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Categories

\ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/categories/index.xml b/aseltmann.github.io-hugo/public/categories/index.xml deleted file mode 100644 index 7cf9044..0000000 --- a/aseltmann.github.io-hugo/public/categories/index.xml +++ /dev/null @@ -1 +0,0 @@ -Categories on Blog & Noteshttp://localhost:1313/categories/Recent content in Categories on Blog & NotesHugoen \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/categories/page/1/index.html b/aseltmann.github.io-hugo/public/categories/page/1/index.html deleted file mode 100644 index adb2000..0000000 --- a/aseltmann.github.io-hugo/public/categories/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/categories/ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de.search-data.min.3c99881476450c58deda692262a83183c22f9e17f0517ec0af7cf57419beebec.json b/aseltmann.github.io-hugo/public/de.search-data.min.3c99881476450c58deda692262a83183c22f9e17f0517ec0af7cf57419beebec.json deleted file mode 100644 index fd4d57f..0000000 --- a/aseltmann.github.io-hugo/public/de.search-data.min.3c99881476450c58deda692262a83183c22f9e17f0517ec0af7cf57419beebec.json +++ /dev/null @@ -1 +0,0 @@ -[{"id":0,"href":"/de/docs/notes/poetry/","title":"Gedichte","section":"Notizen","content":" Table of Contents Erich Fried Lǎozǐ Erich Fried # Kleines Beispiel (1982) Auch ungelebtes Leben\ngeht zu Ende\nzwar vielleicht langsamer\nwie eine Batterie\nin einer Taschenlampe\ndie keiner benutzt\nAber das hilft nicht viel:\nWenn man\n(sagen wir einmal)\ndiese Taschenlampe\nnach so- und sovielen Jahren\nanknipsen will\nkommt kein Atemzug Licht mehr heraus\nund wenn du sie aufmachst\nfindest du nur deine Knochen\nund falls du Pech hast\nauch diese\nschon ganz zerfressen\nDa hättest du\ngenau so gut\nleuchten können\nHerrschaftsfreiheit (1984) Zu sagen\n\u0026ldquo;Hier\nherrscht Freiheit\u0026rdquo;\nist immer\nein Irrtum\noder auch\neine Lüge:\nFreiheit\nherrscht nicht\nHumorlos (1967) Die Jungen\nwerfen\nzum Spaß\nmit Steinen\nnach Fröschen\nDie Frösche\nsterben\nim Ernst\nRückwärtsgewandte Utopie (1981) Angeklagt\nder Unmenschlichkeit\nbehauptet\nder Nichtmehrmensch\nimmer noch\nerst\nein Nochnichtmensch\nzu sein\nZu guter Letzt (1983) Als Kind wusste ich:\nJeder Schmetterling\nden ich rette\njede Schnecke\nund jede Spinne\nund jede Mücke\njeder Ohrwurm\nund jeder Regenwurm\nwird kommen und weinen\nwenn ich begraben werde\nEinmal von mir gerettet\nmuss keines mehr sterben\nAlle werden sie kommen\nzu meinem Begräbnis\nAls ich dann groß wurde\nerkannte ich:\nDas ist großer Unsinn\nKeines wird kommen\nich überlebe sie alle\nJetzt im Alter\nfrage ich: Wenn ich sie aber\nrette bis ganz zuletzt\nkommen doch vielleicht zwei oder drei?\nLǎozǐ # Ich persönlich mag die Übersetzung von Günther Debon.\nDàodéjīng - Kapitel 11 [\u0026hellip;]\nMan knetet Ton zurecht\nZum Trinkgerät:\nEben dort, wo keiner ist,\nIst des Gerätes Brauchbarkeit.\n[\u0026hellip;]\nWahrlich:\nErkennst du das Da-Sein als einen Gewinn,\nErkenne: Das Nicht-Sein macht brauchbar.\nDàodéjīng - Kapitel 15 Wer im Altertum gut war als Meister,\nWar subtil, geheimnisvoll, mystisch, durchdringend;\nSo tief, dass er uns unbegreiflich bleibt.\nWohl! Und weil er unbegreiflich bleibt,\nWill ich lieber dartun sein Gebaren:\nSo zögernd, ach!\nWie wenn man winters quert einen Strom;\nSo ängstlich, ach!\nWie wenn man fürchtet die Nachbarn rings;\nVerhalten, ach!\nAls wäre zu Gast man geladen;\nNachgiebig, ach!\nWie vor der Schmelze das Eis;\nGediegen, ach!\nGleich einem Grobholz;\nWeit, ach!\nGleich einem Flusstal;\nChaotisch, ach!\nGleich einem Strudel.\nWer kann den Strudel stillen,\nAuf dass er mählich werde rein?\nWer kann das Ruhende bewegen,\nAuf dass es mählich Leben gewinne?\nWer diesen \u0026lt;i\u0026gt;Weg\u0026lt;/i\u0026gt; bewahrt,\nWünscht nicht, erfüllt zu sein.\nWohl! Nur was unerfüllt,\nKann auch verschleißen ohne Erneuen.\nDàodéjīng - Kapitel 20 Brich ab das Lernen, so bist du sorgenfrei!\nSind denn \u0026ldquo;Jawohl!\u0026rdquo; und \u0026ldquo;Recht gern!\u0026quot;\nWirklich einander so fern?\nSind denn das Gute, die Schlechtigkeit\nWirklich einander so weit?\n\u0026ldquo;Wem andere Menschen sich beugen,\nDem musst auch du dich beugen\u0026rdquo;:\nWelch Öde doch! Und kein Ende noch!\nDie Menschen alle sind ausgelassen,\nAls säßen sie zechend beim Opferfest,\nAls stiegen sie auf zu den Frühlingsterrassen.\nIch allein liege noch still,\nKein Zeichen hab ich gegeben,\nGleich einem kleinen Kinde,\nDas noch nie gelacht hat im Leben;\nBin schwankend, bin wankend,\nAls hätt ich die Heimat verloren.\nDie Menge der Menschen hat Überfluss;\nNur Ich bin gleichsam von allem entblößt.\nWahrlich, Ich habe das Herz eines Toren,\nSo dunkel und wirr!\nDie gewöhnlichen Menschen sind hell und klar;\nNur Ich bin trübe verhangen.\nDie gewöhnlichen Menschen sind strebig-straff;\nNur Ich bin bang-befangen.\nRuhelos gleich ich dem Meere;\nVerweht, ach, bin gleichsam ich ohne Halt.\nDie Menschen machen sich nützlich all,\nNur Ich bin halsstarr, als ob ich ein Wildling wäre.\nNur Ich bin von den andern Menschen verschieden -\nDer ich die nährende Mutter verehre.\n"},{"id":1,"href":"/de/posts/aoc2024-day02/","title":"Advent of code 2024 - Tag 2: Rotnasige Berichte","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2024 in der auf Statistik fokussierten Programmiersprache R. Dies ist Tag 2 - siehe https:adventofcode.com/2024/day/2\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":2,"href":"/de/posts/aoc2024-day01/","title":"Advent of code 2024 - Tag 1: Hysterie der Historiker:innen","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2024 in der auf Statistik fokussierten Programmiersprache R. Dies ist Tag 1 - siehe https:adventofcode.com/2024/day/1\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":3,"href":"/de/posts/aoc2023-day07/","title":"Advent of code 2023 - Tag 7: Kamelkarten","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 7 - siehe https:adventofcode.com/2023/day/7\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":4,"href":"/de/posts/aoc2023-day06/","title":"Advent of code 2023 - Tag 6: Warte Darauf","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 6 - siehe https:adventofcode.com/2023/day/6\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":5,"href":"/de/posts/aoc2023-day05/","title":"Advent of code 2023 - Tag 5: Wenn Du Einem Samen Dünger Gibst","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 5 - siehe https:adventofcode.com/2023/day/5\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":6,"href":"/de/posts/aoc2023-day04/","title":"Advent of code 2023 - Tag 4: Rubbelkarten","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 4 - siehe https:adventofcode.com/2023/day/4\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":7,"href":"/de/posts/aoc2023-day03/","title":"Advent of code 2023 - Tag 3: Getriebeübersetzungen","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 3 - siehe https:adventofcode.com/2023/day/3\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":8,"href":"/de/posts/aoc2023-day02/","title":"Advent of code 2023 - Tag 2: Würfelrätsel","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 2 - siehe https:adventofcode.com/2023/day/2\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":9,"href":"/de/posts/aoc2023-day01/","title":"Advent of code 2023 - Tag 1: Trebuchet?!","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 1 - siehe https:adventofcode.com/2023/day/1\n"}] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de.search-data.min.4279d5fcce8149553d10e54d8161e2ead2c7fa63e7fe5972a013ef3ea5bcb6e4.json b/aseltmann.github.io-hugo/public/de.search-data.min.4279d5fcce8149553d10e54d8161e2ead2c7fa63e7fe5972a013ef3ea5bcb6e4.json deleted file mode 100644 index 3d2d776..0000000 --- a/aseltmann.github.io-hugo/public/de.search-data.min.4279d5fcce8149553d10e54d8161e2ead2c7fa63e7fe5972a013ef3ea5bcb6e4.json +++ /dev/null @@ -1 +0,0 @@ -[{"id":0,"href":"/de/docs/notes/poetry/","title":"Gedichte","section":"Notizen","content":" Table of Contents Erich Fried Lǎozǐ Erich Fried # Kleines Beispiel (1982) Auch ungelebtes Leben\ngeht zu Ende\nzwar vielleicht langsamer\nwie eine Batterie\nin einer Taschenlampe\ndie keiner benutzt\nAber das hilft nicht viel:\nWenn man\n(sagen wir einmal)\ndiese Taschenlampe\nnach so- und sovielen Jahren\nanknipsen will\nkommt kein Atemzug Licht mehr heraus\nund wenn du sie aufmachst\nfindest du nur deine Knochen\nund falls du Pech hast\nauch diese\nschon ganz zerfressen\nDa hättest du\ngenau so gut\nleuchten können\nHerrschaftsfreiheit (1984) Zu sagen\n\u0026ldquo;Hier\nherrscht Freiheit\u0026rdquo;\nist immer\nein Irrtum\noder auch\neine Lüge:\nFreiheit\nherrscht nicht\nHumorlos (1967) Die Jungen\nwerfen\nzum Spaß\nmit Steinen\nnach Fröschen\nDie Frösche\nsterben\nim Ernst\nRückwärtsgewandte Utopie (1981) Angeklagt\nder Unmenschlichkeit\nbehauptet\nder Nichtmehrmensch\nimmer noch\nerst\nein Nochnichtmensch\nzu sein\nZu guter Letzt (1983) Als Kind wusste ich:\nJeder Schmetterling\nden ich rette\njede Schnecke\nund jede Spinne\nund jede Mücke\njeder Ohrwurm\nund jeder Regenwurm\nwird kommen und weinen\nwenn ich begraben werde\nEinmal von mir gerettet\nmuss keines mehr sterben\nAlle werden sie kommen\nzu meinem Begräbnis\nAls ich dann groß wurde\nerkannte ich:\nDas ist großer Unsinn\nKeines wird kommen\nich überlebe sie alle\nJetzt im Alter\nfrage ich: Wenn ich sie aber\nrette bis ganz zuletzt\nkommen doch vielleicht zwei oder drei?\nLǎozǐ # Ich persönlich mag die Übersetzung von Günther Debon.\nDàodéjīng - Kapitel 11 [\u0026hellip;]\nMan knetet Ton zurecht\nZum Trinkgerät:\nEben dort, wo keiner ist,\nIst des Gerätes Brauchbarkeit.\n[\u0026hellip;]\nWahrlich:\nErkennst du das Da-Sein als einen Gewinn,\nErkenne: Das Nicht-Sein macht brauchbar.\nDàodéjīng - Kapitel 15 Wer im Altertum gut war als Meister,\nWar subtil, geheimnisvoll, mystisch, durchdringend;\nSo tief, dass er uns unbegreiflich bleibt.\nWohl! Und weil er unbegreiflich bleibt,\nWill ich lieber dartun sein Gebaren:\nSo zögernd, ach!\nWie wenn man winters quert einen Strom;\nSo ängstlich, ach!\nWie wenn man fürchtet die Nachbarn rings;\nVerhalten, ach!\nAls wäre zu Gast man geladen;\nNachgiebig, ach!\nWie vor der Schmelze das Eis;\nGediegen, ach!\nGleich einem Grobholz;\nWeit, ach!\nGleich einem Flusstal;\nChaotisch, ach!\nGleich einem Strudel.\nWer kann den Strudel stillen,\nAuf dass er mählich werde rein?\nWer kann das Ruhende bewegen,\nAuf dass es mählich Leben gewinne?\nWer diesen \u0026lt;i\u0026gt;Weg\u0026lt;/i\u0026gt; bewahrt,\nWünscht nicht, erfüllt zu sein.\nWohl! Nur was unerfüllt,\nKann auch verschleißen ohne Erneuen.\nDàodéjīng - Kapitel 20 Brich ab das Lernen, so bist du sorgenfrei!\nSind denn \u0026ldquo;Jawohl!\u0026rdquo; und \u0026ldquo;Recht gern!\u0026quot;\nWirklich einander so fern?\nSind denn das Gute, die Schlechtigkeit\nWirklich einander so weit?\n\u0026ldquo;Wem andere Menschen sich beugen,\nDem musst auch du dich beugen\u0026rdquo;:\nWelch Öde doch! Und kein Ende noch!\nDie Menschen alle sind ausgelassen,\nAls säßen sie zechend beim Opferfest,\nAls stiegen sie auf zu den Frühlingsterrassen.\nIch allein liege noch still,\nKein Zeichen hab ich gegeben,\nGleich einem kleinen Kinde,\nDas noch nie gelacht hat im Leben;\nBin schwankend, bin wankend,\nAls hätt ich die Heimat verloren.\nDie Menge der Menschen hat Überfluss;\nNur Ich bin gleichsam von allem entblößt.\nWahrlich, Ich habe das Herz eines Toren,\nSo dunkel und wirr!\nDie gewöhnlichen Menschen sind hell und klar;\nNur Ich bin trübe verhangen.\nDie gewöhnlichen Menschen sind strebig-straff;\nNur Ich bin bang-befangen.\nRuhelos gleich ich dem Meere;\nVerweht, ach, bin gleichsam ich ohne Halt.\nDie Menschen machen sich nützlich all,\nNur Ich bin halsstarr, als ob ich ein Wildling wäre.\nNur Ich bin von den andern Menschen verschieden -\nDer ich die nährende Mutter verehre.\n"},{"id":1,"href":"/de/posts/aoc2024-day01/","title":"Advent of code 2024 - Tag 1: Hysterie der Historiker:innen","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2024 in der auf Statistik fokussierten Programmiersprache R. Dies ist Tag 1 - siehe https:adventofcode.com/2024/day/1\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":2,"href":"/de/posts/aoc2023-day07/","title":"Advent of code 2023 - Tag 7: Kamelkarten","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 7 - siehe https:adventofcode.com/2023/day/7\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":3,"href":"/de/posts/aoc2023-day06/","title":"Advent of code 2023 - Tag 6: Warte Darauf","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 6 - siehe https:adventofcode.com/2023/day/6\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":4,"href":"/de/posts/aoc2023-day05/","title":"Advent of code 2023 - Tag 5: Wenn Du Einem Samen Dünger Gibst","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 5 - siehe https:adventofcode.com/2023/day/5\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":5,"href":"/de/posts/aoc2023-day04/","title":"Advent of code 2023 - Tag 4: Rubbelkarten","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 4 - siehe https:adventofcode.com/2023/day/4\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":6,"href":"/de/posts/aoc2023-day03/","title":"Advent of code 2023 - Tag 3: Getriebeübersetzungen","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 3 - siehe https:adventofcode.com/2023/day/3\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":7,"href":"/de/posts/aoc2023-day02/","title":"Advent of code 2023 - Tag 2: Würfelrätsel","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 2 - siehe https:adventofcode.com/2023/day/2\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":8,"href":"/de/posts/aoc2023-day01/","title":"Advent of code 2023 - Tag 1: Trebuchet?!","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 1 - siehe https:adventofcode.com/2023/day/1\n"}] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de.search-data.min.4f53cda18c2baa0c0354bb5f9a3ecbe5ed12ab4d8e11ba873c2f11161202b945.json b/aseltmann.github.io-hugo/public/de.search-data.min.4f53cda18c2baa0c0354bb5f9a3ecbe5ed12ab4d8e11ba873c2f11161202b945.json deleted file mode 100644 index 0637a08..0000000 --- a/aseltmann.github.io-hugo/public/de.search-data.min.4f53cda18c2baa0c0354bb5f9a3ecbe5ed12ab4d8e11ba873c2f11161202b945.json +++ /dev/null @@ -1 +0,0 @@ -[] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de.search-data.min.e8bdc17e969113efc5165227f41bcc104a0f5628da42c6f49531c86d99d33fd8.json b/aseltmann.github.io-hugo/public/de.search-data.min.e8bdc17e969113efc5165227f41bcc104a0f5628da42c6f49531c86d99d33fd8.json deleted file mode 100644 index ce2e384..0000000 --- a/aseltmann.github.io-hugo/public/de.search-data.min.e8bdc17e969113efc5165227f41bcc104a0f5628da42c6f49531c86d99d33fd8.json +++ /dev/null @@ -1 +0,0 @@ -[{"id":0,"href":"/de/docs/notes/poetry/","title":"Gedichte","section":"Notizen","content":" Table of Contents Erich Fried Lǎozǐ Erich Fried # Kleines Beispiel (1982) Auch ungelebtes Leben\ngeht zu Ende\nzwar vielleicht langsamer\nwie eine Batterie\nin einer Taschenlampe\ndie keiner benutzt\nAber das hilft nicht viel:\nWenn man\n(sagen wir einmal)\ndiese Taschenlampe\nnach so- und sovielen Jahren\nanknipsen will\nkommt kein Atemzug Licht mehr heraus\nund wenn du sie aufmachst\nfindest du nur deine Knochen\nund falls du Pech hast\nauch diese\nschon ganz zerfressen\nDa hättest du\ngenau so gut\nleuchten können\nHerrschaftsfreiheit (1984) Zu sagen\n\u0026ldquo;Hier\nherrscht Freiheit\u0026rdquo;\nist immer\nein Irrtum\noder auch\neine Lüge:\nFreiheit\nherrscht nicht\nHumorlos (1967) Die Jungen\nwerfen\nzum Spaß\nmit Steinen\nnach Fröschen\nDie Frösche\nsterben\nim Ernst\nRückwärtsgewandte Utopie (1981) Angeklagt\nder Unmenschlichkeit\nbehauptet\nder Nichtmehrmensch\nimmer noch\nerst\nein Nochnichtmensch\nzu sein\nZu guter Letzt (1983) Als Kind wusste ich:\nJeder Schmetterling\nden ich rette\njede Schnecke\nund jede Spinne\nund jede Mücke\njeder Ohrwurm\nund jeder Regenwurm\nwird kommen und weinen\nwenn ich begraben werde\nEinmal von mir gerettet\nmuss keines mehr sterben\nAlle werden sie kommen\nzu meinem Begräbnis\nAls ich dann groß wurde\nerkannte ich:\nDas ist großer Unsinn\nKeines wird kommen\nich überlebe sie alle\nJetzt im Alter\nfrage ich: Wenn ich sie aber\nrette bis ganz zuletzt\nkommen doch vielleicht zwei oder drei?\nLǎozǐ # Ich persönlich mag die Übersetzung von Günther Debon.\nDàodéjīng - Kapitel 11 [\u0026hellip;]\nMan knetet Ton zurecht\nZum Trinkgerät:\nEben dort, wo keiner ist,\nIst des Gerätes Brauchbarkeit.\n[\u0026hellip;]\nWahrlich:\nErkennst du das Da-Sein als einen Gewinn,\nErkenne: Das Nicht-Sein macht brauchbar.\nDàodéjīng - Kapitel 15 Wer im Altertum gut war als Meister,\nWar subtil, geheimnisvoll, mystisch, durchdringend;\nSo tief, dass er uns unbegreiflich bleibt.\nWohl! Und weil er unbegreiflich bleibt,\nWill ich lieber dartun sein Gebaren:\nSo zögernd, ach!\nWie wenn man winters quert einen Strom;\nSo ängstlich, ach!\nWie wenn man fürchtet die Nachbarn rings;\nVerhalten, ach!\nAls wäre zu Gast man geladen;\nNachgiebig, ach!\nWie vor der Schmelze das Eis;\nGediegen, ach!\nGleich einem Grobholz;\nWeit, ach!\nGleich einem Flusstal;\nChaotisch, ach!\nGleich einem Strudel.\nWer kann den Strudel stillen,\nAuf dass er mählich werde rein?\nWer kann das Ruhende bewegen,\nAuf dass es mählich Leben gewinne?\nWer diesen \u0026lt;i\u0026gt;Weg\u0026lt;/i\u0026gt; bewahrt,\nWünscht nicht, erfüllt zu sein.\nWohl! Nur was unerfüllt,\nKann auch verschleißen ohne Erneuen.\nDàodéjīng - Kapitel 20 Brich ab das Lernen, so bist du sorgenfrei!\nSind denn \u0026ldquo;Jawohl!\u0026rdquo; und \u0026ldquo;Recht gern!\u0026quot;\nWirklich einander so fern?\nSind denn das Gute, die Schlechtigkeit\nWirklich einander so weit?\n\u0026ldquo;Wem andere Menschen sich beugen,\nDem musst auch du dich beugen\u0026rdquo;:\nWelch Öde doch! Und kein Ende noch!\nDie Menschen alle sind ausgelassen,\nAls säßen sie zechend beim Opferfest,\nAls stiegen sie auf zu den Frühlingsterrassen.\nIch allein liege noch still,\nKein Zeichen hab ich gegeben,\nGleich einem kleinen Kinde,\nDas noch nie gelacht hat im Leben;\nBin schwankend, bin wankend,\nAls hätt ich die Heimat verloren.\nDie Menge der Menschen hat Überfluss;\nNur Ich bin gleichsam von allem entblößt.\nWahrlich, Ich habe das Herz eines Toren,\nSo dunkel und wirr!\nDie gewöhnlichen Menschen sind hell und klar;\nNur Ich bin trübe verhangen.\nDie gewöhnlichen Menschen sind strebig-straff;\nNur Ich bin bang-befangen.\nRuhelos gleich ich dem Meere;\nVerweht, ach, bin gleichsam ich ohne Halt.\nDie Menschen machen sich nützlich all,\nNur Ich bin halsstarr, als ob ich ein Wildling wäre.\nNur Ich bin von den andern Menschen verschieden -\nDer ich die nährende Mutter verehre.\n"},{"id":1,"href":"/de/posts/aoc2024-day02/","title":"Advent of code 2024 - Tag 2: Rotnasige Berichte","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2024 in der auf Statistik fokussierten Programmiersprache R. Dies ist Tag 2 - siehe https:adventofcode.com/2024/day/2\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":2,"href":"/de/posts/aoc2024-day01/","title":"Advent of code 2024 - Tag 1: Hysterie der Historiker:innen","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2024 in der auf Statistik fokussierten Programmiersprache R. Dies ist Tag 1 - siehe https:adventofcode.com/2024/day/1\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":3,"href":"/de/posts/aoc2023-day07/","title":"Advent of code 2023 - Tag 7: Kamelkarten","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 7 - siehe https:adventofcode.com/2023/day/7\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":4,"href":"/de/posts/aoc2023-day06/","title":"Advent of code 2023 - Tag 6: Warte Darauf","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 6 - siehe https:adventofcode.com/2023/day/6\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":5,"href":"/de/docs/projects/","title":"Projekte","section":"Docs","content":" Aktuelle Projekte # Ich glaube an die Prinzipien von Open Science, Open Data und Open Source. Daher arbeite ich aktuell daran, meine Forschung an Fluoreszenzkorrelationsspektroskopie (FCS) und Maschinellem Lernen in einer offenen Art und Weise verfügbar zu machen. Im Sinne der Open-notebook science ist hier ein Einblick in meine aktuelle Forschung zu finden:\nFluotracify - eine reproduzierbare Doktorarbeit [Englisch]\nPräsentationen auf Konferenzen # Seltmann A, Eggeling C, Waithe D. Automated, User-independent Correction of Artifacts in Fluorescence Correlation Spectroscopy Measurements using Convolutional Neural Networks. Quantitative BioImaging Conference (QBI); 2020 Jan 6-9; Oxford, UK "},{"id":6,"href":"/de/posts/aoc2023-day05/","title":"Advent of code 2023 - Tag 5: Wenn Du Einem Samen Dünger Gibst","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 5 - siehe https:adventofcode.com/2023/day/5\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":7,"href":"/de/posts/aoc2023-day04/","title":"Advent of code 2023 - Tag 4: Rubbelkarten","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 4 - siehe https:adventofcode.com/2023/day/4\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":8,"href":"/de/posts/aoc2023-day03/","title":"Advent of code 2023 - Tag 3: Getriebeübersetzungen","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 3 - siehe https:adventofcode.com/2023/day/3\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":9,"href":"/de/posts/aoc2023-day02/","title":"Advent of code 2023 - Tag 2: Würfelrätsel","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 2 - siehe https:adventofcode.com/2023/day/2\nFür die Lösungen siehe die englische Version dieses Blogbeitrages.\n"},{"id":10,"href":"/de/posts/aoc2023-day01/","title":"Advent of code 2023 - Tag 1: Trebuchet?!","section":"Posts","content":"Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2023. Dies ist Tag 1 - siehe https:adventofcode.com/2023/day/1\n"}] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de.search.min.1d102abf0947cc01ec4686e930a3308334eaffbc6b04471c10189f6a241052ba.js b/aseltmann.github.io-hugo/public/de.search.min.1d102abf0947cc01ec4686e930a3308334eaffbc6b04471c10189f6a241052ba.js deleted file mode 100644 index 3cc464f..0000000 --- a/aseltmann.github.io-hugo/public/de.search.min.1d102abf0947cc01ec4686e930a3308334eaffbc6b04471c10189f6a241052ba.js +++ /dev/null @@ -1 +0,0 @@ -"use strict";(function(){const o="/de.search-data.min.4279d5fcce8149553d10e54d8161e2ead2c7fa63e7fe5972a013ef3ea5bcb6e4.json",i=Object.assign({cache:!0},{includeScore:!0,useExtendedSearch:!0,fieldNormWeight:1.5,threshold:.2,ignoreLocation:!0,keys:[{name:"title",weight:.7},{name:"content",weight:.3}]}),e=document.querySelector("#book-search-input"),t=document.querySelector("#book-search-results");if(!e)return;e.addEventListener("focus",n),e.addEventListener("keyup",s),document.addEventListener("keypress",a);function a(t){if(t.target.value!==void 0)return;if(e===document.activeElement)return;const n=String.fromCharCode(t.charCode);if(!r(n))return;e.focus(),t.preventDefault()}function r(t){const n=e.getAttribute("data-hotkeys")||"";return n.indexOf(t)>=0}function n(){e.removeEventListener("focus",n),e.required=!0,fetch(o).then(e=>e.json()).then(e=>{window.bookSearchIndex=new Fuse(e,i)}).then(()=>e.required=!1).then(s)}function s(){for(;t.firstChild;)t.removeChild(t.firstChild);if(!e.value)return;const n=window.bookSearchIndex.search(e.value).slice(0,10);n.forEach(function(e){const n=c("
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  • "),s=n.querySelector("a"),o=n.querySelector("small");s.href=e.item.href,s.textContent=e.item.title,o.textContent=e.item.section,t.appendChild(n)})}function c(e){const t=document.createElement("div");return t.innerHTML=e,t.firstChild}})() \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/404.html b/aseltmann.github.io-hugo/public/de/404.html deleted file mode 100644 index 7009cff..0000000 --- a/aseltmann.github.io-hugo/public/de/404.html +++ /dev/null @@ -1,2 +0,0 @@ -404 Page not found | Blog & Notes -

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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/categories/index.html b/aseltmann.github.io-hugo/public/de/categories/index.html deleted file mode 100644 index b46341a..0000000 --- a/aseltmann.github.io-hugo/public/de/categories/index.html +++ /dev/null @@ -1,15 +0,0 @@ -Categories | Blog & Notes - -

    Categories

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    Docs

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/docs/index.xml b/aseltmann.github.io-hugo/public/de/docs/index.xml deleted file mode 100644 index 22c119a..0000000 --- a/aseltmann.github.io-hugo/public/de/docs/index.xml +++ /dev/null @@ -1 +0,0 @@ -Docs on Blog & Noteshttp://localhost:1313/de/docs/Recent content in Docs on Blog & NotesHugode \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/docs/notes/index.html b/aseltmann.github.io-hugo/public/de/docs/notes/index.html deleted file mode 100644 index f710a44..0000000 --- a/aseltmann.github.io-hugo/public/de/docs/notes/index.html +++ /dev/null @@ -1,13 +0,0 @@ -Notizen | Blog & Notes - -

    Notizen

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/docs/notes/index.xml b/aseltmann.github.io-hugo/public/de/docs/notes/index.xml deleted file mode 100644 index 3c140a8..0000000 --- a/aseltmann.github.io-hugo/public/de/docs/notes/index.xml +++ /dev/null @@ -1,41 +0,0 @@ -Notizen on Blog & Noteshttp://localhost:1313/de/docs/notes/Recent content in Notizen on Blog & NotesHugodeSun, 22 Dec 2024 01:44:37 +0100Gedichtehttp://localhost:1313/de/docs/notes/poetry/Sun, 22 Dec 2024 01:44:37 +0100http://localhost:1313/de/docs/notes/poetry/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/de/#erich-fried">Erich Fried</a></li> -<li> - <a href="http://localhost:1313/de/#l%c7%8eoz%c7%90">Lǎozǐ</a></li> -</ul> -</div> -<!--endtoc--> -<h2 id="erich-fried"> - Erich Fried - <a class="anchor" href="#erich-fried">#</a> -</h2> -<details> -<summary>Kleines Beispiel (1982)</summary> -<div class="details"> -<div class="verse"> -<p>Auch ungelebtes Leben<br /> -geht zu Ende<br /> -zwar vielleicht langsamer<br /> -wie eine Batterie<br /> -in einer Taschenlampe<br /> -die keiner benutzt<br /> -<br /> -Aber das hilft nicht viel:<br /> -Wenn man<br /> -(sagen wir einmal)<br /> -diese Taschenlampe<br /> -nach so- und sovielen Jahren<br /> -anknipsen will<br /> -kommt kein Atemzug Licht mehr heraus<br /> -und wenn du sie aufmachst<br /> -findest du nur deine Knochen<br /> -und falls du Pech hast<br /> -auch diese<br /> -schon ganz zerfressen<br /> -<br /> -Da hättest du<br /> -genau so gut<br /> -leuchten können<br /></p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/docs/notes/poetry/index.html b/aseltmann.github.io-hugo/public/de/docs/notes/poetry/index.html deleted file mode 100644 index 8e0966a..0000000 --- a/aseltmann.github.io-hugo/public/de/docs/notes/poetry/index.html +++ /dev/null @@ -1,75 +0,0 @@ -Gedichte | Blog & Notes - -

    Gedichte

    Table of Contents

    Erich Fried -#

    Kleines Beispiel (1982)

    Auch ungelebtes Leben
    geht zu Ende
    zwar vielleicht langsamer
    wie eine Batterie
    in einer Taschenlampe
    die keiner benutzt

    Aber das hilft nicht viel:
    Wenn man
    (sagen wir einmal)
    diese Taschenlampe
    nach so- und sovielen Jahren
    anknipsen will
    kommt kein Atemzug Licht mehr heraus
    und wenn du sie aufmachst
    findest du nur deine Knochen
    und falls du Pech hast
    auch diese
    schon ganz zerfressen

    Da hättest du
    genau so gut
    leuchten können

    Herrschaftsfreiheit (1984)

    Zu sagen
    “Hier
    herrscht Freiheit”
    ist immer
    ein Irrtum
    oder auch
    eine Lüge:

    Freiheit
    herrscht nicht

    Humorlos (1967)

    Die Jungen
    werfen
    zum Spaß
    mit Steinen
    nach Fröschen

    Die Frösche
    sterben
    im Ernst

    Rückwärtsgewandte Utopie (1981)

    Angeklagt
    der Unmenschlichkeit
    behauptet
    der Nichtmehrmensch
    immer noch
    erst
    ein Nochnichtmensch
    zu sein

    Zu guter Letzt (1983)

    Als Kind wusste ich:
    Jeder Schmetterling
    den ich rette
    jede Schnecke
    und jede Spinne
    und jede Mücke
    jeder Ohrwurm
    und jeder Regenwurm
    wird kommen und weinen
    wenn ich begraben werde

    Einmal von mir gerettet
    muss keines mehr sterben
    Alle werden sie kommen
    zu meinem Begräbnis

    Als ich dann groß wurde
    erkannte ich:
    Das ist großer Unsinn
    Keines wird kommen
    ich überlebe sie alle

    Jetzt im Alter
    frage ich: Wenn ich sie aber
    rette bis ganz zuletzt
    kommen doch vielleicht zwei oder drei?

    Lǎozǐ -#

    Ich persönlich mag die Übersetzung von Günther Debon.

    Dàodéjīng - Kapitel 11

    […]

    Man knetet Ton zurecht
    Zum Trinkgerät:
    Eben dort, wo keiner ist,
    Ist des Gerätes Brauchbarkeit.

    […]

    Wahrlich:
    Erkennst du das Da-Sein als einen Gewinn,
    Erkenne: Das Nicht-Sein macht brauchbar.

    Dàodéjīng - Kapitel 15

    Wer im Altertum gut war als Meister,
    War subtil, geheimnisvoll, mystisch, durchdringend;
    So tief, dass er uns unbegreiflich bleibt.
    Wohl! Und weil er unbegreiflich bleibt,
    Will ich lieber dartun sein Gebaren:

    So zögernd, ach!
    Wie wenn man winters quert einen Strom;
    So ängstlich, ach!
    Wie wenn man fürchtet die Nachbarn rings;
    Verhalten, ach!
    Als wäre zu Gast man geladen;
    Nachgiebig, ach!
    Wie vor der Schmelze das Eis;
    Gediegen, ach!
    Gleich einem Grobholz;
    Weit, ach!
    Gleich einem Flusstal;
    Chaotisch, ach!
    Gleich einem Strudel.

    Wer kann den Strudel stillen,
    Auf dass er mählich werde rein?
    Wer kann das Ruhende bewegen,
    Auf dass es mählich Leben gewinne?

    Wer diesen <i>Weg</i> bewahrt,
    Wünscht nicht, erfüllt zu sein.
    Wohl! Nur was unerfüllt,
    Kann auch verschleißen ohne Erneuen.

    Dàodéjīng - Kapitel 20

    Brich ab das Lernen, so bist du sorgenfrei!

    Sind denn “Jawohl!” und “Recht gern!"
    Wirklich einander so fern?
    Sind denn das Gute, die Schlechtigkeit
    Wirklich einander so weit?
    “Wem andere Menschen sich beugen,
    Dem musst auch du dich beugen”:
    Welch Öde doch! Und kein Ende noch!

    Die Menschen alle sind ausgelassen,
    Als säßen sie zechend beim Opferfest,
    Als stiegen sie auf zu den Frühlingsterrassen.
    Ich allein liege noch still,
    Kein Zeichen hab ich gegeben,
    Gleich einem kleinen Kinde,
    Das noch nie gelacht hat im Leben;
    Bin schwankend, bin wankend,
    Als hätt ich die Heimat verloren.
    Die Menge der Menschen hat Überfluss;
    Nur Ich bin gleichsam von allem entblößt.
    Wahrlich, Ich habe das Herz eines Toren,
    So dunkel und wirr!
    Die gewöhnlichen Menschen sind hell und klar;
    Nur Ich bin trübe verhangen.
    Die gewöhnlichen Menschen sind strebig-straff;
    Nur Ich bin bang-befangen.
    Ruhelos gleich ich dem Meere;
    Verweht, ach, bin gleichsam ich ohne Halt.

    Die Menschen machen sich nützlich all,
    Nur Ich bin halsstarr, als ob ich ein Wildling wäre.
    Nur Ich bin von den andern Menschen verschieden -
    Der ich die nährende Mutter verehre.

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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/docs/projects/index.html b/aseltmann.github.io-hugo/public/de/docs/projects/index.html deleted file mode 100644 index 4c5ae81..0000000 --- a/aseltmann.github.io-hugo/public/de/docs/projects/index.html +++ /dev/null @@ -1,48 +0,0 @@ -Projekte | Blog & Notes - -

    Projekte

    Aktuelle Projekte -#

    Ich glaube an die Prinzipien von Open Science, Open Data und Open Source. -Daher arbeite ich aktuell daran, meine Forschung an -Fluoreszenzkorrelationsspektroskopie (FCS) und Maschinellem Lernen in einer -offenen Art und Weise verfügbar zu machen. Im Sinne der -Open-notebook -science ist hier ein Einblick in meine aktuelle Forschung zu finden:

    Fluotracify - eine reproduzierbare Doktorarbeit [Englisch]

    Präsentationen auf Konferenzen -#

    • Seltmann A, Eggeling C, Waithe D. Automated, User-independent Correction of -Artifacts in Fluorescence Correlation Spectroscopy Measurements using -Convolutional Neural Networks. -Quantitative BioImaging Conference (QBI); 2020 -Jan 6-9; Oxford, UK
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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/docs/projects/index.xml b/aseltmann.github.io-hugo/public/de/docs/projects/index.xml deleted file mode 100644 index 30ccdaf..0000000 --- a/aseltmann.github.io-hugo/public/de/docs/projects/index.xml +++ /dev/null @@ -1 +0,0 @@ -Projekte on Blog & Noteshttp://localhost:1313/de/docs/projects/Recent content in Projekte on Blog & NotesHugode \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/index.html b/aseltmann.github.io-hugo/public/de/index.html deleted file mode 100644 index 1bca746..0000000 --- a/aseltmann.github.io-hugo/public/de/index.html +++ /dev/null @@ -1,45 +0,0 @@ -Homepage_de | Blog & Notes - -

    Homepage_de

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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/index.xml b/aseltmann.github.io-hugo/public/de/index.xml deleted file mode 100644 index 5d67d17..0000000 --- a/aseltmann.github.io-hugo/public/de/index.xml +++ /dev/null @@ -1,69 +0,0 @@ -Homepage_de on Blog & Noteshttp://localhost:1313/de/Recent content in Homepage_de on Blog & NotesHugodeSun, 22 Dec 2024 01:46:51 +0100Gedichtehttp://localhost:1313/de/docs/notes/poetry/Sun, 22 Dec 2024 01:44:37 +0100http://localhost:1313/de/docs/notes/poetry/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/de/#erich-fried">Erich Fried</a></li> -<li> - <a href="http://localhost:1313/de/#l%c7%8eoz%c7%90">Lǎozǐ</a></li> -</ul> -</div> -<!--endtoc--> -<h2 id="erich-fried"> - Erich Fried - <a class="anchor" href="#erich-fried">#</a> -</h2> -<details> -<summary>Kleines Beispiel (1982)</summary> -<div class="details"> -<div class="verse"> -<p>Auch ungelebtes Leben<br /> -geht zu Ende<br /> -zwar vielleicht langsamer<br /> -wie eine Batterie<br /> -in einer Taschenlampe<br /> -die keiner benutzt<br /> -<br /> -Aber das hilft nicht viel:<br /> -Wenn man<br /> -(sagen wir einmal)<br /> -diese Taschenlampe<br /> -nach so- und sovielen Jahren<br /> -anknipsen will<br /> -kommt kein Atemzug Licht mehr heraus<br /> -und wenn du sie aufmachst<br /> -findest du nur deine Knochen<br /> -und falls du Pech hast<br /> -auch diese<br /> -schon ganz zerfressen<br /> -<br /> -Da hättest du<br /> -genau so gut<br /> -leuchten können<br /></p>Advent of code 2024 - Tag 2: Rotnasige Berichtehttp://localhost:1313/de/posts/aoc2024-day02/Tue, 10 Dec 2024 00:23:00 +0100http://localhost:1313/de/posts/aoc2024-day02/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code 2024</strong> -in der auf Statistik fokussierten Programmiersprache <strong>R</strong>. Dies ist Tag 2 - siehe - - <a href="https:adventofcode.com/2024/day/2">https:adventofcode.com/2024/day/2</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2024 - Tag 1: Hysterie der Historiker:innenhttp://localhost:1313/de/posts/aoc2024-day01/Sat, 07 Dec 2024 00:23:00 +0100http://localhost:1313/de/posts/aoc2024-day01/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code 2024</strong> -in der auf Statistik fokussierten Programmiersprache <strong>R</strong>. Dies ist Tag 1 - siehe - - <a href="https:adventofcode.com/2024/day/1">https:adventofcode.com/2024/day/1</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 7: Kamelkartenhttp://localhost:1313/de/posts/aoc2023-day07/Thu, 04 Jan 2024 16:33:00 +0100http://localhost:1313/de/posts/aoc2023-day07/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 7 - siehe - <a href="https:adventofcode.com/2023/day/7">https:adventofcode.com/2023/day/7</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 6: Warte Daraufhttp://localhost:1313/de/posts/aoc2023-day06/Thu, 04 Jan 2024 00:45:00 +0100http://localhost:1313/de/posts/aoc2023-day06/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 6 - siehe - <a href="https:adventofcode.com/2023/day/6">https:adventofcode.com/2023/day/6</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 5: Wenn Du Einem Samen Dünger Gibsthttp://localhost:1313/de/posts/aoc2023-day05/Sat, 30 Dec 2023 01:10:00 +0100http://localhost:1313/de/posts/aoc2023-day05/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 5 - siehe - <a href="https:adventofcode.com/2023/day/5">https:adventofcode.com/2023/day/5</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 4: Rubbelkartenhttp://localhost:1313/de/posts/aoc2023-day04/Tue, 19 Dec 2023 00:25:00 +0100http://localhost:1313/de/posts/aoc2023-day04/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 4 - siehe - <a href="https:adventofcode.com/2023/day/4">https:adventofcode.com/2023/day/4</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 3: Getriebeübersetzungenhttp://localhost:1313/de/posts/aoc2023-day03/Mon, 18 Dec 2023 01:01:00 +0100http://localhost:1313/de/posts/aoc2023-day03/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 3 - siehe - <a href="https:adventofcode.com/2023/day/3">https:adventofcode.com/2023/day/3</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 2: Würfelrätselhttp://localhost:1313/de/posts/aoc2023-day02/Mon, 11 Dec 2023 23:38:00 +0100http://localhost:1313/de/posts/aoc2023-day02/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 2 - siehe - <a href="https:adventofcode.com/2023/day/2">https:adventofcode.com/2023/day/2</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 1: Trebuchet?!http://localhost:1313/de/posts/aoc2023-day01/Sun, 10 Dec 2023 23:38:00 +0100http://localhost:1313/de/posts/aoc2023-day01/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 1 - siehe - <a href="https:adventofcode.com/2023/day/1">https:adventofcode.com/2023/day/1</a></p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/posts/aoc2023-day01/index.html b/aseltmann.github.io-hugo/public/de/posts/aoc2023-day01/index.html deleted file mode 100644 index 88e846e..0000000 --- a/aseltmann.github.io-hugo/public/de/posts/aoc2023-day01/index.html +++ /dev/null @@ -1,18 +0,0 @@ -Advent of code 2023 - Tag 1: Trebuchet?! | Blog & Notes - -

    Advent of code 2023 - Tag 1: Trebuchet?!

    Advent of code 2023 - Tag 1: Trebuchet?!

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code -2023. Dies ist Tag 1 - siehe -https:adventofcode.com/2023/day/1

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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/posts/aoc2023-day02/index.html b/aseltmann.github.io-hugo/public/de/posts/aoc2023-day02/index.html deleted file mode 100644 index 2130987..0000000 --- a/aseltmann.github.io-hugo/public/de/posts/aoc2023-day02/index.html +++ /dev/null @@ -1,20 +0,0 @@ -Advent of code 2023 - Tag 2: Würfelrätsel | Blog & Notes - -

    Advent of code 2023 - Tag 2: Würfelrätsel

    Advent of code 2023 - Tag 2: Würfelrätsel

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code -2023. Dies ist Tag 2 - siehe -https:adventofcode.com/2023/day/2

    Für die Lösungen siehe die englische Version dieses Blogbeitrages.

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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/posts/aoc2023-day03/index.html b/aseltmann.github.io-hugo/public/de/posts/aoc2023-day03/index.html deleted file mode 100644 index 90690a9..0000000 --- a/aseltmann.github.io-hugo/public/de/posts/aoc2023-day03/index.html +++ /dev/null @@ -1,20 +0,0 @@ -Advent of code 2023 - Tag 3: Getriebeübersetzungen | Blog & Notes - -

    Advent of code 2023 - Tag 3: Getriebeübersetzungen

    Advent of code 2023 - Tag 3: Getriebeübersetzungen

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code -2023. Dies ist Tag 3 - siehe -https:adventofcode.com/2023/day/3

    Für die Lösungen siehe die englische Version dieses Blogbeitrages.

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    Advent of code 2023 - Tag 4: Rubbelkarten

    Advent of code 2023 - Tag 4: Rubbelkarten

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code -2023. Dies ist Tag 4 - siehe -https:adventofcode.com/2023/day/4

    Für die Lösungen siehe die englische Version dieses Blogbeitrages.

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    Advent of code 2023 - Tag 5: Wenn Du Einem Samen Dünger Gibst

    Advent of code 2023 - Tag 5: Wenn Du Einem Samen Dünger Gibst

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code -2023. Dies ist Tag 5 - siehe -https:adventofcode.com/2023/day/5

    Für die Lösungen siehe die englische Version dieses Blogbeitrages.

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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/posts/aoc2023-day06/index.html b/aseltmann.github.io-hugo/public/de/posts/aoc2023-day06/index.html deleted file mode 100644 index f77529d..0000000 --- a/aseltmann.github.io-hugo/public/de/posts/aoc2023-day06/index.html +++ /dev/null @@ -1,20 +0,0 @@ -Advent of code 2023 - Tag 6: Warte Darauf | Blog & Notes - -

    Advent of code 2023 - Tag 6: Warte Darauf

    Advent of code 2023 - Tag 6: Warte Darauf

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code -2023. Dies ist Tag 6 - siehe -https:adventofcode.com/2023/day/6

    Für die Lösungen siehe die englische Version dieses Blogbeitrages.

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    Advent of code 2023 - Tag 7: Kamelkarten

    Advent of code 2023 - Tag 7: Kamelkarten

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code -2023. Dies ist Tag 7 - siehe -https:adventofcode.com/2023/day/7

    Für die Lösungen siehe die englische Version dieses Blogbeitrages.

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    Advent of code 2024 - Tag 1: Hysterie der Historiker:innen

    Advent of code 2024 - Tag 1: Hysterie der Historiker:innen

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2024 -in der auf Statistik fokussierten Programmiersprache R. Dies ist Tag 1 - siehe -https:adventofcode.com/2024/day/1

    Für die Lösungen siehe die englische Version dieses Blogbeitrages.

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    Advent of code 2024 - Tag 2: Rotnasige Berichte

    Advent of code 2024 - Tag 2: Rotnasige Berichte

    Dieses Jahr versuche ich mich an den Herausforderungen des Advent of Code 2024 -in der auf Statistik fokussierten Programmiersprache R. Dies ist Tag 2 - siehe -https:adventofcode.com/2024/day/2

    Für die Lösungen siehe die englische Version dieses Blogbeitrages.

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    Posts

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/posts/index.xml b/aseltmann.github.io-hugo/public/de/posts/index.xml deleted file mode 100644 index 531bcfb..0000000 --- a/aseltmann.github.io-hugo/public/de/posts/index.xml +++ /dev/null @@ -1,29 +0,0 @@ -Posts on Blog & Noteshttp://localhost:1313/de/posts/Recent content in Posts on Blog & NotesHugodeSun, 22 Dec 2024 01:46:51 +0100Advent of code 2024 - Tag 2: Rotnasige Berichtehttp://localhost:1313/de/posts/aoc2024-day02/Tue, 10 Dec 2024 00:23:00 +0100http://localhost:1313/de/posts/aoc2024-day02/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code 2024</strong> -in der auf Statistik fokussierten Programmiersprache <strong>R</strong>. Dies ist Tag 2 - siehe - - <a href="https:adventofcode.com/2024/day/2">https:adventofcode.com/2024/day/2</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2024 - Tag 1: Hysterie der Historiker:innenhttp://localhost:1313/de/posts/aoc2024-day01/Sat, 07 Dec 2024 00:23:00 +0100http://localhost:1313/de/posts/aoc2024-day01/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code 2024</strong> -in der auf Statistik fokussierten Programmiersprache <strong>R</strong>. Dies ist Tag 1 - siehe - - <a href="https:adventofcode.com/2024/day/1">https:adventofcode.com/2024/day/1</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 7: Kamelkartenhttp://localhost:1313/de/posts/aoc2023-day07/Thu, 04 Jan 2024 16:33:00 +0100http://localhost:1313/de/posts/aoc2023-day07/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 7 - siehe - <a href="https:adventofcode.com/2023/day/7">https:adventofcode.com/2023/day/7</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 6: Warte Daraufhttp://localhost:1313/de/posts/aoc2023-day06/Thu, 04 Jan 2024 00:45:00 +0100http://localhost:1313/de/posts/aoc2023-day06/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 6 - siehe - <a href="https:adventofcode.com/2023/day/6">https:adventofcode.com/2023/day/6</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 5: Wenn Du Einem Samen Dünger Gibsthttp://localhost:1313/de/posts/aoc2023-day05/Sat, 30 Dec 2023 01:10:00 +0100http://localhost:1313/de/posts/aoc2023-day05/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 5 - siehe - <a href="https:adventofcode.com/2023/day/5">https:adventofcode.com/2023/day/5</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 4: Rubbelkartenhttp://localhost:1313/de/posts/aoc2023-day04/Tue, 19 Dec 2023 00:25:00 +0100http://localhost:1313/de/posts/aoc2023-day04/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 4 - siehe - <a href="https:adventofcode.com/2023/day/4">https:adventofcode.com/2023/day/4</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 3: Getriebeübersetzungenhttp://localhost:1313/de/posts/aoc2023-day03/Mon, 18 Dec 2023 01:01:00 +0100http://localhost:1313/de/posts/aoc2023-day03/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 3 - siehe - <a href="https:adventofcode.com/2023/day/3">https:adventofcode.com/2023/day/3</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 2: Würfelrätselhttp://localhost:1313/de/posts/aoc2023-day02/Mon, 11 Dec 2023 23:38:00 +0100http://localhost:1313/de/posts/aoc2023-day02/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 2 - siehe - <a href="https:adventofcode.com/2023/day/2">https:adventofcode.com/2023/day/2</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 1: Trebuchet?!http://localhost:1313/de/posts/aoc2023-day01/Sun, 10 Dec 2023 23:38:00 +0100http://localhost:1313/de/posts/aoc2023-day01/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 1 - siehe - <a href="https:adventofcode.com/2023/day/1">https:adventofcode.com/2023/day/1</a></p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/posts/page/1/index.html b/aseltmann.github.io-hugo/public/de/posts/page/1/index.html deleted file mode 100644 index fec448c..0000000 --- a/aseltmann.github.io-hugo/public/de/posts/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/de/posts/ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/sitemap.xml b/aseltmann.github.io-hugo/public/de/sitemap.xml deleted file mode 100644 index 9dda71a..0000000 --- a/aseltmann.github.io-hugo/public/de/sitemap.xml +++ /dev/null @@ -1 +0,0 @@ -http://localhost:1313/de/docs/2024-12-22T01:44:37+01:00http://localhost:1313/de/docs/notes/poetry/2024-12-22T01:44:37+01:00http://localhost:1313/de/docs/notes/2024-12-22T01:44:37+01:00http://localhost:1313/de/2024-12-22T01:44:35+01:00http://localhost:1313/de/posts/2024-12-22T01:46:51+01:00http://localhost:1313/de/posts/aoc2024-day02/2024-12-22T01:46:51+01:00http://localhost:1313/de/tags/R/2024-12-22T01:46:51+01:00http://localhost:1313/de/tags/2024-12-22T01:46:51+01:00http://localhost:1313/de/posts/aoc2024-day01/2024-12-22T01:44:41+01:00http://localhost:1313/de/posts/aoc2023-day07/2024-12-22T01:44:40+01:00http://localhost:1313/de/tags/python/2024-12-22T01:44:40+01:00http://localhost:1313/de/posts/aoc2023-day06/2024-12-22T01:44:40+01:00http://localhost:1313/de/docs/projects/2023-12-31T01:20:57+01:00http://localhost:1313/de/posts/aoc2023-day05/2024-12-22T01:44:40+01:00http://localhost:1313/de/posts/aoc2023-day04/2024-12-22T01:44:39+01:00http://localhost:1313/de/posts/aoc2023-day03/2024-12-22T01:44:39+01:00http://localhost:1313/de/posts/aoc2023-day02/2024-12-22T01:44:38+01:00http://localhost:1313/de/posts/aoc2023-day01/2024-12-22T01:44:38+01:00http://localhost:1313/de/categories/ \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/R/index.html b/aseltmann.github.io-hugo/public/de/tags/R/index.html deleted file mode 100644 index dd0e292..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/R/index.html +++ /dev/null @@ -1,21 +0,0 @@ -R | Blog & Notes - -

    R

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/R/index.xml b/aseltmann.github.io-hugo/public/de/tags/R/index.xml deleted file mode 100644 index 751c841..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/R/index.xml +++ /dev/null @@ -1,9 +0,0 @@ -R on Blog & Noteshttp://localhost:1313/de/tags/R/Recent content in R on Blog & NotesHugodeSun, 22 Dec 2024 01:46:51 +0100Advent of code 2024 - Tag 2: Rotnasige Berichtehttp://localhost:1313/de/posts/aoc2024-day02/Tue, 10 Dec 2024 00:23:00 +0100http://localhost:1313/de/posts/aoc2024-day02/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code 2024</strong> -in der auf Statistik fokussierten Programmiersprache <strong>R</strong>. Dies ist Tag 2 - siehe - - <a href="https:adventofcode.com/2024/day/2">https:adventofcode.com/2024/day/2</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2024 - Tag 1: Hysterie der Historiker:innenhttp://localhost:1313/de/posts/aoc2024-day01/Sat, 07 Dec 2024 00:23:00 +0100http://localhost:1313/de/posts/aoc2024-day01/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code 2024</strong> -in der auf Statistik fokussierten Programmiersprache <strong>R</strong>. Dies ist Tag 1 - siehe - - <a href="https:adventofcode.com/2024/day/1">https:adventofcode.com/2024/day/1</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/R/page/1/index.html b/aseltmann.github.io-hugo/public/de/tags/R/page/1/index.html deleted file mode 100644 index b4417ae..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/R/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/de/tags/R/ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/index.html b/aseltmann.github.io-hugo/public/de/tags/index.html deleted file mode 100644 index afb6e55..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/index.html +++ /dev/null @@ -1,17 +0,0 @@ -Tags | Blog & Notes - -

    Tags

    R

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/index.xml b/aseltmann.github.io-hugo/public/de/tags/index.xml deleted file mode 100644 index 25c5def..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/index.xml +++ /dev/null @@ -1 +0,0 @@ -Tags on Blog & Noteshttp://localhost:1313/de/tags/Recent content in Tags on Blog & NotesHugodeSun, 22 Dec 2024 01:46:51 +0100Rhttp://localhost:1313/de/tags/R/Tue, 10 Dec 2024 00:23:00 +0100http://localhost:1313/de/tags/R/Pythonhttp://localhost:1313/de/tags/python/Thu, 04 Jan 2024 16:33:00 +0100http://localhost:1313/de/tags/python/ \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/page/1/index.html b/aseltmann.github.io-hugo/public/de/tags/page/1/index.html deleted file mode 100644 index 75d85e6..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/de/tags/ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/python/index.html b/aseltmann.github.io-hugo/public/de/tags/python/index.html deleted file mode 100644 index 3626ffe..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/python/index.html +++ /dev/null @@ -1,36 +0,0 @@ -Python | Blog & Notes - -

    Python

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/python/index.xml b/aseltmann.github.io-hugo/public/de/tags/python/index.xml deleted file mode 100644 index b52c526..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/python/index.xml +++ /dev/null @@ -1,21 +0,0 @@ -Python on Blog & Noteshttp://localhost:1313/de/tags/python/Recent content in Python on Blog & NotesHugodeSun, 22 Dec 2024 01:44:40 +0100Advent of code 2023 - Tag 7: Kamelkartenhttp://localhost:1313/de/posts/aoc2023-day07/Thu, 04 Jan 2024 16:33:00 +0100http://localhost:1313/de/posts/aoc2023-day07/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 7 - siehe - <a href="https:adventofcode.com/2023/day/7">https:adventofcode.com/2023/day/7</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 6: Warte Daraufhttp://localhost:1313/de/posts/aoc2023-day06/Thu, 04 Jan 2024 00:45:00 +0100http://localhost:1313/de/posts/aoc2023-day06/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 6 - siehe - <a href="https:adventofcode.com/2023/day/6">https:adventofcode.com/2023/day/6</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 5: Wenn Du Einem Samen Dünger Gibsthttp://localhost:1313/de/posts/aoc2023-day05/Sat, 30 Dec 2023 01:10:00 +0100http://localhost:1313/de/posts/aoc2023-day05/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 5 - siehe - <a href="https:adventofcode.com/2023/day/5">https:adventofcode.com/2023/day/5</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 4: Rubbelkartenhttp://localhost:1313/de/posts/aoc2023-day04/Tue, 19 Dec 2023 00:25:00 +0100http://localhost:1313/de/posts/aoc2023-day04/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 4 - siehe - <a href="https:adventofcode.com/2023/day/4">https:adventofcode.com/2023/day/4</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 3: Getriebeübersetzungenhttp://localhost:1313/de/posts/aoc2023-day03/Mon, 18 Dec 2023 01:01:00 +0100http://localhost:1313/de/posts/aoc2023-day03/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 3 - siehe - <a href="https:adventofcode.com/2023/day/3">https:adventofcode.com/2023/day/3</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 2: Würfelrätselhttp://localhost:1313/de/posts/aoc2023-day02/Mon, 11 Dec 2023 23:38:00 +0100http://localhost:1313/de/posts/aoc2023-day02/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 2 - siehe - <a href="https:adventofcode.com/2023/day/2">https:adventofcode.com/2023/day/2</a></p> -<p>Für die Lösungen siehe die englische Version dieses Blogbeitrages.</p>Advent of code 2023 - Tag 1: Trebuchet?!http://localhost:1313/de/posts/aoc2023-day01/Sun, 10 Dec 2023 23:38:00 +0100http://localhost:1313/de/posts/aoc2023-day01/<p>Dieses Jahr versuche ich mich an den Herausforderungen des <strong>Advent of Code -2023</strong>. Dies ist Tag 1 - siehe - <a href="https:adventofcode.com/2023/day/1">https:adventofcode.com/2023/day/1</a></p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/de/tags/python/page/1/index.html b/aseltmann.github.io-hugo/public/de/tags/python/page/1/index.html deleted file mode 100644 index 6a54d3f..0000000 --- a/aseltmann.github.io-hugo/public/de/tags/python/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/de/tags/python/ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/docs/index.html b/aseltmann.github.io-hugo/public/docs/index.html deleted file mode 100644 index ef0340f..0000000 --- a/aseltmann.github.io-hugo/public/docs/index.html +++ /dev/null @@ -1,11 +0,0 @@ -Docs | Blog & Notes - -

    Docs

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/docs/index.xml b/aseltmann.github.io-hugo/public/docs/index.xml deleted file mode 100644 index 6256831..0000000 --- a/aseltmann.github.io-hugo/public/docs/index.xml +++ /dev/null @@ -1 +0,0 @@ -Docs on Blog & Noteshttp://localhost:1313/docs/Recent content in Docs on Blog & NotesHugoen \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/docs/notes/index.html b/aseltmann.github.io-hugo/public/docs/notes/index.html deleted file mode 100644 index a43b6c8..0000000 --- a/aseltmann.github.io-hugo/public/docs/notes/index.html +++ /dev/null @@ -1,13 +0,0 @@ -Notes | Blog & Notes - -

    Notes

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/docs/notes/index.xml b/aseltmann.github.io-hugo/public/docs/notes/index.xml deleted file mode 100644 index b5ddaec..0000000 --- a/aseltmann.github.io-hugo/public/docs/notes/index.xml +++ /dev/null @@ -1,29 +0,0 @@ -Notes on Blog & Noteshttp://localhost:1313/docs/notes/Recent content in Notes on Blog & NotesHugoenSun, 22 Dec 2024 01:44:35 +0100Poetryhttp://localhost:1313/docs/notes/poetry/Sun, 22 Dec 2024 01:44:35 +0100http://localhost:1313/docs/notes/poetry/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#l%c7%8eoz%c7%90">Lǎozǐ</a></li> -</ul> -</div> -<!--endtoc--> -<h2 id="lǎozǐ"> - Lǎozǐ - <a class="anchor" href="#l%c7%8eoz%c7%90">#</a> -</h2> -<p>I am partial to the - <a href="https://terebess.hu/english/tao/DerekLin.html#Kap20">translation of Derek Lin</a></p> -<details> -<summary>Dàodéjīng - Chapter 11 (The use of what has no substantive existence)</summary> -<div class="details"> -<div class="verse"> -<p>[&hellip;]<br /> -<br /> -Mix clay<br /> -to create a container<br /> -In its emptiness, there is<br /> -the function of a container<br /> -<br /> -[&hellip;]<br /> -Therefore,<br /> -that which exists is used to create benefit<br /> -That which is empty is used to create functionality<br /></p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/docs/notes/poetry/index.html b/aseltmann.github.io-hugo/public/docs/notes/poetry/index.html deleted file mode 100644 index 0bae67e..0000000 --- a/aseltmann.github.io-hugo/public/docs/notes/poetry/index.html +++ /dev/null @@ -1,52 +0,0 @@ -Poetry | Blog & Notes - -

    Poetry

    Table of Contents

    Lǎozǐ -#

    I am partial to the -translation of Derek Lin

    Dàodéjīng - Chapter 11 (The use of what has no substantive existence)

    […]

    Mix clay
    to create a container
    In its emptiness, there is
    the function of a container

    […]
    Therefore,
    that which exists is used to create benefit
    That which is empty is used to create functionality

    Dàodéjīng - Chapter 15 (The exhibition of the qualities of the Dao)

    The Tao masters of antiquity
    Subtle wonders through mystery
    Depths that cannot be discerned
    Because one cannot discern them
    Therefore one is forced to describe the appearance

    Hesitant,
    like crossing a wintry river
    Cautious,
    like fearing four neighbors
    Solemn,
    like a guest
    Loose,
    like ice about to melt
    Genuine,
    like plain wood
    Open,
    like a valley
    Opaque,
    like muddy water

    Who can be muddled yet desist
    In stillness gradually become clear?
    Who can be serene yet persist
    In motion gradually come alive?

    One who holds this <i>Tao</i> does not wish to be overfilled
    Because one is not overfilled
    Therefore one can preserve and not create anew

    Dàodéjīng - Chapter 20 (Being different from ordinary people)

    Cease learning, no more worries

    Respectful response and scornful response
    How much is the difference?
    Goodness and evil
    How much do they differ?
    What the people fear,
    I cannot be unafraid
    So desolate! How limitless it is!

    The people are excited
    As if enjoying a great feast
    As if climbing up to the terrace in spring
    I alone am quiet and uninvolved
    Like an infant
    not yet smiling
    So weary,
    like having no place to return
    The people all have surplus
    While I alone seem lacking
    I have the heart of a fool indeed -
    so ignorant!
    Ordinary people are bright
    I alone am muddled
    Ordinary people are scrutinizing
    I alone am obtuse
    So tranquil, like the ocean
    So moving, as if without limits

    The people all have goals
    And I alone am stubborn and lowly
    I alone am different from them
    And value the nourishing mother

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    Projects

    Current projects -#

    I am a believer in the principles of Open Science, Open Data and Open -Source. Thus, I am currently working on a reproducible way of conducting my -research on simulation of Fluorescence Correlation Spectroscopy (FCS) -measurements and correcting artifacts using neural networks, such as -Convolutional Neural Nets (CNNs). I do not have my workflow fixed yet, but -to have an insight in my current approach - and with the principles of -Open-notebook science in mind:

    Fluotracify - doctoral research project done in a reproducible way

    Description: In a current project, we apply Deep Learning techniques on Fluorescence -Correlation Spectroscopy (FCS) data to correct a variety of hardware- and -sample-related artifacts, such as photobleaching, contamination from additional -slow moving particles, or sudden drops in intensity because of detector -anomalies.

    Conference talks -#

    • Seltmann A, Eggeling C, Waithe D. Automated, User-independent Correction of -Artifacts in Fluorescence Correlation Spectroscopy Measurements using -Convolutional Neural Networks. -Quantitative BioImaging Conference (QBI); 2020 -Jan 6-9; Oxford, UK
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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/docs/projects/index.xml b/aseltmann.github.io-hugo/public/docs/projects/index.xml deleted file mode 100644 index 8760f60..0000000 --- a/aseltmann.github.io-hugo/public/docs/projects/index.xml +++ /dev/null @@ -1 +0,0 @@ -Projects on Blog & Noteshttp://localhost:1313/docs/projects/Recent content in Projects on Blog & NotesHugoen \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/en.search-data.min.4f53cda18c2baa0c0354bb5f9a3ecbe5ed12ab4d8e11ba873c2f11161202b945.json b/aseltmann.github.io-hugo/public/en.search-data.min.4f53cda18c2baa0c0354bb5f9a3ecbe5ed12ab4d8e11ba873c2f11161202b945.json deleted file mode 100644 index 0637a08..0000000 --- a/aseltmann.github.io-hugo/public/en.search-data.min.4f53cda18c2baa0c0354bb5f9a3ecbe5ed12ab4d8e11ba873c2f11161202b945.json +++ /dev/null @@ -1 +0,0 @@ -[] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/en.search-data.min.83957fc8af013f6c1375146741f6476a2787dd36aa372c9f33e7197b01c6f9e2.json b/aseltmann.github.io-hugo/public/en.search-data.min.83957fc8af013f6c1375146741f6476a2787dd36aa372c9f33e7197b01c6f9e2.json deleted file mode 100644 index f930dbf..0000000 --- a/aseltmann.github.io-hugo/public/en.search-data.min.83957fc8af013f6c1375146741f6476a2787dd36aa372c9f33e7197b01c6f9e2.json +++ /dev/null @@ -1 +0,0 @@ -[{"id":0,"href":"/docs/notes/poetry/","title":"Poetry","section":"Notes","content":" Table of Contents Lǎozǐ Lǎozǐ # I am partial to the translation of Derek Lin\nDàodéjīng - Chapter 11 (The use of what has no substantive existence) [\u0026hellip;]\nMix clay\nto create a container\nIn its emptiness, there is\nthe function of a container\n[\u0026hellip;]\nTherefore,\nthat which exists is used to create benefit\nThat which is empty is used to create functionality\nDàodéjīng - Chapter 15 (The exhibition of the qualities of the Dao) The Tao masters of antiquity\nSubtle wonders through mystery\nDepths that cannot be discerned\nBecause one cannot discern them\nTherefore one is forced to describe the appearance\nHesitant,\nlike crossing a wintry river\nCautious,\nlike fearing four neighbors\nSolemn,\nlike a guest\nLoose,\nlike ice about to melt\nGenuine,\nlike plain wood\nOpen,\nlike a valley\nOpaque,\nlike muddy water\nWho can be muddled yet desist\nIn stillness gradually become clear?\nWho can be serene yet persist\nIn motion gradually come alive?\nOne who holds this \u0026lt;i\u0026gt;Tao\u0026lt;/i\u0026gt; does not wish to be overfilled\nBecause one is not overfilled\nTherefore one can preserve and not create anew\nDàodéjīng - Chapter 20 (Being different from ordinary people) Cease learning, no more worries\nRespectful response and scornful response\nHow much is the difference?\nGoodness and evil\nHow much do they differ?\nWhat the people fear,\nI cannot be unafraid\nSo desolate! How limitless it is!\nThe people are excited\nAs if enjoying a great feast\nAs if climbing up to the terrace in spring\nI alone am quiet and uninvolved\nLike an infant\nnot yet smiling\nSo weary,\nlike having no place to return\nThe people all have surplus\nWhile I alone seem lacking\nI have the heart of a fool indeed -\nso ignorant!\nOrdinary people are bright\nI alone am muddled\nOrdinary people are scrutinizing\nI alone am obtuse\nSo tranquil, like the ocean\nSo moving, as if without limits\nThe people all have goals\nAnd I alone am stubborn and lowly\nI alone am different from them\nAnd value the nourishing mother\n"},{"id":1,"href":"/posts/aoc2024-day02/","title":"Advent of code 2024 - Day 2: Red-Nosed Reports","section":"Posts","content":" Table of Contents Part 1: Safe reports Part 2: Problem Dampener I try to solve this year\u0026rsquo;s Advent of Code 2024 riddles in R. This is Day 2 - see https:adventofcode.com/2024/day/2\nPart 1: Safe reports # Lets first read the task: The unusual data (your puzzle input) consists of many reports, one report per line. Each report is a list of numbers called levels that are separated by spaces. For example:\n7 6 4 2 1 1 2 7 8 9 9 7 6 2 1 1 3 2 4 5 8 6 4 4 1 1 3 6 7 9 This example data contains six reports each containing five levels.\nThe engineers are trying to figure out which reports are safe. The Red-Nosed reactor safety systems can only tolerate levels that are either gradually increasing or gradually decreasing. So, a report only counts as safe if both of the following are true:\nThe levels are either all increasing or all decreasing. Any two adjacent levels differ by at least one and at most three. In the example above, the reports can be found safe or unsafe by checking those rules:\n7 6 4 2 1: Safe because the levels are all decreasing by 1 or 2. 1 2 7 8 9: Unsafe because 2 7 is an increase of 5. 9 7 6 2 1: Unsafe because 6 2 is a decrease of 4. 1 3 2 4 5: Unsafe because 1 3 is increasing but 3 2 is decreasing. 8 6 4 4 1: Unsafe because 4 4 is neither an increase or a decrease. 1 3 6 7 9: Safe because the levels are all increasing by 1, 2, or 3. So, in this example, 2 reports are safe.\nAnalyze the unusual data from the engineers. How many reports are safe?\nMy input file: 2024-12-02-1-aoc.txt\nFirst, let\u0026rsquo;s load the data and take a look:\ndata \u0026lt;- data.table::fread(\u0026#34;2024-12-02-1-aoc.txt\u0026#34;, header = FALSE, fill = TRUE) head(data) V1 V2 V3 V4 V5 V6 V7 V8 \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; 74 76 78 79 76 NA NA NA 38 40 43 44 44 NA NA NA 1 2 4 6 8 9 13 NA 65 68 70 72 75 76 81 NA 89 91 92 95 93 94 NA NA 15 17 16 18 19 17 NA NA My core idea here was to compute the differences between each element in a row - luckily, R provides an inbuilt diff() function that provides exactly what I need. Then, we check if all differences are either 1, 2, 3 or -1, -2, -3. Note that each row has a different number of elements, so we have to omit NA values via na.omit when applying the check.\nisalltrue \u0026lt;- function(rep) { return(all(diff(rep) %in% c(1, 2, 3)) | all(diff(rep) %in% c(-1, -2, -3))) } isreport \u0026lt;- function(row) { return(isalltrue(na.omit(row))) } valid.reports \u0026lt;- apply(data, 1, isreport) table(valid.reports) valid.reports FALSE TRUE 743 257 Part 2: Problem Dampener # Let\u0026rsquo;s read the task for part 2! The engineers are surprised by the low number of safe reports until they realize they forgot to tell you about the Problem Dampener.\nThe Problem Dampener is a reactor-mounted module that lets the reactor safety systems tolerate a single bad level in what would otherwise be a safe report. It\u0026rsquo;s like the bad level never happened!\nNow, the same rules apply as before, except if removing a single level from an unsafe report would make it safe, the report instead counts as safe.\nMore of the above example\u0026rsquo;s reports are now safe:\n-7 6 4 2 1: Safe without removing any level. -1 2 7 8 9: Unsafe regardless of which level is removed. -9 7 6 2 1: Unsafe regardless of which level is removed. -1 3 2 4 5: Safe by removing the second level, 3. -8 6 4 4 1: Safe by removing the third level, 4. -1 3 6 7 9: Safe without removing any level. Thanks to the Problem Dampener, 4 reports are actually safe!\nUpdate your analysis by handling situations where the Problem Dampener can remove a single level from unsafe reports. How many reports are now safe?\nSo, this task is a good example on how easy it is to get lost in a overly complicated approach. My first approach was basically to somehow check if the general trend was descending or ascending and based on that to remove opposite differences. This was too complicated and in the end I found 4 less safe reports than were needed\u0026hellip; (Also note the beautiful print debuggin\u0026hellip;)\nisalltrue \u0026lt;- function(reptrue1, reptrue2) { return((all(reptrue1) == TRUE) | (all(reptrue2) == TRUE)) } iscorrtrue \u0026lt;- function(rep, reptrue) { rep \u0026lt;- rep[-which.min(reptrue)] repdiff \u0026lt;- diff(rep) reptrue1 \u0026lt;- !(repdiff \u0026lt; 1 | repdiff \u0026gt; 3) reptrue2 \u0026lt;- !(repdiff \u0026lt; -3 | repdiff \u0026gt; -1) out \u0026lt;- isalltrue(reptrue1, reptrue2) print(rep) print(out) return(out) } isreport2 \u0026lt;- function(row) { print(\u0026#34;\u0026#34;) rep \u0026lt;- na.omit(as.numeric(row)) print(c(rep)) repdiff \u0026lt;- c(1, diff(rep)) reptrue1 \u0026lt;- !(repdiff \u0026lt; 1 | repdiff \u0026gt; 3) repdiff \u0026lt;- c(-1, diff(rep)) reptrue2 \u0026lt;- !(repdiff \u0026lt; -3 | repdiff \u0026gt; -1) if (isalltrue(reptrue1, reptrue2) == TRUE) { out \u0026lt;- isalltrue(reptrue1, reptrue2) print(\u0026#34;----- 1 -----\u0026#34;) print(out) } else if (sum(diff(rep) \u0026gt; 0) \u0026gt; sum(diff(rep) \u0026lt; 0)) { print(\u0026#34;----- 2 -----\u0026#34;) out \u0026lt;- iscorrtrue(rep, reptrue1) } else if (sum(diff(rep) \u0026lt; 0) \u0026gt; sum(diff(rep) \u0026gt; 0)) { print(\u0026#34;----- 3 -----\u0026#34;) out \u0026lt;- iscorrtrue(rep, reptrue2) } else { out \u0026lt;- FALSE } if (out == FALSE) { print(\u0026#34;----- 4 -----\u0026#34;) rep \u0026lt;- na.omit(as.numeric(row))[-1] repdiff \u0026lt;- diff(rep) reptrue1 \u0026lt;- !(repdiff \u0026lt; 1 | repdiff \u0026gt; 3) reptrue2 \u0026lt;- !(repdiff \u0026lt; -3 | repdiff \u0026gt; -1) out \u0026lt;- isalltrue(reptrue1, reptrue2) print(c(rep)) print(out) } return(out) } My actual working approach was in the end brute force. I just threw out each element in a row and checked if the report became positive - much simpler and effective!\nisreport2 \u0026lt;- function(row) { rep \u0026lt;- na.omit(row) out \u0026lt;- isalltrue(rep) if (out == FALSE) { for (i in seq_along(rep)) { # rep \u0026lt;- rep[-i] does not update the variable # with rep.i, a new variable is assigned (rep.1, rep.2, ...) rep.i \u0026lt;- rep[-i] out \u0026lt;- isalltrue(rep.i) if (out == TRUE) { break } } } return(out) } sum(apply(data, 1, isreport2)) 328\n"},{"id":2,"href":"/posts/aoc2024-day01/","title":"Advent of code 2024 - Day 1: Historian Hysteria","section":"Posts","content":" Table of Contents Part 1: Locations Part 2: Similarities I try to solve this year\u0026rsquo;s Advent of Code 2024 riddles in R. This is Day 1 - see https:adventofcode.com/2024/day/1\nPart 1: Locations # Lets first read the task: Throughout the Chief\u0026rsquo;s office, the historically significant locations are listed not by name but by a unique number called the location ID. To make sure they don\u0026rsquo;t miss anything, The Historians split into two groups, each searching the office and trying to create their own complete list of location IDs.\nThere\u0026rsquo;s just one problem: by holding the two lists up side by side (your puzzle input), it quickly becomes clear that the lists aren\u0026rsquo;t very similar. Maybe you can help The Historians reconcile their lists?\nFor example:\n3 4 4 3 2 5 1 3 3 9 3 3 Maybe the lists are only off by a small amount! To find out, pair up the numbers and measure how far apart they are. Pair up the smallest number in the left list with the smallest number in the right list, then the second-smallest left number with the second-smallest right number, and so on.\nWithin each pair, figure out how far apart the two numbers are; you\u0026rsquo;ll need to add up all of those distances. For example, if you pair up a 3 from the left list with a 7 from the right list, the distance apart is 4; if you pair up a 9 with a 3, the distance apart is 6.\nIn the example list above, the pairs and distances would be as follows:\nThe smallest number in the left list is 1, and the smallest number in the right list is 3. The distance between them is 2. The second-smallest number in the left list is 2, and the second-smallest number in the right list is another 3. The distance between them is 1. The third-smallest number in both lists is 3, so the distance between them is 0. The next numbers to pair up are 3 and 4, a distance of 1. The fifth-smallest numbers in each list are 3 and 5, a distance of 2. Finally, the largest number in the left list is 4, while the largest number in the right list is 9; these are a distance 5 apart. To find the total distance between the left list and the right list, add up the distances between all of the pairs you found. In the example above, this is 2 + 1 + 0 + 1 + 2 + 5, a total distance of 11!\nYour actual left and right lists contain many location IDs. What is the total distance between your lists?\nMy input file: 2024-12-01-1-aoc.txt\nFirst, we read the data - the fread command from the data.table package is more versatile than read.delim from base and directly reads the data as a data.table, which has some benefits.\ndata \u0026lt;- data.table::fread(\u0026#34;2024-12-01-1-aoc.txt\u0026#34;, header = FALSE) head(data) V1 V2 \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; 41226 69190 89318 10100 59419 23880 63157 20193 81510 22869 83942 63304 Now, the idea is to get the ordered versions of V1 and V2.\nv1 \u0026lt;- data[order(V1)]$V1 v2 \u0026lt;- data[order(V2)]$V2 head(v1) 10188 10314 10319 10348 10408 10668 Coming from Python, this syntax looks a bit odd. The documentation of data.table::setorder helps:\nsetorder (and setorderv) reorders the rows of a data.table based on the columns (and column order) provided. It reorders the table by reference and is therefore very memory efficient. Note that queries like x[order(.)] are optimised internally to use data.table\u0026#39;s fast order. So, data[order(V1)] is actually short for data.table::setorder(data, V1). Then, we extract the vector by name using the $ operator, which allows to extract elements by name.\nhead(data.table::setorder(data, V1)$V1) 10188 10314 10319 10348 10408 10668 The actual computation is just the sum of the absolute difference:\nsum(abs(v1 - v2)) 3574690\nPart 2: Similarities # Lets first read the task: This time, you\u0026rsquo;ll need to figure out exactly how often each number from the left list appears in the right list. Calculate a total similarity score by adding up each number in the left list after multiplying it by the number of times that number appears in the right list.\nHere are the same example lists again:\n3 4 4 3 2 5 1 3 3 9 3 3 For these example lists, here is the process of finding the similarity score:\nThe first number in the left list is 3. It appears in the right list three times, so the similarity score increases by 3 * 3 = 9. The second number in the left list is 4. It appears in the right list once, so the similarity score increases by 4 * 1 = 4. The third number in the left list is 2. It does not appear in the right list, so the similarity score does not increase (2 * 0 = 0). The fourth number, 1, also does not appear in the right list. The fifth number, 3, appears in the right list three times; the similarity score increases by 9. The last number, 3, appears in the right list three times; the similarity score again increases by 9. So, for these example lists, the similarity score at the end of this process is 31 (9 + 4 + 0 + 0 + 9 + 9).\nOnce again consider your left and right lists. What is their similarity score?\nMy idea here is simple - we first count the occurrences of V2 to be able to check V1 against them. Here, applying the table command works like pandas.value_counts() and achieves this. We convert the output to a data.frame and assign the V1 values as row names. Note: table converts the values which are counted to string, e.g. \u0026quot;10019\u0026quot; instead of 10019.\ntab \u0026lt;- table(v2) v2series \u0026lt;- data.frame(c(tab), row.names = names(tab)) head(v2series) v2series[\u0026#34;10019\u0026#34;, ] c.tab. \u0026lt;int\u0026gt; 10019 1 10100 1 10206 1 10428 1 10645 1 10972 1 1\nNow, we loop through V1 and try to access the frequency in V2 (by converting to string first, via as.character). If nothing is found in V2, an NA value is returned. For everything that is not NA, we compute the similarity score, append it to a list, and sum in the end. Note: a list in R can not directly be summed (I don\u0026rsquo;t know why that is) - so we have to unlist first.\nsim \u0026lt;- list() for (num in v1) { myfreq \u0026lt;- v2series[as.character(num), ] if (!is.na(myfreq)) { score \u0026lt;- myfreq * num sim \u0026lt;- append(sim, score) } } sum(unlist(sim)) 22565391\n"},{"id":3,"href":"/posts/aoc2023-day07/","title":"Advent of code 2023 - Day 7: Camel Cards","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 7 - see https:adventofcode.com/2023/day/7\nPart 1 # Lets first read the task: In Camel Cards, you get a list of hands, and your goal is to order them based on the strength of each hand. A hand consists of five cards labeled one of A, K, Q, J, T, 9, 8, 7, 6, 5, 4, 3, or 2. The relative strength of each card follows this order, where A is the highest and 2 is the lowest.\nEvery hand is exactly one type. From strongest to weakest, they are:\nFive of a kind, where all five cards have the same label: AAAAA Four of a kind, where four cards have the same label and one card has a different label: AA8AA Full house, where three cards have the same label, and the remaining two cards share a different label: 23332 Three of a kind, where three cards have the same label, and the remaining two cards are each different from any other card in the hand: TTT98 Two pair, where two cards share one label, two other cards share a second label, and the remaining card has a third label: 23432 One pair, where two cards share one label, and the other three cards have a different label from the pair and each other: A23A4 High card, where all cards\u0026rsquo; labels are distinct: 23456 Hands are primarily ordered based on type; for example, every full house is stronger than any three of a kind.\nIf two hands have the same type, a second ordering rule takes effect. Start by comparing the first card in each hand. If these cards are different, the hand with the stronger first card is considered stronger. If the first card in each hand have the same label, however, then move on to considering the second card in each hand. If they differ, the hand with the higher second card wins; otherwise, continue with the third card in each hand, then the fourth, then the fifth.\nSo, 33332 and 2AAAA are both four of a kind hands, but 33332 is stronger because its first card is stronger. Similarly, 77888 and 77788 are both a full house, but 77888 is stronger because its third card is stronger (and both hands have the same first and second card).\nTo play Camel Cards, you are given a list of hands and their corresponding bid (your puzzle input). For example:\n32T3K 765 T55J5 684 KK677 28 KTJJT 220 QQQJA 483 This example shows five hands; each hand is followed by its bid amount. Each hand wins an amount equal to its bid multiplied by its rank, where the weakest hand gets rank 1, the second-weakest hand gets rank 2, and so on up to the strongest hand. Because there are five hands in this example, the strongest hand will have rank 5 and its bid will be multiplied by 5.\nSo, the first step is to put the hands in order of strength:\n32T3K is the only one pair and the other hands are all a stronger type, so it gets rank 1. KK677 and KTJJT are both two pair. Their first cards both have the same label, but the second card of KK677 is stronger (K vs T), so KTJJT gets rank 2 and KK677 gets rank 3. T55J5 and QQQJA are both three of a kind. QQQJA has a stronger first card, so it gets rank 5 and T55J5 gets rank 4. Now, you can determine the total winnings of this set of hands by adding up the result of multiplying each hand\u0026rsquo;s bid with its rank (765 * 1 + 220 * 2 + 28 * 3 + 684 * 4 + 483 * 5). So the total winnings in this example are 6440.\nFind the rank of every hand in your set. What are the total winnings?\nMy input file: 2023-12-07-1-aoc.txt\nAs always - let\u0026rsquo;s read in the data.\nimport pandas as pd camel = pd.read_table(\u0026#39;data/2023-12-07-1-aoc.txt\u0026#39;, delim_whitespace=True, header=None, names=[\u0026#39;cards\u0026#39;, \u0026#39;bid\u0026#39;], dtype={\u0026#39;cards\u0026#39;: \u0026#39;string\u0026#39;, \u0026#39;bid\u0026#39;: \u0026#39;Int64\u0026#39;}) camel cards bid 0 A833A 309 1 Q33J3 205 2 55KK5 590 3 K4457 924 4 Q3QT3 11 \u0026hellip; \u0026hellip; \u0026hellip; 995 KQJ53 367 996 57866 537 997 Q94A9 210 998 J448A 903 999 6J66J 114 1000 rows × 2 columns\nFor sorting, we need to compute a type column. The 7 types can be deduced based on how many different cards are there (through set()) and the number of cards.\nAlso, for the next step we prepare one column per card string in the cards.\ndef get_type(cards): cardset = set(cards) cardlen = len(cardset) cardlist = list(cards) cardnum = [cardlist.count(i) for i in cardset] if cardlen == 1: ctype = 1 elif cardlen in (4, 5): ctype = cardlen + 2 elif cardlen == 2: ctype = 2 if set(cardnum) == {1, 4} else 3 else: ctype = 4 if 3 in cardnum else 5 return ctype camel[\u0026#39;type\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: get_type(x)) for i in range(5): camel[f\u0026#39;c{i}\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: list(x)[i]) camel cards bid type c0 c1 c2 c3 c4 0 A833A 309 5 A 8 3 3 A 1 Q33J3 205 4 Q 3 3 J 3 2 55KK5 590 3 5 5 K K 5 3 K4457 924 6 K 4 4 5 7 4 Q3QT3 11 5 Q 3 Q T 3 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 KQJ53 367 7 K Q J 5 3 996 57866 537 6 5 7 8 6 6 997 Q94A9 210 6 Q 9 4 A 9 998 J448A 903 6 J 4 4 8 A 999 6J66J 114 3 6 J 6 6 J 1000 rows × 8 columns\nFor sorting, pandas contributes the sort_values method, which supports sorting by multiple columns (here, we first sort according to type, then c0, c1, \u0026hellip;). For easier sorting by numerical values, we convert the values using the card_dict dictionary, which implements the given sorting order.\ncard_dict = {c: i for i, c in enumerate([7, 6, 5, 4, 3, 2, 1, \u0026#39;2\u0026#39;, \u0026#39;3\u0026#39;, \u0026#39;4\u0026#39;, \u0026#39;5\u0026#39;, \u0026#39;6\u0026#39;, \u0026#39;7\u0026#39;, \u0026#39;8\u0026#39;, \u0026#39;9\u0026#39;, \u0026#39;T\u0026#39;, \u0026#39;J\u0026#39;, \u0026#39;Q\u0026#39;, \u0026#39;K\u0026#39;, \u0026#39;A\u0026#39;])} camel = camel.sort_values(by=[\u0026#39;type\u0026#39;, \u0026#39;c0\u0026#39;, \u0026#39;c1\u0026#39;, \u0026#39;c2\u0026#39;, \u0026#39;c3\u0026#39;, \u0026#39;c4\u0026#39;], ignore_index=True, key=lambda x: x.map(card_dict)) camel.index += 1 camel cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 AA5AA 565 2 A A 5 A A 997 AAATA 704 2 A A A T A 998 AAAA3 145 2 A A A A 3 999 AAAAJ 594 2 A A A A J 1000 JJJJJ 219 1 J J J J J 1000 rows × 8 columns\nLastly, we multiply the bid with the index and sum the result.\ncamel.loc[:, \u0026#39;bid\u0026#39;].mul(camel.index).sum() 246912307 Part 2 # Lets first read the task: To make things a little more interesting, the Elf introduces one additional rule. Now, J cards are jokers - wildcards that can act like whatever card would make the hand the strongest type possible.\nTo balance this, J cards are now the weakest individual cards, weaker even than 2. The other cards stay in the same order: A, K, Q, T, 9, 8, 7, 6, 5, 4, 3, 2, J.\nJ cards can pretend to be whatever card is best for the purpose of determining hand type; for example, QJJQ2 is now considered four of a kind. However, for the purpose of breaking ties between two hands of the same type, J is always treated as J, not the card it\u0026rsquo;s pretending to be: JKKK2 is weaker than QQQQ2 because J is weaker than Q.\nNow, the above example goes very differently:\n32T3K 765 T55J5 684 KK677 28 KTJJT 220 QQQJA 483 32T3K is still the only one pair; it doesn\u0026rsquo;t contain any jokers, so its strength doesn\u0026rsquo;t increase. KK677 is now the only two pair, making it the second-weakest hand. T55J5, KTJJT, and QQQJA are now all four of a kind! T55J5 gets rank 3, QQQJA gets rank 4, and KTJJT gets rank 5. With the new joker rule, the total winnings in this example are 5905.\nUsing the new joker rule, find the rank of every hand in your set. What are the new total winnings?\nWhat changes? We have to update our get_type() function and our card_dict dictionary, otherwise the method stays the same!\nLet\u0026rsquo;s first write get_type_joker(): if no joker is in the cards, we can just refer to get_type(). If we have 1, 2 or 5 different card faces, the answer is trivial. We only have to use the number of jokers and the maximum number of other faces if we have 3 or 4 different faces.\nNote: if a J is present, there can never be high card (type = 7), because we always can get one pair. Also we can never get two pair (type = 5), because we can always construct three of a kind or full house.\ndef get_type_joker(cards): cardset = set(cards) cardlen = len(cardset) cardlist = list(cards) cardnum = [cardlist.count(i) for i in cardset] if not \u0026#39;J\u0026#39; in cardset: ctype = get_type(cards) else: idxj = list(cardset).index(\u0026#39;J\u0026#39;) numj = cardnum[idxj] cardnum.pop(idxj) if cardlen in [1, 2]: ctype = 1 elif cardlen == 5: ctype = 6 else: if (numj in [1, 2]) and (cardlen == 4): ctype = 4 else: ctype = 6 - max(cardnum) - numj return ctype camel[\u0026#39;type\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: get_type_joker(x)) camel cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 AA5AA 565 2 A A 5 A A 997 AAATA 704 2 A A A T A 998 AAAA3 145 2 A A A A 3 999 AAAAJ 594 1 A A A A J 1000 JJJJJ 219 1 J J J J J 1000 rows × 8 columns\nNow, we just move J in our new card_dict_joker to where it belongs, the rest of the solution is the same.\ncard_dict_joker = {c: i for i, c in enumerate( [7, 6, 5, 4, 3, 2, 1, \u0026#39;J\u0026#39;, \u0026#39;2\u0026#39;, \u0026#39;3\u0026#39;, \u0026#39;4\u0026#39;, \u0026#39;5\u0026#39;, \u0026#39;6\u0026#39;, \u0026#39;7\u0026#39;, \u0026#39;8\u0026#39;, \u0026#39;9\u0026#39;, \u0026#39;T\u0026#39;, \u0026#39;Q\u0026#39;, \u0026#39;K\u0026#39;, \u0026#39;A\u0026#39;])} camel = camel.sort_values(by=[\u0026#39;type\u0026#39;, \u0026#39;c0\u0026#39;, \u0026#39;c1\u0026#39;, \u0026#39;c2\u0026#39;, \u0026#39;c3\u0026#39;, \u0026#39;c4\u0026#39;], ignore_index=True, key=lambda x: x.map(card_dict_joker)) camel.index += 1 display(camel) print(\u0026#39;Our new solution: \u0026#39;, camel.loc[:, \u0026#39;bid\u0026#39;].mul(camel.index).sum()) cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 TJTTJ 609 1 T J T T J 997 QQJQJ 131 1 Q Q J Q J 998 QQQJQ 831 1 Q Q Q J Q 999 KKKJK 183 1 K K K J K 1000 AAAAJ 594 1 A A A A J 1000 rows × 8 columns\nOur new solution: 246894760 "},{"id":4,"href":"/posts/aoc2023-day06/","title":"Advent of code 2023 - Day 6: Wait For It","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 6 - see https:adventofcode.com/2023/day/6\nPart 1 # Lets first read the task: You manage to sign up as a competitor in the boat races just in time. The organizer explains that it\u0026rsquo;s not really a traditional race - instead, you will get a fixed amount of time during which your boat has to travel as far as it can, and you win if your boat goes the farthest.\nAs part of signing up, you get a sheet of paper (your puzzle input) that lists the time allowed for each race and also the best distance ever recorded in that race. To guarantee you win the grand prize, you need to make sure you go farther in each race than the current record holder.\nThe organizer brings you over to the area where the boat races are held. The boats are much smaller than you expected - they\u0026rsquo;re actually toy boats, each with a big button on top. Holding down the button charges the boat, and releasing the button allows the boat to move. Boats move faster if their button was held longer, but time spent holding the button counts against the total race time. You can only hold the button at the start of the race, and boats don\u0026rsquo;t move until the button is released.\nFor example:\nTime: 7 15 30 Distance: 9 40 200 This document describes three races:\nThe first race lasts 7 milliseconds. The record distance in this race is 9 millimeters. The second race lasts 15 milliseconds. The record distance in this race is 40 millimeters. The third race lasts 30 milliseconds. The record distance in this race is 200 millimeters. Your toy boat has a starting speed of zero millimeters per millisecond. For each whole millisecond you spend at the beginning of the race holding down the button, the boat\u0026rsquo;s speed increases by one millimeter per millisecond.\nSo, because the first race lasts 7 milliseconds, you only have a few options:\nDon\u0026rsquo;t hold the button at all (that is, hold it for 0 milliseconds) at the start of the race. The boat won\u0026rsquo;t move; it will have traveled 0 millimeters by the end of the race. Hold the button for 1 millisecond at the start of the race. Then, the boat will travel at a speed of 1 millimeter per millisecond for 6 milliseconds, reaching a total distance traveled of 6 millimeters. Hold the button for 2 milliseconds, giving the boat a speed of 2 millimeters per millisecond. It will then get 5 milliseconds to move, reaching a total distance of 10 millimeters. Hold the button for 3 milliseconds. After its remaining 4 milliseconds of travel time, the boat will have gone 12 millimeters. Hold the button for 4 milliseconds. After its remaining 3 milliseconds of travel time, the boat will have gone 12 millimeters. Hold the button for 5 milliseconds, causing the boat to travel a total of 10 millimeters. Hold the button for 6 milliseconds, causing the boat to travel a total of 6 millimeters. Hold the button for 7 milliseconds. That\u0026rsquo;s the entire duration of the race. You never let go of the button. The boat can\u0026rsquo;t move until you let go of the button. Please make sure you let go of the button so the boat gets to move. 0 millimeters. Since the current record for this race is 9 millimeters, there are actually 4 different ways you could win: you could hold the button for 2, 3, 4, or 5 milliseconds at the start of the race.\nIn the second race, you could hold the button for at least 4 milliseconds and at most 11 milliseconds and beat the record, a total of 8 different ways to win.\nIn the third race, you could hold the button for at least 11 milliseconds and no more than 19 milliseconds and still beat the record, a total of 9 ways you could win.\nTo see how much margin of error you have, determine the number of ways you can beat the record in each race; in this example, if you multiply these values together, you get 288 (4 * 8 * 9).\nDetermine the number of ways you could beat the record in each race. What do you get if you multiply these numbers together?\nMy input file: 2023-12-06-1-aoc.txt\nThe solution for this problem is shorter than the task description!\nAs always - let\u0026rsquo;s load the data. Note: we use the delim_whitesspace argument instead of sep=' ' to separate values by spaces of any length.\nimport pandas as pd import numpy as np race = pd.read_table(\u0026#39;data/2023-12-06-1-aoc.txt\u0026#39;, delim_whitespace=True, header=None, index_col=0) race.index = [\u0026#39;time\u0026#39;, \u0026#39;distance\u0026#39;] race 1 2 3 4 time 51 92 68 90 distance 222 2031 1126 1225 The we use Python list comprehensions to quickly calculate all distances for the range of the time value (and keep only the winning distances). Then we get the respective lengths and compute the product.\ndef get_distance(time, distance): return [t * (time - t) for t in range(1, time + 1) if (t * (time - t)) \u0026gt; distance] race.loc[\u0026#39;nwins\u0026#39;, :] = [len(get_distance(int(r.time), int(r.distance))) for c, r in race.items()] nwins_prod = np.prod(race.loc[\u0026#39;nwins\u0026#39;, :].values, dtype=int) display(race) print(f\u0026#39;The product of our nwins is {nwins_prod}\u0026#39;) 1 2 3 4 time 51.0 92.0 68.0 90.0 distance 222.0 2031.0 1126.0 1225.0 nwins 42.0 19.0 11.0 57.0 The product of our nwins is 500346 Part 2 # Lets first read the task: As the race is about to start, you realize the piece of paper with race times and record distances you got earlier actually just has very bad kerning. There\u0026rsquo;s really only one race - ignore the spaces between the numbers on each line.\nSo, the example from before:\nTime: 7 15 30 Distance: 9 40 200 \u0026hellip;now instead means this:\nTime: 71530 Distance: 940200 Now, you have to figure out how many ways there are to win this single race. In this example, the race lasts for 71530 milliseconds and the record distance you need to beat is 940200 millimeters. You could hold the button anywhere from 14 to 71516 milliseconds and beat the record, a total of 71503 ways!\nHow many ways can you beat the record in this one much longer race?\nFor this riddle we can take the same approach as for Part 1. The cell took 18s to evaluate on a consumer laptop from 2015 - no fancy workarounds needed :)\nracetime = int(\u0026#39;\u0026#39;.join(str(r) for r in race.loc[\u0026#39;time\u0026#39;, :4].astype(int))) racedist = int(\u0026#39;\u0026#39;.join(str(r) for r in race.loc[\u0026#39;distance\u0026#39;, :4].astype(int))) racewins = len(get_distance(racetime, racedist)) race[5] = [racetime, racedist, racewins] race 1 2 3 4 5 time 51.0 92.0 68.0 90.0 51926890 distance 222.0 2031.0 1126.0 1225.0 222203111261225 nwins 42.0 19.0 11.0 57.0 42515755 "},{"id":5,"href":"/posts/aoc2023-day05/","title":"Advent of code 2023 - Day 5: If You Give A Seed A Fertilizer","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 5 - see https:adventofcode.com/2023/day/5\nUpdate [2023-12-31 So]:\nsubstitute get_generator() (own implementation) with range() (Python inbuilt) improve grid search so that it goes through all location ranges, still starting with the lowest range Part 1 # Lets first read the task: The almanac (your puzzle input) lists all of the seeds that need to be planted. It also lists what type of soil to use with each kind of seed, what type of fertilizer to use with each kind of soil, what type of water to use with each kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is identified with a number, but numbers are reused by each category - that is, soil 123 and fertilizer 123 aren\u0026rsquo;t necessarily related to each other.\nFor example:\nseeds: 79 14 55 13 seed-to-soil map: 50 98 2 52 50 48 soil-to-fertilizer map: 0 15 37 37 52 2 39 0 15 fertilizer-to-water map: 49 53 8 0 11 42 42 0 7 57 7 4 water-to-light map: 88 18 7 18 25 70 light-to-temperature map: 45 77 23 81 45 19 68 64 13 temperature-to-humidity map: 0 69 1 1 0 69 humidity-to-location map: 60 56 37 56 93 4 The almanac starts by listing which seeds need to be planted: seeds 79, 14, 55, and 13.\nThe rest of the almanac contains a list of maps which describe how to convert numbers from a source category into numbers in a destination category. That is, the section that starts with seed-to-soil map: describes how to convert a seed number (the source) to a soil number (the destination). This lets the gardener and his team know which soil to use with which seeds, which water to use with which fertilizer, and so on.\nRather than list every source number and its corresponding destination number one by one, the maps describe entire ranges of numbers that can be converted. Each line within a map contains three numbers: the destination range start, the source range start, and the range length.\nConsider again the example seed-to-soil map:\n50 98 2 52 50 48\nThe first line has a destination range start of 50, a source range start of 98, and a range length of 2. This line means that the source range starts at 98 and contains two values: 98 and 99. The destination range is the same length, but it starts at 50, so its two values are 50 and 51. With this information, you know that seed number 98 corresponds to soil number 50 and that seed number 99 corresponds to soil number 51. o The second line means that the source range starts at 50 and contains 48 values: 50, 51, \u0026hellip;, 96, 97. This corresponds to a destination range starting at 52 and also containing 48 values: 52, 53, \u0026hellip;, 98, 99. So, seed number 53 corresponds to soil number 55.\nAny source numbers that aren\u0026rsquo;t mapped correspond to the same destination number. So, seed number 10 corresponds to soil number 10.\nSo, the entire list of seed numbers and their corresponding soil numbers looks like this:\nseed soil 0 0 1 1 ... ... 48 48 49 49 50 52 51 53 ... ... 96 98 97 99 98 50 99 51 With this map, you can look up the soil number required for each initial seed number:\nSeed number 79 corresponds to soil number 81. Seed number 14 corresponds to soil number 14. Seed number 55 corresponds to soil number 57. Seed number 13 corresponds to soil number 13. The gardener and his team want to get started as soon as possible, so they\u0026rsquo;d like to know the closest location that needs a seed. Using these maps, find the lowest location number that corresponds to any of the initial seeds. To do this, you\u0026rsquo;ll need to convert each seed number through other categories until you can find its corresponding location number. In this example, the corresponding types are:\nSeed 79, soil 81, fertilizer 81, water 81, light 74, temperature 78, humidity 78, location 82. Seed 14, soil 14, fertilizer 53, water 49, light 42, temperature 42, humidity 43, location 43. Seed 55, soil 57, fertilizer 57, water 53, light 46, temperature 82, humidity 82, location 86. Seed 13, soil 13, fertilizer 52, water 41, light 34, temperature 34, humidity 35, location 35. So, the lowest location number in this example is 35.\nWhat is the lowest location number that corresponds to any of the initial seed numbers?\nMy input file: 2023-12-05-1-aoc.txt\nWow, this task is a mouthful\u0026hellip;\nLet\u0026rsquo;s start slowly and load the data. Our input text document contains several maps, which are clearly separated and have a title (seed-to-soil map etc). So we can tell pandas where each map starts and give each map a dataframe. I got the values for the skiprows and nrows argument by looking at the input file and\u0026hellip; counting :)\nimport pandas as pd import sys seeds = pd.read_table(\u0026#39;data/2023-12-05-1-aoc.txt\u0026#39;, nrows=1, sep=\u0026#39; \u0026#39;, header=None, index_col=0) seeds = seeds.values.flatten() map_opt = {\u0026#39;filepath_or_buffer\u0026#39;: \u0026#39;data/2023-12-05-1-aoc.txt\u0026#39;, \u0026#39;header\u0026#39;: None, \u0026#39;sep\u0026#39;: \u0026#39; \u0026#39;, \u0026#39;dtype\u0026#39;: \u0026#39;Int64\u0026#39;, \u0026#39;names\u0026#39;: [\u0026#39;dest_start\u0026#39;, \u0026#39;src_start\u0026#39;, \u0026#39;range\u0026#39;]} seed_soil = pd.read_table(skiprows=3, nrows=23, **map_opt) soil_fert = pd.read_table(skiprows=28, nrows=9, **map_opt) fert_water = pd.read_table(skiprows=39, nrows=20, **map_opt) water_light = pd.read_table(skiprows=61, nrows=40, **map_opt) light_temp = pd.read_table(skiprows=103, nrows=36, **map_opt) temp_humi = pd.read_table(skiprows=141, nrows=35, **map_opt) humi_loc = pd.read_table(skiprows=178, nrows=26, **map_opt) maps = (seed_soil, soil_fert, fert_water, water_light, light_temp, temp_humi, humi_loc) print(\u0026#39;seeds are just a numpy array:\u0026#39;) display(seeds) print(\u0026#39;The \u0026#34;humidity-to-location\u0026#34; map as an example:\u0026#39;) humi_loc seeds are just a numpy array: array([3169137700, 271717609, 3522125441, 23376095, 1233948799, 811833837, 280549587, 703867355, 166086528, 44766996, 2326968141, 69162222, 2698492851, 14603069, 2755327667, 348999531, 2600461189, 92332846, 1054656969, 169099767])The \u0026#34;humidity-to-location\u0026#34; map as an example: dest_start src_start range 0 3490144003 1623866227 218040905 1 1709610578 1620839197 3027030 2 105449249 586389428 113279526 3 1899604338 2167886292 348178199 4 1712637608 2678624215 186966730 5 218728775 0 245776251 6 2472992580 923734334 143670388 7 2616662968 3670169885 15297294 8 2247782537 1395629154 225210043 9 0 480940179 105449249 10 4113852846 3959729057 159909096 11 3322784653 1067404722 167359350 12 923734334 4119638153 175329143 13 1534964496 2516064491 162559724 14 2631960262 3496028440 140849733 15 2862695906 1234764072 97988355 16 1697524220 3636878173 12086358 17 2985646048 1332752427 62876727 18 1099063477 2970241510 435901019 19 4009202281 2865590945 104650565 20 2960684261 2142924505 24961787 21 2772809995 3406142529 89885911 22 4273761942 3648964531 21205354 23 3708184908 1841907132 301017373 24 464505026 245776251 235163928 25 3048522775 3685467179 274261878 My first attempt was to actually construct ranges like in the example above, mapping out all possible sources and destinations. Python quickly informed me that even constructing one pandas.Series with int64 values mapping seeds to soil would cost 64GB memory - not the best solution.\nSo we take a different approach. For convenience, let\u0026rsquo;s add a src_end and a dest_end column to our maps:\nfor df in maps: df[\u0026#39;src_end\u0026#39;] = df.loc[:, \u0026#39;src_start\u0026#39;] + df.loc[:, \u0026#39;range\u0026#39;] df[\u0026#39;dest_end\u0026#39;] = df.loc[:, \u0026#39;dest_start\u0026#39;] + df.loc[:, \u0026#39;range\u0026#39;] print(\u0026#39;Again the \u0026#34;humidity-to-location\u0026#34; map as an example:\u0026#39;) humi_loc Again the \u0026#34;humidity-to-location\u0026#34; map as an example: dest_start src_start range src_end dest_end 0 3490144003 1623866227 218040905 1841907132 3708184908 1 1709610578 1620839197 3027030 1623866227 1712637608 2 105449249 586389428 113279526 699668954 218728775 3 1899604338 2167886292 348178199 2516064491 2247782537 4 1712637608 2678624215 186966730 2865590945 1899604338 5 218728775 0 245776251 245776251 464505026 6 2472992580 923734334 143670388 1067404722 2616662968 7 2616662968 3670169885 15297294 3685467179 2631960262 8 2247782537 1395629154 225210043 1620839197 2472992580 9 0 480940179 105449249 586389428 105449249 10 4113852846 3959729057 159909096 4119638153 4273761942 11 3322784653 1067404722 167359350 1234764072 3490144003 12 923734334 4119638153 175329143 4294967296 1099063477 13 1534964496 2516064491 162559724 2678624215 1697524220 14 2631960262 3496028440 140849733 3636878173 2772809995 15 2862695906 1234764072 97988355 1332752427 2960684261 16 1697524220 3636878173 12086358 3648964531 1709610578 17 2985646048 1332752427 62876727 1395629154 3048522775 18 1099063477 2970241510 435901019 3406142529 1534964496 19 4009202281 2865590945 104650565 2970241510 4113852846 20 2960684261 2142924505 24961787 2167886292 2985646048 21 2772809995 3406142529 89885911 3496028440 2862695906 22 4273761942 3648964531 21205354 3670169885 4294967296 23 3708184908 1841907132 301017373 2142924505 4009202281 24 464505026 245776251 235163928 480940179 699668954 25 3048522775 3685467179 274261878 3959729057 3322784653 Now we actually compute the mapping. For each seed, we go through all mappings and in each mapping we go through each row. We find the row which contains the mapping and exctract the destination, which is the source for the next map until we reach the last map which gives us the locations.\nApproach 1: df.itertuples() is a convenient way to step through a pandas.DataFrame in this example. It is faster than df.iterrows() and returns the row as a NamedTuple - nice!\nApproach 2: I actually wondered if it would be faster to get all maps in one pd.DataFrame and then iterate through the mappings. To test this let\u0026rsquo;s construct a new DataFrame maps_df which contains all maps. Since the maps have different lengths it is important to cast the datatype to Int64, which is short for pd.Int64Dtype() and keeps values as integers, even if NA values are in the same column.\nApproach 3: A third alternative I tested (not shown here) was to check if a value is in a Python range with the in operator as in: if 3 in range(5):... . This was way too slow.\n# mapping version 1 def get_location(seed): current = seed for df in maps: current_map = [row for row in df.itertuples() if ((current \u0026gt; row.src_start) and (current \u0026lt; row.src_end))] if len(current_map) == 0: pass elif len(current_map) == 1: current = (current_map[0].dest_start + (current - current_map[0].src_start)) else: raise ValueError(\u0026#39;This should not happen!\u0026#39;) return current # mapping version 2 - around 10 times slower # you need to rename the maps_df columns so that they have a unique id # e.g. \u0026#39;src_start_1\u0026#39;, \u0026#39;src_start_2\u0026#39; etc # maps_df = pd.concat(maps, axis=\u0026#39;columns\u0026#39;) # def get_dest(i, src): # return (maps_df[(src \u0026gt; maps_df.loc[:, f\u0026#39;src_start_{i}\u0026#39;]) \u0026amp; # (src \u0026lt; maps_df.loc[:, f\u0026#39;src_end_{i}\u0026#39;])] # .loc[:, f\u0026#39;dest_start_{i}\u0026#39;] # .iloc[0]) # def get_location2(seed): # dest = seed # i = 1 # while i \u0026lt; 7: # dest = get_dest(i, dest) # i += 1 # return dest %timeit get_location(seeds[0]) 1.98 ms ± 72 μs per loop (mean ± std. dev. of 7 runs, 100 loops each) Now let\u0026rsquo;s get a list of locations:\nlocations = [get_location(s) for s in seeds] locations 2493982655 3209845376 3992357533 4163131463 4104485616 1952252479 3218677354 388071289 2181441450 2594336315 4049507670 2084517144 3119633635 428978312 3518771991 3704555655 953918455 2107687768 3448046330 2184454689 Lastly, just get the minimum of all location values.\nmin(locations) 388071289 Part 2 # Let\u0026rsquo;s read the task of part 2! Everyone will starve if you only plant such a small number of seeds. Re-reading the almanac, it looks like the seeds: line actually describes ranges of seed numbers.\nThe values on the initial seeds: line come in pairs. Within each pair, the first value is the start of the range and the second value is the length of the range. So, in the first line of the example above:\nseeds: 79 14 55 13\nThis line describes two ranges of seed numbers to be planted in the garden. The first range starts with seed number 79 and contains 14 values: 79, 80, \u0026hellip;, 91, 92. The second range starts with seed number 55 and contains 13 values: 55, 56, \u0026hellip;, 66, 67.\nNow, rather than considering four seed numbers, you need to consider a total of 27 seed numbers.\nIn the above example, the lowest location number can be obtained from seed number 82, which corresponds to soil 84, fertilizer 84, water 84, light 77, temperature 45, humidity 46, and location 46. So, the lowest location number is 46.\nConsider all of the initial seed numbers listed in the ranges on the first line of the almanac. What is the lowest location number that corresponds to any of the initial seed numbers?\nLet\u0026rsquo;s first construct a dataframe containing the range of seeds:\nseeds_df = pd.DataFrame({\u0026#39;start\u0026#39;: seeds[::2], \u0026#39;range\u0026#39;: seeds[1::2], \u0026#39;end\u0026#39;: seeds[::2] + seeds[1::2]}) print(f\u0026#39;There are {sum(seeds_df.loc[:, \u0026#34;range\u0026#34;]):_} seeds in total\u0026#39;) seeds_df There are 2_549_759_327 seeds in total start range end 0 3169137700 271717609 3440855309 1 3522125441 23376095 3545501536 2 1233948799 811833837 2045782636 3 280549587 703867355 984416942 4 166086528 44766996 210853524 5 2326968141 69162222 2396130363 6 2698492851 14603069 2713095920 7 2755327667 348999531 3104327198 8 2600461189 92332846 2692794035 9 1054656969 169099767 1223756736 Now - I really needed some time to finally realize, that going through all seed values is really unfeasable. So how do we deal with this problem?\nIn the end we need the lowest location number - thus our approach is to take the humi_loc map, start with the lowest location number and go up and get the corresponding seed values. The location of the first seed value which is inside seeds_df is our solution.\nSo first we rebuild the get_location function to a get_seed function (we reverse the maps tuple with maps[::-1] and switch src and dest).\ndef get_seed(location): current = location for df in maps[::-1]: current_map = [row for row in df.itertuples() if ((current \u0026gt;= row.dest_start) and (current \u0026lt; row.dest_end))] if len(current_map) == 0: pass elif len(current_map) == 1: current = (current_map[0].src_start + (current - current_map[0].dest_start)) else: raise ValueError(\u0026#39;This should not happen!\u0026#39;) return current Since the Python 3 range function does not store it\u0026rsquo;s contents in memory (similar to a generator), it is well suited to go through these large ranges.\nLastly, we deal with the sheer amount of possible values by performing a grid search. First, we check every millionth location. After the first match, we stop this search and check the last million locations before the match with a finer grid and so on. The last grid is just 1, so we find our lowest location.\nUpdate: We order our locations from smallest to largest with sort_values and go through them - but the search stops after the first match, since that will be our lowest location.\ndef grid_search(start: int, end: int, grid: list[int]): success = False for g in grid: print(f\u0026#39;Start search with grid={g}\u0026#39;) for l in range(start, end, g): current = get_seed(l) if any((current \u0026gt;= seeds_df.loc[:, \u0026#39;start\u0026#39;]) \u0026amp; (current \u0026lt; seeds_df.loc[:, \u0026#39;end\u0026#39;])): print(f\u0026#39;location {l} is the lowest which contains one of the given seeds ({current})\u0026#39;) start = l - g end = l success = True break return success for row in humi_loc.sort_values(\u0026#39;dest_start\u0026#39;).itertuples(): success = grid_search(start=row.dest_start, end=row.dest_end, grid=[1_000_000, 100_000, 1000, 1]) if success: print(\u0026#39;Finished search\u0026#39;) break Start search with grid=1000000 location 85000000 is the lowest which contains one of the given seeds (2605777210) Start search with grid=100000 location 84300000 is the lowest which contains one of the given seeds (2605077210) Start search with grid=1000 location 84207000 is the lowest which contains one of the given seeds (2604984210) Start search with grid=1 location 84206669 is the lowest which contains one of the given seeds (2604983879) Finished search "},{"id":6,"href":"/posts/aoc2023-day04/","title":"Advent of code 2023 - Day 4: Scratchcards","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 4 - see https:adventofcode.com/2023/day/4\nPart 1 # Lets first read the task: The Elf leads you over to the pile of colorful cards. There, you discover dozens of scratchcards, all with their opaque covering already scratched off. Picking one up, it looks like each card has two lists of numbers separated by a vertical bar (|): a list of winning numbers and then a list of numbers you have. You organize the information into a table (your puzzle input).\nAs far as the Elf has been able to figure out, you have to figure out which of the numbers you have appear in the list of winning numbers. The first match makes the card worth one point and each match after the first doubles the point value of that card.\nFor example:\nCard 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53 Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19 Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1 Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83 Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36 Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11 In the above example, card 1 has five winning numbers (41, 48, 83, 86, and 17) and eight numbers you have (83, 86, 6, 31, 17, 9, 48, and 53). Of the numbers you have, four of them (48, 83, 17, and 86) are winning numbers! That means card 1 is worth 8 points (1 for the first match, then doubled three times for each of the three matches after the first).\nCard 2 has two winning numbers (32 and 61), so it is worth 2 points. Card 3 has two winning numbers (1 and 21), so it is worth 2 points. Card 4 has one winning number (84), so it is worth 1 point. Card 5 has no winning numbers, so it is worth no points. Card 6 has no winning numbers, so it is worth no points. So, in this example, the Elf\u0026rsquo;s pile of scratchcards is worth 13 points.\nTake a seat in the large pile of colorful cards. How many points are they worth in total?\nMy input file: 2023-12-04-1-aoc.txt\nLoading this data is very similar to Day 2 - so let\u0026rsquo;s load the data as we did there. Our goal is to get win and yours columns holding the respective digits which we want to compare. We find the numbers with one or more digits using the regex \\d+. And we want them to be in sets (not lists), as we can logically compare sets in Python.\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-04-1-aoc.txt\u0026#39;, names=[\u0026#39;win\u0026#39;]) txt[\u0026#39;id\u0026#39;] = txt.loc[:, \u0026#39;win\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply( lambda x: int(x[0].strip(\u0026#39;Card \u0026#39;))) txt[\u0026#39;win\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;:\u0026#39;).apply(lambda x: x[1])) txt[\u0026#39;yours\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;|\u0026#39;) .apply(lambda x: x[1]) # get a list of only the numbers / digits .apply(lambda x: re.findall(r\u0026#39;\\d+\u0026#39;, x)) # convert the list of strings to a set of integers .apply(lambda x: set([int(i) for i in x]))) txt[\u0026#39;win\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;|\u0026#39;) .apply(lambda x: x[0]) .apply(lambda x: re.findall(r\u0026#39;\\d+\u0026#39;, x)) .apply(lambda x: set([int(i) for i in x]))) txt win id yours 0 {32, 36, 7, 9, 10, 12, 82, 85, 95, 31} 1 {2, 7, 9, 10, 12, 14, 21, 22, 23, 24, 31, 32, \u0026hellip; 1 {35, 76, 16, 82, 19, 22, 88, 59, 60, 95} 2 {7, 8, 12, 16, 19, 22, 26, 28, 35, 38, 44, 51,\u0026hellip; 2 {1, 70, 11, 78, 48, 19, 52, 88, 28, 94} 3 {3, 4, 8, 17, 18, 19, 24, 31, 34, 45, 52, 54, \u0026hellip; 3 {65, 2, 72, 28, 14, 16, 55, 91, 92, 62} 4 {3, 4, 6, 7, 8, 9, 15, 30, 33, 35, 47, 49, 51,\u0026hellip; 4 {38, 41, 75, 77, 50, 24, 94, 60, 61, 30} 5 {1, 2, 4, 5, 6, 7, 9, 10, 14, 17, 21, 29, 47, \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 213 {97, 98, 39, 41, 43, 12, 13, 19, 93, 95} 214 {5, 10, 17, 20, 28, 29, 33, 34, 36, 50, 51, 52\u0026hellip; 214 {97, 35, 69, 40, 74, 45, 20, 21, 62, 31} 215 {1, 8, 15, 17, 18, 25, 30, 33, 42, 44, 47, 52,\u0026hellip; 215 {33, 70, 71, 12, 78, 17, 51, 86, 60, 94} 216 {7, 8, 9, 10, 22, 29, 37, 39, 41, 43, 46, 47, \u0026hellip; 216 {98, 67, 68, 38, 70, 39, 72, 77, 45, 21} 217 {8, 21, 22, 25, 26, 31, 37, 41, 42, 48, 54, 57\u0026hellip; 217 {34, 9, 44, 78, 79, 16, 17, 19, 55, 92} 218 {1, 4, 20, 21, 27, 38, 39, 40, 41, 45, 46, 52,\u0026hellip; 218 rows × 3 columns\nNow, we get the logical conjunction of win and yours, these are our winning numbers. Then, the number of wins is converted to points - for all number of wins bigger than 1, we can get the points by 2**(n_wins-1).\ntxt[\u0026#39;n_wins\u0026#39;] = txt.apply( lambda row: len(row.loc[\u0026#39;win\u0026#39;] \u0026amp; row.loc[\u0026#39;yours\u0026#39;]), axis=1) txt[\u0026#39;points\u0026#39;] = txt.loc[:, \u0026#39;n_wins\u0026#39;].apply( lambda x: 2**(x-1) if x \u0026gt; 1 else x) txt.loc[:, [\u0026#39;win\u0026#39;, \u0026#39;n_wins\u0026#39;, \u0026#39;points\u0026#39;]] win n_wins points 0 {32, 36, 7, 9, 10, 12, 82, 85, 95, 31} 10 512 1 {35, 76, 16, 82, 19, 22, 88, 59, 60, 95} 10 512 2 {1, 70, 11, 78, 48, 19, 52, 88, 28, 94} 5 16 3 {65, 2, 72, 28, 14, 16, 55, 91, 92, 62} 0 0 4 {38, 41, 75, 77, 50, 24, 94, 60, 61, 30} 0 0 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 213 {97, 98, 39, 41, 43, 12, 13, 19, 93, 95} 0 0 214 {97, 35, 69, 40, 74, 45, 20, 21, 62, 31} 0 0 215 {33, 70, 71, 12, 78, 17, 51, 86, 60, 94} 2 2 216 {98, 67, 68, 38, 70, 39, 72, 77, 45, 21} 1 1 217 {34, 9, 44, 78, 79, 16, 17, 19, 55, 92} 0 0 218 rows × 3 columns\nThen, we just sum up:\nsum(txt.loc[:, \u0026#39;points\u0026#39;]) 25004 Part 2 # Let\u0026rsquo;s read the task of part 2! There\u0026rsquo;s no such thing as \u0026ldquo;points\u0026rdquo;. Instead, scratchcards only cause you to win more scratchcards equal to the number of winning numbers you have.\nSpecifically, you win copies of the scratchcards below the winning card equal to the number of matches. So, if card 10 were to have 5 matching numbers, you would win one copy each of cards 11, 12, 13, 14, and 15.\nCopies of scratchcards are scored like normal scratchcards and have the same card number as the card they copied. So, if you win a copy of card 10 and it has 5 matching numbers, it would then win a copy of the same cards that the original card 10 won: cards 11, 12, 13, 14, and 15. This process repeats until none of the copies cause you to win any more cards. (Cards will never make you copy a card past the end of the table.)\nThis time, the above example goes differently:\nCard 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53 Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19 Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1 Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83 Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36 Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11 Card 1 has four matching numbers, so you win one copy each of the next four cards: cards 2, 3, 4, and 5. Your original card 2 has two matching numbers, so you win one copy each of cards 3 and 4. Your copy of card 2 also wins one copy each of cards 3 and 4. Your four instances of card 3 (one original and three copies) have two matching numbers, so you win four copies each of cards 4 and 5. Your eight instances of card 4 (one original and seven copies) have one matching number, so you win eight copies of card 5. Your fourteen instances of card 5 (one original and thirteen copies) have no matching numbers and win no more cards. Once all of the originals and copies have been processed, you end up with 1 instance of card 1, 2 instances of card 2, 4 instances of card 3, 8 instances of card 4, 14 instances of card 5, and 1 instance of card 6. In total, this example pile of scratchcards causes you to ultimately have 30 scratchcards!\nProcess all of the original and copied scratchcards until no more scratchcards are won. Including the original set of scratchcards, how many total scratchcards do you end up with?\nWe will use the n_wins column we created before and go from there. We step through each Game. Each current game gets +1 card. Then, we step through the number of next games depending on our n_wins. Each of these gets added the card number of the current game.\ntxt[\u0026#39;cards\u0026#39;] = 0 for i, nwin in enumerate(txt.loc[:, \u0026#39;n_wins\u0026#39;]): txt.loc[i, \u0026#39;cards\u0026#39;] += 1 for j in range(1, nwin+1): txt.loc[i+j, \u0026#39;cards\u0026#39;] += txt.loc[i, \u0026#39;cards\u0026#39;] txt.loc[:, [\u0026#39;n_wins\u0026#39;, \u0026#39;cards\u0026#39;]] n_wins cards 0 10 1 1 10 2 2 5 4 3 0 8 4 0 8 \u0026hellip; \u0026hellip; \u0026hellip; 213 0 9608 214 0 8927 215 2 8927 216 1 12636 217 0 21564 218 rows × 2 columns\nNow, we just sum the output again:\nsum(txt[\u0026#39;cards\u0026#39;]) 14427616 "},{"id":7,"href":"/posts/aoc2023-day03/","title":"Advent of code 2023 - Day 3: Gear Ratios","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 3 - see https:adventofcode.com/2023/day/3\nPart 1 # Lets first read the task: The engine schematic (your puzzle input) consists of a visual representation of the engine. There are lots of numbers and symbols you don\u0026rsquo;t really understand, but apparently any number adjacent to a symbol, even diagonally, is a \u0026ldquo;part number\u0026rdquo; and should be included in your sum. (Periods (.) do not count as a symbol.)\nHere is an example engine schematic:\n467..114.. ...*...... ..35..633. ......#... 617*...... .....+.58. ..592..... ......755. ...$.*.... .664.598.. In this schematic, two numbers are not part numbers because they are not adjacent to a symbol: 114 (top right) and 58 (middle right). Every other number is adjacent to a symbol and so is a part number; their sum is 4361.\nOf course, the actual engine schematic is much larger. What is the sum of all of the part numbers in the engine schematic?\nMy input file: 2023-12-03-1-aoc.txt\nOkay, let\u0026rsquo;s first get the input as a numpy character array\nimport numpy as np import pandas as pd import sys import matplotlib.pyplot as plt from scipy import ndimage as ndi np.set_printoptions(threshold=sys.maxsize) txt = pd.read_table(\u0026#39;data/2023-12-03-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) arr = np.chararray((txt.size, txt.size), unicode=True) txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda x: [i for i in x]) for i, code in enumerate(txt[\u0026#39;code\u0026#39;]): arr[i, :] = code print((arr[:15, :15])) [[\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;5\u0026#39; \u0026#39;3\u0026#39; \u0026#39;.\u0026#39; \u0026#39;4\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;2\u0026#39; \u0026#39;6\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;5\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;3\u0026#39; \u0026#39;8\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;4\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;6\u0026#39; \u0026#39;6\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;1\u0026#39; \u0026#39;2\u0026#39; \u0026#39;6\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;3\u0026#39; \u0026#39;1\u0026#39; \u0026#39;/\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;/\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;5\u0026#39; \u0026#39;3\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;5\u0026#39; \u0026#39;5\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;6\u0026#39; \u0026#39;8\u0026#39; \u0026#39;5\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;0\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;3\u0026#39; \u0026#39;8\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39;]] /tmp/ipykernel_18336/87086526.py:9: DeprecationWarning: `np.chararray` is deprecated and will be removed from the main namespace in the future. Use an array with a string or bytes dtype instead. arr = np.chararray((txt.size, txt.size), unicode=True) Now extract symbols, digits and the empty space. We use the numpy character methods for that. This way, we create a binary mask for all digits and a binary mask for all empty space (.). The symbols are then every character which is neither.\ndigits = np.char.isdigit(arr) empty = np.char.endswith(arr, \u0026#39;.\u0026#39;) symbols = ~(digits | empty) # just for visualization plt.figure(figsize=(16, 5)) plt.subplot(131, title=\u0026#39;symbols\u0026#39;).matshow(symbols) plt.subplot(132, title=\u0026#39;digits\u0026#39;).matshow(digits) plt.subplot(133, title=\u0026#39;empty\u0026#39;).matshow(empty) plt.show() Now we use the image processing technique of dilation on the symbols mask. So that we get a new mask which covers the surroundings of all symbols. We use this afterwards to check if the digits are near a symbol.\nstruct = ((1, 1, 1), (1, 1, 1), (1, 1, 1)) dilate = ndi.binary_dilation(symbols, structure=struct) plt.figure(figsize=(10, 6)) plt.subplot(121, title=\u0026#39;symbols\u0026#39;).matshow(symbols[:15, :15]) plt.subplot(122, title=\u0026#39;symbols dilated\u0026#39;).matshow(dilate[:15, :15]) plt.show() Creating this masks as before could be understood as a binary segmentation, as each element in our mask is either True or False. To extract the single digits, we\u0026rsquo;ll convert the binary digits segmentation into a instance segmentation, where each connected segment has an own index.\nmarkers, num_features = ndi.label(digits) plt.figure(figsize=(10, 6)) plt.subplot(121, title=\u0026#39;binary segmentation\u0026#39;).matshow( digits[:15, :15]) plt.subplot(122, title=\u0026#39;instance segmentation\u0026#39;).matshow( markers[:15, :15], cmap=\u0026#39;gnuplot\u0026#39;) plt.show() Now, for each instance, we check if the dilated binary mask overlaps with the instance and if yes, we extract the number.\nnumbers = [int(\u0026#39;\u0026#39;.join(arr[markers == i])) for i in range(1, num_features+1) if np.any((markers == i) \u0026amp; dilate)] Then, we just sum up:\nsum(numbers) 527364 Part 2 # Let\u0026rsquo;s first read the second task! The missing part wasn\u0026rsquo;t the only issue - one of the gears in the engine is wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.\nThis time, you need to find the gear ratio of every gear and add them all up so that the engineer can figure out which gear needs to be replaced.\nConsider the same engine schematic again:The missing part wasn\u0026rsquo;t the only issue - one of the gears in the engine is wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.\nThis time, you need to find the gear ratio of every gear and add them all up so that the engineer can figure out which gear needs to be replaced.\nConsider the same engine schematic again:\n467..114.. ...*...... ..35..633. ......#... 617*...... .....+.58. ..592..... ......755. ...$.*.... .664.598.. In this schematic, there are two gears. The first is in the top left; it has part numbers 467 and 35, so its gear ratio is 16345. The second gear is in the lower right; its gear ratio is 451490. (The * adjacent to 617 is not a gear because it is only adjacent to one part number.) Adding up all of the gear ratios produces 467835.\nWhat is the sum of all of the gear ratios in your engine schematic?\nWe\u0026rsquo;ll use the same method as before, but this time only extract * and create the instance segmentation before the dilation. Why? Because when we dilate first, we could merge two independent gears into one instance.\ngear = np.char.endswith(arr, \u0026#39;*\u0026#39;) gear_markers, gear_num = ndi.label(gear) plt.figure(figsize=(10, 6)) plt.subplot(131, title=\u0026#39;all symbols dilated\u0026#39;).matshow( symbols[:15, :15]) plt.subplot(132, title=\u0026#39;gears\u0026#39;).matshow( gear[:15, :15]) plt.subplot(133, title=\u0026#39;gears instances\u0026#39;).matshow( gear_markers[:15, :15], cmap=\u0026#39;gnuplot\u0026#39;) plt.show() Now, we step through each gear instance, create a mask for that gear, dilate it, and then step through all digits and check if they are in the gear mask. If we get two digits in the end, we multiply them to get the gear ratio and save the ratios to a list.\ngear_ratios = [] for i in range(1, gear_num+1): gear_binary = gear_markers == i gear_dil = ndi.binary_dilation(gear_binary, structure=struct) part_numbers = [int(\u0026#39;\u0026#39;.join(arr[markers == j])) for j in range(1, num_features+1) if np.any((markers == j) \u0026amp; gear_dil)] if len(part_numbers) == 2: gear_ratios.append(part_numbers[0] * part_numbers[1]) Now, we just sum the output again:\nsum(gear_ratios) 79026871 "},{"id":8,"href":"/posts/aoc2023-day02/","title":"Advent of code 2023 - Day 2: Cube Conundrum","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 2 - see https:adventofcode.com/2023/day/2\nPart 1 # Lets first read the task: As you walk, the Elf shows you a small bag and some cubes which are either red, green, or blue. Each time you play this game, he will hide a secret number of cubes of each color in the bag, and your goal is to figure out information about the number of cubes.\nTo get information, once a bag has been loaded with cubes, the Elf will reach into the bag, grab a handful of random cubes, show them to you, and then put them back in the bag. He\u0026rsquo;ll do this a few times per game.\nYou play several games and record the information from each game (your puzzle input). Each game is listed with its ID number (like the 11 in Game 11: ...) followed by a semicolon-separated list of subsets of cubes that were revealed from the bag (like 3 red, 5 green, 4 blue).\nFor example, the record of a few games might look like this:\nGame 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green In game 1, three sets of cubes are revealed from the bag (and then put back again). The first set is 3 blue cubes and 4 red cubes; the second set is 1 red cube, 2 green cubes, and 6 blue cubes; the third set is only 2 green cubes.\nThe Elf would first like to know which games would have been possible if the bag contained only 12 red cubes, 13 green cubes, and 14 blue cubes?\nIn the example above, games 1, 2, and 5 would have been possible if the bag had been loaded with that configuration. However, game 3 would have been impossible because at one point the Elf showed you 20 red cubes at once; similarly, game 4 would also have been impossible because the Elf showed you 15 blue cubes at once. If you add up the IDs of the games that would have been possible, you get 8.\nDetermine which games would have been possible if the bag had been loaded with only 12 red cubes, 13 green cubes, and 14 blue cubes. What is the sum of the IDs of those games?\nMy input file: 2023-12-02-1-aoc.txt\nOkay, let\u0026rsquo;s load our python kernel in emacs-jupyter and get coding! First of all, let\u0026rsquo;s load the input and split the riddle code by colon : to extract the game id and the rest of the code by semicolon ; to get the number of sets played in each game.\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-02-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) txt[\u0026#39;id\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply( lambda x: int(x[0].strip(\u0026#39;Game \u0026#39;))) txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply(lambda x: x[1]) # txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;;\u0026#39;) # txt[\u0026#39;nsets\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda x: len(x)) txt code id 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 100 rows × 2 columns\nNow, let\u0026rsquo;s extract the three colors in different columns with regex. We use the lookahead assertion ?= to find the respective colours and only exctract the digits \\d+ coming before. Then we just keep the max imum drawn number of cubes per color, since this is the only information that matters at the moment.\ntxt[\u0026#39;green\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.green)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt[\u0026#39;red\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.red)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt[\u0026#39;blue\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.blue)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt code id green red blue 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 2 10 10 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 15 10 3 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 4 6 16 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 16 20 18 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 10 6 14 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 4 2 1 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 5 4 8 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 2 5 2 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 2 9 11 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 13 15 11 100 rows × 5 columns\nLastly, we just filter the DataFrame to only include games where all drawn cubes were below or equal the number of cubes in the game and sum the result!\ntxt[\u0026#39;id\u0026#39;][(txt[\u0026#39;green\u0026#39;] \u0026lt; 14) \u0026amp; (txt[\u0026#39;red\u0026#39;] \u0026lt; 13) \u0026amp; (txt[\u0026#39;blue\u0026#39;] \u0026lt; 15)].sum() 3035 Part 2 # First, let\u0026rsquo;s get the instruction from the second part: As you continue your walk, the Elf poses a second question: in each game you played, what is the fewest number of cubes of each color that could have been in the bag to make the game possible?\nAgain consider the example games from earlier:\nGame 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green In game 1, the game could have been played with as few as 4 red, 2 green, and 6 blue cubes. If any color had even one fewer cube, the game would have been impossible. Game 2 could have been played with a minimum of 1 red, 3 green, and 4 blue cubes. Game 3 must have been played with at least 20 red, 13 green, and 6 blue cubes. Game 4 required at least 14 red, 3 green, and 15 blue cubes. Game 5 needed no fewer than 6 red, 3 green, and 2 blue cubes in the bag. The power of a set of cubes is equal to the numbers of red, green, and blue cubes multiplied together. The power of the minimum set of cubes in game 1 is 48. In games 2-5 it was 12, 1560, 630, and 36, respectively. Adding up these five powers produces the sum 2286.\nFor each game, find the minimum set of cubes that must have been present. What is the sum of the power of these sets?\nLuckily, this task is made trivial by the approach we have taken before. We just have to multiply the green, red and blue columns:\ntxt[\u0026#39;power\u0026#39;] = txt.loc[:, \u0026#39;green\u0026#39;] * txt.loc[:, \u0026#39;blue\u0026#39;] * txt.loc[:, \u0026#39;red\u0026#39;] txt code id green red blue power 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 2 10 10 200 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 15 10 3 450 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 4 6 16 384 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 16 20 18 5760 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 10 6 14 840 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 4 2 1 8 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 5 4 8 160 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 2 5 2 20 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 2 9 11 198 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 13 15 11 2145 100 rows × 6 columns\nAnd for this one, the sum is:\ntxt[\u0026#39;power\u0026#39;].sum() 66027 "},{"id":9,"href":"/posts/aoc2023-day01/","title":"Advent of code 2023 - Day 1: Trebuchet?!","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 1 - see https:adventofcode.com/2023/day/1\nPart 1 # Our first task is the following: The newly-improved calibration document consists of lines of text; each line originally contained a specific calibration value that the Elves now need to recover. On each line, the calibration value can be found by combining the first digit and the last digit (in that order) to form a single two-digit number.\nFor example:\n1abc2 pqr3stu8vwx a1b2c3d4e5f treb7uchet In this example, the calibration values of these four lines are 12, 38, 15, and 77. Adding these together produces 142.\nConsider your entire calibration document. What is the sum of all of the calibration values?\nMy input file: 2023-12-01-1-aoc.txt\nLets start jupyter in our shell to start coding!\nconda activate tf jupyter lab --no-browser --port=8888 First, load the test document\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-01-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) txt code 0 jjfvnnlfivejj1 1 6fourfour 2 ninevbmltwo69 3 pcg91vqrfpxxzzzoneightzt 4 jpprthxgjfive3one1qckhrptpqdc \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 996 81s 997 2four3threesxxvlfqfive4 998 nine6eightsevenzx9twoxc 999 hmbfjdfnp989mfivefiverpzrjs 1000 rows × 1 columns\nSecond, extract the digits. I had to wrap my head around regex matching in python first, because I first tried pandas.extract (which only extracts the first match), then pandas.extractall (which extracts all matches but puts them into a multiindex which makes things more difficult in this case). So I settled for the re.findall version, which returns a list. To concatenate the elements in the list, we take use the join function.\ntxt[\u0026#39;digits\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda x: \u0026#39;\u0026#39;.join(re.findall(r\u0026#39;(\\d+)\u0026#39;, x))) txt code digits 0 jjfvnnlfivejj1 1 1 6fourfour 6 2 ninevbmltwo69 69 3 pcg91vqrfpxxzzzoneightzt 91 4 jpprthxgjfive3one1qckhrptpqdc 31 \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 996 81s 81 997 2four3threesxxvlfqfive4 234 998 nine6eightsevenzx9twoxc 69 999 hmbfjdfnp989mfivefiverpzrjs 989 1000 rows × 2 columns\nNext, combine the first and the last digit and convert the result from string to integer\ntxt[\u0026#39;calibration\u0026#39;] = txt.loc[:, \u0026#39;digits\u0026#39;].apply( lambda x: int(x[0] + x[-1])) txt code digits calibration 0 jjfvnnlfivejj1 1 11 1 6fourfour 6 66 2 ninevbmltwo69 69 69 3 pcg91vqrfpxxzzzoneightzt 91 91 4 jpprthxgjfive3one1qckhrptpqdc 31 31 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 996 81s 81 81 997 2four3threesxxvlfqfive4 234 24 998 nine6eightsevenzx9twoxc 69 69 999 hmbfjdfnp989mfivefiverpzrjs 989 99 1000 rows × 3 columns\nLastly, get the sum of our calibration numbers\ntxt.loc[:, \u0026#39;calibration\u0026#39;].sum() 56465 Part 2 # Now follows part two: Your calculation isn\u0026rsquo;t quite right. It looks like some of the digits are actually spelled out with letters: one, two, three, four, five, six, seven, eight, and nine also count as valid \u0026ldquo;digits\u0026rdquo;.\nEquipped with this new information, you now need to find the real first and last digit on each line. For example:\ntwo1nine eightwothree abcone2threexyz xtwone3four 4nineeightseven2 zoneight234 7pqrstsixteen In this example, the calibration values are 29, 83, 13, 24, 42, 14, and 76. Adding these together produces 281.\nWhat is the sum of all of the calibration values?\nOkay, let\u0026rsquo;s see if we can update the pattern matching. To deal with potential overlapping values like oneight which contains one as well as eight, I used the regex positive lookahead ?= as described here. Because this enables capturing overlapping values, I used \\d (one digit) instead of \\d+ (one or more digits), otherwise digits might double. Afterwards, just replace the spelled out digits with their numerical value.\n# for i, r in enumerate(txt.loc[:, \u0026#39;code\u0026#39;]): # matches = re.findall( # r\u0026#39;(?=(\\d|one|two|three|four|five|six|seven|eight|nine))\u0026#39;, r) # result = \u0026#39;\u0026#39;.join([match for match in matches]) # result = result.replace(\u0026#39;one\u0026#39;, \u0026#39;1\u0026#39;).replace(\u0026#39;two\u0026#39;, \u0026#39;2\u0026#39;).replace( # \u0026#39;three\u0026#39;, \u0026#39;3\u0026#39;).replace(\u0026#39;four\u0026#39;, \u0026#39;4\u0026#39;).replace(\u0026#39;five\u0026#39;, \u0026#39;5\u0026#39;).replace( # \u0026#39;six\u0026#39;, \u0026#39;6\u0026#39;).replace(\u0026#39;seven\u0026#39;, \u0026#39;7\u0026#39;).replace(\u0026#39;eight\u0026#39;, \u0026#39;8\u0026#39;).replace( # \u0026#39;nine\u0026#39;, \u0026#39;9\u0026#39;) # txt.loc[i, \u0026#39;digits2\u0026#39;] = result # txt # a very nice alternative suggested by Tomalak: digits = \u0026#39;\\d one two three four five six seven eight nine\u0026#39;.split() txt[\u0026#39;digits2\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda v: \u0026#39;\u0026#39;.join( str(digits.index(m)) if m in digits else m for m in re.findall(rf\u0026#39;(?=({\u0026#34;|\u0026#34;.join(digits)}))\u0026#39;, v) )) txt code digits calibration digits2 0 jjfvnnlfivejj1 1 11 51 1 6fourfour 6 66 644 2 ninevbmltwo69 69 69 9269 3 pcg91vqrfpxxzzzoneightzt 91 91 9118 4 jpprthxgjfive3one1qckhrptpqdc 31 31 5311 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 58375 996 81s 81 81 81 997 2four3threesxxvlfqfive4 234 24 243354 998 nine6eightsevenzx9twoxc 69 69 968792 999 hmbfjdfnp989mfivefiverpzrjs 989 99 98955 1000 rows × 4 columns\nNow, construct the calibration value as before\u0026hellip;\ntxt[\u0026#39;calibration2\u0026#39;] = txt.loc[:, \u0026#39;digits2\u0026#39;].apply(lambda x: int(x[0] + x[-1])) txt code digits calibration digits2 calibration2 0 jjfvnnlfivejj1 1 11 51 51 1 6fourfour 6 66 644 64 2 ninevbmltwo69 69 69 9269 99 3 pcg91vqrfpxxzzzoneightzt 91 91 9118 98 4 jpprthxgjfive3one1qckhrptpqdc 31 31 5311 51 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 58375 55 996 81s 81 81 81 81 997 2four3threesxxvlfqfive4 234 24 243354 24 998 nine6eightsevenzx9twoxc 69 69 968792 92 999 hmbfjdfnp989mfivefiverpzrjs 989 99 98955 95 1000 rows × 5 columns\n\u0026hellip; and get the correct sum!\ntxt.loc[:, \u0026#39;calibration2\u0026#39;].sum() 55902 "}] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/en.search-data.min.8e5342b287ceb737757b551465c44c8ab9c700ea35d479a58e6b93e06004015b.json b/aseltmann.github.io-hugo/public/en.search-data.min.8e5342b287ceb737757b551465c44c8ab9c700ea35d479a58e6b93e06004015b.json deleted file mode 100644 index 78107f3..0000000 --- a/aseltmann.github.io-hugo/public/en.search-data.min.8e5342b287ceb737757b551465c44c8ab9c700ea35d479a58e6b93e06004015b.json +++ /dev/null @@ -1 +0,0 @@ -[{"id":0,"href":"/docs/notes/poetry/","title":"Poetry","section":"Notes","content":" Table of Contents Lǎozǐ Lǎozǐ # I am partial to the translation of Derek Lin\nDàodéjīng - Chapter 11 (The use of what has no substantive existence) [\u0026hellip;]\nMix clay\nto create a container\nIn its emptiness, there is\nthe function of a container\n[\u0026hellip;]\nTherefore,\nthat which exists is used to create benefit\nThat which is empty is used to create functionality\nDàodéjīng - Chapter 15 (The exhibition of the qualities of the Dao) The Tao masters of antiquity\nSubtle wonders through mystery\nDepths that cannot be discerned\nBecause one cannot discern them\nTherefore one is forced to describe the appearance\nHesitant,\nlike crossing a wintry river\nCautious,\nlike fearing four neighbors\nSolemn,\nlike a guest\nLoose,\nlike ice about to melt\nGenuine,\nlike plain wood\nOpen,\nlike a valley\nOpaque,\nlike muddy water\nWho can be muddled yet desist\nIn stillness gradually become clear?\nWho can be serene yet persist\nIn motion gradually come alive?\nOne who holds this \u0026lt;i\u0026gt;Tao\u0026lt;/i\u0026gt; does not wish to be overfilled\nBecause one is not overfilled\nTherefore one can preserve and not create anew\nDàodéjīng - Chapter 20 (Being different from ordinary people) Cease learning, no more worries\nRespectful response and scornful response\nHow much is the difference?\nGoodness and evil\nHow much do they differ?\nWhat the people fear,\nI cannot be unafraid\nSo desolate! How limitless it is!\nThe people are excited\nAs if enjoying a great feast\nAs if climbing up to the terrace in spring\nI alone am quiet and uninvolved\nLike an infant\nnot yet smiling\nSo weary,\nlike having no place to return\nThe people all have surplus\nWhile I alone seem lacking\nI have the heart of a fool indeed -\nso ignorant!\nOrdinary people are bright\nI alone am muddled\nOrdinary people are scrutinizing\nI alone am obtuse\nSo tranquil, like the ocean\nSo moving, as if without limits\nThe people all have goals\nAnd I alone am stubborn and lowly\nI alone am different from them\nAnd value the nourishing mother\n"},{"id":1,"href":"/posts/aoc2024-day02/","title":"Advent of code 2024 - Day 2: Red-Nosed Reports","section":"Posts","content":" Table of Contents Part 1: Safe reports Part 2: Problem Dampener I try to solve this year\u0026rsquo;s Advent of Code 2024 riddles in R. This is Day 2 - see https:adventofcode.com/2024/day/2\nPart 1: Safe reports # Lets first read the task: The unusual data (your puzzle input) consists of many reports, one report per line. Each report is a list of numbers called levels that are separated by spaces. For example:\n7 6 4 2 1 1 2 7 8 9 9 7 6 2 1 1 3 2 4 5 8 6 4 4 1 1 3 6 7 9 This example data contains six reports each containing five levels.\nThe engineers are trying to figure out which reports are safe. The Red-Nosed reactor safety systems can only tolerate levels that are either gradually increasing or gradually decreasing. So, a report only counts as safe if both of the following are true:\nThe levels are either all increasing or all decreasing. Any two adjacent levels differ by at least one and at most three. In the example above, the reports can be found safe or unsafe by checking those rules:\n7 6 4 2 1: Safe because the levels are all decreasing by 1 or 2. 1 2 7 8 9: Unsafe because 2 7 is an increase of 5. 9 7 6 2 1: Unsafe because 6 2 is a decrease of 4. 1 3 2 4 5: Unsafe because 1 3 is increasing but 3 2 is decreasing. 8 6 4 4 1: Unsafe because 4 4 is neither an increase or a decrease. 1 3 6 7 9: Safe because the levels are all increasing by 1, 2, or 3. So, in this example, 2 reports are safe.\nAnalyze the unusual data from the engineers. How many reports are safe?\nMy input file: 2024-12-02-1-aoc.txt\nFirst, let\u0026rsquo;s load the data and take a look:\ndata \u0026lt;- data.table::fread(\u0026#34;2024-12-02-1-aoc.txt\u0026#34;, header = FALSE, fill = TRUE) head(data) V1 V2 V3 V4 V5 V6 V7 V8 \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; 74 76 78 79 76 NA NA NA 38 40 43 44 44 NA NA NA 1 2 4 6 8 9 13 NA 65 68 70 72 75 76 81 NA 89 91 92 95 93 94 NA NA 15 17 16 18 19 17 NA NA My core idea here was to compute the differences between each element in a row - luckily, R provides an inbuilt diff() function that provides exactly what I need. Then, we check if all differences are either 1, 2, 3 or -1, -2, -3. Note that each row has a different number of elements, so we have to omit NA values via na.omit when applying the check.\nisalltrue \u0026lt;- function(rep) { return(all(diff(rep) %in% c(1, 2, 3)) | all(diff(rep) %in% c(-1, -2, -3))) } isreport \u0026lt;- function(row) { return(isalltrue(na.omit(row))) } valid.reports \u0026lt;- apply(data, 1, isreport) table(valid.reports) valid.reports FALSE TRUE 743 257 Part 2: Problem Dampener # Let\u0026rsquo;s read the task for part 2! The engineers are surprised by the low number of safe reports until they realize they forgot to tell you about the Problem Dampener.\nThe Problem Dampener is a reactor-mounted module that lets the reactor safety systems tolerate a single bad level in what would otherwise be a safe report. It\u0026rsquo;s like the bad level never happened!\nNow, the same rules apply as before, except if removing a single level from an unsafe report would make it safe, the report instead counts as safe.\nMore of the above example\u0026rsquo;s reports are now safe:\n-7 6 4 2 1: Safe without removing any level. -1 2 7 8 9: Unsafe regardless of which level is removed. -9 7 6 2 1: Unsafe regardless of which level is removed. -1 3 2 4 5: Safe by removing the second level, 3. -8 6 4 4 1: Safe by removing the third level, 4. -1 3 6 7 9: Safe without removing any level. Thanks to the Problem Dampener, 4 reports are actually safe!\nUpdate your analysis by handling situations where the Problem Dampener can remove a single level from unsafe reports. How many reports are now safe?\nSo, this task is a good example on how easy it is to get lost in a overly complicated approach. My first approach was basically to somehow check if the general trend was descending or ascending and based on that to remove opposite differences. This was too complicated and in the end I found 4 less safe reports than were needed\u0026hellip; (Also note the beautiful print debuggin\u0026hellip;)\nisalltrue \u0026lt;- function(reptrue1, reptrue2) { return((all(reptrue1) == TRUE) | (all(reptrue2) == TRUE)) } iscorrtrue \u0026lt;- function(rep, reptrue) { rep \u0026lt;- rep[-which.min(reptrue)] repdiff \u0026lt;- diff(rep) reptrue1 \u0026lt;- !(repdiff \u0026lt; 1 | repdiff \u0026gt; 3) reptrue2 \u0026lt;- !(repdiff \u0026lt; -3 | repdiff \u0026gt; -1) out \u0026lt;- isalltrue(reptrue1, reptrue2) print(rep) print(out) return(out) } isreport2 \u0026lt;- function(row) { print(\u0026#34;\u0026#34;) rep \u0026lt;- na.omit(as.numeric(row)) print(c(rep)) repdiff \u0026lt;- c(1, diff(rep)) reptrue1 \u0026lt;- !(repdiff \u0026lt; 1 | repdiff \u0026gt; 3) repdiff \u0026lt;- c(-1, diff(rep)) reptrue2 \u0026lt;- !(repdiff \u0026lt; -3 | repdiff \u0026gt; -1) if (isalltrue(reptrue1, reptrue2) == TRUE) { out \u0026lt;- isalltrue(reptrue1, reptrue2) print(\u0026#34;----- 1 -----\u0026#34;) print(out) } else if (sum(diff(rep) \u0026gt; 0) \u0026gt; sum(diff(rep) \u0026lt; 0)) { print(\u0026#34;----- 2 -----\u0026#34;) out \u0026lt;- iscorrtrue(rep, reptrue1) } else if (sum(diff(rep) \u0026lt; 0) \u0026gt; sum(diff(rep) \u0026gt; 0)) { print(\u0026#34;----- 3 -----\u0026#34;) out \u0026lt;- iscorrtrue(rep, reptrue2) } else { out \u0026lt;- FALSE } if (out == FALSE) { print(\u0026#34;----- 4 -----\u0026#34;) rep \u0026lt;- na.omit(as.numeric(row))[-1] repdiff \u0026lt;- diff(rep) reptrue1 \u0026lt;- !(repdiff \u0026lt; 1 | repdiff \u0026gt; 3) reptrue2 \u0026lt;- !(repdiff \u0026lt; -3 | repdiff \u0026gt; -1) out \u0026lt;- isalltrue(reptrue1, reptrue2) print(c(rep)) print(out) } return(out) } My actual working approach was in the end brute force. I just threw out each element in a row and checked if the report became positive - much simpler and effective!\nisreport2 \u0026lt;- function(row) { rep \u0026lt;- na.omit(row) out \u0026lt;- isalltrue(rep) if (out == FALSE) { for (i in seq_along(rep)) { # rep \u0026lt;- rep[-i] does not update the variable # with rep.i, a new variable is assigned (rep.1, rep.2, ...) rep.i \u0026lt;- rep[-i] out \u0026lt;- isalltrue(rep.i) if (out == TRUE) { break } } } return(out) } sum(apply(data, 1, isreport2)) 328\n"},{"id":2,"href":"/posts/aoc2024-day01/","title":"Advent of code 2024 - Day 1: Historian Hysteria","section":"Posts","content":" Table of Contents Part 1: Locations Part 2: Similarities I try to solve this year\u0026rsquo;s Advent of Code 2024 riddles in R. This is Day 1 - see https:adventofcode.com/2024/day/1\nPart 1: Locations # Lets first read the task: Throughout the Chief\u0026rsquo;s office, the historically significant locations are listed not by name but by a unique number called the location ID. To make sure they don\u0026rsquo;t miss anything, The Historians split into two groups, each searching the office and trying to create their own complete list of location IDs.\nThere\u0026rsquo;s just one problem: by holding the two lists up side by side (your puzzle input), it quickly becomes clear that the lists aren\u0026rsquo;t very similar. Maybe you can help The Historians reconcile their lists?\nFor example:\n3 4 4 3 2 5 1 3 3 9 3 3 Maybe the lists are only off by a small amount! To find out, pair up the numbers and measure how far apart they are. Pair up the smallest number in the left list with the smallest number in the right list, then the second-smallest left number with the second-smallest right number, and so on.\nWithin each pair, figure out how far apart the two numbers are; you\u0026rsquo;ll need to add up all of those distances. For example, if you pair up a 3 from the left list with a 7 from the right list, the distance apart is 4; if you pair up a 9 with a 3, the distance apart is 6.\nIn the example list above, the pairs and distances would be as follows:\nThe smallest number in the left list is 1, and the smallest number in the right list is 3. The distance between them is 2. The second-smallest number in the left list is 2, and the second-smallest number in the right list is another 3. The distance between them is 1. The third-smallest number in both lists is 3, so the distance between them is 0. The next numbers to pair up are 3 and 4, a distance of 1. The fifth-smallest numbers in each list are 3 and 5, a distance of 2. Finally, the largest number in the left list is 4, while the largest number in the right list is 9; these are a distance 5 apart. To find the total distance between the left list and the right list, add up the distances between all of the pairs you found. In the example above, this is 2 + 1 + 0 + 1 + 2 + 5, a total distance of 11!\nYour actual left and right lists contain many location IDs. What is the total distance between your lists?\nMy input file: 2024-12-01-1-aoc.txt\nFirst, we read the data - the fread command from the data.table package is more versatile than read.delim from base and directly reads the data as a data.table, which has some benefits.\ndata \u0026lt;- data.table::fread(\u0026#34;2024-12-01-1-aoc.txt\u0026#34;, header = FALSE) head(data) V1 V2 \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; 41226 69190 89318 10100 59419 23880 63157 20193 81510 22869 83942 63304 Now, the idea is to get the ordered versions of V1 and V2.\nv1 \u0026lt;- data[order(V1)]$V1 v2 \u0026lt;- data[order(V2)]$V2 head(v1) 10188 10314 10319 10348 10408 10668 Coming from Python, this syntax looks a bit odd. The documentation of data.table::setorder helps:\nsetorder (and setorderv) reorders the rows of a data.table based on the columns (and column order) provided. It reorders the table by reference and is therefore very memory efficient. Note that queries like x[order(.)] are optimised internally to use data.table\u0026#39;s fast order. So, data[order(V1)] is actually short for data.table::setorder(data, V1). Then, we extract the vector by name using the $ operator, which allows to extract elements by name.\nhead(data.table::setorder(data, V1)$V1) 10188 10314 10319 10348 10408 10668 The actual computation is just the sum of the absolute difference:\nsum(abs(v1 - v2)) 3574690\nPart 2: Similarities # Lets first read the task: This time, you\u0026rsquo;ll need to figure out exactly how often each number from the left list appears in the right list. Calculate a total similarity score by adding up each number in the left list after multiplying it by the number of times that number appears in the right list.\nHere are the same example lists again:\n3 4 4 3 2 5 1 3 3 9 3 3 For these example lists, here is the process of finding the similarity score:\nThe first number in the left list is 3. It appears in the right list three times, so the similarity score increases by 3 * 3 = 9. The second number in the left list is 4. It appears in the right list once, so the similarity score increases by 4 * 1 = 4. The third number in the left list is 2. It does not appear in the right list, so the similarity score does not increase (2 * 0 = 0). The fourth number, 1, also does not appear in the right list. The fifth number, 3, appears in the right list three times; the similarity score increases by 9. The last number, 3, appears in the right list three times; the similarity score again increases by 9. So, for these example lists, the similarity score at the end of this process is 31 (9 + 4 + 0 + 0 + 9 + 9).\nOnce again consider your left and right lists. What is their similarity score?\nMy idea here is simple - we first count the occurrences of V2 to be able to check V1 against them. Here, applying the table command works like pandas.value_counts() and achieves this. We convert the output to a data.frame and assign the V1 values as row names. Note: table converts the values which are counted to string, e.g. \u0026quot;10019\u0026quot; instead of 10019.\ntab \u0026lt;- table(v2) v2series \u0026lt;- data.frame(c(tab), row.names = names(tab)) head(v2series) v2series[\u0026#34;10019\u0026#34;, ] c.tab. \u0026lt;int\u0026gt; 10019 1 10100 1 10206 1 10428 1 10645 1 10972 1 1\nNow, we loop through V1 and try to access the frequency in V2 (by converting to string first, via as.character). If nothing is found in V2, an NA value is returned. For everything that is not NA, we compute the similarity score, append it to a list, and sum in the end. Note: a list in R can not directly be summed (I don\u0026rsquo;t know why that is) - so we have to unlist first.\nsim \u0026lt;- list() for (num in v1) { myfreq \u0026lt;- v2series[as.character(num), ] if (!is.na(myfreq)) { score \u0026lt;- myfreq * num sim \u0026lt;- append(sim, score) } } sum(unlist(sim)) 22565391\n"},{"id":3,"href":"/posts/aoc2023-day07/","title":"Advent of code 2023 - Day 7: Camel Cards","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 7 - see https:adventofcode.com/2023/day/7\nPart 1 # Lets first read the task: In Camel Cards, you get a list of hands, and your goal is to order them based on the strength of each hand. A hand consists of five cards labeled one of A, K, Q, J, T, 9, 8, 7, 6, 5, 4, 3, or 2. The relative strength of each card follows this order, where A is the highest and 2 is the lowest.\nEvery hand is exactly one type. From strongest to weakest, they are:\nFive of a kind, where all five cards have the same label: AAAAA Four of a kind, where four cards have the same label and one card has a different label: AA8AA Full house, where three cards have the same label, and the remaining two cards share a different label: 23332 Three of a kind, where three cards have the same label, and the remaining two cards are each different from any other card in the hand: TTT98 Two pair, where two cards share one label, two other cards share a second label, and the remaining card has a third label: 23432 One pair, where two cards share one label, and the other three cards have a different label from the pair and each other: A23A4 High card, where all cards\u0026rsquo; labels are distinct: 23456 Hands are primarily ordered based on type; for example, every full house is stronger than any three of a kind.\nIf two hands have the same type, a second ordering rule takes effect. Start by comparing the first card in each hand. If these cards are different, the hand with the stronger first card is considered stronger. If the first card in each hand have the same label, however, then move on to considering the second card in each hand. If they differ, the hand with the higher second card wins; otherwise, continue with the third card in each hand, then the fourth, then the fifth.\nSo, 33332 and 2AAAA are both four of a kind hands, but 33332 is stronger because its first card is stronger. Similarly, 77888 and 77788 are both a full house, but 77888 is stronger because its third card is stronger (and both hands have the same first and second card).\nTo play Camel Cards, you are given a list of hands and their corresponding bid (your puzzle input). For example:\n32T3K 765 T55J5 684 KK677 28 KTJJT 220 QQQJA 483 This example shows five hands; each hand is followed by its bid amount. Each hand wins an amount equal to its bid multiplied by its rank, where the weakest hand gets rank 1, the second-weakest hand gets rank 2, and so on up to the strongest hand. Because there are five hands in this example, the strongest hand will have rank 5 and its bid will be multiplied by 5.\nSo, the first step is to put the hands in order of strength:\n32T3K is the only one pair and the other hands are all a stronger type, so it gets rank 1. KK677 and KTJJT are both two pair. Their first cards both have the same label, but the second card of KK677 is stronger (K vs T), so KTJJT gets rank 2 and KK677 gets rank 3. T55J5 and QQQJA are both three of a kind. QQQJA has a stronger first card, so it gets rank 5 and T55J5 gets rank 4. Now, you can determine the total winnings of this set of hands by adding up the result of multiplying each hand\u0026rsquo;s bid with its rank (765 * 1 + 220 * 2 + 28 * 3 + 684 * 4 + 483 * 5). So the total winnings in this example are 6440.\nFind the rank of every hand in your set. What are the total winnings?\nMy input file: 2023-12-07-1-aoc.txt\nAs always - let\u0026rsquo;s read in the data.\nimport pandas as pd camel = pd.read_table(\u0026#39;data/2023-12-07-1-aoc.txt\u0026#39;, delim_whitespace=True, header=None, names=[\u0026#39;cards\u0026#39;, \u0026#39;bid\u0026#39;], dtype={\u0026#39;cards\u0026#39;: \u0026#39;string\u0026#39;, \u0026#39;bid\u0026#39;: \u0026#39;Int64\u0026#39;}) camel cards bid 0 A833A 309 1 Q33J3 205 2 55KK5 590 3 K4457 924 4 Q3QT3 11 \u0026hellip; \u0026hellip; \u0026hellip; 995 KQJ53 367 996 57866 537 997 Q94A9 210 998 J448A 903 999 6J66J 114 1000 rows × 2 columns\nFor sorting, we need to compute a type column. The 7 types can be deduced based on how many different cards are there (through set()) and the number of cards.\nAlso, for the next step we prepare one column per card string in the cards.\ndef get_type(cards): cardset = set(cards) cardlen = len(cardset) cardlist = list(cards) cardnum = [cardlist.count(i) for i in cardset] if cardlen == 1: ctype = 1 elif cardlen in (4, 5): ctype = cardlen + 2 elif cardlen == 2: ctype = 2 if set(cardnum) == {1, 4} else 3 else: ctype = 4 if 3 in cardnum else 5 return ctype camel[\u0026#39;type\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: get_type(x)) for i in range(5): camel[f\u0026#39;c{i}\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: list(x)[i]) camel cards bid type c0 c1 c2 c3 c4 0 A833A 309 5 A 8 3 3 A 1 Q33J3 205 4 Q 3 3 J 3 2 55KK5 590 3 5 5 K K 5 3 K4457 924 6 K 4 4 5 7 4 Q3QT3 11 5 Q 3 Q T 3 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 KQJ53 367 7 K Q J 5 3 996 57866 537 6 5 7 8 6 6 997 Q94A9 210 6 Q 9 4 A 9 998 J448A 903 6 J 4 4 8 A 999 6J66J 114 3 6 J 6 6 J 1000 rows × 8 columns\nFor sorting, pandas contributes the sort_values method, which supports sorting by multiple columns (here, we first sort according to type, then c0, c1, \u0026hellip;). For easier sorting by numerical values, we convert the values using the card_dict dictionary, which implements the given sorting order.\ncard_dict = {c: i for i, c in enumerate([7, 6, 5, 4, 3, 2, 1, \u0026#39;2\u0026#39;, \u0026#39;3\u0026#39;, \u0026#39;4\u0026#39;, \u0026#39;5\u0026#39;, \u0026#39;6\u0026#39;, \u0026#39;7\u0026#39;, \u0026#39;8\u0026#39;, \u0026#39;9\u0026#39;, \u0026#39;T\u0026#39;, \u0026#39;J\u0026#39;, \u0026#39;Q\u0026#39;, \u0026#39;K\u0026#39;, \u0026#39;A\u0026#39;])} camel = camel.sort_values(by=[\u0026#39;type\u0026#39;, \u0026#39;c0\u0026#39;, \u0026#39;c1\u0026#39;, \u0026#39;c2\u0026#39;, \u0026#39;c3\u0026#39;, \u0026#39;c4\u0026#39;], ignore_index=True, key=lambda x: x.map(card_dict)) camel.index += 1 camel cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 AA5AA 565 2 A A 5 A A 997 AAATA 704 2 A A A T A 998 AAAA3 145 2 A A A A 3 999 AAAAJ 594 2 A A A A J 1000 JJJJJ 219 1 J J J J J 1000 rows × 8 columns\nLastly, we multiply the bid with the index and sum the result.\ncamel.loc[:, \u0026#39;bid\u0026#39;].mul(camel.index).sum() 246912307 Part 2 # Lets first read the task: To make things a little more interesting, the Elf introduces one additional rule. Now, J cards are jokers - wildcards that can act like whatever card would make the hand the strongest type possible.\nTo balance this, J cards are now the weakest individual cards, weaker even than 2. The other cards stay in the same order: A, K, Q, T, 9, 8, 7, 6, 5, 4, 3, 2, J.\nJ cards can pretend to be whatever card is best for the purpose of determining hand type; for example, QJJQ2 is now considered four of a kind. However, for the purpose of breaking ties between two hands of the same type, J is always treated as J, not the card it\u0026rsquo;s pretending to be: JKKK2 is weaker than QQQQ2 because J is weaker than Q.\nNow, the above example goes very differently:\n32T3K 765 T55J5 684 KK677 28 KTJJT 220 QQQJA 483 32T3K is still the only one pair; it doesn\u0026rsquo;t contain any jokers, so its strength doesn\u0026rsquo;t increase. KK677 is now the only two pair, making it the second-weakest hand. T55J5, KTJJT, and QQQJA are now all four of a kind! T55J5 gets rank 3, QQQJA gets rank 4, and KTJJT gets rank 5. With the new joker rule, the total winnings in this example are 5905.\nUsing the new joker rule, find the rank of every hand in your set. What are the new total winnings?\nWhat changes? We have to update our get_type() function and our card_dict dictionary, otherwise the method stays the same!\nLet\u0026rsquo;s first write get_type_joker(): if no joker is in the cards, we can just refer to get_type(). If we have 1, 2 or 5 different card faces, the answer is trivial. We only have to use the number of jokers and the maximum number of other faces if we have 3 or 4 different faces.\nNote: if a J is present, there can never be high card (type = 7), because we always can get one pair. Also we can never get two pair (type = 5), because we can always construct three of a kind or full house.\ndef get_type_joker(cards): cardset = set(cards) cardlen = len(cardset) cardlist = list(cards) cardnum = [cardlist.count(i) for i in cardset] if not \u0026#39;J\u0026#39; in cardset: ctype = get_type(cards) else: idxj = list(cardset).index(\u0026#39;J\u0026#39;) numj = cardnum[idxj] cardnum.pop(idxj) if cardlen in [1, 2]: ctype = 1 elif cardlen == 5: ctype = 6 else: if (numj in [1, 2]) and (cardlen == 4): ctype = 4 else: ctype = 6 - max(cardnum) - numj return ctype camel[\u0026#39;type\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: get_type_joker(x)) camel cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 AA5AA 565 2 A A 5 A A 997 AAATA 704 2 A A A T A 998 AAAA3 145 2 A A A A 3 999 AAAAJ 594 1 A A A A J 1000 JJJJJ 219 1 J J J J J 1000 rows × 8 columns\nNow, we just move J in our new card_dict_joker to where it belongs, the rest of the solution is the same.\ncard_dict_joker = {c: i for i, c in enumerate( [7, 6, 5, 4, 3, 2, 1, \u0026#39;J\u0026#39;, \u0026#39;2\u0026#39;, \u0026#39;3\u0026#39;, \u0026#39;4\u0026#39;, \u0026#39;5\u0026#39;, \u0026#39;6\u0026#39;, \u0026#39;7\u0026#39;, \u0026#39;8\u0026#39;, \u0026#39;9\u0026#39;, \u0026#39;T\u0026#39;, \u0026#39;Q\u0026#39;, \u0026#39;K\u0026#39;, \u0026#39;A\u0026#39;])} camel = camel.sort_values(by=[\u0026#39;type\u0026#39;, \u0026#39;c0\u0026#39;, \u0026#39;c1\u0026#39;, \u0026#39;c2\u0026#39;, \u0026#39;c3\u0026#39;, \u0026#39;c4\u0026#39;], ignore_index=True, key=lambda x: x.map(card_dict_joker)) camel.index += 1 display(camel) print(\u0026#39;Our new solution: \u0026#39;, camel.loc[:, \u0026#39;bid\u0026#39;].mul(camel.index).sum()) cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 TJTTJ 609 1 T J T T J 997 QQJQJ 131 1 Q Q J Q J 998 QQQJQ 831 1 Q Q Q J Q 999 KKKJK 183 1 K K K J K 1000 AAAAJ 594 1 A A A A J 1000 rows × 8 columns\nOur new solution: 246894760 "},{"id":4,"href":"/posts/aoc2023-day06/","title":"Advent of code 2023 - Day 6: Wait For It","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 6 - see https:adventofcode.com/2023/day/6\nPart 1 # Lets first read the task: You manage to sign up as a competitor in the boat races just in time. The organizer explains that it\u0026rsquo;s not really a traditional race - instead, you will get a fixed amount of time during which your boat has to travel as far as it can, and you win if your boat goes the farthest.\nAs part of signing up, you get a sheet of paper (your puzzle input) that lists the time allowed for each race and also the best distance ever recorded in that race. To guarantee you win the grand prize, you need to make sure you go farther in each race than the current record holder.\nThe organizer brings you over to the area where the boat races are held. The boats are much smaller than you expected - they\u0026rsquo;re actually toy boats, each with a big button on top. Holding down the button charges the boat, and releasing the button allows the boat to move. Boats move faster if their button was held longer, but time spent holding the button counts against the total race time. You can only hold the button at the start of the race, and boats don\u0026rsquo;t move until the button is released.\nFor example:\nTime: 7 15 30 Distance: 9 40 200 This document describes three races:\nThe first race lasts 7 milliseconds. The record distance in this race is 9 millimeters. The second race lasts 15 milliseconds. The record distance in this race is 40 millimeters. The third race lasts 30 milliseconds. The record distance in this race is 200 millimeters. Your toy boat has a starting speed of zero millimeters per millisecond. For each whole millisecond you spend at the beginning of the race holding down the button, the boat\u0026rsquo;s speed increases by one millimeter per millisecond.\nSo, because the first race lasts 7 milliseconds, you only have a few options:\nDon\u0026rsquo;t hold the button at all (that is, hold it for 0 milliseconds) at the start of the race. The boat won\u0026rsquo;t move; it will have traveled 0 millimeters by the end of the race. Hold the button for 1 millisecond at the start of the race. Then, the boat will travel at a speed of 1 millimeter per millisecond for 6 milliseconds, reaching a total distance traveled of 6 millimeters. Hold the button for 2 milliseconds, giving the boat a speed of 2 millimeters per millisecond. It will then get 5 milliseconds to move, reaching a total distance of 10 millimeters. Hold the button for 3 milliseconds. After its remaining 4 milliseconds of travel time, the boat will have gone 12 millimeters. Hold the button for 4 milliseconds. After its remaining 3 milliseconds of travel time, the boat will have gone 12 millimeters. Hold the button for 5 milliseconds, causing the boat to travel a total of 10 millimeters. Hold the button for 6 milliseconds, causing the boat to travel a total of 6 millimeters. Hold the button for 7 milliseconds. That\u0026rsquo;s the entire duration of the race. You never let go of the button. The boat can\u0026rsquo;t move until you let go of the button. Please make sure you let go of the button so the boat gets to move. 0 millimeters. Since the current record for this race is 9 millimeters, there are actually 4 different ways you could win: you could hold the button for 2, 3, 4, or 5 milliseconds at the start of the race.\nIn the second race, you could hold the button for at least 4 milliseconds and at most 11 milliseconds and beat the record, a total of 8 different ways to win.\nIn the third race, you could hold the button for at least 11 milliseconds and no more than 19 milliseconds and still beat the record, a total of 9 ways you could win.\nTo see how much margin of error you have, determine the number of ways you can beat the record in each race; in this example, if you multiply these values together, you get 288 (4 * 8 * 9).\nDetermine the number of ways you could beat the record in each race. What do you get if you multiply these numbers together?\nMy input file: 2023-12-06-1-aoc.txt\nThe solution for this problem is shorter than the task description!\nAs always - let\u0026rsquo;s load the data. Note: we use the delim_whitesspace argument instead of sep=' ' to separate values by spaces of any length.\nimport pandas as pd import numpy as np race = pd.read_table(\u0026#39;data/2023-12-06-1-aoc.txt\u0026#39;, delim_whitespace=True, header=None, index_col=0) race.index = [\u0026#39;time\u0026#39;, \u0026#39;distance\u0026#39;] race 1 2 3 4 time 51 92 68 90 distance 222 2031 1126 1225 The we use Python list comprehensions to quickly calculate all distances for the range of the time value (and keep only the winning distances). Then we get the respective lengths and compute the product.\ndef get_distance(time, distance): return [t * (time - t) for t in range(1, time + 1) if (t * (time - t)) \u0026gt; distance] race.loc[\u0026#39;nwins\u0026#39;, :] = [len(get_distance(int(r.time), int(r.distance))) for c, r in race.items()] nwins_prod = np.prod(race.loc[\u0026#39;nwins\u0026#39;, :].values, dtype=int) display(race) print(f\u0026#39;The product of our nwins is {nwins_prod}\u0026#39;) 1 2 3 4 time 51.0 92.0 68.0 90.0 distance 222.0 2031.0 1126.0 1225.0 nwins 42.0 19.0 11.0 57.0 The product of our nwins is 500346 Part 2 # Lets first read the task: As the race is about to start, you realize the piece of paper with race times and record distances you got earlier actually just has very bad kerning. There\u0026rsquo;s really only one race - ignore the spaces between the numbers on each line.\nSo, the example from before:\nTime: 7 15 30 Distance: 9 40 200 \u0026hellip;now instead means this:\nTime: 71530 Distance: 940200 Now, you have to figure out how many ways there are to win this single race. In this example, the race lasts for 71530 milliseconds and the record distance you need to beat is 940200 millimeters. You could hold the button anywhere from 14 to 71516 milliseconds and beat the record, a total of 71503 ways!\nHow many ways can you beat the record in this one much longer race?\nFor this riddle we can take the same approach as for Part 1. The cell took 18s to evaluate on a consumer laptop from 2015 - no fancy workarounds needed :)\nracetime = int(\u0026#39;\u0026#39;.join(str(r) for r in race.loc[\u0026#39;time\u0026#39;, :4].astype(int))) racedist = int(\u0026#39;\u0026#39;.join(str(r) for r in race.loc[\u0026#39;distance\u0026#39;, :4].astype(int))) racewins = len(get_distance(racetime, racedist)) race[5] = [racetime, racedist, racewins] race 1 2 3 4 5 time 51.0 92.0 68.0 90.0 51926890 distance 222.0 2031.0 1126.0 1225.0 222203111261225 nwins 42.0 19.0 11.0 57.0 42515755 "},{"id":5,"href":"/docs/projects/","title":"Projects","section":"Docs","content":" Current projects # I am a believer in the principles of Open Science, Open Data and Open Source. Thus, I am currently working on a reproducible way of conducting my research on simulation of Fluorescence Correlation Spectroscopy (FCS) measurements and correcting artifacts using neural networks, such as Convolutional Neural Nets (CNNs). I do not have my workflow fixed yet, but to have an insight in my current approach - and with the principles of Open-notebook science in mind:\nFluotracify - doctoral research project done in a reproducible way\nDescription: In a current project, we apply Deep Learning techniques on Fluorescence Correlation Spectroscopy (FCS) data to correct a variety of hardware- and sample-related artifacts, such as photobleaching, contamination from additional slow moving particles, or sudden drops in intensity because of detector anomalies.\nConference talks # Seltmann A, Eggeling C, Waithe D. Automated, User-independent Correction of Artifacts in Fluorescence Correlation Spectroscopy Measurements using Convolutional Neural Networks. Quantitative BioImaging Conference (QBI); 2020 Jan 6-9; Oxford, UK "},{"id":6,"href":"/posts/aoc2023-day05/","title":"Advent of code 2023 - Day 5: If You Give A Seed A Fertilizer","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 5 - see https:adventofcode.com/2023/day/5\nUpdate [2023-12-31 So]:\nsubstitute get_generator() (own implementation) with range() (Python inbuilt) improve grid search so that it goes through all location ranges, still starting with the lowest range Part 1 # Lets first read the task: The almanac (your puzzle input) lists all of the seeds that need to be planted. It also lists what type of soil to use with each kind of seed, what type of fertilizer to use with each kind of soil, what type of water to use with each kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is identified with a number, but numbers are reused by each category - that is, soil 123 and fertilizer 123 aren\u0026rsquo;t necessarily related to each other.\nFor example:\nseeds: 79 14 55 13 seed-to-soil map: 50 98 2 52 50 48 soil-to-fertilizer map: 0 15 37 37 52 2 39 0 15 fertilizer-to-water map: 49 53 8 0 11 42 42 0 7 57 7 4 water-to-light map: 88 18 7 18 25 70 light-to-temperature map: 45 77 23 81 45 19 68 64 13 temperature-to-humidity map: 0 69 1 1 0 69 humidity-to-location map: 60 56 37 56 93 4 The almanac starts by listing which seeds need to be planted: seeds 79, 14, 55, and 13.\nThe rest of the almanac contains a list of maps which describe how to convert numbers from a source category into numbers in a destination category. That is, the section that starts with seed-to-soil map: describes how to convert a seed number (the source) to a soil number (the destination). This lets the gardener and his team know which soil to use with which seeds, which water to use with which fertilizer, and so on.\nRather than list every source number and its corresponding destination number one by one, the maps describe entire ranges of numbers that can be converted. Each line within a map contains three numbers: the destination range start, the source range start, and the range length.\nConsider again the example seed-to-soil map:\n50 98 2 52 50 48\nThe first line has a destination range start of 50, a source range start of 98, and a range length of 2. This line means that the source range starts at 98 and contains two values: 98 and 99. The destination range is the same length, but it starts at 50, so its two values are 50 and 51. With this information, you know that seed number 98 corresponds to soil number 50 and that seed number 99 corresponds to soil number 51. o The second line means that the source range starts at 50 and contains 48 values: 50, 51, \u0026hellip;, 96, 97. This corresponds to a destination range starting at 52 and also containing 48 values: 52, 53, \u0026hellip;, 98, 99. So, seed number 53 corresponds to soil number 55.\nAny source numbers that aren\u0026rsquo;t mapped correspond to the same destination number. So, seed number 10 corresponds to soil number 10.\nSo, the entire list of seed numbers and their corresponding soil numbers looks like this:\nseed soil 0 0 1 1 ... ... 48 48 49 49 50 52 51 53 ... ... 96 98 97 99 98 50 99 51 With this map, you can look up the soil number required for each initial seed number:\nSeed number 79 corresponds to soil number 81. Seed number 14 corresponds to soil number 14. Seed number 55 corresponds to soil number 57. Seed number 13 corresponds to soil number 13. The gardener and his team want to get started as soon as possible, so they\u0026rsquo;d like to know the closest location that needs a seed. Using these maps, find the lowest location number that corresponds to any of the initial seeds. To do this, you\u0026rsquo;ll need to convert each seed number through other categories until you can find its corresponding location number. In this example, the corresponding types are:\nSeed 79, soil 81, fertilizer 81, water 81, light 74, temperature 78, humidity 78, location 82. Seed 14, soil 14, fertilizer 53, water 49, light 42, temperature 42, humidity 43, location 43. Seed 55, soil 57, fertilizer 57, water 53, light 46, temperature 82, humidity 82, location 86. Seed 13, soil 13, fertilizer 52, water 41, light 34, temperature 34, humidity 35, location 35. So, the lowest location number in this example is 35.\nWhat is the lowest location number that corresponds to any of the initial seed numbers?\nMy input file: 2023-12-05-1-aoc.txt\nWow, this task is a mouthful\u0026hellip;\nLet\u0026rsquo;s start slowly and load the data. Our input text document contains several maps, which are clearly separated and have a title (seed-to-soil map etc). So we can tell pandas where each map starts and give each map a dataframe. I got the values for the skiprows and nrows argument by looking at the input file and\u0026hellip; counting :)\nimport pandas as pd import sys seeds = pd.read_table(\u0026#39;data/2023-12-05-1-aoc.txt\u0026#39;, nrows=1, sep=\u0026#39; \u0026#39;, header=None, index_col=0) seeds = seeds.values.flatten() map_opt = {\u0026#39;filepath_or_buffer\u0026#39;: \u0026#39;data/2023-12-05-1-aoc.txt\u0026#39;, \u0026#39;header\u0026#39;: None, \u0026#39;sep\u0026#39;: \u0026#39; \u0026#39;, \u0026#39;dtype\u0026#39;: \u0026#39;Int64\u0026#39;, \u0026#39;names\u0026#39;: [\u0026#39;dest_start\u0026#39;, \u0026#39;src_start\u0026#39;, \u0026#39;range\u0026#39;]} seed_soil = pd.read_table(skiprows=3, nrows=23, **map_opt) soil_fert = pd.read_table(skiprows=28, nrows=9, **map_opt) fert_water = pd.read_table(skiprows=39, nrows=20, **map_opt) water_light = pd.read_table(skiprows=61, nrows=40, **map_opt) light_temp = pd.read_table(skiprows=103, nrows=36, **map_opt) temp_humi = pd.read_table(skiprows=141, nrows=35, **map_opt) humi_loc = pd.read_table(skiprows=178, nrows=26, **map_opt) maps = (seed_soil, soil_fert, fert_water, water_light, light_temp, temp_humi, humi_loc) print(\u0026#39;seeds are just a numpy array:\u0026#39;) display(seeds) print(\u0026#39;The \u0026#34;humidity-to-location\u0026#34; map as an example:\u0026#39;) humi_loc seeds are just a numpy array: array([3169137700, 271717609, 3522125441, 23376095, 1233948799, 811833837, 280549587, 703867355, 166086528, 44766996, 2326968141, 69162222, 2698492851, 14603069, 2755327667, 348999531, 2600461189, 92332846, 1054656969, 169099767])The \u0026#34;humidity-to-location\u0026#34; map as an example: dest_start src_start range 0 3490144003 1623866227 218040905 1 1709610578 1620839197 3027030 2 105449249 586389428 113279526 3 1899604338 2167886292 348178199 4 1712637608 2678624215 186966730 5 218728775 0 245776251 6 2472992580 923734334 143670388 7 2616662968 3670169885 15297294 8 2247782537 1395629154 225210043 9 0 480940179 105449249 10 4113852846 3959729057 159909096 11 3322784653 1067404722 167359350 12 923734334 4119638153 175329143 13 1534964496 2516064491 162559724 14 2631960262 3496028440 140849733 15 2862695906 1234764072 97988355 16 1697524220 3636878173 12086358 17 2985646048 1332752427 62876727 18 1099063477 2970241510 435901019 19 4009202281 2865590945 104650565 20 2960684261 2142924505 24961787 21 2772809995 3406142529 89885911 22 4273761942 3648964531 21205354 23 3708184908 1841907132 301017373 24 464505026 245776251 235163928 25 3048522775 3685467179 274261878 My first attempt was to actually construct ranges like in the example above, mapping out all possible sources and destinations. Python quickly informed me that even constructing one pandas.Series with int64 values mapping seeds to soil would cost 64GB memory - not the best solution.\nSo we take a different approach. For convenience, let\u0026rsquo;s add a src_end and a dest_end column to our maps:\nfor df in maps: df[\u0026#39;src_end\u0026#39;] = df.loc[:, \u0026#39;src_start\u0026#39;] + df.loc[:, \u0026#39;range\u0026#39;] df[\u0026#39;dest_end\u0026#39;] = df.loc[:, \u0026#39;dest_start\u0026#39;] + df.loc[:, \u0026#39;range\u0026#39;] print(\u0026#39;Again the \u0026#34;humidity-to-location\u0026#34; map as an example:\u0026#39;) humi_loc Again the \u0026#34;humidity-to-location\u0026#34; map as an example: dest_start src_start range src_end dest_end 0 3490144003 1623866227 218040905 1841907132 3708184908 1 1709610578 1620839197 3027030 1623866227 1712637608 2 105449249 586389428 113279526 699668954 218728775 3 1899604338 2167886292 348178199 2516064491 2247782537 4 1712637608 2678624215 186966730 2865590945 1899604338 5 218728775 0 245776251 245776251 464505026 6 2472992580 923734334 143670388 1067404722 2616662968 7 2616662968 3670169885 15297294 3685467179 2631960262 8 2247782537 1395629154 225210043 1620839197 2472992580 9 0 480940179 105449249 586389428 105449249 10 4113852846 3959729057 159909096 4119638153 4273761942 11 3322784653 1067404722 167359350 1234764072 3490144003 12 923734334 4119638153 175329143 4294967296 1099063477 13 1534964496 2516064491 162559724 2678624215 1697524220 14 2631960262 3496028440 140849733 3636878173 2772809995 15 2862695906 1234764072 97988355 1332752427 2960684261 16 1697524220 3636878173 12086358 3648964531 1709610578 17 2985646048 1332752427 62876727 1395629154 3048522775 18 1099063477 2970241510 435901019 3406142529 1534964496 19 4009202281 2865590945 104650565 2970241510 4113852846 20 2960684261 2142924505 24961787 2167886292 2985646048 21 2772809995 3406142529 89885911 3496028440 2862695906 22 4273761942 3648964531 21205354 3670169885 4294967296 23 3708184908 1841907132 301017373 2142924505 4009202281 24 464505026 245776251 235163928 480940179 699668954 25 3048522775 3685467179 274261878 3959729057 3322784653 Now we actually compute the mapping. For each seed, we go through all mappings and in each mapping we go through each row. We find the row which contains the mapping and exctract the destination, which is the source for the next map until we reach the last map which gives us the locations.\nApproach 1: df.itertuples() is a convenient way to step through a pandas.DataFrame in this example. It is faster than df.iterrows() and returns the row as a NamedTuple - nice!\nApproach 2: I actually wondered if it would be faster to get all maps in one pd.DataFrame and then iterate through the mappings. To test this let\u0026rsquo;s construct a new DataFrame maps_df which contains all maps. Since the maps have different lengths it is important to cast the datatype to Int64, which is short for pd.Int64Dtype() and keeps values as integers, even if NA values are in the same column.\nApproach 3: A third alternative I tested (not shown here) was to check if a value is in a Python range with the in operator as in: if 3 in range(5):... . This was way too slow.\n# mapping version 1 def get_location(seed): current = seed for df in maps: current_map = [row for row in df.itertuples() if ((current \u0026gt; row.src_start) and (current \u0026lt; row.src_end))] if len(current_map) == 0: pass elif len(current_map) == 1: current = (current_map[0].dest_start + (current - current_map[0].src_start)) else: raise ValueError(\u0026#39;This should not happen!\u0026#39;) return current # mapping version 2 - around 10 times slower # you need to rename the maps_df columns so that they have a unique id # e.g. \u0026#39;src_start_1\u0026#39;, \u0026#39;src_start_2\u0026#39; etc # maps_df = pd.concat(maps, axis=\u0026#39;columns\u0026#39;) # def get_dest(i, src): # return (maps_df[(src \u0026gt; maps_df.loc[:, f\u0026#39;src_start_{i}\u0026#39;]) \u0026amp; # (src \u0026lt; maps_df.loc[:, f\u0026#39;src_end_{i}\u0026#39;])] # .loc[:, f\u0026#39;dest_start_{i}\u0026#39;] # .iloc[0]) # def get_location2(seed): # dest = seed # i = 1 # while i \u0026lt; 7: # dest = get_dest(i, dest) # i += 1 # return dest %timeit get_location(seeds[0]) 1.98 ms ± 72 μs per loop (mean ± std. dev. of 7 runs, 100 loops each) Now let\u0026rsquo;s get a list of locations:\nlocations = [get_location(s) for s in seeds] locations 2493982655 3209845376 3992357533 4163131463 4104485616 1952252479 3218677354 388071289 2181441450 2594336315 4049507670 2084517144 3119633635 428978312 3518771991 3704555655 953918455 2107687768 3448046330 2184454689 Lastly, just get the minimum of all location values.\nmin(locations) 388071289 Part 2 # Let\u0026rsquo;s read the task of part 2! Everyone will starve if you only plant such a small number of seeds. Re-reading the almanac, it looks like the seeds: line actually describes ranges of seed numbers.\nThe values on the initial seeds: line come in pairs. Within each pair, the first value is the start of the range and the second value is the length of the range. So, in the first line of the example above:\nseeds: 79 14 55 13\nThis line describes two ranges of seed numbers to be planted in the garden. The first range starts with seed number 79 and contains 14 values: 79, 80, \u0026hellip;, 91, 92. The second range starts with seed number 55 and contains 13 values: 55, 56, \u0026hellip;, 66, 67.\nNow, rather than considering four seed numbers, you need to consider a total of 27 seed numbers.\nIn the above example, the lowest location number can be obtained from seed number 82, which corresponds to soil 84, fertilizer 84, water 84, light 77, temperature 45, humidity 46, and location 46. So, the lowest location number is 46.\nConsider all of the initial seed numbers listed in the ranges on the first line of the almanac. What is the lowest location number that corresponds to any of the initial seed numbers?\nLet\u0026rsquo;s first construct a dataframe containing the range of seeds:\nseeds_df = pd.DataFrame({\u0026#39;start\u0026#39;: seeds[::2], \u0026#39;range\u0026#39;: seeds[1::2], \u0026#39;end\u0026#39;: seeds[::2] + seeds[1::2]}) print(f\u0026#39;There are {sum(seeds_df.loc[:, \u0026#34;range\u0026#34;]):_} seeds in total\u0026#39;) seeds_df There are 2_549_759_327 seeds in total start range end 0 3169137700 271717609 3440855309 1 3522125441 23376095 3545501536 2 1233948799 811833837 2045782636 3 280549587 703867355 984416942 4 166086528 44766996 210853524 5 2326968141 69162222 2396130363 6 2698492851 14603069 2713095920 7 2755327667 348999531 3104327198 8 2600461189 92332846 2692794035 9 1054656969 169099767 1223756736 Now - I really needed some time to finally realize, that going through all seed values is really unfeasable. So how do we deal with this problem?\nIn the end we need the lowest location number - thus our approach is to take the humi_loc map, start with the lowest location number and go up and get the corresponding seed values. The location of the first seed value which is inside seeds_df is our solution.\nSo first we rebuild the get_location function to a get_seed function (we reverse the maps tuple with maps[::-1] and switch src and dest).\ndef get_seed(location): current = location for df in maps[::-1]: current_map = [row for row in df.itertuples() if ((current \u0026gt;= row.dest_start) and (current \u0026lt; row.dest_end))] if len(current_map) == 0: pass elif len(current_map) == 1: current = (current_map[0].src_start + (current - current_map[0].dest_start)) else: raise ValueError(\u0026#39;This should not happen!\u0026#39;) return current Since the Python 3 range function does not store it\u0026rsquo;s contents in memory (similar to a generator), it is well suited to go through these large ranges.\nLastly, we deal with the sheer amount of possible values by performing a grid search. First, we check every millionth location. After the first match, we stop this search and check the last million locations before the match with a finer grid and so on. The last grid is just 1, so we find our lowest location.\nUpdate: We order our locations from smallest to largest with sort_values and go through them - but the search stops after the first match, since that will be our lowest location.\ndef grid_search(start: int, end: int, grid: list[int]): success = False for g in grid: print(f\u0026#39;Start search with grid={g}\u0026#39;) for l in range(start, end, g): current = get_seed(l) if any((current \u0026gt;= seeds_df.loc[:, \u0026#39;start\u0026#39;]) \u0026amp; (current \u0026lt; seeds_df.loc[:, \u0026#39;end\u0026#39;])): print(f\u0026#39;location {l} is the lowest which contains one of the given seeds ({current})\u0026#39;) start = l - g end = l success = True break return success for row in humi_loc.sort_values(\u0026#39;dest_start\u0026#39;).itertuples(): success = grid_search(start=row.dest_start, end=row.dest_end, grid=[1_000_000, 100_000, 1000, 1]) if success: print(\u0026#39;Finished search\u0026#39;) break Start search with grid=1000000 location 85000000 is the lowest which contains one of the given seeds (2605777210) Start search with grid=100000 location 84300000 is the lowest which contains one of the given seeds (2605077210) Start search with grid=1000 location 84207000 is the lowest which contains one of the given seeds (2604984210) Start search with grid=1 location 84206669 is the lowest which contains one of the given seeds (2604983879) Finished search "},{"id":7,"href":"/posts/aoc2023-day04/","title":"Advent of code 2023 - Day 4: Scratchcards","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 4 - see https:adventofcode.com/2023/day/4\nPart 1 # Lets first read the task: The Elf leads you over to the pile of colorful cards. There, you discover dozens of scratchcards, all with their opaque covering already scratched off. Picking one up, it looks like each card has two lists of numbers separated by a vertical bar (|): a list of winning numbers and then a list of numbers you have. You organize the information into a table (your puzzle input).\nAs far as the Elf has been able to figure out, you have to figure out which of the numbers you have appear in the list of winning numbers. The first match makes the card worth one point and each match after the first doubles the point value of that card.\nFor example:\nCard 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53 Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19 Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1 Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83 Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36 Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11 In the above example, card 1 has five winning numbers (41, 48, 83, 86, and 17) and eight numbers you have (83, 86, 6, 31, 17, 9, 48, and 53). Of the numbers you have, four of them (48, 83, 17, and 86) are winning numbers! That means card 1 is worth 8 points (1 for the first match, then doubled three times for each of the three matches after the first).\nCard 2 has two winning numbers (32 and 61), so it is worth 2 points. Card 3 has two winning numbers (1 and 21), so it is worth 2 points. Card 4 has one winning number (84), so it is worth 1 point. Card 5 has no winning numbers, so it is worth no points. Card 6 has no winning numbers, so it is worth no points. So, in this example, the Elf\u0026rsquo;s pile of scratchcards is worth 13 points.\nTake a seat in the large pile of colorful cards. How many points are they worth in total?\nMy input file: 2023-12-04-1-aoc.txt\nLoading this data is very similar to Day 2 - so let\u0026rsquo;s load the data as we did there. Our goal is to get win and yours columns holding the respective digits which we want to compare. We find the numbers with one or more digits using the regex \\d+. And we want them to be in sets (not lists), as we can logically compare sets in Python.\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-04-1-aoc.txt\u0026#39;, names=[\u0026#39;win\u0026#39;]) txt[\u0026#39;id\u0026#39;] = txt.loc[:, \u0026#39;win\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply( lambda x: int(x[0].strip(\u0026#39;Card \u0026#39;))) txt[\u0026#39;win\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;:\u0026#39;).apply(lambda x: x[1])) txt[\u0026#39;yours\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;|\u0026#39;) .apply(lambda x: x[1]) # get a list of only the numbers / digits .apply(lambda x: re.findall(r\u0026#39;\\d+\u0026#39;, x)) # convert the list of strings to a set of integers .apply(lambda x: set([int(i) for i in x]))) txt[\u0026#39;win\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;|\u0026#39;) .apply(lambda x: x[0]) .apply(lambda x: re.findall(r\u0026#39;\\d+\u0026#39;, x)) .apply(lambda x: set([int(i) for i in x]))) txt win id yours 0 {32, 36, 7, 9, 10, 12, 82, 85, 95, 31} 1 {2, 7, 9, 10, 12, 14, 21, 22, 23, 24, 31, 32, \u0026hellip; 1 {35, 76, 16, 82, 19, 22, 88, 59, 60, 95} 2 {7, 8, 12, 16, 19, 22, 26, 28, 35, 38, 44, 51,\u0026hellip; 2 {1, 70, 11, 78, 48, 19, 52, 88, 28, 94} 3 {3, 4, 8, 17, 18, 19, 24, 31, 34, 45, 52, 54, \u0026hellip; 3 {65, 2, 72, 28, 14, 16, 55, 91, 92, 62} 4 {3, 4, 6, 7, 8, 9, 15, 30, 33, 35, 47, 49, 51,\u0026hellip; 4 {38, 41, 75, 77, 50, 24, 94, 60, 61, 30} 5 {1, 2, 4, 5, 6, 7, 9, 10, 14, 17, 21, 29, 47, \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 213 {97, 98, 39, 41, 43, 12, 13, 19, 93, 95} 214 {5, 10, 17, 20, 28, 29, 33, 34, 36, 50, 51, 52\u0026hellip; 214 {97, 35, 69, 40, 74, 45, 20, 21, 62, 31} 215 {1, 8, 15, 17, 18, 25, 30, 33, 42, 44, 47, 52,\u0026hellip; 215 {33, 70, 71, 12, 78, 17, 51, 86, 60, 94} 216 {7, 8, 9, 10, 22, 29, 37, 39, 41, 43, 46, 47, \u0026hellip; 216 {98, 67, 68, 38, 70, 39, 72, 77, 45, 21} 217 {8, 21, 22, 25, 26, 31, 37, 41, 42, 48, 54, 57\u0026hellip; 217 {34, 9, 44, 78, 79, 16, 17, 19, 55, 92} 218 {1, 4, 20, 21, 27, 38, 39, 40, 41, 45, 46, 52,\u0026hellip; 218 rows × 3 columns\nNow, we get the logical conjunction of win and yours, these are our winning numbers. Then, the number of wins is converted to points - for all number of wins bigger than 1, we can get the points by 2**(n_wins-1).\ntxt[\u0026#39;n_wins\u0026#39;] = txt.apply( lambda row: len(row.loc[\u0026#39;win\u0026#39;] \u0026amp; row.loc[\u0026#39;yours\u0026#39;]), axis=1) txt[\u0026#39;points\u0026#39;] = txt.loc[:, \u0026#39;n_wins\u0026#39;].apply( lambda x: 2**(x-1) if x \u0026gt; 1 else x) txt.loc[:, [\u0026#39;win\u0026#39;, \u0026#39;n_wins\u0026#39;, \u0026#39;points\u0026#39;]] win n_wins points 0 {32, 36, 7, 9, 10, 12, 82, 85, 95, 31} 10 512 1 {35, 76, 16, 82, 19, 22, 88, 59, 60, 95} 10 512 2 {1, 70, 11, 78, 48, 19, 52, 88, 28, 94} 5 16 3 {65, 2, 72, 28, 14, 16, 55, 91, 92, 62} 0 0 4 {38, 41, 75, 77, 50, 24, 94, 60, 61, 30} 0 0 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 213 {97, 98, 39, 41, 43, 12, 13, 19, 93, 95} 0 0 214 {97, 35, 69, 40, 74, 45, 20, 21, 62, 31} 0 0 215 {33, 70, 71, 12, 78, 17, 51, 86, 60, 94} 2 2 216 {98, 67, 68, 38, 70, 39, 72, 77, 45, 21} 1 1 217 {34, 9, 44, 78, 79, 16, 17, 19, 55, 92} 0 0 218 rows × 3 columns\nThen, we just sum up:\nsum(txt.loc[:, \u0026#39;points\u0026#39;]) 25004 Part 2 # Let\u0026rsquo;s read the task of part 2! There\u0026rsquo;s no such thing as \u0026ldquo;points\u0026rdquo;. Instead, scratchcards only cause you to win more scratchcards equal to the number of winning numbers you have.\nSpecifically, you win copies of the scratchcards below the winning card equal to the number of matches. So, if card 10 were to have 5 matching numbers, you would win one copy each of cards 11, 12, 13, 14, and 15.\nCopies of scratchcards are scored like normal scratchcards and have the same card number as the card they copied. So, if you win a copy of card 10 and it has 5 matching numbers, it would then win a copy of the same cards that the original card 10 won: cards 11, 12, 13, 14, and 15. This process repeats until none of the copies cause you to win any more cards. (Cards will never make you copy a card past the end of the table.)\nThis time, the above example goes differently:\nCard 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53 Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19 Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1 Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83 Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36 Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11 Card 1 has four matching numbers, so you win one copy each of the next four cards: cards 2, 3, 4, and 5. Your original card 2 has two matching numbers, so you win one copy each of cards 3 and 4. Your copy of card 2 also wins one copy each of cards 3 and 4. Your four instances of card 3 (one original and three copies) have two matching numbers, so you win four copies each of cards 4 and 5. Your eight instances of card 4 (one original and seven copies) have one matching number, so you win eight copies of card 5. Your fourteen instances of card 5 (one original and thirteen copies) have no matching numbers and win no more cards. Once all of the originals and copies have been processed, you end up with 1 instance of card 1, 2 instances of card 2, 4 instances of card 3, 8 instances of card 4, 14 instances of card 5, and 1 instance of card 6. In total, this example pile of scratchcards causes you to ultimately have 30 scratchcards!\nProcess all of the original and copied scratchcards until no more scratchcards are won. Including the original set of scratchcards, how many total scratchcards do you end up with?\nWe will use the n_wins column we created before and go from there. We step through each Game. Each current game gets +1 card. Then, we step through the number of next games depending on our n_wins. Each of these gets added the card number of the current game.\ntxt[\u0026#39;cards\u0026#39;] = 0 for i, nwin in enumerate(txt.loc[:, \u0026#39;n_wins\u0026#39;]): txt.loc[i, \u0026#39;cards\u0026#39;] += 1 for j in range(1, nwin+1): txt.loc[i+j, \u0026#39;cards\u0026#39;] += txt.loc[i, \u0026#39;cards\u0026#39;] txt.loc[:, [\u0026#39;n_wins\u0026#39;, \u0026#39;cards\u0026#39;]] n_wins cards 0 10 1 1 10 2 2 5 4 3 0 8 4 0 8 \u0026hellip; \u0026hellip; \u0026hellip; 213 0 9608 214 0 8927 215 2 8927 216 1 12636 217 0 21564 218 rows × 2 columns\nNow, we just sum the output again:\nsum(txt[\u0026#39;cards\u0026#39;]) 14427616 "},{"id":8,"href":"/posts/aoc2023-day03/","title":"Advent of code 2023 - Day 3: Gear Ratios","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 3 - see https:adventofcode.com/2023/day/3\nPart 1 # Lets first read the task: The engine schematic (your puzzle input) consists of a visual representation of the engine. There are lots of numbers and symbols you don\u0026rsquo;t really understand, but apparently any number adjacent to a symbol, even diagonally, is a \u0026ldquo;part number\u0026rdquo; and should be included in your sum. (Periods (.) do not count as a symbol.)\nHere is an example engine schematic:\n467..114.. ...*...... ..35..633. ......#... 617*...... .....+.58. ..592..... ......755. ...$.*.... .664.598.. In this schematic, two numbers are not part numbers because they are not adjacent to a symbol: 114 (top right) and 58 (middle right). Every other number is adjacent to a symbol and so is a part number; their sum is 4361.\nOf course, the actual engine schematic is much larger. What is the sum of all of the part numbers in the engine schematic?\nMy input file: 2023-12-03-1-aoc.txt\nOkay, let\u0026rsquo;s first get the input as a numpy character array\nimport numpy as np import pandas as pd import sys import matplotlib.pyplot as plt from scipy import ndimage as ndi np.set_printoptions(threshold=sys.maxsize) txt = pd.read_table(\u0026#39;data/2023-12-03-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) arr = np.chararray((txt.size, txt.size), unicode=True) txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda x: [i for i in x]) for i, code in enumerate(txt[\u0026#39;code\u0026#39;]): arr[i, :] = code print((arr[:15, :15])) [[\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;5\u0026#39; \u0026#39;3\u0026#39; \u0026#39;.\u0026#39; \u0026#39;4\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;2\u0026#39; \u0026#39;6\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;5\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;3\u0026#39; \u0026#39;8\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;4\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;6\u0026#39; \u0026#39;6\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;1\u0026#39; \u0026#39;2\u0026#39; \u0026#39;6\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;3\u0026#39; \u0026#39;1\u0026#39; \u0026#39;/\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;/\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;5\u0026#39; \u0026#39;3\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;5\u0026#39; \u0026#39;5\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;6\u0026#39; \u0026#39;8\u0026#39; \u0026#39;5\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;0\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;3\u0026#39; \u0026#39;8\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39;]] /tmp/ipykernel_18336/87086526.py:9: DeprecationWarning: `np.chararray` is deprecated and will be removed from the main namespace in the future. Use an array with a string or bytes dtype instead. arr = np.chararray((txt.size, txt.size), unicode=True) Now extract symbols, digits and the empty space. We use the numpy character methods for that. This way, we create a binary mask for all digits and a binary mask for all empty space (.). The symbols are then every character which is neither.\ndigits = np.char.isdigit(arr) empty = np.char.endswith(arr, \u0026#39;.\u0026#39;) symbols = ~(digits | empty) # just for visualization plt.figure(figsize=(16, 5)) plt.subplot(131, title=\u0026#39;symbols\u0026#39;).matshow(symbols) plt.subplot(132, title=\u0026#39;digits\u0026#39;).matshow(digits) plt.subplot(133, title=\u0026#39;empty\u0026#39;).matshow(empty) plt.show() Now we use the image processing technique of dilation on the symbols mask. So that we get a new mask which covers the surroundings of all symbols. We use this afterwards to check if the digits are near a symbol.\nstruct = ((1, 1, 1), (1, 1, 1), (1, 1, 1)) dilate = ndi.binary_dilation(symbols, structure=struct) plt.figure(figsize=(10, 6)) plt.subplot(121, title=\u0026#39;symbols\u0026#39;).matshow(symbols[:15, :15]) plt.subplot(122, title=\u0026#39;symbols dilated\u0026#39;).matshow(dilate[:15, :15]) plt.show() Creating this masks as before could be understood as a binary segmentation, as each element in our mask is either True or False. To extract the single digits, we\u0026rsquo;ll convert the binary digits segmentation into a instance segmentation, where each connected segment has an own index.\nmarkers, num_features = ndi.label(digits) plt.figure(figsize=(10, 6)) plt.subplot(121, title=\u0026#39;binary segmentation\u0026#39;).matshow( digits[:15, :15]) plt.subplot(122, title=\u0026#39;instance segmentation\u0026#39;).matshow( markers[:15, :15], cmap=\u0026#39;gnuplot\u0026#39;) plt.show() Now, for each instance, we check if the dilated binary mask overlaps with the instance and if yes, we extract the number.\nnumbers = [int(\u0026#39;\u0026#39;.join(arr[markers == i])) for i in range(1, num_features+1) if np.any((markers == i) \u0026amp; dilate)] Then, we just sum up:\nsum(numbers) 527364 Part 2 # Let\u0026rsquo;s first read the second task! The missing part wasn\u0026rsquo;t the only issue - one of the gears in the engine is wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.\nThis time, you need to find the gear ratio of every gear and add them all up so that the engineer can figure out which gear needs to be replaced.\nConsider the same engine schematic again:The missing part wasn\u0026rsquo;t the only issue - one of the gears in the engine is wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.\nThis time, you need to find the gear ratio of every gear and add them all up so that the engineer can figure out which gear needs to be replaced.\nConsider the same engine schematic again:\n467..114.. ...*...... ..35..633. ......#... 617*...... .....+.58. ..592..... ......755. ...$.*.... .664.598.. In this schematic, there are two gears. The first is in the top left; it has part numbers 467 and 35, so its gear ratio is 16345. The second gear is in the lower right; its gear ratio is 451490. (The * adjacent to 617 is not a gear because it is only adjacent to one part number.) Adding up all of the gear ratios produces 467835.\nWhat is the sum of all of the gear ratios in your engine schematic?\nWe\u0026rsquo;ll use the same method as before, but this time only extract * and create the instance segmentation before the dilation. Why? Because when we dilate first, we could merge two independent gears into one instance.\ngear = np.char.endswith(arr, \u0026#39;*\u0026#39;) gear_markers, gear_num = ndi.label(gear) plt.figure(figsize=(10, 6)) plt.subplot(131, title=\u0026#39;all symbols dilated\u0026#39;).matshow( symbols[:15, :15]) plt.subplot(132, title=\u0026#39;gears\u0026#39;).matshow( gear[:15, :15]) plt.subplot(133, title=\u0026#39;gears instances\u0026#39;).matshow( gear_markers[:15, :15], cmap=\u0026#39;gnuplot\u0026#39;) plt.show() Now, we step through each gear instance, create a mask for that gear, dilate it, and then step through all digits and check if they are in the gear mask. If we get two digits in the end, we multiply them to get the gear ratio and save the ratios to a list.\ngear_ratios = [] for i in range(1, gear_num+1): gear_binary = gear_markers == i gear_dil = ndi.binary_dilation(gear_binary, structure=struct) part_numbers = [int(\u0026#39;\u0026#39;.join(arr[markers == j])) for j in range(1, num_features+1) if np.any((markers == j) \u0026amp; gear_dil)] if len(part_numbers) == 2: gear_ratios.append(part_numbers[0] * part_numbers[1]) Now, we just sum the output again:\nsum(gear_ratios) 79026871 "},{"id":9,"href":"/posts/aoc2023-day02/","title":"Advent of code 2023 - Day 2: Cube Conundrum","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 2 - see https:adventofcode.com/2023/day/2\nPart 1 # Lets first read the task: As you walk, the Elf shows you a small bag and some cubes which are either red, green, or blue. Each time you play this game, he will hide a secret number of cubes of each color in the bag, and your goal is to figure out information about the number of cubes.\nTo get information, once a bag has been loaded with cubes, the Elf will reach into the bag, grab a handful of random cubes, show them to you, and then put them back in the bag. He\u0026rsquo;ll do this a few times per game.\nYou play several games and record the information from each game (your puzzle input). Each game is listed with its ID number (like the 11 in Game 11: ...) followed by a semicolon-separated list of subsets of cubes that were revealed from the bag (like 3 red, 5 green, 4 blue).\nFor example, the record of a few games might look like this:\nGame 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green In game 1, three sets of cubes are revealed from the bag (and then put back again). The first set is 3 blue cubes and 4 red cubes; the second set is 1 red cube, 2 green cubes, and 6 blue cubes; the third set is only 2 green cubes.\nThe Elf would first like to know which games would have been possible if the bag contained only 12 red cubes, 13 green cubes, and 14 blue cubes?\nIn the example above, games 1, 2, and 5 would have been possible if the bag had been loaded with that configuration. However, game 3 would have been impossible because at one point the Elf showed you 20 red cubes at once; similarly, game 4 would also have been impossible because the Elf showed you 15 blue cubes at once. If you add up the IDs of the games that would have been possible, you get 8.\nDetermine which games would have been possible if the bag had been loaded with only 12 red cubes, 13 green cubes, and 14 blue cubes. What is the sum of the IDs of those games?\nMy input file: 2023-12-02-1-aoc.txt\nOkay, let\u0026rsquo;s load our python kernel in emacs-jupyter and get coding! First of all, let\u0026rsquo;s load the input and split the riddle code by colon : to extract the game id and the rest of the code by semicolon ; to get the number of sets played in each game.\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-02-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) txt[\u0026#39;id\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply( lambda x: int(x[0].strip(\u0026#39;Game \u0026#39;))) txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply(lambda x: x[1]) # txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;;\u0026#39;) # txt[\u0026#39;nsets\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda x: len(x)) txt code id 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 100 rows × 2 columns\nNow, let\u0026rsquo;s extract the three colors in different columns with regex. We use the lookahead assertion ?= to find the respective colours and only exctract the digits \\d+ coming before. Then we just keep the max imum drawn number of cubes per color, since this is the only information that matters at the moment.\ntxt[\u0026#39;green\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.green)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt[\u0026#39;red\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.red)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt[\u0026#39;blue\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.blue)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt code id green red blue 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 2 10 10 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 15 10 3 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 4 6 16 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 16 20 18 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 10 6 14 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 4 2 1 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 5 4 8 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 2 5 2 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 2 9 11 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 13 15 11 100 rows × 5 columns\nLastly, we just filter the DataFrame to only include games where all drawn cubes were below or equal the number of cubes in the game and sum the result!\ntxt[\u0026#39;id\u0026#39;][(txt[\u0026#39;green\u0026#39;] \u0026lt; 14) \u0026amp; (txt[\u0026#39;red\u0026#39;] \u0026lt; 13) \u0026amp; (txt[\u0026#39;blue\u0026#39;] \u0026lt; 15)].sum() 3035 Part 2 # First, let\u0026rsquo;s get the instruction from the second part: As you continue your walk, the Elf poses a second question: in each game you played, what is the fewest number of cubes of each color that could have been in the bag to make the game possible?\nAgain consider the example games from earlier:\nGame 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green In game 1, the game could have been played with as few as 4 red, 2 green, and 6 blue cubes. If any color had even one fewer cube, the game would have been impossible. Game 2 could have been played with a minimum of 1 red, 3 green, and 4 blue cubes. Game 3 must have been played with at least 20 red, 13 green, and 6 blue cubes. Game 4 required at least 14 red, 3 green, and 15 blue cubes. Game 5 needed no fewer than 6 red, 3 green, and 2 blue cubes in the bag. The power of a set of cubes is equal to the numbers of red, green, and blue cubes multiplied together. The power of the minimum set of cubes in game 1 is 48. In games 2-5 it was 12, 1560, 630, and 36, respectively. Adding up these five powers produces the sum 2286.\nFor each game, find the minimum set of cubes that must have been present. What is the sum of the power of these sets?\nLuckily, this task is made trivial by the approach we have taken before. We just have to multiply the green, red and blue columns:\ntxt[\u0026#39;power\u0026#39;] = txt.loc[:, \u0026#39;green\u0026#39;] * txt.loc[:, \u0026#39;blue\u0026#39;] * txt.loc[:, \u0026#39;red\u0026#39;] txt code id green red blue power 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 2 10 10 200 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 15 10 3 450 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 4 6 16 384 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 16 20 18 5760 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 10 6 14 840 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 4 2 1 8 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 5 4 8 160 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 2 5 2 20 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 2 9 11 198 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 13 15 11 2145 100 rows × 6 columns\nAnd for this one, the sum is:\ntxt[\u0026#39;power\u0026#39;].sum() 66027 "},{"id":10,"href":"/posts/aoc2023-day01/","title":"Advent of code 2023 - Day 1: Trebuchet?!","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 1 - see https:adventofcode.com/2023/day/1\nPart 1 # Our first task is the following: The newly-improved calibration document consists of lines of text; each line originally contained a specific calibration value that the Elves now need to recover. On each line, the calibration value can be found by combining the first digit and the last digit (in that order) to form a single two-digit number.\nFor example:\n1abc2 pqr3stu8vwx a1b2c3d4e5f treb7uchet In this example, the calibration values of these four lines are 12, 38, 15, and 77. Adding these together produces 142.\nConsider your entire calibration document. What is the sum of all of the calibration values?\nMy input file: 2023-12-01-1-aoc.txt\nLets start jupyter in our shell to start coding!\nconda activate tf jupyter lab --no-browser --port=8888 First, load the test document\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-01-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) txt code 0 jjfvnnlfivejj1 1 6fourfour 2 ninevbmltwo69 3 pcg91vqrfpxxzzzoneightzt 4 jpprthxgjfive3one1qckhrptpqdc \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 996 81s 997 2four3threesxxvlfqfive4 998 nine6eightsevenzx9twoxc 999 hmbfjdfnp989mfivefiverpzrjs 1000 rows × 1 columns\nSecond, extract the digits. I had to wrap my head around regex matching in python first, because I first tried pandas.extract (which only extracts the first match), then pandas.extractall (which extracts all matches but puts them into a multiindex which makes things more difficult in this case). So I settled for the re.findall version, which returns a list. To concatenate the elements in the list, we take use the join function.\ntxt[\u0026#39;digits\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda x: \u0026#39;\u0026#39;.join(re.findall(r\u0026#39;(\\d+)\u0026#39;, x))) txt code digits 0 jjfvnnlfivejj1 1 1 6fourfour 6 2 ninevbmltwo69 69 3 pcg91vqrfpxxzzzoneightzt 91 4 jpprthxgjfive3one1qckhrptpqdc 31 \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 996 81s 81 997 2four3threesxxvlfqfive4 234 998 nine6eightsevenzx9twoxc 69 999 hmbfjdfnp989mfivefiverpzrjs 989 1000 rows × 2 columns\nNext, combine the first and the last digit and convert the result from string to integer\ntxt[\u0026#39;calibration\u0026#39;] = txt.loc[:, \u0026#39;digits\u0026#39;].apply( lambda x: int(x[0] + x[-1])) txt code digits calibration 0 jjfvnnlfivejj1 1 11 1 6fourfour 6 66 2 ninevbmltwo69 69 69 3 pcg91vqrfpxxzzzoneightzt 91 91 4 jpprthxgjfive3one1qckhrptpqdc 31 31 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 996 81s 81 81 997 2four3threesxxvlfqfive4 234 24 998 nine6eightsevenzx9twoxc 69 69 999 hmbfjdfnp989mfivefiverpzrjs 989 99 1000 rows × 3 columns\nLastly, get the sum of our calibration numbers\ntxt.loc[:, \u0026#39;calibration\u0026#39;].sum() 56465 Part 2 # Now follows part two: Your calculation isn\u0026rsquo;t quite right. It looks like some of the digits are actually spelled out with letters: one, two, three, four, five, six, seven, eight, and nine also count as valid \u0026ldquo;digits\u0026rdquo;.\nEquipped with this new information, you now need to find the real first and last digit on each line. For example:\ntwo1nine eightwothree abcone2threexyz xtwone3four 4nineeightseven2 zoneight234 7pqrstsixteen In this example, the calibration values are 29, 83, 13, 24, 42, 14, and 76. Adding these together produces 281.\nWhat is the sum of all of the calibration values?\nOkay, let\u0026rsquo;s see if we can update the pattern matching. To deal with potential overlapping values like oneight which contains one as well as eight, I used the regex positive lookahead ?= as described here. Because this enables capturing overlapping values, I used \\d (one digit) instead of \\d+ (one or more digits), otherwise digits might double. Afterwards, just replace the spelled out digits with their numerical value.\n# for i, r in enumerate(txt.loc[:, \u0026#39;code\u0026#39;]): # matches = re.findall( # r\u0026#39;(?=(\\d|one|two|three|four|five|six|seven|eight|nine))\u0026#39;, r) # result = \u0026#39;\u0026#39;.join([match for match in matches]) # result = result.replace(\u0026#39;one\u0026#39;, \u0026#39;1\u0026#39;).replace(\u0026#39;two\u0026#39;, \u0026#39;2\u0026#39;).replace( # \u0026#39;three\u0026#39;, \u0026#39;3\u0026#39;).replace(\u0026#39;four\u0026#39;, \u0026#39;4\u0026#39;).replace(\u0026#39;five\u0026#39;, \u0026#39;5\u0026#39;).replace( # \u0026#39;six\u0026#39;, \u0026#39;6\u0026#39;).replace(\u0026#39;seven\u0026#39;, \u0026#39;7\u0026#39;).replace(\u0026#39;eight\u0026#39;, \u0026#39;8\u0026#39;).replace( # \u0026#39;nine\u0026#39;, \u0026#39;9\u0026#39;) # txt.loc[i, \u0026#39;digits2\u0026#39;] = result # txt # a very nice alternative suggested by Tomalak: digits = \u0026#39;\\d one two three four five six seven eight nine\u0026#39;.split() txt[\u0026#39;digits2\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda v: \u0026#39;\u0026#39;.join( str(digits.index(m)) if m in digits else m for m in re.findall(rf\u0026#39;(?=({\u0026#34;|\u0026#34;.join(digits)}))\u0026#39;, v) )) txt code digits calibration digits2 0 jjfvnnlfivejj1 1 11 51 1 6fourfour 6 66 644 2 ninevbmltwo69 69 69 9269 3 pcg91vqrfpxxzzzoneightzt 91 91 9118 4 jpprthxgjfive3one1qckhrptpqdc 31 31 5311 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 58375 996 81s 81 81 81 997 2four3threesxxvlfqfive4 234 24 243354 998 nine6eightsevenzx9twoxc 69 69 968792 999 hmbfjdfnp989mfivefiverpzrjs 989 99 98955 1000 rows × 4 columns\nNow, construct the calibration value as before\u0026hellip;\ntxt[\u0026#39;calibration2\u0026#39;] = txt.loc[:, \u0026#39;digits2\u0026#39;].apply(lambda x: int(x[0] + x[-1])) txt code digits calibration digits2 calibration2 0 jjfvnnlfivejj1 1 11 51 51 1 6fourfour 6 66 644 64 2 ninevbmltwo69 69 69 9269 99 3 pcg91vqrfpxxzzzoneightzt 91 91 9118 98 4 jpprthxgjfive3one1qckhrptpqdc 31 31 5311 51 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 58375 55 996 81s 81 81 81 81 997 2four3threesxxvlfqfive4 234 24 243354 24 998 nine6eightsevenzx9twoxc 69 69 968792 92 999 hmbfjdfnp989mfivefiverpzrjs 989 99 98955 95 1000 rows × 5 columns\n\u0026hellip; and get the correct sum!\ntxt.loc[:, \u0026#39;calibration2\u0026#39;].sum() 55902 "}] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/en.search-data.min.a906b147eddae07740dc484d51a62d0c44a622ead6b0e0755e9b5ffccefdea9b.json b/aseltmann.github.io-hugo/public/en.search-data.min.a906b147eddae07740dc484d51a62d0c44a622ead6b0e0755e9b5ffccefdea9b.json deleted file mode 100644 index ba5bbd9..0000000 --- a/aseltmann.github.io-hugo/public/en.search-data.min.a906b147eddae07740dc484d51a62d0c44a622ead6b0e0755e9b5ffccefdea9b.json +++ /dev/null @@ -1 +0,0 @@ -[{"id":0,"href":"/docs/notes/poetry/","title":"Poetry","section":"Notes","content":" Table of Contents Lǎozǐ Lǎozǐ # I am partial to the translation of Derek Lin\nDàodéjīng - Chapter 11 (The use of what has no substantive existence) [\u0026hellip;]\nMix clay\nto create a container\nIn its emptiness, there is\nthe function of a container\n[\u0026hellip;]\nTherefore,\nthat which exists is used to create benefit\nThat which is empty is used to create functionality\nDàodéjīng - Chapter 15 (The exhibition of the qualities of the Dao) The Tao masters of antiquity\nSubtle wonders through mystery\nDepths that cannot be discerned\nBecause one cannot discern them\nTherefore one is forced to describe the appearance\nHesitant,\nlike crossing a wintry river\nCautious,\nlike fearing four neighbors\nSolemn,\nlike a guest\nLoose,\nlike ice about to melt\nGenuine,\nlike plain wood\nOpen,\nlike a valley\nOpaque,\nlike muddy water\nWho can be muddled yet desist\nIn stillness gradually become clear?\nWho can be serene yet persist\nIn motion gradually come alive?\nOne who holds this \u0026lt;i\u0026gt;Tao\u0026lt;/i\u0026gt; does not wish to be overfilled\nBecause one is not overfilled\nTherefore one can preserve and not create anew\nDàodéjīng - Chapter 20 (Being different from ordinary people) Cease learning, no more worries\nRespectful response and scornful response\nHow much is the difference?\nGoodness and evil\nHow much do they differ?\nWhat the people fear,\nI cannot be unafraid\nSo desolate! How limitless it is!\nThe people are excited\nAs if enjoying a great feast\nAs if climbing up to the terrace in spring\nI alone am quiet and uninvolved\nLike an infant\nnot yet smiling\nSo weary,\nlike having no place to return\nThe people all have surplus\nWhile I alone seem lacking\nI have the heart of a fool indeed -\nso ignorant!\nOrdinary people are bright\nI alone am muddled\nOrdinary people are scrutinizing\nI alone am obtuse\nSo tranquil, like the ocean\nSo moving, as if without limits\nThe people all have goals\nAnd I alone am stubborn and lowly\nI alone am different from them\nAnd value the nourishing mother\n"},{"id":1,"href":"/posts/aoc2024-day01/","title":"Advent of code 2024 - Day 1: Historian Hysteria","section":"Posts","content":" Table of Contents Part 1: Locations Part 2: Similarities I try to solve this year\u0026rsquo;s Advent of Code 2024 riddles in R. This is Day 1 - see https:adventofcode.com/2024/day/1\nPart 1: Locations # Lets first read the task: Throughout the Chief\u0026rsquo;s office, the historically significant locations are listed not by name but by a unique number called the location ID. To make sure they don\u0026rsquo;t miss anything, The Historians split into two groups, each searching the office and trying to create their own complete list of location IDs.\nThere\u0026rsquo;s just one problem: by holding the two lists up side by side (your puzzle input), it quickly becomes clear that the lists aren\u0026rsquo;t very similar. Maybe you can help The Historians reconcile their lists?\nFor example:\n3 4 4 3 2 5 1 3 3 9 3 3 Maybe the lists are only off by a small amount! To find out, pair up the numbers and measure how far apart they are. Pair up the smallest number in the left list with the smallest number in the right list, then the second-smallest left number with the second-smallest right number, and so on.\nWithin each pair, figure out how far apart the two numbers are; you\u0026rsquo;ll need to add up all of those distances. For example, if you pair up a 3 from the left list with a 7 from the right list, the distance apart is 4; if you pair up a 9 with a 3, the distance apart is 6.\nIn the example list above, the pairs and distances would be as follows:\nThe smallest number in the left list is 1, and the smallest number in the right list is 3. The distance between them is 2. The second-smallest number in the left list is 2, and the second-smallest number in the right list is another 3. The distance between them is 1. The third-smallest number in both lists is 3, so the distance between them is 0. The next numbers to pair up are 3 and 4, a distance of 1. The fifth-smallest numbers in each list are 3 and 5, a distance of 2. Finally, the largest number in the left list is 4, while the largest number in the right list is 9; these are a distance 5 apart. To find the total distance between the left list and the right list, add up the distances between all of the pairs you found. In the example above, this is 2 + 1 + 0 + 1 + 2 + 5, a total distance of 11!\nYour actual left and right lists contain many location IDs. What is the total distance between your lists?\nFirst, we read the data - the fread command from the data.table package is more versatile than read.delim from base and directly reads the data as a data.table, which has some benefits.\ndata \u0026lt;- data.table::fread(\u0026#34;2024-12-01-1-aoc.txt\u0026#34;, header = FALSE) head(data) V1 V2 \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; 41226 69190 89318 10100 59419 23880 63157 20193 81510 22869 83942 63304 Now, the idea is to get the ordered versions of V1 and V2.\nv1 \u0026lt;- data[order(V1)]$V1 v2 \u0026lt;- data[order(V2)]$V2 head(v1) 10188 10314 10319 10348 10408 10668 Coming from Python, this syntax looks a bit odd. The documentation of data.table::setorder helps:\nsetorder (and setorderv) reorders the rows of a data.table based on the columns (and column order) provided. It reorders the table by reference and is therefore very memory efficient. Note that queries like x[order(.)] are optimised internally to use data.table\u0026#39;s fast order. So, data[order(V1)] is actually short for data.table::setorder(data, V1). Then, we extract the vector by name using the $ operator, which allows to extract elements by name.\nhead(data.table::setorder(data, V1)$V1) 10188 10314 10319 10348 10408 10668 The actual computation is just the sum of the absolute difference:\nsum(abs(v1 - v2)) 3574690\nPart 2: Similarities # Lets first read the task: This time, you\u0026rsquo;ll need to figure out exactly how often each number from the left list appears in the right list. Calculate a total similarity score by adding up each number in the left list after multiplying it by the number of times that number appears in the right list.\nHere are the same example lists again:\n3 4 4 3 2 5 1 3 3 9 3 3 For these example lists, here is the process of finding the similarity score:\nThe first number in the left list is 3. It appears in the right list three times, so the similarity score increases by 3 * 3 = 9. The second number in the left list is 4. It appears in the right list once, so the similarity score increases by 4 * 1 = 4. The third number in the left list is 2. It does not appear in the right list, so the similarity score does not increase (2 * 0 = 0). The fourth number, 1, also does not appear in the right list. The fifth number, 3, appears in the right list three times; the similarity score increases by 9. The last number, 3, appears in the right list three times; the similarity score again increases by 9. So, for these example lists, the similarity score at the end of this process is 31 (9 + 4 + 0 + 0 + 9 + 9).\nOnce again consider your left and right lists. What is their similarity score?\nMy idea here is simple - we first count the occurrences of V2 to be able to check V1 against them. Here, applying the table command works like pandas.value_counts() and achieves this. We convert the output to a data.frame and assign the V1 values as row names. Note: table converts the values which are counted to string, e.g. \u0026quot;10019\u0026quot; instead of 10019.\ntab \u0026lt;- table(v2) v2series \u0026lt;- data.frame(c(tab), row.names = names(tab)) head(v2series) v2series[\u0026#34;10019\u0026#34;, ] c.tab. \u0026lt;int\u0026gt; 10019 1 10100 1 10206 1 10428 1 10645 1 10972 1 1\nNow, we loop through V1 and try to access the frequency in V2 (by converting to string first, via as.character). If nothing is found in V2, an NA value is returned. For everything that is not NA, we compute the similarity score, append it to a list, and sum in the end. Note: a list in R can not directly be summed (I don\u0026rsquo;t know why that is) - so we have to unlist first.\nsim \u0026lt;- list() for (num in v1) { myfreq \u0026lt;- v2series[as.character(num), ] if (!is.na(myfreq)) { score \u0026lt;- myfreq * num sim \u0026lt;- append(sim, score) } } sum(unlist(sim)) 22565391\n"},{"id":2,"href":"/posts/aoc2023-day07/","title":"Advent of code 2023 - Day 7: Camel Cards","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 7 - see https:adventofcode.com/2023/day/7\nPart 1 # Lets first read the task: In Camel Cards, you get a list of hands, and your goal is to order them based on the strength of each hand. A hand consists of five cards labeled one of A, K, Q, J, T, 9, 8, 7, 6, 5, 4, 3, or 2. The relative strength of each card follows this order, where A is the highest and 2 is the lowest.\nEvery hand is exactly one type. From strongest to weakest, they are:\nFive of a kind, where all five cards have the same label: AAAAA Four of a kind, where four cards have the same label and one card has a different label: AA8AA Full house, where three cards have the same label, and the remaining two cards share a different label: 23332 Three of a kind, where three cards have the same label, and the remaining two cards are each different from any other card in the hand: TTT98 Two pair, where two cards share one label, two other cards share a second label, and the remaining card has a third label: 23432 One pair, where two cards share one label, and the other three cards have a different label from the pair and each other: A23A4 High card, where all cards\u0026rsquo; labels are distinct: 23456 Hands are primarily ordered based on type; for example, every full house is stronger than any three of a kind.\nIf two hands have the same type, a second ordering rule takes effect. Start by comparing the first card in each hand. If these cards are different, the hand with the stronger first card is considered stronger. If the first card in each hand have the same label, however, then move on to considering the second card in each hand. If they differ, the hand with the higher second card wins; otherwise, continue with the third card in each hand, then the fourth, then the fifth.\nSo, 33332 and 2AAAA are both four of a kind hands, but 33332 is stronger because its first card is stronger. Similarly, 77888 and 77788 are both a full house, but 77888 is stronger because its third card is stronger (and both hands have the same first and second card).\nTo play Camel Cards, you are given a list of hands and their corresponding bid (your puzzle input). For example:\n32T3K 765 T55J5 684 KK677 28 KTJJT 220 QQQJA 483 This example shows five hands; each hand is followed by its bid amount. Each hand wins an amount equal to its bid multiplied by its rank, where the weakest hand gets rank 1, the second-weakest hand gets rank 2, and so on up to the strongest hand. Because there are five hands in this example, the strongest hand will have rank 5 and its bid will be multiplied by 5.\nSo, the first step is to put the hands in order of strength:\n32T3K is the only one pair and the other hands are all a stronger type, so it gets rank 1. KK677 and KTJJT are both two pair. Their first cards both have the same label, but the second card of KK677 is stronger (K vs T), so KTJJT gets rank 2 and KK677 gets rank 3. T55J5 and QQQJA are both three of a kind. QQQJA has a stronger first card, so it gets rank 5 and T55J5 gets rank 4. Now, you can determine the total winnings of this set of hands by adding up the result of multiplying each hand\u0026rsquo;s bid with its rank (765 * 1 + 220 * 2 + 28 * 3 + 684 * 4 + 483 * 5). So the total winnings in this example are 6440.\nFind the rank of every hand in your set. What are the total winnings?\nAs always - let\u0026rsquo;s read in the data.\nimport pandas as pd camel = pd.read_table(\u0026#39;data/2023-12-07-1-aoc.txt\u0026#39;, delim_whitespace=True, header=None, names=[\u0026#39;cards\u0026#39;, \u0026#39;bid\u0026#39;], dtype={\u0026#39;cards\u0026#39;: \u0026#39;string\u0026#39;, \u0026#39;bid\u0026#39;: \u0026#39;Int64\u0026#39;}) camel cards bid 0 A833A 309 1 Q33J3 205 2 55KK5 590 3 K4457 924 4 Q3QT3 11 \u0026hellip; \u0026hellip; \u0026hellip; 995 KQJ53 367 996 57866 537 997 Q94A9 210 998 J448A 903 999 6J66J 114 1000 rows × 2 columns\nFor sorting, we need to compute a type column. The 7 types can be deduced based on how many different cards are there (through set()) and the number of cards.\nAlso, for the next step we prepare one column per card string in the cards.\ndef get_type(cards): cardset = set(cards) cardlen = len(cardset) cardlist = list(cards) cardnum = [cardlist.count(i) for i in cardset] if cardlen == 1: ctype = 1 elif cardlen in (4, 5): ctype = cardlen + 2 elif cardlen == 2: ctype = 2 if set(cardnum) == {1, 4} else 3 else: ctype = 4 if 3 in cardnum else 5 return ctype camel[\u0026#39;type\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: get_type(x)) for i in range(5): camel[f\u0026#39;c{i}\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: list(x)[i]) camel cards bid type c0 c1 c2 c3 c4 0 A833A 309 5 A 8 3 3 A 1 Q33J3 205 4 Q 3 3 J 3 2 55KK5 590 3 5 5 K K 5 3 K4457 924 6 K 4 4 5 7 4 Q3QT3 11 5 Q 3 Q T 3 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 KQJ53 367 7 K Q J 5 3 996 57866 537 6 5 7 8 6 6 997 Q94A9 210 6 Q 9 4 A 9 998 J448A 903 6 J 4 4 8 A 999 6J66J 114 3 6 J 6 6 J 1000 rows × 8 columns\nFor sorting, pandas contributes the sort_values method, which supports sorting by multiple columns (here, we first sort according to type, then c0, c1, \u0026hellip;). For easier sorting by numerical values, we convert the values using the card_dict dictionary, which implements the given sorting order.\ncard_dict = {c: i for i, c in enumerate([7, 6, 5, 4, 3, 2, 1, \u0026#39;2\u0026#39;, \u0026#39;3\u0026#39;, \u0026#39;4\u0026#39;, \u0026#39;5\u0026#39;, \u0026#39;6\u0026#39;, \u0026#39;7\u0026#39;, \u0026#39;8\u0026#39;, \u0026#39;9\u0026#39;, \u0026#39;T\u0026#39;, \u0026#39;J\u0026#39;, \u0026#39;Q\u0026#39;, \u0026#39;K\u0026#39;, \u0026#39;A\u0026#39;])} camel = camel.sort_values(by=[\u0026#39;type\u0026#39;, \u0026#39;c0\u0026#39;, \u0026#39;c1\u0026#39;, \u0026#39;c2\u0026#39;, \u0026#39;c3\u0026#39;, \u0026#39;c4\u0026#39;], ignore_index=True, key=lambda x: x.map(card_dict)) camel.index += 1 camel cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 AA5AA 565 2 A A 5 A A 997 AAATA 704 2 A A A T A 998 AAAA3 145 2 A A A A 3 999 AAAAJ 594 2 A A A A J 1000 JJJJJ 219 1 J J J J J 1000 rows × 8 columns\nLastly, we multiply the bid with the index and sum the result.\ncamel.loc[:, \u0026#39;bid\u0026#39;].mul(camel.index).sum() 246912307 Part 2 # Lets first read the task: To make things a little more interesting, the Elf introduces one additional rule. Now, J cards are jokers - wildcards that can act like whatever card would make the hand the strongest type possible.\nTo balance this, J cards are now the weakest individual cards, weaker even than 2. The other cards stay in the same order: A, K, Q, T, 9, 8, 7, 6, 5, 4, 3, 2, J.\nJ cards can pretend to be whatever card is best for the purpose of determining hand type; for example, QJJQ2 is now considered four of a kind. However, for the purpose of breaking ties between two hands of the same type, J is always treated as J, not the card it\u0026rsquo;s pretending to be: JKKK2 is weaker than QQQQ2 because J is weaker than Q.\nNow, the above example goes very differently:\n32T3K 765 T55J5 684 KK677 28 KTJJT 220 QQQJA 483 32T3K is still the only one pair; it doesn\u0026rsquo;t contain any jokers, so its strength doesn\u0026rsquo;t increase. KK677 is now the only two pair, making it the second-weakest hand. T55J5, KTJJT, and QQQJA are now all four of a kind! T55J5 gets rank 3, QQQJA gets rank 4, and KTJJT gets rank 5. With the new joker rule, the total winnings in this example are 5905.\nUsing the new joker rule, find the rank of every hand in your set. What are the new total winnings?\nWhat changes? We have to update our get_type() function and our card_dict dictionary, otherwise the method stays the same!\nLet\u0026rsquo;s first write get_type_joker(): if no joker is in the cards, we can just refer to get_type(). If we have 1, 2 or 5 different card faces, the answer is trivial. We only have to use the number of jokers and the maximum number of other faces if we have 3 or 4 different faces.\nNote: if a J is present, there can never be high card (type = 7), because we always can get one pair. Also we can never get two pair (type = 5), because we can always construct three of a kind or full house.\ndef get_type_joker(cards): cardset = set(cards) cardlen = len(cardset) cardlist = list(cards) cardnum = [cardlist.count(i) for i in cardset] if not \u0026#39;J\u0026#39; in cardset: ctype = get_type(cards) else: idxj = list(cardset).index(\u0026#39;J\u0026#39;) numj = cardnum[idxj] cardnum.pop(idxj) if cardlen in [1, 2]: ctype = 1 elif cardlen == 5: ctype = 6 else: if (numj in [1, 2]) and (cardlen == 4): ctype = 4 else: ctype = 6 - max(cardnum) - numj return ctype camel[\u0026#39;type\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: get_type_joker(x)) camel cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 AA5AA 565 2 A A 5 A A 997 AAATA 704 2 A A A T A 998 AAAA3 145 2 A A A A 3 999 AAAAJ 594 1 A A A A J 1000 JJJJJ 219 1 J J J J J 1000 rows × 8 columns\nNow, we just move J in our new card_dict_joker to where it belongs, the rest of the solution is the same.\ncard_dict_joker = {c: i for i, c in enumerate( [7, 6, 5, 4, 3, 2, 1, \u0026#39;J\u0026#39;, \u0026#39;2\u0026#39;, \u0026#39;3\u0026#39;, \u0026#39;4\u0026#39;, \u0026#39;5\u0026#39;, \u0026#39;6\u0026#39;, \u0026#39;7\u0026#39;, \u0026#39;8\u0026#39;, \u0026#39;9\u0026#39;, \u0026#39;T\u0026#39;, \u0026#39;Q\u0026#39;, \u0026#39;K\u0026#39;, \u0026#39;A\u0026#39;])} camel = camel.sort_values(by=[\u0026#39;type\u0026#39;, \u0026#39;c0\u0026#39;, \u0026#39;c1\u0026#39;, \u0026#39;c2\u0026#39;, \u0026#39;c3\u0026#39;, \u0026#39;c4\u0026#39;], ignore_index=True, key=lambda x: x.map(card_dict_joker)) camel.index += 1 display(camel) print(\u0026#39;Our new solution: \u0026#39;, camel.loc[:, \u0026#39;bid\u0026#39;].mul(camel.index).sum()) cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 TJTTJ 609 1 T J T T J 997 QQJQJ 131 1 Q Q J Q J 998 QQQJQ 831 1 Q Q Q J Q 999 KKKJK 183 1 K K K J K 1000 AAAAJ 594 1 A A A A J 1000 rows × 8 columns\nOur new solution: 246894760 "},{"id":3,"href":"/posts/aoc2023-day06/","title":"Advent of code 2023 - Day 6: Wait For It","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 6 - see https:adventofcode.com/2023/day/6\nPart 1 # Lets first read the task: You manage to sign up as a competitor in the boat races just in time. The organizer explains that it\u0026rsquo;s not really a traditional race - instead, you will get a fixed amount of time during which your boat has to travel as far as it can, and you win if your boat goes the farthest.\nAs part of signing up, you get a sheet of paper (your puzzle input) that lists the time allowed for each race and also the best distance ever recorded in that race. To guarantee you win the grand prize, you need to make sure you go farther in each race than the current record holder.\nThe organizer brings you over to the area where the boat races are held. The boats are much smaller than you expected - they\u0026rsquo;re actually toy boats, each with a big button on top. Holding down the button charges the boat, and releasing the button allows the boat to move. Boats move faster if their button was held longer, but time spent holding the button counts against the total race time. You can only hold the button at the start of the race, and boats don\u0026rsquo;t move until the button is released.\nFor example:\nTime: 7 15 30 Distance: 9 40 200 This document describes three races:\nThe first race lasts 7 milliseconds. The record distance in this race is 9 millimeters. The second race lasts 15 milliseconds. The record distance in this race is 40 millimeters. The third race lasts 30 milliseconds. The record distance in this race is 200 millimeters. Your toy boat has a starting speed of zero millimeters per millisecond. For each whole millisecond you spend at the beginning of the race holding down the button, the boat\u0026rsquo;s speed increases by one millimeter per millisecond.\nSo, because the first race lasts 7 milliseconds, you only have a few options:\nDon\u0026rsquo;t hold the button at all (that is, hold it for 0 milliseconds) at the start of the race. The boat won\u0026rsquo;t move; it will have traveled 0 millimeters by the end of the race. Hold the button for 1 millisecond at the start of the race. Then, the boat will travel at a speed of 1 millimeter per millisecond for 6 milliseconds, reaching a total distance traveled of 6 millimeters. Hold the button for 2 milliseconds, giving the boat a speed of 2 millimeters per millisecond. It will then get 5 milliseconds to move, reaching a total distance of 10 millimeters. Hold the button for 3 milliseconds. After its remaining 4 milliseconds of travel time, the boat will have gone 12 millimeters. Hold the button for 4 milliseconds. After its remaining 3 milliseconds of travel time, the boat will have gone 12 millimeters. Hold the button for 5 milliseconds, causing the boat to travel a total of 10 millimeters. Hold the button for 6 milliseconds, causing the boat to travel a total of 6 millimeters. Hold the button for 7 milliseconds. That\u0026rsquo;s the entire duration of the race. You never let go of the button. The boat can\u0026rsquo;t move until you let go of the button. Please make sure you let go of the button so the boat gets to move. 0 millimeters. Since the current record for this race is 9 millimeters, there are actually 4 different ways you could win: you could hold the button for 2, 3, 4, or 5 milliseconds at the start of the race.\nIn the second race, you could hold the button for at least 4 milliseconds and at most 11 milliseconds and beat the record, a total of 8 different ways to win.\nIn the third race, you could hold the button for at least 11 milliseconds and no more than 19 milliseconds and still beat the record, a total of 9 ways you could win.\nTo see how much margin of error you have, determine the number of ways you can beat the record in each race; in this example, if you multiply these values together, you get 288 (4 * 8 * 9).\nDetermine the number of ways you could beat the record in each race. What do you get if you multiply these numbers together?\nThe solution for this problem is shorter than the task description!\nAs always - let\u0026rsquo;s load the data. Note: we use the delim_whitesspace argument instead of sep=' ' to separate values by spaces of any length.\nimport pandas as pd import numpy as np race = pd.read_table(\u0026#39;data/2023-12-06-1-aoc.txt\u0026#39;, delim_whitespace=True, header=None, index_col=0) race.index = [\u0026#39;time\u0026#39;, \u0026#39;distance\u0026#39;] race 1 2 3 4 time 51 92 68 90 distance 222 2031 1126 1225 The we use Python list comprehensions to quickly calculate all distances for the range of the time value (and keep only the winning distances). Then we get the respective lengths and compute the product.\ndef get_distance(time, distance): return [t * (time - t) for t in range(1, time + 1) if (t * (time - t)) \u0026gt; distance] race.loc[\u0026#39;nwins\u0026#39;, :] = [len(get_distance(int(r.time), int(r.distance))) for c, r in race.items()] nwins_prod = np.prod(race.loc[\u0026#39;nwins\u0026#39;, :].values, dtype=int) display(race) print(f\u0026#39;The product of our nwins is {nwins_prod}\u0026#39;) 1 2 3 4 time 51.0 92.0 68.0 90.0 distance 222.0 2031.0 1126.0 1225.0 nwins 42.0 19.0 11.0 57.0 The product of our nwins is 500346 Part 2 # Lets first read the task: As the race is about to start, you realize the piece of paper with race times and record distances you got earlier actually just has very bad kerning. There\u0026rsquo;s really only one race - ignore the spaces between the numbers on each line.\nSo, the example from before:\nTime: 7 15 30 Distance: 9 40 200 \u0026hellip;now instead means this:\nTime: 71530 Distance: 940200 Now, you have to figure out how many ways there are to win this single race. In this example, the race lasts for 71530 milliseconds and the record distance you need to beat is 940200 millimeters. You could hold the button anywhere from 14 to 71516 milliseconds and beat the record, a total of 71503 ways!\nHow many ways can you beat the record in this one much longer race?\nFor this riddle we can take the same approach as for Part 1. The cell took 18s to evaluate on a consumer laptop from 2015 - no fancy workarounds needed :)\nracetime = int(\u0026#39;\u0026#39;.join(str(r) for r in race.loc[\u0026#39;time\u0026#39;, :4].astype(int))) racedist = int(\u0026#39;\u0026#39;.join(str(r) for r in race.loc[\u0026#39;distance\u0026#39;, :4].astype(int))) racewins = len(get_distance(racetime, racedist)) race[5] = [racetime, racedist, racewins] race 1 2 3 4 5 time 51.0 92.0 68.0 90.0 51926890 distance 222.0 2031.0 1126.0 1225.0 222203111261225 nwins 42.0 19.0 11.0 57.0 42515755 "},{"id":4,"href":"/posts/aoc2023-day05/","title":"Advent of code 2023 - Day 5: If You Give A Seed A Fertilizer","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 5 - see https:adventofcode.com/2023/day/5\nUpdate [2023-12-31 So]:\nsubstitute get_generator() (own implementation) with range() (Python inbuilt) improve grid search so that it goes through all location ranges, still starting with the lowest range Part 1 # Lets first read the task: The almanac (your puzzle input) lists all of the seeds that need to be planted. It also lists what type of soil to use with each kind of seed, what type of fertilizer to use with each kind of soil, what type of water to use with each kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is identified with a number, but numbers are reused by each category - that is, soil 123 and fertilizer 123 aren\u0026rsquo;t necessarily related to each other.\nFor example:\nseeds: 79 14 55 13 seed-to-soil map: 50 98 2 52 50 48 soil-to-fertilizer map: 0 15 37 37 52 2 39 0 15 fertilizer-to-water map: 49 53 8 0 11 42 42 0 7 57 7 4 water-to-light map: 88 18 7 18 25 70 light-to-temperature map: 45 77 23 81 45 19 68 64 13 temperature-to-humidity map: 0 69 1 1 0 69 humidity-to-location map: 60 56 37 56 93 4 The almanac starts by listing which seeds need to be planted: seeds 79, 14, 55, and 13.\nThe rest of the almanac contains a list of maps which describe how to convert numbers from a source category into numbers in a destination category. That is, the section that starts with seed-to-soil map: describes how to convert a seed number (the source) to a soil number (the destination). This lets the gardener and his team know which soil to use with which seeds, which water to use with which fertilizer, and so on.\nRather than list every source number and its corresponding destination number one by one, the maps describe entire ranges of numbers that can be converted. Each line within a map contains three numbers: the destination range start, the source range start, and the range length.\nConsider again the example seed-to-soil map:\n50 98 2 52 50 48\nThe first line has a destination range start of 50, a source range start of 98, and a range length of 2. This line means that the source range starts at 98 and contains two values: 98 and 99. The destination range is the same length, but it starts at 50, so its two values are 50 and 51. With this information, you know that seed number 98 corresponds to soil number 50 and that seed number 99 corresponds to soil number 51. o The second line means that the source range starts at 50 and contains 48 values: 50, 51, \u0026hellip;, 96, 97. This corresponds to a destination range starting at 52 and also containing 48 values: 52, 53, \u0026hellip;, 98, 99. So, seed number 53 corresponds to soil number 55.\nAny source numbers that aren\u0026rsquo;t mapped correspond to the same destination number. So, seed number 10 corresponds to soil number 10.\nSo, the entire list of seed numbers and their corresponding soil numbers looks like this:\nseed soil 0 0 1 1 ... ... 48 48 49 49 50 52 51 53 ... ... 96 98 97 99 98 50 99 51 With this map, you can look up the soil number required for each initial seed number:\nSeed number 79 corresponds to soil number 81. Seed number 14 corresponds to soil number 14. Seed number 55 corresponds to soil number 57. Seed number 13 corresponds to soil number 13. The gardener and his team want to get started as soon as possible, so they\u0026rsquo;d like to know the closest location that needs a seed. Using these maps, find the lowest location number that corresponds to any of the initial seeds. To do this, you\u0026rsquo;ll need to convert each seed number through other categories until you can find its corresponding location number. In this example, the corresponding types are:\nSeed 79, soil 81, fertilizer 81, water 81, light 74, temperature 78, humidity 78, location 82. Seed 14, soil 14, fertilizer 53, water 49, light 42, temperature 42, humidity 43, location 43. Seed 55, soil 57, fertilizer 57, water 53, light 46, temperature 82, humidity 82, location 86. Seed 13, soil 13, fertilizer 52, water 41, light 34, temperature 34, humidity 35, location 35. So, the lowest location number in this example is 35.\nWhat is the lowest location number that corresponds to any of the initial seed numbers?\nWow, this task is a mouthful\u0026hellip;\nLet\u0026rsquo;s start slowly and load the data. Our input text document contains several maps, which are clearly separated and have a title (seed-to-soil map etc). So we can tell pandas where each map starts and give each map a dataframe. I got the values for the skiprows and nrows argument by looking at the input file and\u0026hellip; counting :)\nimport pandas as pd import sys seeds = pd.read_table(\u0026#39;data/2023-12-05-1-aoc.txt\u0026#39;, nrows=1, sep=\u0026#39; \u0026#39;, header=None, index_col=0) seeds = seeds.values.flatten() map_opt = {\u0026#39;filepath_or_buffer\u0026#39;: \u0026#39;data/2023-12-05-1-aoc.txt\u0026#39;, \u0026#39;header\u0026#39;: None, \u0026#39;sep\u0026#39;: \u0026#39; \u0026#39;, \u0026#39;dtype\u0026#39;: \u0026#39;Int64\u0026#39;, \u0026#39;names\u0026#39;: [\u0026#39;dest_start\u0026#39;, \u0026#39;src_start\u0026#39;, \u0026#39;range\u0026#39;]} seed_soil = pd.read_table(skiprows=3, nrows=23, **map_opt) soil_fert = pd.read_table(skiprows=28, nrows=9, **map_opt) fert_water = pd.read_table(skiprows=39, nrows=20, **map_opt) water_light = pd.read_table(skiprows=61, nrows=40, **map_opt) light_temp = pd.read_table(skiprows=103, nrows=36, **map_opt) temp_humi = pd.read_table(skiprows=141, nrows=35, **map_opt) humi_loc = pd.read_table(skiprows=178, nrows=26, **map_opt) maps = (seed_soil, soil_fert, fert_water, water_light, light_temp, temp_humi, humi_loc) print(\u0026#39;seeds are just a numpy array:\u0026#39;) display(seeds) print(\u0026#39;The \u0026#34;humidity-to-location\u0026#34; map as an example:\u0026#39;) humi_loc seeds are just a numpy array: array([3169137700, 271717609, 3522125441, 23376095, 1233948799, 811833837, 280549587, 703867355, 166086528, 44766996, 2326968141, 69162222, 2698492851, 14603069, 2755327667, 348999531, 2600461189, 92332846, 1054656969, 169099767])The \u0026#34;humidity-to-location\u0026#34; map as an example: dest_start src_start range 0 3490144003 1623866227 218040905 1 1709610578 1620839197 3027030 2 105449249 586389428 113279526 3 1899604338 2167886292 348178199 4 1712637608 2678624215 186966730 5 218728775 0 245776251 6 2472992580 923734334 143670388 7 2616662968 3670169885 15297294 8 2247782537 1395629154 225210043 9 0 480940179 105449249 10 4113852846 3959729057 159909096 11 3322784653 1067404722 167359350 12 923734334 4119638153 175329143 13 1534964496 2516064491 162559724 14 2631960262 3496028440 140849733 15 2862695906 1234764072 97988355 16 1697524220 3636878173 12086358 17 2985646048 1332752427 62876727 18 1099063477 2970241510 435901019 19 4009202281 2865590945 104650565 20 2960684261 2142924505 24961787 21 2772809995 3406142529 89885911 22 4273761942 3648964531 21205354 23 3708184908 1841907132 301017373 24 464505026 245776251 235163928 25 3048522775 3685467179 274261878 My first attempt was to actually construct ranges like in the example above, mapping out all possible sources and destinations. Python quickly informed me that even constructing one pandas.Series with int64 values mapping seeds to soil would cost 64GB memory - not the best solution.\nSo we take a different approach. For convenience, let\u0026rsquo;s add a src_end and a dest_end column to our maps:\nfor df in maps: df[\u0026#39;src_end\u0026#39;] = df.loc[:, \u0026#39;src_start\u0026#39;] + df.loc[:, \u0026#39;range\u0026#39;] df[\u0026#39;dest_end\u0026#39;] = df.loc[:, \u0026#39;dest_start\u0026#39;] + df.loc[:, \u0026#39;range\u0026#39;] print(\u0026#39;Again the \u0026#34;humidity-to-location\u0026#34; map as an example:\u0026#39;) humi_loc Again the \u0026#34;humidity-to-location\u0026#34; map as an example: dest_start src_start range src_end dest_end 0 3490144003 1623866227 218040905 1841907132 3708184908 1 1709610578 1620839197 3027030 1623866227 1712637608 2 105449249 586389428 113279526 699668954 218728775 3 1899604338 2167886292 348178199 2516064491 2247782537 4 1712637608 2678624215 186966730 2865590945 1899604338 5 218728775 0 245776251 245776251 464505026 6 2472992580 923734334 143670388 1067404722 2616662968 7 2616662968 3670169885 15297294 3685467179 2631960262 8 2247782537 1395629154 225210043 1620839197 2472992580 9 0 480940179 105449249 586389428 105449249 10 4113852846 3959729057 159909096 4119638153 4273761942 11 3322784653 1067404722 167359350 1234764072 3490144003 12 923734334 4119638153 175329143 4294967296 1099063477 13 1534964496 2516064491 162559724 2678624215 1697524220 14 2631960262 3496028440 140849733 3636878173 2772809995 15 2862695906 1234764072 97988355 1332752427 2960684261 16 1697524220 3636878173 12086358 3648964531 1709610578 17 2985646048 1332752427 62876727 1395629154 3048522775 18 1099063477 2970241510 435901019 3406142529 1534964496 19 4009202281 2865590945 104650565 2970241510 4113852846 20 2960684261 2142924505 24961787 2167886292 2985646048 21 2772809995 3406142529 89885911 3496028440 2862695906 22 4273761942 3648964531 21205354 3670169885 4294967296 23 3708184908 1841907132 301017373 2142924505 4009202281 24 464505026 245776251 235163928 480940179 699668954 25 3048522775 3685467179 274261878 3959729057 3322784653 Now we actually compute the mapping. For each seed, we go through all mappings and in each mapping we go through each row. We find the row which contains the mapping and exctract the destination, which is the source for the next map until we reach the last map which gives us the locations.\nApproach 1: df.itertuples() is a convenient way to step through a pandas.DataFrame in this example. It is faster than df.iterrows() and returns the row as a NamedTuple - nice!\nApproach 2: I actually wondered if it would be faster to get all maps in one pd.DataFrame and then iterate through the mappings. To test this let\u0026rsquo;s construct a new DataFrame maps_df which contains all maps. Since the maps have different lengths it is important to cast the datatype to Int64, which is short for pd.Int64Dtype() and keeps values as integers, even if NA values are in the same column.\nApproach 3: A third alternative I tested (not shown here) was to check if a value is in a Python range with the in operator as in: if 3 in range(5):... . This was way too slow.\n# mapping version 1 def get_location(seed): current = seed for df in maps: current_map = [row for row in df.itertuples() if ((current \u0026gt; row.src_start) and (current \u0026lt; row.src_end))] if len(current_map) == 0: pass elif len(current_map) == 1: current = (current_map[0].dest_start + (current - current_map[0].src_start)) else: raise ValueError(\u0026#39;This should not happen!\u0026#39;) return current # mapping version 2 - around 10 times slower # you need to rename the maps_df columns so that they have a unique id # e.g. \u0026#39;src_start_1\u0026#39;, \u0026#39;src_start_2\u0026#39; etc # maps_df = pd.concat(maps, axis=\u0026#39;columns\u0026#39;) # def get_dest(i, src): # return (maps_df[(src \u0026gt; maps_df.loc[:, f\u0026#39;src_start_{i}\u0026#39;]) \u0026amp; # (src \u0026lt; maps_df.loc[:, f\u0026#39;src_end_{i}\u0026#39;])] # .loc[:, f\u0026#39;dest_start_{i}\u0026#39;] # .iloc[0]) # def get_location2(seed): # dest = seed # i = 1 # while i \u0026lt; 7: # dest = get_dest(i, dest) # i += 1 # return dest %timeit get_location(seeds[0]) 1.98 ms ± 72 μs per loop (mean ± std. dev. of 7 runs, 100 loops each) Now let\u0026rsquo;s get a list of locations:\nlocations = [get_location(s) for s in seeds] locations 2493982655 3209845376 3992357533 4163131463 4104485616 1952252479 3218677354 388071289 2181441450 2594336315 4049507670 2084517144 3119633635 428978312 3518771991 3704555655 953918455 2107687768 3448046330 2184454689 Lastly, just get the minimum of all location values.\nmin(locations) 388071289 Part 2 # Let\u0026rsquo;s read the task of part 2! Everyone will starve if you only plant such a small number of seeds. Re-reading the almanac, it looks like the seeds: line actually describes ranges of seed numbers.\nThe values on the initial seeds: line come in pairs. Within each pair, the first value is the start of the range and the second value is the length of the range. So, in the first line of the example above:\nseeds: 79 14 55 13\nThis line describes two ranges of seed numbers to be planted in the garden. The first range starts with seed number 79 and contains 14 values: 79, 80, \u0026hellip;, 91, 92. The second range starts with seed number 55 and contains 13 values: 55, 56, \u0026hellip;, 66, 67.\nNow, rather than considering four seed numbers, you need to consider a total of 27 seed numbers.\nIn the above example, the lowest location number can be obtained from seed number 82, which corresponds to soil 84, fertilizer 84, water 84, light 77, temperature 45, humidity 46, and location 46. So, the lowest location number is 46.\nConsider all of the initial seed numbers listed in the ranges on the first line of the almanac. What is the lowest location number that corresponds to any of the initial seed numbers?\nLet\u0026rsquo;s first construct a dataframe containing the range of seeds:\nseeds_df = pd.DataFrame({\u0026#39;start\u0026#39;: seeds[::2], \u0026#39;range\u0026#39;: seeds[1::2], \u0026#39;end\u0026#39;: seeds[::2] + seeds[1::2]}) print(f\u0026#39;There are {sum(seeds_df.loc[:, \u0026#34;range\u0026#34;]):_} seeds in total\u0026#39;) seeds_df There are 2_549_759_327 seeds in total start range end 0 3169137700 271717609 3440855309 1 3522125441 23376095 3545501536 2 1233948799 811833837 2045782636 3 280549587 703867355 984416942 4 166086528 44766996 210853524 5 2326968141 69162222 2396130363 6 2698492851 14603069 2713095920 7 2755327667 348999531 3104327198 8 2600461189 92332846 2692794035 9 1054656969 169099767 1223756736 Now - I really needed some time to finally realize, that going through all seed values is really unfeasable. So how do we deal with this problem?\nIn the end we need the lowest location number - thus our approach is to take the humi_loc map, start with the lowest location number and go up and get the corresponding seed values. The location of the first seed value which is inside seeds_df is our solution.\nSo first we rebuild the get_location function to a get_seed function (we reverse the maps tuple with maps[::-1] and switch src and dest).\ndef get_seed(location): current = location for df in maps[::-1]: current_map = [row for row in df.itertuples() if ((current \u0026gt;= row.dest_start) and (current \u0026lt; row.dest_end))] if len(current_map) == 0: pass elif len(current_map) == 1: current = (current_map[0].src_start + (current - current_map[0].dest_start)) else: raise ValueError(\u0026#39;This should not happen!\u0026#39;) return current Since the Python 3 range function does not store it\u0026rsquo;s contents in memory (similar to a generator), it is well suited to go through these large ranges.\nLastly, we deal with the sheer amount of possible values by performing a grid search. First, we check every millionth location. After the first match, we stop this search and check the last million locations before the match with a finer grid and so on. The last grid is just 1, so we find our lowest location.\nUpdate: We order our locations from smallest to largest with sort_values and go through them - but the search stops after the first match, since that will be our lowest location.\ndef grid_search(start: int, end: int, grid: list[int]): success = False for g in grid: print(f\u0026#39;Start search with grid={g}\u0026#39;) for l in range(start, end, g): current = get_seed(l) if any((current \u0026gt;= seeds_df.loc[:, \u0026#39;start\u0026#39;]) \u0026amp; (current \u0026lt; seeds_df.loc[:, \u0026#39;end\u0026#39;])): print(f\u0026#39;location {l} is the lowest which contains one of the given seeds ({current})\u0026#39;) start = l - g end = l success = True break return success for row in humi_loc.sort_values(\u0026#39;dest_start\u0026#39;).itertuples(): success = grid_search(start=row.dest_start, end=row.dest_end, grid=[1_000_000, 100_000, 1000, 1]) if success: print(\u0026#39;Finished search\u0026#39;) break Start search with grid=1000000 location 85000000 is the lowest which contains one of the given seeds (2605777210) Start search with grid=100000 location 84300000 is the lowest which contains one of the given seeds (2605077210) Start search with grid=1000 location 84207000 is the lowest which contains one of the given seeds (2604984210) Start search with grid=1 location 84206669 is the lowest which contains one of the given seeds (2604983879) Finished search "},{"id":5,"href":"/posts/aoc2023-day04/","title":"Advent of code 2023 - Day 4: Scratchcards","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 4 - see https:adventofcode.com/2023/day/4\nPart 1 # Lets first read the task: The Elf leads you over to the pile of colorful cards. There, you discover dozens of scratchcards, all with their opaque covering already scratched off. Picking one up, it looks like each card has two lists of numbers separated by a vertical bar (|): a list of winning numbers and then a list of numbers you have. You organize the information into a table (your puzzle input).\nAs far as the Elf has been able to figure out, you have to figure out which of the numbers you have appear in the list of winning numbers. The first match makes the card worth one point and each match after the first doubles the point value of that card.\nFor example:\nCard 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53 Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19 Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1 Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83 Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36 Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11 In the above example, card 1 has five winning numbers (41, 48, 83, 86, and 17) and eight numbers you have (83, 86, 6, 31, 17, 9, 48, and 53). Of the numbers you have, four of them (48, 83, 17, and 86) are winning numbers! That means card 1 is worth 8 points (1 for the first match, then doubled three times for each of the three matches after the first).\nCard 2 has two winning numbers (32 and 61), so it is worth 2 points. Card 3 has two winning numbers (1 and 21), so it is worth 2 points. Card 4 has one winning number (84), so it is worth 1 point. Card 5 has no winning numbers, so it is worth no points. Card 6 has no winning numbers, so it is worth no points. So, in this example, the Elf\u0026rsquo;s pile of scratchcards is worth 13 points.\nTake a seat in the large pile of colorful cards. How many points are they worth in total?\nLoading this data is very similar to Day 2 - so let\u0026rsquo;s load the data as we did there. Our goal is to get win and yours columns holding the respective digits which we want to compare. We find the numbers with one or more digits using the regex \\d+. And we want them to be in sets (not lists), as we can logically compare sets in Python.\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-04-1-aoc.txt\u0026#39;, names=[\u0026#39;win\u0026#39;]) txt[\u0026#39;id\u0026#39;] = txt.loc[:, \u0026#39;win\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply( lambda x: int(x[0].strip(\u0026#39;Card \u0026#39;))) txt[\u0026#39;win\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;:\u0026#39;).apply(lambda x: x[1])) txt[\u0026#39;yours\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;|\u0026#39;) .apply(lambda x: x[1]) # get a list of only the numbers / digits .apply(lambda x: re.findall(r\u0026#39;\\d+\u0026#39;, x)) # convert the list of strings to a set of integers .apply(lambda x: set([int(i) for i in x]))) txt[\u0026#39;win\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;|\u0026#39;) .apply(lambda x: x[0]) .apply(lambda x: re.findall(r\u0026#39;\\d+\u0026#39;, x)) .apply(lambda x: set([int(i) for i in x]))) txt win id yours 0 {32, 36, 7, 9, 10, 12, 82, 85, 95, 31} 1 {2, 7, 9, 10, 12, 14, 21, 22, 23, 24, 31, 32, \u0026hellip; 1 {35, 76, 16, 82, 19, 22, 88, 59, 60, 95} 2 {7, 8, 12, 16, 19, 22, 26, 28, 35, 38, 44, 51,\u0026hellip; 2 {1, 70, 11, 78, 48, 19, 52, 88, 28, 94} 3 {3, 4, 8, 17, 18, 19, 24, 31, 34, 45, 52, 54, \u0026hellip; 3 {65, 2, 72, 28, 14, 16, 55, 91, 92, 62} 4 {3, 4, 6, 7, 8, 9, 15, 30, 33, 35, 47, 49, 51,\u0026hellip; 4 {38, 41, 75, 77, 50, 24, 94, 60, 61, 30} 5 {1, 2, 4, 5, 6, 7, 9, 10, 14, 17, 21, 29, 47, \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 213 {97, 98, 39, 41, 43, 12, 13, 19, 93, 95} 214 {5, 10, 17, 20, 28, 29, 33, 34, 36, 50, 51, 52\u0026hellip; 214 {97, 35, 69, 40, 74, 45, 20, 21, 62, 31} 215 {1, 8, 15, 17, 18, 25, 30, 33, 42, 44, 47, 52,\u0026hellip; 215 {33, 70, 71, 12, 78, 17, 51, 86, 60, 94} 216 {7, 8, 9, 10, 22, 29, 37, 39, 41, 43, 46, 47, \u0026hellip; 216 {98, 67, 68, 38, 70, 39, 72, 77, 45, 21} 217 {8, 21, 22, 25, 26, 31, 37, 41, 42, 48, 54, 57\u0026hellip; 217 {34, 9, 44, 78, 79, 16, 17, 19, 55, 92} 218 {1, 4, 20, 21, 27, 38, 39, 40, 41, 45, 46, 52,\u0026hellip; 218 rows × 3 columns\nNow, we get the logical conjunction of win and yours, these are our winning numbers. Then, the number of wins is converted to points - for all number of wins bigger than 1, we can get the points by 2**(n_wins-1).\ntxt[\u0026#39;n_wins\u0026#39;] = txt.apply( lambda row: len(row.loc[\u0026#39;win\u0026#39;] \u0026amp; row.loc[\u0026#39;yours\u0026#39;]), axis=1) txt[\u0026#39;points\u0026#39;] = txt.loc[:, \u0026#39;n_wins\u0026#39;].apply( lambda x: 2**(x-1) if x \u0026gt; 1 else x) txt.loc[:, [\u0026#39;win\u0026#39;, \u0026#39;n_wins\u0026#39;, \u0026#39;points\u0026#39;]] win n_wins points 0 {32, 36, 7, 9, 10, 12, 82, 85, 95, 31} 10 512 1 {35, 76, 16, 82, 19, 22, 88, 59, 60, 95} 10 512 2 {1, 70, 11, 78, 48, 19, 52, 88, 28, 94} 5 16 3 {65, 2, 72, 28, 14, 16, 55, 91, 92, 62} 0 0 4 {38, 41, 75, 77, 50, 24, 94, 60, 61, 30} 0 0 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 213 {97, 98, 39, 41, 43, 12, 13, 19, 93, 95} 0 0 214 {97, 35, 69, 40, 74, 45, 20, 21, 62, 31} 0 0 215 {33, 70, 71, 12, 78, 17, 51, 86, 60, 94} 2 2 216 {98, 67, 68, 38, 70, 39, 72, 77, 45, 21} 1 1 217 {34, 9, 44, 78, 79, 16, 17, 19, 55, 92} 0 0 218 rows × 3 columns\nThen, we just sum up:\nsum(txt.loc[:, \u0026#39;points\u0026#39;]) 25004 Part 2 # Let\u0026rsquo;s read the task of part 2! There\u0026rsquo;s no such thing as \u0026ldquo;points\u0026rdquo;. Instead, scratchcards only cause you to win more scratchcards equal to the number of winning numbers you have.\nSpecifically, you win copies of the scratchcards below the winning card equal to the number of matches. So, if card 10 were to have 5 matching numbers, you would win one copy each of cards 11, 12, 13, 14, and 15.\nCopies of scratchcards are scored like normal scratchcards and have the same card number as the card they copied. So, if you win a copy of card 10 and it has 5 matching numbers, it would then win a copy of the same cards that the original card 10 won: cards 11, 12, 13, 14, and 15. This process repeats until none of the copies cause you to win any more cards. (Cards will never make you copy a card past the end of the table.)\nThis time, the above example goes differently:\nCard 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53 Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19 Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1 Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83 Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36 Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11 Card 1 has four matching numbers, so you win one copy each of the next four cards: cards 2, 3, 4, and 5. Your original card 2 has two matching numbers, so you win one copy each of cards 3 and 4. Your copy of card 2 also wins one copy each of cards 3 and 4. Your four instances of card 3 (one original and three copies) have two matching numbers, so you win four copies each of cards 4 and 5. Your eight instances of card 4 (one original and seven copies) have one matching number, so you win eight copies of card 5. Your fourteen instances of card 5 (one original and thirteen copies) have no matching numbers and win no more cards. Once all of the originals and copies have been processed, you end up with 1 instance of card 1, 2 instances of card 2, 4 instances of card 3, 8 instances of card 4, 14 instances of card 5, and 1 instance of card 6. In total, this example pile of scratchcards causes you to ultimately have 30 scratchcards!\nProcess all of the original and copied scratchcards until no more scratchcards are won. Including the original set of scratchcards, how many total scratchcards do you end up with?\nWe will use the n_wins column we created before and go from there. We step through each Game. Each current game gets +1 card. Then, we step through the number of next games depending on our n_wins. Each of these gets added the card number of the current game.\ntxt[\u0026#39;cards\u0026#39;] = 0 for i, nwin in enumerate(txt.loc[:, \u0026#39;n_wins\u0026#39;]): txt.loc[i, \u0026#39;cards\u0026#39;] += 1 for j in range(1, nwin+1): txt.loc[i+j, \u0026#39;cards\u0026#39;] += txt.loc[i, \u0026#39;cards\u0026#39;] txt.loc[:, [\u0026#39;n_wins\u0026#39;, \u0026#39;cards\u0026#39;]] n_wins cards 0 10 1 1 10 2 2 5 4 3 0 8 4 0 8 \u0026hellip; \u0026hellip; \u0026hellip; 213 0 9608 214 0 8927 215 2 8927 216 1 12636 217 0 21564 218 rows × 2 columns\nNow, we just sum the output again:\nsum(txt[\u0026#39;cards\u0026#39;]) 14427616 "},{"id":6,"href":"/posts/aoc2023-day03/","title":"Advent of code 2023 - Day 3: Gear Ratios","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 3 - see https:adventofcode.com/2023/day/3\nPart 1 # Lets first read the task: The engine schematic (your puzzle input) consists of a visual representation of the engine. There are lots of numbers and symbols you don\u0026rsquo;t really understand, but apparently any number adjacent to a symbol, even diagonally, is a \u0026ldquo;part number\u0026rdquo; and should be included in your sum. (Periods (.) do not count as a symbol.)\nHere is an example engine schematic:\n467..114.. ...*...... ..35..633. ......#... 617*...... .....+.58. ..592..... ......755. ...$.*.... .664.598.. In this schematic, two numbers are not part numbers because they are not adjacent to a symbol: 114 (top right) and 58 (middle right). Every other number is adjacent to a symbol and so is a part number; their sum is 4361.\nOf course, the actual engine schematic is much larger. What is the sum of all of the part numbers in the engine schematic?\nOkay, let\u0026rsquo;s first get the input as a numpy character array\nimport numpy as np import pandas as pd import sys import matplotlib.pyplot as plt from scipy import ndimage as ndi np.set_printoptions(threshold=sys.maxsize) txt = pd.read_table(\u0026#39;data/2023-12-03-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) arr = np.chararray((txt.size, txt.size), unicode=True) txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda x: [i for i in x]) for i, code in enumerate(txt[\u0026#39;code\u0026#39;]): arr[i, :] = code print((arr[:15, :15])) [[\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;5\u0026#39; \u0026#39;3\u0026#39; \u0026#39;.\u0026#39; \u0026#39;4\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;2\u0026#39; \u0026#39;6\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;5\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;3\u0026#39; \u0026#39;8\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;4\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;6\u0026#39; \u0026#39;6\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;1\u0026#39; \u0026#39;2\u0026#39; \u0026#39;6\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;3\u0026#39; \u0026#39;1\u0026#39; \u0026#39;/\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;/\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;5\u0026#39; \u0026#39;3\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;5\u0026#39; \u0026#39;5\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;6\u0026#39; \u0026#39;8\u0026#39; \u0026#39;5\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;0\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;3\u0026#39; \u0026#39;8\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39;]] /tmp/ipykernel_18336/87086526.py:9: DeprecationWarning: `np.chararray` is deprecated and will be removed from the main namespace in the future. Use an array with a string or bytes dtype instead. arr = np.chararray((txt.size, txt.size), unicode=True) Now extract symbols, digits and the empty space. We use the numpy character methods for that. This way, we create a binary mask for all digits and a binary mask for all empty space (.). The symbols are then every character which is neither.\ndigits = np.char.isdigit(arr) empty = np.char.endswith(arr, \u0026#39;.\u0026#39;) symbols = ~(digits | empty) # just for visualization plt.figure(figsize=(16, 5)) plt.subplot(131, title=\u0026#39;symbols\u0026#39;).matshow(symbols) plt.subplot(132, title=\u0026#39;digits\u0026#39;).matshow(digits) plt.subplot(133, title=\u0026#39;empty\u0026#39;).matshow(empty) plt.show() Now we use the image processing technique of dilation on the symbols mask. So that we get a new mask which covers the surroundings of all symbols. We use this afterwards to check if the digits are near a symbol.\nstruct = ((1, 1, 1), (1, 1, 1), (1, 1, 1)) dilate = ndi.binary_dilation(symbols, structure=struct) plt.figure(figsize=(10, 6)) plt.subplot(121, title=\u0026#39;symbols\u0026#39;).matshow(symbols[:15, :15]) plt.subplot(122, title=\u0026#39;symbols dilated\u0026#39;).matshow(dilate[:15, :15]) plt.show() Creating this masks as before could be understood as a binary segmentation, as each element in our mask is either True or False. To extract the single digits, we\u0026rsquo;ll convert the binary digits segmentation into a instance segmentation, where each connected segment has an own index.\nmarkers, num_features = ndi.label(digits) plt.figure(figsize=(10, 6)) plt.subplot(121, title=\u0026#39;binary segmentation\u0026#39;).matshow( digits[:15, :15]) plt.subplot(122, title=\u0026#39;instance segmentation\u0026#39;).matshow( markers[:15, :15], cmap=\u0026#39;gnuplot\u0026#39;) plt.show() Now, for each instance, we check if the dilated binary mask overlaps with the instance and if yes, we extract the number.\nnumbers = [int(\u0026#39;\u0026#39;.join(arr[markers == i])) for i in range(1, num_features+1) if np.any((markers == i) \u0026amp; dilate)] Then, we just sum up:\nsum(numbers) 527364 Part 2 # Let\u0026rsquo;s first read the second task! The missing part wasn\u0026rsquo;t the only issue - one of the gears in the engine is wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.\nThis time, you need to find the gear ratio of every gear and add them all up so that the engineer can figure out which gear needs to be replaced.\nConsider the same engine schematic again:The missing part wasn\u0026rsquo;t the only issue - one of the gears in the engine is wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.\nThis time, you need to find the gear ratio of every gear and add them all up so that the engineer can figure out which gear needs to be replaced.\nConsider the same engine schematic again:\n467..114.. ...*...... ..35..633. ......#... 617*...... .....+.58. ..592..... ......755. ...$.*.... .664.598.. In this schematic, there are two gears. The first is in the top left; it has part numbers 467 and 35, so its gear ratio is 16345. The second gear is in the lower right; its gear ratio is 451490. (The * adjacent to 617 is not a gear because it is only adjacent to one part number.) Adding up all of the gear ratios produces 467835.\nWhat is the sum of all of the gear ratios in your engine schematic?\nWe\u0026rsquo;ll use the same method as before, but this time only extract * and create the instance segmentation before the dilation. Why? Because when we dilate first, we could merge two independent gears into one instance.\ngear = np.char.endswith(arr, \u0026#39;*\u0026#39;) gear_markers, gear_num = ndi.label(gear) plt.figure(figsize=(10, 6)) plt.subplot(131, title=\u0026#39;all symbols dilated\u0026#39;).matshow( symbols[:15, :15]) plt.subplot(132, title=\u0026#39;gears\u0026#39;).matshow( gear[:15, :15]) plt.subplot(133, title=\u0026#39;gears instances\u0026#39;).matshow( gear_markers[:15, :15], cmap=\u0026#39;gnuplot\u0026#39;) plt.show() Now, we step through each gear instance, create a mask for that gear, dilate it, and then step through all digits and check if they are in the gear mask. If we get two digits in the end, we multiply them to get the gear ratio and save the ratios to a list.\ngear_ratios = [] for i in range(1, gear_num+1): gear_binary = gear_markers == i gear_dil = ndi.binary_dilation(gear_binary, structure=struct) part_numbers = [int(\u0026#39;\u0026#39;.join(arr[markers == j])) for j in range(1, num_features+1) if np.any((markers == j) \u0026amp; gear_dil)] if len(part_numbers) == 2: gear_ratios.append(part_numbers[0] * part_numbers[1]) Now, we just sum the output again:\nsum(gear_ratios) 79026871 "},{"id":7,"href":"/posts/aoc2023-day02/","title":"Advent of code 2023 - Day 2: Cube Conundrum","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 2 - see https:adventofcode.com/2023/day/2\nPart 1 # Lets first read the task: As you walk, the Elf shows you a small bag and some cubes which are either red, green, or blue. Each time you play this game, he will hide a secret number of cubes of each color in the bag, and your goal is to figure out information about the number of cubes.\nTo get information, once a bag has been loaded with cubes, the Elf will reach into the bag, grab a handful of random cubes, show them to you, and then put them back in the bag. He\u0026rsquo;ll do this a few times per game.\nYou play several games and record the information from each game (your puzzle input). Each game is listed with its ID number (like the 11 in Game 11: ...) followed by a semicolon-separated list of subsets of cubes that were revealed from the bag (like 3 red, 5 green, 4 blue).\nFor example, the record of a few games might look like this:\nGame 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green In game 1, three sets of cubes are revealed from the bag (and then put back again). The first set is 3 blue cubes and 4 red cubes; the second set is 1 red cube, 2 green cubes, and 6 blue cubes; the third set is only 2 green cubes.\nThe Elf would first like to know which games would have been possible if the bag contained only 12 red cubes, 13 green cubes, and 14 blue cubes?\nIn the example above, games 1, 2, and 5 would have been possible if the bag had been loaded with that configuration. However, game 3 would have been impossible because at one point the Elf showed you 20 red cubes at once; similarly, game 4 would also have been impossible because the Elf showed you 15 blue cubes at once. If you add up the IDs of the games that would have been possible, you get 8.\nDetermine which games would have been possible if the bag had been loaded with only 12 red cubes, 13 green cubes, and 14 blue cubes. What is the sum of the IDs of those games?\nOkay, let\u0026rsquo;s load our python kernel in emacs-jupyter and get coding! First of all, let\u0026rsquo;s load the input and split the riddle code by colon : to extract the game id and the rest of the code by semicolon ; to get the number of sets played in each game.\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-02-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) txt[\u0026#39;id\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply( lambda x: int(x[0].strip(\u0026#39;Game \u0026#39;))) txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply(lambda x: x[1]) # txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;;\u0026#39;) # txt[\u0026#39;nsets\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda x: len(x)) txt code id 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 100 rows × 2 columns\nNow, let\u0026rsquo;s extract the three colors in different columns with regex. We use the lookahead assertion ?= to find the respective colours and only exctract the digits \\d+ coming before. Then we just keep the max imum drawn number of cubes per color, since this is the only information that matters at the moment.\ntxt[\u0026#39;green\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.green)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt[\u0026#39;red\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.red)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt[\u0026#39;blue\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.blue)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt code id green red blue 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 2 10 10 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 15 10 3 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 4 6 16 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 16 20 18 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 10 6 14 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 4 2 1 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 5 4 8 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 2 5 2 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 2 9 11 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 13 15 11 100 rows × 5 columns\nLastly, we just filter the DataFrame to only include games where all drawn cubes were below or equal the number of cubes in the game and sum the result!\ntxt[\u0026#39;id\u0026#39;][(txt[\u0026#39;green\u0026#39;] \u0026lt; 14) \u0026amp; (txt[\u0026#39;red\u0026#39;] \u0026lt; 13) \u0026amp; (txt[\u0026#39;blue\u0026#39;] \u0026lt; 15)].sum() 3035 Part 2 # First, let\u0026rsquo;s get the instruction from the second part: As you continue your walk, the Elf poses a second question: in each game you played, what is the fewest number of cubes of each color that could have been in the bag to make the game possible?\nAgain consider the example games from earlier:\nGame 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green In game 1, the game could have been played with as few as 4 red, 2 green, and 6 blue cubes. If any color had even one fewer cube, the game would have been impossible. Game 2 could have been played with a minimum of 1 red, 3 green, and 4 blue cubes. Game 3 must have been played with at least 20 red, 13 green, and 6 blue cubes. Game 4 required at least 14 red, 3 green, and 15 blue cubes. Game 5 needed no fewer than 6 red, 3 green, and 2 blue cubes in the bag. The power of a set of cubes is equal to the numbers of red, green, and blue cubes multiplied together. The power of the minimum set of cubes in game 1 is 48. In games 2-5 it was 12, 1560, 630, and 36, respectively. Adding up these five powers produces the sum 2286.\nFor each game, find the minimum set of cubes that must have been present. What is the sum of the power of these sets?\nLuckily, this task is made trivial by the approach we have taken before. We just have to multiply the green, red and blue columns:\ntxt[\u0026#39;power\u0026#39;] = txt.loc[:, \u0026#39;green\u0026#39;] * txt.loc[:, \u0026#39;blue\u0026#39;] * txt.loc[:, \u0026#39;red\u0026#39;] txt code id green red blue power 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 2 10 10 200 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 15 10 3 450 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 4 6 16 384 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 16 20 18 5760 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 10 6 14 840 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 4 2 1 8 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 5 4 8 160 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 2 5 2 20 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 2 9 11 198 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 13 15 11 2145 100 rows × 6 columns\nAnd for this one, the sum is:\ntxt[\u0026#39;power\u0026#39;].sum() 66027 "},{"id":8,"href":"/posts/aoc2023-day01/","title":"Advent of code 2023 - Day 1: Trebuchet?!","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 1 - see https:adventofcode.com/2023/day/1\nPart 1 # Our first task is the following: The newly-improved calibration document consists of lines of text; each line originally contained a specific calibration value that the Elves now need to recover. On each line, the calibration value can be found by combining the first digit and the last digit (in that order) to form a single two-digit number.\nFor example:\n1abc2 pqr3stu8vwx a1b2c3d4e5f treb7uchet In this example, the calibration values of these four lines are 12, 38, 15, and 77. Adding these together produces 142.\nConsider your entire calibration document. What is the sum of all of the calibration values?\nLets start jupyter in our shell to start coding!\nconda activate tf jupyter lab --no-browser --port=8888 First, load the test document\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-01-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) txt code 0 jjfvnnlfivejj1 1 6fourfour 2 ninevbmltwo69 3 pcg91vqrfpxxzzzoneightzt 4 jpprthxgjfive3one1qckhrptpqdc \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 996 81s 997 2four3threesxxvlfqfive4 998 nine6eightsevenzx9twoxc 999 hmbfjdfnp989mfivefiverpzrjs 1000 rows × 1 columns\nSecond, extract the digits. I had to wrap my head around regex matching in python first, because I first tried pandas.extract (which only extracts the first match), then pandas.extractall (which extracts all matches but puts them into a multiindex which makes things more difficult in this case). So I settled for the re.findall version, which returns a list. To concatenate the elements in the list, we take use the join function.\ntxt[\u0026#39;digits\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda x: \u0026#39;\u0026#39;.join(re.findall(r\u0026#39;(\\d+)\u0026#39;, x))) txt code digits 0 jjfvnnlfivejj1 1 1 6fourfour 6 2 ninevbmltwo69 69 3 pcg91vqrfpxxzzzoneightzt 91 4 jpprthxgjfive3one1qckhrptpqdc 31 \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 996 81s 81 997 2four3threesxxvlfqfive4 234 998 nine6eightsevenzx9twoxc 69 999 hmbfjdfnp989mfivefiverpzrjs 989 1000 rows × 2 columns\nNext, combine the first and the last digit and convert the result from string to integer\ntxt[\u0026#39;calibration\u0026#39;] = txt.loc[:, \u0026#39;digits\u0026#39;].apply( lambda x: int(x[0] + x[-1])) txt code digits calibration 0 jjfvnnlfivejj1 1 11 1 6fourfour 6 66 2 ninevbmltwo69 69 69 3 pcg91vqrfpxxzzzoneightzt 91 91 4 jpprthxgjfive3one1qckhrptpqdc 31 31 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 996 81s 81 81 997 2four3threesxxvlfqfive4 234 24 998 nine6eightsevenzx9twoxc 69 69 999 hmbfjdfnp989mfivefiverpzrjs 989 99 1000 rows × 3 columns\nLastly, get the sum of our calibration numbers\ntxt.loc[:, \u0026#39;calibration\u0026#39;].sum() 56465 Part 2 # Now follows part two: Your calculation isn\u0026rsquo;t quite right. It looks like some of the digits are actually spelled out with letters: one, two, three, four, five, six, seven, eight, and nine also count as valid \u0026ldquo;digits\u0026rdquo;.\nEquipped with this new information, you now need to find the real first and last digit on each line. For example:\ntwo1nine eightwothree abcone2threexyz xtwone3four 4nineeightseven2 zoneight234 7pqrstsixteen In this example, the calibration values are 29, 83, 13, 24, 42, 14, and 76. Adding these together produces 281.\nWhat is the sum of all of the calibration values?\nOkay, let\u0026rsquo;s see if we can update the pattern matching. To deal with potential overlapping values like oneight which contains one as well as eight, I used the regex positive lookahead ?= as described here. Because this enables capturing overlapping values, I used \\d (one digit) instead of \\d+ (one or more digits), otherwise digits might double. Afterwards, just replace the spelled out digits with their numerical value.\n# for i, r in enumerate(txt.loc[:, \u0026#39;code\u0026#39;]): # matches = re.findall( # r\u0026#39;(?=(\\d|one|two|three|four|five|six|seven|eight|nine))\u0026#39;, r) # result = \u0026#39;\u0026#39;.join([match for match in matches]) # result = result.replace(\u0026#39;one\u0026#39;, \u0026#39;1\u0026#39;).replace(\u0026#39;two\u0026#39;, \u0026#39;2\u0026#39;).replace( # \u0026#39;three\u0026#39;, \u0026#39;3\u0026#39;).replace(\u0026#39;four\u0026#39;, \u0026#39;4\u0026#39;).replace(\u0026#39;five\u0026#39;, \u0026#39;5\u0026#39;).replace( # \u0026#39;six\u0026#39;, \u0026#39;6\u0026#39;).replace(\u0026#39;seven\u0026#39;, \u0026#39;7\u0026#39;).replace(\u0026#39;eight\u0026#39;, \u0026#39;8\u0026#39;).replace( # \u0026#39;nine\u0026#39;, \u0026#39;9\u0026#39;) # txt.loc[i, \u0026#39;digits2\u0026#39;] = result # txt # a very nice alternative suggested by Tomalak: digits = \u0026#39;\\d one two three four five six seven eight nine\u0026#39;.split() txt[\u0026#39;digits2\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda v: \u0026#39;\u0026#39;.join( str(digits.index(m)) if m in digits else m for m in re.findall(rf\u0026#39;(?=({\u0026#34;|\u0026#34;.join(digits)}))\u0026#39;, v) )) txt code digits calibration digits2 0 jjfvnnlfivejj1 1 11 51 1 6fourfour 6 66 644 2 ninevbmltwo69 69 69 9269 3 pcg91vqrfpxxzzzoneightzt 91 91 9118 4 jpprthxgjfive3one1qckhrptpqdc 31 31 5311 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 58375 996 81s 81 81 81 997 2four3threesxxvlfqfive4 234 24 243354 998 nine6eightsevenzx9twoxc 69 69 968792 999 hmbfjdfnp989mfivefiverpzrjs 989 99 98955 1000 rows × 4 columns\nNow, construct the calibration value as before\u0026hellip;\ntxt[\u0026#39;calibration2\u0026#39;] = txt.loc[:, \u0026#39;digits2\u0026#39;].apply(lambda x: int(x[0] + x[-1])) txt code digits calibration digits2 calibration2 0 jjfvnnlfivejj1 1 11 51 51 1 6fourfour 6 66 644 64 2 ninevbmltwo69 69 69 9269 99 3 pcg91vqrfpxxzzzoneightzt 91 91 9118 98 4 jpprthxgjfive3one1qckhrptpqdc 31 31 5311 51 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 58375 55 996 81s 81 81 81 81 997 2four3threesxxvlfqfive4 234 24 243354 24 998 nine6eightsevenzx9twoxc 69 69 968792 92 999 hmbfjdfnp989mfivefiverpzrjs 989 99 98955 95 1000 rows × 5 columns\n\u0026hellip; and get the correct sum!\ntxt.loc[:, \u0026#39;calibration2\u0026#39;].sum() 55902 "}] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/en.search-data.min.fd07bc3fbf8c5c6b0d51c093e137b6631f07588fcb9dcedbf19c3224be5a9392.json b/aseltmann.github.io-hugo/public/en.search-data.min.fd07bc3fbf8c5c6b0d51c093e137b6631f07588fcb9dcedbf19c3224be5a9392.json deleted file mode 100644 index 0bccb20..0000000 --- a/aseltmann.github.io-hugo/public/en.search-data.min.fd07bc3fbf8c5c6b0d51c093e137b6631f07588fcb9dcedbf19c3224be5a9392.json +++ /dev/null @@ -1 +0,0 @@ -[{"id":0,"href":"/docs/notes/poetry/","title":"Poetry","section":"Notes","content":" Table of Contents Lǎozǐ Lǎozǐ # I am partial to the translation of Derek Lin\nDàodéjīng - Chapter 11 (The use of what has no substantive existence) [\u0026hellip;]\nMix clay\nto create a container\nIn its emptiness, there is\nthe function of a container\n[\u0026hellip;]\nTherefore,\nthat which exists is used to create benefit\nThat which is empty is used to create functionality\nDàodéjīng - Chapter 15 (The exhibition of the qualities of the Dao) The Tao masters of antiquity\nSubtle wonders through mystery\nDepths that cannot be discerned\nBecause one cannot discern them\nTherefore one is forced to describe the appearance\nHesitant,\nlike crossing a wintry river\nCautious,\nlike fearing four neighbors\nSolemn,\nlike a guest\nLoose,\nlike ice about to melt\nGenuine,\nlike plain wood\nOpen,\nlike a valley\nOpaque,\nlike muddy water\nWho can be muddled yet desist\nIn stillness gradually become clear?\nWho can be serene yet persist\nIn motion gradually come alive?\nOne who holds this \u0026lt;i\u0026gt;Tao\u0026lt;/i\u0026gt; does not wish to be overfilled\nBecause one is not overfilled\nTherefore one can preserve and not create anew\nDàodéjīng - Chapter 20 (Being different from ordinary people) Cease learning, no more worries\nRespectful response and scornful response\nHow much is the difference?\nGoodness and evil\nHow much do they differ?\nWhat the people fear,\nI cannot be unafraid\nSo desolate! How limitless it is!\nThe people are excited\nAs if enjoying a great feast\nAs if climbing up to the terrace in spring\nI alone am quiet and uninvolved\nLike an infant\nnot yet smiling\nSo weary,\nlike having no place to return\nThe people all have surplus\nWhile I alone seem lacking\nI have the heart of a fool indeed -\nso ignorant!\nOrdinary people are bright\nI alone am muddled\nOrdinary people are scrutinizing\nI alone am obtuse\nSo tranquil, like the ocean\nSo moving, as if without limits\nThe people all have goals\nAnd I alone am stubborn and lowly\nI alone am different from them\nAnd value the nourishing mother\n"},{"id":1,"href":"/posts/aoc2024-day01/","title":"Advent of code 2024 - Day 1: Historian Hysteria","section":"Posts","content":" Table of Contents Part 1: Locations Part 2: Similarities I try to solve this year\u0026rsquo;s Advent of Code 2024 riddles in R. This is Day 1 - see https:adventofcode.com/2024/day/1\nPart 1: Locations # Lets first read the task: Throughout the Chief\u0026rsquo;s office, the historically significant locations are listed not by name but by a unique number called the location ID. To make sure they don\u0026rsquo;t miss anything, The Historians split into two groups, each searching the office and trying to create their own complete list of location IDs.\nThere\u0026rsquo;s just one problem: by holding the two lists up side by side (your puzzle input), it quickly becomes clear that the lists aren\u0026rsquo;t very similar. Maybe you can help The Historians reconcile their lists?\nFor example:\n3 4 4 3 2 5 1 3 3 9 3 3 Maybe the lists are only off by a small amount! To find out, pair up the numbers and measure how far apart they are. Pair up the smallest number in the left list with the smallest number in the right list, then the second-smallest left number with the second-smallest right number, and so on.\nWithin each pair, figure out how far apart the two numbers are; you\u0026rsquo;ll need to add up all of those distances. For example, if you pair up a 3 from the left list with a 7 from the right list, the distance apart is 4; if you pair up a 9 with a 3, the distance apart is 6.\nIn the example list above, the pairs and distances would be as follows:\nThe smallest number in the left list is 1, and the smallest number in the right list is 3. The distance between them is 2. The second-smallest number in the left list is 2, and the second-smallest number in the right list is another 3. The distance between them is 1. The third-smallest number in both lists is 3, so the distance between them is 0. The next numbers to pair up are 3 and 4, a distance of 1. The fifth-smallest numbers in each list are 3 and 5, a distance of 2. Finally, the largest number in the left list is 4, while the largest number in the right list is 9; these are a distance 5 apart. To find the total distance between the left list and the right list, add up the distances between all of the pairs you found. In the example above, this is 2 + 1 + 0 + 1 + 2 + 5, a total distance of 11!\nYour actual left and right lists contain many location IDs. What is the total distance between your lists?\nFirst, we read the data - the fread command from the data.table package is more versatile than read.delim from base and directly reads the data as a data.table, which has some benefits.\ndata \u0026lt;- data.table::fread(\u0026#34;2024-12-01-1-aoc.txt\u0026#34;, header = FALSE) head(data) V1 V2 \u0026lt;int\u0026gt; \u0026lt;int\u0026gt; 41226 69190 89318 10100 59419 23880 63157 20193 81510 22869 83942 63304 Now, the idea is to get the ordered versions of V1 and V2.\nv1 \u0026lt;- data[order(V1)]$V1 v2 \u0026lt;- data[order(V2)]$V2 head(v1) 10188 10314 10319 10348 10408 10668 Coming from Python, this syntax looks a bit odd. The documentation of data.table::setorder helps:\nsetorder (and setorderv) reorders the rows of a data.table based on the columns (and column order) provided. It reorders the table by reference and is therefore very memory efficient. Note that queries like x[order(.)] are optimised internally to use data.table\u0026#39;s fast order. So, data[order(V1)] is actually short for data.table::setorder(data, V1). Then, we extract the vector by name using the $ operator, which allows to extract elements by name.\nhead(data.table::setorder(data, V1)$V1) 10188 10314 10319 10348 10408 10668 The actual computation is just the sum of the absolute difference:\nsum(abs(v1 - v2)) 3574690\nPart 2: Similarities # Lets first read the task: This time, you\u0026rsquo;ll need to figure out exactly how often each number from the left list appears in the right list. Calculate a total similarity score by adding up each number in the left list after multiplying it by the number of times that number appears in the right list.\nHere are the same example lists again:\n3 4 4 3 2 5 1 3 3 9 3 3 For these example lists, here is the process of finding the similarity score:\nThe first number in the left list is 3. It appears in the right list three times, so the similarity score increases by 3 * 3 = 9. The second number in the left list is 4. It appears in the right list once, so the similarity score increases by 4 * 1 = 4. The third number in the left list is 2. It does not appear in the right list, so the similarity score does not increase (2 * 0 = 0). The fourth number, 1, also does not appear in the right list. The fifth number, 3, appears in the right list three times; the similarity score increases by 9. The last number, 3, appears in the right list three times; the similarity score again increases by 9. So, for these example lists, the similarity score at the end of this process is 31 (9 + 4 + 0 + 0 + 9 + 9).\nOnce again consider your left and right lists. What is their similarity score?\nMy idea here is simple - we first count the occurrences of V2 to be able to check V1 against them. Here, applying the table command works like pandas.value_counts() and achieves this. We convert the output to a data.frame and assign the V1 values as row names. Note: table converts the values which are counted to string, e.g. \u0026quot;10019\u0026quot; instead of 10019.\ntab \u0026lt;- table(v2) v2series \u0026lt;- data.frame(c(tab), row.names = names(tab)) head(v2series) v2series[\u0026#34;10019\u0026#34;, ] c.tab. \u0026lt;int\u0026gt; 10019 1 10100 1 10206 1 10428 1 10645 1 10972 1 1\nNow, we loop through V1 and try to access the frequency in V2 (by converting to string first, via as.character). If nothing is found in V2, an NA value is returned. For everything that is not NA, we compute the similarity score, append it to a list, and sum in the end. Note: a list in R can not directly be summed (I don\u0026rsquo;t know why that is) - so we have to unlist first.\nsim \u0026lt;- list() for (num in v1) { myfreq \u0026lt;- v2series[as.character(num), ] if (!is.na(myfreq)) { score \u0026lt;- myfreq * num sim \u0026lt;- append(sim, score) } } sum(unlist(sim)) 22565391\n"},{"id":2,"href":"/posts/aoc2023-day07/","title":"Advent of code 2023 - Day 7: Camel Cards","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 7 - see https:adventofcode.com/2023/day/7\nPart 1 # Lets first read the task: In Camel Cards, you get a list of hands, and your goal is to order them based on the strength of each hand. A hand consists of five cards labeled one of A, K, Q, J, T, 9, 8, 7, 6, 5, 4, 3, or 2. The relative strength of each card follows this order, where A is the highest and 2 is the lowest.\nEvery hand is exactly one type. From strongest to weakest, they are:\nFive of a kind, where all five cards have the same label: AAAAA Four of a kind, where four cards have the same label and one card has a different label: AA8AA Full house, where three cards have the same label, and the remaining two cards share a different label: 23332 Three of a kind, where three cards have the same label, and the remaining two cards are each different from any other card in the hand: TTT98 Two pair, where two cards share one label, two other cards share a second label, and the remaining card has a third label: 23432 One pair, where two cards share one label, and the other three cards have a different label from the pair and each other: A23A4 High card, where all cards\u0026rsquo; labels are distinct: 23456 Hands are primarily ordered based on type; for example, every full house is stronger than any three of a kind.\nIf two hands have the same type, a second ordering rule takes effect. Start by comparing the first card in each hand. If these cards are different, the hand with the stronger first card is considered stronger. If the first card in each hand have the same label, however, then move on to considering the second card in each hand. If they differ, the hand with the higher second card wins; otherwise, continue with the third card in each hand, then the fourth, then the fifth.\nSo, 33332 and 2AAAA are both four of a kind hands, but 33332 is stronger because its first card is stronger. Similarly, 77888 and 77788 are both a full house, but 77888 is stronger because its third card is stronger (and both hands have the same first and second card).\nTo play Camel Cards, you are given a list of hands and their corresponding bid (your puzzle input). For example:\n32T3K 765 T55J5 684 KK677 28 KTJJT 220 QQQJA 483 This example shows five hands; each hand is followed by its bid amount. Each hand wins an amount equal to its bid multiplied by its rank, where the weakest hand gets rank 1, the second-weakest hand gets rank 2, and so on up to the strongest hand. Because there are five hands in this example, the strongest hand will have rank 5 and its bid will be multiplied by 5.\nSo, the first step is to put the hands in order of strength:\n32T3K is the only one pair and the other hands are all a stronger type, so it gets rank 1. KK677 and KTJJT are both two pair. Their first cards both have the same label, but the second card of KK677 is stronger (K vs T), so KTJJT gets rank 2 and KK677 gets rank 3. T55J5 and QQQJA are both three of a kind. QQQJA has a stronger first card, so it gets rank 5 and T55J5 gets rank 4. Now, you can determine the total winnings of this set of hands by adding up the result of multiplying each hand\u0026rsquo;s bid with its rank (765 * 1 + 220 * 2 + 28 * 3 + 684 * 4 + 483 * 5). So the total winnings in this example are 6440.\nFind the rank of every hand in your set. What are the total winnings?\nAs always - let\u0026rsquo;s read in the data.\nimport pandas as pd camel = pd.read_table(\u0026#39;data/2023-12-07-1-aoc.txt\u0026#39;, delim_whitespace=True, header=None, names=[\u0026#39;cards\u0026#39;, \u0026#39;bid\u0026#39;], dtype={\u0026#39;cards\u0026#39;: \u0026#39;string\u0026#39;, \u0026#39;bid\u0026#39;: \u0026#39;Int64\u0026#39;}) camel cards bid 0 A833A 309 1 Q33J3 205 2 55KK5 590 3 K4457 924 4 Q3QT3 11 \u0026hellip; \u0026hellip; \u0026hellip; 995 KQJ53 367 996 57866 537 997 Q94A9 210 998 J448A 903 999 6J66J 114 1000 rows × 2 columns\nFor sorting, we need to compute a type column. The 7 types can be deduced based on how many different cards are there (through set()) and the number of cards.\nAlso, for the next step we prepare one column per card string in the cards.\ndef get_type(cards): cardset = set(cards) cardlen = len(cardset) cardlist = list(cards) cardnum = [cardlist.count(i) for i in cardset] if cardlen == 1: ctype = 1 elif cardlen in (4, 5): ctype = cardlen + 2 elif cardlen == 2: ctype = 2 if set(cardnum) == {1, 4} else 3 else: ctype = 4 if 3 in cardnum else 5 return ctype camel[\u0026#39;type\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: get_type(x)) for i in range(5): camel[f\u0026#39;c{i}\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: list(x)[i]) camel cards bid type c0 c1 c2 c3 c4 0 A833A 309 5 A 8 3 3 A 1 Q33J3 205 4 Q 3 3 J 3 2 55KK5 590 3 5 5 K K 5 3 K4457 924 6 K 4 4 5 7 4 Q3QT3 11 5 Q 3 Q T 3 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 KQJ53 367 7 K Q J 5 3 996 57866 537 6 5 7 8 6 6 997 Q94A9 210 6 Q 9 4 A 9 998 J448A 903 6 J 4 4 8 A 999 6J66J 114 3 6 J 6 6 J 1000 rows × 8 columns\nFor sorting, pandas contributes the sort_values method, which supports sorting by multiple columns (here, we first sort according to type, then c0, c1, \u0026hellip;). For easier sorting by numerical values, we convert the values using the card_dict dictionary, which implements the given sorting order.\ncard_dict = {c: i for i, c in enumerate([7, 6, 5, 4, 3, 2, 1, \u0026#39;2\u0026#39;, \u0026#39;3\u0026#39;, \u0026#39;4\u0026#39;, \u0026#39;5\u0026#39;, \u0026#39;6\u0026#39;, \u0026#39;7\u0026#39;, \u0026#39;8\u0026#39;, \u0026#39;9\u0026#39;, \u0026#39;T\u0026#39;, \u0026#39;J\u0026#39;, \u0026#39;Q\u0026#39;, \u0026#39;K\u0026#39;, \u0026#39;A\u0026#39;])} camel = camel.sort_values(by=[\u0026#39;type\u0026#39;, \u0026#39;c0\u0026#39;, \u0026#39;c1\u0026#39;, \u0026#39;c2\u0026#39;, \u0026#39;c3\u0026#39;, \u0026#39;c4\u0026#39;], ignore_index=True, key=lambda x: x.map(card_dict)) camel.index += 1 camel cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 AA5AA 565 2 A A 5 A A 997 AAATA 704 2 A A A T A 998 AAAA3 145 2 A A A A 3 999 AAAAJ 594 2 A A A A J 1000 JJJJJ 219 1 J J J J J 1000 rows × 8 columns\nLastly, we multiply the bid with the index and sum the result.\ncamel.loc[:, \u0026#39;bid\u0026#39;].mul(camel.index).sum() 246912307 Part 2 # Lets first read the task: To make things a little more interesting, the Elf introduces one additional rule. Now, J cards are jokers - wildcards that can act like whatever card would make the hand the strongest type possible.\nTo balance this, J cards are now the weakest individual cards, weaker even than 2. The other cards stay in the same order: A, K, Q, T, 9, 8, 7, 6, 5, 4, 3, 2, J.\nJ cards can pretend to be whatever card is best for the purpose of determining hand type; for example, QJJQ2 is now considered four of a kind. However, for the purpose of breaking ties between two hands of the same type, J is always treated as J, not the card it\u0026rsquo;s pretending to be: JKKK2 is weaker than QQQQ2 because J is weaker than Q.\nNow, the above example goes very differently:\n32T3K 765 T55J5 684 KK677 28 KTJJT 220 QQQJA 483 32T3K is still the only one pair; it doesn\u0026rsquo;t contain any jokers, so its strength doesn\u0026rsquo;t increase. KK677 is now the only two pair, making it the second-weakest hand. T55J5, KTJJT, and QQQJA are now all four of a kind! T55J5 gets rank 3, QQQJA gets rank 4, and KTJJT gets rank 5. With the new joker rule, the total winnings in this example are 5905.\nUsing the new joker rule, find the rank of every hand in your set. What are the new total winnings?\nWhat changes? We have to update our get_type() function and our card_dict dictionary, otherwise the method stays the same!\nLet\u0026rsquo;s first write get_type_joker(): if no joker is in the cards, we can just refer to get_type(). If we have 1, 2 or 5 different card faces, the answer is trivial. We only have to use the number of jokers and the maximum number of other faces if we have 3 or 4 different faces.\nNote: if a J is present, there can never be high card (type = 7), because we always can get one pair. Also we can never get two pair (type = 5), because we can always construct three of a kind or full house.\ndef get_type_joker(cards): cardset = set(cards) cardlen = len(cardset) cardlist = list(cards) cardnum = [cardlist.count(i) for i in cardset] if not \u0026#39;J\u0026#39; in cardset: ctype = get_type(cards) else: idxj = list(cardset).index(\u0026#39;J\u0026#39;) numj = cardnum[idxj] cardnum.pop(idxj) if cardlen in [1, 2]: ctype = 1 elif cardlen == 5: ctype = 6 else: if (numj in [1, 2]) and (cardlen == 4): ctype = 4 else: ctype = 6 - max(cardnum) - numj return ctype camel[\u0026#39;type\u0026#39;] = camel.loc[:, \u0026#39;cards\u0026#39;].apply(lambda x: get_type_joker(x)) camel cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 AA5AA 565 2 A A 5 A A 997 AAATA 704 2 A A A T A 998 AAAA3 145 2 A A A A 3 999 AAAAJ 594 1 A A A A J 1000 JJJJJ 219 1 J J J J J 1000 rows × 8 columns\nNow, we just move J in our new card_dict_joker to where it belongs, the rest of the solution is the same.\ncard_dict_joker = {c: i for i, c in enumerate( [7, 6, 5, 4, 3, 2, 1, \u0026#39;J\u0026#39;, \u0026#39;2\u0026#39;, \u0026#39;3\u0026#39;, \u0026#39;4\u0026#39;, \u0026#39;5\u0026#39;, \u0026#39;6\u0026#39;, \u0026#39;7\u0026#39;, \u0026#39;8\u0026#39;, \u0026#39;9\u0026#39;, \u0026#39;T\u0026#39;, \u0026#39;Q\u0026#39;, \u0026#39;K\u0026#39;, \u0026#39;A\u0026#39;])} camel = camel.sort_values(by=[\u0026#39;type\u0026#39;, \u0026#39;c0\u0026#39;, \u0026#39;c1\u0026#39;, \u0026#39;c2\u0026#39;, \u0026#39;c3\u0026#39;, \u0026#39;c4\u0026#39;], ignore_index=True, key=lambda x: x.map(card_dict_joker)) camel.index += 1 display(camel) print(\u0026#39;Our new solution: \u0026#39;, camel.loc[:, \u0026#39;bid\u0026#39;].mul(camel.index).sum()) cards bid type c0 c1 c2 c3 c4 1 237AQ 157 7 2 3 7 A Q 2 23K47 759 7 2 3 K 4 7 3 249K8 612 7 2 4 9 K 8 4 264A7 341 7 2 6 4 A 7 5 26578 10 7 2 6 5 7 8 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 996 TJTTJ 609 1 T J T T J 997 QQJQJ 131 1 Q Q J Q J 998 QQQJQ 831 1 Q Q Q J Q 999 KKKJK 183 1 K K K J K 1000 AAAAJ 594 1 A A A A J 1000 rows × 8 columns\nOur new solution: 246894760 "},{"id":3,"href":"/posts/aoc2023-day06/","title":"Advent of code 2023 - Day 6: Wait For It","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 6 - see https:adventofcode.com/2023/day/6\nPart 1 # Lets first read the task: You manage to sign up as a competitor in the boat races just in time. The organizer explains that it\u0026rsquo;s not really a traditional race - instead, you will get a fixed amount of time during which your boat has to travel as far as it can, and you win if your boat goes the farthest.\nAs part of signing up, you get a sheet of paper (your puzzle input) that lists the time allowed for each race and also the best distance ever recorded in that race. To guarantee you win the grand prize, you need to make sure you go farther in each race than the current record holder.\nThe organizer brings you over to the area where the boat races are held. The boats are much smaller than you expected - they\u0026rsquo;re actually toy boats, each with a big button on top. Holding down the button charges the boat, and releasing the button allows the boat to move. Boats move faster if their button was held longer, but time spent holding the button counts against the total race time. You can only hold the button at the start of the race, and boats don\u0026rsquo;t move until the button is released.\nFor example:\nTime: 7 15 30 Distance: 9 40 200 This document describes three races:\nThe first race lasts 7 milliseconds. The record distance in this race is 9 millimeters. The second race lasts 15 milliseconds. The record distance in this race is 40 millimeters. The third race lasts 30 milliseconds. The record distance in this race is 200 millimeters. Your toy boat has a starting speed of zero millimeters per millisecond. For each whole millisecond you spend at the beginning of the race holding down the button, the boat\u0026rsquo;s speed increases by one millimeter per millisecond.\nSo, because the first race lasts 7 milliseconds, you only have a few options:\nDon\u0026rsquo;t hold the button at all (that is, hold it for 0 milliseconds) at the start of the race. The boat won\u0026rsquo;t move; it will have traveled 0 millimeters by the end of the race. Hold the button for 1 millisecond at the start of the race. Then, the boat will travel at a speed of 1 millimeter per millisecond for 6 milliseconds, reaching a total distance traveled of 6 millimeters. Hold the button for 2 milliseconds, giving the boat a speed of 2 millimeters per millisecond. It will then get 5 milliseconds to move, reaching a total distance of 10 millimeters. Hold the button for 3 milliseconds. After its remaining 4 milliseconds of travel time, the boat will have gone 12 millimeters. Hold the button for 4 milliseconds. After its remaining 3 milliseconds of travel time, the boat will have gone 12 millimeters. Hold the button for 5 milliseconds, causing the boat to travel a total of 10 millimeters. Hold the button for 6 milliseconds, causing the boat to travel a total of 6 millimeters. Hold the button for 7 milliseconds. That\u0026rsquo;s the entire duration of the race. You never let go of the button. The boat can\u0026rsquo;t move until you let go of the button. Please make sure you let go of the button so the boat gets to move. 0 millimeters. Since the current record for this race is 9 millimeters, there are actually 4 different ways you could win: you could hold the button for 2, 3, 4, or 5 milliseconds at the start of the race.\nIn the second race, you could hold the button for at least 4 milliseconds and at most 11 milliseconds and beat the record, a total of 8 different ways to win.\nIn the third race, you could hold the button for at least 11 milliseconds and no more than 19 milliseconds and still beat the record, a total of 9 ways you could win.\nTo see how much margin of error you have, determine the number of ways you can beat the record in each race; in this example, if you multiply these values together, you get 288 (4 * 8 * 9).\nDetermine the number of ways you could beat the record in each race. What do you get if you multiply these numbers together?\nThe solution for this problem is shorter than the task description!\nAs always - let\u0026rsquo;s load the data. Note: we use the delim_whitesspace argument instead of sep=' ' to separate values by spaces of any length.\nimport pandas as pd import numpy as np race = pd.read_table(\u0026#39;data/2023-12-06-1-aoc.txt\u0026#39;, delim_whitespace=True, header=None, index_col=0) race.index = [\u0026#39;time\u0026#39;, \u0026#39;distance\u0026#39;] race 1 2 3 4 time 51 92 68 90 distance 222 2031 1126 1225 The we use Python list comprehensions to quickly calculate all distances for the range of the time value (and keep only the winning distances). Then we get the respective lengths and compute the product.\ndef get_distance(time, distance): return [t * (time - t) for t in range(1, time + 1) if (t * (time - t)) \u0026gt; distance] race.loc[\u0026#39;nwins\u0026#39;, :] = [len(get_distance(int(r.time), int(r.distance))) for c, r in race.items()] nwins_prod = np.prod(race.loc[\u0026#39;nwins\u0026#39;, :].values, dtype=int) display(race) print(f\u0026#39;The product of our nwins is {nwins_prod}\u0026#39;) 1 2 3 4 time 51.0 92.0 68.0 90.0 distance 222.0 2031.0 1126.0 1225.0 nwins 42.0 19.0 11.0 57.0 The product of our nwins is 500346 Part 2 # Lets first read the task: As the race is about to start, you realize the piece of paper with race times and record distances you got earlier actually just has very bad kerning. There\u0026rsquo;s really only one race - ignore the spaces between the numbers on each line.\nSo, the example from before:\nTime: 7 15 30 Distance: 9 40 200 \u0026hellip;now instead means this:\nTime: 71530 Distance: 940200 Now, you have to figure out how many ways there are to win this single race. In this example, the race lasts for 71530 milliseconds and the record distance you need to beat is 940200 millimeters. You could hold the button anywhere from 14 to 71516 milliseconds and beat the record, a total of 71503 ways!\nHow many ways can you beat the record in this one much longer race?\nFor this riddle we can take the same approach as for Part 1. The cell took 18s to evaluate on a consumer laptop from 2015 - no fancy workarounds needed :)\nracetime = int(\u0026#39;\u0026#39;.join(str(r) for r in race.loc[\u0026#39;time\u0026#39;, :4].astype(int))) racedist = int(\u0026#39;\u0026#39;.join(str(r) for r in race.loc[\u0026#39;distance\u0026#39;, :4].astype(int))) racewins = len(get_distance(racetime, racedist)) race[5] = [racetime, racedist, racewins] race 1 2 3 4 5 time 51.0 92.0 68.0 90.0 51926890 distance 222.0 2031.0 1126.0 1225.0 222203111261225 nwins 42.0 19.0 11.0 57.0 42515755 "},{"id":4,"href":"/posts/aoc2023-day05/","title":"Advent of code 2023 - Day 5: If You Give A Seed A Fertilizer","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 5 - see https:adventofcode.com/2023/day/5\nUpdate [2023-12-31 So]:\nsubstitute get_generator() (own implementation) with range() (Python inbuilt) improve grid search so that it goes through all location ranges, still starting with the lowest range Part 1 # Lets first read the task: The almanac (your puzzle input) lists all of the seeds that need to be planted. It also lists what type of soil to use with each kind of seed, what type of fertilizer to use with each kind of soil, what type of water to use with each kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is identified with a number, but numbers are reused by each category - that is, soil 123 and fertilizer 123 aren\u0026rsquo;t necessarily related to each other.\nFor example:\nseeds: 79 14 55 13 seed-to-soil map: 50 98 2 52 50 48 soil-to-fertilizer map: 0 15 37 37 52 2 39 0 15 fertilizer-to-water map: 49 53 8 0 11 42 42 0 7 57 7 4 water-to-light map: 88 18 7 18 25 70 light-to-temperature map: 45 77 23 81 45 19 68 64 13 temperature-to-humidity map: 0 69 1 1 0 69 humidity-to-location map: 60 56 37 56 93 4 The almanac starts by listing which seeds need to be planted: seeds 79, 14, 55, and 13.\nThe rest of the almanac contains a list of maps which describe how to convert numbers from a source category into numbers in a destination category. That is, the section that starts with seed-to-soil map: describes how to convert a seed number (the source) to a soil number (the destination). This lets the gardener and his team know which soil to use with which seeds, which water to use with which fertilizer, and so on.\nRather than list every source number and its corresponding destination number one by one, the maps describe entire ranges of numbers that can be converted. Each line within a map contains three numbers: the destination range start, the source range start, and the range length.\nConsider again the example seed-to-soil map:\n50 98 2 52 50 48\nThe first line has a destination range start of 50, a source range start of 98, and a range length of 2. This line means that the source range starts at 98 and contains two values: 98 and 99. The destination range is the same length, but it starts at 50, so its two values are 50 and 51. With this information, you know that seed number 98 corresponds to soil number 50 and that seed number 99 corresponds to soil number 51. o The second line means that the source range starts at 50 and contains 48 values: 50, 51, \u0026hellip;, 96, 97. This corresponds to a destination range starting at 52 and also containing 48 values: 52, 53, \u0026hellip;, 98, 99. So, seed number 53 corresponds to soil number 55.\nAny source numbers that aren\u0026rsquo;t mapped correspond to the same destination number. So, seed number 10 corresponds to soil number 10.\nSo, the entire list of seed numbers and their corresponding soil numbers looks like this:\nseed soil 0 0 1 1 ... ... 48 48 49 49 50 52 51 53 ... ... 96 98 97 99 98 50 99 51 With this map, you can look up the soil number required for each initial seed number:\nSeed number 79 corresponds to soil number 81. Seed number 14 corresponds to soil number 14. Seed number 55 corresponds to soil number 57. Seed number 13 corresponds to soil number 13. The gardener and his team want to get started as soon as possible, so they\u0026rsquo;d like to know the closest location that needs a seed. Using these maps, find the lowest location number that corresponds to any of the initial seeds. To do this, you\u0026rsquo;ll need to convert each seed number through other categories until you can find its corresponding location number. In this example, the corresponding types are:\nSeed 79, soil 81, fertilizer 81, water 81, light 74, temperature 78, humidity 78, location 82. Seed 14, soil 14, fertilizer 53, water 49, light 42, temperature 42, humidity 43, location 43. Seed 55, soil 57, fertilizer 57, water 53, light 46, temperature 82, humidity 82, location 86. Seed 13, soil 13, fertilizer 52, water 41, light 34, temperature 34, humidity 35, location 35. So, the lowest location number in this example is 35.\nWhat is the lowest location number that corresponds to any of the initial seed numbers?\nWow, this task is a mouthful\u0026hellip;\nLet\u0026rsquo;s start slowly and load the data. Our input text document contains several maps, which are clearly separated and have a title (seed-to-soil map etc). So we can tell pandas where each map starts and give each map a dataframe. I got the values for the skiprows and nrows argument by looking at the input file and\u0026hellip; counting :)\nimport pandas as pd import sys seeds = pd.read_table(\u0026#39;data/2023-12-05-1-aoc.txt\u0026#39;, nrows=1, sep=\u0026#39; \u0026#39;, header=None, index_col=0) seeds = seeds.values.flatten() map_opt = {\u0026#39;filepath_or_buffer\u0026#39;: \u0026#39;data/2023-12-05-1-aoc.txt\u0026#39;, \u0026#39;header\u0026#39;: None, \u0026#39;sep\u0026#39;: \u0026#39; \u0026#39;, \u0026#39;dtype\u0026#39;: \u0026#39;Int64\u0026#39;, \u0026#39;names\u0026#39;: [\u0026#39;dest_start\u0026#39;, \u0026#39;src_start\u0026#39;, \u0026#39;range\u0026#39;]} seed_soil = pd.read_table(skiprows=3, nrows=23, **map_opt) soil_fert = pd.read_table(skiprows=28, nrows=9, **map_opt) fert_water = pd.read_table(skiprows=39, nrows=20, **map_opt) water_light = pd.read_table(skiprows=61, nrows=40, **map_opt) light_temp = pd.read_table(skiprows=103, nrows=36, **map_opt) temp_humi = pd.read_table(skiprows=141, nrows=35, **map_opt) humi_loc = pd.read_table(skiprows=178, nrows=26, **map_opt) maps = (seed_soil, soil_fert, fert_water, water_light, light_temp, temp_humi, humi_loc) print(\u0026#39;seeds are just a numpy array:\u0026#39;) display(seeds) print(\u0026#39;The \u0026#34;humidity-to-location\u0026#34; map as an example:\u0026#39;) humi_loc seeds are just a numpy array: array([3169137700, 271717609, 3522125441, 23376095, 1233948799, 811833837, 280549587, 703867355, 166086528, 44766996, 2326968141, 69162222, 2698492851, 14603069, 2755327667, 348999531, 2600461189, 92332846, 1054656969, 169099767])The \u0026#34;humidity-to-location\u0026#34; map as an example: dest_start src_start range 0 3490144003 1623866227 218040905 1 1709610578 1620839197 3027030 2 105449249 586389428 113279526 3 1899604338 2167886292 348178199 4 1712637608 2678624215 186966730 5 218728775 0 245776251 6 2472992580 923734334 143670388 7 2616662968 3670169885 15297294 8 2247782537 1395629154 225210043 9 0 480940179 105449249 10 4113852846 3959729057 159909096 11 3322784653 1067404722 167359350 12 923734334 4119638153 175329143 13 1534964496 2516064491 162559724 14 2631960262 3496028440 140849733 15 2862695906 1234764072 97988355 16 1697524220 3636878173 12086358 17 2985646048 1332752427 62876727 18 1099063477 2970241510 435901019 19 4009202281 2865590945 104650565 20 2960684261 2142924505 24961787 21 2772809995 3406142529 89885911 22 4273761942 3648964531 21205354 23 3708184908 1841907132 301017373 24 464505026 245776251 235163928 25 3048522775 3685467179 274261878 My first attempt was to actually construct ranges like in the example above, mapping out all possible sources and destinations. Python quickly informed me that even constructing one pandas.Series with int64 values mapping seeds to soil would cost 64GB memory - not the best solution.\nSo we take a different approach. For convenience, let\u0026rsquo;s add a src_end and a dest_end column to our maps:\nfor df in maps: df[\u0026#39;src_end\u0026#39;] = df.loc[:, \u0026#39;src_start\u0026#39;] + df.loc[:, \u0026#39;range\u0026#39;] df[\u0026#39;dest_end\u0026#39;] = df.loc[:, \u0026#39;dest_start\u0026#39;] + df.loc[:, \u0026#39;range\u0026#39;] print(\u0026#39;Again the \u0026#34;humidity-to-location\u0026#34; map as an example:\u0026#39;) humi_loc Again the \u0026#34;humidity-to-location\u0026#34; map as an example: dest_start src_start range src_end dest_end 0 3490144003 1623866227 218040905 1841907132 3708184908 1 1709610578 1620839197 3027030 1623866227 1712637608 2 105449249 586389428 113279526 699668954 218728775 3 1899604338 2167886292 348178199 2516064491 2247782537 4 1712637608 2678624215 186966730 2865590945 1899604338 5 218728775 0 245776251 245776251 464505026 6 2472992580 923734334 143670388 1067404722 2616662968 7 2616662968 3670169885 15297294 3685467179 2631960262 8 2247782537 1395629154 225210043 1620839197 2472992580 9 0 480940179 105449249 586389428 105449249 10 4113852846 3959729057 159909096 4119638153 4273761942 11 3322784653 1067404722 167359350 1234764072 3490144003 12 923734334 4119638153 175329143 4294967296 1099063477 13 1534964496 2516064491 162559724 2678624215 1697524220 14 2631960262 3496028440 140849733 3636878173 2772809995 15 2862695906 1234764072 97988355 1332752427 2960684261 16 1697524220 3636878173 12086358 3648964531 1709610578 17 2985646048 1332752427 62876727 1395629154 3048522775 18 1099063477 2970241510 435901019 3406142529 1534964496 19 4009202281 2865590945 104650565 2970241510 4113852846 20 2960684261 2142924505 24961787 2167886292 2985646048 21 2772809995 3406142529 89885911 3496028440 2862695906 22 4273761942 3648964531 21205354 3670169885 4294967296 23 3708184908 1841907132 301017373 2142924505 4009202281 24 464505026 245776251 235163928 480940179 699668954 25 3048522775 3685467179 274261878 3959729057 3322784653 Now we actually compute the mapping. For each seed, we go through all mappings and in each mapping we go through each row. We find the row which contains the mapping and exctract the destination, which is the source for the next map until we reach the last map which gives us the locations.\nApproach 1: df.itertuples() is a convenient way to step through a pandas.DataFrame in this example. It is faster than df.iterrows() and returns the row as a NamedTuple - nice!\nApproach 2: I actually wondered if it would be faster to get all maps in one pd.DataFrame and then iterate through the mappings. To test this let\u0026rsquo;s construct a new DataFrame maps_df which contains all maps. Since the maps have different lengths it is important to cast the datatype to Int64, which is short for pd.Int64Dtype() and keeps values as integers, even if NA values are in the same column.\nApproach 3: A third alternative I tested (not shown here) was to check if a value is in a Python range with the in operator as in: if 3 in range(5):... . This was way too slow.\n# mapping version 1 def get_location(seed): current = seed for df in maps: current_map = [row for row in df.itertuples() if ((current \u0026gt; row.src_start) and (current \u0026lt; row.src_end))] if len(current_map) == 0: pass elif len(current_map) == 1: current = (current_map[0].dest_start + (current - current_map[0].src_start)) else: raise ValueError(\u0026#39;This should not happen!\u0026#39;) return current # mapping version 2 - around 10 times slower # you need to rename the maps_df columns so that they have a unique id # e.g. \u0026#39;src_start_1\u0026#39;, \u0026#39;src_start_2\u0026#39; etc # maps_df = pd.concat(maps, axis=\u0026#39;columns\u0026#39;) # def get_dest(i, src): # return (maps_df[(src \u0026gt; maps_df.loc[:, f\u0026#39;src_start_{i}\u0026#39;]) \u0026amp; # (src \u0026lt; maps_df.loc[:, f\u0026#39;src_end_{i}\u0026#39;])] # .loc[:, f\u0026#39;dest_start_{i}\u0026#39;] # .iloc[0]) # def get_location2(seed): # dest = seed # i = 1 # while i \u0026lt; 7: # dest = get_dest(i, dest) # i += 1 # return dest %timeit get_location(seeds[0]) 1.98 ms ± 72 μs per loop (mean ± std. dev. of 7 runs, 100 loops each) Now let\u0026rsquo;s get a list of locations:\nlocations = [get_location(s) for s in seeds] locations 2493982655 3209845376 3992357533 4163131463 4104485616 1952252479 3218677354 388071289 2181441450 2594336315 4049507670 2084517144 3119633635 428978312 3518771991 3704555655 953918455 2107687768 3448046330 2184454689 Lastly, just get the minimum of all location values.\nmin(locations) 388071289 Part 2 # Let\u0026rsquo;s read the task of part 2! Everyone will starve if you only plant such a small number of seeds. Re-reading the almanac, it looks like the seeds: line actually describes ranges of seed numbers.\nThe values on the initial seeds: line come in pairs. Within each pair, the first value is the start of the range and the second value is the length of the range. So, in the first line of the example above:\nseeds: 79 14 55 13\nThis line describes two ranges of seed numbers to be planted in the garden. The first range starts with seed number 79 and contains 14 values: 79, 80, \u0026hellip;, 91, 92. The second range starts with seed number 55 and contains 13 values: 55, 56, \u0026hellip;, 66, 67.\nNow, rather than considering four seed numbers, you need to consider a total of 27 seed numbers.\nIn the above example, the lowest location number can be obtained from seed number 82, which corresponds to soil 84, fertilizer 84, water 84, light 77, temperature 45, humidity 46, and location 46. So, the lowest location number is 46.\nConsider all of the initial seed numbers listed in the ranges on the first line of the almanac. What is the lowest location number that corresponds to any of the initial seed numbers?\nLet\u0026rsquo;s first construct a dataframe containing the range of seeds:\nseeds_df = pd.DataFrame({\u0026#39;start\u0026#39;: seeds[::2], \u0026#39;range\u0026#39;: seeds[1::2], \u0026#39;end\u0026#39;: seeds[::2] + seeds[1::2]}) print(f\u0026#39;There are {sum(seeds_df.loc[:, \u0026#34;range\u0026#34;]):_} seeds in total\u0026#39;) seeds_df There are 2_549_759_327 seeds in total start range end 0 3169137700 271717609 3440855309 1 3522125441 23376095 3545501536 2 1233948799 811833837 2045782636 3 280549587 703867355 984416942 4 166086528 44766996 210853524 5 2326968141 69162222 2396130363 6 2698492851 14603069 2713095920 7 2755327667 348999531 3104327198 8 2600461189 92332846 2692794035 9 1054656969 169099767 1223756736 Now - I really needed some time to finally realize, that going through all seed values is really unfeasable. So how do we deal with this problem?\nIn the end we need the lowest location number - thus our approach is to take the humi_loc map, start with the lowest location number and go up and get the corresponding seed values. The location of the first seed value which is inside seeds_df is our solution.\nSo first we rebuild the get_location function to a get_seed function (we reverse the maps tuple with maps[::-1] and switch src and dest).\ndef get_seed(location): current = location for df in maps[::-1]: current_map = [row for row in df.itertuples() if ((current \u0026gt;= row.dest_start) and (current \u0026lt; row.dest_end))] if len(current_map) == 0: pass elif len(current_map) == 1: current = (current_map[0].src_start + (current - current_map[0].dest_start)) else: raise ValueError(\u0026#39;This should not happen!\u0026#39;) return current Since the Python 3 range function does not store it\u0026rsquo;s contents in memory (similar to a generator), it is well suited to go through these large ranges.\nLastly, we deal with the sheer amount of possible values by performing a grid search. First, we check every millionth location. After the first match, we stop this search and check the last million locations before the match with a finer grid and so on. The last grid is just 1, so we find our lowest location.\nUpdate: We order our locations from smallest to largest with sort_values and go through them - but the search stops after the first match, since that will be our lowest location.\ndef grid_search(start: int, end: int, grid: list[int]): success = False for g in grid: print(f\u0026#39;Start search with grid={g}\u0026#39;) for l in range(start, end, g): current = get_seed(l) if any((current \u0026gt;= seeds_df.loc[:, \u0026#39;start\u0026#39;]) \u0026amp; (current \u0026lt; seeds_df.loc[:, \u0026#39;end\u0026#39;])): print(f\u0026#39;location {l} is the lowest which contains one of the given seeds ({current})\u0026#39;) start = l - g end = l success = True break return success for row in humi_loc.sort_values(\u0026#39;dest_start\u0026#39;).itertuples(): success = grid_search(start=row.dest_start, end=row.dest_end, grid=[1_000_000, 100_000, 1000, 1]) if success: print(\u0026#39;Finished search\u0026#39;) break Start search with grid=1000000 location 85000000 is the lowest which contains one of the given seeds (2605777210) Start search with grid=100000 location 84300000 is the lowest which contains one of the given seeds (2605077210) Start search with grid=1000 location 84207000 is the lowest which contains one of the given seeds (2604984210) Start search with grid=1 location 84206669 is the lowest which contains one of the given seeds (2604983879) Finished search "},{"id":5,"href":"/posts/aoc2023-day04/","title":"Advent of code 2023 - Day 4: Scratchcards","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 4 - see https:adventofcode.com/2023/day/4\nPart 1 # Lets first read the task: The Elf leads you over to the pile of colorful cards. There, you discover dozens of scratchcards, all with their opaque covering already scratched off. Picking one up, it looks like each card has two lists of numbers separated by a vertical bar (|): a list of winning numbers and then a list of numbers you have. You organize the information into a table (your puzzle input).\nAs far as the Elf has been able to figure out, you have to figure out which of the numbers you have appear in the list of winning numbers. The first match makes the card worth one point and each match after the first doubles the point value of that card.\nFor example:\nCard 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53 Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19 Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1 Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83 Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36 Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11 In the above example, card 1 has five winning numbers (41, 48, 83, 86, and 17) and eight numbers you have (83, 86, 6, 31, 17, 9, 48, and 53). Of the numbers you have, four of them (48, 83, 17, and 86) are winning numbers! That means card 1 is worth 8 points (1 for the first match, then doubled three times for each of the three matches after the first).\nCard 2 has two winning numbers (32 and 61), so it is worth 2 points. Card 3 has two winning numbers (1 and 21), so it is worth 2 points. Card 4 has one winning number (84), so it is worth 1 point. Card 5 has no winning numbers, so it is worth no points. Card 6 has no winning numbers, so it is worth no points. So, in this example, the Elf\u0026rsquo;s pile of scratchcards is worth 13 points.\nTake a seat in the large pile of colorful cards. How many points are they worth in total?\nLoading this data is very similar to Day 2 - so let\u0026rsquo;s load the data as we did there. Our goal is to get win and yours columns holding the respective digits which we want to compare. We find the numbers with one or more digits using the regex \\d+. And we want them to be in sets (not lists), as we can logically compare sets in Python.\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-04-1-aoc.txt\u0026#39;, names=[\u0026#39;win\u0026#39;]) txt[\u0026#39;id\u0026#39;] = txt.loc[:, \u0026#39;win\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply( lambda x: int(x[0].strip(\u0026#39;Card \u0026#39;))) txt[\u0026#39;win\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;:\u0026#39;).apply(lambda x: x[1])) txt[\u0026#39;yours\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;|\u0026#39;) .apply(lambda x: x[1]) # get a list of only the numbers / digits .apply(lambda x: re.findall(r\u0026#39;\\d+\u0026#39;, x)) # convert the list of strings to a set of integers .apply(lambda x: set([int(i) for i in x]))) txt[\u0026#39;win\u0026#39;] = (txt.loc[:, \u0026#39;win\u0026#39;] .str.split(\u0026#39;|\u0026#39;) .apply(lambda x: x[0]) .apply(lambda x: re.findall(r\u0026#39;\\d+\u0026#39;, x)) .apply(lambda x: set([int(i) for i in x]))) txt win id yours 0 {32, 36, 7, 9, 10, 12, 82, 85, 95, 31} 1 {2, 7, 9, 10, 12, 14, 21, 22, 23, 24, 31, 32, \u0026hellip; 1 {35, 76, 16, 82, 19, 22, 88, 59, 60, 95} 2 {7, 8, 12, 16, 19, 22, 26, 28, 35, 38, 44, 51,\u0026hellip; 2 {1, 70, 11, 78, 48, 19, 52, 88, 28, 94} 3 {3, 4, 8, 17, 18, 19, 24, 31, 34, 45, 52, 54, \u0026hellip; 3 {65, 2, 72, 28, 14, 16, 55, 91, 92, 62} 4 {3, 4, 6, 7, 8, 9, 15, 30, 33, 35, 47, 49, 51,\u0026hellip; 4 {38, 41, 75, 77, 50, 24, 94, 60, 61, 30} 5 {1, 2, 4, 5, 6, 7, 9, 10, 14, 17, 21, 29, 47, \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 213 {97, 98, 39, 41, 43, 12, 13, 19, 93, 95} 214 {5, 10, 17, 20, 28, 29, 33, 34, 36, 50, 51, 52\u0026hellip; 214 {97, 35, 69, 40, 74, 45, 20, 21, 62, 31} 215 {1, 8, 15, 17, 18, 25, 30, 33, 42, 44, 47, 52,\u0026hellip; 215 {33, 70, 71, 12, 78, 17, 51, 86, 60, 94} 216 {7, 8, 9, 10, 22, 29, 37, 39, 41, 43, 46, 47, \u0026hellip; 216 {98, 67, 68, 38, 70, 39, 72, 77, 45, 21} 217 {8, 21, 22, 25, 26, 31, 37, 41, 42, 48, 54, 57\u0026hellip; 217 {34, 9, 44, 78, 79, 16, 17, 19, 55, 92} 218 {1, 4, 20, 21, 27, 38, 39, 40, 41, 45, 46, 52,\u0026hellip; 218 rows × 3 columns\nNow, we get the logical conjunction of win and yours, these are our winning numbers. Then, the number of wins is converted to points - for all number of wins bigger than 1, we can get the points by 2**(n_wins-1).\ntxt[\u0026#39;n_wins\u0026#39;] = txt.apply( lambda row: len(row.loc[\u0026#39;win\u0026#39;] \u0026amp; row.loc[\u0026#39;yours\u0026#39;]), axis=1) txt[\u0026#39;points\u0026#39;] = txt.loc[:, \u0026#39;n_wins\u0026#39;].apply( lambda x: 2**(x-1) if x \u0026gt; 1 else x) txt.loc[:, [\u0026#39;win\u0026#39;, \u0026#39;n_wins\u0026#39;, \u0026#39;points\u0026#39;]] win n_wins points 0 {32, 36, 7, 9, 10, 12, 82, 85, 95, 31} 10 512 1 {35, 76, 16, 82, 19, 22, 88, 59, 60, 95} 10 512 2 {1, 70, 11, 78, 48, 19, 52, 88, 28, 94} 5 16 3 {65, 2, 72, 28, 14, 16, 55, 91, 92, 62} 0 0 4 {38, 41, 75, 77, 50, 24, 94, 60, 61, 30} 0 0 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 213 {97, 98, 39, 41, 43, 12, 13, 19, 93, 95} 0 0 214 {97, 35, 69, 40, 74, 45, 20, 21, 62, 31} 0 0 215 {33, 70, 71, 12, 78, 17, 51, 86, 60, 94} 2 2 216 {98, 67, 68, 38, 70, 39, 72, 77, 45, 21} 1 1 217 {34, 9, 44, 78, 79, 16, 17, 19, 55, 92} 0 0 218 rows × 3 columns\nThen, we just sum up:\nsum(txt.loc[:, \u0026#39;points\u0026#39;]) 25004 Part 2 # Let\u0026rsquo;s read the task of part 2! There\u0026rsquo;s no such thing as \u0026ldquo;points\u0026rdquo;. Instead, scratchcards only cause you to win more scratchcards equal to the number of winning numbers you have.\nSpecifically, you win copies of the scratchcards below the winning card equal to the number of matches. So, if card 10 were to have 5 matching numbers, you would win one copy each of cards 11, 12, 13, 14, and 15.\nCopies of scratchcards are scored like normal scratchcards and have the same card number as the card they copied. So, if you win a copy of card 10 and it has 5 matching numbers, it would then win a copy of the same cards that the original card 10 won: cards 11, 12, 13, 14, and 15. This process repeats until none of the copies cause you to win any more cards. (Cards will never make you copy a card past the end of the table.)\nThis time, the above example goes differently:\nCard 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53 Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19 Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1 Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83 Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36 Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11 Card 1 has four matching numbers, so you win one copy each of the next four cards: cards 2, 3, 4, and 5. Your original card 2 has two matching numbers, so you win one copy each of cards 3 and 4. Your copy of card 2 also wins one copy each of cards 3 and 4. Your four instances of card 3 (one original and three copies) have two matching numbers, so you win four copies each of cards 4 and 5. Your eight instances of card 4 (one original and seven copies) have one matching number, so you win eight copies of card 5. Your fourteen instances of card 5 (one original and thirteen copies) have no matching numbers and win no more cards. Once all of the originals and copies have been processed, you end up with 1 instance of card 1, 2 instances of card 2, 4 instances of card 3, 8 instances of card 4, 14 instances of card 5, and 1 instance of card 6. In total, this example pile of scratchcards causes you to ultimately have 30 scratchcards!\nProcess all of the original and copied scratchcards until no more scratchcards are won. Including the original set of scratchcards, how many total scratchcards do you end up with?\nWe will use the n_wins column we created before and go from there. We step through each Game. Each current game gets +1 card. Then, we step through the number of next games depending on our n_wins. Each of these gets added the card number of the current game.\ntxt[\u0026#39;cards\u0026#39;] = 0 for i, nwin in enumerate(txt.loc[:, \u0026#39;n_wins\u0026#39;]): txt.loc[i, \u0026#39;cards\u0026#39;] += 1 for j in range(1, nwin+1): txt.loc[i+j, \u0026#39;cards\u0026#39;] += txt.loc[i, \u0026#39;cards\u0026#39;] txt.loc[:, [\u0026#39;n_wins\u0026#39;, \u0026#39;cards\u0026#39;]] n_wins cards 0 10 1 1 10 2 2 5 4 3 0 8 4 0 8 \u0026hellip; \u0026hellip; \u0026hellip; 213 0 9608 214 0 8927 215 2 8927 216 1 12636 217 0 21564 218 rows × 2 columns\nNow, we just sum the output again:\nsum(txt[\u0026#39;cards\u0026#39;]) 14427616 "},{"id":6,"href":"/posts/aoc2023-day03/","title":"Advent of code 2023 - Day 3: Gear Ratios","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 3 - see https:adventofcode.com/2023/day/3\nPart 1 # Lets first read the task: The engine schematic (your puzzle input) consists of a visual representation of the engine. There are lots of numbers and symbols you don\u0026rsquo;t really understand, but apparently any number adjacent to a symbol, even diagonally, is a \u0026ldquo;part number\u0026rdquo; and should be included in your sum. (Periods (.) do not count as a symbol.)\nHere is an example engine schematic:\n467..114.. ...*...... ..35..633. ......#... 617*...... .....+.58. ..592..... ......755. ...$.*.... .664.598.. In this schematic, two numbers are not part numbers because they are not adjacent to a symbol: 114 (top right) and 58 (middle right). Every other number is adjacent to a symbol and so is a part number; their sum is 4361.\nOf course, the actual engine schematic is much larger. What is the sum of all of the part numbers in the engine schematic?\nOkay, let\u0026rsquo;s first get the input as a numpy character array\nimport numpy as np import pandas as pd import sys import matplotlib.pyplot as plt from scipy import ndimage as ndi np.set_printoptions(threshold=sys.maxsize) txt = pd.read_table(\u0026#39;data/2023-12-03-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) arr = np.chararray((txt.size, txt.size), unicode=True) txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda x: [i for i in x]) for i, code in enumerate(txt[\u0026#39;code\u0026#39;]): arr[i, :] = code print((arr[:15, :15])) [[\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;5\u0026#39; \u0026#39;3\u0026#39; \u0026#39;.\u0026#39; \u0026#39;4\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;2\u0026#39; \u0026#39;6\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;5\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;3\u0026#39; \u0026#39;8\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;4\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;6\u0026#39; \u0026#39;6\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;1\u0026#39; \u0026#39;2\u0026#39; \u0026#39;6\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;3\u0026#39; \u0026#39;1\u0026#39; \u0026#39;/\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;/\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;5\u0026#39; \u0026#39;3\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;3\u0026#39; \u0026#39;5\u0026#39; \u0026#39;5\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;6\u0026#39; \u0026#39;8\u0026#39; \u0026#39;5\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;7\u0026#39; \u0026#39;0\u0026#39;] [\u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;9\u0026#39; \u0026#39;3\u0026#39; \u0026#39;8\u0026#39; \u0026#39;.\u0026#39; \u0026#39;.\u0026#39; \u0026#39;*\u0026#39;]] /tmp/ipykernel_18336/87086526.py:9: DeprecationWarning: `np.chararray` is deprecated and will be removed from the main namespace in the future. Use an array with a string or bytes dtype instead. arr = np.chararray((txt.size, txt.size), unicode=True) Now extract symbols, digits and the empty space. We use the numpy character methods for that. This way, we create a binary mask for all digits and a binary mask for all empty space (.). The symbols are then every character which is neither.\ndigits = np.char.isdigit(arr) empty = np.char.endswith(arr, \u0026#39;.\u0026#39;) symbols = ~(digits | empty) # just for visualization plt.figure(figsize=(16, 5)) plt.subplot(131, title=\u0026#39;symbols\u0026#39;).matshow(symbols) plt.subplot(132, title=\u0026#39;digits\u0026#39;).matshow(digits) plt.subplot(133, title=\u0026#39;empty\u0026#39;).matshow(empty) plt.show() Now we use the image processing technique of dilation on the symbols mask. So that we get a new mask which covers the surroundings of all symbols. We use this afterwards to check if the digits are near a symbol.\nstruct = ((1, 1, 1), (1, 1, 1), (1, 1, 1)) dilate = ndi.binary_dilation(symbols, structure=struct) plt.figure(figsize=(10, 6)) plt.subplot(121, title=\u0026#39;symbols\u0026#39;).matshow(symbols[:15, :15]) plt.subplot(122, title=\u0026#39;symbols dilated\u0026#39;).matshow(dilate[:15, :15]) plt.show() Creating this masks as before could be understood as a binary segmentation, as each element in our mask is either True or False. To extract the single digits, we\u0026rsquo;ll convert the binary digits segmentation into a instance segmentation, where each connected segment has an own index.\nmarkers, num_features = ndi.label(digits) plt.figure(figsize=(10, 6)) plt.subplot(121, title=\u0026#39;binary segmentation\u0026#39;).matshow( digits[:15, :15]) plt.subplot(122, title=\u0026#39;instance segmentation\u0026#39;).matshow( markers[:15, :15], cmap=\u0026#39;gnuplot\u0026#39;) plt.show() Now, for each instance, we check if the dilated binary mask overlaps with the instance and if yes, we extract the number.\nnumbers = [int(\u0026#39;\u0026#39;.join(arr[markers == i])) for i in range(1, num_features+1) if np.any((markers == i) \u0026amp; dilate)] Then, we just sum up:\nsum(numbers) 527364 Part 2 # Let\u0026rsquo;s first read the second task! The missing part wasn\u0026rsquo;t the only issue - one of the gears in the engine is wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.\nThis time, you need to find the gear ratio of every gear and add them all up so that the engineer can figure out which gear needs to be replaced.\nConsider the same engine schematic again:The missing part wasn\u0026rsquo;t the only issue - one of the gears in the engine is wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.\nThis time, you need to find the gear ratio of every gear and add them all up so that the engineer can figure out which gear needs to be replaced.\nConsider the same engine schematic again:\n467..114.. ...*...... ..35..633. ......#... 617*...... .....+.58. ..592..... ......755. ...$.*.... .664.598.. In this schematic, there are two gears. The first is in the top left; it has part numbers 467 and 35, so its gear ratio is 16345. The second gear is in the lower right; its gear ratio is 451490. (The * adjacent to 617 is not a gear because it is only adjacent to one part number.) Adding up all of the gear ratios produces 467835.\nWhat is the sum of all of the gear ratios in your engine schematic?\nWe\u0026rsquo;ll use the same method as before, but this time only extract * and create the instance segmentation before the dilation. Why? Because when we dilate first, we could merge two independent gears into one instance.\nprint(markers[2], arr[markers == 17]) [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 17 17 17 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 18 18 18 0 0 0 0 19 19 19 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 20 20 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 21 21 21 0 22 22 22 0 0 0 23 23 23 0 0 0 0 0 0 0 0] [\u0026#39;1\u0026#39; \u0026#39;3\u0026#39; \u0026#39;2\u0026#39;] gear = np.char.endswith(arr, \u0026#39;*\u0026#39;) gear_markers, gear_num = ndi.label(gear) plt.figure(figsize=(10, 6)) plt.subplot(131, title=\u0026#39;all symbols dilated\u0026#39;).matshow( symbols[:15, :15]) plt.subplot(132, title=\u0026#39;gears\u0026#39;).matshow( gear[:15, :15]) plt.subplot(133, title=\u0026#39;gears instances\u0026#39;).matshow( gear_markers[:15, :15], cmap=\u0026#39;gnuplot\u0026#39;) plt.show() Now, we step through each gear instance, create a mask for that gear, dilate it, and then step through all digits and check if they are in the gear mask. If we get two digits in the end, we multiply them to get the gear ratio and save the ratios to a list.\ngear_ratios = [] for i in range(1, gear_num+1): gear_binary = gear_markers == i gear_dil = ndi.binary_dilation(gear_binary, structure=struct) part_numbers = [int(\u0026#39;\u0026#39;.join(arr[markers == j])) for j in range(1, num_features+1) if np.any((markers == j) \u0026amp; gear_dil)] if len(part_numbers) == 2: gear_ratios.append(part_numbers[0] * part_numbers[1]) Now, we just sum the output again:\nsum(gear_ratios) 79026871 "},{"id":7,"href":"/posts/aoc2023-day02/","title":"Advent of code 2023 - Day 2: Cube Conundrum","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 2 - see https:adventofcode.com/2023/day/2\nPart 1 # Lets first read the task: As you walk, the Elf shows you a small bag and some cubes which are either red, green, or blue. Each time you play this game, he will hide a secret number of cubes of each color in the bag, and your goal is to figure out information about the number of cubes.\nTo get information, once a bag has been loaded with cubes, the Elf will reach into the bag, grab a handful of random cubes, show them to you, and then put them back in the bag. He\u0026rsquo;ll do this a few times per game.\nYou play several games and record the information from each game (your puzzle input). Each game is listed with its ID number (like the 11 in Game 11: ...) followed by a semicolon-separated list of subsets of cubes that were revealed from the bag (like 3 red, 5 green, 4 blue).\nFor example, the record of a few games might look like this:\nGame 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green In game 1, three sets of cubes are revealed from the bag (and then put back again). The first set is 3 blue cubes and 4 red cubes; the second set is 1 red cube, 2 green cubes, and 6 blue cubes; the third set is only 2 green cubes.\nThe Elf would first like to know which games would have been possible if the bag contained only 12 red cubes, 13 green cubes, and 14 blue cubes?\nIn the example above, games 1, 2, and 5 would have been possible if the bag had been loaded with that configuration. However, game 3 would have been impossible because at one point the Elf showed you 20 red cubes at once; similarly, game 4 would also have been impossible because the Elf showed you 15 blue cubes at once. If you add up the IDs of the games that would have been possible, you get 8.\nDetermine which games would have been possible if the bag had been loaded with only 12 red cubes, 13 green cubes, and 14 blue cubes. What is the sum of the IDs of those games?\nOkay, let\u0026rsquo;s load our python kernel in emacs-jupyter and get coding! First of all, let\u0026rsquo;s load the input and split the riddle code by colon : to extract the game id and the rest of the code by semicolon ; to get the number of sets played in each game.\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-02-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) txt[\u0026#39;id\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply( lambda x: int(x[0].strip(\u0026#39;Game \u0026#39;))) txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;:\u0026#39;).apply(lambda x: x[1]) # txt[\u0026#39;code\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].str.split(\u0026#39;;\u0026#39;) # txt[\u0026#39;nsets\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda x: len(x)) txt code id 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 100 rows × 2 columns\nNow, let\u0026rsquo;s extract the three colors in different columns with regex. We use the lookahead assertion ?= to find the respective colours and only exctract the digits \\d+ coming before. Then we just keep the max imum drawn number of cubes per color, since this is the only information that matters at the moment.\ntxt[\u0026#39;green\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.green)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt[\u0026#39;red\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.red)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt[\u0026#39;blue\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda code: re.findall(r\u0026#39;\\d+(?=.blue)\u0026#39;, code)).apply( lambda list: max([int(i) for i in list])) txt code id green red blue 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 2 10 10 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 15 10 3 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 4 6 16 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 16 20 18 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 10 6 14 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 4 2 1 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 5 4 8 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 2 5 2 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 2 9 11 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 13 15 11 100 rows × 5 columns\nLastly, we just filter the DataFrame to only include games where all drawn cubes were below or equal the number of cubes in the game and sum the result!\ntxt[\u0026#39;id\u0026#39;][(txt[\u0026#39;green\u0026#39;] \u0026lt; 14) \u0026amp; (txt[\u0026#39;red\u0026#39;] \u0026lt; 13) \u0026amp; (txt[\u0026#39;blue\u0026#39;] \u0026lt; 15)].sum() 3035 Part 2 # First, let\u0026rsquo;s get the instruction from the second part: As you continue your walk, the Elf poses a second question: in each game you played, what is the fewest number of cubes of each color that could have been in the bag to make the game possible?\nAgain consider the example games from earlier:\nGame 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green In game 1, the game could have been played with as few as 4 red, 2 green, and 6 blue cubes. If any color had even one fewer cube, the game would have been impossible. Game 2 could have been played with a minimum of 1 red, 3 green, and 4 blue cubes. Game 3 must have been played with at least 20 red, 13 green, and 6 blue cubes. Game 4 required at least 14 red, 3 green, and 15 blue cubes. Game 5 needed no fewer than 6 red, 3 green, and 2 blue cubes in the bag. The power of a set of cubes is equal to the numbers of red, green, and blue cubes multiplied together. The power of the minimum set of cubes in game 1 is 48. In games 2-5 it was 12, 1560, 630, and 36, respectively. Adding up these five powers produces the sum 2286.\nFor each game, find the minimum set of cubes that must have been present. What is the sum of the power of these sets?\nLuckily, this task is made trivial by the approach we have taken before. We just have to multiply the green, red and blue columns:\ntxt[\u0026#39;power\u0026#39;] = txt.loc[:, \u0026#39;green\u0026#39;] * txt.loc[:, \u0026#39;blue\u0026#39;] * txt.loc[:, \u0026#39;red\u0026#39;] txt code id green red blue power 0 1 green, 1 blue, 1 red; 1 green, 8 red, 7 blu\u0026hellip; 1 2 10 10 200 1 9 red, 7 green, 3 blue; 15 green, 2 blue, 5 r\u0026hellip; 2 15 10 3 450 2 3 red, 1 blue, 4 green; 6 red, 3 green, 2 blu\u0026hellip; 3 4 6 16 384 3 2 blue, 2 green, 19 red; 3 blue, 11 red, 16 g\u0026hellip; 4 16 20 18 5760 4 8 green, 1 red, 12 blue; 10 green, 6 red, 13 \u0026hellip; 5 10 6 14 840 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 95 2 red, 2 green, 1 blue; 1 red, 4 green; 1 green 96 4 2 1 8 96 4 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree\u0026hellip; 97 5 4 8 160 97 1 blue; 2 green, 1 red; 5 red, 2 green; 4 red\u0026hellip; 98 2 5 2 20 98 6 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree\u0026hellip; 99 2 9 11 198 99 1 blue, 13 green, 14 red; 11 green, 11 blue, \u0026hellip; 100 13 15 11 2145 100 rows × 6 columns\nAnd for this one, the sum is:\ntxt[\u0026#39;power\u0026#39;].sum() 66027 "},{"id":8,"href":"/posts/aoc2023-day01/","title":"Advent of code 2023 - Day 1: Trebuchet?!","section":"Posts","content":" Table of Contents Part 1 Part 2 This year I try to record my attempt at solving the Advent of Code 2023 riddles. This is Day 1 - see https:adventofcode.com/2023/day/1\nPart 1 # Our first task is the following: The newly-improved calibration document consists of lines of text; each line originally contained a specific calibration value that the Elves now need to recover. On each line, the calibration value can be found by combining the first digit and the last digit (in that order) to form a single two-digit number.\nFor example:\n1abc2 pqr3stu8vwx a1b2c3d4e5f treb7uchet In this example, the calibration values of these four lines are 12, 38, 15, and 77. Adding these together produces 142.\nConsider your entire calibration document. What is the sum of all of the calibration values?\nLets start jupyter in our shell to start coding!\nconda activate tf jupyter lab --no-browser --port=8888 First, load the test document\nimport pandas as pd import re txt = pd.read_table(\u0026#39;data/2023-12-01-1-aoc.txt\u0026#39;, names=[\u0026#39;code\u0026#39;]) txt code 0 jjfvnnlfivejj1 1 6fourfour 2 ninevbmltwo69 3 pcg91vqrfpxxzzzoneightzt 4 jpprthxgjfive3one1qckhrptpqdc \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 996 81s 997 2four3threesxxvlfqfive4 998 nine6eightsevenzx9twoxc 999 hmbfjdfnp989mfivefiverpzrjs 1000 rows × 1 columns\nSecond, extract the digits. I had to wrap my head around regex matching in python first, because I first tried pandas.extract (which only extracts the first match), then pandas.extractall (which extracts all matches but puts them into a multiindex which makes things more difficult in this case). So I settled for the re.findall version, which returns a list. To concatenate the elements in the list, we take use the join function.\ntxt[\u0026#39;digits\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply( lambda x: \u0026#39;\u0026#39;.join(re.findall(r\u0026#39;(\\d+)\u0026#39;, x))) txt code digits 0 jjfvnnlfivejj1 1 1 6fourfour 6 2 ninevbmltwo69 69 3 pcg91vqrfpxxzzzoneightzt 91 4 jpprthxgjfive3one1qckhrptpqdc 31 \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 996 81s 81 997 2four3threesxxvlfqfive4 234 998 nine6eightsevenzx9twoxc 69 999 hmbfjdfnp989mfivefiverpzrjs 989 1000 rows × 2 columns\nNext, combine the first and the last digit and convert the result from string to integer\ntxt[\u0026#39;calibration\u0026#39;] = txt.loc[:, \u0026#39;digits\u0026#39;].apply( lambda x: int(x[0] + x[-1])) txt code digits calibration 0 jjfvnnlfivejj1 1 11 1 6fourfour 6 66 2 ninevbmltwo69 69 69 3 pcg91vqrfpxxzzzoneightzt 91 91 4 jpprthxgjfive3one1qckhrptpqdc 31 31 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 996 81s 81 81 997 2four3threesxxvlfqfive4 234 24 998 nine6eightsevenzx9twoxc 69 69 999 hmbfjdfnp989mfivefiverpzrjs 989 99 1000 rows × 3 columns\nLastly, get the sum of our calibration numbers\ntxt.loc[:, \u0026#39;calibration\u0026#39;].sum() 56465 Part 2 # Now follows part two: Your calculation isn\u0026rsquo;t quite right. It looks like some of the digits are actually spelled out with letters: one, two, three, four, five, six, seven, eight, and nine also count as valid \u0026ldquo;digits\u0026rdquo;.\nEquipped with this new information, you now need to find the real first and last digit on each line. For example:\ntwo1nine eightwothree abcone2threexyz xtwone3four 4nineeightseven2 zoneight234 7pqrstsixteen In this example, the calibration values are 29, 83, 13, 24, 42, 14, and 76. Adding these together produces 281.\nWhat is the sum of all of the calibration values?\nOkay, let\u0026rsquo;s see if we can update the pattern matching. To deal with potential overlapping values like oneight which contains one as well as eight, I used the regex positive lookahead ?= as described here. Because this enables capturing overlapping values, I used \\d (one digit) instead of \\d+ (one or more digits), otherwise digits might double. Afterwards, just replace the spelled out digits with their numerical value.\n# for i, r in enumerate(txt.loc[:, \u0026#39;code\u0026#39;]): # matches = re.findall( # r\u0026#39;(?=(\\d|one|two|three|four|five|six|seven|eight|nine))\u0026#39;, r) # result = \u0026#39;\u0026#39;.join([match for match in matches]) # result = result.replace(\u0026#39;one\u0026#39;, \u0026#39;1\u0026#39;).replace(\u0026#39;two\u0026#39;, \u0026#39;2\u0026#39;).replace( # \u0026#39;three\u0026#39;, \u0026#39;3\u0026#39;).replace(\u0026#39;four\u0026#39;, \u0026#39;4\u0026#39;).replace(\u0026#39;five\u0026#39;, \u0026#39;5\u0026#39;).replace( # \u0026#39;six\u0026#39;, \u0026#39;6\u0026#39;).replace(\u0026#39;seven\u0026#39;, \u0026#39;7\u0026#39;).replace(\u0026#39;eight\u0026#39;, \u0026#39;8\u0026#39;).replace( # \u0026#39;nine\u0026#39;, \u0026#39;9\u0026#39;) # txt.loc[i, \u0026#39;digits2\u0026#39;] = result # txt # a very nice alternative suggested by Tomalak: digits = \u0026#39;\\d one two three four five six seven eight nine\u0026#39;.split() txt[\u0026#39;digits2\u0026#39;] = txt.loc[:, \u0026#39;code\u0026#39;].apply(lambda v: \u0026#39;\u0026#39;.join( str(digits.index(m)) if m in digits else m for m in re.findall(rf\u0026#39;(?=({\u0026#34;|\u0026#34;.join(digits)}))\u0026#39;, v) )) txt code digits calibration digits2 0 jjfvnnlfivejj1 1 11 51 1 6fourfour 6 66 644 2 ninevbmltwo69 69 69 9269 3 pcg91vqrfpxxzzzoneightzt 91 91 9118 4 jpprthxgjfive3one1qckhrptpqdc 31 31 5311 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 58375 996 81s 81 81 81 997 2four3threesxxvlfqfive4 234 24 243354 998 nine6eightsevenzx9twoxc 69 69 968792 999 hmbfjdfnp989mfivefiverpzrjs 989 99 98955 1000 rows × 4 columns\nNow, construct the calibration value as before\u0026hellip;\ntxt[\u0026#39;calibration2\u0026#39;] = txt.loc[:, \u0026#39;digits2\u0026#39;].apply(lambda x: int(x[0] + x[-1])) txt code digits calibration digits2 calibration2 0 jjfvnnlfivejj1 1 11 51 51 1 6fourfour 6 66 644 64 2 ninevbmltwo69 69 69 9269 99 3 pcg91vqrfpxxzzzoneightzt 91 91 9118 98 4 jpprthxgjfive3one1qckhrptpqdc 31 31 5311 51 \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; \u0026hellip; 995 583sevenhjxlqzjgbzxhkcl5 5835 55 58375 55 996 81s 81 81 81 81 997 2four3threesxxvlfqfive4 234 24 243354 24 998 nine6eightsevenzx9twoxc 69 69 968792 92 999 hmbfjdfnp989mfivefiverpzrjs 989 99 98955 95 1000 rows × 5 columns\n\u0026hellip; and get the correct sum!\ntxt.loc[:, \u0026#39;calibration2\u0026#39;].sum() 55902 "}] \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/en.search.min.2b6333591953211fc501497937cbcdc6fcf29ad8ccca8f8e9dfa3e18bbb5848a.js b/aseltmann.github.io-hugo/public/en.search.min.2b6333591953211fc501497937cbcdc6fcf29ad8ccca8f8e9dfa3e18bbb5848a.js deleted file mode 100644 index e32ccd9..0000000 --- a/aseltmann.github.io-hugo/public/en.search.min.2b6333591953211fc501497937cbcdc6fcf29ad8ccca8f8e9dfa3e18bbb5848a.js +++ /dev/null @@ -1 +0,0 @@ -"use strict";(function(){const o="/en.search-data.min.4f53cda18c2baa0c0354bb5f9a3ecbe5ed12ab4d8e11ba873c2f11161202b945.json",i=Object.assign({cache:!0},{includeScore:!0,useExtendedSearch:!0,fieldNormWeight:1.5,threshold:.2,ignoreLocation:!0,keys:[{name:"title",weight:.7},{name:"content",weight:.3}]}),e=document.querySelector("#book-search-input"),t=document.querySelector("#book-search-results");if(!e)return;e.addEventListener("focus",n),e.addEventListener("keyup",s),document.addEventListener("keypress",a);function a(t){if(t.target.value!==void 0)return;if(e===document.activeElement)return;const n=String.fromCharCode(t.charCode);if(!r(n))return;e.focus(),t.preventDefault()}function r(t){const n=e.getAttribute("data-hotkeys")||"";return n.indexOf(t)>=0}function n(){e.removeEventListener("focus",n),e.required=!0,fetch(o).then(e=>e.json()).then(e=>{window.bookSearchIndex=new Fuse(e,i)}).then(()=>e.required=!1).then(s)}function s(){for(;t.firstChild;)t.removeChild(t.firstChild);if(!e.value)return;const n=window.bookSearchIndex.search(e.value).slice(0,10);n.forEach(function(e){const n=c("
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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/index.xml b/aseltmann.github.io-hugo/public/index.xml deleted file mode 100644 index 83cc691..0000000 --- a/aseltmann.github.io-hugo/public/index.xml +++ /dev/null @@ -1,253 +0,0 @@ -Homepage on Blog & Noteshttp://localhost:1313/Recent content in Homepage on Blog & NotesHugoenSun, 22 Dec 2024 01:46:31 +0100Poetryhttp://localhost:1313/docs/notes/poetry/Sun, 22 Dec 2024 01:44:35 +0100http://localhost:1313/docs/notes/poetry/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#l%c7%8eoz%c7%90">Lǎozǐ</a></li> -</ul> -</div> -<!--endtoc--> -<h2 id="lǎozǐ"> - Lǎozǐ - <a class="anchor" href="#l%c7%8eoz%c7%90">#</a> -</h2> -<p>I am partial to the - <a href="https://terebess.hu/english/tao/DerekLin.html#Kap20">translation of Derek Lin</a></p> -<details> -<summary>Dàodéjīng - Chapter 11 (The use of what has no substantive existence)</summary> -<div class="details"> -<div class="verse"> -<p>[&hellip;]<br /> -<br /> -Mix clay<br /> -to create a container<br /> -In its emptiness, there is<br /> -the function of a container<br /> -<br /> -[&hellip;]<br /> -Therefore,<br /> -that which exists is used to create benefit<br /> -That which is empty is used to create functionality<br /></p>Advent of code 2024 - Day 2: Red-Nosed Reportshttp://localhost:1313/posts/aoc2024-day02/Tue, 10 Dec 2024 01:43:00 +0100http://localhost:1313/posts/aoc2024-day02/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1-safe-reports">Part 1: Safe reports</a></li> -<li> - <a href="http://localhost:1313/#part-2-problem-dampener">Part 2: Problem Dampener</a></li> -</ul> -</div> -<!--endtoc--> -<p>I try to solve this year&rsquo;s <strong>Advent of Code 2024</strong> riddles in R. This is Day 2 - see - <a href="https:adventofcode.com/2024/day/2">https:adventofcode.com/2024/day/2</a></p> -<h2 id="part-1-safe-reports"> - Part 1: Safe reports - <a class="anchor" href="#part-1-safe-reports">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The unusual data (your puzzle input) consists of many <strong>reports</strong>, -one report per line. Each report is a list of numbers called <strong>levels</strong> that are -separated by spaces. For example:</p>Advent of code 2024 - Day 1: Historian Hysteriahttp://localhost:1313/posts/aoc2024-day01/Sat, 07 Dec 2024 00:22:00 +0100http://localhost:1313/posts/aoc2024-day01/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1-locations">Part 1: Locations</a></li> -<li> - <a href="http://localhost:1313/#part-2-similarities">Part 2: Similarities</a></li> -</ul> -</div> -<!--endtoc--> -<p>I try to solve this year&rsquo;s <strong>Advent of Code 2024</strong> riddles in R. This is Day 1 - see - <a href="https:adventofcode.com/2024/day/1">https:adventofcode.com/2024/day/1</a></p> -<h2 id="part-1-locations"> - Part 1: Locations - <a class="anchor" href="#part-1-locations">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>Throughout the Chief&rsquo;s office, the historically significant locations are listed -not by name but by a unique number called the <strong>location ID</strong>. To make sure they -don&rsquo;t miss anything, The Historians split into two groups, each searching the -office and trying to create their own complete list of location IDs.</p>Advent of code 2023 - Day 7: Camel Cardshttp://localhost:1313/posts/aoc2023-day07/Thu, 04 Jan 2024 16:32:00 +0100http://localhost:1313/posts/aoc2023-day07/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 7 - see - <a href="https:adventofcode.com/2023/day/7">https:adventofcode.com/2023/day/7</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>In Camel Cards, you get a list of <strong>hands</strong>, and your goal is to order them based -on the <strong>strength</strong> of each hand. A hand consists of <strong>five cards</strong> labeled one of -<code>A</code>, <code>K</code>, <code>Q</code>, <code>J</code>, <code>T</code>, <code>9</code>, <code>8</code>, <code>7</code>, <code>6</code>, <code>5</code>, <code>4</code>, <code>3</code>, or <code>2</code>. The relative -strength of each card follows this order, where <code>A</code> is the highest and <code>2</code> is -the lowest.</p>Advent of code 2023 - Day 6: Wait For Ithttp://localhost:1313/posts/aoc2023-day06/Thu, 04 Jan 2024 00:44:00 +0100http://localhost:1313/posts/aoc2023-day06/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 6 - see - <a href="https:adventofcode.com/2023/day/6">https:adventofcode.com/2023/day/6</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>You manage to sign up as a competitor in the boat races just in time. The -organizer explains that it&rsquo;s not really a traditional race - instead, you will -get a fixed amount of time during which your boat has to travel as far as it -can, and you win if your boat goes the farthest.</p>Advent of code 2023 - Day 5: If You Give A Seed A Fertilizerhttp://localhost:1313/posts/aoc2023-day05/Sat, 30 Dec 2023 01:10:00 +0100http://localhost:1313/posts/aoc2023-day05/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 5 - see - <a href="https:adventofcode.com/2023/day/5">https:adventofcode.com/2023/day/5</a></p> -<p>Update <span class="timestamp-wrapper"><span class="timestamp">[2023-12-31 So]</span></span>:</p> -<ul> -<li>substitute <code>get_generator()</code> (own implementation) with <code>range()</code> (Python inbuilt)</li> -<li>improve grid search so that it goes through all location ranges, still -starting with the lowest range</li> -</ul> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The almanac (your puzzle input) lists all of the seeds that need to be planted. -It also lists what type of soil to use with each kind of seed, what type of -fertilizer to use with each kind of soil, what type of water to use with each -kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is -identified with a number, but numbers are reused by each category - that is, -soil <code>123</code> and fertilizer <code>123</code> aren&rsquo;t necessarily related to each other.</p>Advent of code 2023 - Day 4: Scratchcardshttp://localhost:1313/posts/aoc2023-day04/Tue, 19 Dec 2023 00:25:00 +0100http://localhost:1313/posts/aoc2023-day04/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 4 - see - <a href="https:adventofcode.com/2023/day/4">https:adventofcode.com/2023/day/4</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The Elf leads you over to the pile of colorful cards. There, you discover dozens -of scratchcards, all with their opaque covering already scratched off. Picking -one up, it looks like each card has two lists of numbers separated by a vertical -bar (<code>|</code>): a list of <strong>winning numbers</strong> and then a list of <strong>numbers you have</strong>. -You organize the information into a table (your puzzle input).</p>Advent of code 2023 - Day 3: Gear Ratioshttp://localhost:1313/posts/aoc2023-day03/Mon, 18 Dec 2023 01:01:00 +0100http://localhost:1313/posts/aoc2023-day03/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 3 - see - <a href="https:adventofcode.com/2023/day/3">https:adventofcode.com/2023/day/3</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The engine schematic (your puzzle input) consists of a visual representation of -the engine. There are lots of numbers and symbols you don&rsquo;t really understand, -but apparently <strong>any number adjacent to a symbol</strong>, even diagonally, is a &ldquo;part -number&rdquo; and should be included in your sum. (Periods (<code>.</code>) do not count as a -symbol.)</p>Advent of code 2023 - Day 2: Cube Conundrumhttp://localhost:1313/posts/aoc2023-day02/Mon, 11 Dec 2023 23:38:00 +0100http://localhost:1313/posts/aoc2023-day02/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 2 - see - <a href="https:adventofcode.com/2023/day/2">https:adventofcode.com/2023/day/2</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>As you walk, the Elf shows you a small bag and some cubes which are either red, -green, or blue. Each time you play this game, he will hide a secret number of -cubes of each color in the bag, and your goal is to figure out information about -the number of cubes.</p>Advent of code 2023 - Day 1: Trebuchet?!http://localhost:1313/posts/aoc2023-day01/Sun, 10 Dec 2023 03:05:00 +0100http://localhost:1313/posts/aoc2023-day01/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. 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cScale7||Oe(this.primaryColor,{h:150}),this.cScale8=this.cScale8||Oe(this.primaryColor,{h:210,l:150}),this.cScale9=this.cScale9||Oe(this.primaryColor,{h:270}),this.cScale10=this.cScale10||Oe(this.primaryColor,{h:300}),this.cScale11=this.cScale11||Oe(this.primaryColor,{h:330}),this.darkMode)for(let r=0;r{this[n]=e[n]}),this.updateColors(),r.forEach(n=>{this[n]=e[n]})}},Nz=o(t=>{let e=new QC;return e.calculate(t),e},"getThemeVariables")});var ZC,Iz,Oz=R(()=>{"use strict";al();up();ZC=class{static{o(this,"Theme")}constructor(){this.background="#333",this.primaryColor="#1f2020",this.secondaryColor=Et(this.primaryColor,16),this.tertiaryColor=Oe(this.primaryColor,{h:-160}),this.primaryBorderColor=ot(this.background),this.secondaryBorderColor=yi(this.secondaryColor,this.darkMode),this.tertiaryBorderColor=yi(this.tertiaryColor,this.darkMode),this.primaryTextColor=ot(this.primaryColor),this.secondaryTextColor=ot(this.secondaryColor),this.tertiaryTextColor=ot(this.tertiaryColor),this.lineColor=ot(this.background),this.textColor=ot(this.background),this.mainBkg="#1f2020",this.secondBkg="calculated",this.mainContrastColor="lightgrey",this.darkTextColor=Et(ot("#323D47"),10),this.lineColor="calculated",this.border1="#ccc",this.border2=Ws(255,255,255,.25),this.arrowheadColor="calculated",this.fontFamily='"trebuchet ms", verdana, arial, sans-serif',this.fontSize="16px",this.labelBackground="#181818",this.textColor="#ccc",this.THEME_COLOR_LIMIT=12,this.nodeBkg="calculated",this.nodeBorder="calculated",this.clusterBkg="calculated",this.clusterBorder="calculated",this.defaultLinkColor="calculated",this.titleColor="#F9FFFE",this.edgeLabelBackground="calculated",this.actorBorder="calculated",this.actorBkg="calculated",this.actorTextColor="calculated",this.actorLineColor="calculated",this.signalColor="calculated",this.signalTextColor="calculated",this.labelBoxBkgColor="calculated",this.labelBoxBorderColor="calculated",this.labelTextColor="calculated",this.loopTextColor="calculated",this.noteBorderColor="calculated",this.noteBkgColor="#fff5ad",this.noteTextColor="calculated",this.activationBorderColor="calculated",this.activationBkgColor="calculated",this.sequenceNumberColor="black",this.sectionBkgColor=Dt("#EAE8D9",30),this.altSectionBkgColor="calculated",this.sectionBkgColor2="#EAE8D9",this.excludeBkgColor=Dt(this.sectionBkgColor,10),this.taskBorderColor=Ws(255,255,255,70),this.taskBkgColor="calculated",this.taskTextColor="calculated",this.taskTextLightColor="calculated",this.taskTextOutsideColor="calculated",this.taskTextClickableColor="#003163",this.activeTaskBorderColor=Ws(255,255,255,50),this.activeTaskBkgColor="#81B1DB",this.gridColor="calculated",this.doneTaskBkgColor="calculated",this.doneTaskBorderColor="grey",this.critBorderColor="#E83737",this.critBkgColor="#E83737",this.taskTextDarkColor="calculated",this.todayLineColor="#DB5757",this.personBorder=this.primaryBorderColor,this.personBkg=this.mainBkg,this.archEdgeColor="calculated",this.archEdgeArrowColor="calculated",this.archEdgeWidth="3",this.archGroupBorderColor=this.primaryBorderColor,this.archGroupBorderWidth="2px",this.labelColor="calculated",this.errorBkgColor="#a44141",this.errorTextColor="#ddd"}updateColors(){this.secondBkg=Et(this.mainBkg,16),this.lineColor=this.mainContrastColor,this.arrowheadColor=this.mainContrastColor,this.nodeBkg=this.mainBkg,this.nodeBorder=this.border1,this.clusterBkg=this.secondBkg,this.clusterBorder=this.border2,this.defaultLinkColor=this.lineColor,this.edgeLabelBackground=Et(this.labelBackground,25),this.actorBorder=this.border1,this.actorBkg=this.mainBkg,this.actorTextColor=this.mainContrastColor,this.actorLineColor=this.actorBorder,this.signalColor=this.mainContrastColor,this.signalTextColor=this.mainContrastColor,this.labelBoxBkgColor=this.actorBkg,this.labelBoxBorderColor=this.actorBorder,this.labelTextColor=this.mainContrastColor,this.loopTextColor=this.mainContrastColor,this.noteBorderColor=this.secondaryBorderColor,this.noteBkgColor=this.secondBkg,this.noteTextColor=this.secondaryTextColor,this.activationBorderColor=this.border1,this.activationBkgColor=this.secondBkg,this.altSectionBkgColor=this.background,this.taskBkgColor=Et(this.mainBkg,23),this.taskTextColor=this.darkTextColor,this.taskTextLightColor=this.mainContrastColor,this.taskTextOutsideColor=this.taskTextLightColor,this.gridColor=this.mainContrastColor,this.doneTaskBkgColor=this.mainContrastColor,this.taskTextDarkColor=this.darkTextColor,this.archEdgeColor=this.lineColor,this.archEdgeArrowColor=this.lineColor,this.transitionColor=this.transitionColor||this.lineColor,this.transitionLabelColor=this.transitionLabelColor||this.textColor,this.stateLabelColor=this.stateLabelColor||this.stateBkg||this.primaryTextColor,this.stateBkg=this.stateBkg||this.mainBkg,this.labelBackgroundColor=this.labelBackgroundColor||this.stateBkg,this.compositeBackground=this.compositeBackground||this.background||this.tertiaryColor,this.altBackground=this.altBackground||"#555",this.compositeTitleBackground=this.compositeTitleBackground||this.mainBkg,this.compositeBorder=this.compositeBorder||this.nodeBorder,this.innerEndBackground=this.primaryBorderColor,this.specialStateColor="#f4f4f4",this.errorBkgColor=this.errorBkgColor||this.tertiaryColor,this.errorTextColor=this.errorTextColor||this.tertiaryTextColor,this.fillType0=this.primaryColor,this.fillType1=this.secondaryColor,this.fillType2=Oe(this.primaryColor,{h:64}),this.fillType3=Oe(this.secondaryColor,{h:64}),this.fillType4=Oe(this.primaryColor,{h:-64}),this.fillType5=Oe(this.secondaryColor,{h:-64}),this.fillType6=Oe(this.primaryColor,{h:128}),this.fillType7=Oe(this.secondaryColor,{h:128}),this.cScale1=this.cScale1||"#0b0000",this.cScale2=this.cScale2||"#4d1037",this.cScale3=this.cScale3||"#3f5258",this.cScale4=this.cScale4||"#4f2f1b",this.cScale5=this.cScale5||"#6e0a0a",this.cScale6=this.cScale6||"#3b0048",this.cScale7=this.cScale7||"#995a01",this.cScale8=this.cScale8||"#154706",this.cScale9=this.cScale9||"#161722",this.cScale10=this.cScale10||"#00296f",this.cScale11=this.cScale11||"#01629c",this.cScale12=this.cScale12||"#010029",this.cScale0=this.cScale0||this.primaryColor,this.cScale1=this.cScale1||this.secondaryColor,this.cScale2=this.cScale2||this.tertiaryColor,this.cScale3=this.cScale3||Oe(this.primaryColor,{h:30}),this.cScale4=this.cScale4||Oe(this.primaryColor,{h:60}),this.cScale5=this.cScale5||Oe(this.primaryColor,{h:90}),this.cScale6=this.cScale6||Oe(this.primaryColor,{h:120}),this.cScale7=this.cScale7||Oe(this.primaryColor,{h:150}),this.cScale8=this.cScale8||Oe(this.primaryColor,{h:210}),this.cScale9=this.cScale9||Oe(this.primaryColor,{h:270}),this.cScale10=this.cScale10||Oe(this.primaryColor,{h:300}),this.cScale11=this.cScale11||Oe(this.primaryColor,{h:330});for(let e=0;e{this[n]=e[n]}),this.updateColors(),r.forEach(n=>{this[n]=e[n]})}},Iz=o(t=>{let e=new ZC;return e.calculate(t),e},"getThemeVariables")});var JC,hp,jb=R(()=>{"use strict";al();up();j1();JC=class{static{o(this,"Theme")}constructor(){this.background="#f4f4f4",this.primaryColor="#ECECFF",this.secondaryColor=Oe(this.primaryColor,{h:120}),this.secondaryColor="#ffffde",this.tertiaryColor=Oe(this.primaryColor,{h:-160}),this.primaryBorderColor=yi(this.primaryColor,this.darkMode),this.secondaryBorderColor=yi(this.secondaryColor,this.darkMode),this.tertiaryBorderColor=yi(this.tertiaryColor,this.darkMode),this.primaryTextColor=ot(this.primaryColor),this.secondaryTextColor=ot(this.secondaryColor),this.tertiaryTextColor=ot(this.tertiaryColor),this.lineColor=ot(this.background),this.textColor=ot(this.background),this.background="white",this.mainBkg="#ECECFF",this.secondBkg="#ffffde",this.lineColor="#333333",this.border1="#9370DB",this.border2="#aaaa33",this.arrowheadColor="#333333",this.fontFamily='"trebuchet ms", verdana, arial, sans-serif',this.fontSize="16px",this.labelBackground="rgba(232,232,232, 0.8)",this.textColor="#333",this.THEME_COLOR_LIMIT=12,this.nodeBkg="calculated",this.nodeBorder="calculated",this.clusterBkg="calculated",this.clusterBorder="calculated",this.defaultLinkColor="calculated",this.titleColor="calculated",this.edgeLabelBackground="calculated",this.actorBorder="calculated",this.actorBkg="calculated",this.actorTextColor="black",this.actorLineColor="calculated",this.signalColor="calculated",this.signalTextColor="calculated",this.labelBoxBkgColor="calculated",this.labelBoxBorderColor="calculated",this.labelTextColor="calculated",this.loopTextColor="calculated",this.noteBorderColor="calculated",this.noteBkgColor="#fff5ad",this.noteTextColor="calculated",this.activationBorderColor="#666",this.activationBkgColor="#f4f4f4",this.sequenceNumberColor="white",this.sectionBkgColor="calculated",this.altSectionBkgColor="calculated",this.sectionBkgColor2="calculated",this.excludeBkgColor="#eeeeee",this.taskBorderColor="calculated",this.taskBkgColor="calculated",this.taskTextLightColor="calculated",this.taskTextColor=this.taskTextLightColor,this.taskTextDarkColor="calculated",this.taskTextOutsideColor=this.taskTextDarkColor,this.taskTextClickableColor="calculated",this.activeTaskBorderColor="calculated",this.activeTaskBkgColor="calculated",this.gridColor="calculated",this.doneTaskBkgColor="calculated",this.doneTaskBorderColor="calculated",this.critBorderColor="calculated",this.critBkgColor="calculated",this.todayLineColor="calculated",this.sectionBkgColor=Ws(102,102,255,.49),this.altSectionBkgColor="white",this.sectionBkgColor2="#fff400",this.taskBorderColor="#534fbc",this.taskBkgColor="#8a90dd",this.taskTextLightColor="white",this.taskTextColor="calculated",this.taskTextDarkColor="black",this.taskTextOutsideColor="calculated",this.taskTextClickableColor="#003163",this.activeTaskBorderColor="#534fbc",this.activeTaskBkgColor="#bfc7ff",this.gridColor="lightgrey",this.doneTaskBkgColor="lightgrey",this.doneTaskBorderColor="grey",this.critBorderColor="#ff8888",this.critBkgColor="red",this.todayLineColor="red",this.personBorder=this.primaryBorderColor,this.personBkg=this.mainBkg,this.archEdgeColor="calculated",this.archEdgeArrowColor="calculated",this.archEdgeWidth="3",this.archGroupBorderColor=this.primaryBorderColor,this.archGroupBorderWidth="2px",this.labelColor="black",this.errorBkgColor="#552222",this.errorTextColor="#552222",this.updateColors()}updateColors(){this.cScale0=this.cScale0||this.primaryColor,this.cScale1=this.cScale1||this.secondaryColor,this.cScale2=this.cScale2||this.tertiaryColor,this.cScale3=this.cScale3||Oe(this.primaryColor,{h:30}),this.cScale4=this.cScale4||Oe(this.primaryColor,{h:60}),this.cScale5=this.cScale5||Oe(this.primaryColor,{h:90}),this.cScale6=this.cScale6||Oe(this.primaryColor,{h:120}),this.cScale7=this.cScale7||Oe(this.primaryColor,{h:150}),this.cScale8=this.cScale8||Oe(this.primaryColor,{h:210}),this.cScale9=this.cScale9||Oe(this.primaryColor,{h:270}),this.cScale10=this.cScale10||Oe(this.primaryColor,{h:300}),this.cScale11=this.cScale11||Oe(this.primaryColor,{h:330}),this.cScalePeer1=this.cScalePeer1||Dt(this.secondaryColor,45),this.cScalePeer2=this.cScalePeer2||Dt(this.tertiaryColor,40);for(let e=0;e{this[n]=e[n]}),this.updateColors(),r.forEach(n=>{this[n]=e[n]})}},hp=o(t=>{let e=new JC;return e.calculate(t),e},"getThemeVariables")});var e7,Pz,Bz=R(()=>{"use strict";al();j1();up();e7=class{static{o(this,"Theme")}constructor(){this.background="#f4f4f4",this.primaryColor="#cde498",this.secondaryColor="#cdffb2",this.background="white",this.mainBkg="#cde498",this.secondBkg="#cdffb2",this.lineColor="green",this.border1="#13540c",this.border2="#6eaa49",this.arrowheadColor="green",this.fontFamily='"trebuchet ms", verdana, arial, sans-serif',this.fontSize="16px",this.tertiaryColor=Et("#cde498",10),this.primaryBorderColor=yi(this.primaryColor,this.darkMode),this.secondaryBorderColor=yi(this.secondaryColor,this.darkMode),this.tertiaryBorderColor=yi(this.tertiaryColor,this.darkMode),this.primaryTextColor=ot(this.primaryColor),this.secondaryTextColor=ot(this.secondaryColor),this.tertiaryTextColor=ot(this.primaryColor),this.lineColor=ot(this.background),this.textColor=ot(this.background),this.THEME_COLOR_LIMIT=12,this.nodeBkg="calculated",this.nodeBorder="calculated",this.clusterBkg="calculated",this.clusterBorder="calculated",this.defaultLinkColor="calculated",this.titleColor="#333",this.edgeLabelBackground="#e8e8e8",this.actorBorder="calculated",this.actorBkg="calculated",this.actorTextColor="black",this.actorLineColor="calculated",this.signalColor="#333",this.signalTextColor="#333",this.labelBoxBkgColor="calculated",this.labelBoxBorderColor="#326932",this.labelTextColor="calculated",this.loopTextColor="calculated",this.noteBorderColor="calculated",this.noteBkgColor="#fff5ad",this.noteTextColor="calculated",this.activationBorderColor="#666",this.activationBkgColor="#f4f4f4",this.sequenceNumberColor="white",this.sectionBkgColor="#6eaa49",this.altSectionBkgColor="white",this.sectionBkgColor2="#6eaa49",this.excludeBkgColor="#eeeeee",this.taskBorderColor="calculated",this.taskBkgColor="#487e3a",this.taskTextLightColor="white",this.taskTextColor="calculated",this.taskTextDarkColor="black",this.taskTextOutsideColor="calculated",this.taskTextClickableColor="#003163",this.activeTaskBorderColor="calculated",this.activeTaskBkgColor="calculated",this.gridColor="lightgrey",this.doneTaskBkgColor="lightgrey",this.doneTaskBorderColor="grey",this.critBorderColor="#ff8888",this.critBkgColor="red",this.todayLineColor="red",this.personBorder=this.primaryBorderColor,this.personBkg=this.mainBkg,this.archEdgeColor="calculated",this.archEdgeArrowColor="calculated",this.archEdgeWidth="3",this.archGroupBorderColor=this.primaryBorderColor,this.archGroupBorderWidth="2px",this.labelColor="black",this.errorBkgColor="#552222",this.errorTextColor="#552222"}updateColors(){this.actorBorder=Dt(this.mainBkg,20),this.actorBkg=this.mainBkg,this.labelBoxBkgColor=this.actorBkg,this.labelTextColor=this.actorTextColor,this.loopTextColor=this.actorTextColor,this.noteBorderColor=this.border2,this.noteTextColor=this.actorTextColor,this.actorLineColor=this.actorBorder,this.cScale0=this.cScale0||this.primaryColor,this.cScale1=this.cScale1||this.secondaryColor,this.cScale2=this.cScale2||this.tertiaryColor,this.cScale3=this.cScale3||Oe(this.primaryColor,{h:30}),this.cScale4=this.cScale4||Oe(this.primaryColor,{h:60}),this.cScale5=this.cScale5||Oe(this.primaryColor,{h:90}),this.cScale6=this.cScale6||Oe(this.primaryColor,{h:120}),this.cScale7=this.cScale7||Oe(this.primaryColor,{h:150}),this.cScale8=this.cScale8||Oe(this.primaryColor,{h:210}),this.cScale9=this.cScale9||Oe(this.primaryColor,{h:270}),this.cScale10=this.cScale10||Oe(this.primaryColor,{h:300}),this.cScale11=this.cScale11||Oe(this.primaryColor,{h:330}),this.cScalePeer1=this.cScalePeer1||Dt(this.secondaryColor,45),this.cScalePeer2=this.cScalePeer2||Dt(this.tertiaryColor,40);for(let e=0;e{this[n]=e[n]}),this.updateColors(),r.forEach(n=>{this[n]=e[n]})}},Pz=o(t=>{let e=new e7;return e.calculate(t),e},"getThemeVariables")});var t7,Fz,zz=R(()=>{"use strict";al();up();j1();t7=class{static{o(this,"Theme")}constructor(){this.primaryColor="#eee",this.contrast="#707070",this.secondaryColor=Et(this.contrast,55),this.background="#ffffff",this.tertiaryColor=Oe(this.primaryColor,{h:-160}),this.primaryBorderColor=yi(this.primaryColor,this.darkMode),this.secondaryBorderColor=yi(this.secondaryColor,this.darkMode),this.tertiaryBorderColor=yi(this.tertiaryColor,this.darkMode),this.primaryTextColor=ot(this.primaryColor),this.secondaryTextColor=ot(this.secondaryColor),this.tertiaryTextColor=ot(this.tertiaryColor),this.lineColor=ot(this.background),this.textColor=ot(this.background),this.mainBkg="#eee",this.secondBkg="calculated",this.lineColor="#666",this.border1="#999",this.border2="calculated",this.note="#ffa",this.text="#333",this.critical="#d42",this.done="#bbb",this.arrowheadColor="#333333",this.fontFamily='"trebuchet ms", verdana, arial, sans-serif',this.fontSize="16px",this.THEME_COLOR_LIMIT=12,this.nodeBkg="calculated",this.nodeBorder="calculated",this.clusterBkg="calculated",this.clusterBorder="calculated",this.defaultLinkColor="calculated",this.titleColor="calculated",this.edgeLabelBackground="white",this.actorBorder="calculated",this.actorBkg="calculated",this.actorTextColor="calculated",this.actorLineColor=this.actorBorder,this.signalColor="calculated",this.signalTextColor="calculated",this.labelBoxBkgColor="calculated",this.labelBoxBorderColor="calculated",this.labelTextColor="calculated",this.loopTextColor="calculated",this.noteBorderColor="calculated",this.noteBkgColor="calculated",this.noteTextColor="calculated",this.activationBorderColor="#666",this.activationBkgColor="#f4f4f4",this.sequenceNumberColor="white",this.sectionBkgColor="calculated",this.altSectionBkgColor="white",this.sectionBkgColor2="calculated",this.excludeBkgColor="#eeeeee",this.taskBorderColor="calculated",this.taskBkgColor="calculated",this.taskTextLightColor="white",this.taskTextColor="calculated",this.taskTextDarkColor="calculated",this.taskTextOutsideColor="calculated",this.taskTextClickableColor="#003163",this.activeTaskBorderColor="calculated",this.activeTaskBkgColor="calculated",this.gridColor="calculated",this.doneTaskBkgColor="calculated",this.doneTaskBorderColor="calculated",this.critBkgColor="calculated",this.critBorderColor="calculated",this.todayLineColor="calculated",this.personBorder=this.primaryBorderColor,this.personBkg=this.mainBkg,this.archEdgeColor="calculated",this.archEdgeArrowColor="calculated",this.archEdgeWidth="3",this.archGroupBorderColor=this.primaryBorderColor,this.archGroupBorderWidth="2px",this.labelColor="black",this.errorBkgColor="#552222",this.errorTextColor="#552222"}updateColors(){this.secondBkg=Et(this.contrast,55),this.border2=this.contrast,this.actorBorder=Et(this.border1,23),this.actorBkg=this.mainBkg,this.actorTextColor=this.text,this.actorLineColor=this.actorBorder,this.signalColor=this.text,this.signalTextColor=this.text,this.labelBoxBkgColor=this.actorBkg,this.labelBoxBorderColor=this.actorBorder,this.labelTextColor=this.text,this.loopTextColor=this.text,this.noteBorderColor="#999",this.noteBkgColor="#666",this.noteTextColor="#fff",this.cScale0=this.cScale0||"#555",this.cScale1=this.cScale1||"#F4F4F4",this.cScale2=this.cScale2||"#555",this.cScale3=this.cScale3||"#BBB",this.cScale4=this.cScale4||"#777",this.cScale5=this.cScale5||"#999",this.cScale6=this.cScale6||"#DDD",this.cScale7=this.cScale7||"#FFF",this.cScale8=this.cScale8||"#DDD",this.cScale9=this.cScale9||"#BBB",this.cScale10=this.cScale10||"#999",this.cScale11=this.cScale11||"#777";for(let 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r=="function"||!r.unexpandable:fh.hasOwnProperty(e)&&!fh[e].primitive}},SG=/^[₊₋₌₍₎₀₁₂₃₄₅₆₇₈₉ₐₑₕᵢⱼₖₗₘₙₒₚᵣₛₜᵤᵥₓᵦᵧᵨᵩᵪ]/,c4=Object.freeze({"\u208A":"+","\u208B":"-","\u208C":"=","\u208D":"(","\u208E":")","\u2080":"0","\u2081":"1","\u2082":"2","\u2083":"3","\u2084":"4","\u2085":"5","\u2086":"6","\u2087":"7","\u2088":"8","\u2089":"9","\u2090":"a","\u2091":"e","\u2095":"h","\u1D62":"i","\u2C7C":"j","\u2096":"k","\u2097":"l","\u2098":"m","\u2099":"n","\u2092":"o","\u209A":"p","\u1D63":"r","\u209B":"s","\u209C":"t","\u1D64":"u","\u1D65":"v","\u2093":"x","\u1D66":"\u03B2","\u1D67":"\u03B3","\u1D68":"\u03C1","\u1D69":"\u03D5","\u1D6A":"\u03C7","\u207A":"+","\u207B":"-","\u207C":"=","\u207D":"(","\u207E":")","\u2070":"0","\xB9":"1","\xB2":"2","\xB3":"3","\u2074":"4","\u2075":"5","\u2076":"6","\u2077":"7","\u2078":"8","\u2079":"9","\u1D2C":"A","\u1D2E":"B","\u1D30":"D","\u1D31":"E","\u1D33":"G","\u1D34":"H","\u1D35":"I","\u1D36":"J","\u1D37":"K","\u1D38":"L","\u1D39":"M","\u1D3A":"N","\u1D3C":"O","\u1D3E":"P","\u1D3F":"R","\u1D40":"T","\u1D41":"U","\u2C7D":"V","\u1D42":"W","\u1D43":"a","\u1D47":"b","\u1D9C":"c","\u1D48":"d","\u1D49":"e","\u1DA0":"f","\u1D4D":"g",\u02B0:"h","\u2071":"i",\u02B2:"j","\u1D4F":"k",\u02E1:"l","\u1D50":"m",\u207F:"n","\u1D52":"o","\u1D56":"p",\u02B3:"r",\u02E2:"s","\u1D57":"t","\u1D58":"u","\u1D5B":"v",\u02B7:"w",\u02E3:"x",\u02B8:"y","\u1DBB":"z","\u1D5D":"\u03B2","\u1D5E":"\u03B3","\u1D5F":"\u03B4","\u1D60":"\u03D5","\u1D61":"\u03C7","\u1DBF":"\u03B8"}),x7={"\u0301":{text:"\\'",math:"\\acute"},"\u0300":{text:"\\`",math:"\\grave"},"\u0308":{text:'\\"',math:"\\ddot"},"\u0303":{text:"\\~",math:"\\tilde"},"\u0304":{text:"\\=",math:"\\bar"},"\u0306":{text:"\\u",math:"\\breve"},"\u030C":{text:"\\v",math:"\\check"},"\u0302":{text:"\\^",math:"\\hat"},"\u0307":{text:"\\.",math:"\\dot"},"\u030A":{text:"\\r",math:"\\mathring"},"\u030B":{text:"\\H"},"\u0327":{text:"\\c"}},AG={\u00E1:"a\u0301",\u00E0:"a\u0300",\u00E4:"a\u0308",\u01DF:"a\u0308\u0304",\u00E3:"a\u0303",\u0101:"a\u0304",\u0103:"a\u0306",\u1EAF:"a\u0306\u0301",\u1EB1:"a\u0306\u0300",\u1EB5:"a\u0306\u0303",\u01CE:"a\u030C",\u00E2:"a\u0302",\u1EA5:"a\u0302\u0301",\u1EA7:"a\u0302\u0300",\u1EAB:"a\u0302\u0303",\u0227:"a\u0307",\u01E1:"a\u0307\u0304",\u00E5:"a\u030A",\u01FB:"a\u030A\u0301",\u1E03:"b\u0307",\u0107:"c\u0301",\u1E09:"c\u0327\u0301",\u010D:"c\u030C",\u0109:"c\u0302",\u010B:"c\u0307",\u00E7:"c\u0327",\u010F:"d\u030C",\u1E0B:"d\u0307",\u1E11:"d\u0327",\u00E9:"e\u0301",\u00E8:"e\u0300",\u00EB:"e\u0308",\u1EBD:"e\u0303",\u0113:"e\u0304",\u1E17:"e\u0304\u0301",\u1E15:"e\u0304\u0300",\u0115:"e\u0306",\u1E1D:"e\u0327\u0306",\u011B:"e\u030C",\u00EA:"e\u0302",\u1EBF:"e\u0302\u0301",\u1EC1:"e\u0302\u0300",\u1EC5:"e\u0302\u0303",\u0117:"e\u0307",\u0229:"e\u0327",\u1E1F:"f\u0307",\u01F5:"g\u0301",\u1E21:"g\u0304",\u011F:"g\u0306",\u01E7:"g\u030C",\u011D:"g\u0302",\u0121:"g\u0307",\u0123:"g\u0327",\u1E27:"h\u0308",\u021F:"h\u030C",\u0125:"h\u0302",\u1E23:"h\u0307",\u1E29:"h\u0327",\u00ED:"i\u0301",\u00EC:"i\u0300",\u00EF:"i\u0308",\u1E2F:"i\u0308\u0301",\u0129:"i\u0303",\u012B:"i\u0304",\u012D:"i\u0306",\u01D0:"i\u030C",\u00EE:"i\u0302",\u01F0:"j\u030C",\u0135:"j\u0302",\u1E31:"k\u0301",\u01E9:"k\u030C",\u0137:"k\u0327",\u013A:"l\u0301",\u013E:"l\u030C",\u013C:"l\u0327",\u1E3F:"m\u0301",\u1E41:"m\u0307",\u0144:"n\u0301",\u01F9:"n\u0300",\u00F1:"n\u0303",\u0148:"n\u030C",\u1E45:"n\u0307",\u0146:"n\u0327",\u00F3:"o\u0301",\u00F2:"o\u0300",\u00F6:"o\u0308",\u022B:"o\u0308\u0304",\u00F5:"o\u0303",\u1E4D:"o\u0303\u0301",\u1E4F:"o\u0303\u0308",\u022D:"o\u0303\u0304",\u014D:"o\u0304",\u1E53:"o\u0304\u0301",\u1E51:"o\u0304\u0300",\u014F:"o\u0306",\u01D2:"o\u030C",\u00F4:"o\u0302",\u1ED1:"o\u0302\u0301",\u1ED3:"o\u0302\u0300",\u1ED7:"o\u0302\u0303",\u022F:"o\u0307",\u0231:"o\u0307\u0304",\u0151:"o\u030B",\u1E55:"p\u0301",\u1E57:"p\u0307",\u0155:"r\u0301",\u0159:"r\u030C",\u1E59:"r\u0307",\u0157:"r\u0327",\u015B:"s\u0301",\u1E65:"s\u0301\u0307",\u0161:"s\u030C",\u1E67:"s\u030C\u0307",\u015D:"s\u0302",\u1E61:"s\u0307",\u015F:"s\u0327",\u1E97:"t\u0308",\u0165:"t\u030C",\u1E6B:"t\u0307",\u0163:"t\u0327",\u00FA:"u\u0301",\u00F9:"u\u0300",\u00FC:"u\u0308",\u01D8:"u\u0308\u0301",\u01DC:"u\u0308\u0300",\u01D6:"u\u0308\u0304",\u01DA:"u\u0308\u030C",\u0169:"u\u0303",\u1E79:"u\u0303\u0301",\u016B:"u\u0304",\u1E7B:"u\u0304\u0308",\u016D:"u\u0306",\u01D4:"u\u030C",\u00FB:"u\u0302",\u016F:"u\u030A",\u0171:"u\u030B",\u1E7D:"v\u0303",\u1E83:"w\u0301",\u1E81:"w\u0300",\u1E85:"w\u0308",\u0175:"w\u0302",\u1E87:"w\u0307",\u1E98:"w\u030A",\u1E8D:"x\u0308",\u1E8B:"x\u0307",\u00FD:"y\u0301",\u1EF3:"y\u0300",\u00FF:"y\u0308",\u1EF9:"y\u0303",\u0233:"y\u0304",\u0177:"y\u0302",\u1E8F:"y\u0307",\u1E99:"y\u030A",\u017A:"z\u0301",\u017E:"z\u030C",\u1E91:"z\u0302",\u017C:"z\u0307",\u00C1:"A\u0301",\u00C0:"A\u0300",\u00C4:"A\u0308",\u01DE:"A\u0308\u0304",\u00C3:"A\u0303",\u0100:"A\u0304",\u0102:"A\u0306",\u1EAE:"A\u0306\u0301",\u1EB0:"A\u0306\u0300",\u1EB4:"A\u0306\u0303",\u01CD:"A\u03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this.expect("}"),this.nextToken=r,n}parseExpression(e,r){for(var n=[];;){this.mode==="math"&&this.consumeSpaces();var i=this.fetch();if(t.endOfExpression.indexOf(i.text)!==-1||r&&i.text===r||e&&fh[i.text]&&fh[i.text].infix)break;var a=this.parseAtom(r);if(a){if(a.type==="internal")continue}else break;n.push(a)}return this.mode==="text"&&this.formLigatures(n),this.handleInfixNodes(n)}handleInfixNodes(e){for(var r=-1,n,i=0;i=0&&this.settings.reportNonstrict("unicodeTextInMathMode",'Latin-1/Unicode text character "'+r[0]+'" used in math mode',e);var l=wn[this.mode][r].group,u=Xs.range(e),h;if(hxe.hasOwnProperty(l)){var f=l;h={type:"atom",mode:this.mode,family:f,loc:u,text:r}}else h={type:l,mode:this.mode,loc:u,text:r};s=h}else if(r.charCodeAt(0)>=128)this.settings.strict&&(LG(r.charCodeAt(0))?this.mode==="math"&&this.settings.reportNonstrict("unicodeTextInMathMode",'Unicode text character "'+r[0]+'" used in math mode',e):this.settings.reportNonstrict("unknownSymbol",'Unrecognized Unicode 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2:if(Hs=this.productions_[us[1]][1],sh.$=Ot[Ot.length-Hs],sh._$={first_line:pe[pe.length-(Hs||1)].first_line,last_line:pe[pe.length-1].last_line,first_column:pe[pe.length-(Hs||1)].first_column,last_column:pe[pe.length-1].last_column},ah&&(sh._$.range=[pe[pe.length-(Hs||1)].range[0],pe[pe.length-1].range[1]]),Wl=this.performAction.apply(sh,[be,Xc,Ir,ko.yy,us[1],Ot,pe].concat(O1)),typeof Wl<"u")return Wl;Hs&&($t=$t.slice(0,-1*Hs*2),Ot=Ot.slice(0,-1*Hs),pe=pe.slice(0,-1*Hs)),$t.push(this.productions_[us[1]][0]),Ot.push(sh.$),pe.push(sh._$),B1=ur[$t[$t.length-2]][$t[$t.length-1]],$t.push(B1);break;case 3:return!0}}return!0},"parse")},_a=function(){var qi={EOF:1,parseError:o(function(At,$t){if(this.yy.parser)this.yy.parser.parseError(At,$t);else throw new Error(At)},"parseError"),setInput:o(function(ht,At){return this.yy=At||this.yy||{},this._input=ht,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},"setInput"),input:o(function(){var ht=this._input[0];this.yytext+=ht,this.yyleng++,this.offset++,this.match+=ht,this.matched+=ht;var At=ht.match(/(?:\r\n?|\n).*/g);return At?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),ht},"input"),unput:o(function(ht){var At=ht.length,$t=ht.split(/(?:\r\n?|\n)/g);this._input=ht+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-At),this.offset-=At;var rt=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),$t.length-1&&(this.yylineno-=$t.length-1);var Ot=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:$t?($t.length===rt.length?this.yylloc.first_column:0)+rt[rt.length-$t.length].length-$t[0].length:this.yylloc.first_column-At},this.options.ranges&&(this.yylloc.range=[Ot[0],Ot[0]+this.yyleng-At]),this.yyleng=this.yytext.length,this},"unput"),more:o(function(){return this._more=!0,this},"more"),reject:o(function(){if(this.options.backtrack_lexer)this._backtrack=!0;else return this.parseError("Lexical error on line "+(this.yylineno+1)+`. You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true). -`+this.showPosition(),{text:"",token:null,line:this.yylineno});return this},"reject"),less:o(function(ht){this.unput(this.match.slice(ht))},"less"),pastInput:o(function(){var ht=this.matched.substr(0,this.matched.length-this.match.length);return(ht.length>20?"...":"")+ht.substr(-20).replace(/\n/g,"")},"pastInput"),upcomingInput:o(function(){var ht=this.match;return ht.length<20&&(ht+=this._input.substr(0,20-ht.length)),(ht.substr(0,20)+(ht.length>20?"...":"")).replace(/\n/g,"")},"upcomingInput"),showPosition:o(function(){var ht=this.pastInput(),At=new Array(ht.length+1).join("-");return ht+this.upcomingInput()+` -`+At+"^"},"showPosition"),test_match:o(function(ht,At){var $t,rt,Ot;if(this.options.backtrack_lexer&&(Ot={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(Ot.yylloc.range=this.yylloc.range.slice(0))),rt=ht[0].match(/(?:\r\n?|\n).*/g),rt&&(this.yylineno+=rt.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:rt?rt[rt.length-1].length-rt[rt.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+ht[0].length},this.yytext+=ht[0],this.match+=ht[0],this.matches=ht,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(ht[0].length),this.matched+=ht[0],$t=this.performAction.call(this,this.yy,this,At,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),$t)return $t;if(this._backtrack){for(var pe in Ot)this[pe]=Ot[pe];return!1}return!1},"test_match"),next:o(function(){if(this.done)return this.EOF;this._input||(this.done=!0);var ht,At,$t,rt;this._more||(this.yytext="",this.match="");for(var Ot=this._currentRules(),pe=0;peAt[0].length)){if(At=$t,rt=pe,this.options.backtrack_lexer){if(ht=this.test_match($t,Ot[pe]),ht!==!1)return ht;if(this._backtrack){At=!1;continue}else return!1}else if(!this.options.flex)break}return At?(ht=this.test_match(At,Ot[rt]),ht!==!1?ht:!1):this._input===""?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+`. Unrecognized text. -`+this.showPosition(),{text:"",token:null,line:this.yylineno})},"next"),lex:o(function(){var At=this.next();return At||this.lex()},"lex"),begin:o(function(At){this.conditionStack.push(At)},"begin"),popState:o(function(){var At=this.conditionStack.length-1;return At>0?this.conditionStack.pop():this.conditionStack[0]},"popState"),_currentRules:o(function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},"_currentRules"),topState:o(function(At){return At=this.conditionStack.length-1-Math.abs(At||0),At>=0?this.conditionStack[At]:"INITIAL"},"topState"),pushState:o(function(At){this.begin(At)},"pushState"),stateStackSize:o(function(){return this.conditionStack.length},"stateStackSize"),options:{},performAction:o(function(At,$t,rt,Ot){var pe=Ot;switch(rt){case 0:return this.begin("acc_title"),34;break;case 1:return this.popState(),"acc_title_value";break;case 2:return this.begin("acc_descr"),36;break;case 3:return this.popState(),"acc_descr_value";break;case 4:this.begin("acc_descr_multiline");break;case 5:this.popState();break;case 6:return"acc_descr_multiline_value";case 7:this.begin("callbackname");break;case 8:this.popState();break;case 9:this.popState(),this.begin("callbackargs");break;case 10:return 92;case 11:this.popState();break;case 12:return 93;case 13:return"MD_STR";case 14:this.popState();break;case 15:this.begin("md_string");break;case 16:return"STR";case 17:this.popState();break;case 18:this.pushState("string");break;case 19:return 81;case 20:return 99;case 21:return 82;case 22:return 101;case 23:return 83;case 24:return 84;case 25:return 94;case 26:this.begin("click");break;case 27:this.popState();break;case 28:return 85;case 29:return At.lex.firstGraph()&&this.begin("dir"),12;break;case 30:return At.lex.firstGraph()&&this.begin("dir"),12;break;case 31:return At.lex.firstGraph()&&this.begin("dir"),12;break;case 32:return 27;case 33:return 32;case 34:return 95;case 35:return 95;case 36:return 95;case 37:return 95;case 38:return this.popState(),13;break;case 39:return this.popState(),14;break;case 40:return this.popState(),14;break;case 41:return this.popState(),14;break;case 42:return this.popState(),14;break;case 43:return this.popState(),14;break;case 44:return this.popState(),14;break;case 45:return this.popState(),14;break;case 46:return this.popState(),14;break;case 47:return this.popState(),14;break;case 48:return this.popState(),14;break;case 49:return 118;case 50:return 119;case 51:return 120;case 52:return 121;case 53:return 102;case 54:return 108;case 55:return 44;case 56:return 58;case 57:return 42;case 58:return 8;case 59:return 103;case 60:return 112;case 61:return this.popState(),75;break;case 62:return this.pushState("edgeText"),73;break;case 63:return 116;case 64:return this.popState(),75;break;case 65:return this.pushState("thickEdgeText"),73;break;case 66:return 116;case 67:return this.popState(),75;break;case 68:return this.pushState("dottedEdgeText"),73;break;case 69:return 116;case 70:return 75;case 71:return this.popState(),51;break;case 72:return"TEXT";case 73:return this.pushState("ellipseText"),50;break;case 74:return this.popState(),53;break;case 75:return this.pushState("text"),52;break;case 76:return this.popState(),55;break;case 77:return this.pushState("text"),54;break;case 78:return 56;case 79:return this.pushState("text"),65;break;case 80:return this.popState(),62;break;case 81:return this.pushState("text"),61;break;case 82:return this.popState(),47;break;case 83:return this.pushState("text"),46;break;case 84:return this.popState(),67;break;case 85:return this.popState(),69;break;case 86:return 114;case 87:return this.pushState("trapText"),66;break;case 88:return this.pushState("trapText"),68;break;case 89:return 115;case 90:return 65;case 91:return 87;case 92:return"SEP";case 93:return 86;case 94:return 112;case 95:return 108;case 96:return 42;case 97:return 106;case 98:return 111;case 99:return 113;case 100:return this.popState(),60;break;case 101:return this.pushState("text"),60;break;case 102:return this.popState(),49;break;case 103:return this.pushState("text"),48;break;case 104:return this.popState(),31;break;case 105:return this.pushState("text"),29;break;case 106:return this.popState(),64;break;case 107:return this.pushState("text"),63;break;case 108:return"TEXT";case 109:return"QUOTE";case 110:return 9;case 111:return 10;case 112:return 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24:this.$={attributeType:C[D-2],attributeName:C[D-1],attributeKeyTypeList:C[D]};break;case 25:this.$={attributeType:C[D-2],attributeName:C[D-1],attributeComment:C[D]};break;case 26:this.$={attributeType:C[D-3],attributeName:C[D-2],attributeKeyTypeList:C[D-1],attributeComment:C[D]};break;case 27:case 28:case 31:this.$=C[D];break;case 30:C[D-2].push(C[D]),this.$=C[D-2];break;case 32:this.$=C[D].replace(/"/g,"");break;case 33:this.$={cardA:C[D],relType:C[D-1],cardB:C[D-2]};break;case 34:this.$=k.Cardinality.ZERO_OR_ONE;break;case 35:this.$=k.Cardinality.ZERO_OR_MORE;break;case 36:this.$=k.Cardinality.ONE_OR_MORE;break;case 37:this.$=k.Cardinality.ONLY_ONE;break;case 38:this.$=k.Cardinality.MD_PARENT;break;case 39:this.$=k.Identification.NON_IDENTIFYING;break;case 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N=new Error(L);throw N.hash=M,N}},"parseError"),parse:o(function(L){var M=this,N=[0],k=[],I=[null],C=[],O=this.table,D="",P=0,F=0,B=0,$=2,z=1,Y=C.slice.call(arguments,1),Q=Object.create(this.lexer),X={yy:{}};for(var ie in this.yy)Object.prototype.hasOwnProperty.call(this.yy,ie)&&(X.yy[ie]=this.yy[ie]);Q.setInput(L,X.yy),X.yy.lexer=Q,X.yy.parser=this,typeof Q.yylloc>"u"&&(Q.yylloc={});var j=Q.yylloc;C.push(j);var J=Q.options&&Q.options.ranges;typeof X.yy.parseError=="function"?this.parseError=X.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;function Z(Pe){N.length=N.length-2*Pe,I.length=I.length-Pe,C.length=C.length-Pe}o(Z,"popStack");function H(){var Pe;return Pe=k.pop()||Q.lex()||z,typeof Pe!="number"&&(Pe instanceof Array&&(k=Pe,Pe=k.pop()),Pe=M.symbols_[Pe]||Pe),Pe}o(H,"lex");for(var q,K,se,ce,ue,te,De={},oe,ke,Ie,Se;;){if(se=N[N.length-1],this.defaultActions[se]?ce=this.defaultActions[se]:((q===null||typeof q>"u")&&(q=H()),ce=O[se]&&O[se][q]),typeof 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I=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:N?(N.length===k.length?this.yylloc.first_column:0)+k[k.length-N.length].length-N[0].length:this.yylloc.first_column-M},this.options.ranges&&(this.yylloc.range=[I[0],I[0]+this.yyleng-M]),this.yyleng=this.yytext.length,this},"unput"),more:o(function(){return this._more=!0,this},"more"),reject:o(function(){if(this.options.backtrack_lexer)this._backtrack=!0;else return this.parseError("Lexical error on line "+(this.yylineno+1)+`. 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this.popState(),"acc_title_value";break;case 2:return this.begin("acc_descr"),24;break;case 3:return this.popState(),"acc_descr_value";break;case 4:this.begin("acc_descr_multiline");break;case 5:this.popState();break;case 6:return"acc_descr_multiline_value";case 7:return 10;case 8:break;case 9:return 8;case 10:return 28;case 11:return 48;case 12:return 4;case 13:return this.begin("block"),15;break;case 14:return 36;case 15:break;case 16:return 37;case 17:return 34;case 18:return 34;case 19:return 38;case 20:break;case 21:return this.popState(),17;break;case 22:return N.yytext[0];case 23:return 18;case 24:return 19;case 25:return 41;case 26:return 43;case 27:return 43;case 28:return 43;case 29:return 41;case 30:return 41;case 31:return 42;case 32:return 42;case 33:return 42;case 34:return 42;case 35:return 42;case 36:return 43;case 37:return 42;case 38:return 43;case 39:return 44;case 40:return 44;case 41:return 44;case 42:return 44;case 43:return 41;case 44:return 42;case 45:return 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n=this.config.traceInitPerf;n===!0?(this.traceInitMaxIdent=1/0,this.traceInitPerf=!0):typeof n=="number"&&(this.traceInitMaxIdent=n,this.traceInitPerf=!0),this.traceInitIndent=-1,this.TRACE_INIT("Lexer Constructor",()=>{let i,a=!0;this.TRACE_INIT("Lexer Config handling",()=>{if(this.config.lineTerminatorsPattern===a2.lineTerminatorsPattern)this.config.lineTerminatorsPattern=kie;else if(this.config.lineTerminatorCharacters===a2.lineTerminatorCharacters)throw Error(`Error: Missing property on the Lexer config. - For details See: https://chevrotain.io/docs/guide/resolving_lexer_errors.html#MISSING_LINE_TERM_CHARS`);if(r.safeMode&&r.ensureOptimizations)throw Error('"safeMode" and "ensureOptimizations" flags are mutually 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h;this.TRACE_INIT("analyzeTokenTypes",()=>{h=yie(l,{lineTerminatorCharacters:this.config.lineTerminatorCharacters,positionTracking:r.positionTracking,ensureOptimizations:r.ensureOptimizations,safeMode:r.safeMode,tracer:this.TRACE_INIT})}),this.patternIdxToConfig[u]=h.patternIdxToConfig,this.charCodeToPatternIdxToConfig[u]=h.charCodeToPatternIdxToConfig,this.emptyGroups=pa({},this.emptyGroups,h.emptyGroups),this.hasCustom=h.hasCustom||this.hasCustom,this.canModeBeOptimized[u]=h.canBeOptimized}})}),this.defaultMode=i.defaultMode,!Qt(this.lexerDefinitionErrors)&&!this.config.deferDefinitionErrorsHandling){let u=qe(this.lexerDefinitionErrors,h=>h.message).join(`----------------------- -`);throw new Error(`Errors detected in definition of Lexer: -`+u)}Ee(this.lexerDefinitionWarning,l=>{e2(l.message)}),this.TRACE_INIT("Choosing sub-methods implementations",()=>{if(fN?(this.chopInput=ea,this.match=this.matchWithTest):(this.updateLastIndex=qn,this.match=this.matchWithExec),a&&(this.handleModes=qn),this.trackStartLines===!1&&(this.computeNewColumn=ea),this.trackEndLines===!1&&(this.updateTokenEndLineColumnLocation=qn),/full/i.test(this.config.positionTracking))this.createTokenInstance=this.createFullToken;else if(/onlyStart/i.test(this.config.positionTracking))this.createTokenInstance=this.createStartOnlyToken;else if(/onlyOffset/i.test(this.config.positionTracking))this.createTokenInstance=this.createOffsetOnlyToken;else throw Error(`Invalid config option: "${this.config.positionTracking}"`);this.hasCustom?(this.addToken=this.addTokenUsingPush,this.handlePayload=this.handlePayloadWithCustom):(this.addToken=this.addTokenUsingMemberAccess,this.handlePayload=this.handlePayloadNoCustom)}),this.TRACE_INIT("Failed Optimization Warnings",()=>{let l=Vr(this.canModeBeOptimized,(u,h,f)=>(h===!1&&u.push(f),u),[]);if(r.ensureOptimizations&&!Qt(l))throw Error(`Lexer Modes: < ${l.join(", ")} > cannot be optimized. - Disable the "ensureOptimizations" lexer config flag to silently ignore this and run the lexer in an un-optimized mode. - Or inspect the console log for details on how to resolve these issues.`)}),this.TRACE_INIT("clearRegExpParserCache",()=>{uie()}),this.TRACE_INIT("toFastProperties",()=>{r2(this)})})}tokenize(e,r=this.defaultMode){if(!Qt(this.lexerDefinitionErrors)){let i=qe(this.lexerDefinitionErrors,a=>a.message).join(`----------------------- -`);throw new Error(`Unable to Tokenize because Errors detected in definition of Lexer: -`+i)}return this.tokenizeInternal(e,r)}tokenizeInternal(e,r){let n,i,a,s,l,u,h,f,d,p,m,g,y,v,x,b,w=e,S=w.length,T=0,E=0,_=this.hasCustom?0:Math.floor(e.length/10),A=new Array(_),L=[],M=this.trackStartLines?1:void 0,N=this.trackStartLines?1:void 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ie,j=this.config.recoveryEnabled;for(;Tu.length){u=s,h=f,ie=ce;break}}}break}}if(u!==null){if(d=u.length,p=ie.group,p!==void 0&&(m=ie.tokenTypeIdx,g=this.createTokenInstance(u,T,m,ie.tokenType,M,N,d),this.handlePayload(g,h),p===!1?E=this.addToken(A,E,g):k[p].push(g)),e=this.chopInput(e,d),T=T+d,N=this.computeNewColumn(N,d),I===!0&&ie.canLineTerminator===!0){let q=0,K,se;C.lastIndex=0;do K=C.test(u),K===!0&&(se=C.lastIndex-1,q++);while(K===!0);q!==0&&(M=M+q,N=d-se,this.updateTokenEndLineColumnLocation(g,p,se,q,M,N,d))}this.handleModes(ie,Q,X,g)}else{let q=T,K=M,se=N,ce=j===!1;for(;ce===!1&&T{"use strict";Pt();i2();s0();o(zu,"tokenLabel");o(gN,"hasTokenLabel");EIe="parent",Rie="categories",Nie="label",Mie="group",Iie="push_mode",Oie="pop_mode",Pie="longer_alt",Bie="line_breaks",Fie="start_chars_hint";o(VT,"createToken");o(CIe,"createTokenInternal");fo=VT({name:"EOF",pattern:ni.NA});Fu([fo]);o(o0,"createTokenInstance");o(s2,"tokenMatcher")});var Gu,zie,Ol,Jm=R(()=>{"use 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easy to accomplish by using the convention that Terminal names start with an uppercase letter -and Non-Terminal names start with a lower case letter.`},buildAlternationPrefixAmbiguityError(t){let e=qe(t.prefixPath,i=>zu(i)).join(", "),r=t.alternation.idx===0?"":t.alternation.idx;return`Ambiguous alternatives: <${t.ambiguityIndices.join(" ,")}> due to common lookahead prefix -in inside <${t.topLevelRule.name}> Rule, -<${e}> may appears as a prefix path in all these alternatives. -See: https://chevrotain.io/docs/guide/resolving_grammar_errors.html#COMMON_PREFIX -For Further details.`},buildAlternationAmbiguityError(t){let e=qe(t.prefixPath,i=>zu(i)).join(", "),r=t.alternation.idx===0?"":t.alternation.idx,n=`Ambiguous Alternatives Detected: <${t.ambiguityIndices.join(" ,")}> in inside <${t.topLevelRule.name}> Rule, -<${e}> may appears as a prefix path in all these alternatives. -`;return n=n+`See: https://chevrotain.io/docs/guide/resolving_grammar_errors.html#AMBIGUOUS_ALTERNATIVES -For Further details.`,n},buildEmptyRepetitionError(t){let e=Rs(t.repetition);return t.repetition.idx!==0&&(e+=t.repetition.idx),`The repetition <${e}> within Rule <${t.topLevelRule.name}> can never consume any tokens. -This could lead to an infinite loop.`},buildTokenNameError(t){return"deprecated"},buildEmptyAlternationError(t){return`Ambiguous empty alternative: <${t.emptyChoiceIdx+1}> in inside <${t.topLevelRule.name}> Rule. -Only the last alternative may be an empty alternative.`},buildTooManyAlternativesError(t){return`An Alternation cannot have more than 256 alternatives: - inside <${t.topLevelRule.name}> Rule. - has ${t.alternation.definition.length+1} alternatives.`},buildLeftRecursionError(t){let e=t.topLevelRule.name,r=qe(t.leftRecursionPath,a=>a.name),n=`${e} --> ${r.concat([e]).join(" --> ")}`;return`Left Recursion found in grammar. -rule: <${e}> can be invoked from itself (directly or indirectly) -without consuming any Tokens. The grammar path that causes this is: - ${n} - To fix this refactor your grammar to remove the left recursion. -see: https://en.wikipedia.org/wiki/LL_parser#Left_factoring.`},buildInvalidRuleNameError(t){return"deprecated"},buildDuplicateRuleNameError(t){let e;return t.topLevelRule instanceof ts?e=t.topLevelRule.name:e=t.topLevelRule,`Duplicate definition, rule: ->${e}<- is already defined in the grammar: ->${t.grammarName}<-`}}});function Gie(t,e){let r=new yN(t,e);return r.resolveRefs(),r.errors}var yN,$ie=R(()=>{"use strict";Ns();Pt();ns();o(Gie,"resolveGrammar");yN=class extends rs{static{o(this,"GastRefResolverVisitor")}constructor(e,r){super(),this.nameToTopRule=e,this.errMsgProvider=r,this.errors=[]}resolveRefs(){Ee(or(this.nameToTopRule),e=>{this.currTopLevel=e,e.accept(this)})}visitNonTerminal(e){let r=this.nameToTopRule[e.nonTerminalName];if(r)e.referencedRule=r;else{let n=this.errMsgProvider.buildRuleNotFoundError(this.currTopLevel,e);this.errors.push({message:n,type:Pi.UNRESOLVED_SUBRULE_REF,ruleName:this.currTopLevel.name,unresolvedRefName:e.nonTerminalName})}}}});function WT(t,e,r=[]){r=Qr(r);let n=[],i=0;function a(l){return l.concat(fi(t,i+1))}o(a,"remainingPathWith");function s(l){let u=WT(a(l),e,r);return n.concat(u)}for(o(s,"getAlternativesForProd");r.length{Qt(u.definition)===!1&&(n=s(u.definition))}),n;if(l instanceof fr)r.push(l.terminalType);else throw Error("non exhaustive match")}i++}return n.push({partialPath:r,suffixDef:fi(t,i)}),n}function qT(t,e,r,n){let i="EXIT_NONE_TERMINAL",a=[i],s="EXIT_ALTERNATIVE",l=!1,u=e.length,h=u-n-1,f=[],d=[];for(d.push({idx:-1,def:t,ruleStack:[],occurrenceStack:[]});!Qt(d);){let p=d.pop();if(p===s){l&&ma(d).idx<=h&&d.pop();continue}let m=p.def,g=p.idx,y=p.ruleStack,v=p.occurrenceStack;if(Qt(m))continue;let x=m[0];if(x===i){let b={idx:g,def:fi(m),ruleStack:Ru(y),occurrenceStack:Ru(v)};d.push(b)}else if(x 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this.ruleStack=Qr(this.path.ruleStack).reverse(),this.occurrenceStack=Qr(this.path.occurrenceStack).reverse(),this.ruleStack.pop(),this.occurrenceStack.pop(),this.updateExpectedNext(),this.walk(this.topProd),this.possibleTokTypes}walk(e,r=[]){this.found||super.walk(e,r)}walkProdRef(e,r,n){if(e.referencedRule.name===this.nextProductionName&&e.idx===this.nextProductionOccurrence){let i=r.concat(n);this.updateExpectedNext(),this.walk(e.referencedRule,i)}}updateExpectedNext(){Qt(this.ruleStack)?(this.nextProductionName="",this.nextProductionOccurrence=0,this.isAtEndOfPath=!0):(this.nextProductionName=this.ruleStack.pop(),this.nextProductionOccurrence=this.occurrenceStack.pop())}},UT=class extends vN{static{o(this,"NextAfterTokenWalker")}constructor(e,r){super(e,r),this.path=r,this.nextTerminalName="",this.nextTerminalOccurrence=0,this.nextTerminalName=this.path.lastTok.name,this.nextTerminalOccurrence=this.path.lastTokOccurrence}walkTerminal(e,r,n){if(this.isAtEndOfPath&&e.terminalType.name===this.nextTerminalName&&e.idx===this.nextTerminalOccurrence&&!this.found){let i=r.concat(n),a=new Sn({definition:i});this.possibleTokTypes=i0(a),this.found=!0}}},eg=class extends Pu{static{o(this,"AbstractNextTerminalAfterProductionWalker")}constructor(e,r){super(),this.topRule=e,this.occurrence=r,this.result={token:void 0,occurrence:void 0,isEndOfRule:void 0}}startWalking(){return this.walk(this.topRule),this.result}},HT=class extends eg{static{o(this,"NextTerminalAfterManyWalker")}walkMany(e,r,n){if(e.idx===this.occurrence){let i=na(r.concat(n));this.result.isEndOfRule=i===void 0,i instanceof fr&&(this.result.token=i.terminalType,this.result.occurrence=i.idx)}else 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i=[],a=Vr(t,(l,u,h)=>(e.definition[h].ignoreAmbiguities===!0||Ee(u,f=>{let d=[h];Ee(t,(p,m)=>{h!==m&&KT(p,f)&&e.definition[m].ignoreAmbiguities!==!0&&d.push(m)}),d.length>1&&!KT(i,f)&&(i.push(f),l.push({alts:d,path:f}))}),l),[]);return qe(a,l=>{let u=qe(l.alts,f=>f+1);return{message:n.buildAlternationAmbiguityError({topLevelRule:r,alternation:e,ambiguityIndices:u,prefixPath:l.path}),type:Pi.AMBIGUOUS_ALTS,ruleName:r.name,occurrence:e.idx,alternatives:l.alts}})}function MIe(t,e,r,n){let i=Vr(t,(s,l,u)=>{let h=qe(l,f=>({idx:u,path:f}));return s.concat(h)},[]);return wc(ga(i,s=>{if(e.definition[s.idx].ignoreAmbiguities===!0)return[];let u=s.idx,h=s.path,f=$r(i,p=>e.definition[p.idx].ignoreAmbiguities!==!0&&p.idx{let m=[p.idx+1,u+1],g=e.idx===0?"":e.idx;return{message:n.buildAlternationPrefixAmbiguityError({topLevelRule:r,alternation:e,ambiguityIndices:m,prefixPath:p.path}),type:Pi.AMBIGUOUS_PREFIX_ALTS,ruleName:r.name,occurrence:g,alternatives:m}})}))}function IIe(t,e,r){let n=[],i=qe(e,a=>a.name);return Ee(t,a=>{let s=a.name;if(Fn(i,s)){let l=r.buildNamespaceConflictError(a);n.push({message:l,type:Pi.CONFLICT_TOKENS_RULES_NAMESPACE,ruleName:s})}}),n}var wN,h2,TN,f2=R(()=>{"use strict";Pt();Ns();ns();ng();c2();s0();o(Kie,"validateLookahead");o(Qie,"validateGrammar");o(_Ie,"validateDuplicateProductions");o(LIe,"identifyProductionForDuplicates");o(Zie,"getExtraProductionArgument");wN=class extends 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l=this.getKeyForAutomaticLookahead(n,i),u=this.firstAfterRepMap[l];if(u===void 0){let p=this.getCurrRuleFullName(),m=this.getGAstProductions()[p];u=new a(m,i).startWalking(),this.firstAfterRepMap[l]=u}let h=u.token,f=u.occurrence,d=u.isEndOfRule;this.RULE_STACK.length===1&&d&&h===void 0&&(h=fo,f=1),!(h===void 0||f===void 0)&&this.shouldInRepetitionRecoveryBeTried(h,f,s)&&this.tryInRepetitionRecovery(t,e,r,h)}var EN,SN,CN,ZT,AN=R(()=>{"use strict";l0();Pt();ag();aN();Ns();EN={},SN="InRuleRecoveryException",CN=class extends Error{static{o(this,"InRuleRecoveryException")}constructor(e){super(e),this.name=SN}},ZT=class{static{o(this,"Recoverable")}initRecoverable(e){this.firstAfterRepMap={},this.resyncFollows={},this.recoveryEnabled=Xe(e,"recoveryEnabled")?e.recoveryEnabled:is.recoveryEnabled,this.recoveryEnabled&&(this.attemptInRepetitionRecovery=OIe)}getTokenToInsert(e){let r=o0(e,"",NaN,NaN,NaN,NaN,NaN,NaN);return 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strict";Pt();mae();o(FIe,"defaultVisit");o(gae,"createBaseSemanticVisitorConstructor");o(yae,"createBaseVisitorConstructorWithDefaults");(function(t){t[t.REDUNDANT_METHOD=0]="REDUNDANT_METHOD",t[t.MISSING_METHOD=1]="MISSING_METHOD"})(ON||(ON={}));o(zIe,"validateVisitor");o(GIe,"validateMissingCstMethods")});var sk,xae=R(()=>{"use strict";pae();Pt();vae();Ns();sk=class{static{o(this,"TreeBuilder")}initTreeBuilder(e){if(this.CST_STACK=[],this.outputCst=e.outputCst,this.nodeLocationTracking=Xe(e,"nodeLocationTracking")?e.nodeLocationTracking:is.nodeLocationTracking,!this.outputCst)this.cstInvocationStateUpdate=qn,this.cstFinallyStateUpdate=qn,this.cstPostTerminal=qn,this.cstPostNonTerminal=qn,this.cstPostRule=qn;else if(/full/i.test(this.nodeLocationTracking))this.recoveryEnabled?(this.setNodeLocationFromToken=MN,this.setNodeLocationFromNode=MN,this.cstPostRule=qn,this.setInitialNodeLocation=this.setInitialNodeLocationFullRecovery):(this.setNodeLocationFromToken=qn,this.setNodeLocationFromNode=qn,this.cstPostRule=this.cstPostRuleFull,this.setInitialNodeLocation=this.setInitialNodeLocationFullRegular);else if(/onlyOffset/i.test(this.nodeLocationTracking))this.recoveryEnabled?(this.setNodeLocationFromToken=NN,this.setNodeLocationFromNode=NN,this.cstPostRule=qn,this.setInitialNodeLocation=this.setInitialNodeLocationOnlyOffsetRecovery):(this.setNodeLocationFromToken=qn,this.setNodeLocationFromNode=qn,this.cstPostRule=this.cstPostRuleOnlyOffset,this.setInitialNodeLocation=this.setInitialNodeLocationOnlyOffsetRegular);else if(/none/i.test(this.nodeLocationTracking))this.setNodeLocationFromToken=qn,this.setNodeLocationFromNode=qn,this.cstPostRule=qn,this.setInitialNodeLocation=qn;else throw Error(`Invalid config option: "${e.nodeLocationTracking}"`)}setInitialNodeLocationOnlyOffsetRecovery(e){e.location={startOffset:NaN,endOffset:NaN}}setInitialNodeLocationOnlyOffsetRegular(e){e.location={startOffset:this.LA(1).startOffset,endOffset:NaN}}setInitialNodeLocationFullRecovery(e){e.location={startOffset:NaN,startLine:NaN,startColumn:NaN,endOffset:NaN,endLine:NaN,endColumn:NaN}}setInitialNodeLocationFullRegular(e){let r=this.LA(1);e.location={startOffset:r.startOffset,startLine:r.startLine,startColumn:r.startColumn,endOffset:NaN,endLine:NaN,endColumn:NaN}}cstInvocationStateUpdate(e){let r={name:e,children:Object.create(null)};this.setInitialNodeLocation(r),this.CST_STACK.push(r)}cstFinallyStateUpdate(){this.CST_STACK.pop()}cstPostRuleFull(e){let r=this.LA(0),n=e.location;n.startOffset<=r.startOffset?(n.endOffset=r.endOffset,n.endLine=r.endLine,n.endColumn=r.endColumn):(n.startOffset=NaN,n.startLine=NaN,n.startColumn=NaN)}cstPostRuleOnlyOffset(e){let r=this.LA(0),n=e.location;n.startOffset<=r.startOffset?n.endOffset=r.endOffset:n.startOffset=NaN}cstPostTerminal(e,r){let n=this.CST_STACK[this.CST_STACK.length-1];fae(n,r,e),this.setNodeLocationFromToken(n.location,r)}cstPostNonTerminal(e,r){let n=this.CST_STACK[this.CST_STACK.length-1];dae(n,r,e),this.setNodeLocationFromNode(n.location,e.location)}getBaseCstVisitorConstructor(){if(er(this.baseCstVisitorConstructor)){let e=gae(this.className,Dr(this.gastProductionsCache));return this.baseCstVisitorConstructor=e,e}return this.baseCstVisitorConstructor}getBaseCstVisitorConstructorWithDefaults(){if(er(this.baseCstVisitorWithDefaultsConstructor)){let e=yae(this.className,Dr(this.gastProductionsCache),this.getBaseCstVisitorConstructor());return this.baseCstVisitorWithDefaultsConstructor=e,e}return this.baseCstVisitorWithDefaultsConstructor}getLastExplicitRuleShortName(){let e=this.RULE_STACK;return e[e.length-1]}getPreviousExplicitRuleShortName(){let e=this.RULE_STACK;return e[e.length-2]}getLastExplicitRuleOccurrenceIndex(){let e=this.RULE_OCCURRENCE_STACK;return e[e.length-1]}}});var ok,bae=R(()=>{"use strict";Ns();ok=class{static{o(this,"LexerAdapter")}initLexerAdapter(){this.tokVector=[],this.tokVectorLength=0,this.currIdx=-1}set input(e){if(this.selfAnalysisDone!==!0)throw Error("Missing invocation at the end of the Parser's constructor.");this.reset(),this.tokVector=e,this.tokVectorLength=e.length}get input(){return this.tokVector}SKIP_TOKEN(){return this.currIdx<=this.tokVector.length-2?(this.consumeToken(),this.LA(1)):sg}LA(e){let r=this.currIdx+e;return r<0||this.tokVectorLength<=r?sg:this.tokVector[r]}consumeToken(){this.currIdx++}exportLexerState(){return this.currIdx}importLexerState(e){this.currIdx=e}resetLexerState(){this.currIdx=-1}moveToTerminatedState(){this.currIdx=this.tokVector.length-1}getLexerPosition(){return this.exportLexerState()}}});var lk,wae=R(()=>{"use strict";Pt();ag();Ns();Jm();f2();ns();lk=class{static{o(this,"RecognizerApi")}ACTION(e){return e.call(this)}consume(e,r,n){return this.consumeInternal(r,e,n)}subrule(e,r,n){return this.subruleInternal(r,e,n)}option(e,r){return this.optionInternal(r,e)}or(e,r){return this.orInternal(r,e)}many(e,r){return this.manyInternal(e,r)}atLeastOne(e,r){return this.atLeastOneInternal(e,r)}CONSUME(e,r){return this.consumeInternal(e,0,r)}CONSUME1(e,r){return this.consumeInternal(e,1,r)}CONSUME2(e,r){return this.consumeInternal(e,2,r)}CONSUME3(e,r){return this.consumeInternal(e,3,r)}CONSUME4(e,r){return this.consumeInternal(e,4,r)}CONSUME5(e,r){return this.consumeInternal(e,5,r)}CONSUME6(e,r){return this.consumeInternal(e,6,r)}CONSUME7(e,r){return this.consumeInternal(e,7,r)}CONSUME8(e,r){return this.consumeInternal(e,8,r)}CONSUME9(e,r){return this.consumeInternal(e,9,r)}SUBRULE(e,r){return this.subruleInternal(e,0,r)}SUBRULE1(e,r){return this.subruleInternal(e,1,r)}SUBRULE2(e,r){return this.subruleInternal(e,2,r)}SUBRULE3(e,r){return this.subruleInternal(e,3,r)}SUBRULE4(e,r){return this.subruleInternal(e,4,r)}SUBRULE5(e,r){return this.subruleInternal(e,5,r)}SUBRULE6(e,r){return this.subruleInternal(e,6,r)}SUBRULE7(e,r){return this.subruleInternal(e,7,r)}SUBRULE8(e,r){return this.subruleInternal(e,8,r)}SUBRULE9(e,r){return this.subruleInternal(e,9,r)}OPTION(e){return this.optionInternal(e,0)}OPTION1(e){return this.optionInternal(e,1)}OPTION2(e){return this.optionInternal(e,2)}OPTION3(e){return this.optionInternal(e,3)}OPTION4(e){return this.optionInternal(e,4)}OPTION5(e){return this.optionInternal(e,5)}OPTION6(e){return this.optionInternal(e,6)}OPTION7(e){return this.optionInternal(e,7)}OPTION8(e){return this.optionInternal(e,8)}OPTION9(e){return this.optionInternal(e,9)}OR(e){return this.orInternal(e,0)}OR1(e){return this.orInternal(e,1)}OR2(e){return this.orInternal(e,2)}OR3(e){return this.orInternal(e,3)}OR4(e){return this.orInternal(e,4)}OR5(e){return this.orInternal(e,5)}OR6(e){return this.orInternal(e,6)}OR7(e){return this.orInternal(e,7)}OR8(e){return this.orInternal(e,8)}OR9(e){return this.orInternal(e,9)}MANY(e){this.manyInternal(0,e)}MANY1(e){this.manyInternal(1,e)}MANY2(e){this.manyInternal(2,e)}MANY3(e){this.manyInternal(3,e)}MANY4(e){this.manyInternal(4,e)}MANY5(e){this.manyInternal(5,e)}MANY6(e){this.manyInternal(6,e)}MANY7(e){this.manyInternal(7,e)}MANY8(e){this.manyInternal(8,e)}MANY9(e){this.manyInternal(9,e)}MANY_SEP(e){this.manySepFirstInternal(0,e)}MANY_SEP1(e){this.manySepFirstInternal(1,e)}MANY_SEP2(e){this.manySepFirstInternal(2,e)}MANY_SEP3(e){this.manySepFirstInternal(3,e)}MANY_SEP4(e){this.manySepFirstInternal(4,e)}MANY_SEP5(e){this.manySepFirstInternal(5,e)}MANY_SEP6(e){this.manySepFirstInternal(6,e)}MANY_SEP7(e){this.manySepFirstInternal(7,e)}MANY_SEP8(e){this.manySepFirstInternal(8,e)}MANY_SEP9(e){this.manySepFirstInternal(9,e)}AT_LEAST_ONE(e){this.atLeastOneInternal(0,e)}AT_LEAST_ONE1(e){return this.atLeastOneInternal(1,e)}AT_LEAST_ONE2(e){this.atLeastOneInternal(2,e)}AT_LEAST_ONE3(e){this.atLeastOneInternal(3,e)}AT_LEAST_ONE4(e){this.atLeastOneInternal(4,e)}AT_LEAST_ONE5(e){this.atLeastOneInternal(5,e)}AT_LEAST_ONE6(e){this.atLeastOneInternal(6,e)}AT_LEAST_ONE7(e){this.atLeastOneInternal(7,e)}AT_LEAST_ONE8(e){this.atLeastOneInternal(8,e)}AT_LEAST_ONE9(e){this.atLeastOneInternal(9,e)}AT_LEAST_ONE_SEP(e){this.atLeastOneSepFirstInternal(0,e)}AT_LEAST_ONE_SEP1(e){this.atLeastOneSepFirstInternal(1,e)}AT_LEAST_ONE_SEP2(e){this.atLeastOneSepFirstInternal(2,e)}AT_LEAST_ONE_SEP3(e){this.atLeastOneSepFirstInternal(3,e)}AT_LEAST_ONE_SEP4(e){this.atLeastOneSepFirstInternal(4,e)}AT_LEAST_ONE_SEP5(e){this.atLeastOneSepFirstInternal(5,e)}AT_LEAST_ONE_SEP6(e){this.atLeastOneSepFirstInternal(6,e)}AT_LEAST_ONE_SEP7(e){this.atLeastOneSepFirstInternal(7,e)}AT_LEAST_ONE_SEP8(e){this.atLeastOneSepFirstInternal(8,e)}AT_LEAST_ONE_SEP9(e){this.atLeastOneSepFirstInternal(9,e)}RULE(e,r,n=og){if(Fn(this.definedRulesNames,e)){let s={message:Ol.buildDuplicateRuleNameError({topLevelRule:e,grammarName:this.className}),type:Pi.DUPLICATE_RULE_NAME,ruleName:e};this.definitionErrors.push(s)}this.definedRulesNames.push(e);let i=this.defineRule(e,r,n);return this[e]=i,i}OVERRIDE_RULE(e,r,n=og){let i=Jie(e,this.definedRulesNames,this.className);this.definitionErrors=this.definitionErrors.concat(i);let a=this.defineRule(e,r,n);return this[e]=a,a}BACKTRACK(e,r){return function(){this.isBackTrackingStack.push(1);let n=this.saveRecogState();try{return e.apply(this,r),!0}catch(i){if(nf(i))return!1;throw i}finally{this.reloadRecogState(n),this.isBackTrackingStack.pop()}}}getGAstProductions(){return this.gastProductionsCache}getSerializedGastProductions(){return NT(or(this.gastProductionsCache))}}});var ck,Tae=R(()=>{"use strict";Pt();ek();ag();ng();c2();Ns();AN();l0();s0();ck=class{static{o(this,"RecognizerEngine")}initRecognizerEngine(e,r){if(this.className=this.constructor.name,this.shortRuleNameToFull={},this.fullRuleNameToShort={},this.ruleShortNameIdx=256,this.tokenMatcher=Zm,this.subruleIdx=0,this.definedRulesNames=[],this.tokensMap={},this.isBackTrackingStack=[],this.RULE_STACK=[],this.RULE_OCCURRENCE_STACK=[],this.gastProductionsCache={},Xe(r,"serializedGrammar"))throw Error(`The Parser's configuration can no longer contain a property. - See: https://chevrotain.io/docs/changes/BREAKING_CHANGES.html#_6-0-0 - For Further details.`);if(wt(e)){if(Qt(e))throw Error(`A Token Vocabulary cannot be empty. - Note that the first argument for the parser constructor - is no longer a Token vector (since v4.0).`);if(typeof e[0].startOffset=="number")throw Error(`The Parser constructor no longer accepts a token vector as the first argument. - See: https://chevrotain.io/docs/changes/BREAKING_CHANGES.html#_4-0-0 - For Further details.`)}if(wt(e))this.tokensMap=Vr(e,(a,s)=>(a[s.name]=s,a),{});else if(Xe(e,"modes")&&Ia(Gr(or(e.modes)),Die)){let a=Gr(or(e.modes)),s=Pm(a);this.tokensMap=Vr(s,(l,u)=>(l[u.name]=u,l),{})}else if(pn(e))this.tokensMap=Qr(e);else throw new Error(" argument must be An Array of Token constructors, A dictionary of Token constructors or an IMultiModeLexerDefinition");this.tokensMap.EOF=fo;let n=Xe(e,"modes")?Gr(or(e.modes)):or(e),i=Ia(n,a=>Qt(a.categoryMatches));this.tokenMatcher=i?Zm:Bu,Fu(or(this.tokensMap))}defineRule(e,r,n){if(this.selfAnalysisDone)throw Error(`Grammar rule <${e}> may not be defined after the 'performSelfAnalysis' method has been called' -Make sure that all grammar rule definitions are done before 'performSelfAnalysis' is called.`);let i=Xe(n,"resyncEnabled")?n.resyncEnabled:og.resyncEnabled,a=Xe(n,"recoveryValueFunc")?n.recoveryValueFunc:og.recoveryValueFunc,s=this.ruleShortNameIdx<<12;this.ruleShortNameIdx++,this.shortRuleNameToFull[s]=e,this.fullRuleNameToShort[e]=s;let l;return this.outputCst===!0?l=o(function(...f){try{this.ruleInvocationStateUpdate(s,e,this.subruleIdx),r.apply(this,f);let d=this.CST_STACK[this.CST_STACK.length-1];return this.cstPostRule(d),d}catch(d){return this.invokeRuleCatch(d,i,a)}finally{this.ruleFinallyStateUpdate()}},"invokeRuleWithTry"):l=o(function(...f){try{return this.ruleInvocationStateUpdate(s,e,this.subruleIdx),r.apply(this,f)}catch(d){return this.invokeRuleCatch(d,i,a)}finally{this.ruleFinallyStateUpdate()}},"invokeRuleWithTryCst"),Object.assign(l,{ruleName:e,originalGrammarAction:r})}invokeRuleCatch(e,r,n){let i=this.RULE_STACK.length===1,a=r&&!this.isBackTracking()&&this.recoveryEnabled;if(nf(e)){let s=e;if(a){let l=this.findReSyncTokenType();if(this.isInCurrentRuleReSyncSet(l))if(s.resyncedTokens=this.reSyncTo(l),this.outputCst){let u=this.CST_STACK[this.CST_STACK.length-1];return u.recoveredNode=!0,u}else return n(e);else{if(this.outputCst){let u=this.CST_STACK[this.CST_STACK.length-1];u.recoveredNode=!0,s.partialCstResult=u}throw s}}else{if(i)return this.moveToTerminatedState(),n(e);throw s}}else throw e}optionInternal(e,r){let n=this.getKeyForAutomaticLookahead(512,r);return this.optionInternalLogic(e,r,n)}optionInternalLogic(e,r,n){let i=this.getLaFuncFromCache(n),a;if(typeof e!="function"){a=e.DEF;let s=e.GATE;if(s!==void 0){let l=i;i=o(()=>s.call(this)&&l.call(this),"lookAheadFunc")}}else a=e;if(i.call(this)===!0)return a.call(this)}atLeastOneInternal(e,r){let n=this.getKeyForAutomaticLookahead(1024,e);return this.atLeastOneInternalLogic(e,r,n)}atLeastOneInternalLogic(e,r,n){let i=this.getLaFuncFromCache(n),a;if(typeof r!="function"){a=r.DEF;let s=r.GATE;if(s!==void 0){let l=i;i=o(()=>s.call(this)&&l.call(this),"lookAheadFunc")}}else a=r;if(i.call(this)===!0){let s=this.doSingleRepetition(a);for(;i.call(this)===!0&&s===!0;)s=this.doSingleRepetition(a)}else throw this.raiseEarlyExitException(e,$n.REPETITION_MANDATORY,r.ERR_MSG);this.attemptInRepetitionRecovery(this.atLeastOneInternal,[e,r],i,1024,e,YT)}atLeastOneSepFirstInternal(e,r){let n=this.getKeyForAutomaticLookahead(1536,e);this.atLeastOneSepFirstInternalLogic(e,r,n)}atLeastOneSepFirstInternalLogic(e,r,n){let i=r.DEF,a=r.SEP;if(this.getLaFuncFromCache(n).call(this)===!0){i.call(this);let l=o(()=>this.tokenMatcher(this.LA(1),a),"separatorLookAheadFunc");for(;this.tokenMatcher(this.LA(1),a)===!0;)this.CONSUME(a),i.call(this);this.attemptInRepetitionRecovery(this.repetitionSepSecondInternal,[e,a,l,i,l2],l,1536,e,l2)}else throw this.raiseEarlyExitException(e,$n.REPETITION_MANDATORY_WITH_SEPARATOR,r.ERR_MSG)}manyInternal(e,r){let n=this.getKeyForAutomaticLookahead(768,e);return this.manyInternalLogic(e,r,n)}manyInternalLogic(e,r,n){let i=this.getLaFuncFromCache(n),a;if(typeof r!="function"){a=r.DEF;let l=r.GATE;if(l!==void 0){let u=i;i=o(()=>l.call(this)&&u.call(this),"lookaheadFunction")}}else a=r;let s=!0;for(;i.call(this)===!0&&s===!0;)s=this.doSingleRepetition(a);this.attemptInRepetitionRecovery(this.manyInternal,[e,r],i,768,e,HT,s)}manySepFirstInternal(e,r){let n=this.getKeyForAutomaticLookahead(1280,e);this.manySepFirstInternalLogic(e,r,n)}manySepFirstInternalLogic(e,r,n){let i=r.DEF,a=r.SEP;if(this.getLaFuncFromCache(n).call(this)===!0){i.call(this);let l=o(()=>this.tokenMatcher(this.LA(1),a),"separatorLookAheadFunc");for(;this.tokenMatcher(this.LA(1),a)===!0;)this.CONSUME(a),i.call(this);this.attemptInRepetitionRecovery(this.repetitionSepSecondInternal,[e,a,l,i,o2],l,1280,e,o2)}}repetitionSepSecondInternal(e,r,n,i,a){for(;n();)this.CONSUME(r),i.call(this);this.attemptInRepetitionRecovery(this.repetitionSepSecondInternal,[e,r,n,i,a],n,1536,e,a)}doSingleRepetition(e){let r=this.getLexerPosition();return e.call(this),this.getLexerPosition()>r}orInternal(e,r){let n=this.getKeyForAutomaticLookahead(256,r),i=wt(e)?e:e.DEF,s=this.getLaFuncFromCache(n).call(this,i);if(s!==void 0)return i[s].ALT.call(this);this.raiseNoAltException(r,e.ERR_MSG)}ruleFinallyStateUpdate(){if(this.RULE_STACK.pop(),this.RULE_OCCURRENCE_STACK.pop(),this.cstFinallyStateUpdate(),this.RULE_STACK.length===0&&this.isAtEndOfInput()===!1){let e=this.LA(1),r=this.errorMessageProvider.buildNotAllInputParsedMessage({firstRedundant:e,ruleName:this.getCurrRuleFullName()});this.SAVE_ERROR(new p2(r,e))}}subruleInternal(e,r,n){let i;try{let a=n!==void 0?n.ARGS:void 0;return this.subruleIdx=r,i=e.apply(this,a),this.cstPostNonTerminal(i,n!==void 0&&n.LABEL!==void 0?n.LABEL:e.ruleName),i}catch(a){throw this.subruleInternalError(a,n,e.ruleName)}}subruleInternalError(e,r,n){throw nf(e)&&e.partialCstResult!==void 0&&(this.cstPostNonTerminal(e.partialCstResult,r!==void 0&&r.LABEL!==void 0?r.LABEL:n),delete e.partialCstResult),e}consumeInternal(e,r,n){let i;try{let a=this.LA(1);this.tokenMatcher(a,e)===!0?(this.consumeToken(),i=a):this.consumeInternalError(e,a,n)}catch(a){i=this.consumeInternalRecovery(e,r,a)}return this.cstPostTerminal(n!==void 0&&n.LABEL!==void 0?n.LABEL:e.name,i),i}consumeInternalError(e,r,n){let i,a=this.LA(0);throw n!==void 0&&n.ERR_MSG?i=n.ERR_MSG:i=this.errorMessageProvider.buildMismatchTokenMessage({expected:e,actual:r,previous:a,ruleName:this.getCurrRuleFullName()}),this.SAVE_ERROR(new c0(i,r,a))}consumeInternalRecovery(e,r,n){if(this.recoveryEnabled&&n.name==="MismatchedTokenException"&&!this.isBackTracking()){let i=this.getFollowsForInRuleRecovery(e,r);try{return this.tryInRuleRecovery(e,i)}catch(a){throw a.name===SN?n:a}}else throw n}saveRecogState(){let e=this.errors,r=Qr(this.RULE_STACK);return{errors:e,lexerState:this.exportLexerState(),RULE_STACK:r,CST_STACK:this.CST_STACK}}reloadRecogState(e){this.errors=e.errors,this.importLexerState(e.lexerState),this.RULE_STACK=e.RULE_STACK}ruleInvocationStateUpdate(e,r,n){this.RULE_OCCURRENCE_STACK.push(n),this.RULE_STACK.push(e),this.cstInvocationStateUpdate(r)}isBackTracking(){return this.isBackTrackingStack.length!==0}getCurrRuleFullName(){let e=this.getLastExplicitRuleShortName();return this.shortRuleNameToFull[e]}shortRuleNameToFullName(e){return this.shortRuleNameToFull[e]}isAtEndOfInput(){return this.tokenMatcher(this.LA(1),fo)}reset(){this.resetLexerState(),this.subruleIdx=0,this.isBackTrackingStack=[],this.errors=[],this.RULE_STACK=[],this.CST_STACK=[],this.RULE_OCCURRENCE_STACK=[]}}});var uk,kae=R(()=>{"use strict";ag();Pt();ng();Ns();uk=class{static{o(this,"ErrorHandler")}initErrorHandler(e){this._errors=[],this.errorMessageProvider=Xe(e,"errorMessageProvider")?e.errorMessageProvider:is.errorMessageProvider}SAVE_ERROR(e){if(nf(e))return e.context={ruleStack:this.getHumanReadableRuleStack(),ruleOccurrenceStack:Qr(this.RULE_OCCURRENCE_STACK)},this._errors.push(e),e;throw Error("Trying to save an Error which is not a RecognitionException")}get errors(){return Qr(this._errors)}set errors(e){this._errors=e}raiseEarlyExitException(e,r,n){let i=this.getCurrRuleFullName(),a=this.getGAstProductions()[i],l=rg(e,a,r,this.maxLookahead)[0],u=[];for(let f=1;f<=this.maxLookahead;f++)u.push(this.LA(f));let h=this.errorMessageProvider.buildEarlyExitMessage({expectedIterationPaths:l,actual:u,previous:this.LA(0),customUserDescription:n,ruleName:i});throw this.SAVE_ERROR(new m2(h,this.LA(1),this.LA(0)))}raiseNoAltException(e,r){let n=this.getCurrRuleFullName(),i=this.getGAstProductions()[n],a=tg(e,i,this.maxLookahead),s=[];for(let h=1;h<=this.maxLookahead;h++)s.push(this.LA(h));let l=this.LA(0),u=this.errorMessageProvider.buildNoViableAltMessage({expectedPathsPerAlt:a,actual:s,previous:l,customUserDescription:r,ruleName:this.getCurrRuleFullName()});throw this.SAVE_ERROR(new d2(u,this.LA(1),l))}}});var hk,Eae=R(()=>{"use strict";c2();Pt();hk=class{static{o(this,"ContentAssist")}initContentAssist(){}computeContentAssist(e,r){let n=this.gastProductionsCache[e];if(er(n))throw Error(`Rule ->${e}<- does not exist in this grammar.`);return qT([n],r,this.tokenMatcher,this.maxLookahead)}getNextPossibleTokenTypes(e){let r=na(e.ruleStack),i=this.getGAstProductions()[r];return new UT(i,e).startWalking()}}});function y2(t,e,r,n=!1){dk(r);let i=ma(this.recordingProdStack),a=wi(e)?e:e.DEF,s=new t({definition:[],idx:r});return n&&(s.separator=e.SEP),Xe(e,"MAX_LOOKAHEAD")&&(s.maxLookahead=e.MAX_LOOKAHEAD),this.recordingProdStack.push(s),a.call(this),i.definition.push(s),this.recordingProdStack.pop(),pk}function UIe(t,e){dk(e);let r=ma(this.recordingProdStack),n=wt(t)===!1,i=n===!1?t:t.DEF,a=new gn({definition:[],idx:e,ignoreAmbiguities:n&&t.IGNORE_AMBIGUITIES===!0});Xe(t,"MAX_LOOKAHEAD")&&(a.maxLookahead=t.MAX_LOOKAHEAD);let s=Nv(i,l=>wi(l.GATE));return a.hasPredicates=s,r.definition.push(a),Ee(i,l=>{let u=new Sn({definition:[]});a.definition.push(u),Xe(l,"IGNORE_AMBIGUITIES")?u.ignoreAmbiguities=l.IGNORE_AMBIGUITIES:Xe(l,"GATE")&&(u.ignoreAmbiguities=!0),this.recordingProdStack.push(u),l.ALT.call(this),this.recordingProdStack.pop()}),pk}function Aae(t){return t===0?"":`${t}`}function dk(t){if(t<0||t>Sae){let e=new Error(`Invalid DSL Method idx value: <${t}> - Idx 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Ge=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),Ae.length-1&&(this.yylineno-=Ae.length-1);var Me=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:Ae?(Ae.length===Ge.length?this.yylloc.first_column:0)+Ge[Ge.length-Ae.length].length-Ae[0].length:this.yylloc.first_column-Ce},this.options.ranges&&(this.yylloc.range=[Me[0],Me[0]+this.yyleng-Ce]),this.yyleng=this.yytext.length,this},"unput"),more:o(function(){return this._more=!0,this},"more"),reject:o(function(){if(this.options.backtrack_lexer)this._backtrack=!0;else return this.parseError("Lexical error on line "+(this.yylineno+1)+`. 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32:this.popState(),this.begin("point_y");break;case 33:return this.popState(),46;break;case 34:return 28;case 35:return 4;case 36:return 11;case 37:return 64;case 38:return 10;case 39:return 65;case 40:return 65;case 41:return 14;case 42:return 13;case 43:return 67;case 44:return 66;case 45:return 12;case 46:return 8;case 47:return 5;case 48:return 18;case 49:return 56;case 50:return 63;case 51:return 57}},"anonymous"),rules:[/^(?:%%(?!\{)[^\n]*)/i,/^(?:[^\}]%%[^\n]*)/i,/^(?:[\n\r]+)/i,/^(?:%%[^\n]*)/i,/^(?:title\b)/i,/^(?:(?!\n||)*[^\n]*)/i,/^(?:accTitle\s*:\s*)/i,/^(?:(?!\n||)*[^\n]*)/i,/^(?:accDescr\s*:\s*)/i,/^(?:(?!\n||)*[^\n]*)/i,/^(?:accDescr\s*\{\s*)/i,/^(?:[\}])/i,/^(?:[^\}]*)/i,/^(?: *x-axis *)/i,/^(?: *y-axis *)/i,/^(?: *--+> *)/i,/^(?: *quadrant-1 *)/i,/^(?: *quadrant-2 *)/i,/^(?: *quadrant-3 *)/i,/^(?: *quadrant-4 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Map,V.info("clear called")}setData(e){this.data={...this.data,...e}}addPoints(e){this.data.points=[...e,...this.data.points]}addClass(e,r){this.classes.set(e,r)}setConfig(e){V.trace("setConfig called with: ",e),this.config={...this.config,...e}}setThemeConfig(e){V.trace("setThemeConfig called with: ",e),this.themeConfig={...this.themeConfig,...e}}calculateSpace(e,r,n,i){let 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this.textDimensionCalculator.getMaxDimension(this.getTickValues().map(e=>e.toString()),this.axisConfig.labelFontSize)}recalculateOuterPaddingToDrawBar(){.7*this.getTickDistance()>this.outerPadding*2&&(this.outerPadding=Math.floor(.7*this.getTickDistance()/2)),this.recalculateScale()}calculateSpaceIfDrawnHorizontally(e){let r=e.height;if(this.axisConfig.showAxisLine&&r>this.axisConfig.axisLineWidth&&(r-=this.axisConfig.axisLineWidth,this.showAxisLine=!0),this.axisConfig.showLabel){let n=this.getLabelDimension(),i=.2*e.width;this.outerPadding=Math.min(n.width/2,i);let a=n.height+this.axisConfig.labelPadding*2;this.labelTextHeight=n.height,a<=r&&(r-=a,this.showLabel=!0)}if(this.axisConfig.showTick&&r>=this.axisConfig.tickLength&&(this.showTick=!0,r-=this.axisConfig.tickLength),this.axisConfig.showTitle&&this.title){let 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49:te[oe-1].draw="participant",te[oe-1].type="addParticipant",this.$=te[oe-1];break;case 50:te[oe-3].draw="actor",te[oe-3].type="addParticipant",te[oe-3].description=ce.parseMessage(te[oe-1]),this.$=te[oe-3];break;case 51:te[oe-1].draw="actor",te[oe-1].type="addParticipant",this.$=te[oe-1];break;case 52:te[oe-1].type="destroyParticipant",this.$=te[oe-1];break;case 53:this.$=[te[oe-1],{type:"addNote",placement:te[oe-2],actor:te[oe-1].actor,text:te[oe]}];break;case 54:te[oe-2]=[].concat(te[oe-1],te[oe-1]).slice(0,2),te[oe-2][0]=te[oe-2][0].actor,te[oe-2][1]=te[oe-2][1].actor,this.$=[te[oe-1],{type:"addNote",placement:ce.PLACEMENT.OVER,actor:te[oe-2].slice(0,2),text:te[oe]}];break;case 55:this.$=[te[oe-1],{type:"addLinks",actor:te[oe-1].actor,text:te[oe]}];break;case 56:this.$=[te[oe-1],{type:"addALink",actor:te[oe-1].actor,text:te[oe]}];break;case 57:this.$=[te[oe-1],{type:"addProperties",actor:te[oe-1].actor,text:te[oe]}];break;case 58:this.$=[te[oe-1],{type:"addDetails",actor:te[oe-1].actor,text:te[oe]}];break;case 61:this.$=[te[oe-2],te[oe]];break;case 62:this.$=te[oe];break;case 63:this.$=ce.PLACEMENT.LEFTOF;break;case 64:this.$=ce.PLACEMENT.RIGHTOF;break;case 65:this.$=[te[oe-4],te[oe-1],{type:"addMessage",from:te[oe-4].actor,to:te[oe-1].actor,signalType:te[oe-3],msg:te[oe],activate:!0},{type:"activeStart",signalType:ce.LINETYPE.ACTIVE_START,actor:te[oe-1].actor}];break;case 66:this.$=[te[oe-4],te[oe-1],{type:"addMessage",from:te[oe-4].actor,to:te[oe-1].actor,signalType:te[oe-3],msg:te[oe]},{type:"activeEnd",signalType:ce.LINETYPE.ACTIVE_END,actor:te[oe-4].actor}];break;case 67:this.$=[te[oe-3],te[oe-1],{type:"addMessage",from:te[oe-3].actor,to:te[oe-1].actor,signalType:te[oe-2],msg:te[oe]}];break;case 68:this.$={type:"addParticipant",actor:te[oe]};break;case 69:this.$=ce.LINETYPE.SOLID_OPEN;break;case 70:this.$=ce.LINETYPE.DOTTED_OPEN;break;case 71:this.$=ce.LINETYPE.SOLID;break;case 72:this.$=ce.LINETYPE.BIDIRECTIONAL_SOLID;break;case 73:this.$=ce.LINETYPE.DOTTED;break;case 74:this.$=ce.LINETYPE.BIDIRECTIONAL_DOTTED;break;case 75:this.$=ce.LINETYPE.SOLID_CROSS;break;case 76:this.$=ce.LINETYPE.DOTTED_CROSS;break;case 77:this.$=ce.LINETYPE.SOLID_POINT;break;case 78:this.$=ce.LINETYPE.DOTTED_POINT;break;case 79:this.$=ce.parseMessage(te[oe].trim().substring(1));break}},"anonymous"),table:[{3:1,4:e,5:r,6:n},{1:[3]},{3:5,4:e,5:r,6:n},{3:6,4:e,5:r,6:n},t([1,4,5,13,14,18,21,23,29,30,31,33,35,36,37,38,39,41,43,44,46,50,52,53,54,59,60,61,62,70],i,{7:7}),{1:[2,1]},{1:[2,2]},{1:[2,3],4:a,5:s,8:8,9:10,12:12,13:l,14:u,17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},t(F,[2,5]),{9:47,12:12,13:l,14:u,17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},t(F,[2,7]),t(F,[2,8]),t(F,[2,14]),{12:48,50:L,52:M,53:N},{15:[1,49]},{5:[1,50]},{5:[1,53],19:[1,51],20:[1,52]},{22:54,70:P},{22:55,70:P},{5:[1,56]},{5:[1,57]},{5:[1,58]},{5:[1,59]},{5:[1,60]},t(F,[2,29]),t(F,[2,30]),{32:[1,61]},{34:[1,62]},t(F,[2,33]),{15:[1,63]},{15:[1,64]},{15:[1,65]},{15:[1,66]},{15:[1,67]},{15:[1,68]},{15:[1,69]},{15:[1,70]},{22:71,70:P},{22:72,70:P},{22:73,70:P},{67:74,71:[1,75],72:[1,76],73:[1,77],74:[1,78],75:[1,79],76:[1,80],77:[1,81],78:[1,82],79:[1,83],80:[1,84]},{55:85,57:[1,86],65:[1,87],66:[1,88]},{22:89,70:P},{22:90,70:P},{22:91,70:P},{22:92,70:P},t([5,51,64,71,72,73,74,75,76,77,78,79,80,81],[2,68]),t(F,[2,6]),t(F,[2,15]),t(B,[2,9],{10:93}),t(F,[2,17]),{5:[1,95],19:[1,94]},{5:[1,96]},t(F,[2,21]),{5:[1,97]},{5:[1,98]},t(F,[2,24]),t(F,[2,25]),t(F,[2,26]),t(F,[2,27]),t(F,[2,28]),t(F,[2,31]),t(F,[2,32]),t($,i,{7:99}),t($,i,{7:100}),t($,i,{7:101}),t(z,i,{40:102,7:103}),t(Y,i,{42:104,7:105}),t(Y,i,{7:105,42:106}),t(Q,i,{45:107,7:108}),t($,i,{7:109}),{5:[1,111],51:[1,110]},{5:[1,113],51:[1,112]},{5:[1,114]},{22:117,68:[1,115],69:[1,116],70:P},t(X,[2,69]),t(X,[2,70]),t(X,[2,71]),t(X,[2,72]),t(X,[2,73]),t(X,[2,74]),t(X,[2,75]),t(X,[2,76]),t(X,[2,77]),t(X,[2,78]),{22:118,70:P},{22:120,58:119,70:P},{70:[2,63]},{70:[2,64]},{56:121,81:ie},{56:123,81:ie},{56:124,81:ie},{56:125,81:ie},{4:[1,128],5:[1,130],11:127,12:129,16:[1,126],50:L,52:M,53:N},{5:[1,131]},t(F,[2,19]),t(F,[2,20]),t(F,[2,22]),t(F,[2,23]),{4:a,5:s,8:8,9:10,12:12,13:l,14:u,16:[1,132],17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},{4:a,5:s,8:8,9:10,12:12,13:l,14:u,16:[1,133],17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},{4:a,5:s,8:8,9:10,12:12,13:l,14:u,16:[1,134],17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},{16:[1,135]},{4:a,5:s,8:8,9:10,12:12,13:l,14:u,16:[2,46],17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,49:[1,136],50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},{16:[1,137]},{4:a,5:s,8:8,9:10,12:12,13:l,14:u,16:[2,44],17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,48:[1,138],50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},{16:[1,139]},{16:[1,140]},{4:a,5:s,8:8,9:10,12:12,13:l,14:u,16:[2,42],17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,47:[1,141],50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},{4:a,5:s,8:8,9:10,12:12,13:l,14:u,16:[1,142],17:15,18:h,21:f,22:40,23:d,24:19,25:20,26:21,27:22,28:23,29:p,30:m,31:g,33:y,35:v,36:x,37:b,38:w,39:S,41:T,43:E,44:_,46:A,50:L,52:M,53:N,54:k,59:I,60:C,61:O,62:D,70:P},{15:[1,143]},t(F,[2,49]),{15:[1,144]},t(F,[2,51]),t(F,[2,52]),{22:145,70:P},{22:146,70:P},{56:147,81:ie},{56:148,81:ie},{56:149,81:ie},{64:[1,150],81:[2,62]},{5:[2,55]},{5:[2,79]},{5:[2,56]},{5:[2,57]},{5:[2,58]},t(F,[2,16]),t(B,[2,10]),{12:151,50:L,52:M,53:N},t(B,[2,12]),t(B,[2,13]),t(F,[2,18]),t(F,[2,34]),t(F,[2,35]),t(F,[2,36]),t(F,[2,37]),{15:[1,152]},t(F,[2,38]),{15:[1,153]},t(F,[2,39]),t(F,[2,40]),{15:[1,154]},t(F,[2,41]),{5:[1,155]},{5:[1,156]},{56:157,81:ie},{56:158,81:ie},{5:[2,67]},{5:[2,53]},{5:[2,54]},{22:159,70:P},t(B,[2,11]),t(z,i,{7:103,40:160}),t(Y,i,{7:105,42:161}),t(Q,i,{7:108,45:162}),t(F,[2,48]),t(F,[2,50]),{5:[2,65]},{5:[2,66]},{81:[2,61]},{16:[2,47]},{16:[2,45]},{16:[2,43]}],defaultActions:{5:[2,1],6:[2,2],87:[2,63],88:[2,64],121:[2,55],122:[2,79],123:[2,56],124:[2,57],125:[2,58],147:[2,67],148:[2,53],149:[2,54],157:[2,65],158:[2,66],159:[2,61],160:[2,47],161:[2,45],162:[2,43]},parseError:o(function(q,K){if(K.recoverable)this.trace(q);else{var se=new Error(q);throw se.hash=K,se}},"parseError"),parse:o(function(q){var K=this,se=[0],ce=[],ue=[null],te=[],De=this.table,oe="",ke=0,Ie=0,Se=0,Ue=2,Pe=1,_e=te.slice.call(arguments,1),me=Object.create(this.lexer),W={yy:{}};for(var fe in this.yy)Object.prototype.hasOwnProperty.call(this.yy,fe)&&(W.yy[fe]=this.yy[fe]);me.setInput(q,W.yy),W.yy.lexer=me,W.yy.parser=this,typeof me.yylloc>"u"&&(me.yylloc={});var ge=me.yylloc;te.push(ge);var re=me.options&&me.options.ranges;typeof W.yy.parseError=="function"?this.parseError=W.yy.parseError:this.parseError=Object.getPrototypeOf(this).parseError;function he(yt){se.length=se.length-2*yt,ue.length=ue.length-yt,te.length=te.length-yt}o(he,"popStack");function ne(){var yt;return yt=ce.pop()||me.lex()||Pe,typeof yt!="number"&&(yt instanceof Array&&(ce=yt,yt=ce.pop()),yt=K.symbols_[yt]||yt),yt}o(ne,"lex");for(var ae,we,Te,Ce,Ae,Ge,Me={},ye,He,ze,Ze;;){if(Te=se[se.length-1],this.defaultActions[Te]?Ce=this.defaultActions[Te]:((ae===null||typeof ae>"u")&&(ae=ne()),Ce=De[Te]&&De[Te][ae]),typeof Ce>"u"||!Ce.length||!Ce[0]){var gt="";Ze=[];for(ye in De[Te])this.terminals_[ye]&&ye>Ue&&Ze.push("'"+this.terminals_[ye]+"'");me.showPosition?gt="Parse error on line "+(ke+1)+`: -`+me.showPosition()+` -Expecting `+Ze.join(", ")+", got '"+(this.terminals_[ae]||ae)+"'":gt="Parse error on line "+(ke+1)+": Unexpected "+(ae==Pe?"end of input":"'"+(this.terminals_[ae]||ae)+"'"),this.parseError(gt,{text:me.match,token:this.terminals_[ae]||ae,line:me.yylineno,loc:ge,expected:Ze})}if(Ce[0]instanceof Array&&Ce.length>1)throw new Error("Parse Error: multiple actions possible at state: "+Te+", token: "+ae);switch(Ce[0]){case 1:se.push(ae),ue.push(me.yytext),te.push(me.yylloc),se.push(Ce[1]),ae=null,we?(ae=we,we=null):(Ie=me.yyleng,oe=me.yytext,ke=me.yylineno,ge=me.yylloc,Se>0&&Se--);break;case 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this.yy=K||this.yy||{},this._input=q,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},"setInput"),input:o(function(){var q=this._input[0];this.yytext+=q,this.yyleng++,this.offset++,this.match+=q,this.matched+=q;var K=q.match(/(?:\r\n?|\n).*/g);return K?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),q},"input"),unput:o(function(q){var K=q.length,se=q.split(/(?:\r\n?|\n)/g);this._input=q+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-K),this.offset-=K;var ce=this.match.split(/(?:\r\n?|\n)/g);this.match=this.match.substr(0,this.match.length-1),this.matched=this.matched.substr(0,this.matched.length-1),se.length-1&&(this.yylineno-=se.length-1);var ue=this.yylloc.range;return this.yylloc={first_line:this.yylloc.first_line,last_line:this.yylineno+1,first_column:this.yylloc.first_column,last_column:se?(se.length===ce.length?this.yylloc.first_column:0)+ce[ce.length-se.length].length-se[0].length:this.yylloc.first_column-K},this.options.ranges&&(this.yylloc.range=[ue[0],ue[0]+this.yyleng-K]),this.yyleng=this.yytext.length,this},"unput"),more:o(function(){return this._more=!0,this},"more"),reject:o(function(){if(this.options.backtrack_lexer)this._backtrack=!0;else return this.parseError("Lexical error on line "+(this.yylineno+1)+`. You can only invoke reject() in the lexer when the lexer is of the backtracking persuasion (options.backtrack_lexer = true). -`+this.showPosition(),{text:"",token:null,line:this.yylineno});return this},"reject"),less:o(function(q){this.unput(this.match.slice(q))},"less"),pastInput:o(function(){var q=this.matched.substr(0,this.matched.length-this.match.length);return(q.length>20?"...":"")+q.substr(-20).replace(/\n/g,"")},"pastInput"),upcomingInput:o(function(){var q=this.match;return q.length<20&&(q+=this._input.substr(0,20-q.length)),(q.substr(0,20)+(q.length>20?"...":"")).replace(/\n/g,"")},"upcomingInput"),showPosition:o(function(){var q=this.pastInput(),K=new Array(q.length+1).join("-");return q+this.upcomingInput()+` -`+K+"^"},"showPosition"),test_match:o(function(q,K){var se,ce,ue;if(this.options.backtrack_lexer&&(ue={yylineno:this.yylineno,yylloc:{first_line:this.yylloc.first_line,last_line:this.last_line,first_column:this.yylloc.first_column,last_column:this.yylloc.last_column},yytext:this.yytext,match:this.match,matches:this.matches,matched:this.matched,yyleng:this.yyleng,offset:this.offset,_more:this._more,_input:this._input,yy:this.yy,conditionStack:this.conditionStack.slice(0),done:this.done},this.options.ranges&&(ue.yylloc.range=this.yylloc.range.slice(0))),ce=q[0].match(/(?:\r\n?|\n).*/g),ce&&(this.yylineno+=ce.length),this.yylloc={first_line:this.yylloc.last_line,last_line:this.yylineno+1,first_column:this.yylloc.last_column,last_column:ce?ce[ce.length-1].length-ce[ce.length-1].match(/\r?\n?/)[0].length:this.yylloc.last_column+q[0].length},this.yytext+=q[0],this.match+=q[0],this.matches=q,this.yyleng=this.yytext.length,this.options.ranges&&(this.yylloc.range=[this.offset,this.offset+=this.yyleng]),this._more=!1,this._backtrack=!1,this._input=this._input.slice(q[0].length),this.matched+=q[0],se=this.performAction.call(this,this.yy,this,K,this.conditionStack[this.conditionStack.length-1]),this.done&&this._input&&(this.done=!1),se)return se;if(this._backtrack){for(var te in ue)this[te]=ue[te];return!1}return!1},"test_match"),next:o(function(){if(this.done)return this.EOF;this._input||(this.done=!0);var q,K,se,ce;this._more||(this.yytext="",this.match="");for(var ue=this._currentRules(),te=0;teK[0].length)){if(K=se,ce=te,this.options.backtrack_lexer){if(q=this.test_match(se,ue[te]),q!==!1)return q;if(this._backtrack){K=!1;continue}else return!1}else if(!this.options.flex)break}return K?(q=this.test_match(K,ue[ce]),q!==!1?q:!1):this._input===""?this.EOF:this.parseError("Lexical error on line "+(this.yylineno+1)+`. Unrecognized text. -`+this.showPosition(),{text:"",token:null,line:this.yylineno})},"next"),lex:o(function(){var K=this.next();return K||this.lex()},"lex"),begin:o(function(K){this.conditionStack.push(K)},"begin"),popState:o(function(){var K=this.conditionStack.length-1;return K>0?this.conditionStack.pop():this.conditionStack[0]},"popState"),_currentRules:o(function(){return this.conditionStack.length&&this.conditionStack[this.conditionStack.length-1]?this.conditions[this.conditionStack[this.conditionStack.length-1]].rules:this.conditions.INITIAL.rules},"_currentRules"),topState:o(function(K){return K=this.conditionStack.length-1-Math.abs(K||0),K>=0?this.conditionStack[K]:"INITIAL"},"topState"),pushState:o(function(K){this.begin(K)},"pushState"),stateStackSize:o(function(){return this.conditionStack.length},"stateStackSize"),options:{"case-insensitive":!0},performAction:o(function(K,se,ce,ue){var te=ue;switch(ce){case 0:return 5;case 1:break;case 2:break;case 3:break;case 4:break;case 5:break;case 6:return 19;case 7:return this.begin("LINE"),14;break;case 8:return this.begin("ID"),50;break;case 9:return this.begin("ID"),52;break;case 10:return 13;case 11:return this.begin("ID"),53;break;case 12:return se.yytext=se.yytext.trim(),this.begin("ALIAS"),70;break;case 13:return this.popState(),this.popState(),this.begin("LINE"),51;break;case 14:return this.popState(),this.popState(),5;break;case 15:return this.begin("LINE"),36;break;case 16:return this.begin("LINE"),37;break;case 17:return this.begin("LINE"),38;break;case 18:return this.begin("LINE"),39;break;case 19:return this.begin("LINE"),49;break;case 20:return this.begin("LINE"),41;break;case 21:return this.begin("LINE"),43;break;case 22:return this.begin("LINE"),48;break;case 23:return this.begin("LINE"),44;break;case 24:return this.begin("LINE"),47;break;case 25:return this.begin("LINE"),46;break;case 26:return this.popState(),15;break;case 27:return 16;case 28:return 65;case 29:return 66;case 30:return 59;case 31:return 60;case 32:return 61;case 33:return 62;case 34:return 57;case 35:return 54;case 36:return this.begin("ID"),21;break;case 37:return this.begin("ID"),23;break;case 38:return 29;case 39:return 30;case 40:return this.begin("acc_title"),31;break;case 41:return this.popState(),"acc_title_value";break;case 42:return this.begin("acc_descr"),33;break;case 43:return this.popState(),"acc_descr_value";break;case 44:this.begin("acc_descr_multiline");break;case 45:this.popState();break;case 46:return"acc_descr_multiline_value";case 47:return 6;case 48:return 18;case 49:return 20;case 50:return 64;case 51:return 5;case 52:return se.yytext=se.yytext.trim(),70;break;case 53:return 73;case 54:return 74;case 55:return 75;case 56:return 76;case 57:return 71;case 58:return 72;case 59:return 77;case 60:return 78;case 61:return 79;case 62:return 80;case 63:return 81;case 64:return 68;case 65:return 69;case 66:return 5;case 67:return"INVALID"}},"anonymous"),rules:[/^(?:[\n]+)/i,/^(?:\s+)/i,/^(?:((?!\n)\s)+)/i,/^(?:#[^\n]*)/i,/^(?:%(?!\{)[^\n]*)/i,/^(?:[^\}]%%[^\n]*)/i,/^(?:[0-9]+(?=[ \n]+))/i,/^(?:box\b)/i,/^(?:participant\b)/i,/^(?:actor\b)/i,/^(?:create\b)/i,/^(?:destroy\b)/i,/^(?:[^\<->\->:\n,;]+?([\-]*[^\<->\->:\n,;]+?)*?(?=((?!\n)\s)+as(?!\n)\s|[#\n;]|$))/i,/^(?:as\b)/i,/^(?:(?:))/i,/^(?:loop\b)/i,/^(?:rect\b)/i,/^(?:opt\b)/i,/^(?:alt\b)/i,/^(?:else\b)/i,/^(?:par\b)/i,/^(?:par_over\b)/i,/^(?:and\b)/i,/^(?:critical\b)/i,/^(?:option\b)/i,/^(?:break\b)/i,/^(?:(?:[:]?(?:no)?wrap)?[^#\n;]*)/i,/^(?:end\b)/i,/^(?:left of\b)/i,/^(?:right of\b)/i,/^(?:links\b)/i,/^(?:link\b)/i,/^(?:properties\b)/i,/^(?:details\b)/i,/^(?:over\b)/i,/^(?:note\b)/i,/^(?:activate\b)/i,/^(?:deactivate\b)/i,/^(?:title\s[^#\n;]+)/i,/^(?:title:\s[^#\n;]+)/i,/^(?:accTitle\s*:\s*)/i,/^(?:(?!\n||)*[^\n]*)/i,/^(?:accDescr\s*:\s*)/i,/^(?:(?!\n||)*[^\n]*)/i,/^(?:accDescr\s*\{\s*)/i,/^(?:[\}])/i,/^(?:[^\}]*)/i,/^(?:sequenceDiagram\b)/i,/^(?:autonumber\b)/i,/^(?:off\b)/i,/^(?:,)/i,/^(?:;)/i,/^(?:[^\+\<->\->:\n,;]+((?!(-x|--x|-\)|--\)))[\-]*[^\+\<->\->:\n,;]+)*)/i,/^(?:->>)/i,/^(?:<<->>)/i,/^(?:-->>)/i,/^(?:<<-->>)/i,/^(?:->)/i,/^(?:-->)/i,/^(?:-[x])/i,/^(?:--[x])/i,/^(?:-[\)])/i,/^(?:--[\)])/i,/^(?::(?:(?:no)?wrap)?[^#\n;]+)/i,/^(?:\+)/i,/^(?:-)/i,/^(?:$)/i,/^(?:.)/i],conditions:{acc_descr_multiline:{rules:[45,46],inclusive:!1},acc_descr:{rules:[43],inclusive:!1},acc_title:{rules:[41],inclusive:!1},ID:{rules:[2,3,12],inclusive:!1},ALIAS:{rules:[2,3,13,14],inclusive:!1},LINE:{rules:[2,3,26],inclusive:!1},INITIAL:{rules:[0,1,3,4,5,6,7,8,9,10,11,15,16,17,18,19,20,21,22,23,24,25,27,28,29,30,31,32,33,34,35,36,37,38,39,40,42,44,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67],inclusive:!0}}};return H}();j.lexer=J;function Z(){this.yy={}}return o(Z,"Parser"),Z.prototype=j,j.Parser=Z,new Z}();hO.parser=hO;Wue=hO});function dO(t,e){if(t.links==null)t.links=e;else for(let r in e)t.links[r]=e[r]}function Zue(t,e){if(t.properties==null)t.properties=e;else for(let r in e)t.properties[r]=e[r]}function SGe(){Mt.records.currentBox=void 0}var Mt,aGe,fO,sGe,oGe,pi,lGe,cGe,uGe,hGe,fGe,dGe,pGe,xx,mGe,gGe,yGe,vGe,xGe,Xue,A0,bGe,wGe,TGe,vx,kGe,EGe,jue,Kue,CGe,Que,Jue,AGe,ehe,pO,the=R(()=>{"use strict";_t();ut();Jk();rr();bi();Mt=new uf(()=>({prevActor:void 0,actors:new Map,createdActors:new Map,destroyedActors:new Map,boxes:[],messages:[],notes:[],sequenceNumbersEnabled:!1,wrapEnabled:void 0,currentBox:void 0,lastCreated:void 0,lastDestroyed:void 0})),aGe=o(function(t){Mt.records.boxes.push({name:t.text,wrap:t.wrap??A0(),fill:t.color,actorKeys:[]}),Mt.records.currentBox=Mt.records.boxes.slice(-1)[0]},"addBox"),fO=o(function(t,e,r,n){let i=Mt.records.currentBox,a=Mt.records.actors.get(t);if(a){if(Mt.records.currentBox&&a.box&&Mt.records.currentBox!==a.box)throw new Error(`A same participant should only be defined in one Box: ${a.name} can't be in '${a.box.name}' and in '${Mt.records.currentBox.name}' at the same time.`);if(i=a.box?a.box:Mt.records.currentBox,a.box=i,a&&e===a.name&&r==null)return}if(r?.text==null&&(r={text:e,type:n}),(n==null||r.text==null)&&(r={text:e,type:n}),Mt.records.actors.set(t,{box:i,name:e,description:r.text,wrap:r.wrap??A0(),prevActor:Mt.records.prevActor,links:{},properties:{},actorCnt:null,rectData:null,type:n??"participant"}),Mt.records.prevActor){let s=Mt.records.actors.get(Mt.records.prevActor);s&&(s.nextActor=t)}Mt.records.currentBox&&Mt.records.currentBox.actorKeys.push(t),Mt.records.prevActor=t},"addActor"),sGe=o(t=>{let e,r=0;if(!t)return 0;for(e=0;e>-",token:"->>-",line:"1",loc:{first_line:1,last_line:1,first_column:1,last_column:1},expected:["'ACTIVE_PARTICIPANT'"]},s}return Mt.records.messages.push({from:t,to:e,message:r?.text??"",wrap:r?.wrap??A0(),type:n,activate:i}),!0},"addSignal"),lGe=o(function(){return Mt.records.boxes.length>0},"hasAtLeastOneBox"),cGe=o(function(){return Mt.records.boxes.some(t=>t.name)},"hasAtLeastOneBoxWithTitle"),uGe=o(function(){return Mt.records.messages},"getMessages"),hGe=o(function(){return Mt.records.boxes},"getBoxes"),fGe=o(function(){return Mt.records.actors},"getActors"),dGe=o(function(){return Mt.records.createdActors},"getCreatedActors"),pGe=o(function(){return Mt.records.destroyedActors},"getDestroyedActors"),xx=o(function(t){return Mt.records.actors.get(t)},"getActor"),mGe=o(function(){return[...Mt.records.actors.keys()]},"getActorKeys"),gGe=o(function(){Mt.records.sequenceNumbersEnabled=!0},"enableSequenceNumbers"),yGe=o(function(){Mt.records.sequenceNumbersEnabled=!1},"disableSequenceNumbers"),vGe=o(()=>Mt.records.sequenceNumbersEnabled,"showSequenceNumbers"),xGe=o(function(t){Mt.records.wrapEnabled=t},"setWrap"),Xue=o(t=>{if(t===void 0)return{};t=t.trim();let e=/^:?wrap:/.exec(t)!==null?!0:/^:?nowrap:/.exec(t)!==null?!1:void 0;return{cleanedText:(e===void 0?t:t.replace(/^:?(?:no)?wrap:/,"")).trim(),wrap:e}},"extractWrap"),A0=o(()=>Mt.records.wrapEnabled!==void 0?Mt.records.wrapEnabled:de().sequence?.wrap??!1,"autoWrap"),bGe=o(function(){Mt.reset(),vr()},"clear"),wGe=o(function(t){let e=t.trim(),{wrap:r,cleanedText:n}=Xue(e),i={text:n,wrap:r};return V.debug(`parseMessage: ${JSON.stringify(i)}`),i},"parseMessage"),TGe=o(function(t){let e=/^((?:rgba?|hsla?)\s*\(.*\)|\w*)(.*)$/.exec(t),r=e?.[1]?e[1].trim():"transparent",n=e?.[2]?e[2].trim():void 0;if(window?.CSS)window.CSS.supports("color",r)||(r="transparent",n=t.trim());else{let s=new Option().style;s.color=r,s.color!==r&&(r="transparent",n=t.trim())}let{wrap:i,cleanedText:a}=Xue(n);return{text:a?qr(a,de()):void 0,color:r,wrap:i}},"parseBoxData"),vx={SOLID:0,DOTTED:1,NOTE:2,SOLID_CROSS:3,DOTTED_CROSS:4,SOLID_OPEN:5,DOTTED_OPEN:6,LOOP_START:10,LOOP_END:11,ALT_START:12,ALT_ELSE:13,ALT_END:14,OPT_START:15,OPT_END:16,ACTIVE_START:17,ACTIVE_END:18,PAR_START:19,PAR_AND:20,PAR_END:21,RECT_START:22,RECT_END:23,SOLID_POINT:24,DOTTED_POINT:25,AUTONUMBER:26,CRITICAL_START:27,CRITICAL_OPTION:28,CRITICAL_END:29,BREAK_START:30,BREAK_END:31,PAR_OVER_START:32,BIDIRECTIONAL_SOLID:33,BIDIRECTIONAL_DOTTED:34},kGe={FILLED:0,OPEN:1},EGe={LEFTOF:0,RIGHTOF:1,OVER:2},jue=o(function(t,e,r){let n={actor:t,placement:e,message:r.text,wrap:r.wrap??A0()},i=[].concat(t,t);Mt.records.notes.push(n),Mt.records.messages.push({from:i[0],to:i[1],message:r.text,wrap:r.wrap??A0(),type:vx.NOTE,placement:e})},"addNote"),Kue=o(function(t,e){let r=xx(t);try{let n=qr(e.text,de());n=n.replace(/&/g,"&"),n=n.replace(/=/g,"=");let i=JSON.parse(n);dO(r,i)}catch(n){V.error("error while parsing actor link text",n)}},"addLinks"),CGe=o(function(t,e){let r=xx(t);try{let n={},i=qr(e.text,de()),a=i.indexOf("@");i=i.replace(/&/g,"&"),i=i.replace(/=/g,"=");let s=i.slice(0,a-1).trim(),l=i.slice(a+1).trim();n[s]=l,dO(r,n)}catch(n){V.error("error while parsing actor link text",n)}},"addALink");o(dO,"insertLinks");Que=o(function(t,e){let r=xx(t);try{let n=qr(e.text,de()),i=JSON.parse(n);Zue(r,i)}catch(n){V.error("error while parsing actor properties text",n)}},"addProperties");o(Zue,"insertProperties");o(SGe,"boxEnd");Jue=o(function(t,e){let r=xx(t),n=document.getElementById(e.text);try{let i=n.innerHTML,a=JSON.parse(i);a.properties&&Zue(r,a.properties),a.links&&dO(r,a.links)}catch(i){V.error("error while parsing actor details text",i)}},"addDetails"),AGe=o(function(t,e){if(t?.properties!==void 0)return t.properties[e]},"getActorProperty"),ehe=o(function(t){if(Array.isArray(t))t.forEach(function(e){ehe(e)});else switch(t.type){case"sequenceIndex":Mt.records.messages.push({from:void 0,to:void 0,message:{start:t.sequenceIndex,step:t.sequenceIndexStep,visible:t.sequenceVisible},wrap:!1,type:t.signalType});break;case"addParticipant":fO(t.actor,t.actor,t.description,t.draw);break;case"createParticipant":if(Mt.records.actors.has(t.actor))throw new Error("It is not possible to have actors with the same id, even if one is destroyed before the next is created. Use 'AS' aliases to simulate the behavior");Mt.records.lastCreated=t.actor,fO(t.actor,t.actor,t.description,t.draw),Mt.records.createdActors.set(t.actor,Mt.records.messages.length);break;case"destroyParticipant":Mt.records.lastDestroyed=t.actor,Mt.records.destroyedActors.set(t.actor,Mt.records.messages.length);break;case"activeStart":pi(t.actor,void 0,void 0,t.signalType);break;case"activeEnd":pi(t.actor,void 0,void 0,t.signalType);break;case"addNote":jue(t.actor,t.placement,t.text);break;case"addLinks":Kue(t.actor,t.text);break;case"addALink":CGe(t.actor,t.text);break;case"addProperties":Que(t.actor,t.text);break;case"addDetails":Jue(t.actor,t.text);break;case"addMessage":if(Mt.records.lastCreated){if(t.to!==Mt.records.lastCreated)throw new Error("The created participant "+Mt.records.lastCreated.name+" does not have an associated creating message after its declaration. Please check the sequence diagram.");Mt.records.lastCreated=void 0}else if(Mt.records.lastDestroyed){if(t.to!==Mt.records.lastDestroyed&&t.from!==Mt.records.lastDestroyed)throw new Error("The destroyed participant "+Mt.records.lastDestroyed.name+" does not have an associated destroying message after its declaration. Please check the sequence diagram.");Mt.records.lastDestroyed=void 0}pi(t.from,t.to,t.msg,t.signalType,t.activate);break;case"boxStart":aGe(t.boxData);break;case"boxEnd":SGe();break;case"loopStart":pi(void 0,void 0,t.loopText,t.signalType);break;case"loopEnd":pi(void 0,void 0,void 0,t.signalType);break;case"rectStart":pi(void 0,void 0,t.color,t.signalType);break;case"rectEnd":pi(void 0,void 0,void 0,t.signalType);break;case"optStart":pi(void 0,void 0,t.optText,t.signalType);break;case"optEnd":pi(void 0,void 0,void 0,t.signalType);break;case"altStart":pi(void 0,void 0,t.altText,t.signalType);break;case"else":pi(void 0,void 0,t.altText,t.signalType);break;case"altEnd":pi(void 0,void 0,void 0,t.signalType);break;case"setAccTitle":kr(t.text);break;case"parStart":pi(void 0,void 0,t.parText,t.signalType);break;case"and":pi(void 0,void 0,t.parText,t.signalType);break;case"parEnd":pi(void 0,void 0,void 0,t.signalType);break;case"criticalStart":pi(void 0,void 0,t.criticalText,t.signalType);break;case"option":pi(void 0,void 0,t.optionText,t.signalType);break;case"criticalEnd":pi(void 0,void 0,void 0,t.signalType);break;case"breakStart":pi(void 0,void 0,t.breakText,t.signalType);break;case"breakEnd":pi(void 0,void 0,void 0,t.signalType);break}},"apply"),pO={addActor:fO,addMessage:oGe,addSignal:pi,addLinks:Kue,addDetails:Jue,addProperties:Que,autoWrap:A0,setWrap:xGe,enableSequenceNumbers:gGe,disableSequenceNumbers:yGe,showSequenceNumbers:vGe,getMessages:uGe,getActors:fGe,getCreatedActors:dGe,getDestroyedActors:pGe,getActor:xx,getActorKeys:mGe,getActorProperty:AGe,getAccTitle:Ar,getBoxes:hGe,getDiagramTitle:Xr,setDiagramTitle:nn,getConfig:o(()=>de().sequence,"getConfig"),clear:bGe,parseMessage:wGe,parseBoxData:TGe,LINETYPE:vx,ARROWTYPE:kGe,PLACEMENT:EGe,addNote:jue,setAccTitle:kr,apply:ehe,setAccDescription:_r,getAccDescription:Lr,hasAtLeastOneBox:lGe,hasAtLeastOneBoxWithTitle:cGe}});var _Ge,rhe,nhe=R(()=>{"use strict";_Ge=o(t=>`.actor { - stroke: ${t.actorBorder}; - fill: ${t.actorBkg}; - } - - text.actor > tspan { - fill: ${t.actorTextColor}; - stroke: none; - } - - .actor-line { - stroke: ${t.actorLineColor}; - } - - .messageLine0 { - stroke-width: 1.5; - stroke-dasharray: none; - stroke: ${t.signalColor}; - } - - .messageLine1 { - stroke-width: 1.5; - stroke-dasharray: 2, 2; - stroke: ${t.signalColor}; - } - - #arrowhead path { - fill: ${t.signalColor}; - stroke: ${t.signalColor}; - } - - .sequenceNumber { - fill: ${t.sequenceNumberColor}; - } - - #sequencenumber { - fill: ${t.signalColor}; - } - - #crosshead path { - fill: ${t.signalColor}; - stroke: ${t.signalColor}; - } - - .messageText { - fill: ${t.signalTextColor}; - stroke: none; - } - - .labelBox { - stroke: ${t.labelBoxBorderColor}; - fill: ${t.labelBoxBkgColor}; - } - - .labelText, .labelText > tspan { - fill: ${t.labelTextColor}; - stroke: none; - } - - .loopText, .loopText > tspan { - fill: ${t.loopTextColor}; - stroke: none; - } - - .loopLine { - stroke-width: 2px; - stroke-dasharray: 2, 2; 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function(t,e,r,n){let{startx:i,stopx:a,starty:s,message:l,type:u,sequenceIndex:h,sequenceVisible:f}=e,d=Lt.calculateTextDimensions(l,L0(Ne)),p=Ky();p.x=i,p.y=s+10,p.width=a-i,p.class="messageText",p.dy="1em",p.text=l,p.fontFamily=Ne.messageFontFamily,p.fontSize=Ne.messageFontSize,p.fontWeight=Ne.messageFontWeight,p.anchor=Ne.messageAlign,p.valign="center",p.textMargin=Ne.wrapPadding,p.tspan=!1,Ni(p.text)?await bx(t,p,{startx:i,stopx:a,starty:r}):_0(t,p);let m=d.width,g;i===a?Ne.rightAngles?g=t.append("path").attr("d",`M ${i},${r} H ${i+We.getMax(Ne.width/2,m/2)} V ${r+25} H ${i}`):g=t.append("path").attr("d","M "+i+","+r+" C "+(i+60)+","+(r-10)+" "+(i+60)+","+(r+30)+" "+i+","+(r+20)):(g=t.append("line"),g.attr("x1",i),g.attr("y1",r),g.attr("x2",a),g.attr("y2",r)),u===n.db.LINETYPE.DOTTED||u===n.db.LINETYPE.DOTTED_CROSS||u===n.db.LINETYPE.DOTTED_POINT||u===n.db.LINETYPE.DOTTED_OPEN||u===n.db.LINETYPE.BIDIRECTIONAL_DOTTED?(g.style("stroke-dasharray","3, 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y="";Ne.arrowMarkerAbsolute&&(y=window.location.protocol+"//"+window.location.host+window.location.pathname+window.location.search,y=y.replace(/\(/g,"\\("),y=y.replace(/\)/g,"\\)")),g.attr("stroke-width",2),g.attr("stroke","none"),g.style("fill","none"),(u===n.db.LINETYPE.SOLID||u===n.db.LINETYPE.DOTTED)&&g.attr("marker-end","url("+y+"#arrowhead)"),(u===n.db.LINETYPE.BIDIRECTIONAL_SOLID||u===n.db.LINETYPE.BIDIRECTIONAL_DOTTED)&&(g.attr("marker-start","url("+y+"#arrowhead)"),g.attr("marker-end","url("+y+"#arrowhead)")),(u===n.db.LINETYPE.SOLID_POINT||u===n.db.LINETYPE.DOTTED_POINT)&&g.attr("marker-end","url("+y+"#filled-head)"),(u===n.db.LINETYPE.SOLID_CROSS||u===n.db.LINETYPE.DOTTED_CROSS)&&g.attr("marker-end","url("+y+"#crosshead)"),(f||Ne.showSequenceNumbers)&&(g.attr("marker-start","url("+y+"#sequencenumber)"),t.append("text").attr("x",i).attr("y",r+4).attr("font-family","sans-serif").attr("font-size","12px").attr("text-anchor","middle").attr("class","sequenceNumber").text(h))},"drawMessage"),QGe=o(function(t,e,r,n,i,a,s){let l=0,u=0,h,f=0;for(let d of n){let p=e.get(d),m=p.box;h&&h!=m&&(s||Ke.models.addBox(h),u+=Ne.boxMargin+h.margin),m&&m!=h&&(s||(m.x=l+u,m.y=i),u+=m.margin),p.width=p.width||Ne.width,p.height=We.getMax(p.height||Ne.height,Ne.height),p.margin=p.margin||Ne.actorMargin,f=We.getMax(f,p.height),r.get(p.name)&&(u+=p.width/2),p.x=l+u,p.starty=Ke.getVerticalPos(),Ke.insert(p.x,i,p.x+p.width,p.height),l+=p.width+u,p.box&&(p.box.width=l+m.margin-p.box.x),u=p.margin,h=p.box,Ke.models.addActor(p)}h&&!s&&Ke.models.addBox(h),Ke.bumpVerticalPos(f)},"addActorRenderingData"),bO=o(async function(t,e,r,n){if(n){let i=0;Ke.bumpVerticalPos(Ne.boxMargin*2);for(let a of r){let s=e.get(a);s.stopy||(s.stopy=Ke.getVerticalPos());let l=await si.drawActor(t,s,Ne,!0);i=We.getMax(i,l)}Ke.bumpVerticalPos(i+Ne.boxMargin)}else for(let i of r){let a=e.get(i);await si.drawActor(t,a,Ne,!1)}},"drawActors"),fhe=o(function(t,e,r,n){let i=0,a=0;for(let s of r){let l=e.get(s),u=t$e(l),h=si.drawPopup(t,l,u,Ne,Ne.forceMenus,n);h.height>i&&(i=h.height),h.width+l.x>a&&(a=h.width+l.x)}return{maxHeight:i,maxWidth:a}},"drawActorsPopup"),dhe=o(function(t){On(Ne,t),t.fontFamily&&(Ne.actorFontFamily=Ne.noteFontFamily=Ne.messageFontFamily=t.fontFamily),t.fontSize&&(Ne.actorFontSize=Ne.noteFontSize=Ne.messageFontSize=t.fontSize),t.fontWeight&&(Ne.actorFontWeight=Ne.noteFontWeight=Ne.messageFontWeight=t.fontWeight)},"setConf"),bE=o(function(t){return Ke.activations.filter(function(e){return e.actor===t})},"actorActivations"),hhe=o(function(t,e){let r=e.get(t),n=bE(t),i=n.reduce(function(s,l){return We.getMin(s,l.startx)},r.x+r.width/2-1),a=n.reduce(function(s,l){return We.getMax(s,l.stopx)},r.x+r.width/2+1);return[i,a]},"activationBounds");o(Fc,"adjustLoopHeightForWrap");o(ZGe,"adjustCreatedDestroyedData");JGe=o(async function(t,e,r,n){let{securityLevel:i,sequence:a}=de();Ne=a;let s;i==="sandbox"&&(s=$e("#i"+e));let l=i==="sandbox"?$e(s.nodes()[0].contentDocument.body):$e("body"),u=i==="sandbox"?s.nodes()[0].contentDocument:document;Ke.init(),V.debug(n.db);let h=i==="sandbox"?l.select(`[id="${e}"]`):$e(`[id="${e}"]`),f=n.db.getActors(),d=n.db.getCreatedActors(),p=n.db.getDestroyedActors(),m=n.db.getBoxes(),g=n.db.getActorKeys(),y=n.db.getMessages(),v=n.db.getDiagramTitle(),x=n.db.hasAtLeastOneBox(),b=n.db.hasAtLeastOneBoxWithTitle(),w=await e$e(f,y,n);if(Ne.height=await r$e(f,w,m),si.insertComputerIcon(h),si.insertDatabaseIcon(h),si.insertClockIcon(h),x&&(Ke.bumpVerticalPos(Ne.boxMargin),b&&Ke.bumpVerticalPos(m[0].textMaxHeight)),Ne.hideUnusedParticipants===!0){let F=new Set;y.forEach(B=>{F.add(B.from),F.add(B.to)}),g=g.filter(B=>F.has(B))}QGe(h,f,d,g,0,y,!1);let S=await a$e(y,f,w,n);si.insertArrowHead(h),si.insertArrowCrossHead(h),si.insertArrowFilledHead(h),si.insertSequenceNumber(h);function T(F,B){let $=Ke.endActivation(F);$.starty+18>B&&($.starty=B-6,B+=12),si.drawActivation(h,$,B,Ne,bE(F.from).length),Ke.insert($.startx,B-10,$.stopx,B)}o(T,"activeEnd");let E=1,_=1,A=[],L=[],M=0;for(let F of y){let B,$,z;switch(F.type){case n.db.LINETYPE.NOTE:Ke.resetVerticalPos(),$=F.noteModel,await XGe(h,$);break;case n.db.LINETYPE.ACTIVE_START:Ke.newActivation(F,h,f);break;case n.db.LINETYPE.ACTIVE_END:T(F,Ke.getVerticalPos());break;case n.db.LINETYPE.LOOP_START:Fc(S,F,Ne.boxMargin,Ne.boxMargin+Ne.boxTextMargin,Y=>Ke.newLoop(Y));break;case n.db.LINETYPE.LOOP_END:B=Ke.endLoop(),await si.drawLoop(h,B,"loop",Ne),Ke.bumpVerticalPos(B.stopy-Ke.getVerticalPos()),Ke.models.addLoop(B);break;case n.db.LINETYPE.RECT_START:Fc(S,F,Ne.boxMargin,Ne.boxMargin,Y=>Ke.newLoop(void 0,Y.message));break;case n.db.LINETYPE.RECT_END:B=Ke.endLoop(),L.push(B),Ke.models.addLoop(B),Ke.bumpVerticalPos(B.stopy-Ke.getVerticalPos());break;case n.db.LINETYPE.OPT_START:Fc(S,F,Ne.boxMargin,Ne.boxMargin+Ne.boxTextMargin,Y=>Ke.newLoop(Y));break;case n.db.LINETYPE.OPT_END:B=Ke.endLoop(),await si.drawLoop(h,B,"opt",Ne),Ke.bumpVerticalPos(B.stopy-Ke.getVerticalPos()),Ke.models.addLoop(B);break;case n.db.LINETYPE.ALT_START:Fc(S,F,Ne.boxMargin,Ne.boxMargin+Ne.boxTextMargin,Y=>Ke.newLoop(Y));break;case n.db.LINETYPE.ALT_ELSE:Fc(S,F,Ne.boxMargin+Ne.boxTextMargin,Ne.boxMargin,Y=>Ke.addSectionToLoop(Y));break;case n.db.LINETYPE.ALT_END:B=Ke.endLoop(),await si.drawLoop(h,B,"alt",Ne),Ke.bumpVerticalPos(B.stopy-Ke.getVerticalPos()),Ke.models.addLoop(B);break;case n.db.LINETYPE.PAR_START:case n.db.LINETYPE.PAR_OVER_START:Fc(S,F,Ne.boxMargin,Ne.boxMargin+Ne.boxTextMargin,Y=>Ke.newLoop(Y)),Ke.saveVerticalPos();break;case n.db.LINETYPE.PAR_AND:Fc(S,F,Ne.boxMargin+Ne.boxTextMargin,Ne.boxMargin,Y=>Ke.addSectionToLoop(Y));break;case n.db.LINETYPE.PAR_END:B=Ke.endLoop(),await si.drawLoop(h,B,"par",Ne),Ke.bumpVerticalPos(B.stopy-Ke.getVerticalPos()),Ke.models.addLoop(B);break;case n.db.LINETYPE.AUTONUMBER:E=F.message.start||E,_=F.message.step||_,F.message.visible?n.db.enableSequenceNumbers():n.db.disableSequenceNumbers();break;case n.db.LINETYPE.CRITICAL_START:Fc(S,F,Ne.boxMargin,Ne.boxMargin+Ne.boxTextMargin,Y=>Ke.newLoop(Y));break;case n.db.LINETYPE.CRITICAL_OPTION:Fc(S,F,Ne.boxMargin+Ne.boxTextMargin,Ne.boxMargin,Y=>Ke.addSectionToLoop(Y));break;case n.db.LINETYPE.CRITICAL_END:B=Ke.endLoop(),await si.drawLoop(h,B,"critical",Ne),Ke.bumpVerticalPos(B.stopy-Ke.getVerticalPos()),Ke.models.addLoop(B);break;case n.db.LINETYPE.BREAK_START:Fc(S,F,Ne.boxMargin,Ne.boxMargin+Ne.boxTextMargin,Y=>Ke.newLoop(Y));break;case n.db.LINETYPE.BREAK_END:B=Ke.endLoop(),await si.drawLoop(h,B,"break",Ne),Ke.bumpVerticalPos(B.stopy-Ke.getVerticalPos()),Ke.models.addLoop(B);break;default:try{z=F.msgModel,z.starty=Ke.getVerticalPos(),z.sequenceIndex=E,z.sequenceVisible=n.db.showSequenceNumbers();let Y=await jGe(h,z);ZGe(F,z,Y,M,f,d,p),A.push({messageModel:z,lineStartY:Y}),Ke.models.addMessage(z)}catch(Y){V.error("error while drawing message",Y)}}[n.db.LINETYPE.SOLID_OPEN,n.db.LINETYPE.DOTTED_OPEN,n.db.LINETYPE.SOLID,n.db.LINETYPE.DOTTED,n.db.LINETYPE.SOLID_CROSS,n.db.LINETYPE.DOTTED_CROSS,n.db.LINETYPE.SOLID_POINT,n.db.LINETYPE.DOTTED_POINT,n.db.LINETYPE.BIDIRECTIONAL_SOLID,n.db.LINETYPE.BIDIRECTIONAL_DOTTED].includes(F.type)&&(E=E+_),M++}V.debug("createdActors",d),V.debug("destroyedActors",p),await bO(h,f,g,!1);for(let F of A)await KGe(h,F.messageModel,F.lineStartY,n);Ne.mirrorActors&&await bO(h,f,g,!0),L.forEach(F=>si.drawBackgroundRect(h,F)),yO(h,f,g,Ne);for(let F of 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this.yy=me||this.yy||{},this._input=_e,this._more=this._backtrack=this.done=!1,this.yylineno=this.yyleng=0,this.yytext=this.matched=this.match="",this.conditionStack=["INITIAL"],this.yylloc={first_line:1,first_column:0,last_line:1,last_column:0},this.options.ranges&&(this.yylloc.range=[0,0]),this.offset=0,this},"setInput"),input:o(function(){var _e=this._input[0];this.yytext+=_e,this.yyleng++,this.offset++,this.match+=_e,this.matched+=_e;var me=_e.match(/(?:\r\n?|\n).*/g);return me?(this.yylineno++,this.yylloc.last_line++):this.yylloc.last_column++,this.options.ranges&&this.yylloc.range[1]++,this._input=this._input.slice(1),_e},"input"),unput:o(function(_e){var me=_e.length,W=_e.split(/(?:\r\n?|\n)/g);this._input=_e+this._input,this.yytext=this.yytext.substr(0,this.yytext.length-me),this.offset-=me;var 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v=g.value,x=y.value,b=v[0]===x[0]&&v[1]===x[1]&&v[2]===x[2]&&(v[3]===x[3]||(v[3]==null||v[3]===1)&&(x[3]==null||x[3]===1));if(b)return!1}return{name:t,value:d,strValue:""+e,mapped:m,field:d[1],fieldMin:parseFloat(d[2]),fieldMax:parseFloat(d[3]),valueMin:g.value,valueMax:y.value,bypass:r}}}if(h.multiple&&n!=="multiple"){var w;if(u?w=e.split(/\s+/):vn(e)?w=e:w=[e],h.evenMultiple&&w.length%2!==0)return null;for(var S=[],T=[],E=[],_="",A=!1,L=0;L0?" 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i=e.boundingBox(),a=this.width(),s=this.height();r=r===void 0?this._private.zoom:r;var l={x:(a-r*(i.x1+i.x2))/2,y:(s-r*(i.y1+i.y2))/2};return l}}},"getCenterPan"),reset:o(function(){return!this._private.panningEnabled||!this._private.zoomingEnabled?this:(this.viewport({pan:{x:0,y:0},zoom:1}),this)},"reset"),invalidateSize:o(function(){this._private.sizeCache=null},"invalidateSize"),size:o(function(){var e=this._private,r=e.container,n=this;return e.sizeCache=e.sizeCache||(r?function(){var i=n.window().getComputedStyle(r),a=o(function(l){return parseFloat(i.getPropertyValue(l))},"val");return{width:r.clientWidth-a("padding-left")-a("padding-right"),height:r.clientHeight-a("padding-top")-a("padding-bottom")}}():{width:1,height:1})},"size"),width:o(function(){return this.size().width},"width"),height:o(function(){return this.size().height},"height"),extent:o(function(){var e=this._private.pan,r=this._private.zoom,n=this.renderedExtent(),i={x1:(n.x1-e.x)/r,x2:(n.x2-e.x)/r,y1:(n.y1-e.y)/r,y2:(n.y2-e.y)/r};return i.w=i.x2-i.x1,i.h=i.y2-i.y1,i},"extent"),renderedExtent:o(function(){var e=this.width(),r=this.height();return{x1:0,y1:0,x2:e,y2:r,w:e,h:r}},"renderedExtent"),multiClickDebounceTime:o(function(e){if(e)this._private.multiClickDebounceTime=e;else return this._private.multiClickDebounceTime;return this},"multiClickDebounceTime")};q0.centre=q0.center;q0.autolockNodes=q0.autolock;q0.autoungrabifyNodes=q0.autoungrabify;jx={data:en.data({field:"data",bindingEvent:"data",allowBinding:!0,allowSetting:!0,settingEvent:"data",settingTriggersEvent:!0,triggerFnName:"trigger",allowGetting:!0,updateStyle:!0}),removeData:en.removeData({field:"data",event:"data",triggerFnName:"trigger",triggerEvent:!0,updateStyle:!0}),scratch:en.data({field:"scratch",bindingEvent:"scratch",allowBinding:!0,allowSetting:!0,settingEvent:"scratch",settingTriggersEvent:!0,triggerFnName:"trigger",allowGetting:!0,updateStyle:!0}),removeScratch:en.removeData({field:"scratch",event:"scratch",triggerFnName:"trigger",triggerEvent:!0,updateStyle:!0})};jx.attr=jx.data;jx.removeAttr=jx.removeData;Kx=o(function(e){var r=this;e=Wt({},e);var n=e.container;n&&!k6(n)&&k6(n[0])&&(n=n[0]);var i=n?n._cyreg:null;i=i||{},i&&i.cy&&(i.cy.destroy(),i={});var a=i.readies=i.readies||[];n&&(n._cyreg=i),i.cy=r;var s=Vi!==void 0&&n!==void 0&&!e.headless,l=e;l.layout=Wt({name:s?"grid":"null"},l.layout),l.renderer=Wt({name:s?"canvas":"null"},l.renderer);var u=o(function(g,y,v){return y!==void 0?y:v!==void 0?v:g},"defVal"),h=this._private={container:n,ready:!1,options:l,elements:new Ca(this),listeners:[],aniEles:new Ca(this),data:l.data||{},scratch:{},layout:null,renderer:null,destroyed:!1,notificationsEnabled:!0,minZoom:1e-50,maxZoom:1e50,zoomingEnabled:u(!0,l.zoomingEnabled),userZoomingEnabled:u(!0,l.userZoomingEnabled),panningEnabled:u(!0,l.panningEnabled),userPanningEnabled:u(!0,l.userPanningEnabled),boxSelectionEnabled:u(!0,l.boxSelectionEnabled),autolock:u(!1,l.autolock,l.autolockNodes),autoungrabify:u(!1,l.autoungrabify,l.autoungrabifyNodes),autounselectify:u(!1,l.autounselectify),styleEnabled:l.styleEnabled===void 0?s:l.styleEnabled,zoom:ft(l.zoom)?l.zoom:1,pan:{x:Mr(l.pan)&&ft(l.pan.x)?l.pan.x:0,y:Mr(l.pan)&&ft(l.pan.y)?l.pan.y:0},animation:{current:[],queue:[]},hasCompoundNodes:!1,multiClickDebounceTime:u(250,l.multiClickDebounceTime)};this.createEmitter(),this.selectionType(l.selectionType),this.zoomRange({min:l.minZoom,max:l.maxZoom});var f=o(function(g,y){var v=g.some(zHe);if(v)return u1.all(g).then(y);y(g)},"loadExtData");h.styleEnabled&&r.setStyle([]);var d=Wt({},l,l.renderer);r.initRenderer(d);var p=o(function(g,y,v){r.notifications(!1);var x=r.mutableElements();x.length>0&&x.remove(),g!=null&&(Mr(g)||vn(g))&&r.add(g),r.one("layoutready",function(w){r.notifications(!0),r.emit(w),r.one("load",y),r.emitAndNotify("load")}).one("layoutstop",function(){r.one("done",v),r.emit("done")});var b=Wt({},r._private.options.layout);b.eles=r.elements(),r.layout(b).run()},"setElesAndLayout");f([l.style,l.elements],function(m){var g=m[0],y=m[1];h.styleEnabled&&r.style().append(g),p(y,function(){r.startAnimationLoop(),h.ready=!0,jn(l.ready)&&r.on("ready",l.ready);for(var v=0;v0,u=Gs(e.boundingBox?e.boundingBox:{x1:0,y1:0,w:r.width(),h:r.height()}),h;if(xo(e.roots))h=e.roots;else if(vn(e.roots)){for(var f=[],d=0;d0;){var O=C(),D=M(O,k);if(D)O.outgoers().filter(function(ue){return ue.isNode()&&n.has(ue)}).forEach(I);else if(D===null){tn("Detected double maximal shift for node `"+O.id()+"`. 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s=Gs(e.boundingBox?e.boundingBox:{x1:0,y1:0,w:r.width(),h:r.height()}),l={x:s.x1+s.w/2,y:s.y1+s.h/2},u=e.sweep===void 0?2*Math.PI-2*Math.PI/a.length:e.sweep,h=u/Math.max(1,a.length-1),f,d=0,p=0;p1&&e.avoidOverlap){d*=1.75;var x=Math.cos(h)-Math.cos(0),b=Math.sin(h)-Math.sin(0),w=Math.sqrt(d*d/(x*x+b*b));f=Math.max(w,f)}var S=o(function(E,_){var A=e.startAngle+_*h*(i?1:-1),L=f*Math.cos(A),M=f*Math.sin(A),N={x:l.x+L,y:l.y+M};return N},"getPos");return n.nodes().layoutPositions(this,e,S),this};GKe={fit:!0,padding:30,startAngle:3/2*Math.PI,sweep:void 0,clockwise:!0,equidistant:!1,minNodeSpacing:10,boundingBox:void 0,avoidOverlap:!0,nodeDimensionsIncludeLabels:!1,height:void 0,width:void 0,spacingFactor:void 0,concentric:o(function(e){return e.degree()},"concentric"),levelWidth:o(function(e){return e.maxDegree()/4},"levelWidth"),animate:!1,animationDuration:500,animationEasing:void 0,animateFilter:o(function(e,r){return!0},"animateFilter"),ready:void 0,stop:void 0,transform:o(function(e,r){return r},"transform")};o(Lme,"ConcentricLayout");Lme.prototype.run=function(){for(var t=this.options,e=t,r=e.counterclockwise!==void 0?!e.counterclockwise:e.clockwise,n=t.cy,i=e.eles,a=i.nodes().not(":parent"),s=Gs(e.boundingBox?e.boundingBox:{x1:0,y1:0,w:n.width(),h:n.height()}),l={x:s.x1+s.w/2,y:s.y1+s.h/2},u=[],h=0,f=0;f0){var T=Math.abs(b[0].value-S.value);T>=v&&(b=[],x.push(b))}b.push(S)}var E=h+e.minNodeSpacing;if(!e.avoidOverlap){var _=x.length>0&&x[0].length>1,A=Math.min(s.w,s.h)/2-E,L=A/(x.length+_?1:0);E=Math.min(E,L)}for(var M=0,N=0;N1&&e.avoidOverlap){var O=Math.cos(C)-Math.cos(0),D=Math.sin(C)-Math.sin(0),P=Math.sqrt(E*E/(O*O+D*D));M=Math.max(P,M)}k.r=M,M+=E}if(e.equidistant){for(var F=0,B=0,$=0;$=t.numIter||(XKe(n,t),n.temperature=n.temperature*t.coolingFactor,n.temperature=t.animationThreshold&&a(),E6(d)}},"frame");f()}else{for(;h;)h=s(u),u++;rpe(n,t),l()}return this};j6.prototype.stop=function(){return this.stopped=!0,this.thread&&this.thread.stop(),this.emit("layoutstop"),this};j6.prototype.destroy=function(){return this.thread&&this.thread.stop(),this};VKe=o(function(e,r,n){for(var i=n.eles.edges(),a=n.eles.nodes(),s=Gs(n.boundingBox?n.boundingBox:{x1:0,y1:0,w:e.width(),h:e.height()}),l={isCompound:e.hasCompoundNodes(),layoutNodes:[],idToIndex:{},nodeSize:a.size(),graphSet:[],indexToGraph:[],layoutEdges:[],edgeSize:i.size(),temperature:n.initialTemp,clientWidth:s.w,clientHeight:s.h,boundingBox:s},u=n.eles.components(),h={},f=0;f0){l.graphSet.push(A);for(var f=0;fi.count?0:i.graph},"findLCA"),HKe=o(function t(e,r,n,i){var a=i.graphSet[n];if(-10)var d=i.nodeOverlap*f,p=Math.sqrt(l*l+u*u),m=d*l/p,g=d*u/p;else var y=R6(e,l,u),v=R6(r,-1*l,-1*u),x=v.x-y.x,b=v.y-y.y,w=x*x+b*b,p=Math.sqrt(w),d=(e.nodeRepulsion+r.nodeRepulsion)/w,m=d*x/p,g=d*b/p;e.isLocked||(e.offsetX-=m,e.offsetY-=g),r.isLocked||(r.offsetX+=m,r.offsetY+=g)}},"nodeRepulsion"),QKe=o(function(e,r,n,i){if(n>0)var a=e.maxX-r.minX;else var a=r.maxX-e.minX;if(i>0)var s=e.maxY-r.minY;else var s=r.maxY-e.minY;return a>=0&&s>=0?Math.sqrt(a*a+s*s):0},"nodesOverlap"),R6=o(function(e,r,n){var i=e.positionX,a=e.positionY,s=e.height||1,l=e.width||1,u=n/r,h=s/l,f={};return r===0&&0n?(f.x=i,f.y=a+s/2,f):0r&&-1*h<=u&&u<=h?(f.x=i-l/2,f.y=a-l*n/2/r,f):0=h)?(f.x=i+s*r/2/n,f.y=a+s/2,f):(0>n&&(u<=-1*h||u>=h)&&(f.x=i-s*r/2/n,f.y=a-s/2),f)},"findClippingPoint"),ZKe=o(function(e,r){for(var n=0;nn){var v=r.gravity*m/y,x=r.gravity*g/y;p.offsetX+=v,p.offsetY+=x}}}}},"calculateGravityForces"),eQe=o(function(e,r){var n=[],i=0,a=-1;for(n.push.apply(n,e.graphSet[0]),a+=e.graphSet[0].length;i<=a;){var s=n[i++],l=e.idToIndex[s],u=e.layoutNodes[l],h=u.children;if(0n)var a={x:n*e/i,y:n*r/i};else var a={x:e,y:r};return a},"limitForce"),nQe=o(function t(e,r){var n=e.parentId;if(n!=null){var i=r.layoutNodes[r.idToIndex[n]],a=!1;if((i.maxX==null||e.maxX+i.padRight>i.maxX)&&(i.maxX=e.maxX+i.padRight,a=!0),(i.minX==null||e.minX-i.padLefti.maxY)&&(i.maxY=e.maxY+i.padBottom,a=!0),(i.minY==null||e.minY-i.padTopx&&(g+=v+r.componentSpacing,m=0,y=0,v=0)}}},"separateComponents"),iQe={fit:!0,padding:30,boundingBox:void 0,avoidOverlap:!0,avoidOverlapPadding:10,nodeDimensionsIncludeLabels:!1,spacingFactor:void 0,condense:!1,rows:void 0,cols:void 0,position:o(function(e){},"position"),sort:void 0,animate:!1,animationDuration:500,animationEasing:void 0,animateFilter:o(function(e,r){return!0},"animateFilter"),ready:void 0,stop:void 0,transform:o(function(e,r){return r},"transform")};o(Rme,"GridLayout");Rme.prototype.run=function(){var t=this.options,e=t,r=t.cy,n=e.eles,i=n.nodes().not(":parent");e.sort&&(i=i.sort(e.sort));var a=Gs(e.boundingBox?e.boundingBox:{x1:0,y1:0,w:r.width(),h:r.height()});if(a.h===0||a.w===0)n.nodes().layoutPositions(this,e,function(Q){return{x:a.x1,y:a.y1}});else{var s=i.size(),l=Math.sqrt(s*a.h/a.w),u=Math.round(l),h=Math.round(a.w/a.h*l),f=o(function(X){if(X==null)return Math.min(u,h);var ie=Math.min(u,h);ie==u?u=X:h=X},"small"),d=o(function(X){if(X==null)return Math.max(u,h);var ie=Math.max(u,h);ie==u?u=X:h=X},"large"),p=e.rows,m=e.cols!=null?e.cols:e.columns;if(p!=null&&m!=null)u=p,h=m;else if(p!=null&&m==null)u=p,h=Math.ceil(s/u);else if(p==null&&m!=null)h=m,u=Math.ceil(s/h);else if(h*u>s){var g=f(),y=d();(g-1)*y>=s?f(g-1):(y-1)*g>=s&&d(y-1)}else for(;h*u=s?d(x+1):f(v+1)}var b=a.w/h,w=a.h/u;if(e.condense&&(b=0,w=0),e.avoidOverlap)for(var S=0;S=h&&(O=0,C++)},"moveToNextCell"),P={},F=0;F(O=_We(t,e,D[P],D[P+1],D[P+2],D[P+3])))return v(_,O),!0}else if(L.edgeType==="bezier"||L.edgeType==="multibezier"||L.edgeType==="self"||L.edgeType==="compound"){for(var D=L.allpts,P=0;P+5(O=AWe(t,e,D[P],D[P+1],D[P+2],D[P+3],D[P+4],D[P+5])))return v(_,O),!0}for(var F=F||A.source,B=B||A.target,$=i.getArrowWidth(M,N),z=[{name:"source",x:L.arrowStartX,y:L.arrowStartY,angle:L.srcArrowAngle},{name:"target",x:L.arrowEndX,y:L.arrowEndY,angle:L.tgtArrowAngle},{name:"mid-source",x:L.midX,y:L.midY,angle:L.midsrcArrowAngle},{name:"mid-target",x:L.midX,y:L.midY,angle:L.midtgtArrowAngle}],P=0;P0&&(x(F),x(B))}o(b,"checkEdge");function w(_,A,L){return Ul(_,A,L)}o(w,"preprop");function S(_,A){var L=_._private,M=p,N;A?N=A+"-":N="",_.boundingBox();var k=L.labelBounds[A||"main"],I=_.pstyle(N+"label").value,C=_.pstyle("text-events").strValue==="yes";if(!(!C||!I)){var O=w(L.rscratch,"labelX",A),D=w(L.rscratch,"labelY",A),P=w(L.rscratch,"labelAngle",A),F=_.pstyle(N+"text-margin-x").pfValue,B=_.pstyle(N+"text-margin-y").pfValue,$=k.x1-M-F,z=k.x2+M-F,Y=k.y1-M-B,Q=k.y2+M-B;if(P){var X=Math.cos(P),ie=Math.sin(P),j=o(function(ce,ue){return ce=ce-O,ue=ue-D,{x:ce*X-ue*ie+O,y:ce*ie+ue*X+D}},"rotate"),J=j($,Y),Z=j($,Q),H=j(z,Y),q=j(z,Q),K=[J.x+F,J.y+B,H.x+F,H.y+B,q.x+F,q.y+B,Z.x+F,Z.y+B];if(zs(t,e,K))return v(_),!0}else if(s1(k,t,e))return v(_),!0}}o(S,"checkLabel");for(var T=s.length-1;T>=0;T--){var E=s[T];E.isNode()?x(E)||S(E):b(E)||S(E)||S(E,"source")||S(E,"target")}return l};j0.getAllInBox=function(t,e,r,n){var i=this.getCachedZSortedEles().interactive,a=[],s=Math.min(t,r),l=Math.max(t,r),u=Math.min(e,n),h=Math.max(e,n);t=s,r=l,e=u,n=h;for(var f=Gs({x1:t,y1:e,x2:r,y2:n}),d=0;d0?-(Math.PI-e.ang):Math.PI+e.ang},"invertVec"),uQe=o(function(e,r,n,i,a){if(e!==ope?lpe(r,e,Gc):cQe(Jo,Gc),lpe(r,n,Jo),ape=Gc.nx*Jo.ny-Gc.ny*Jo.nx,spe=Gc.nx*Jo.nx-Gc.ny*-Jo.ny,ju=Math.asin(Math.max(-1,Math.min(1,ape))),Math.abs(ju)<1e-6){GP=r.x,$P=r.y,z0=Qg=0;return}G0=1,w6=!1,spe<0?ju<0?ju=Math.PI+ju:(ju=Math.PI-ju,G0=-1,w6=!0):ju>0&&(G0=-1,w6=!0),r.radius!==void 0?Qg=r.radius:Qg=i,O0=ju/2,f6=Math.min(Gc.len/2,Jo.len/2),a?(zc=Math.abs(Math.cos(O0)*Qg/Math.sin(O0)),zc>f6?(zc=f6,z0=Math.abs(zc*Math.sin(O0)/Math.cos(O0))):z0=Qg):(zc=Math.min(f6,Qg),z0=Math.abs(zc*Math.sin(O0)/Math.cos(O0))),VP=r.x+Jo.nx*zc,UP=r.y+Jo.ny*zc,GP=VP-Jo.ny*z0*G0,$P=UP+Jo.nx*z0*G0,Ome=r.x+Gc.nx*zc,Pme=r.y+Gc.ny*zc,ope=r},"calcCornerArc");o(Bme,"drawPreparedRoundCorner");o(mB,"getRoundCorner");$a={};$a.findMidptPtsEtc=function(t,e){var r=e.posPts,n=e.intersectionPts,i=e.vectorNormInverse,a,s=t.pstyle("source-endpoint"),l=t.pstyle("target-endpoint"),u=s.units!=null&&l.units!=null,h=o(function(T,E,_,A){var L=A-E,M=_-T,N=Math.sqrt(M*M+L*L);return{x:-L/N,y:M/N}},"recalcVectorNormInverse"),f=t.pstyle("edge-distances").value;switch(f){case"node-position":a=r;break;case"intersection":a=n;break;case"endpoints":{if(u){var d=this.manualEndptToPx(t.source()[0],s),p=$l(d,2),m=p[0],g=p[1],y=this.manualEndptToPx(t.target()[0],l),v=$l(y,2),x=v[0],b=v[1],w={x1:m,y1:g,x2:x,y2:b};i=h(m,g,x,b),a=w}else tn("Edge ".concat(t.id()," has edge-distances:endpoints specified without manual endpoints specified via source-endpoint and target-endpoint. Falling back on edge-distances:intersection (default).")),a=n;break}}return{midptPts:a,vectorNormInverse:i}};$a.findHaystackPoints=function(t){for(var e=0;e0?Math.max(Te-Ce,0):Math.min(Te+Ce,0)},"subDWH"),I=k(M,A),C=k(N,L),O=!1;b===h?x=Math.abs(I)>Math.abs(C)?i:n:b===u||b===l?(x=n,O=!0):(b===a||b===s)&&(x=i,O=!0);var D=x===n,P=D?C:I,F=D?N:M,B=$pe(F),$=!1;!(O&&(S||E))&&(b===l&&F<0||b===u&&F>0||b===a&&F>0||b===s&&F<0)&&(B*=-1,P=B*Math.abs(P),$=!0);var z;if(S){var Y=T<0?1+T:T;z=Y*P}else{var Q=T<0?P:0;z=Q+T*B}var X=o(function(Te){return Math.abs(Te)<_||Math.abs(Te)>=Math.abs(P)},"getIsTooClose"),ie=X(z),j=X(Math.abs(P)-Math.abs(z)),J=ie||j;if(J&&!$)if(D){var Z=Math.abs(F)<=p/2,H=Math.abs(M)<=m/2;if(Z){var q=(f.x1+f.x2)/2,K=f.y1,se=f.y2;r.segpts=[q,K,q,se]}else if(H){var ce=(f.y1+f.y2)/2,ue=f.x1,te=f.x2;r.segpts=[ue,ce,te,ce]}else r.segpts=[f.x1,f.y2]}else{var De=Math.abs(F)<=d/2,oe=Math.abs(N)<=g/2;if(De){var ke=(f.y1+f.y2)/2,Ie=f.x1,Se=f.x2;r.segpts=[Ie,ke,Se,ke]}else if(oe){var Ue=(f.x1+f.x2)/2,Pe=f.y1,_e=f.y2;r.segpts=[Ue,Pe,Ue,_e]}else r.segpts=[f.x2,f.y1]}else if(D){var me=f.y1+z+(v?p/2*B:0),W=f.x1,fe=f.x2;r.segpts=[W,me,fe,me]}else{var ge=f.x1+z+(v?d/2*B:0),re=f.y1,he=f.y2;r.segpts=[ge,re,ge,he]}if(r.isRound){var ne=t.pstyle("taxi-radius").value,ae=t.pstyle("radius-type").value[0]==="arc-radius";r.radii=new Array(r.segpts.length/2).fill(ne),r.isArcRadius=new Array(r.segpts.length/2).fill(ae)}};$a.tryToCorrectInvalidPoints=function(t,e){var r=t._private.rscratch;if(r.edgeType==="bezier"){var n=e.srcPos,i=e.tgtPos,a=e.srcW,s=e.srcH,l=e.tgtW,u=e.tgtH,h=e.srcShape,f=e.tgtShape,d=e.srcCornerRadius,p=e.tgtCornerRadius,m=e.srcRs,g=e.tgtRs,y=!ft(r.startX)||!ft(r.startY),v=!ft(r.arrowStartX)||!ft(r.arrowStartY),x=!ft(r.endX)||!ft(r.endY),b=!ft(r.arrowEndX)||!ft(r.arrowEndY),w=3,S=this.getArrowWidth(t.pstyle("width").pfValue,t.pstyle("arrow-scale").value)*this.arrowShapeWidth,T=w*S,E=H0({x:r.ctrlpts[0],y:r.ctrlpts[1]},{x:r.startX,y:r.startY}),_=EC.poolIndex()){var O=I;I=C,C=O}var D=L.srcPos=I.position(),P=L.tgtPos=C.position(),F=L.srcW=I.outerWidth(),B=L.srcH=I.outerHeight(),$=L.tgtW=C.outerWidth(),z=L.tgtH=C.outerHeight(),Y=L.srcShape=r.nodeShapes[e.getNodeShape(I)],Q=L.tgtShape=r.nodeShapes[e.getNodeShape(C)],X=L.srcCornerRadius=I.pstyle("corner-radius").value==="auto"?"auto":I.pstyle("corner-radius").pfValue,ie=L.tgtCornerRadius=C.pstyle("corner-radius").value==="auto"?"auto":C.pstyle("corner-radius").pfValue,j=L.tgtRs=C._private.rscratch,J=L.srcRs=I._private.rscratch;L.dirCounts={north:0,west:0,south:0,east:0,northwest:0,southwest:0,northeast:0,southeast:0};for(var Z=0;Z0){var se=a,ce=B0(se,Jg(r)),ue=B0(se,Jg(K)),te=ce;if(ue2){var De=B0(se,{x:K[2],y:K[3]});De0){var he=s,ne=B0(he,Jg(r)),ae=B0(he,Jg(re)),we=ne;if(ae2){var Te=B0(he,{x:re[2],y:re[3]});Te=g||_){v={cp:S,segment:E};break}}if(v)break}var A=v.cp,L=v.segment,M=(g-x)/L.length,N=L.t1-L.t0,k=m?L.t0+N*M:L.t1-N*M;k=Hx(0,k,1),e=t1(A.p0,A.p1,A.p2,k),p=fQe(A.p0,A.p1,A.p2,k);break}case"straight":case"segments":case"haystack":{for(var I=0,C,O,D,P,F=n.allpts.length,B=0;B+3=g));B+=2);var $=g-O,z=$/C;z=Hx(0,z,1),e=yWe(D,P,z),p=Gme(D,P);break}}s("labelX",d,e.x),s("labelY",d,e.y),s("labelAutoAngle",d,p)}},"calculateEndProjection");h("source"),h("target"),this.applyLabelDimensions(t)}};Hc.applyLabelDimensions=function(t){this.applyPrefixedLabelDimensions(t),t.isEdge()&&(this.applyPrefixedLabelDimensions(t,"source"),this.applyPrefixedLabelDimensions(t,"target"))};Hc.applyPrefixedLabelDimensions=function(t,e){var r=t._private,n=this.getLabelText(t,e),i=this.calculateLabelDimensions(t,n),a=t.pstyle("line-height").pfValue,s=t.pstyle("text-wrap").strValue,l=Ul(r.rscratch,"labelWrapCachedLines",e)||[],u=s!=="wrap"?1:Math.max(l.length,1),h=i.height/u,f=h*a,d=i.width,p=i.height+(u-1)*(a-1)*h;Tf(r.rstyle,"labelWidth",e,d),Tf(r.rscratch,"labelWidth",e,d),Tf(r.rstyle,"labelHeight",e,p),Tf(r.rscratch,"labelHeight",e,p),Tf(r.rscratch,"labelLineHeight",e,f)};Hc.getLabelText=function(t,e){var r=t._private,n=e?e+"-":"",i=t.pstyle(n+"label").strValue,a=t.pstyle("text-transform").value,s=o(function(Q,X){return X?(Tf(r.rscratch,Q,e,X),X):Ul(r.rscratch,Q,e)},"rscratch");if(!i)return"";a=="none"||(a=="uppercase"?i=i.toUpperCase():a=="lowercase"&&(i=i.toLowerCase()));var l=t.pstyle("text-wrap").value;if(l==="wrap"){var u=s("labelKey");if(u!=null&&s("labelWrapKey")===u)return s("labelWrapCachedText");for(var h="\u200B",f=i.split(` -`),d=t.pstyle("text-max-width").pfValue,p=t.pstyle("text-overflow-wrap").value,m=p==="anywhere",g=[],y=/[\s\u200b]+|$/g,v=0;vd){var T=x.matchAll(y),E="",_=0,A=Tpe(T),L;try{for(A.s();!(L=A.n()).done;){var M=L.value,N=M[0],k=x.substring(_,M.index);_=M.index+N.length;var I=E.length===0?k:E+k+N,C=this.calculateLabelDimensions(t,I),O=C.width;O<=d?E+=k+N:(E&&g.push(E),E=k+N)}}catch(Y){A.e(Y)}finally{A.f()}E.match(/^[\s\u200b]+$/)||g.push(E)}else g.push(x)}s("labelWrapCachedLines",g),i=s("labelWrapCachedText",g.join(` -`)),s("labelWrapKey",u)}else if(l==="ellipsis"){var D=t.pstyle("text-max-width").pfValue,P="",F="\u2026",B=!1;if(this.calculateLabelDimensions(t,i).widthD)break;P+=i[$],$===i.length-1&&(B=!0)}return B||(P+=F),P}return i};Hc.getLabelJustification=function(t){var e=t.pstyle("text-justification").strValue,r=t.pstyle("text-halign").strValue;if(e==="auto")if(t.isNode())switch(r){case"left":return"right";case"right":return"left";default:return"center"}else return"center";else return e};Hc.calculateLabelDimensions=function(t,e){var r=this,n=r.cy.window(),i=n.document,a=U0(e,t._private.labelDimsKey),s=r.labelDimCache||(r.labelDimCache=[]),l=s[a];if(l!=null)return l;var u=0,h=t.pstyle("font-style").strValue,f=t.pstyle("font-size").pfValue,d=t.pstyle("font-family").strValue,p=t.pstyle("font-weight").strValue,m=this.labelCalcCanvas,g=this.labelCalcCanvasContext;if(!m){m=this.labelCalcCanvas=i.createElement("canvas"),g=this.labelCalcCanvasContext=m.getContext("2d");var y=m.style;y.position="absolute",y.left="-9999px",y.top="-9999px",y.zIndex="-1",y.visibility="hidden",y.pointerEvents="none"}g.font="".concat(h," ").concat(p," ").concat(f,"px ").concat(d);for(var v=0,x=0,b=e.split(` -`),w=0;w1&&arguments[1]!==void 0?arguments[1]:!0;if(e.merge(s),l)for(var u=0;u=t.desktopTapThreshold2}var Je=i(W);ze&&(t.hoverData.tapholdCancelled=!0);var Ve=o(function(){var St=t.hoverData.dragDelta=t.hoverData.dragDelta||[];St.length===0?(St.push(ye[0]),St.push(ye[1])):(St[0]+=ye[0],St[1]+=ye[1])},"updateDragDelta");ge=!0,n(Ae,["mousemove","vmousemove","tapdrag"],W,{x:ae[0],y:ae[1]});var je=o(function(){t.data.bgActivePosistion=void 0,t.hoverData.selecting||re.emit({originalEvent:W,type:"boxstart",position:{x:ae[0],y:ae[1]}}),Ce[4]=1,t.hoverData.selecting=!0,t.redrawHint("select",!0),t.redraw()},"goIntoBoxMode");if(t.hoverData.which===3){if(ze){var kt={originalEvent:W,type:"cxtdrag",position:{x:ae[0],y:ae[1]}};Me?Me.emit(kt):re.emit(kt),t.hoverData.cxtDragged=!0,(!t.hoverData.cxtOver||Ae!==t.hoverData.cxtOver)&&(t.hoverData.cxtOver&&t.hoverData.cxtOver.emit({originalEvent:W,type:"cxtdragout",position:{x:ae[0],y:ae[1]}}),t.hoverData.cxtOver=Ae,Ae&&Ae.emit({originalEvent:W,type:"cxtdragover",position:{x:ae[0],y:ae[1]}}))}}else if(t.hoverData.dragging){if(ge=!0,re.panningEnabled()&&re.userPanningEnabled()){var at;if(t.hoverData.justStartedPan){var xt=t.hoverData.mdownPos;at={x:(ae[0]-xt[0])*he,y:(ae[1]-xt[1])*he},t.hoverData.justStartedPan=!1}else at={x:ye[0]*he,y:ye[1]*he};re.panBy(at),re.emit("dragpan"),t.hoverData.dragged=!0}ae=t.projectIntoViewport(W.clientX,W.clientY)}else if(Ce[4]==1&&(Me==null||Me.pannable())){if(ze){if(!t.hoverData.dragging&&re.boxSelectionEnabled()&&(Je||!re.panningEnabled()||!re.userPanningEnabled()))je();else if(!t.hoverData.selecting&&re.panningEnabled()&&re.userPanningEnabled()){var it=a(Me,t.hoverData.downs);it&&(t.hoverData.dragging=!0,t.hoverData.justStartedPan=!0,Ce[4]=0,t.data.bgActivePosistion=Jg(we),t.redrawHint("select",!0),t.redraw())}Me&&Me.pannable()&&Me.active()&&Me.unactivate()}}else{if(Me&&Me.pannable()&&Me.active()&&Me.unactivate(),(!Me||!Me.grabbed())&&Ae!=Ge&&(Ge&&n(Ge,["mouseout","tapdragout"],W,{x:ae[0],y:ae[1]}),Ae&&n(Ae,["mouseover","tapdragover"],W,{x:ae[0],y:ae[1]}),t.hoverData.last=Ae),Me)if(ze){if(re.boxSelectionEnabled()&&Je)Me&&Me.grabbed()&&(v(He),Me.emit("freeon"),He.emit("free"),t.dragData.didDrag&&(Me.emit("dragfreeon"),He.emit("dragfree"))),je();else if(Me&&Me.grabbed()&&t.nodeIsDraggable(Me)){var dt=!t.dragData.didDrag;dt&&t.redrawHint("eles",!0),t.dragData.didDrag=!0,t.hoverData.draggingEles||g(He,{inDragLayer:!0});var lt={x:0,y:0};if(ft(ye[0])&&ft(ye[1])&&(lt.x+=ye[0],lt.y+=ye[1],dt)){var It=t.hoverData.dragDelta;It&&ft(It[0])&&ft(It[1])&&(lt.x+=It[0],lt.y+=It[1])}t.hoverData.draggingEles=!0,He.silentShift(lt).emit("position drag"),t.redrawHint("drag",!0),t.redraw()}}else Ve();ge=!0}if(Ce[2]=ae[0],Ce[3]=ae[1],ge)return W.stopPropagation&&W.stopPropagation(),W.preventDefault&&W.preventDefault(),!1}},"mousemoveHandler"),!1);var M,N,k;t.registerBinding(e,"mouseup",o(function(W){if(!(t.hoverData.which===1&&W.which!==1&&t.hoverData.capture)){var fe=t.hoverData.capture;if(fe){t.hoverData.capture=!1;var ge=t.cy,re=t.projectIntoViewport(W.clientX,W.clientY),he=t.selection,ne=t.findNearestElement(re[0],re[1],!0,!1),ae=t.dragData.possibleDragElements,we=t.hoverData.down,Te=i(W);if(t.data.bgActivePosistion&&(t.redrawHint("select",!0),t.redraw()),t.hoverData.tapholdCancelled=!0,t.data.bgActivePosistion=void 0,we&&we.unactivate(),t.hoverData.which===3){var Ce={originalEvent:W,type:"cxttapend",position:{x:re[0],y:re[1]}};if(we?we.emit(Ce):ge.emit(Ce),!t.hoverData.cxtDragged){var Ae={originalEvent:W,type:"cxttap",position:{x:re[0],y:re[1]}};we?we.emit(Ae):ge.emit(Ae)}t.hoverData.cxtDragged=!1,t.hoverData.which=null}else if(t.hoverData.which===1){if(n(ne,["mouseup","tapend","vmouseup"],W,{x:re[0],y:re[1]}),!t.dragData.didDrag&&!t.hoverData.dragged&&!t.hoverData.selecting&&!t.hoverData.isOverThresholdDrag&&(n(we,["click","tap","vclick"],W,{x:re[0],y:re[1]}),N=!1,W.timeStamp-k<=ge.multiClickDebounceTime()?(M&&clearTimeout(M),N=!0,k=null,n(we,["dblclick","dbltap","vdblclick"],W,{x:re[0],y:re[1]})):(M=setTimeout(function(){N||n(we,["oneclick","onetap","voneclick"],W,{x:re[0],y:re[1]})},ge.multiClickDebounceTime()),k=W.timeStamp)),we==null&&!t.dragData.didDrag&&!t.hoverData.selecting&&!t.hoverData.dragged&&!i(W)&&(ge.$(r).unselect(["tapunselect"]),ae.length>0&&t.redrawHint("eles",!0),t.dragData.possibleDragElements=ae=ge.collection()),ne==we&&!t.dragData.didDrag&&!t.hoverData.selecting&&ne!=null&&ne._private.selectable&&(t.hoverData.dragging||(ge.selectionType()==="additive"||Te?ne.selected()?ne.unselect(["tapunselect"]):ne.select(["tapselect"]):Te||(ge.$(r).unmerge(ne).unselect(["tapunselect"]),ne.select(["tapselect"]))),t.redrawHint("eles",!0)),t.hoverData.selecting){var Ge=ge.collection(t.getAllInBox(he[0],he[1],he[2],he[3]));t.redrawHint("select",!0),Ge.length>0&&t.redrawHint("eles",!0),ge.emit({type:"boxend",originalEvent:W,position:{x:re[0],y:re[1]}});var Me=o(function(ze){return ze.selectable()&&!ze.selected()},"eleWouldBeSelected");ge.selectionType()==="additive"||Te||ge.$(r).unmerge(Ge).unselect(),Ge.emit("box").stdFilter(Me).select().emit("boxselect"),t.redraw()}if(t.hoverData.dragging&&(t.hoverData.dragging=!1,t.redrawHint("select",!0),t.redrawHint("eles",!0),t.redraw()),!he[4]){t.redrawHint("drag",!0),t.redrawHint("eles",!0);var ye=we&&we.grabbed();v(ae),ye&&(we.emit("freeon"),ae.emit("free"),t.dragData.didDrag&&(we.emit("dragfreeon"),ae.emit("dragfree")))}}he[4]=0,t.hoverData.down=null,t.hoverData.cxtStarted=!1,t.hoverData.draggingEles=!1,t.hoverData.selecting=!1,t.hoverData.isOverThresholdDrag=!1,t.dragData.didDrag=!1,t.hoverData.dragged=!1,t.hoverData.dragDelta=[],t.hoverData.mdownPos=null,t.hoverData.mdownGPos=null}}},"mouseupHandler"),!1);var I=o(function(W){if(!t.scrollingPage){var fe=t.cy,ge=fe.zoom(),re=fe.pan(),he=t.projectIntoViewport(W.clientX,W.clientY),ne=[he[0]*ge+re.x,he[1]*ge+re.y];if(t.hoverData.draggingEles||t.hoverData.dragging||t.hoverData.cxtStarted||A()){W.preventDefault();return}if(fe.panningEnabled()&&fe.userPanningEnabled()&&fe.zoomingEnabled()&&fe.userZoomingEnabled()){W.preventDefault(),t.data.wheelZooming=!0,clearTimeout(t.data.wheelTimeout),t.data.wheelTimeout=setTimeout(function(){t.data.wheelZooming=!1,t.redrawHint("eles",!0),t.redraw()},150);var ae;W.deltaY!=null?ae=W.deltaY/-250:W.wheelDeltaY!=null?ae=W.wheelDeltaY/1e3:ae=W.wheelDelta/1e3,ae=ae*t.wheelSensitivity;var we=W.deltaMode===1;we&&(ae*=33);var Te=fe.zoom()*Math.pow(10,ae);W.type==="gesturechange"&&(Te=t.gestureStartZoom*W.scale),fe.zoom({level:Te,renderedPosition:{x:ne[0],y:ne[1]}}),fe.emit(W.type==="gesturechange"?"pinchzoom":"scrollzoom")}}},"wheelHandler");t.registerBinding(t.container,"wheel",I,!0),t.registerBinding(e,"scroll",o(function(W){t.scrollingPage=!0,clearTimeout(t.scrollingPageTimeout),t.scrollingPageTimeout=setTimeout(function(){t.scrollingPage=!1},250)},"scrollHandler"),!0),t.registerBinding(t.container,"gesturestart",o(function(W){t.gestureStartZoom=t.cy.zoom(),t.hasTouchStarted||W.preventDefault()},"gestureStartHandler"),!0),t.registerBinding(t.container,"gesturechange",function(me){t.hasTouchStarted||I(me)},!0),t.registerBinding(t.container,"mouseout",o(function(W){var fe=t.projectIntoViewport(W.clientX,W.clientY);t.cy.emit({originalEvent:W,type:"mouseout",position:{x:fe[0],y:fe[1]}})},"mouseOutHandler"),!1),t.registerBinding(t.container,"mouseover",o(function(W){var fe=t.projectIntoViewport(W.clientX,W.clientY);t.cy.emit({originalEvent:W,type:"mouseover",position:{x:fe[0],y:fe[1]}})},"mouseOverHandler"),!1);var C,O,D,P,F,B,$,z,Y,Q,X,ie,j,J=o(function(W,fe,ge,re){return Math.sqrt((ge-W)*(ge-W)+(re-fe)*(re-fe))},"distance"),Z=o(function(W,fe,ge,re){return(ge-W)*(ge-W)+(re-fe)*(re-fe)},"distanceSq"),H;t.registerBinding(t.container,"touchstart",H=o(function(W){if(t.hasTouchStarted=!0,!!L(W)){b(),t.touchData.capture=!0,t.data.bgActivePosistion=void 0;var fe=t.cy,ge=t.touchData.now,re=t.touchData.earlier;if(W.touches[0]){var he=t.projectIntoViewport(W.touches[0].clientX,W.touches[0].clientY);ge[0]=he[0],ge[1]=he[1]}if(W.touches[1]){var he=t.projectIntoViewport(W.touches[1].clientX,W.touches[1].clientY);ge[2]=he[0],ge[3]=he[1]}if(W.touches[2]){var he=t.projectIntoViewport(W.touches[2].clientX,W.touches[2].clientY);ge[4]=he[0],ge[5]=he[1]}if(W.touches[1]){t.touchData.singleTouchMoved=!0,v(t.dragData.touchDragEles);var ne=t.findContainerClientCoords();Y=ne[0],Q=ne[1],X=ne[2],ie=ne[3],C=W.touches[0].clientX-Y,O=W.touches[0].clientY-Q,D=W.touches[1].clientX-Y,P=W.touches[1].clientY-Q,j=0<=C&&C<=X&&0<=D&&D<=X&&0<=O&&O<=ie&&0<=P&&P<=ie;var ae=fe.pan(),we=fe.zoom();F=J(C,O,D,P),B=Z(C,O,D,P),$=[(C+D)/2,(O+P)/2],z=[($[0]-ae.x)/we,($[1]-ae.y)/we];var Te=200,Ce=Te*Te;if(B=1){for(var gt=t.touchData.startPosition=[null,null,null,null,null,null],yt=0;yt=t.touchTapThreshold2}if(fe&&t.touchData.cxt){W.preventDefault();var gt=W.touches[0].clientX-Y,yt=W.touches[0].clientY-Q,tt=W.touches[1].clientX-Y,Ye=W.touches[1].clientY-Q,Je=Z(gt,yt,tt,Ye),Ve=Je/B,je=150,kt=je*je,at=1.5,xt=at*at;if(Ve>=xt||Je>=kt){t.touchData.cxt=!1,t.data.bgActivePosistion=void 0,t.redrawHint("select",!0);var it={originalEvent:W,type:"cxttapend",position:{x:he[0],y:he[1]}};t.touchData.start?(t.touchData.start.unactivate().emit(it),t.touchData.start=null):re.emit(it)}}if(fe&&t.touchData.cxt){var it={originalEvent:W,type:"cxtdrag",position:{x:he[0],y:he[1]}};t.data.bgActivePosistion=void 0,t.redrawHint("select",!0),t.touchData.start?t.touchData.start.emit(it):re.emit(it),t.touchData.start&&(t.touchData.start._private.grabbed=!1),t.touchData.cxtDragged=!0;var dt=t.findNearestElement(he[0],he[1],!0,!0);(!t.touchData.cxtOver||dt!==t.touchData.cxtOver)&&(t.touchData.cxtOver&&t.touchData.cxtOver.emit({originalEvent:W,type:"cxtdragout",position:{x:he[0],y:he[1]}}),t.touchData.cxtOver=dt,dt&&dt.emit({originalEvent:W,type:"cxtdragover",position:{x:he[0],y:he[1]}}))}else if(fe&&W.touches[2]&&re.boxSelectionEnabled())W.preventDefault(),t.data.bgActivePosistion=void 0,this.lastThreeTouch=+new Date,t.touchData.selecting||re.emit({originalEvent:W,type:"boxstart",position:{x:he[0],y:he[1]}}),t.touchData.selecting=!0,t.touchData.didSelect=!0,ge[4]=1,!ge||ge.length===0||ge[0]===void 0?(ge[0]=(he[0]+he[2]+he[4])/3,ge[1]=(he[1]+he[3]+he[5])/3,ge[2]=(he[0]+he[2]+he[4])/3+1,ge[3]=(he[1]+he[3]+he[5])/3+1):(ge[2]=(he[0]+he[2]+he[4])/3,ge[3]=(he[1]+he[3]+he[5])/3),t.redrawHint("select",!0),t.redraw();else if(fe&&W.touches[1]&&!t.touchData.didSelect&&re.zoomingEnabled()&&re.panningEnabled()&&re.userZoomingEnabled()&&re.userPanningEnabled()){W.preventDefault(),t.data.bgActivePosistion=void 0,t.redrawHint("select",!0);var lt=t.dragData.touchDragEles;if(lt){t.redrawHint("drag",!0);for(var It=0;It0&&!t.hoverData.draggingEles&&!t.swipePanning&&t.data.bgActivePosistion!=null&&(t.data.bgActivePosistion=void 0,t.redrawHint("select",!0),t.redraw())}},"touchmoveHandler"),!1);var K;t.registerBinding(e,"touchcancel",K=o(function(W){var fe=t.touchData.start;t.touchData.capture=!1,fe&&fe.unactivate()},"touchcancelHandler"));var se,ce,ue,te;if(t.registerBinding(e,"touchend",se=o(function(W){var fe=t.touchData.start,ge=t.touchData.capture;if(ge)W.touches.length===0&&(t.touchData.capture=!1),W.preventDefault();else return;var re=t.selection;t.swipePanning=!1,t.hoverData.draggingEles=!1;var he=t.cy,ne=he.zoom(),ae=t.touchData.now,we=t.touchData.earlier;if(W.touches[0]){var Te=t.projectIntoViewport(W.touches[0].clientX,W.touches[0].clientY);ae[0]=Te[0],ae[1]=Te[1]}if(W.touches[1]){var Te=t.projectIntoViewport(W.touches[1].clientX,W.touches[1].clientY);ae[2]=Te[0],ae[3]=Te[1]}if(W.touches[2]){var Te=t.projectIntoViewport(W.touches[2].clientX,W.touches[2].clientY);ae[4]=Te[0],ae[5]=Te[1]}fe&&fe.unactivate();var Ce;if(t.touchData.cxt){if(Ce={originalEvent:W,type:"cxttapend",position:{x:ae[0],y:ae[1]}},fe?fe.emit(Ce):he.emit(Ce),!t.touchData.cxtDragged){var Ae={originalEvent:W,type:"cxttap",position:{x:ae[0],y:ae[1]}};fe?fe.emit(Ae):he.emit(Ae)}t.touchData.start&&(t.touchData.start._private.grabbed=!1),t.touchData.cxt=!1,t.touchData.start=null,t.redraw();return}if(!W.touches[2]&&he.boxSelectionEnabled()&&t.touchData.selecting){t.touchData.selecting=!1;var Ge=he.collection(t.getAllInBox(re[0],re[1],re[2],re[3]));re[0]=void 0,re[1]=void 0,re[2]=void 0,re[3]=void 0,re[4]=0,t.redrawHint("select",!0),he.emit({type:"boxend",originalEvent:W,position:{x:ae[0],y:ae[1]}});var Me=o(function(kt){return kt.selectable()&&!kt.selected()},"eleWouldBeSelected");Ge.emit("box").stdFilter(Me).select().emit("boxselect"),Ge.nonempty()&&t.redrawHint("eles",!0),t.redraw()}if(fe?.unactivate(),W.touches[2])t.data.bgActivePosistion=void 0,t.redrawHint("select",!0);else if(!W.touches[1]){if(!W.touches[0]){if(!W.touches[0]){t.data.bgActivePosistion=void 0,t.redrawHint("select",!0);var ye=t.dragData.touchDragEles;if(fe!=null){var He=fe._private.grabbed;v(ye),t.redrawHint("drag",!0),t.redrawHint("eles",!0),He&&(fe.emit("freeon"),ye.emit("free"),t.dragData.didDrag&&(fe.emit("dragfreeon"),ye.emit("dragfree"))),n(fe,["touchend","tapend","vmouseup","tapdragout"],W,{x:ae[0],y:ae[1]}),fe.unactivate(),t.touchData.start=null}else{var ze=t.findNearestElement(ae[0],ae[1],!0,!0);n(ze,["touchend","tapend","vmouseup","tapdragout"],W,{x:ae[0],y:ae[1]})}var Ze=t.touchData.startPosition[0]-ae[0],gt=Ze*Ze,yt=t.touchData.startPosition[1]-ae[1],tt=yt*yt,Ye=gt+tt,Je=Ye*ne*ne;t.touchData.singleTouchMoved||(fe||he.$(":selected").unselect(["tapunselect"]),n(fe,["tap","vclick"],W,{x:ae[0],y:ae[1]}),ce=!1,W.timeStamp-te<=he.multiClickDebounceTime()?(ue&&clearTimeout(ue),ce=!0,te=null,n(fe,["dbltap","vdblclick"],W,{x:ae[0],y:ae[1]})):(ue=setTimeout(function(){ce||n(fe,["onetap","voneclick"],W,{x:ae[0],y:ae[1]})},he.multiClickDebounceTime()),te=W.timeStamp)),fe!=null&&!t.dragData.didDrag&&fe._private.selectable&&Je"u"){var De=[],oe=o(function(W){return{clientX:W.clientX,clientY:W.clientY,force:1,identifier:W.pointerId,pageX:W.pageX,pageY:W.pageY,radiusX:W.width/2,radiusY:W.height/2,screenX:W.screenX,screenY:W.screenY,target:W.target}},"makeTouch"),ke=o(function(W){return{event:W,touch:oe(W)}},"makePointer"),Ie=o(function(W){De.push(ke(W))},"addPointer"),Se=o(function(W){for(var fe=0;fe0)return Y[0]}return null},"getCurveT"),g=Object.keys(p),y=0;y0?m:Hpe(a,s,e,r,n,i,l,u)},"intersectLine"),checkPoint:o(function(e,r,n,i,a,s,l,u){u=u==="auto"?Y0(i,a):u;var h=2*u;if(Qu(e,r,this.points,s,l,i,a-h,[0,-1],n)||Qu(e,r,this.points,s,l,i-h,a,[0,-1],n))return!0;var f=i/2+2*n,d=a/2+2*n,p=[s-f,l-d,s-f,l,s+f,l,s+f,l-d];return!!(zs(e,r,p)||$0(e,r,h,h,s+i/2-u,l+a/2-u,n)||$0(e,r,h,h,s-i/2+u,l+a/2-u,n))},"checkPoint")}};Ju.registerNodeShapes=function(){var t=this.nodeShapes={},e=this;this.generateEllipse(),this.generatePolygon("triangle",ls(3,0)),this.generateRoundPolygon("round-triangle",ls(3,0)),this.generatePolygon("rectangle",ls(4,0)),t.square=t.rectangle,this.generateRoundRectangle(),this.generateCutRectangle(),this.generateBarrel(),this.generateBottomRoundrectangle();{var r=[0,1,1,0,0,-1,-1,0];this.generatePolygon("diamond",r),this.generateRoundPolygon("round-diamond",r)}this.generatePolygon("pentagon",ls(5,0)),this.generateRoundPolygon("round-pentagon",ls(5,0)),this.generatePolygon("hexagon",ls(6,0)),this.generateRoundPolygon("round-hexagon",ls(6,0)),this.generatePolygon("heptagon",ls(7,0)),this.generateRoundPolygon("round-heptagon",ls(7,0)),this.generatePolygon("octagon",ls(8,0)),this.generateRoundPolygon("round-octagon",ls(8,0));var n=new Array(20);{var i=NP(5,0),a=NP(5,Math.PI/5),s=.5*(3-Math.sqrt(5));s*=1.57;for(var l=0;l=e.deqFastCost*S)break}else if(h){if(b>=e.deqCost*m||b>=e.deqAvgCost*p)break}else if(w>=e.deqNoDrawCost*LP)break;var T=e.deq(n,v,y);if(T.length>0)for(var E=0;E0&&(e.onDeqd(n,g),!h&&e.shouldRedraw(n,g,v,y)&&a())},"dequeue"),l=e.priority||JP;i.beforeRender(s,l(n))}},"setupDequeueingImpl")},"setupDequeueing")},pQe=function(){function t(e){var r=arguments.length>1&&arguments[1]!==void 0?arguments[1]:C6;XP(this,t),this.idsByKey=new Vc,this.keyForId=new Vc,this.cachesByLvl=new Vc,this.lvls=[],this.getKey=e,this.doesEleInvalidateKey=r}return o(t,"ElementTextureCacheLookup"),jP(t,[{key:"getIdsFor",value:o(function(r){r==null&&oi("Can not get id list for null key");var n=this.idsByKey,i=this.idsByKey.get(r);return i||(i=new c1,n.set(r,i)),i},"getIdsFor")},{key:"addIdForKey",value:o(function(r,n){r!=null&&this.getIdsFor(r).add(n)},"addIdForKey")},{key:"deleteIdForKey",value:o(function(r,n){r!=null&&this.getIdsFor(r).delete(n)},"deleteIdForKey")},{key:"getNumberOfIdsForKey",value:o(function(r){return r==null?0:this.getIdsFor(r).size},"getNumberOfIdsForKey")},{key:"updateKeyMappingFor",value:o(function(r){var n=r.id(),i=this.keyForId.get(n),a=this.getKey(r);this.deleteIdForKey(i,n),this.addIdForKey(a,n),this.keyForId.set(n,a)},"updateKeyMappingFor")},{key:"deleteKeyMappingFor",value:o(function(r){var n=r.id(),i=this.keyForId.get(n);this.deleteIdForKey(i,n),this.keyForId.delete(n)},"deleteKeyMappingFor")},{key:"keyHasChangedFor",value:o(function(r){var n=r.id(),i=this.keyForId.get(n),a=this.getKey(r);return i!==a},"keyHasChangedFor")},{key:"isInvalid",value:o(function(r){return this.keyHasChangedFor(r)||this.doesEleInvalidateKey(r)},"isInvalid")},{key:"getCachesAt",value:o(function(r){var n=this.cachesByLvl,i=this.lvls,a=n.get(r);return a||(a=new Vc,n.set(r,a),i.push(r)),a},"getCachesAt")},{key:"getCache",value:o(function(r,n){return this.getCachesAt(n).get(r)},"getCache")},{key:"get",value:o(function(r,n){var i=this.getKey(r),a=this.getCache(i,n);return 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this.rect.x+this.rect.width/2},h.prototype.getCenterY=function(){return this.rect.y+this.rect.height/2},h.prototype.getCenter=function(){return new u(this.rect.x+this.rect.width/2,this.rect.y+this.rect.height/2)},h.prototype.getLocation=function(){return new u(this.rect.x,this.rect.y)},h.prototype.getRect=function(){return this.rect},h.prototype.getDiagonal=function(){return Math.sqrt(this.rect.width*this.rect.width+this.rect.height*this.rect.height)},h.prototype.getHalfTheDiagonal=function(){return Math.sqrt(this.rect.height*this.rect.height+this.rect.width*this.rect.width)/2},h.prototype.setRect=function(d,p){this.rect.x=d.x,this.rect.y=d.y,this.rect.width=p.width,this.rect.height=p.height},h.prototype.setCenter=function(d,p){this.rect.x=d-this.rect.width/2,this.rect.y=p-this.rect.height/2},h.prototype.setLocation=function(d,p){this.rect.x=d,this.rect.y=p},h.prototype.moveBy=function(d,p){this.rect.x+=d,this.rect.y+=p},h.prototype.getEdgeListToNode=function(d){var p=[],m,g=this;return g.edges.forEach(function(y){if(y.target==d){if(y.source!=g)throw"Incorrect edge source!";p.push(y)}}),p},h.prototype.getEdgesBetween=function(d){var p=[],m,g=this;return g.edges.forEach(function(y){if(!(y.source==g||y.target==g))throw"Incorrect edge source and/or target";(y.target==d||y.source==d)&&p.push(y)}),p},h.prototype.getNeighborsList=function(){var d=new Set,p=this;return p.edges.forEach(function(m){if(m.source==p)d.add(m.target);else{if(m.target!=p)throw"Incorrect incidency!";d.add(m.source)}}),d},h.prototype.withChildren=function(){var d=new Set,p,m;if(d.add(this),this.child!=null)for(var g=this.child.getNodes(),y=0;yp&&(this.rect.x-=(this.labelWidth-p)/2,this.setWidth(this.labelWidth)),this.labelHeight>m&&(this.labelPos=="center"?this.rect.y-=(this.labelHeight-m)/2:this.labelPos=="top"&&(this.rect.y-=this.labelHeight-m),this.setHeight(this.labelHeight))}}},h.prototype.getInclusionTreeDepth=function(){if(this.inclusionTreeDepth==i.MAX_VALUE)throw"assert failed";return this.inclusionTreeDepth},h.prototype.transform=function(d){var p=this.rect.x;p>s.WORLD_BOUNDARY?p=s.WORLD_BOUNDARY:p<-s.WORLD_BOUNDARY&&(p=-s.WORLD_BOUNDARY);var m=this.rect.y;m>s.WORLD_BOUNDARY?m=s.WORLD_BOUNDARY:m<-s.WORLD_BOUNDARY&&(m=-s.WORLD_BOUNDARY);var g=new u(p,m),y=d.inverseTransformPoint(g);this.setLocation(y.x,y.y)},h.prototype.getLeft=function(){return this.rect.x},h.prototype.getRight=function(){return this.rect.x+this.rect.width},h.prototype.getTop=function(){return this.rect.y},h.prototype.getBottom=function(){return this.rect.y+this.rect.height},h.prototype.getParent=function(){return this.owner==null?null:this.owner.getParent()},t.exports=h},function(t,e,r){"use strict";function n(i,a){i==null&&a==null?(this.x=0,this.y=0):(this.x=i,this.y=a)}o(n,"PointD"),n.prototype.getX=function(){return this.x},n.prototype.getY=function(){return this.y},n.prototype.setX=function(i){this.x=i},n.prototype.setY=function(i){this.y=i},n.prototype.getDifference=function(i){return new DimensionD(this.x-i.x,this.y-i.y)},n.prototype.getCopy=function(){return new n(this.x,this.y)},n.prototype.translate=function(i){return this.x+=i.width,this.y+=i.height,this},t.exports=n},function(t,e,r){"use strict";var n=r(2),i=r(10),a=r(0),s=r(6),l=r(3),u=r(1),h=r(13),f=r(12),d=r(11);function p(g,y,v){n.call(this,v),this.estimatedSize=i.MIN_VALUE,this.margin=a.DEFAULT_GRAPH_MARGIN,this.edges=[],this.nodes=[],this.isConnected=!1,this.parent=g,y!=null&&y instanceof s?this.graphManager=y:y!=null&&y instanceof Layout&&(this.graphManager=y.graphManager)}o(p,"LGraph"),p.prototype=Object.create(n.prototype);for(var m in n)p[m]=n[m];p.prototype.getNodes=function(){return this.nodes},p.prototype.getEdges=function(){return this.edges},p.prototype.getGraphManager=function(){return this.graphManager},p.prototype.getParent=function(){return this.parent},p.prototype.getLeft=function(){return this.left},p.prototype.getRight=function(){return this.right},p.prototype.getTop=function(){return this.top},p.prototype.getBottom=function(){return this.bottom},p.prototype.isConnected=function(){return this.isConnected},p.prototype.add=function(g,y,v){if(y==null&&v==null){var x=g;if(this.graphManager==null)throw"Graph has no graph mgr!";if(this.getNodes().indexOf(x)>-1)throw"Node already in graph!";return x.owner=this,this.getNodes().push(x),x}else{var b=g;if(!(this.getNodes().indexOf(y)>-1&&this.getNodes().indexOf(v)>-1))throw"Source or target not in graph!";if(!(y.owner==v.owner&&y.owner==this))throw"Both owners must be this graph!";return y.owner!=v.owner?null:(b.source=y,b.target=v,b.isInterGraph=!1,this.getEdges().push(b),y.edges.push(b),v!=y&&v.edges.push(b),b)}},p.prototype.remove=function(g){var y=g;if(g instanceof l){if(y==null)throw"Node is null!";if(!(y.owner!=null&&y.owner==this))throw"Owner graph is invalid!";if(this.graphManager==null)throw"Owner graph manager is invalid!";for(var v=y.edges.slice(),x,b=v.length,w=0;w-1&&E>-1))throw"Source and/or target doesn't know this edge!";x.source.edges.splice(T,1),x.target!=x.source&&x.target.edges.splice(E,1);var S=x.source.owner.getEdges().indexOf(x);if(S==-1)throw"Not in owner's edge list!";x.source.owner.getEdges().splice(S,1)}},p.prototype.updateLeftTop=function(){for(var g=i.MAX_VALUE,y=i.MAX_VALUE,v,x,b,w=this.getNodes(),S=w.length,T=0;Tv&&(g=v),y>x&&(y=x)}return g==i.MAX_VALUE?null:(w[0].getParent().paddingLeft!=null?b=w[0].getParent().paddingLeft:b=this.margin,this.left=y-b,this.top=g-b,new f(this.left,this.top))},p.prototype.updateBounds=function(g){for(var y=i.MAX_VALUE,v=-i.MAX_VALUE,x=i.MAX_VALUE,b=-i.MAX_VALUE,w,S,T,E,_,A=this.nodes,L=A.length,M=0;Mw&&(y=w),vT&&(x=T),bw&&(y=w),vT&&(x=T),b=this.nodes.length){var L=0;v.forEach(function(M){M.owner==g&&L++}),L==this.nodes.length&&(this.isConnected=!0)}},t.exports=p},function(t,e,r){"use strict";var n,i=r(1);function a(s){n=r(5),this.layout=s,this.graphs=[],this.edges=[]}o(a,"LGraphManager"),a.prototype.addRoot=function(){var s=this.layout.newGraph(),l=this.layout.newNode(null),u=this.add(s,l);return this.setRootGraph(u),this.rootGraph},a.prototype.add=function(s,l,u,h,f){if(u==null&&h==null&&f==null){if(s==null)throw"Graph is null!";if(l==null)throw"Parent node is null!";if(this.graphs.indexOf(s)>-1)throw"Graph already in this graph mgr!";if(this.graphs.push(s),s.parent!=null)throw"Already has a parent!";if(l.child!=null)throw"Already has a child!";return s.parent=l,l.child=s,s}else{f=u,h=l,u=s;var d=h.getOwner(),p=f.getOwner();if(!(d!=null&&d.getGraphManager()==this))throw"Source not in this graph mgr!";if(!(p!=null&&p.getGraphManager()==this))throw"Target not in this graph mgr!";if(d==p)return u.isInterGraph=!1,d.add(u,h,f);if(u.isInterGraph=!0,u.source=h,u.target=f,this.edges.indexOf(u)>-1)throw"Edge already in inter-graph edge list!";if(this.edges.push(u),!(u.source!=null&&u.target!=null))throw"Edge source and/or target is null!";if(!(u.source.edges.indexOf(u)==-1&&u.target.edges.indexOf(u)==-1))throw"Edge already in source and/or target incidency list!";return u.source.edges.push(u),u.target.edges.push(u),u}},a.prototype.remove=function(s){if(s instanceof n){var l=s;if(l.getGraphManager()!=this)throw"Graph not in this graph mgr";if(!(l==this.rootGraph||l.parent!=null&&l.parent.graphManager==this))throw"Invalid parent node!";var u=[];u=u.concat(l.getEdges());for(var h,f=u.length,d=0;d=s.getRight()?l[0]+=Math.min(s.getX()-a.getX(),a.getRight()-s.getRight()):s.getX()<=a.getX()&&s.getRight()>=a.getRight()&&(l[0]+=Math.min(a.getX()-s.getX(),s.getRight()-a.getRight())),a.getY()<=s.getY()&&a.getBottom()>=s.getBottom()?l[1]+=Math.min(s.getY()-a.getY(),a.getBottom()-s.getBottom()):s.getY()<=a.getY()&&s.getBottom()>=a.getBottom()&&(l[1]+=Math.min(a.getY()-s.getY(),s.getBottom()-a.getBottom()));var f=Math.abs((s.getCenterY()-a.getCenterY())/(s.getCenterX()-a.getCenterX()));s.getCenterY()===a.getCenterY()&&s.getCenterX()===a.getCenterX()&&(f=1);var d=f*l[0],p=l[1]/f;l[0]d)return l[0]=u,l[1]=m,l[2]=f,l[3]=A,!1;if(hf)return l[0]=p,l[1]=h,l[2]=E,l[3]=d,!1;if(uf?(l[0]=y,l[1]=v,k=!0):(l[0]=g,l[1]=m,k=!0):C===D&&(u>f?(l[0]=p,l[1]=m,k=!0):(l[0]=x,l[1]=v,k=!0)),-O===D?f>u?(l[2]=_,l[3]=A,I=!0):(l[2]=E,l[3]=T,I=!0):O===D&&(f>u?(l[2]=S,l[3]=T,I=!0):(l[2]=L,l[3]=A,I=!0)),k&&I)return!1;if(u>f?h>d?(P=this.getCardinalDirection(C,D,4),F=this.getCardinalDirection(O,D,2)):(P=this.getCardinalDirection(-C,D,3),F=this.getCardinalDirection(-O,D,1)):h>d?(P=this.getCardinalDirection(-C,D,1),F=this.getCardinalDirection(-O,D,3)):(P=this.getCardinalDirection(C,D,2),F=this.getCardinalDirection(O,D,4)),!k)switch(P){case 1:$=m,B=u+-w/D,l[0]=B,l[1]=$;break;case 2:B=x,$=h+b*D,l[0]=B,l[1]=$;break;case 3:$=v,B=u+w/D,l[0]=B,l[1]=$;break;case 4:B=y,$=h+-b*D,l[0]=B,l[1]=$;break}if(!I)switch(F){case 1:Y=T,z=f+-N/D,l[2]=z,l[3]=Y;break;case 2:z=L,Y=d+M*D,l[2]=z,l[3]=Y;break;case 3:Y=A,z=f+N/D,l[2]=z,l[3]=Y;break;case 4:z=_,Y=d+-M*D,l[2]=z,l[3]=Y;break}}return!1},i.getCardinalDirection=function(a,s,l){return a>s?l:1+l%4},i.getIntersection=function(a,s,l,u){if(u==null)return this.getIntersection2(a,s,l);var h=a.x,f=a.y,d=s.x,p=s.y,m=l.x,g=l.y,y=u.x,v=u.y,x=void 0,b=void 0,w=void 0,S=void 0,T=void 0,E=void 0,_=void 0,A=void 0,L=void 0;return w=p-f,T=h-d,_=d*f-h*p,S=v-g,E=m-y,A=y*g-m*v,L=w*E-S*T,L===0?null:(x=(T*A-E*_)/L,b=(S*_-w*A)/L,new n(x,b))},i.angleOfVector=function(a,s,l,u){var h=void 0;return a!==l?(h=Math.atan((u-s)/(l-a)),l0?1:i<0?-1:0},n.floor=function(i){return i<0?Math.ceil(i):Math.floor(i)},n.ceil=function(i){return i<0?Math.floor(i):Math.ceil(i)},t.exports=n},function(t,e,r){"use strict";function n(){}o(n,"Integer"),n.MAX_VALUE=2147483647,n.MIN_VALUE=-2147483648,t.exports=n},function(t,e,r){"use strict";var n=function(){function h(f,d){for(var p=0;p"u"?"undefined":n(a);return a==null||s!="object"&&s!="function"},t.exports=i},function(t,e,r){"use strict";function n(m){if(Array.isArray(m)){for(var g=0,y=Array(m.length);g0&&g;){for(w.push(T[0]);w.length>0&&g;){var E=w[0];w.splice(0,1),b.add(E);for(var _=E.getEdges(),x=0;x<_.length;x++){var A=_[x].getOtherEnd(E);if(S.get(E)!=A)if(!b.has(A))w.push(A),S.set(A,E);else{g=!1;break}}}if(!g)m=[];else{var L=[].concat(n(b));m.push(L);for(var x=0;x-1&&T.splice(N,1)}b=new Set,S=new Map}}return m},p.prototype.createDummyNodesForBendpoints=function(m){for(var g=[],y=m.source,v=this.graphManager.calcLowestCommonAncestor(m.source,m.target),x=0;x0){for(var v=this.edgeToDummyNodes.get(y),x=0;x=0&&g.splice(A,1);var L=S.getNeighborsList();L.forEach(function(k){if(y.indexOf(k)<0){var I=v.get(k),C=I-1;C==1&&E.push(k),v.set(k,C)}})}y=y.concat(E),(g.length==1||g.length==2)&&(x=!0,b=g[0])}return b},p.prototype.setGraphManager=function(m){this.graphManager=m},t.exports=p},function(t,e,r){"use strict";function n(){}o(n,"RandomSeed"),n.seed=1,n.x=0,n.nextDouble=function(){return n.x=Math.sin(n.seed++)*1e4,n.x-Math.floor(n.x)},t.exports=n},function(t,e,r){"use strict";var n=r(4);function i(a,s){this.lworldOrgX=0,this.lworldOrgY=0,this.ldeviceOrgX=0,this.ldeviceOrgY=0,this.lworldExtX=1,this.lworldExtY=1,this.ldeviceExtX=1,this.ldeviceExtY=1}o(i,"Transform"),i.prototype.getWorldOrgX=function(){return this.lworldOrgX},i.prototype.setWorldOrgX=function(a){this.lworldOrgX=a},i.prototype.getWorldOrgY=function(){return this.lworldOrgY},i.prototype.setWorldOrgY=function(a){this.lworldOrgY=a},i.prototype.getWorldExtX=function(){return this.lworldExtX},i.prototype.setWorldExtX=function(a){this.lworldExtX=a},i.prototype.getWorldExtY=function(){return this.lworldExtY},i.prototype.setWorldExtY=function(a){this.lworldExtY=a},i.prototype.getDeviceOrgX=function(){return this.ldeviceOrgX},i.prototype.setDeviceOrgX=function(a){this.ldeviceOrgX=a},i.prototype.getDeviceOrgY=function(){return this.ldeviceOrgY},i.prototype.setDeviceOrgY=function(a){this.ldeviceOrgY=a},i.prototype.getDeviceExtX=function(){return this.ldeviceExtX},i.prototype.setDeviceExtX=function(a){this.ldeviceExtX=a},i.prototype.getDeviceExtY=function(){return this.ldeviceExtY},i.prototype.setDeviceExtY=function(a){this.ldeviceExtY=a},i.prototype.transformX=function(a){var s=0,l=this.lworldExtX;return l!=0&&(s=this.ldeviceOrgX+(a-this.lworldOrgX)*this.ldeviceExtX/l),s},i.prototype.transformY=function(a){var s=0,l=this.lworldExtY;return l!=0&&(s=this.ldeviceOrgY+(a-this.lworldOrgY)*this.ldeviceExtY/l),s},i.prototype.inverseTransformX=function(a){var s=0,l=this.ldeviceExtX;return l!=0&&(s=this.lworldOrgX+(a-this.ldeviceOrgX)*this.lworldExtX/l),s},i.prototype.inverseTransformY=function(a){var s=0,l=this.ldeviceExtY;return l!=0&&(s=this.lworldOrgY+(a-this.ldeviceOrgY)*this.lworldExtY/l),s},i.prototype.inverseTransformPoint=function(a){var s=new n(this.inverseTransformX(a.x),this.inverseTransformY(a.y));return s},t.exports=i},function(t,e,r){"use strict";function n(d){if(Array.isArray(d)){for(var p=0,m=Array(d.length);pa.ADAPTATION_LOWER_NODE_LIMIT&&(this.coolingFactor=Math.max(this.coolingFactor*a.COOLING_ADAPTATION_FACTOR,this.coolingFactor-(d-a.ADAPTATION_LOWER_NODE_LIMIT)/(a.ADAPTATION_UPPER_NODE_LIMIT-a.ADAPTATION_LOWER_NODE_LIMIT)*this.coolingFactor*(1-a.COOLING_ADAPTATION_FACTOR))),this.maxNodeDisplacement=a.MAX_NODE_DISPLACEMENT_INCREMENTAL):(d>a.ADAPTATION_LOWER_NODE_LIMIT?this.coolingFactor=Math.max(a.COOLING_ADAPTATION_FACTOR,1-(d-a.ADAPTATION_LOWER_NODE_LIMIT)/(a.ADAPTATION_UPPER_NODE_LIMIT-a.ADAPTATION_LOWER_NODE_LIMIT)*(1-a.COOLING_ADAPTATION_FACTOR)):this.coolingFactor=1,this.initialCoolingFactor=this.coolingFactor,this.maxNodeDisplacement=a.MAX_NODE_DISPLACEMENT),this.maxIterations=Math.max(this.getAllNodes().length*5,this.maxIterations),this.totalDisplacementThreshold=this.displacementThresholdPerNode*this.getAllNodes().length,this.repulsionRange=this.calcRepulsionRange()},h.prototype.calcSpringForces=function(){for(var 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g=this.child.getNodes(),y=0;yp?(this.rect.x-=(this.labelWidth-p)/2,this.setWidth(this.labelWidth)):this.labelPosHorizontal=="right"&&this.setWidth(p+this.labelWidth)),this.labelHeight&&(this.labelPosVertical=="top"?(this.rect.y-=this.labelHeight,this.setHeight(m+this.labelHeight)):this.labelPosVertical=="center"&&this.labelHeight>m?(this.rect.y-=(this.labelHeight-m)/2,this.setHeight(this.labelHeight)):this.labelPosVertical=="bottom"&&this.setHeight(m+this.labelHeight))}}},h.prototype.getInclusionTreeDepth=function(){if(this.inclusionTreeDepth==i.MAX_VALUE)throw"assert failed";return this.inclusionTreeDepth},h.prototype.transform=function(d){var p=this.rect.x;p>s.WORLD_BOUNDARY?p=s.WORLD_BOUNDARY:p<-s.WORLD_BOUNDARY&&(p=-s.WORLD_BOUNDARY);var m=this.rect.y;m>s.WORLD_BOUNDARY?m=s.WORLD_BOUNDARY:m<-s.WORLD_BOUNDARY&&(m=-s.WORLD_BOUNDARY);var g=new u(p,m),y=d.inverseTransformPoint(g);this.setLocation(y.x,y.y)},h.prototype.getLeft=function(){return this.rect.x},h.prototype.getRight=function(){return this.rect.x+this.rect.width},h.prototype.getTop=function(){return this.rect.y},h.prototype.getBottom=function(){return this.rect.y+this.rect.height},h.prototype.getParent=function(){return this.owner==null?null:this.owner.getParent()},t.exports=h},function(t,e,r){"use strict";var n=r(0);function i(){}o(i,"FDLayoutConstants");for(var a in n)i[a]=n[a];i.MAX_ITERATIONS=2500,i.DEFAULT_EDGE_LENGTH=50,i.DEFAULT_SPRING_STRENGTH=.45,i.DEFAULT_REPULSION_STRENGTH=4500,i.DEFAULT_GRAVITY_STRENGTH=.4,i.DEFAULT_COMPOUND_GRAVITY_STRENGTH=1,i.DEFAULT_GRAVITY_RANGE_FACTOR=3.8,i.DEFAULT_COMPOUND_GRAVITY_RANGE_FACTOR=1.5,i.DEFAULT_USE_SMART_IDEAL_EDGE_LENGTH_CALCULATION=!0,i.DEFAULT_USE_SMART_REPULSION_RANGE_CALCULATION=!0,i.DEFAULT_COOLING_FACTOR_INCREMENTAL=.3,i.COOLING_ADAPTATION_FACTOR=.33,i.ADAPTATION_LOWER_NODE_LIMIT=1e3,i.ADAPTATION_UPPER_NODE_LIMIT=5e3,i.MAX_NODE_DISPLACEMENT_INCREMENTAL=100,i.MAX_NODE_DISPLACEMENT=i.MAX_NODE_DISPLACEMENT_INCREMENTAL*3,i.MIN_REPULSION_DIST=i.DEFAULT_EDGE_LENGTH/10,i.CONVERGENCE_CHECK_PERIOD=100,i.PER_LEVEL_IDEAL_EDGE_LENGTH_FACTOR=.1,i.MIN_EDGE_LENGTH=1,i.GRID_CALCULATION_CHECK_PERIOD=10,t.exports=i},function(t,e,r){"use strict";function n(i,a){i==null&&a==null?(this.x=0,this.y=0):(this.x=i,this.y=a)}o(n,"PointD"),n.prototype.getX=function(){return this.x},n.prototype.getY=function(){return this.y},n.prototype.setX=function(i){this.x=i},n.prototype.setY=function(i){this.y=i},n.prototype.getDifference=function(i){return new DimensionD(this.x-i.x,this.y-i.y)},n.prototype.getCopy=function(){return new n(this.x,this.y)},n.prototype.translate=function(i){return this.x+=i.width,this.y+=i.height,this},t.exports=n},function(t,e,r){"use strict";var n=r(2),i=r(10),a=r(0),s=r(7),l=r(3),u=r(1),h=r(13),f=r(12),d=r(11);function p(g,y,v){n.call(this,v),this.estimatedSize=i.MIN_VALUE,this.margin=a.DEFAULT_GRAPH_MARGIN,this.edges=[],this.nodes=[],this.isConnected=!1,this.parent=g,y!=null&&y instanceof s?this.graphManager=y:y!=null&&y instanceof Layout&&(this.graphManager=y.graphManager)}o(p,"LGraph"),p.prototype=Object.create(n.prototype);for(var m in n)p[m]=n[m];p.prototype.getNodes=function(){return this.nodes},p.prototype.getEdges=function(){return this.edges},p.prototype.getGraphManager=function(){return this.graphManager},p.prototype.getParent=function(){return this.parent},p.prototype.getLeft=function(){return this.left},p.prototype.getRight=function(){return this.right},p.prototype.getTop=function(){return this.top},p.prototype.getBottom=function(){return this.bottom},p.prototype.isConnected=function(){return this.isConnected},p.prototype.add=function(g,y,v){if(y==null&&v==null){var x=g;if(this.graphManager==null)throw"Graph has no graph mgr!";if(this.getNodes().indexOf(x)>-1)throw"Node already in graph!";return x.owner=this,this.getNodes().push(x),x}else{var b=g;if(!(this.getNodes().indexOf(y)>-1&&this.getNodes().indexOf(v)>-1))throw"Source or target not in graph!";if(!(y.owner==v.owner&&y.owner==this))throw"Both owners must be this graph!";return y.owner!=v.owner?null:(b.source=y,b.target=v,b.isInterGraph=!1,this.getEdges().push(b),y.edges.push(b),v!=y&&v.edges.push(b),b)}},p.prototype.remove=function(g){var y=g;if(g instanceof l){if(y==null)throw"Node is null!";if(!(y.owner!=null&&y.owner==this))throw"Owner graph is invalid!";if(this.graphManager==null)throw"Owner graph manager is invalid!";for(var v=y.edges.slice(),x,b=v.length,w=0;w-1&&E>-1))throw"Source and/or target doesn't know this edge!";x.source.edges.splice(T,1),x.target!=x.source&&x.target.edges.splice(E,1);var S=x.source.owner.getEdges().indexOf(x);if(S==-1)throw"Not in owner's edge list!";x.source.owner.getEdges().splice(S,1)}},p.prototype.updateLeftTop=function(){for(var g=i.MAX_VALUE,y=i.MAX_VALUE,v,x,b,w=this.getNodes(),S=w.length,T=0;Tv&&(g=v),y>x&&(y=x)}return g==i.MAX_VALUE?null:(w[0].getParent().paddingLeft!=null?b=w[0].getParent().paddingLeft:b=this.margin,this.left=y-b,this.top=g-b,new f(this.left,this.top))},p.prototype.updateBounds=function(g){for(var y=i.MAX_VALUE,v=-i.MAX_VALUE,x=i.MAX_VALUE,b=-i.MAX_VALUE,w,S,T,E,_,A=this.nodes,L=A.length,M=0;Mw&&(y=w),vT&&(x=T),bw&&(y=w),vT&&(x=T),b=this.nodes.length){var L=0;v.forEach(function(M){M.owner==g&&L++}),L==this.nodes.length&&(this.isConnected=!0)}},t.exports=p},function(t,e,r){"use strict";var n,i=r(1);function a(s){n=r(6),this.layout=s,this.graphs=[],this.edges=[]}o(a,"LGraphManager"),a.prototype.addRoot=function(){var s=this.layout.newGraph(),l=this.layout.newNode(null),u=this.add(s,l);return this.setRootGraph(u),this.rootGraph},a.prototype.add=function(s,l,u,h,f){if(u==null&&h==null&&f==null){if(s==null)throw"Graph is null!";if(l==null)throw"Parent node is null!";if(this.graphs.indexOf(s)>-1)throw"Graph already in this graph mgr!";if(this.graphs.push(s),s.parent!=null)throw"Already has a parent!";if(l.child!=null)throw"Already has a child!";return s.parent=l,l.child=s,s}else{f=u,h=l,u=s;var d=h.getOwner(),p=f.getOwner();if(!(d!=null&&d.getGraphManager()==this))throw"Source not in this graph mgr!";if(!(p!=null&&p.getGraphManager()==this))throw"Target not in this graph mgr!";if(d==p)return u.isInterGraph=!1,d.add(u,h,f);if(u.isInterGraph=!0,u.source=h,u.target=f,this.edges.indexOf(u)>-1)throw"Edge already in inter-graph edge list!";if(this.edges.push(u),!(u.source!=null&&u.target!=null))throw"Edge source and/or target is null!";if(!(u.source.edges.indexOf(u)==-1&&u.target.edges.indexOf(u)==-1))throw"Edge already in source and/or target incidency list!";return u.source.edges.push(u),u.target.edges.push(u),u}},a.prototype.remove=function(s){if(s instanceof n){var l=s;if(l.getGraphManager()!=this)throw"Graph not in this graph mgr";if(!(l==this.rootGraph||l.parent!=null&&l.parent.graphManager==this))throw"Invalid parent node!";var u=[];u=u.concat(l.getEdges());for(var h,f=u.length,d=0;d=s.getRight()?l[0]+=Math.min(s.getX()-a.getX(),a.getRight()-s.getRight()):s.getX()<=a.getX()&&s.getRight()>=a.getRight()&&(l[0]+=Math.min(a.getX()-s.getX(),s.getRight()-a.getRight())),a.getY()<=s.getY()&&a.getBottom()>=s.getBottom()?l[1]+=Math.min(s.getY()-a.getY(),a.getBottom()-s.getBottom()):s.getY()<=a.getY()&&s.getBottom()>=a.getBottom()&&(l[1]+=Math.min(a.getY()-s.getY(),s.getBottom()-a.getBottom()));var f=Math.abs((s.getCenterY()-a.getCenterY())/(s.getCenterX()-a.getCenterX()));s.getCenterY()===a.getCenterY()&&s.getCenterX()===a.getCenterX()&&(f=1);var d=f*l[0],p=l[1]/f;l[0]d)return l[0]=u,l[1]=m,l[2]=f,l[3]=A,!1;if(hf)return l[0]=p,l[1]=h,l[2]=E,l[3]=d,!1;if(uf?(l[0]=y,l[1]=v,k=!0):(l[0]=g,l[1]=m,k=!0):C===D&&(u>f?(l[0]=p,l[1]=m,k=!0):(l[0]=x,l[1]=v,k=!0)),-O===D?f>u?(l[2]=_,l[3]=A,I=!0):(l[2]=E,l[3]=T,I=!0):O===D&&(f>u?(l[2]=S,l[3]=T,I=!0):(l[2]=L,l[3]=A,I=!0)),k&&I)return!1;if(u>f?h>d?(P=this.getCardinalDirection(C,D,4),F=this.getCardinalDirection(O,D,2)):(P=this.getCardinalDirection(-C,D,3),F=this.getCardinalDirection(-O,D,1)):h>d?(P=this.getCardinalDirection(-C,D,1),F=this.getCardinalDirection(-O,D,3)):(P=this.getCardinalDirection(C,D,2),F=this.getCardinalDirection(O,D,4)),!k)switch(P){case 1:$=m,B=u+-w/D,l[0]=B,l[1]=$;break;case 2:B=x,$=h+b*D,l[0]=B,l[1]=$;break;case 3:$=v,B=u+w/D,l[0]=B,l[1]=$;break;case 4:B=y,$=h+-b*D,l[0]=B,l[1]=$;break}if(!I)switch(F){case 1:Y=T,z=f+-N/D,l[2]=z,l[3]=Y;break;case 2:z=L,Y=d+M*D,l[2]=z,l[3]=Y;break;case 3:Y=A,z=f+N/D,l[2]=z,l[3]=Y;break;case 4:z=_,Y=d+-M*D,l[2]=z,l[3]=Y;break}}return!1},i.getCardinalDirection=function(a,s,l){return a>s?l:1+l%4},i.getIntersection=function(a,s,l,u){if(u==null)return this.getIntersection2(a,s,l);var h=a.x,f=a.y,d=s.x,p=s.y,m=l.x,g=l.y,y=u.x,v=u.y,x=void 0,b=void 0,w=void 0,S=void 0,T=void 0,E=void 0,_=void 0,A=void 0,L=void 0;return w=p-f,T=h-d,_=d*f-h*p,S=v-g,E=m-y,A=y*g-m*v,L=w*E-S*T,L===0?null:(x=(T*A-E*_)/L,b=(S*_-w*A)/L,new n(x,b))},i.angleOfVector=function(a,s,l,u){var h=void 0;return a!==l?(h=Math.atan((u-s)/(l-a)),l=0){var v=(-m+Math.sqrt(m*m-4*p*g))/(2*p),x=(-m-Math.sqrt(m*m-4*p*g))/(2*p),b=null;return v>=0&&v<=1?[v]:x>=0&&x<=1?[x]:b}else return null},i.HALF_PI=.5*Math.PI,i.ONE_AND_HALF_PI=1.5*Math.PI,i.TWO_PI=2*Math.PI,i.THREE_PI=3*Math.PI,t.exports=i},function(t,e,r){"use strict";function n(){}o(n,"IMath"),n.sign=function(i){return i>0?1:i<0?-1:0},n.floor=function(i){return i<0?Math.ceil(i):Math.floor(i)},n.ceil=function(i){return i<0?Math.floor(i):Math.ceil(i)},t.exports=n},function(t,e,r){"use strict";function n(){}o(n,"Integer"),n.MAX_VALUE=2147483647,n.MIN_VALUE=-2147483648,t.exports=n},function(t,e,r){"use strict";var n=function(){function h(f,d){for(var p=0;p"u"?"undefined":n(a);return a==null||s!="object"&&s!="function"},t.exports=i},function(t,e,r){"use strict";function n(m){if(Array.isArray(m)){for(var g=0,y=Array(m.length);g0&&g;){for(w.push(T[0]);w.length>0&&g;){var E=w[0];w.splice(0,1),b.add(E);for(var _=E.getEdges(),x=0;x<_.length;x++){var A=_[x].getOtherEnd(E);if(S.get(E)!=A)if(!b.has(A))w.push(A),S.set(A,E);else{g=!1;break}}}if(!g)m=[];else{var L=[].concat(n(b));m.push(L);for(var x=0;x-1&&T.splice(N,1)}b=new Set,S=new Map}}return m},p.prototype.createDummyNodesForBendpoints=function(m){for(var g=[],y=m.source,v=this.graphManager.calcLowestCommonAncestor(m.source,m.target),x=0;x0){for(var v=this.edgeToDummyNodes.get(y),x=0;x=0&&g.splice(A,1);var L=S.getNeighborsList();L.forEach(function(k){if(y.indexOf(k)<0){var I=v.get(k),C=I-1;C==1&&E.push(k),v.set(k,C)}})}y=y.concat(E),(g.length==1||g.length==2)&&(x=!0,b=g[0])}return b},p.prototype.setGraphManager=function(m){this.graphManager=m},t.exports=p},function(t,e,r){"use strict";function n(){}o(n,"RandomSeed"),n.seed=1,n.x=0,n.nextDouble=function(){return n.x=Math.sin(n.seed++)*1e4,n.x-Math.floor(n.x)},t.exports=n},function(t,e,r){"use strict";var n=r(5);function i(a,s){this.lworldOrgX=0,this.lworldOrgY=0,this.ldeviceOrgX=0,this.ldeviceOrgY=0,this.lworldExtX=1,this.lworldExtY=1,this.ldeviceExtX=1,this.ldeviceExtY=1}o(i,"Transform"),i.prototype.getWorldOrgX=function(){return this.lworldOrgX},i.prototype.setWorldOrgX=function(a){this.lworldOrgX=a},i.prototype.getWorldOrgY=function(){return this.lworldOrgY},i.prototype.setWorldOrgY=function(a){this.lworldOrgY=a},i.prototype.getWorldExtX=function(){return this.lworldExtX},i.prototype.setWorldExtX=function(a){this.lworldExtX=a},i.prototype.getWorldExtY=function(){return this.lworldExtY},i.prototype.setWorldExtY=function(a){this.lworldExtY=a},i.prototype.getDeviceOrgX=function(){return this.ldeviceOrgX},i.prototype.setDeviceOrgX=function(a){this.ldeviceOrgX=a},i.prototype.getDeviceOrgY=function(){return this.ldeviceOrgY},i.prototype.setDeviceOrgY=function(a){this.ldeviceOrgY=a},i.prototype.getDeviceExtX=function(){return this.ldeviceExtX},i.prototype.setDeviceExtX=function(a){this.ldeviceExtX=a},i.prototype.getDeviceExtY=function(){return this.ldeviceExtY},i.prototype.setDeviceExtY=function(a){this.ldeviceExtY=a},i.prototype.transformX=function(a){var s=0,l=this.lworldExtX;return l!=0&&(s=this.ldeviceOrgX+(a-this.lworldOrgX)*this.ldeviceExtX/l),s},i.prototype.transformY=function(a){var s=0,l=this.lworldExtY;return l!=0&&(s=this.ldeviceOrgY+(a-this.lworldOrgY)*this.ldeviceExtY/l),s},i.prototype.inverseTransformX=function(a){var s=0,l=this.ldeviceExtX;return l!=0&&(s=this.lworldOrgX+(a-this.ldeviceOrgX)*this.lworldExtX/l),s},i.prototype.inverseTransformY=function(a){var s=0,l=this.ldeviceExtY;return l!=0&&(s=this.lworldOrgY+(a-this.ldeviceOrgY)*this.lworldExtY/l),s},i.prototype.inverseTransformPoint=function(a){var s=new n(this.inverseTransformX(a.x),this.inverseTransformY(a.y));return s},t.exports=i},function(t,e,r){"use strict";function n(d){if(Array.isArray(d)){for(var p=0,m=Array(d.length);pa.ADAPTATION_LOWER_NODE_LIMIT&&(this.coolingFactor=Math.max(this.coolingFactor*a.COOLING_ADAPTATION_FACTOR,this.coolingFactor-(d-a.ADAPTATION_LOWER_NODE_LIMIT)/(a.ADAPTATION_UPPER_NODE_LIMIT-a.ADAPTATION_LOWER_NODE_LIMIT)*this.coolingFactor*(1-a.COOLING_ADAPTATION_FACTOR))),this.maxNodeDisplacement=a.MAX_NODE_DISPLACEMENT_INCREMENTAL):(d>a.ADAPTATION_LOWER_NODE_LIMIT?this.coolingFactor=Math.max(a.COOLING_ADAPTATION_FACTOR,1-(d-a.ADAPTATION_LOWER_NODE_LIMIT)/(a.ADAPTATION_UPPER_NODE_LIMIT-a.ADAPTATION_LOWER_NODE_LIMIT)*(1-a.COOLING_ADAPTATION_FACTOR)):this.coolingFactor=1,this.initialCoolingFactor=this.coolingFactor,this.maxNodeDisplacement=a.MAX_NODE_DISPLACEMENT),this.maxIterations=Math.max(this.getAllNodes().length*5,this.maxIterations),this.displacementThresholdPerNode=3*a.DEFAULT_EDGE_LENGTH/100,this.totalDisplacementThreshold=this.displacementThresholdPerNode*this.getAllNodes().length,this.repulsionRange=this.calcRepulsionRange()},h.prototype.calcSpringForces=function(){for(var d=this.getAllEdges(),p,m=0;m0&&arguments[0]!==void 0?arguments[0]:!0,p=arguments.length>1&&arguments[1]!==void 0?arguments[1]:!1,m,g,y,v,x=this.getAllNodes(),b;if(this.useFRGridVariant)for(this.totalIterations%a.GRID_CALCULATION_CHECK_PERIOD==1&&d&&this.updateGrid(),b=new Set,m=0;mw||b>w)&&(d.gravitationForceX=-this.gravityConstant*y,d.gravitationForceY=-this.gravityConstant*v)):(w=p.getEstimatedSize()*this.compoundGravityRangeFactor,(x>w||b>w)&&(d.gravitationForceX=-this.gravityConstant*y*this.compoundGravityConstant,d.gravitationForceY=-this.gravityConstant*v*this.compoundGravityConstant))},h.prototype.isConverged=function(){var d,p=!1;return this.totalIterations>this.maxIterations/3&&(p=Math.abs(this.totalDisplacement-this.oldTotalDisplacement)<2),d=this.totalDisplacement=x.length||w>=x[0].length)){for(var S=0;Sh},"_defaultCompareFunction")}]),l}();t.exports=s},function(t,e,r){"use strict";function n(){}o(n,"SVD"),n.svd=function(i){this.U=null,this.V=null,this.s=null,this.m=0,this.n=0,this.m=i.length,this.n=i[0].length;var a=Math.min(this.m,this.n);this.s=function(it){for(var dt=[];it-- >0;)dt.push(0);return dt}(Math.min(this.m+1,this.n)),this.U=function(it){var dt=o(function lt(It){if(It.length==0)return 0;for(var mt=[],St=0;St0;)dt.push(0);return dt}(this.n),l=function(it){for(var dt=[];it-- >0;)dt.push(0);return dt}(this.m),u=!0,h=!0,f=Math.min(this.m-1,this.n),d=Math.max(0,Math.min(this.n-2,this.m)),p=0;p=0;D--)if(this.s[D]!==0){for(var P=D+1;P=0;X--){if(function(it,dt){return it&&dt}(X0;){var ue=void 0,te=void 0;for(ue=I-2;ue>=-1&&ue!==-1;ue--)if(Math.abs(s[ue])<=ce+se*(Math.abs(this.s[ue])+Math.abs(this.s[ue+1]))){s[ue]=0;break}if(ue===I-2)te=4;else{var De=void 0;for(De=I-1;De>=ue&&De!==ue;De--){var oe=(De!==I?Math.abs(s[De]):0)+(De!==ue+1?Math.abs(s[De-1]):0);if(Math.abs(this.s[De])<=ce+se*oe){this.s[De]=0;break}}De===ue?te=3:De===I-1?te=1:(te=2,ue=De)}switch(ue++,te){case 1:{var ke=s[I-2];s[I-2]=0;for(var Ie=I-2;Ie>=ue;Ie--){var Se=n.hypot(this.s[Ie],ke),Ue=this.s[Ie]/Se,Pe=ke/Se;if(this.s[Ie]=Se,Ie!==ue&&(ke=-Pe*s[Ie-1],s[Ie-1]=Ue*s[Ie-1]),h)for(var _e=0;_e=this.s[ue+1]);){var je=this.s[ue];if(this.s[ue]=this.s[ue+1],this.s[ue+1]=je,h&&ueMath.abs(a)?(s=a/i,s=Math.abs(i)*Math.sqrt(1+s*s)):a!=0?(s=i/a,s=Math.abs(a)*Math.sqrt(1+s*s)):s=0,s},t.exports=n},function(t,e,r){"use strict";var n=function(){function s(l,u){for(var h=0;h2&&arguments[2]!==void 0?arguments[2]:1,f=arguments.length>3&&arguments[3]!==void 0?arguments[3]:-1,d=arguments.length>4&&arguments[4]!==void 0?arguments[4]:-1;i(this,s),this.sequence1=l,this.sequence2=u,this.match_score=h,this.mismatch_penalty=f,this.gap_penalty=d,this.iMax=l.length+1,this.jMax=u.length+1,this.grid=new Array(this.iMax);for(var p=0;p=0;l--){var u=this.listeners[l];u.event===a&&u.callback===s&&this.listeners.splice(l,1)}},i.emit=function(a,s){for(var l=0;l{"use strict";o(function(e,r){typeof vb=="object"&&typeof QB=="object"?QB.exports=r(KB()):typeof define=="function"&&define.amd?define(["layout-base"],r):typeof vb=="object"?vb.coseBase=r(KB()):e.coseBase=r(e.layoutBase)},"webpackUniversalModuleDefinition")(vb,function(t){return(()=>{"use strict";var e={45:(a,s,l)=>{var u={};u.layoutBase=l(551),u.CoSEConstants=l(806),u.CoSEEdge=l(767),u.CoSEGraph=l(880),u.CoSEGraphManager=l(578),u.CoSELayout=l(765),u.CoSENode=l(991),u.ConstraintHandler=l(902),a.exports=u},806:(a,s,l)=>{var u=l(551).FDLayoutConstants;function h(){}o(h,"CoSEConstants");for(var f in u)h[f]=u[f];h.DEFAULT_USE_MULTI_LEVEL_SCALING=!1,h.DEFAULT_RADIAL_SEPARATION=u.DEFAULT_EDGE_LENGTH,h.DEFAULT_COMPONENT_SEPERATION=60,h.TILE=!0,h.TILING_PADDING_VERTICAL=10,h.TILING_PADDING_HORIZONTAL=10,h.TRANSFORM_ON_CONSTRAINT_HANDLING=!0,h.ENFORCE_CONSTRAINTS=!0,h.APPLY_LAYOUT=!0,h.RELAX_MOVEMENT_ON_CONSTRAINTS=!0,h.TREE_REDUCTION_ON_INCREMENTAL=!0,h.PURE_INCREMENTAL=h.DEFAULT_INCREMENTAL,a.exports=h},767:(a,s,l)=>{var u=l(551).FDLayoutEdge;function h(d,p,m){u.call(this,d,p,m)}o(h,"CoSEEdge"),h.prototype=Object.create(u.prototype);for(var f in u)h[f]=u[f];a.exports=h},880:(a,s,l)=>{var u=l(551).LGraph;function h(d,p,m){u.call(this,d,p,m)}o(h,"CoSEGraph"),h.prototype=Object.create(u.prototype);for(var f in u)h[f]=u[f];a.exports=h},578:(a,s,l)=>{var u=l(551).LGraphManager;function h(d){u.call(this,d)}o(h,"CoSEGraphManager"),h.prototype=Object.create(u.prototype);for(var f in u)h[f]=u[f];a.exports=h},765:(a,s,l)=>{var u=l(551).FDLayout,h=l(578),f=l(880),d=l(991),p=l(767),m=l(806),g=l(902),y=l(551).FDLayoutConstants,v=l(551).LayoutConstants,x=l(551).Point,b=l(551).PointD,w=l(551).DimensionD,S=l(551).Layout,T=l(551).Integer,E=l(551).IGeometry,_=l(551).LGraph,A=l(551).Transform,L=l(551).LinkedList;function M(){u.call(this),this.toBeTiled={},this.constraints={}}o(M,"CoSELayout"),M.prototype=Object.create(u.prototype);for(var N in u)M[N]=u[N];M.prototype.newGraphManager=function(){var k=new h(this);return this.graphManager=k,k},M.prototype.newGraph=function(k){return new f(null,this.graphManager,k)},M.prototype.newNode=function(k){return new d(this.graphManager,k)},M.prototype.newEdge=function(k){return new p(null,null,k)},M.prototype.initParameters=function(){u.prototype.initParameters.call(this,arguments),this.isSubLayout||(m.DEFAULT_EDGE_LENGTH<10?this.idealEdgeLength=10:this.idealEdgeLength=m.DEFAULT_EDGE_LENGTH,this.useSmartIdealEdgeLengthCalculation=m.DEFAULT_USE_SMART_IDEAL_EDGE_LENGTH_CALCULATION,this.gravityConstant=y.DEFAULT_GRAVITY_STRENGTH,this.compoundGravityConstant=y.DEFAULT_COMPOUND_GRAVITY_STRENGTH,this.gravityRangeFactor=y.DEFAULT_GRAVITY_RANGE_FACTOR,this.compoundGravityRangeFactor=y.DEFAULT_COMPOUND_GRAVITY_RANGE_FACTOR,this.prunedNodesAll=[],this.growTreeIterations=0,this.afterGrowthIterations=0,this.isTreeGrowing=!1,this.isGrowthFinished=!1)},M.prototype.initSpringEmbedder=function(){u.prototype.initSpringEmbedder.call(this),this.coolingCycle=0,this.maxCoolingCycle=this.maxIterations/y.CONVERGENCE_CHECK_PERIOD,this.finalTemperature=.04,this.coolingAdjuster=1},M.prototype.layout=function(){var k=v.DEFAULT_CREATE_BENDS_AS_NEEDED;return k&&(this.createBendpoints(),this.graphManager.resetAllEdges()),this.level=0,this.classicLayout()},M.prototype.classicLayout=function(){if(this.nodesWithGravity=this.calculateNodesToApplyGravitationTo(),this.graphManager.setAllNodesToApplyGravitation(this.nodesWithGravity),this.calcNoOfChildrenForAllNodes(),this.graphManager.calcLowestCommonAncestors(),this.graphManager.calcInclusionTreeDepths(),this.graphManager.getRoot().calcEstimatedSize(),this.calcIdealEdgeLengths(),this.incremental){if(m.TREE_REDUCTION_ON_INCREMENTAL){this.reduceTrees(),this.graphManager.resetAllNodesToApplyGravitation();var I=new Set(this.getAllNodes()),C=this.nodesWithGravity.filter(function(P){return I.has(P)});this.graphManager.setAllNodesToApplyGravitation(C)}}else{var k=this.getFlatForest();if(k.length>0)this.positionNodesRadially(k);else{this.reduceTrees(),this.graphManager.resetAllNodesToApplyGravitation();var I=new Set(this.getAllNodes()),C=this.nodesWithGravity.filter(function(O){return I.has(O)});this.graphManager.setAllNodesToApplyGravitation(C),this.positionNodesRandomly()}}return Object.keys(this.constraints).length>0&&(g.handleConstraints(this),this.initConstraintVariables()),this.initSpringEmbedder(),m.APPLY_LAYOUT&&this.runSpringEmbedder(),!0},M.prototype.tick=function(){if(this.totalIterations++,this.totalIterations===this.maxIterations&&!this.isTreeGrowing&&!this.isGrowthFinished)if(this.prunedNodesAll.length>0)this.isTreeGrowing=!0;else return!0;if(this.totalIterations%y.CONVERGENCE_CHECK_PERIOD==0&&!this.isTreeGrowing&&!this.isGrowthFinished){if(this.isConverged())if(this.prunedNodesAll.length>0)this.isTreeGrowing=!0;else return!0;this.coolingCycle++,this.layoutQuality==0?this.coolingAdjuster=this.coolingCycle:this.layoutQuality==1&&(this.coolingAdjuster=this.coolingCycle/3),this.coolingFactor=Math.max(this.initialCoolingFactor-Math.pow(this.coolingCycle,Math.log(100*(this.initialCoolingFactor-this.finalTemperature))/Math.log(this.maxCoolingCycle))/100*this.coolingAdjuster,this.finalTemperature),this.animationPeriod=Math.ceil(this.initialAnimationPeriod*Math.sqrt(this.coolingFactor))}if(this.isTreeGrowing){if(this.growTreeIterations%10==0)if(this.prunedNodesAll.length>0){this.graphManager.updateBounds(),this.updateGrid(),this.growTree(this.prunedNodesAll),this.graphManager.resetAllNodesToApplyGravitation();var k=new Set(this.getAllNodes()),I=this.nodesWithGravity.filter(function(D){return k.has(D)});this.graphManager.setAllNodesToApplyGravitation(I),this.graphManager.updateBounds(),this.updateGrid(),m.PURE_INCREMENTAL?this.coolingFactor=y.DEFAULT_COOLING_FACTOR_INCREMENTAL/2:this.coolingFactor=y.DEFAULT_COOLING_FACTOR_INCREMENTAL}else this.isTreeGrowing=!1,this.isGrowthFinished=!0;this.growTreeIterations++}if(this.isGrowthFinished){if(this.isConverged())return!0;this.afterGrowthIterations%10==0&&(this.graphManager.updateBounds(),this.updateGrid()),m.PURE_INCREMENTAL?this.coolingFactor=y.DEFAULT_COOLING_FACTOR_INCREMENTAL/2*((100-this.afterGrowthIterations)/100):this.coolingFactor=y.DEFAULT_COOLING_FACTOR_INCREMENTAL*((100-this.afterGrowthIterations)/100),this.afterGrowthIterations++}var C=!this.isTreeGrowing&&!this.isGrowthFinished,O=this.growTreeIterations%10==1&&this.isTreeGrowing||this.afterGrowthIterations%10==1&&this.isGrowthFinished;return this.totalDisplacement=0,this.graphManager.updateBounds(),this.calcSpringForces(),this.calcRepulsionForces(C,O),this.calcGravitationalForces(),this.moveNodes(),this.animate(),!1},M.prototype.getPositionsData=function(){for(var k=this.graphManager.getAllNodes(),I={},C=0;C0&&this.updateDisplacements();for(var C=0;C0&&(O.fixedNodeWeight=P)}}if(this.constraints.relativePlacementConstraint){var F=new Map,B=new Map;if(this.dummyToNodeForVerticalAlignment=new Map,this.dummyToNodeForHorizontalAlignment=new Map,this.fixedNodesOnHorizontal=new Set,this.fixedNodesOnVertical=new Set,this.fixedNodeSet.forEach(function(J){k.fixedNodesOnHorizontal.add(J),k.fixedNodesOnVertical.add(J)}),this.constraints.alignmentConstraint){if(this.constraints.alignmentConstraint.vertical)for(var $=this.constraints.alignmentConstraint.vertical,C=0;C<$.length;C++)this.dummyToNodeForVerticalAlignment.set("dummy"+C,[]),$[C].forEach(function(Z){F.set(Z,"dummy"+C),k.dummyToNodeForVerticalAlignment.get("dummy"+C).push(Z),k.fixedNodeSet.has(Z)&&k.fixedNodesOnHorizontal.add("dummy"+C)});if(this.constraints.alignmentConstraint.horizontal)for(var z=this.constraints.alignmentConstraint.horizontal,C=0;C=2*J.length/3;q--)Z=Math.floor(Math.random()*(q+1)),H=J[q],J[q]=J[Z],J[Z]=H;return J},this.nodesInRelativeHorizontal=[],this.nodesInRelativeVertical=[],this.nodeToRelativeConstraintMapHorizontal=new Map,this.nodeToRelativeConstraintMapVertical=new Map,this.nodeToTempPositionMapHorizontal=new Map,this.nodeToTempPositionMapVertical=new Map,this.constraints.relativePlacementConstraint.forEach(function(J){if(J.left){var Z=F.has(J.left)?F.get(J.left):J.left,H=F.has(J.right)?F.get(J.right):J.right;k.nodesInRelativeHorizontal.includes(Z)||(k.nodesInRelativeHorizontal.push(Z),k.nodeToRelativeConstraintMapHorizontal.set(Z,[]),k.dummyToNodeForVerticalAlignment.has(Z)?k.nodeToTempPositionMapHorizontal.set(Z,k.idToNodeMap.get(k.dummyToNodeForVerticalAlignment.get(Z)[0]).getCenterX()):k.nodeToTempPositionMapHorizontal.set(Z,k.idToNodeMap.get(Z).getCenterX())),k.nodesInRelativeHorizontal.includes(H)||(k.nodesInRelativeHorizontal.push(H),k.nodeToRelativeConstraintMapHorizontal.set(H,[]),k.dummyToNodeForVerticalAlignment.has(H)?k.nodeToTempPositionMapHorizontal.set(H,k.idToNodeMap.get(k.dummyToNodeForVerticalAlignment.get(H)[0]).getCenterX()):k.nodeToTempPositionMapHorizontal.set(H,k.idToNodeMap.get(H).getCenterX())),k.nodeToRelativeConstraintMapHorizontal.get(Z).push({right:H,gap:J.gap}),k.nodeToRelativeConstraintMapHorizontal.get(H).push({left:Z,gap:J.gap})}else{var q=B.has(J.top)?B.get(J.top):J.top,K=B.has(J.bottom)?B.get(J.bottom):J.bottom;k.nodesInRelativeVertical.includes(q)||(k.nodesInRelativeVertical.push(q),k.nodeToRelativeConstraintMapVertical.set(q,[]),k.dummyToNodeForHorizontalAlignment.has(q)?k.nodeToTempPositionMapVertical.set(q,k.idToNodeMap.get(k.dummyToNodeForHorizontalAlignment.get(q)[0]).getCenterY()):k.nodeToTempPositionMapVertical.set(q,k.idToNodeMap.get(q).getCenterY())),k.nodesInRelativeVertical.includes(K)||(k.nodesInRelativeVertical.push(K),k.nodeToRelativeConstraintMapVertical.set(K,[]),k.dummyToNodeForHorizontalAlignment.has(K)?k.nodeToTempPositionMapVertical.set(K,k.idToNodeMap.get(k.dummyToNodeForHorizontalAlignment.get(K)[0]).getCenterY()):k.nodeToTempPositionMapVertical.set(K,k.idToNodeMap.get(K).getCenterY())),k.nodeToRelativeConstraintMapVertical.get(q).push({bottom:K,gap:J.gap}),k.nodeToRelativeConstraintMapVertical.get(K).push({top:q,gap:J.gap})}});else{var Y=new Map,Q=new Map;this.constraints.relativePlacementConstraint.forEach(function(J){if(J.left){var Z=F.has(J.left)?F.get(J.left):J.left,H=F.has(J.right)?F.get(J.right):J.right;Y.has(Z)?Y.get(Z).push(H):Y.set(Z,[H]),Y.has(H)?Y.get(H).push(Z):Y.set(H,[Z])}else{var q=B.has(J.top)?B.get(J.top):J.top,K=B.has(J.bottom)?B.get(J.bottom):J.bottom;Q.has(q)?Q.get(q).push(K):Q.set(q,[K]),Q.has(K)?Q.get(K).push(q):Q.set(K,[q])}});var X=o(function(Z,H){var q=[],K=[],se=new L,ce=new Set,ue=0;return Z.forEach(function(te,De){if(!ce.has(De)){q[ue]=[],K[ue]=!1;var oe=De;for(se.push(oe),ce.add(oe),q[ue].push(oe);se.length!=0;){oe=se.shift(),H.has(oe)&&(K[ue]=!0);var ke=Z.get(oe);ke.forEach(function(Ie){ce.has(Ie)||(se.push(Ie),ce.add(Ie),q[ue].push(Ie))})}ue++}}),{components:q,isFixed:K}},"constructComponents"),ie=X(Y,k.fixedNodesOnHorizontal);this.componentsOnHorizontal=ie.components,this.fixedComponentsOnHorizontal=ie.isFixed;var j=X(Q,k.fixedNodesOnVertical);this.componentsOnVertical=j.components,this.fixedComponentsOnVertical=j.isFixed}}},M.prototype.updateDisplacements=function(){var k=this;if(this.constraints.fixedNodeConstraint&&this.constraints.fixedNodeConstraint.forEach(function(j){var J=k.idToNodeMap.get(j.nodeId);J.displacementX=0,J.displacementY=0}),this.constraints.alignmentConstraint){if(this.constraints.alignmentConstraint.vertical)for(var I=this.constraints.alignmentConstraint.vertical,C=0;C1){var B;for(B=0;BO&&(O=Math.floor(F.y)),P=Math.floor(F.x+m.DEFAULT_COMPONENT_SEPERATION)}this.transform(new b(v.WORLD_CENTER_X-F.x/2,v.WORLD_CENTER_Y-F.y/2))},M.radialLayout=function(k,I,C){var O=Math.max(this.maxDiagonalInTree(k),m.DEFAULT_RADIAL_SEPARATION);M.branchRadialLayout(I,null,0,359,0,O);var D=_.calculateBounds(k),P=new A;P.setDeviceOrgX(D.getMinX()),P.setDeviceOrgY(D.getMinY()),P.setWorldOrgX(C.x),P.setWorldOrgY(C.y);for(var F=0;F1;){var q=H[0];H.splice(0,1);var K=X.indexOf(q);K>=0&&X.splice(K,1),J--,ie--}I!=null?Z=(X.indexOf(H[0])+1)%J:Z=0;for(var se=Math.abs(O-C)/ie,ce=Z;j!=ie;ce=++ce%J){var ue=X[ce].getOtherEnd(k);if(ue!=I){var te=(C+j*se)%360,De=(te+se)%360;M.branchRadialLayout(ue,k,te,De,D+P,P),j++}}},M.maxDiagonalInTree=function(k){for(var I=T.MIN_VALUE,C=0;CI&&(I=D)}return I},M.prototype.calcRepulsionRange=function(){return 2*(this.level+1)*this.idealEdgeLength},M.prototype.groupZeroDegreeMembers=function(){var k=this,I={};this.memberGroups={},this.idToDummyNode={};for(var C=[],O=this.graphManager.getAllNodes(),D=0;D"u"&&(I[B]=[]),I[B]=I[B].concat(P)}Object.keys(I).forEach(function($){if(I[$].length>1){var z="DummyCompound_"+$;k.memberGroups[z]=I[$];var Y=I[$][0].getParent(),Q=new d(k.graphManager);Q.id=z,Q.paddingLeft=Y.paddingLeft||0,Q.paddingRight=Y.paddingRight||0,Q.paddingBottom=Y.paddingBottom||0,Q.paddingTop=Y.paddingTop||0,k.idToDummyNode[z]=Q;var X=k.getGraphManager().add(k.newGraph(),Q),ie=Y.getChild();ie.add(Q);for(var j=0;jD?(O.rect.x-=(O.labelWidth-D)/2,O.setWidth(O.labelWidth),O.labelMarginLeft=(O.labelWidth-D)/2):O.labelPosHorizontal=="right"&&O.setWidth(D+O.labelWidth)),O.labelHeight&&(O.labelPosVertical=="top"?(O.rect.y-=O.labelHeight,O.setHeight(P+O.labelHeight),O.labelMarginTop=O.labelHeight):O.labelPosVertical=="center"&&O.labelHeight>P?(O.rect.y-=(O.labelHeight-P)/2,O.setHeight(O.labelHeight),O.labelMarginTop=(O.labelHeight-P)/2):O.labelPosVertical=="bottom"&&O.setHeight(P+O.labelHeight))}})},M.prototype.repopulateCompounds=function(){for(var k=this.compoundOrder.length-1;k>=0;k--){var I=this.compoundOrder[k],C=I.id,O=I.paddingLeft,D=I.paddingTop,P=I.labelMarginLeft,F=I.labelMarginTop;this.adjustLocations(this.tiledMemberPack[C],I.rect.x,I.rect.y,O,D,P,F)}},M.prototype.repopulateZeroDegreeMembers=function(){var k=this,I=this.tiledZeroDegreePack;Object.keys(I).forEach(function(C){var O=k.idToDummyNode[C],D=O.paddingLeft,P=O.paddingTop,F=O.labelMarginLeft,B=O.labelMarginTop;k.adjustLocations(I[C],O.rect.x,O.rect.y,D,P,F,B)})},M.prototype.getToBeTiled=function(k){var I=k.id;if(this.toBeTiled[I]!=null)return this.toBeTiled[I];var C=k.getChild();if(C==null)return this.toBeTiled[I]=!1,!1;for(var O=C.getNodes(),D=0;D0)return this.toBeTiled[I]=!1,!1;if(P.getChild()==null){this.toBeTiled[P.id]=!1;continue}if(!this.getToBeTiled(P))return this.toBeTiled[I]=!1,!1}return this.toBeTiled[I]=!0,!0},M.prototype.getNodeDegree=function(k){for(var I=k.id,C=k.getEdges(),O=0,D=0;DY&&(Y=X.rect.height)}C+=Y+k.verticalPadding}},M.prototype.tileCompoundMembers=function(k,I){var C=this;this.tiledMemberPack=[],Object.keys(k).forEach(function(O){var D=I[O];if(C.tiledMemberPack[O]=C.tileNodes(k[O],D.paddingLeft+D.paddingRight),D.rect.width=C.tiledMemberPack[O].width,D.rect.height=C.tiledMemberPack[O].height,D.setCenter(C.tiledMemberPack[O].centerX,C.tiledMemberPack[O].centerY),D.labelMarginLeft=0,D.labelMarginTop=0,m.NODE_DIMENSIONS_INCLUDE_LABELS){var P=D.rect.width,F=D.rect.height;D.labelWidth&&(D.labelPosHorizontal=="left"?(D.rect.x-=D.labelWidth,D.setWidth(P+D.labelWidth),D.labelMarginLeft=D.labelWidth):D.labelPosHorizontal=="center"&&D.labelWidth>P?(D.rect.x-=(D.labelWidth-P)/2,D.setWidth(D.labelWidth),D.labelMarginLeft=(D.labelWidth-P)/2):D.labelPosHorizontal=="right"&&D.setWidth(P+D.labelWidth)),D.labelHeight&&(D.labelPosVertical=="top"?(D.rect.y-=D.labelHeight,D.setHeight(F+D.labelHeight),D.labelMarginTop=D.labelHeight):D.labelPosVertical=="center"&&D.labelHeight>F?(D.rect.y-=(D.labelHeight-F)/2,D.setHeight(D.labelHeight),D.labelMarginTop=(D.labelHeight-F)/2):D.labelPosVertical=="bottom"&&D.setHeight(F+D.labelHeight))}})},M.prototype.tileNodes=function(k,I){var C=this.tileNodesByFavoringDim(k,I,!0),O=this.tileNodesByFavoringDim(k,I,!1),D=this.getOrgRatio(C),P=this.getOrgRatio(O),F;return PB&&(B=j.getWidth())});var $=P/D,z=F/D,Y=Math.pow(C-O,2)+4*($+O)*(z+C)*D,Q=(O-C+Math.sqrt(Y))/(2*($+O)),X;I?(X=Math.ceil(Q),X==Q&&X++):X=Math.floor(Q);var ie=X*($+O)-O;return B>ie&&(ie=B),ie+=O*2,ie},M.prototype.tileNodesByFavoringDim=function(k,I,C){var O=m.TILING_PADDING_VERTICAL,D=m.TILING_PADDING_HORIZONTAL,P=m.TILING_COMPARE_BY,F={rows:[],rowWidth:[],rowHeight:[],width:0,height:I,verticalPadding:O,horizontalPadding:D,centerX:0,centerY:0};P&&(F.idealRowWidth=this.calcIdealRowWidth(k,C));var B=o(function(J){return J.rect.width*J.rect.height},"getNodeArea"),$=o(function(J,Z){return B(Z)-B(J)},"areaCompareFcn");k.sort(function(j,J){var Z=$;return F.idealRowWidth?(Z=P,Z(j.id,J.id)):Z(j,J)});for(var z=0,Y=0,Q=0;Q0&&(F+=k.horizontalPadding),k.rowWidth[C]=F,k.width0&&(B+=k.verticalPadding);var $=0;B>k.rowHeight[C]&&($=k.rowHeight[C],k.rowHeight[C]=B,$=k.rowHeight[C]-$),k.height+=$,k.rows[C].push(I)},M.prototype.getShortestRowIndex=function(k){for(var I=-1,C=Number.MAX_VALUE,O=0;OC&&(I=O,C=k.rowWidth[O]);return 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u=l(551).FDLayoutNode,h=l(551).IMath;function f(p,m,g,y){u.call(this,p,m,g,y)}o(f,"CoSENode"),f.prototype=Object.create(u.prototype);for(var d in u)f[d]=u[d];f.prototype.calculateDisplacement=function(){var p=this.graphManager.getLayout();this.getChild()!=null&&this.fixedNodeWeight?(this.displacementX+=p.coolingFactor*(this.springForceX+this.repulsionForceX+this.gravitationForceX)/this.fixedNodeWeight,this.displacementY+=p.coolingFactor*(this.springForceY+this.repulsionForceY+this.gravitationForceY)/this.fixedNodeWeight):(this.displacementX+=p.coolingFactor*(this.springForceX+this.repulsionForceX+this.gravitationForceX)/this.noOfChildren,this.displacementY+=p.coolingFactor*(this.springForceY+this.repulsionForceY+this.gravitationForceY)/this.noOfChildren),Math.abs(this.displacementX)>p.coolingFactor*p.maxNodeDisplacement&&(this.displacementX=p.coolingFactor*p.maxNodeDisplacement*h.sign(this.displacementX)),Math.abs(this.displacementY)>p.coolingFactor*p.maxNodeDisplacement&&(this.displacementY=p.coolingFactor*p.maxNodeDisplacement*h.sign(this.displacementY)),this.child&&this.child.getNodes().length>0&&this.propogateDisplacementToChildren(this.displacementX,this.displacementY)},f.prototype.propogateDisplacementToChildren=function(p,m){for(var g=this.getChild().getNodes(),y,v=0;v{function u(g){if(Array.isArray(g)){for(var y=0,v=Array(g.length);y0){var Je=0;Ye.forEach(function(je){we=="horizontal"?(ye.set(je,x.has(je)?b[x.get(je)]:Ce.get(je)),Je+=ye.get(je)):(ye.set(je,x.has(je)?w[x.get(je)]:Ce.get(je)),Je+=ye.get(je))}),Je=Je/Ye.length,tt.forEach(function(je){Te.has(je)||ye.set(je,Je)})}else{var Ve=0;tt.forEach(function(je){we=="horizontal"?Ve+=x.has(je)?b[x.get(je)]:Ce.get(je):Ve+=x.has(je)?w[x.get(je)]:Ce.get(je)}),Ve=Ve/tt.length,tt.forEach(function(je){ye.set(je,Ve)})}});for(var Ze=o(function(){var Ye=ze.shift(),Je=ae.get(Ye);Je.forEach(function(Ve){if(ye.get(Ve.id)je&&(je=mt),Stkt&&(kt=St)}}catch(Qn){xt=!0,it=Qn}finally{try{!at&&dt.return&&dt.return()}finally{if(xt)throw it}}var gr=(Je+je)/2-(Ve+kt)/2,xn=!0,jt=!1,rn=void 0;try{for(var Er=tt[Symbol.iterator](),Kn;!(xn=(Kn=Er.next()).done);xn=!0){var hn=Kn.value;ye.set(hn,ye.get(hn)+gr)}}catch(Qn){jt=!0,rn=Qn}finally{try{!xn&&Er.return&&Er.return()}finally{if(jt)throw rn}}})}return ye},"findAppropriatePositionForRelativePlacement"),N=o(function(ae){var we=0,Te=0,Ce=0,Ae=0;if(ae.forEach(function(He){He.left?b[x.get(He.left)]-b[x.get(He.right)]>=0?we++:Te++:w[x.get(He.top)]-w[x.get(He.bottom)]>=0?Ce++:Ae++}),we>Te&&Ce>Ae)for(var Ge=0;GeTe)for(var Me=0;MeAe)for(var ye=0;ye1)y.fixedNodeConstraint.forEach(function(ne,ae){O[ae]=[ne.position.x,ne.position.y],D[ae]=[b[x.get(ne.nodeId)],w[x.get(ne.nodeId)]]}),P=!0;else if(y.alignmentConstraint)(function(){var ne=0;if(y.alignmentConstraint.vertical){for(var ae=y.alignmentConstraint.vertical,we=o(function(ye){var He=new Set;ae[ye].forEach(function(gt){He.add(gt)});var ze=new Set([].concat(u(He)).filter(function(gt){return B.has(gt)})),Ze=void 0;ze.size>0?Ze=b[x.get(ze.values().next().value)]:Ze=L(He).x,ae[ye].forEach(function(gt){O[ne]=[Ze,w[x.get(gt)]],D[ne]=[b[x.get(gt)],w[x.get(gt)]],ne++})},"_loop2"),Te=0;Te0?Ze=b[x.get(ze.values().next().value)]:Ze=L(He).y,Ce[ye].forEach(function(gt){O[ne]=[b[x.get(gt)],Ze],D[ne]=[b[x.get(gt)],w[x.get(gt)]],ne++})},"_loop3"),Ge=0;GeQ&&(Q=Y[ie].length,X=ie);if(Q0){var Ue={x:0,y:0};y.fixedNodeConstraint.forEach(function(ne,ae){var we={x:b[x.get(ne.nodeId)],y:w[x.get(ne.nodeId)]},Te=ne.position,Ce=A(Te,we);Ue.x+=Ce.x,Ue.y+=Ce.y}),Ue.x/=y.fixedNodeConstraint.length,Ue.y/=y.fixedNodeConstraint.length,b.forEach(function(ne,ae){b[ae]+=Ue.x}),w.forEach(function(ne,ae){w[ae]+=Ue.y}),y.fixedNodeConstraint.forEach(function(ne){b[x.get(ne.nodeId)]=ne.position.x,w[x.get(ne.nodeId)]=ne.position.y})}if(y.alignmentConstraint){if(y.alignmentConstraint.vertical)for(var Pe=y.alignmentConstraint.vertical,_e=o(function(ae){var we=new Set;Pe[ae].forEach(function(Ae){we.add(Ae)});var Te=new Set([].concat(u(we)).filter(function(Ae){return B.has(Ae)})),Ce=void 0;Te.size>0?Ce=b[x.get(Te.values().next().value)]:Ce=L(we).x,we.forEach(function(Ae){B.has(Ae)||(b[x.get(Ae)]=Ce)})},"_loop4"),me=0;me0?Ce=w[x.get(Te.values().next().value)]:Ce=L(we).y,we.forEach(function(Ae){B.has(Ae)||(w[x.get(Ae)]=Ce)})},"_loop5"),ge=0;ge{a.exports=t}},r={};function n(a){var s=r[a];if(s!==void 0)return s.exports;var l=r[a]={exports:{}};return e[a](l,l.exports,n),l.exports}o(n,"__webpack_require__");var i=n(45);return i})()})});var X1e=gi((xb,JB)=>{"use strict";o(function(e,r){typeof xb=="object"&&typeof JB=="object"?JB.exports=r(ZB()):typeof define=="function"&&define.amd?define(["cose-base"],r):typeof xb=="object"?xb.cytoscapeFcose=r(ZB()):e.cytoscapeFcose=r(e.coseBase)},"webpackUniversalModuleDefinition")(xb,function(t){return(()=>{"use strict";var e={658:a=>{a.exports=Object.assign!=null?Object.assign.bind(Object):function(s){for(var l=arguments.length,u=Array(l>1?l-1:0),h=1;h{var u=function(){function d(p,m){var g=[],y=!0,v=!1,x=void 0;try{for(var b=p[Symbol.iterator](),w;!(y=(w=b.next()).done)&&(g.push(w.value),!(m&&g.length===m));y=!0);}catch(S){v=!0,x=S}finally{try{!y&&b.return&&b.return()}finally{if(v)throw x}}return g}return o(d,"sliceIterator"),function(p,m){if(Array.isArray(p))return p;if(Symbol.iterator in Object(p))return d(p,m);throw new TypeError("Invalid attempt to destructure non-iterable instance")}}(),h=l(140).layoutBase.LinkedList,f={};f.getTopMostNodes=function(d){for(var p={},m=0;m0&&P.merge(z)});for(var F=0;F1){w=x[0],S=w.connectedEdges().length,x.forEach(function(D){D.connectedEdges().length0&&g.set("dummy"+(g.size+1),_),A},f.relocateComponent=function(d,p,m){if(!m.fixedNodeConstraint){var g=Number.POSITIVE_INFINITY,y=Number.NEGATIVE_INFINITY,v=Number.POSITIVE_INFINITY,x=Number.NEGATIVE_INFINITY;if(m.quality=="draft"){var b=!0,w=!1,S=void 0;try{for(var T=p.nodeIndexes[Symbol.iterator](),E;!(b=(E=T.next()).done);b=!0){var _=E.value,A=u(_,2),L=A[0],M=A[1],N=m.cy.getElementById(L);if(N){var k=N.boundingBox(),I=p.xCoords[M]-k.w/2,C=p.xCoords[M]+k.w/2,O=p.yCoords[M]-k.h/2,D=p.yCoords[M]+k.h/2;Iy&&(y=C),Ox&&(x=D)}}}catch(z){w=!0,S=z}finally{try{!b&&T.return&&T.return()}finally{if(w)throw S}}var P=d.x-(y+g)/2,F=d.y-(x+v)/2;p.xCoords=p.xCoords.map(function(z){return z+P}),p.yCoords=p.yCoords.map(function(z){return z+F})}else{Object.keys(p).forEach(function(z){var Y=p[z],Q=Y.getRect().x,X=Y.getRect().x+Y.getRect().width,ie=Y.getRect().y,j=Y.getRect().y+Y.getRect().height;Qy&&(y=X),iex&&(x=j)});var B=d.x-(y+g)/2,$=d.y-(x+v)/2;Object.keys(p).forEach(function(z){var Y=p[z];Y.setCenter(Y.getCenterX()+B,Y.getCenterY()+$)})}}},f.calcBoundingBox=function(d,p,m,g){for(var y=Number.MAX_SAFE_INTEGER,v=Number.MIN_SAFE_INTEGER,x=Number.MAX_SAFE_INTEGER,b=Number.MIN_SAFE_INTEGER,w=void 0,S=void 0,T=void 0,E=void 0,_=d.descendants().not(":parent"),A=_.length,L=0;Lw&&(y=w),vT&&(x=T),b{var u=l(548),h=l(140).CoSELayout,f=l(140).CoSENode,d=l(140).layoutBase.PointD,p=l(140).layoutBase.DimensionD,m=l(140).layoutBase.LayoutConstants,g=l(140).layoutBase.FDLayoutConstants,y=l(140).CoSEConstants,v=o(function(b,w){var S=b.cy,T=b.eles,E=T.nodes(),_=T.edges(),A=void 0,L=void 0,M=void 0,N={};b.randomize&&(A=w.nodeIndexes,L=w.xCoords,M=w.yCoords);var k=o(function(z){return typeof z=="function"},"isFn"),I=o(function(z,Y){return k(z)?z(Y):z},"optFn"),C=u.calcParentsWithoutChildren(S,T),O=o(function $(z,Y,Q,X){for(var ie=Y.length,j=0;j0){var se=void 0;se=Q.getGraphManager().add(Q.newGraph(),H),$(se,Z,Q,X)}}},"processChildrenList"),D=o(function(z,Y,Q){for(var X=0,ie=0,j=0;j0?y.DEFAULT_EDGE_LENGTH=g.DEFAULT_EDGE_LENGTH=X/ie:k(b.idealEdgeLength)?y.DEFAULT_EDGE_LENGTH=g.DEFAULT_EDGE_LENGTH=50:y.DEFAULT_EDGE_LENGTH=g.DEFAULT_EDGE_LENGTH=b.idealEdgeLength,y.MIN_REPULSION_DIST=g.MIN_REPULSION_DIST=g.DEFAULT_EDGE_LENGTH/10,y.DEFAULT_RADIAL_SEPARATION=g.DEFAULT_EDGE_LENGTH)},"processEdges"),P=o(function(z,Y){Y.fixedNodeConstraint&&(z.constraints.fixedNodeConstraint=Y.fixedNodeConstraint),Y.alignmentConstraint&&(z.constraints.alignmentConstraint=Y.alignmentConstraint),Y.relativePlacementConstraint&&(z.constraints.relativePlacementConstraint=Y.relativePlacementConstraint)},"processConstraints");b.nestingFactor!=null&&(y.PER_LEVEL_IDEAL_EDGE_LENGTH_FACTOR=g.PER_LEVEL_IDEAL_EDGE_LENGTH_FACTOR=b.nestingFactor),b.gravity!=null&&(y.DEFAULT_GRAVITY_STRENGTH=g.DEFAULT_GRAVITY_STRENGTH=b.gravity),b.numIter!=null&&(y.MAX_ITERATIONS=g.MAX_ITERATIONS=b.numIter),b.gravityRange!=null&&(y.DEFAULT_GRAVITY_RANGE_FACTOR=g.DEFAULT_GRAVITY_RANGE_FACTOR=b.gravityRange),b.gravityCompound!=null&&(y.DEFAULT_COMPOUND_GRAVITY_STRENGTH=g.DEFAULT_COMPOUND_GRAVITY_STRENGTH=b.gravityCompound),b.gravityRangeCompound!=null&&(y.DEFAULT_COMPOUND_GRAVITY_RANGE_FACTOR=g.DEFAULT_COMPOUND_GRAVITY_RANGE_FACTOR=b.gravityRangeCompound),b.initialEnergyOnIncremental!=null&&(y.DEFAULT_COOLING_FACTOR_INCREMENTAL=g.DEFAULT_COOLING_FACTOR_INCREMENTAL=b.initialEnergyOnIncremental),b.tilingCompareBy!=null&&(y.TILING_COMPARE_BY=b.tilingCompareBy),b.quality=="proof"?m.QUALITY=2:m.QUALITY=0,y.NODE_DIMENSIONS_INCLUDE_LABELS=g.NODE_DIMENSIONS_INCLUDE_LABELS=m.NODE_DIMENSIONS_INCLUDE_LABELS=b.nodeDimensionsIncludeLabels,y.DEFAULT_INCREMENTAL=g.DEFAULT_INCREMENTAL=m.DEFAULT_INCREMENTAL=!b.randomize,y.ANIMATE=g.ANIMATE=m.ANIMATE=b.animate,y.TILE=b.tile,y.TILING_PADDING_VERTICAL=typeof b.tilingPaddingVertical=="function"?b.tilingPaddingVertical.call():b.tilingPaddingVertical,y.TILING_PADDING_HORIZONTAL=typeof b.tilingPaddingHorizontal=="function"?b.tilingPaddingHorizontal.call():b.tilingPaddingHorizontal,y.DEFAULT_INCREMENTAL=g.DEFAULT_INCREMENTAL=m.DEFAULT_INCREMENTAL=!0,y.PURE_INCREMENTAL=!b.randomize,m.DEFAULT_UNIFORM_LEAF_NODE_SIZES=b.uniformNodeDimensions,b.step=="transformed"&&(y.TRANSFORM_ON_CONSTRAINT_HANDLING=!0,y.ENFORCE_CONSTRAINTS=!1,y.APPLY_LAYOUT=!1),b.step=="enforced"&&(y.TRANSFORM_ON_CONSTRAINT_HANDLING=!1,y.ENFORCE_CONSTRAINTS=!0,y.APPLY_LAYOUT=!1),b.step=="cose"&&(y.TRANSFORM_ON_CONSTRAINT_HANDLING=!1,y.ENFORCE_CONSTRAINTS=!1,y.APPLY_LAYOUT=!0),b.step=="all"&&(b.randomize?y.TRANSFORM_ON_CONSTRAINT_HANDLING=!0:y.TRANSFORM_ON_CONSTRAINT_HANDLING=!1,y.ENFORCE_CONSTRAINTS=!0,y.APPLY_LAYOUT=!0),b.fixedNodeConstraint||b.alignmentConstraint||b.relativePlacementConstraint?y.TREE_REDUCTION_ON_INCREMENTAL=!1:y.TREE_REDUCTION_ON_INCREMENTAL=!0;var F=new h,B=F.newGraphManager();return O(B.addRoot(),u.getTopMostNodes(E),F,b),D(F,B,_),P(F,b),F.runLayout(),N},"coseLayout");a.exports={coseLayout:v}},212:(a,s,l)=>{var u=function(){function b(w,S){for(var T=0;T0)if(D){var B=d.getTopMostNodes(T.eles.nodes());if(k=d.connectComponents(E,T.eles,B),k.forEach(function(oe){var ke=oe.boundingBox();I.push({x:ke.x1+ke.w/2,y:ke.y1+ke.h/2})}),T.randomize&&k.forEach(function(oe){T.eles=oe,A.push(m(T))}),T.quality=="default"||T.quality=="proof"){var $=E.collection();if(T.tile){var z=new Map,Y=[],Q=[],X=0,ie={nodeIndexes:z,xCoords:Y,yCoords:Q},j=[];if(k.forEach(function(oe,ke){oe.edges().length==0&&(oe.nodes().forEach(function(Ie,Se){$.merge(oe.nodes()[Se]),Ie.isParent()||(ie.nodeIndexes.set(oe.nodes()[Se].id(),X++),ie.xCoords.push(oe.nodes()[0].position().x),ie.yCoords.push(oe.nodes()[0].position().y))}),j.push(ke))}),$.length>1){var J=$.boundingBox();I.push({x:J.x1+J.w/2,y:J.y1+J.h/2}),k.push($),A.push(ie);for(var Z=j.length-1;Z>=0;Z--)k.splice(j[Z],1),A.splice(j[Z],1),I.splice(j[Z],1)}}k.forEach(function(oe,ke){T.eles=oe,N.push(y(T,A[ke])),d.relocateComponent(I[ke],N[ke],T)})}else k.forEach(function(oe,ke){d.relocateComponent(I[ke],A[ke],T)});var H=new Set;if(k.length>1){var q=[],K=_.filter(function(oe){return oe.css("display")=="none"});k.forEach(function(oe,ke){var Ie=void 0;if(T.quality=="draft"&&(Ie=A[ke].nodeIndexes),oe.nodes().not(K).length>0){var Se={};Se.edges=[],Se.nodes=[];var Ue=void 0;oe.nodes().not(K).forEach(function(Pe){if(T.quality=="draft")if(!Pe.isParent())Ue=Ie.get(Pe.id()),Se.nodes.push({x:A[ke].xCoords[Ue]-Pe.boundingbox().w/2,y:A[ke].yCoords[Ue]-Pe.boundingbox().h/2,width:Pe.boundingbox().w,height:Pe.boundingbox().h});else{var _e=d.calcBoundingBox(Pe,A[ke].xCoords,A[ke].yCoords,Ie);Se.nodes.push({x:_e.topLeftX,y:_e.topLeftY,width:_e.width,height:_e.height})}else N[ke][Pe.id()]&&Se.nodes.push({x:N[ke][Pe.id()].getLeft(),y:N[ke][Pe.id()].getTop(),width:N[ke][Pe.id()].getWidth(),height:N[ke][Pe.id()].getHeight()})}),oe.edges().forEach(function(Pe){var _e=Pe.source(),me=Pe.target();if(_e.css("display")!="none"&&me.css("display")!="none")if(T.quality=="draft"){var W=Ie.get(_e.id()),fe=Ie.get(me.id()),ge=[],re=[];if(_e.isParent()){var he=d.calcBoundingBox(_e,A[ke].xCoords,A[ke].yCoords,Ie);ge.push(he.topLeftX+he.width/2),ge.push(he.topLeftY+he.height/2)}else ge.push(A[ke].xCoords[W]),ge.push(A[ke].yCoords[W]);if(me.isParent()){var ne=d.calcBoundingBox(me,A[ke].xCoords,A[ke].yCoords,Ie);re.push(ne.topLeftX+ne.width/2),re.push(ne.topLeftY+ne.height/2)}else re.push(A[ke].xCoords[fe]),re.push(A[ke].yCoords[fe]);Se.edges.push({startX:ge[0],startY:ge[1],endX:re[0],endY:re[1]})}else 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r={name:t.filename,buffer:t.input.slice(0,-1),position:t.position,line:t.line,column:t.position-t.lineStart};return r.snippet=Met(r),new $s(e,r)}o(ave,"generateError");function Gt(t,e){throw ave(t,e)}o(Gt,"throwError");function TC(t,e){t.onWarning&&t.onWarning.call(null,ave(t,e))}o(TC,"throwWarning");var Pye={YAML:o(function(e,r,n){var i,a,s;e.version!==null&&Gt(e,"duplication of %YAML directive"),n.length!==1&&Gt(e,"YAML directive accepts exactly one argument"),i=/^([0-9]+)\.([0-9]+)$/.exec(n[0]),i===null&&Gt(e,"ill-formed argument of the YAML directive"),a=parseInt(i[1],10),s=parseInt(i[2],10),a!==1&&Gt(e,"unacceptable YAML version of the document"),e.version=n[0],e.checkLineBreaks=s<2,s!==1&&s!==2&&TC(e,"unsupported YAML version of the document")},"handleYamlDirective"),TAG:o(function(e,r,n){var i,a;n.length!==2&&Gt(e,"TAG directive accepts exactly two arguments"),i=n[0],a=n[1],tve.test(i)||Gt(e,"ill-formed tag handle (first argument) of the TAG 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s=t.input.charCodeAt(++t.position);while(s!==0&&!qc(s));break}if(qc(s))break;for(r=t.position;s!==0&&!Vs(s);)s=t.input.charCodeAt(++t.position);i.push(t.input.slice(r,t.position))}s!==0&&fF(t),Bf.call(Pye,n)?Pye[n](t,n,i):TC(t,'unknown document directive "'+n+'"')}if(Ai(t,!0,-1),t.lineIndent===0&&t.input.charCodeAt(t.position)===45&&t.input.charCodeAt(t.position+1)===45&&t.input.charCodeAt(t.position+2)===45?(t.position+=3,Ai(t,!0,-1)):a&&Gt(t,"directives end mark is expected"),N1(t,t.lineIndent-1,wC,!1,!0),Ai(t,!0,-1),t.checkLineBreaks&&ztt.test(t.input.slice(e,t.position))&&TC(t,"non-ASCII line breaks are interpreted as content"),t.documents.push(t.result),t.position===t.lineStart&&CC(t)){t.input.charCodeAt(t.position)===46&&(t.position+=3,Ai(t,!0,-1));return}if(t.position"u"&&(r=e,e=null);var n=sve(t,r);if(typeof e!="function")return n;for(var i=0,a=n.length;i=55296&&r<=56319&&e+1=56320&&n<=57343)?(r-55296)*1024+n-56320+65536:r}o(kb,"codePointAt");function mve(t){var e=/^\n* /;return e.test(t)}o(mve,"needIndentIndicator");var gve=1,cF=2,yve=3,vve=4,L1=5;function Rrt(t,e,r,n,i,a,s,l){var u,h=0,f=null,d=!1,p=!1,m=n!==-1,g=-1,y=Lrt(kb(t,0))&&Drt(kb(t,t.length-1));if(e||s)for(u=0;u=65536?u+=2:u++){if(h=kb(t,u),!Ab(h))return L1;y=y&&$ye(h,f,l),f=h}else{for(u=0;u=65536?u+=2:u++){if(h=kb(t,u),h===Cb)d=!0,m&&(p=p||u-g-1>n&&t[g+1]!==" ",g=u);else if(!Ab(h))return L1;y=y&&$ye(h,f,l),f=h}p=p||m&&u-g-1>n&&t[g+1]!==" "}return!d&&!p?y&&!s&&!i(t)?gve:a===Sb?L1:cF:r>9&&mve(t)?L1:s?a===Sb?L1:cF:p?vve:yve}o(Rrt,"chooseScalarStyle");function Nrt(t,e,r,n,i){t.dump=function(){if(e.length===0)return t.quotingType===Sb?'""':"''";if(!t.noCompatMode&&(Trt.indexOf(e)!==-1||krt.test(e)))return t.quotingType===Sb?'"'+e+'"':"'"+e+"'";var a=t.indent*Math.max(1,r),s=t.lineWidth===-1?-1:Math.max(Math.min(t.lineWidth,40),t.lineWidth-a),l=n||t.flowLevel>-1&&r>=t.flowLevel;function u(h){return _rt(t,h)}switch(o(u,"testAmbiguity"),Rrt(e,l,t.indent,s,u,t.quotingType,t.forceQuotes&&!n,i)){case gve:return e;case cF:return"'"+e.replace(/'/g,"''")+"'";case yve:return"|"+Vye(e,t.indent)+Uye(zye(e,a));case vve:return">"+Vye(e,t.indent)+Uye(zye(Mrt(e,s),a));case L1:return'"'+Irt(e)+'"';default:throw new $s("impossible error: invalid scalar style")}}()}o(Nrt,"writeScalar");function Vye(t,e){var r=mve(t)?String(e):"",n=t[t.length-1]===` -`,i=n&&(t[t.length-2]===` -`||t===` -`),a=i?"+":n?"":"-";return r+a+` -`}o(Vye,"blockHeader");function Uye(t){return t[t.length-1]===` -`?t.slice(0,-1):t}o(Uye,"dropEndingNewline");function Mrt(t,e){for(var r=/(\n+)([^\n]*)/g,n=function(){var h=t.indexOf(` -`);return h=h!==-1?h:t.length,r.lastIndex=h,Hye(t.slice(0,h),e)}(),i=t[0]===` -`||t[0]===" ",a,s;s=r.exec(t);){var l=s[1],u=s[2];a=u[0]===" ",n+=l+(!i&&!a&&u!==""?` -`:"")+Hye(u,e),i=a}return n}o(Mrt,"foldString");function Hye(t,e){if(t===""||t[0]===" ")return t;for(var r=/ [^ ]/g,n,i=0,a,s=0,l=0,u="";n=r.exec(t);)l=n.index,l-i>e&&(a=s>i?s:l,u+=` -`+t.slice(i,a),i=a+1),s=l;return u+=` 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    Advent of code 2023 - Day 1: Trebuchet?!

    Advent of code 2023 - Day 1: Trebuchet?!

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 1 - see -https:adventofcode.com/2023/day/1

    Part 1 -#

    Our first task is the following:

    The newly-improved calibration document consists of lines of text; each line -originally contained a specific calibration value that the Elves now need to -recover. On each line, the calibration value can be found by combining the first -digit and the last digit (in that order) to form a single two-digit number.

    For example:

    1abc2
    -pqr3stu8vwx
    -a1b2c3d4e5f
    -treb7uchet
    -

    In this example, the calibration values of these four lines are 12, 38, 15, -and 77. Adding these together produces 142.

    Consider your entire calibration document. What is the sum of all of the -calibration values?

    My input file: -2023-12-01-1-aoc.txt

    Lets start jupyter in our shell to start coding!

    conda activate tf
    -jupyter lab --no-browser --port=8888
    -

    First, load the test document

    import pandas as pd
    -import re
    -
    -txt = pd.read_table('data/2023-12-01-1-aoc.txt', names=['code'])
    -txt
    -
    code
    0jjfvnnlfivejj1
    16fourfour
    2ninevbmltwo69
    3pcg91vqrfpxxzzzoneightzt
    4jpprthxgjfive3one1qckhrptpqdc
    995583sevenhjxlqzjgbzxhkcl5
    99681s
    9972four3threesxxvlfqfive4
    998nine6eightsevenzx9twoxc
    999hmbfjdfnp989mfivefiverpzrjs

    1000 rows × 1 columns

    Second, extract the digits. I had to wrap my head around regex matching in -python first, because I first tried pandas.extract (which only extracts the -first match), then pandas.extractall (which extracts all matches but puts them -into a multiindex which makes things more difficult in this case). So I settled -for the re.findall version, which returns a list. To concatenate the elements -in the list, we take use the join function.

    txt['digits'] = txt.loc[:, 'code'].apply(
    -    lambda x: ''.join(re.findall(r'(\d+)', x)))
    -txt
    -
    codedigits
    0jjfvnnlfivejj11
    16fourfour6
    2ninevbmltwo6969
    3pcg91vqrfpxxzzzoneightzt91
    4jpprthxgjfive3one1qckhrptpqdc31
    995583sevenhjxlqzjgbzxhkcl55835
    99681s81
    9972four3threesxxvlfqfive4234
    998nine6eightsevenzx9twoxc69
    999hmbfjdfnp989mfivefiverpzrjs989

    1000 rows × 2 columns

    Next, combine the first and the last digit and convert the result from string to integer

    txt['calibration'] = txt.loc[:, 'digits'].apply(
    -    lambda x: int(x[0] + x[-1]))
    -txt
    -
    codedigitscalibration
    0jjfvnnlfivejj1111
    16fourfour666
    2ninevbmltwo696969
    3pcg91vqrfpxxzzzoneightzt9191
    4jpprthxgjfive3one1qckhrptpqdc3131
    995583sevenhjxlqzjgbzxhkcl5583555
    99681s8181
    9972four3threesxxvlfqfive423424
    998nine6eightsevenzx9twoxc6969
    999hmbfjdfnp989mfivefiverpzrjs98999

    1000 rows × 3 columns

    Lastly, get the sum of our calibration numbers

    txt.loc[:, 'calibration'].sum()
    -
    56465
    -

    Part 2 -#

    Now follows part two:

    Your calculation isn’t quite right. It looks like some of the digits are -actually spelled out with letters: one, two, three, four, five, six, seven, -eight, and nine also count as valid “digits”.

    Equipped with this new information, you now need to find the real first and last -digit on each line. For example:

    two1nine
    -eightwothree
    -abcone2threexyz
    -xtwone3four
    -4nineeightseven2
    -zoneight234
    -7pqrstsixteen
    -

    In this example, the calibration values are 29, 83, 13, 24, 42, 14, and 76. -Adding these together produces 281.

    What is the sum of all of the calibration values?

    Okay, let’s see if we can update the pattern matching. To deal with potential -overlapping values like oneight which contains one as well as eight, I -used the regex positive lookahead ?= as described -here. Because this enables -capturing overlapping values, I used \d (one digit) instead of \d+ (one or -more digits), otherwise digits might double. Afterwards, just replace the -spelled out digits with their numerical value.

    # for i, r in enumerate(txt.loc[:, 'code']):
    -#     matches = re.findall(
    -#         r'(?=(\d|one|two|three|four|five|six|seven|eight|nine))', r)
    -#     result = ''.join([match for match in matches])
    -#     result = result.replace('one', '1').replace('two', '2').replace(
    -#         'three', '3').replace('four', '4').replace('five', '5').replace(
    -#         'six', '6').replace('seven', '7').replace('eight', '8').replace(
    -#         'nine', '9')
    -#     txt.loc[i, 'digits2'] = result
    -# txt
    -
    -# a very nice alternative suggested by Tomalak:
    -digits = '\d one two three four five six seven eight nine'.split()
    -
    -
    -txt['digits2'] = txt.loc[:, 'code'].apply(lambda v: ''.join(
    -    str(digits.index(m)) if m in digits else m
    -    for m in re.findall(rf'(?=({"|".join(digits)}))', v)
    -))
    -txt
    -
    codedigitscalibrationdigits2
    0jjfvnnlfivejj111151
    16fourfour666644
    2ninevbmltwo6969699269
    3pcg91vqrfpxxzzzoneightzt91919118
    4jpprthxgjfive3one1qckhrptpqdc31315311
    995583sevenhjxlqzjgbzxhkcl558355558375
    99681s818181
    9972four3threesxxvlfqfive423424243354
    998nine6eightsevenzx9twoxc6969968792
    999hmbfjdfnp989mfivefiverpzrjs9899998955

    1000 rows × 4 columns

    Now, construct the calibration value as before…

    txt['calibration2'] = txt.loc[:, 'digits2'].apply(lambda x: int(x[0] + x[-1]))
    -txt
    -
    codedigitscalibrationdigits2calibration2
    0jjfvnnlfivejj11115151
    16fourfour66664464
    2ninevbmltwo696969926999
    3pcg91vqrfpxxzzzoneightzt9191911898
    4jpprthxgjfive3one1qckhrptpqdc3131531151
    995583sevenhjxlqzjgbzxhkcl55835555837555
    99681s81818181
    9972four3threesxxvlfqfive42342424335424
    998nine6eightsevenzx9twoxc696996879292
    999hmbfjdfnp989mfivefiverpzrjs989999895595

    1000 rows × 5 columns

    … and get the correct sum!

    txt.loc[:, 'calibration2'].sum()
    -
    55902
    -
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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/posts/aoc2023-day02/index.html b/aseltmann.github.io-hugo/public/posts/aoc2023-day02/index.html deleted file mode 100644 index d669347..0000000 --- a/aseltmann.github.io-hugo/public/posts/aoc2023-day02/index.html +++ /dev/null @@ -1,115 +0,0 @@ -Advent of code 2023 - Day 2: Cube Conundrum | Blog & Notes - -

    Advent of code 2023 - Day 2: Cube Conundrum

    Advent of code 2023 - Day 2: Cube Conundrum

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 2 - see -https:adventofcode.com/2023/day/2

    Part 1 -#

    Lets first read the task:

    As you walk, the Elf shows you a small bag and some cubes which are either red, -green, or blue. Each time you play this game, he will hide a secret number of -cubes of each color in the bag, and your goal is to figure out information about -the number of cubes.

    To get information, once a bag has been loaded with cubes, the Elf will reach -into the bag, grab a handful of random cubes, show them to you, and then put -them back in the bag. He’ll do this a few times per game.

    You play several games and record the information from each game (your puzzle -input). Each game is listed with its ID number (like the 11 in Game 11: ...) -followed by a semicolon-separated list of subsets of cubes that were revealed -from the bag (like 3 red, 5 green, 4 blue).

    For example, the record of a few games might look like this:

    Game 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green
    -Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue
    -Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red
    -Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red
    -Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green
    -

    In game 1, three sets of cubes are revealed from the bag (and then put back -again). The first set is 3 blue cubes and 4 red cubes; the second set is 1 red -cube, 2 green cubes, and 6 blue cubes; the third set is only 2 green cubes.

    The Elf would first like to know which games would have been possible if the bag -contained only 12 red cubes, 13 green cubes, and 14 blue cubes?

    In the example above, games 1, 2, and 5 would have been possible if the bag had -been loaded with that configuration. However, game 3 would have been impossible -because at one point the Elf showed you 20 red cubes at once; similarly, game 4 -would also have been impossible because the Elf showed you 15 blue cubes at -once. If you add up the IDs of the games that would have been possible, you -get 8.

    Determine which games would have been possible if the bag had been loaded with -only 12 red cubes, 13 green cubes, and 14 blue cubes. What is the sum of the IDs -of those games?

    My input file: -2023-12-02-1-aoc.txt

    Okay, let’s load our python kernel in emacs-jupyter and get coding! First of -all, let’s load the input and split the riddle code by colon : to extract the -game id and the rest of the code by semicolon ; to get the number of sets -played in each game.

    import pandas as pd
    -import re
    -
    -txt = pd.read_table('data/2023-12-02-1-aoc.txt', names=['code'])
    -txt['id'] = txt.loc[:, 'code'].str.split(':').apply(
    -    lambda x: int(x[0].strip('Game ')))
    -txt['code'] = txt.loc[:, 'code'].str.split(':').apply(lambda x: x[1])
    -# txt['code'] = txt.loc[:, 'code'].str.split(';')
    -# txt['nsets'] = txt.loc[:, 'code'].apply(lambda x: len(x))
    -txt
    -
    codeid
    01 green, 1 blue, 1 red; 1 green, 8 red, 7 blu…1
    19 red, 7 green, 3 blue; 15 green, 2 blue, 5 r…2
    23 red, 1 blue, 4 green; 6 red, 3 green, 2 blu…3
    32 blue, 2 green, 19 red; 3 blue, 11 red, 16 g…4
    48 green, 1 red, 12 blue; 10 green, 6 red, 13 …5
    952 red, 2 green, 1 blue; 1 red, 4 green; 1 green96
    964 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree…97
    971 blue; 2 green, 1 red; 5 red, 2 green; 4 red…98
    986 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree…99
    991 blue, 13 green, 14 red; 11 green, 11 blue, …100

    100 rows × 2 columns

    Now, let’s extract the three colors in different columns with regex. We use the -lookahead assertion ?= to find the respective colours and only exctract the -digits \d+ coming before. Then we just keep the max imum drawn number of cubes -per color, since this is the only information that matters at the moment.

    txt['green'] = txt.loc[:, 'code'].apply(
    -    lambda code: re.findall(r'\d+(?=.green)', code)).apply(
    -        lambda list: max([int(i) for i in list]))
    -txt['red'] = txt.loc[:, 'code'].apply(
    -    lambda code: re.findall(r'\d+(?=.red)', code)).apply(
    -        lambda list: max([int(i) for i in list]))
    -txt['blue'] = txt.loc[:, 'code'].apply(
    -    lambda code: re.findall(r'\d+(?=.blue)', code)).apply(
    -        lambda list: max([int(i) for i in list]))
    -txt
    -
    codeidgreenredblue
    01 green, 1 blue, 1 red; 1 green, 8 red, 7 blu…121010
    19 red, 7 green, 3 blue; 15 green, 2 blue, 5 r…215103
    23 red, 1 blue, 4 green; 6 red, 3 green, 2 blu…34616
    32 blue, 2 green, 19 red; 3 blue, 11 red, 16 g…4162018
    48 green, 1 red, 12 blue; 10 green, 6 red, 13 …510614
    952 red, 2 green, 1 blue; 1 red, 4 green; 1 green96421
    964 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree…97548
    971 blue; 2 green, 1 red; 5 red, 2 green; 4 red…98252
    986 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree…992911
    991 blue, 13 green, 14 red; 11 green, 11 blue, …100131511

    100 rows × 5 columns

    Lastly, we just filter the DataFrame to only include games where all drawn cubes -were below or equal the number of cubes in the game and sum the result!

    txt['id'][(txt['green'] < 14) & (txt['red'] < 13) & (txt['blue'] < 15)].sum()
    -
    3035
    -

    Part 2 -#

    First, let’s get the instruction from the second part:

    As you continue your walk, the Elf poses a second question: in each game you -played, what is the fewest number of cubes of each color that could have been in -the bag to make the game possible?

    Again consider the example games from earlier:

    Game 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green
    -Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue
    -Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red
    -Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red
    -Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green
    -
    • In game 1, the game could have been played with as few as 4 red, 2 green, and -6 blue cubes. If any color had even one fewer cube, the game would have been -impossible.
    • Game 2 could have been played with a minimum of 1 red, 3 green, and 4 blue -cubes.
    • Game 3 must have been played with at least 20 red, 13 green, and 6 blue cubes.
    • Game 4 required at least 14 red, 3 green, and 15 blue cubes.
    • Game 5 needed no fewer than 6 red, 3 green, and 2 blue cubes in the bag.

    The power of a set of cubes is equal to the numbers of red, green, and blue -cubes multiplied together. The power of the minimum set of cubes in game 1 -is 48. In games 2-5 it was 12, 1560, 630, and 36, respectively. Adding up these -five powers produces the sum 2286.

    For each game, find the minimum set of cubes that must have been present. What -is the sum of the power of these sets?

    Luckily, this task is made trivial by the approach we have taken before. We just -have to multiply the green, red and blue columns:

    txt['power'] = txt.loc[:, 'green'] * txt.loc[:, 'blue'] * txt.loc[:, 'red']
    -txt
    -
    codeidgreenredbluepower
    01 green, 1 blue, 1 red; 1 green, 8 red, 7 blu…121010200
    19 red, 7 green, 3 blue; 15 green, 2 blue, 5 r…215103450
    23 red, 1 blue, 4 green; 6 red, 3 green, 2 blu…34616384
    32 blue, 2 green, 19 red; 3 blue, 11 red, 16 g…41620185760
    48 green, 1 red, 12 blue; 10 green, 6 red, 13 …510614840
    952 red, 2 green, 1 blue; 1 red, 4 green; 1 green964218
    964 red, 5 green; 5 blue, 3 red; 8 blue, 2 gree…97548160
    971 blue; 2 green, 1 red; 5 red, 2 green; 4 red…9825220
    986 blue, 5 red, 2 green; 9 red, 1 blue; 2 gree…992911198
    991 blue, 13 green, 14 red; 11 green, 11 blue, …1001315112145

    100 rows × 6 columns

    And for this one, the sum is:

    txt['power'].sum()
    -
    66027
    -
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    Advent of code 2023 - Day 3: Gear Ratios

    Advent of code 2023 - Day 3: Gear Ratios

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 3 - see -https:adventofcode.com/2023/day/3

    Part 1 -#

    Lets first read the task:

    The engine schematic (your puzzle input) consists of a visual representation of -the engine. There are lots of numbers and symbols you don’t really understand, -but apparently any number adjacent to a symbol, even diagonally, is a “part -number” and should be included in your sum. (Periods (.) do not count as a -symbol.)

    Here is an example engine schematic:

    467..114..
    -...*......
    -..35..633.
    -......#...
    -617*......
    -.....+.58.
    -..592.....
    -......755.
    -...$.*....
    -.664.598..
    -

    In this schematic, two numbers are not part numbers because they are not -adjacent to a symbol: 114 (top right) and 58 (middle right). Every other -number is adjacent to a symbol and so is a part number; their sum is 4361.

    Of course, the actual engine schematic is much larger. What is the sum of all of -the part numbers in the engine schematic?

    My input file: -2023-12-03-1-aoc.txt

    Okay, let’s first get the input as a numpy character array

    import numpy as np
    -import pandas as pd
    -import sys
    -import matplotlib.pyplot as plt
    -from scipy import ndimage as ndi
    -np.set_printoptions(threshold=sys.maxsize)
    -
    -txt = pd.read_table('data/2023-12-03-1-aoc.txt', names=['code'])
    -arr = np.chararray((txt.size, txt.size), unicode=True)
    -
    -txt['code'] = txt.loc[:, 'code'].apply(lambda x: [i for i in x])
    -
    -for i, code in enumerate(txt['code']):
    -    arr[i, :] = code
    -
    -print((arr[:15, :15]))
    -
    [['.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.']
    - ['.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '5' '3' '.' '4']
    - ['.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '*' '.' '.' '.' '.']
    - ['.' '.' '.' '.' '.' '.' '.' '7' '2' '6' '.' '.' '.' '.' '.']
    - ['.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '9' '5']
    - ['.' '.' '.' '.' '.' '.' '.' '7' '3' '8' '.' '.' '.' '*' '.']
    - ['.' '7' '4' '.' '.' '.' '.' '.' '.' '.' '.' '.' '3' '6' '6']
    - ['.' '.' '.' '*' '1' '2' '6' '.' '.' '.' '.' '.' '.' '.' '.']
    - ['.' '.' '.' '.' '.' '.' '.' '.' '3' '3' '1' '/' '.' '.' '9']
    - ['.' '.' '.' '.' '/' '.' '.' '.' '.' '.' '.' '.' '.' '.' '*']
    - ['.' '.' '.' '.' '9' '5' '3' '.' '.' '.' '.' '3' '5' '5' '.']
    - ['.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.']
    - ['.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.' '.']
    - ['.' '6' '8' '5' '.' '.' '.' '.' '*' '.' '.' '.' '.' '7' '0']
    - ['.' '.' '.' '.' '.' '.' '.' '.' '.' '9' '3' '8' '.' '.' '*']]
    -/tmp/ipykernel_18336/87086526.py:9: DeprecationWarning: `np.chararray` is deprecated and will be removed from the main namespace in the future. Use an array with a string or bytes dtype instead.
    -  arr = np.chararray((txt.size, txt.size), unicode=True)
    -

    Now extract symbols, digits and the empty space. We use the numpy character -methods for that. This way, we create a binary mask for all digits and a -binary mask for all empty space (.). The symbols are then every character -which is neither.

    digits = np.char.isdigit(arr)
    -empty = np.char.endswith(arr, '.')
    -symbols = ~(digits | empty)
    -
    -# just for visualization
    -plt.figure(figsize=(16, 5))
    -plt.subplot(131, title='symbols').matshow(symbols)
    -plt.subplot(132, title='digits').matshow(digits)
    -plt.subplot(133, title='empty').matshow(empty)
    -plt.show()
    -

    Now we use the image processing technique of dilation on the symbols mask. So -that we get a new mask which covers the surroundings of all symbols. We use this -afterwards to check if the digits are near a symbol.

    struct = ((1, 1, 1), (1, 1, 1), (1, 1, 1))
    -dilate = ndi.binary_dilation(symbols, structure=struct)
    -plt.figure(figsize=(10, 6))
    -plt.subplot(121, title='symbols').matshow(symbols[:15, :15])
    -plt.subplot(122, title='symbols dilated').matshow(dilate[:15, :15])
    -plt.show()
    -

    Creating this masks as before could be understood as a binary segmentation, as -each element in our mask is either True or False. To extract the single -digits, we’ll convert the binary digits segmentation into a instance -segmentation, where each connected segment has an own index.

    markers, num_features = ndi.label(digits)
    -plt.figure(figsize=(10, 6))
    -plt.subplot(121, title='binary segmentation').matshow(
    -    digits[:15, :15])
    -plt.subplot(122, title='instance segmentation').matshow(
    -    markers[:15, :15], cmap='gnuplot')
    -plt.show()
    -

    Now, for each instance, we check if the dilated binary mask overlaps with the -instance and if yes, we extract the number.

    numbers = [int(''.join(arr[markers == i]))
    -           for i in range(1, num_features+1)
    -           if np.any((markers == i) & dilate)]
    -

    Then, we just sum up:

    sum(numbers)
    -
    527364
    -

    Part 2 -#

    Let’s first read the second task!

    The missing part wasn’t the only issue - one of the gears in the engine is -wrong. A gear is any * symbol that is adjacent to exactly two part numbers. Its -gear ratio is the result of multiplying those two numbers together.

    This time, you need to find the gear ratio of every gear and add them all up so -that the engineer can figure out which gear needs to be replaced.

    Consider the same engine schematic again:The missing part wasn’t the only -issue - one of the gears in the engine is wrong. A gear is any * symbol that is -adjacent to exactly two part numbers. Its gear ratio is the result of -multiplying those two numbers together.

    This time, you need to find the gear ratio of every gear and add them all up so -that the engineer can figure out which gear needs to be replaced.

    Consider the same engine schematic again:

    467..114..
    -...*......
    -..35..633.
    -......#...
    -617*......
    -.....+.58.
    -..592.....
    -......755.
    -...$.*....
    -.664.598..
    -

    In this schematic, there are two gears. The first is in the top left; it has -part numbers 467 and 35, so its gear ratio is 16345. The second gear is in -the lower right; its gear ratio is 451490. (The * adjacent to 617 is not a -gear because it is only adjacent to one part number.) Adding up all of the gear -ratios produces 467835.

    What is the sum of all of the gear ratios in your engine schematic?

    We’ll use the same method as before, but this time only extract * and create -the instance segmentation before the dilation. Why? Because when we dilate -first, we could merge two independent gears into one instance.

    gear = np.char.endswith(arr, '*')
    -gear_markers, gear_num = ndi.label(gear)
    -
    -plt.figure(figsize=(10, 6))
    -plt.subplot(131, title='all symbols dilated').matshow(
    -    symbols[:15, :15])
    -plt.subplot(132, title='gears').matshow(
    -    gear[:15, :15])
    -plt.subplot(133, title='gears instances').matshow(
    -    gear_markers[:15, :15], cmap='gnuplot')
    -plt.show()
    -

    Now, we step through each gear instance, create a mask for that gear, dilate it, -and then step through all digits and check if they are in the gear mask. If we -get two digits in the end, we multiply them to get the gear ratio and save the -ratios to a list.

    gear_ratios = []
    -for i in range(1, gear_num+1):
    -    gear_binary = gear_markers == i
    -    gear_dil = ndi.binary_dilation(gear_binary, structure=struct)
    -    part_numbers = [int(''.join(arr[markers == j]))
    -                    for j in range(1, num_features+1)
    -                    if np.any((markers == j) & gear_dil)]
    -    if len(part_numbers) == 2:
    -        gear_ratios.append(part_numbers[0] * part_numbers[1])
    -

    Now, we just sum the output again:

    sum(gear_ratios)
    -
    79026871
    -
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    Advent of code 2023 - Day 4: Scratchcards

    Advent of code 2023 - Day 4: Scratchcards

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 4 - see -https:adventofcode.com/2023/day/4

    Part 1 -#

    Lets first read the task:

    The Elf leads you over to the pile of colorful cards. There, you discover dozens -of scratchcards, all with their opaque covering already scratched off. Picking -one up, it looks like each card has two lists of numbers separated by a vertical -bar (|): a list of winning numbers and then a list of numbers you have. -You organize the information into a table (your puzzle input).

    As far as the Elf has been able to figure out, you have to figure out which of -the numbers you have appear in the list of winning numbers. The first match -makes the card worth one point and each match after the first doubles the -point value of that card.

    For example:

    Card 1: 41 48 83 86 17 | 83 86  6 31 17  9 48 53
    -Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19
    -Card 3:  1 21 53 59 44 | 69 82 63 72 16 21 14  1
    -Card 4: 41 92 73 84 69 | 59 84 76 51 58  5 54 83
    -Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36
    -Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11
    -

    In the above example, card 1 has five winning numbers (41, 48, 83, 86, -and 17) and eight numbers you have (83, 86, 6, 31, 17, 9, 48, -and 53). Of the numbers you have, four of them (48, 83, 17, and 86) -are winning numbers! That means card 1 is worth 8 points (1 for the first -match, then doubled three times for each of the three matches after the first).

    • Card 2 has two winning numbers (32 and 61), so it is worth 2 points.
    • Card 3 has two winning numbers (1 and 21), so it is worth 2 points.
    • Card 4 has one winning number (84), so it is worth 1 point.
    • Card 5 has no winning numbers, so it is worth no points.
    • Card 6 has no winning numbers, so it is worth no points.

    So, in this example, the Elf’s pile of scratchcards is worth 13 points.

    Take a seat in the large pile of colorful cards. How many points are they worth -in total?

    My input file: -2023-12-04-1-aoc.txt

    Loading this data is very similar to Day 2 - so let’s load the data as we did -there. Our goal is to get win and yours columns holding the respective -digits which we want to compare. We find the numbers with one or more digits -using the regex \d+. And we want them to be in sets (not lists), as we can -logically compare sets in Python.

    import pandas as pd
    -import re
    -
    -txt = pd.read_table('data/2023-12-04-1-aoc.txt', names=['win'])
    -txt['id'] = txt.loc[:, 'win'].str.split(':').apply(
    -    lambda x: int(x[0].strip('Card ')))
    -txt['win'] = (txt.loc[:, 'win']
    -              .str.split(':').apply(lambda x: x[1]))
    -txt['yours'] = (txt.loc[:, 'win']
    -                .str.split('|')
    -                .apply(lambda x: x[1])
    -                # get a list of only the numbers / digits
    -                .apply(lambda x: re.findall(r'\d+', x))
    -                # convert the list of strings to a set of integers
    -                .apply(lambda x: set([int(i) for i in x])))
    -txt['win'] = (txt.loc[:, 'win']
    -              .str.split('|')
    -              .apply(lambda x: x[0])
    -              .apply(lambda x: re.findall(r'\d+', x))
    -              .apply(lambda x: set([int(i) for i in x])))
    -
    -txt
    -
    winidyours
    0{32, 36, 7, 9, 10, 12, 82, 85, 95, 31}1{2, 7, 9, 10, 12, 14, 21, 22, 23, 24, 31, 32, …
    1{35, 76, 16, 82, 19, 22, 88, 59, 60, 95}2{7, 8, 12, 16, 19, 22, 26, 28, 35, 38, 44, 51,…
    2{1, 70, 11, 78, 48, 19, 52, 88, 28, 94}3{3, 4, 8, 17, 18, 19, 24, 31, 34, 45, 52, 54, …
    3{65, 2, 72, 28, 14, 16, 55, 91, 92, 62}4{3, 4, 6, 7, 8, 9, 15, 30, 33, 35, 47, 49, 51,…
    4{38, 41, 75, 77, 50, 24, 94, 60, 61, 30}5{1, 2, 4, 5, 6, 7, 9, 10, 14, 17, 21, 29, 47, …
    213{97, 98, 39, 41, 43, 12, 13, 19, 93, 95}214{5, 10, 17, 20, 28, 29, 33, 34, 36, 50, 51, 52…
    214{97, 35, 69, 40, 74, 45, 20, 21, 62, 31}215{1, 8, 15, 17, 18, 25, 30, 33, 42, 44, 47, 52,…
    215{33, 70, 71, 12, 78, 17, 51, 86, 60, 94}216{7, 8, 9, 10, 22, 29, 37, 39, 41, 43, 46, 47, …
    216{98, 67, 68, 38, 70, 39, 72, 77, 45, 21}217{8, 21, 22, 25, 26, 31, 37, 41, 42, 48, 54, 57…
    217{34, 9, 44, 78, 79, 16, 17, 19, 55, 92}218{1, 4, 20, 21, 27, 38, 39, 40, 41, 45, 46, 52,…

    218 rows × 3 columns

    Now, we get the -logical conjunction of win and yours, these are our winning -numbers. Then, the number of wins is converted to points - for all number of -wins bigger than 1, we can get the points by 2**(n_wins-1).

    txt['n_wins'] = txt.apply(
    -    lambda row: len(row.loc['win'] & row.loc['yours']), axis=1)
    -txt['points'] = txt.loc[:, 'n_wins'].apply(
    -    lambda x: 2**(x-1) if x > 1 else x)
    -
    -txt.loc[:, ['win', 'n_wins', 'points']]
    -
    winn_winspoints
    0{32, 36, 7, 9, 10, 12, 82, 85, 95, 31}10512
    1{35, 76, 16, 82, 19, 22, 88, 59, 60, 95}10512
    2{1, 70, 11, 78, 48, 19, 52, 88, 28, 94}516
    3{65, 2, 72, 28, 14, 16, 55, 91, 92, 62}00
    4{38, 41, 75, 77, 50, 24, 94, 60, 61, 30}00
    213{97, 98, 39, 41, 43, 12, 13, 19, 93, 95}00
    214{97, 35, 69, 40, 74, 45, 20, 21, 62, 31}00
    215{33, 70, 71, 12, 78, 17, 51, 86, 60, 94}22
    216{98, 67, 68, 38, 70, 39, 72, 77, 45, 21}11
    217{34, 9, 44, 78, 79, 16, 17, 19, 55, 92}00

    218 rows × 3 columns

    Then, we just sum up:

    sum(txt.loc[:, 'points'])
    -
    25004
    -

    Part 2 -#

    Let’s read the task of part 2!

    There’s no such thing as “points”. Instead, scratchcards only cause you to win -more scratchcards equal to the number of winning numbers you have.

    Specifically, you win copies of the scratchcards below the winning card equal to -the number of matches. So, if card 10 were to have 5 matching numbers, you would -win one copy each of cards 11, 12, 13, 14, and 15.

    Copies of scratchcards are scored like normal scratchcards and have the same -card number as the card they copied. So, if you win a copy of card 10 and it -has 5 matching numbers, it would then win a copy of the same cards that the -original card 10 won: cards 11, 12, 13, 14, and 15. This process repeats until -none of the copies cause you to win any more cards. (Cards will never make you -copy a card past the end of the table.)

    This time, the above example goes differently:

    Card 1: 41 48 83 86 17 | 83 86  6 31 17  9 48 53
    -Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19
    -Card 3:  1 21 53 59 44 | 69 82 63 72 16 21 14  1
    -Card 4: 41 92 73 84 69 | 59 84 76 51 58  5 54 83
    -Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36
    -Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11
    -
    • Card 1 has four matching numbers, so you win one copy each of the next four -cards: cards 2, 3, 4, and 5.
    • Your original card 2 has two matching numbers, so you win one copy each of -cards 3 and 4.
    • Your copy of card 2 also wins one copy each of cards 3 and 4.
    • Your four instances of card 3 (one original and three copies) have two -matching numbers, so you win four copies each of cards 4 and 5.
    • Your eight instances of card 4 (one original and seven copies) have one -matching number, so you win eight copies of card 5.
    • Your fourteen instances of card 5 (one original and thirteen copies) have no -matching numbers and win no more cards.

    Once all of the originals and copies have been processed, you end up with 1 -instance of card 1, 2 instances of card 2, 4 instances of card 3, 8 -instances of card 4, 14 instances of card 5, and 1 instance of card 6. In -total, this example pile of scratchcards causes you to ultimately have 30 -scratchcards!

    Process all of the original and copied scratchcards until no more scratchcards -are won. Including the original set of scratchcards, how many total scratchcards -do you end up with?

    We will use the n_wins column we created before and go from there. We step -through each Game. Each current game gets +1 card. Then, we step through the -number of next games depending on our n_wins. Each of these gets added the -card number of the current game.

    txt['cards'] = 0
    -
    -for i, nwin in enumerate(txt.loc[:, 'n_wins']):
    -    txt.loc[i, 'cards'] += 1
    -    for j in range(1, nwin+1):
    -        txt.loc[i+j, 'cards'] += txt.loc[i, 'cards']
    -
    -txt.loc[:, ['n_wins', 'cards']]
    -
    n_winscards
    0101
    1102
    254
    308
    408
    21309608
    21408927
    21528927
    216112636
    217021564

    218 rows × 2 columns

    Now, we just sum the output again:

    sum(txt['cards'])
    -
    14427616
    -
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    Advent of code 2023 - Day 5: If You Give A Seed A Fertilizer

    Advent of code 2023 - Day 5: If You Give A Seed A Fertilizer

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 5 - see -https:adventofcode.com/2023/day/5

    Update [2023-12-31 So]:

    • substitute get_generator() (own implementation) with range() (Python inbuilt)
    • improve grid search so that it goes through all location ranges, still -starting with the lowest range

    Part 1 -#

    Lets first read the task:

    The almanac (your puzzle input) lists all of the seeds that need to be planted. -It also lists what type of soil to use with each kind of seed, what type of -fertilizer to use with each kind of soil, what type of water to use with each -kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is -identified with a number, but numbers are reused by each category - that is, -soil 123 and fertilizer 123 aren’t necessarily related to each other.

    For example:

    seeds: 79 14 55 13
    -
    -seed-to-soil map:
    -50 98 2
    -52 50 48
    -
    -soil-to-fertilizer map:
    -0 15 37
    -37 52 2
    -39 0 15
    -
    -fertilizer-to-water map:
    -49 53 8
    -0 11 42
    -42 0 7
    -57 7 4
    -
    -water-to-light map:
    -88 18 7
    -18 25 70
    -
    -light-to-temperature map:
    -45 77 23
    -81 45 19
    -68 64 13
    -
    -temperature-to-humidity map:
    -0 69 1
    -1 0 69
    -
    -humidity-to-location map:
    -60 56 37
    -56 93 4
    -

    The almanac starts by listing which seeds need to be planted: seeds 79, 14, -55, and 13.

    The rest of the almanac contains a list of maps which describe how to convert -numbers from a source category into numbers in a destination category. That -is, the section that starts with seed-to-soil map: describes how to convert a -seed number (the source) to a soil number (the destination). This lets the -gardener and his team know which soil to use with which seeds, which water to -use with which fertilizer, and so on.

    Rather than list every source number and its corresponding destination number -one by one, the maps describe entire ranges of numbers that can be converted. -Each line within a map contains three numbers: the destination range start, -the source range start, and the range length.

    Consider again the example seed-to-soil map:

    50 98 2 52 50 48

    The first line has a destination range start of 50, a source range start -of 98, and a range length of 2. This line means that the source range -starts at 98 and contains two values: 98 and 99. The destination range is -the same length, but it starts at 50, so its two values are 50 and 51. -With this information, you know that seed number 98 corresponds to soil number -50 and that seed number 99 corresponds to soil number 51. -o -The second line means that the source range starts at 50 and contains 48 -values: 50, 51, …, 96, 97. This corresponds to a destination range -starting at 52 and also containing 48 values: 52, 53, …, 98, 99. -So, seed number 53 corresponds to soil number 55.

    Any source numbers that aren’t mapped correspond to the same destination -number. So, seed number 10 corresponds to soil number 10.

    So, the entire list of seed numbers and their corresponding soil numbers looks -like this:

    seed  soil
    -0     0
    -1     1
    -...   ...
    -48    48
    -49    49
    -50    52
    -51    53
    -...   ...
    -96    98
    -97    99
    -98    50
    -99    51
    -

    With this map, you can look up the soil number required for each initial seed -number:

    • Seed number 79 corresponds to soil number 81.
    • Seed number 14 corresponds to soil number 14.
    • Seed number 55 corresponds to soil number 57.
    • Seed number 13 corresponds to soil number 13.

    The gardener and his team want to get started as soon as possible, so they’d -like to know the closest location that needs a seed. Using these maps, find the -lowest location number that corresponds to any of the initial seeds. To do -this, you’ll need to convert each seed number through other categories until you -can find its corresponding location number. In this example, the corresponding -types are:

    • Seed 79, soil 81, fertilizer 81, water 81, light 74, temperature -78, humidity 78, location 82.
    • Seed 14, soil 14, fertilizer 53, water 49, light 42, temperature -42, humidity 43, location 43.
    • Seed 55, soil 57, fertilizer 57, water 53, light 46, temperature -82, humidity 82, location 86.
    • Seed 13, soil 13, fertilizer 52, water 41, light 34, temperature -34, humidity 35, location 35.

    So, the lowest location number in this example is 35.

    What is the lowest location number that corresponds to any of the initial seed -numbers?

    My input file: -2023-12-05-1-aoc.txt

    Wow, this task is a mouthful…

    Let’s start slowly and load the data. Our input text document contains several -maps, which are clearly separated and have a title (seed-to-soil map etc). So -we can tell pandas where each map starts and give each map a dataframe. I got -the values for the skiprows and nrows argument by looking at the input file -and… counting -:)

    import pandas as pd
    -import sys
    -
    -seeds = pd.read_table('data/2023-12-05-1-aoc.txt', nrows=1, sep=' ',
    -                      header=None, index_col=0)
    -seeds = seeds.values.flatten()
    -
    -map_opt = {'filepath_or_buffer': 'data/2023-12-05-1-aoc.txt',
    -           'header': None, 'sep': ' ', 'dtype': 'Int64',
    -           'names': ['dest_start', 'src_start', 'range']}
    -
    -seed_soil = pd.read_table(skiprows=3, nrows=23, **map_opt)
    -soil_fert = pd.read_table(skiprows=28, nrows=9, **map_opt)
    -fert_water = pd.read_table(skiprows=39, nrows=20, **map_opt)
    -water_light = pd.read_table(skiprows=61, nrows=40, **map_opt)
    -light_temp = pd.read_table(skiprows=103, nrows=36, **map_opt)
    -temp_humi = pd.read_table(skiprows=141, nrows=35, **map_opt)
    -humi_loc = pd.read_table(skiprows=178, nrows=26, **map_opt)
    -
    -maps = (seed_soil, soil_fert, fert_water, water_light, light_temp,
    -        temp_humi, humi_loc)
    -
    -print('seeds are just a numpy array:')
    -display(seeds)
    -print('The "humidity-to-location" map as an example:')
    -humi_loc
    -
    seeds are just a numpy array:
    -array([3169137700,  271717609, 3522125441,   23376095, 1233948799,
    -        811833837,  280549587,  703867355,  166086528,   44766996,
    -       2326968141,   69162222, 2698492851,   14603069, 2755327667,
    -        348999531, 2600461189,   92332846, 1054656969,  169099767])The "humidity-to-location" map as an example:
    -
    dest_startsrc_startrange
    034901440031623866227218040905
    1170961057816208391973027030
    2105449249586389428113279526
    318996043382167886292348178199
    417126376082678624215186966730
    52187287750245776251
    62472992580923734334143670388
    72616662968367016988515297294
    822477825371395629154225210043
    90480940179105449249
    1041138528463959729057159909096
    1133227846531067404722167359350
    129237343344119638153175329143
    1315349644962516064491162559724
    1426319602623496028440140849733
    152862695906123476407297988355
    161697524220363687817312086358
    172985646048133275242762876727
    1810990634772970241510435901019
    1940092022812865590945104650565
    202960684261214292450524961787
    212772809995340614252989885911
    224273761942364896453121205354
    2337081849081841907132301017373
    24464505026245776251235163928
    2530485227753685467179274261878

    My first attempt was to actually construct ranges like in the example above, -mapping out all possible sources and destinations. Python quickly informed me -that even constructing one pandas.Series with int64 values mapping seeds to -soil would cost 64GB memory - not the best solution.

    So we take a different approach. For convenience, let’s add a src_end and a -dest_end column to our maps:

    for df in maps:
    -    df['src_end'] = df.loc[:, 'src_start'] + df.loc[:, 'range']
    -    df['dest_end'] = df.loc[:, 'dest_start'] + df.loc[:, 'range']
    -
    -print('Again the "humidity-to-location" map as an example:')
    -humi_loc
    -
    Again the "humidity-to-location" map as an example:
    -
    dest_startsrc_startrangesrc_enddest_end
    03490144003162386622721804090518419071323708184908
    117096105781620839197302703016238662271712637608
    2105449249586389428113279526699668954218728775
    31899604338216788629234817819925160644912247782537
    41712637608267862421518696673028655909451899604338
    52187287750245776251245776251464505026
    6247299258092373433414367038810674047222616662968
    7261666296836701698851529729436854671792631960262
    82247782537139562915422521004316208391972472992580
    90480940179105449249586389428105449249
    104113852846395972905715990909641196381534273761942
    113322784653106740472216735935012347640723490144003
    12923734334411963815317532914342949672961099063477
    131534964496251606449116255972426786242151697524220
    142631960262349602844014084973336368781732772809995
    15286269590612347640729798835513327524272960684261
    16169752422036368781731208635836489645311709610578
    17298564604813327524276287672713956291543048522775
    181099063477297024151043590101934061425291534964496
    194009202281286559094510465056529702415104113852846
    20296068426121429245052496178721678862922985646048
    21277280999534061425298988591134960284402862695906
    22427376194236489645312120535436701698854294967296
    233708184908184190713230101737321429245054009202281
    24464505026245776251235163928480940179699668954
    253048522775368546717927426187839597290573322784653

    Now we actually compute the mapping. For each seed, we go through all mappings -and in each mapping we go through each row. We find the row which contains the -mapping and exctract the destination, which is the source for the next map until -we reach the last map which gives us the locations.

    • Approach 1: df.itertuples() is a convenient way to step through a -pandas.DataFrame in this example. It is faster than df.iterrows() and -returns the row as a NamedTuple - nice!

    • Approach 2: I actually wondered if it would be faster to get all maps in one -pd.DataFrame and then iterate through the mappings. To test this let’s -construct a new DataFrame maps_df which contains all maps. Since the maps -have different lengths it is important to cast the datatype to Int64, which -is short for pd.Int64Dtype() and keeps values as integers, even if NA -values are in the same column.

    • Approach 3: A third alternative I tested (not shown here) was to check if a -value is in a Python range with the in operator as in: if 3 in range(5):... . This was way too slow.

    # mapping version 1
    -def get_location(seed):
    -    current = seed
    -    for df in maps:
    -        current_map = [row
    -                       for row in df.itertuples()
    -                       if ((current > row.src_start)
    -                           and (current < row.src_end))]
    -        if len(current_map) == 0:
    -            pass
    -        elif len(current_map) == 1:
    -            current = (current_map[0].dest_start
    -                       + (current - current_map[0].src_start))
    -        else:
    -            raise ValueError('This should not happen!')
    -    return current
    -
    -# mapping version 2 - around 10 times slower
    -# you need to rename the maps_df columns so that they have a unique id
    -# e.g. 'src_start_1', 'src_start_2' etc
    -
    -# maps_df = pd.concat(maps, axis='columns')
    -
    -# def get_dest(i, src):
    -#     return (maps_df[(src > maps_df.loc[:, f'src_start_{i}']) &
    -#                     (src < maps_df.loc[:, f'src_end_{i}'])]
    -#             .loc[:, f'dest_start_{i}']
    -#             .iloc[0])
    -
    -# def get_location2(seed):
    -#     dest = seed
    -#     i = 1
    -#     while i < 7:
    -#         dest = get_dest(i, dest)
    -#         i += 1
    -#     return dest
    -
    -%timeit get_location(seeds[0])
    -
    1.98 ms ± 72 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
    -

    Now let’s get a list of locations:

    locations = [get_location(s) for s in seeds]
    -locations
    -
    24939826553209845376399235753341631314634104485616195225247932186773543880712892181441450259433631540495076702084517144311963363542897831235187719913704555655953918455210768776834480463302184454689

    Lastly, just get the minimum of all location values.

      min(locations)
    -
    388071289
    -

    Part 2 -#

    Let’s read the task of part 2!

    Everyone will starve if you only plant such a small number of seeds. Re-reading -the almanac, it looks like the seeds: line actually describes ranges of seed -numbers.

    The values on the initial seeds: line come in pairs. Within each pair, the -first value is the start of the range and the second value is the length of -the range. So, in the first line of the example above:

    seeds: 79 14 55 13

    This line describes two ranges of seed numbers to be planted in the garden. The -first range starts with seed number 79 and contains 14 values: 79, 80, -…, 91, 92. The second range starts with seed number 55 and contains 13 -values: 55, 56, …, 66, 67.

    Now, rather than considering four seed numbers, you need to consider a total of -27 seed numbers.

    In the above example, the lowest location number can be obtained from seed -number 82, which corresponds to soil 84, fertilizer 84, water 84, light -77, temperature 45, humidity 46, and location 46. So, the lowest -location number is 46.

    Consider all of the initial seed numbers listed in the ranges on the first line -of the almanac. What is the lowest location number that corresponds to any of -the initial seed numbers?

    Let’s first construct a dataframe containing the range of seeds:

    seeds_df = pd.DataFrame({'start': seeds[::2],
    -                         'range': seeds[1::2],
    -                         'end': seeds[::2] + seeds[1::2]})
    -print(f'There are {sum(seeds_df.loc[:, "range"]):_} seeds in total')
    -seeds_df
    -
    There are 2_549_759_327 seeds in total
    -
    startrangeend
    031691377002717176093440855309
    13522125441233760953545501536
    212339487998118338372045782636
    3280549587703867355984416942
    416608652844766996210853524
    52326968141691622222396130363
    62698492851146030692713095920
    727553276673489995313104327198
    82600461189923328462692794035
    910546569691690997671223756736

    Now - I really needed some time to finally realize, that going through all seed -values is really unfeasable. So how do we deal with this problem?

    In the end we need the lowest location number - thus our approach is to take -the humi_loc map, start with the lowest location number and go up and get the -corresponding seed values. The location of the first seed value which is inside -seeds_df is our solution.

    So first we rebuild the get_location function to a get_seed function (we -reverse the maps tuple with maps[::-1] and switch src and dest).

    def get_seed(location):
    -    current = location
    -    for df in maps[::-1]:
    -        current_map = [row
    -                       for row in df.itertuples()
    -                       if ((current >= row.dest_start)
    -                           and (current < row.dest_end))]
    -        if len(current_map) == 0:
    -            pass
    -        elif len(current_map) == 1:
    -            current = (current_map[0].src_start
    -                       + (current - current_map[0].dest_start))
    -        else:
    -            raise ValueError('This should not happen!')
    -    return current
    -

    Since the Python 3 range function does not store it’s contents in memory -(similar to a generator), it is well suited to go through these large ranges.

    Lastly, we deal with the sheer amount of possible values by performing a grid -search. First, we check every millionth location. After the first match, we -stop this search and check the last million locations before the match with a -finer grid and so on. The last grid is just 1, so we find our lowest location.

    Update: We order our locations from smallest to largest with sort_values and -go through them - but the search stops after the first match, since that will be -our lowest location.

    def grid_search(start: int, end: int, grid: list[int]):
    -    success = False
    -    for g in grid:
    -        print(f'Start search with grid={g}')
    -        for l in range(start, end, g):
    -            current = get_seed(l)
    -            if any((current >= seeds_df.loc[:, 'start']) & (current < seeds_df.loc[:, 'end'])):
    -                print(f'location {l} is the lowest which contains one of the given seeds ({current})')
    -                start = l - g
    -                end = l
    -                success = True
    -                break
    -    return success
    -
    -for row in humi_loc.sort_values('dest_start').itertuples():
    -    success = grid_search(start=row.dest_start, end=row.dest_end,
    -                          grid=[1_000_000, 100_000, 1000, 1])
    -    if success:
    -        print('Finished search')
    -        break
    -
    Start search with grid=1000000
    -location 85000000 is the lowest which contains one of the given seeds (2605777210)
    -Start search with grid=100000
    -location 84300000 is the lowest which contains one of the given seeds (2605077210)
    -Start search with grid=1000
    -location 84207000 is the lowest which contains one of the given seeds (2604984210)
    -Start search with grid=1
    -location 84206669 is the lowest which contains one of the given seeds (2604983879)
    -Finished search
    -
    comments powered by Disqus
    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/posts/aoc2023-day06/index.html b/aseltmann.github.io-hugo/public/posts/aoc2023-day06/index.html deleted file mode 100644 index 716c2bc..0000000 --- a/aseltmann.github.io-hugo/public/posts/aoc2023-day06/index.html +++ /dev/null @@ -1,123 +0,0 @@ -Advent of code 2023 - Day 6: Wait For It | Blog & Notes - -

    Advent of code 2023 - Day 6: Wait For It

    Advent of code 2023 - Day 6: Wait For It

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 6 - see -https:adventofcode.com/2023/day/6

    Part 1 -#

    Lets first read the task:

    You manage to sign up as a competitor in the boat races just in time. The -organizer explains that it’s not really a traditional race - instead, you will -get a fixed amount of time during which your boat has to travel as far as it -can, and you win if your boat goes the farthest.

    As part of signing up, you get a sheet of paper (your puzzle input) that lists -the time allowed for each race and also the best distance ever recorded in that -race. To guarantee you win the grand prize, you need to make sure you go farther -in each race than the current record holder.

    The organizer brings you over to the area where the boat races are held. The -boats are much smaller than you expected - they’re actually toy boats, each with -a big button on top. Holding down the button charges the boat, and releasing the -button allows the boat to move. Boats move faster if their button was held -longer, but time spent holding the button counts against the total race time. -You can only hold the button at the start of the race, and boats don’t move -until the button is released.

    For example:

    Time:      7  15   30
    -Distance:  9  40  200
    -

    This document describes three races:

    • The first race lasts 7 milliseconds. The record distance in this race is 9 -millimeters.
    • The second race lasts 15 milliseconds. The record distance in this race is 40 -millimeters.
    • The third race lasts 30 milliseconds. The record distance in this race is 200 -millimeters.

    Your toy boat has a starting speed of zero millimeters per millisecond. For -each whole millisecond you spend at the beginning of the race holding down the -button, the boat’s speed increases by one millimeter per millisecond.

    So, because the first race lasts 7 milliseconds, you only have a few options:

    • Don’t hold the button at all (that is, hold it for 0 milliseconds) at the -start of the race. The boat won’t move; it will have traveled 0 -millimeters by the end of the race.
    • Hold the button for 1 millisecond at the start of the race. Then, the boat -will travel at a speed of 1 millimeter per millisecond for 6 milliseconds, -reaching a total distance traveled of 6 millimeters.
    • Hold the button for 2 milliseconds, giving the boat a speed of 2 -millimeters per millisecond. It will then get 5 milliseconds to move, -reaching a total distance of 10 millimeters.
    • Hold the button for 3 milliseconds. After its remaining 4 milliseconds of -travel time, the boat will have gone 12 millimeters.
    • Hold the button for 4 milliseconds. After its remaining 3 milliseconds of -travel time, the boat will have gone 12 millimeters.
    • Hold the button for 5 milliseconds, causing the boat to travel a total of -10 millimeters.
    • Hold the button for 6 milliseconds, causing the boat to travel a total of -6 millimeters.
    • Hold the button for 7 milliseconds. That’s the entire duration of the -race. You never let go of the button. The boat can’t move until you let go of -the button. Please make sure you let go of the button so the boat gets to -move. 0 millimeters.

    Since the current record for this race is 9 millimeters, there are actually -4 different ways you could win: you could hold the button for 2, 3, 4, -or 5 milliseconds at the start of the race.

    In the second race, you could hold the button for at least 4 milliseconds and -at most 11 milliseconds and beat the record, a total of 8 different ways -to win.

    In the third race, you could hold the button for at least 11 milliseconds and -no more than 19 milliseconds and still beat the record, a total of 9 ways -you could win.

    To see how much margin of error you have, determine the number of ways you can -beat the record in each race; in this example, if you multiply these values -together, you get 288 (4 * 8 * 9).

    Determine the number of ways you could beat the record in each race. What do -you get if you multiply these numbers together?

    My input file: -2023-12-06-1-aoc.txt

    The solution for this problem is shorter than the task description!

    As always - let’s load the data. Note: we use the delim_whitesspace argument -instead of sep=' ' to separate values by spaces of any length.

    import pandas as pd
    -import numpy as np
    -
    -race = pd.read_table('data/2023-12-06-1-aoc.txt', delim_whitespace=True,
    -                     header=None, index_col=0)
    -race.index = ['time', 'distance']
    -race
    -
    1234
    time51926890
    distance222203111261225

    The we use Python list comprehensions to quickly calculate all distances for the -range of the time value (and keep only the winning distances). Then we get the -respective lengths and compute the product.

    def get_distance(time, distance):
    -    return [t * (time - t) for t in range(1, time + 1)
    -            if (t * (time - t)) > distance]
    -
    -race.loc['nwins', :] = [len(get_distance(int(r.time), int(r.distance)))
    -                        for c, r in race.items()]
    -nwins_prod = np.prod(race.loc['nwins', :].values, dtype=int)
    -display(race)
    -print(f'The product of our nwins is {nwins_prod}')
    -
    1234
    time51.092.068.090.0
    distance222.02031.01126.01225.0
    nwins42.019.011.057.0
    The product of our nwins is 500346
    -

    Part 2 -#

    Lets first read the task:

    As the race is about to start, you realize the piece of paper with race times -and record distances you got earlier actually just has very bad -kerning. There’s -really only one race - ignore the spaces between the numbers on each line.

    So, the example from before:

    Time:      7  15   30
    -Distance:  9  40  200
    -

    …now instead means this:

    Time:      71530
    -Distance:  940200
    -

    Now, you have to figure out how many ways there are to win this single race. In -this example, the race lasts for 71530 milliseconds and the record distance -you need to beat is 940200 millimeters. You could hold the button anywhere -from 14 to 71516 milliseconds and beat the record, a total of 71503 -ways!

    How many ways can you beat the record in this one much longer race?

    For this riddle we can take the same approach as for Part 1. The cell took 18s -to evaluate on a consumer laptop from 2015 - no fancy workarounds needed :)

    racetime = int(''.join(str(r) for r in race.loc['time', :4].astype(int)))
    -racedist = int(''.join(str(r) for r in race.loc['distance', :4].astype(int)))
    -racewins = len(get_distance(racetime, racedist))
    -
    -race[5] = [racetime, racedist, racewins]
    -race
    -
    12345
    time51.092.068.090.051926890
    distance222.02031.01126.01225.0222203111261225
    nwins42.019.011.057.042515755
    comments powered by Disqus
    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/posts/aoc2023-day07/index.html b/aseltmann.github.io-hugo/public/posts/aoc2023-day07/index.html deleted file mode 100644 index 286021d..0000000 --- a/aseltmann.github.io-hugo/public/posts/aoc2023-day07/index.html +++ /dev/null @@ -1,181 +0,0 @@ -Advent of code 2023 - Day 7: Camel Cards | Blog & Notes - -

    Advent of code 2023 - Day 7: Camel Cards

    Advent of code 2023 - Day 7: Camel Cards

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 7 - see -https:adventofcode.com/2023/day/7

    Part 1 -#

    Lets first read the task:

    In Camel Cards, you get a list of hands, and your goal is to order them based -on the strength of each hand. A hand consists of five cards labeled one of -A, K, Q, J, T, 9, 8, 7, 6, 5, 4, 3, or 2. The relative -strength of each card follows this order, where A is the highest and 2 is -the lowest.

    Every hand is exactly one type. From strongest to weakest, they are:

    • Five of a kind, where all five cards have the same label: AAAAA
    • Four of a kind, where four cards have the same label and one card has a -different label: AA8AA
    • Full house, where three cards have the same label, and the remaining two -cards share a different label: 23332
    • Three of a kind, where three cards have the same label, and the remaining -two cards are each different from any other card in the hand: TTT98
    • Two pair, where two cards share one label, two other cards share a second -label, and the remaining card has a third label: 23432
    • One pair, where two cards share one label, and the other three cards have a -different label from the pair and each other: A23A4
    • High card, where all cards’ labels are distinct: 23456

    Hands are primarily ordered based on type; for example, every full house is -stronger than any three of a kind.

    If two hands have the same type, a second ordering rule takes effect. Start by -comparing the first card in each hand. If these cards are different, the hand -with the stronger first card is considered stronger. If the first card in each -hand have the same label, however, then move on to considering the second card -in each hand. If they differ, the hand with the higher second card wins; -otherwise, continue with the third card in each hand, then the fourth, then the -fifth.

    So, 33332 and 2AAAA are both four of a kind hands, but 33332 is stronger -because its first card is stronger. Similarly, 77888 and 77788 are both a -full house, but 77888 is stronger because its third card is stronger (and -both hands have the same first and second card).

    To play Camel Cards, you are given a list of hands and their corresponding bid -(your puzzle input). For example:

    32T3K 765
    -T55J5 684
    -KK677 28
    -KTJJT 220
    -QQQJA 483
    -

    This example shows five hands; each hand is followed by its bid amount. Each -hand wins an amount equal to its bid multiplied by its rank, where the weakest -hand gets rank 1, the second-weakest hand gets rank 2, and so on up to the -strongest hand. Because there are five hands in this example, the strongest hand -will have rank 5 and its bid will be multiplied by 5.

    So, the first step is to put the hands in order of strength:

    • 32T3K is the only one pair and the other hands are all a stronger type, so -it gets rank 1.
    • KK677 and KTJJT are both two pair. Their first cards both have the same -label, but the second card of KK677 is stronger (K vs T), so KTJJT -gets rank 2 and KK677 gets rank 3.
    • T55J5 and QQQJA are both three of a kind. QQQJA has a stronger first -card, so it gets rank 5 and T55J5 gets rank 4.

    Now, you can determine the total winnings of this set of hands by adding up the -result of multiplying each hand’s bid with its rank (765 * 1 + 220 * 2 + -28 * 3 + 684 * 4 + 483 * 5). So the total winnings in this example are -6440.

    Find the rank of every hand in your set. What are the total winnings?

    My input file: -2023-12-07-1-aoc.txt

    As always - let’s read in the data.

    import pandas as pd
    -
    -camel = pd.read_table('data/2023-12-07-1-aoc.txt', delim_whitespace=True,
    -                      header=None, names=['cards', 'bid'],
    -                      dtype={'cards': 'string', 'bid': 'Int64'})
    -camel
    -
    cardsbid
    0A833A309
    1Q33J3205
    255KK5590
    3K4457924
    4Q3QT311
    995KQJ53367
    99657866537
    997Q94A9210
    998J448A903
    9996J66J114

    1000 rows × 2 columns

    For sorting, we need to compute a type column. The 7 types can be deduced -based on how many different cards are there (through set()) and the number of -cards.

    Also, for the next step we prepare one column per card string in the cards.

    def get_type(cards):
    -    cardset = set(cards)
    -    cardlen = len(cardset)
    -    cardlist = list(cards)
    -    cardnum = [cardlist.count(i) for i in cardset]
    -    if cardlen == 1:
    -        ctype = 1
    -    elif cardlen in (4, 5):
    -        ctype = cardlen + 2
    -    elif cardlen == 2:
    -        ctype = 2 if set(cardnum) == {1, 4} else 3
    -    else:
    -        ctype = 4 if 3 in cardnum else 5
    -    return ctype
    -
    -camel['type'] = camel.loc[:, 'cards'].apply(lambda x: get_type(x))
    -for i in range(5):
    -    camel[f'c{i}'] = camel.loc[:, 'cards'].apply(lambda x: list(x)[i])
    -camel
    -
    cardsbidtypec0c1c2c3c4
    0A833A3095A833A
    1Q33J32054Q33J3
    255KK5590355KK5
    3K44579246K4457
    4Q3QT3115Q3QT3
    995KQJ533677KQJ53
    99657866537657866
    997Q94A92106Q94A9
    998J448A9036J448A
    9996J66J11436J66J

    1000 rows × 8 columns

    For sorting, pandas contributes the sort_values method, which supports -sorting by multiple columns (here, we first sort according to type, then c0, -c1, …). For easier sorting by numerical values, we convert the values using -the card_dict dictionary, which implements the given sorting order.

    card_dict = {c: i for i, c in enumerate([7, 6, 5, 4, 3, 2, 1, '2', '3',
    -                                         '4', '5', '6', '7', '8', '9',
    -                                         'T', 'J', 'Q', 'K', 'A'])}
    -
    -camel = camel.sort_values(by=['type', 'c0', 'c1', 'c2', 'c3', 'c4'],
    -                          ignore_index=True,
    -                          key=lambda x: x.map(card_dict))
    -camel.index += 1
    -camel
    -
    cardsbidtypec0c1c2c3c4
    1237AQ1577237AQ
    223K47759723K47
    3249K86127249K8
    4264A73417264A7
    52657810726578
    996AA5AA5652AA5AA
    997AAATA7042AAATA
    998AAAA31452AAAA3
    999AAAAJ5942AAAAJ
    1000JJJJJ2191JJJJJ

    1000 rows × 8 columns

    Lastly, we multiply the bid with the index and sum the result.

    camel.loc[:, 'bid'].mul(camel.index).sum()
    -
    246912307
    -

    Part 2 -#

    Lets first read the task:

    To make things a little more interesting, the Elf introduces one additional -rule. Now, J cards are jokers - wildcards that can act like whatever card -would make the hand the strongest type possible.

    To balance this, J cards are now the weakest individual cards, weaker even -than 2. The other cards stay in the same order: A, K, Q, T, 9, 8, -7, 6, 5, 4, 3, 2, J.

    J cards can pretend to be whatever card is best for the purpose of determining -hand type; for example, QJJQ2 is now considered four of a kind. However, for -the purpose of breaking ties between two hands of the same type, J is always -treated as J, not the card it’s pretending to be: JKKK2 is weaker than -QQQQ2 because J is weaker than Q.

    Now, the above example goes very differently:

    32T3K 765
    -T55J5 684
    -KK677 28
    -KTJJT 220
    -QQQJA 483
    -
    • 32T3K is still the only one pair; it doesn’t contain any jokers, so its -strength doesn’t increase.
    • KK677 is now the only two pair, making it the second-weakest hand.
    • T55J5, KTJJT, and QQQJA are now all four of a kind! T55J5 gets rank -3, QQQJA gets rank 4, and KTJJT gets rank 5.

    With the new joker rule, the total winnings in this example are 5905.

    Using the new joker rule, find the rank of every hand in your set. What are the -new total winnings?

    What changes? We have to update our get_type() function and our card_dict -dictionary, otherwise the method stays the same!

    Let’s first write get_type_joker(): if no joker is in the cards, we can just -refer to get_type(). If we have 1, 2 or 5 different card faces, the answer is -trivial. We only have to use the number of jokers and the maximum number of -other faces if we have 3 or 4 different faces.

    Note: if a J is present, there can never be high card (type = 7), because -we always can get one pair. Also we can never get two pair (type = 5), -because we can always construct three of a kind or full house.

    def get_type_joker(cards):
    -    cardset = set(cards)
    -    cardlen = len(cardset)
    -    cardlist = list(cards)
    -    cardnum = [cardlist.count(i) for i in cardset]
    -    if not 'J' in cardset:
    -        ctype = get_type(cards)
    -    else:
    -        idxj = list(cardset).index('J')
    -        numj = cardnum[idxj]
    -        cardnum.pop(idxj)
    -        if cardlen in [1, 2]:
    -            ctype = 1
    -        elif cardlen == 5:
    -            ctype = 6
    -        else:
    -            if (numj in [1, 2]) and (cardlen == 4):
    -                ctype = 4
    -            else:
    -                ctype = 6 - max(cardnum) - numj
    -    return ctype
    -
    -camel['type'] = camel.loc[:, 'cards'].apply(lambda x: get_type_joker(x))
    -
    -camel
    -
    cardsbidtypec0c1c2c3c4
    1237AQ1577237AQ
    223K47759723K47
    3249K86127249K8
    4264A73417264A7
    52657810726578
    996AA5AA5652AA5AA
    997AAATA7042AAATA
    998AAAA31452AAAA3
    999AAAAJ5941AAAAJ
    1000JJJJJ2191JJJJJ

    1000 rows × 8 columns

    Now, we just move J in our new card_dict_joker to where it belongs, the rest -of the solution is the same.

    card_dict_joker = {c: i for i, c in enumerate(
    -    [7, 6, 5, 4, 3, 2, 1, 'J', '2', '3', '4', '5', '6', '7', '8', '9',
    -     'T', 'Q', 'K', 'A'])}
    -
    -camel = camel.sort_values(by=['type', 'c0', 'c1', 'c2', 'c3', 'c4'],
    -                          ignore_index=True,
    -                          key=lambda x: x.map(card_dict_joker))
    -camel.index += 1
    -display(camel)
    -print('Our new solution: ', camel.loc[:, 'bid'].mul(camel.index).sum())
    -
    cardsbidtypec0c1c2c3c4
    1237AQ1577237AQ
    223K47759723K47
    3249K86127249K8
    4264A73417264A7
    52657810726578
    996TJTTJ6091TJTTJ
    997QQJQJ1311QQJQJ
    998QQQJQ8311QQQJQ
    999KKKJK1831KKKJK
    1000AAAAJ5941AAAAJ

    1000 rows × 8 columns

    Our new solution:  246894760
    -
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    Advent of code 2024 - Day 1: Historian Hysteria

    Advent of code 2024 - Day 1: Historian Hysteria

    I try to solve this year’s Advent of Code 2024 riddles in R. This is Day 1 - see -https:adventofcode.com/2024/day/1

    Part 1: Locations -#

    Lets first read the task:

    Throughout the Chief’s office, the historically significant locations are listed -not by name but by a unique number called the location ID. To make sure they -don’t miss anything, The Historians split into two groups, each searching the -office and trying to create their own complete list of location IDs.

    There’s just one problem: by holding the two lists up side by side (your -puzzle input), it quickly becomes clear that the lists aren’t very similar. -Maybe you can help The Historians reconcile their lists?

    For example:

    3   4
    -4   3
    -2   5
    -1   3
    -3   9
    -3   3
    -

    Maybe the lists are only off by a small amount! To find out, pair up the numbers -and measure how far apart they are. Pair up the smallest number in the left -list with the smallest number in the right list, then the second-smallest -left number with the second-smallest right number, and so on.

    Within each pair, figure out how far apart the two numbers are; you’ll need to -add up all of those distances. For example, if you pair up a 3 from the left -list with a 7 from the right list, the distance apart is 4; if you pair up a -9 with a 3, the distance apart is 6.

    In the example list above, the pairs and distances would be as follows:

    • The smallest number in the left list is 1, and the smallest number in the -right list is 3. The distance between them is 2.
    • The second-smallest number in the left list is 2, and the second-smallest -number in the right list is another 3. The distance between them is 1.
    • The third-smallest number in both lists is 3, so the distance between them -is 0.
    • The next numbers to pair up are 3 and 4, a distance of 1.
    • The fifth-smallest numbers in each list are 3 and 5, a distance of 2.
    • Finally, the largest number in the left list is 4, while the largest number -in the right list is 9; these are a distance 5 apart.

    To find the total distance between the left list and the right list, add up -the distances between all of the pairs you found. In the example above, this is -2 + 1 + 0 + 1 + 2 + 5, a total distance of 11!

    Your actual left and right lists contain many location IDs. What is the total distance between your lists?

    My input file: -2024-12-01-1-aoc.txt

    First, we read the data - the fread command from the data.table package is -more versatile than read.delim from base and directly reads the data as a -data.table, which has -some benefits.

    data <- data.table::fread("2024-12-01-1-aoc.txt", header = FALSE)
    -head(data)
    -
    V1V2
    <int><int>
    4122669190
    8931810100
    5941923880
    6315720193
    8151022869
    8394263304

    Now, the idea is to get the ordered versions of V1 and V2.

    v1 <- data[order(V1)]$V1
    -v2 <- data[order(V2)]$V2
    -head(v1)
    -
    1. 10188
    2. 10314
    3. 10319
    4. 10348
    5. 10408
    6. 10668

    Coming from Python, this syntax looks a bit odd. The documentation of -data.table::setorder helps:

    setorder (and setorderv) reorders the rows of a data.table based on the columns (and column
    -order) provided. It reorders the table by reference and is therefore very memory efficient.
    -
    -Note that queries like x[order(.)] are optimised internally to use data.table's fast order.
    -

    So, data[order(V1)] is actually short for data.table::setorder(data, V1). -Then, we extract the vector by name using the $ operator, which allows to -extract elements by name.

    head(data.table::setorder(data, V1)$V1)
    -
    1. 10188
    2. 10314
    3. 10319
    4. 10348
    5. 10408
    6. 10668

    The actual computation is just the sum of the absolute difference:

    sum(abs(v1 - v2))
    -

    3574690

    Part 2: Similarities -#

    Lets first read the task:

    This time, you’ll need to figure out exactly how often each number from the left -list appears in the right list. Calculate a total similarity score by adding -up each number in the left list after multiplying it by the number of times that -number appears in the right list.

    Here are the same example lists again:

    3   4
    -4   3
    -2   5
    -1   3
    -3   9
    -3   3
    -

    For these example lists, here is the process of finding the similarity score:

    • The first number in the left list is 3. It appears in the right list three -times, so the similarity score increases by 3 * 3 = 9.
    • The second number in the left list is 4. It appears in the right list once, -so the similarity score increases by 4 * 1 = 4.
    • The third number in the left list is 2. It does not appear in the right -list, so the similarity score does not increase (2 * 0 = 0).
    • The fourth number, 1, also does not appear in the right list.
    • The fifth number, 3, appears in the right list three times; the similarity -score increases by 9.
    • The last number, 3, appears in the right list three times; the similarity -score again increases by 9.

    So, for these example lists, the similarity score at the end of this process is -31 (9 + 4 + 0 + 0 + 9 + 9).

    Once again consider your left and right lists. What is their similarity score?

    My idea here is simple - we first count the occurrences of V2 to be able to -check V1 against them. Here, applying the table command works like -pandas.value_counts() and achieves this. We convert the output to a -data.frame and assign the V1 values as row names. Note: table converts the -values which are counted to string, e.g. "10019" instead of 10019.

    tab <- table(v2)
    -v2series <- data.frame(c(tab), row.names = names(tab))
    -head(v2series)
    -v2series["10019", ]
    -
    c.tab.
    <int>
    100191
    101001
    102061
    104281
    106451
    109721

    1

    Now, we loop through V1 and try to access the frequency in V2 (by converting -to string first, via as.character). If nothing is found in V2, an NA value -is returned. For everything that is not NA, we compute the similarity score, -append it to a list, and sum in the end. Note: a list in R can not directly be -summed (I don’t know why that is) - so we have to unlist first.

    sim <- list()
    -for (num in v1) {
    -    myfreq <- v2series[as.character(num), ]
    -    if (!is.na(myfreq)) {
    -        score <- myfreq * num
    -        sim <- append(sim, score)
    -    }
    -
    -}
    -sum(unlist(sim))
    -

    22565391

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    Advent of code 2024 - Day 2: Red-Nosed Reports

    Advent of code 2024 - Day 2: Red-Nosed Reports

    I try to solve this year’s Advent of Code 2024 riddles in R. This is Day 2 - see -https:adventofcode.com/2024/day/2

    Part 1: Safe reports -#

    Lets first read the task:

    The unusual data (your puzzle input) consists of many reports, -one report per line. Each report is a list of numbers called levels that are -separated by spaces. For example:

    7 6 4 2 1
    -1 2 7 8 9
    -9 7 6 2 1
    -1 3 2 4 5
    -8 6 4 4 1
    -1 3 6 7 9
    -

    This example data contains six reports each containing five levels.

    The engineers are trying to figure out which reports are safe. The Red-Nosed -reactor safety systems can only tolerate levels that are either gradually -increasing or gradually decreasing. So, a report only counts as safe if both of -the following are true:

    • The levels are either all increasing or all decreasing.
    • Any two adjacent levels differ by at least one and at most three.

    In the example above, the reports can be found safe or unsafe by checking those -rules:

    • 7 6 4 2 1: Safe because the levels are all decreasing by 1 or 2.
    • 1 2 7 8 9: Unsafe because 2 7 is an increase of 5.
    • 9 7 6 2 1: Unsafe because 6 2 is a decrease of 4.
    • 1 3 2 4 5: Unsafe because 1 3 is increasing but 3 2 is decreasing.
    • 8 6 4 4 1: Unsafe because 4 4 is neither an increase or a decrease.
    • 1 3 6 7 9: Safe because the levels are all increasing by 1, 2, or 3.

    So, in this example, 2 reports are safe.

    Analyze the unusual data from the engineers. How many reports are safe?

    My input file: -2024-12-02-1-aoc.txt

    First, let’s load the data and take a look:

    data <- data.table::fread("2024-12-02-1-aoc.txt", header = FALSE, fill = TRUE)
    -head(data)
    -
    V1V2V3V4V5V6V7V8
    <int><int><int><int><int><int><int><int>
    7476787976NANANA
    3840434444NANANA
    12468913NA
    65687072757681NA
    899192959394NANA
    151716181917NANA

    My core idea here was to compute the differences between each element in a row - -luckily, R provides an inbuilt diff() function that provides exactly what I -need. Then, we check if all differences are either 1, 2, 3 or -1, -2, -3. Note -that each row has a different number of elements, so we have to omit NA values -via na.omit when applying the check.

    isalltrue <- function(rep) {
    -    return(all(diff(rep) %in% c(1, 2, 3)) |
    -           all(diff(rep) %in% c(-1, -2, -3)))
    -}
    -
    -isreport <- function(row) {
    -    return(isalltrue(na.omit(row)))
    -}
    -valid.reports <- apply(data, 1, isreport)
    -table(valid.reports)
    -
    valid.reports
    -FALSE  TRUE
    -  743   257
    -

    Part 2: Problem Dampener -#

    Let’s read the task for part 2!

    The engineers are surprised by the low number of safe reports until they realize -they forgot to tell you about the Problem Dampener.

    The Problem Dampener is a reactor-mounted module that lets the reactor safety -systems tolerate a single bad level in what would otherwise be a safe report. -It’s like the bad level never happened!

    Now, the same rules apply as before, except if removing a single level from an -unsafe report would make it safe, the report instead counts as safe.

    More of the above example’s reports are now safe:

    • -7 6 4 2 1: Safe without removing any level.
    • -1 2 7 8 9: Unsafe regardless of which level is removed.
    • -9 7 6 2 1: Unsafe regardless of which level is removed.
    • -1 3 2 4 5: Safe by removing the second level, 3.
    • -8 6 4 4 1: Safe by removing the third level, 4.
    • -1 3 6 7 9: Safe without removing any level.

    Thanks to the Problem Dampener, 4 reports are actually safe!

    Update your analysis by handling situations where the Problem Dampener can -remove a single level from unsafe reports. How many reports are now safe?

    So, this task is a good example on how easy it is to get lost in a overly -complicated approach. My first approach was basically to somehow check if the -general trend was descending or ascending and based on that to remove opposite -differences. This was too complicated and in the end I found 4 less safe reports -than were needed… (Also note the beautiful print debuggin…)

    
    -isalltrue <- function(reptrue1, reptrue2) {
    -    return((all(reptrue1) == TRUE) | (all(reptrue2) == TRUE))
    -}
    -
    -iscorrtrue <- function(rep, reptrue) {
    -    rep <- rep[-which.min(reptrue)]
    -    repdiff <- diff(rep)
    -    reptrue1 <- !(repdiff < 1 | repdiff > 3)
    -    reptrue2 <- !(repdiff < -3 | repdiff > -1)
    -    out <- isalltrue(reptrue1, reptrue2)
    -    print(rep)
    -    print(out)
    -    return(out)
    -}
    -
    -isreport2 <- function(row) {
    -    print("")
    -    rep <- na.omit(as.numeric(row))
    -    print(c(rep))
    -    repdiff <- c(1, diff(rep))
    -    reptrue1 <- !(repdiff < 1 | repdiff > 3)
    -    repdiff <- c(-1, diff(rep))
    -    reptrue2 <- !(repdiff < -3 | repdiff > -1)
    -    if (isalltrue(reptrue1, reptrue2) == TRUE) {
    -        out <- isalltrue(reptrue1, reptrue2)
    -        print("----- 1 -----")
    -        print(out)
    -    } else if (sum(diff(rep) > 0) > sum(diff(rep) < 0)) {
    -        print("----- 2 -----")
    -        out <- iscorrtrue(rep, reptrue1)
    -    } else if (sum(diff(rep) < 0) > sum(diff(rep) > 0)) {
    -        print("----- 3 -----")
    -        out <- iscorrtrue(rep, reptrue2)
    -    } else {
    -        out <- FALSE
    -    }
    -    if (out == FALSE) {
    -        print("----- 4 -----")
    -        rep <- na.omit(as.numeric(row))[-1]
    -        repdiff <- diff(rep)
    -        reptrue1 <- !(repdiff < 1 | repdiff > 3)
    -        reptrue2 <- !(repdiff < -3 | repdiff > -1)
    -        out <- isalltrue(reptrue1, reptrue2)
    -        print(c(rep))
    -        print(out)
    -    }
    -    return(out)
    -}
    -

    My actual working approach was in the end brute force. I just threw out each -element in a row and checked if the report became positive - much simpler and effective!

    isreport2 <- function(row) {
    -    rep <- na.omit(row)
    -    out <- isalltrue(rep)
    -    if (out == FALSE) {
    -        for (i in seq_along(rep)) {
    -            # rep <- rep[-i] does not update the variable
    -            # with rep.i, a new variable is assigned (rep.1, rep.2, ...)
    -            rep.i <- rep[-i]
    -            out <- isalltrue(rep.i)
    -            if (out == TRUE) {
    -                break
    -            }
    -        }
    -    }
    -    return(out)
    -}
    -sum(apply(data, 1, isreport2))
    -

    328

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    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/posts/index.html b/aseltmann.github.io-hugo/public/posts/index.html deleted file mode 100644 index ea70d76..0000000 --- a/aseltmann.github.io-hugo/public/posts/index.html +++ /dev/null @@ -1,81 +0,0 @@ -Posts | Blog & Notes - -

    Posts

    Advent of code 2024 - Day 1: Historian Hysteria

    I try to solve this year’s Advent of Code 2024 riddles in R. This is Day 1 - see -https:adventofcode.com/2024/day/1

    Part 1: Locations -#

    Lets first read the task:

    Throughout the Chief’s office, the historically significant locations are listed -not by name but by a unique number called the location ID. To make sure they -don’t miss anything, The Historians split into two groups, each searching the -office and trying to create their own complete list of location IDs.

    ...

    Advent of code 2023 - Day 7: Camel Cards

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 7 - see -https:adventofcode.com/2023/day/7

    Part 1 -#

    Lets first read the task:

    In Camel Cards, you get a list of hands, and your goal is to order them based -on the strength of each hand. A hand consists of five cards labeled one of -A, K, Q, J, T, 9, 8, 7, 6, 5, 4, 3, or 2. The relative -strength of each card follows this order, where A is the highest and 2 is -the lowest.

    ...

    Advent of code 2023 - Day 6: Wait For It

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 6 - see -https:adventofcode.com/2023/day/6

    Part 1 -#

    Lets first read the task:

    You manage to sign up as a competitor in the boat races just in time. The -organizer explains that it’s not really a traditional race - instead, you will -get a fixed amount of time during which your boat has to travel as far as it -can, and you win if your boat goes the farthest.

    ...

    Advent of code 2023 - Day 5: If You Give A Seed A Fertilizer

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 5 - see -https:adventofcode.com/2023/day/5

    Update [2023-12-31 So]:

    • substitute get_generator() (own implementation) with range() (Python inbuilt)
    • improve grid search so that it goes through all location ranges, still -starting with the lowest range

    Part 1 -#

    Lets first read the task:

    The almanac (your puzzle input) lists all of the seeds that need to be planted. -It also lists what type of soil to use with each kind of seed, what type of -fertilizer to use with each kind of soil, what type of water to use with each -kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is -identified with a number, but numbers are reused by each category - that is, -soil 123 and fertilizer 123 aren’t necessarily related to each other.

    ...

    Advent of code 2023 - Day 4: Scratchcards

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 4 - see -https:adventofcode.com/2023/day/4

    Part 1 -#

    Lets first read the task:

    The Elf leads you over to the pile of colorful cards. There, you discover dozens -of scratchcards, all with their opaque covering already scratched off. Picking -one up, it looks like each card has two lists of numbers separated by a vertical -bar (|): a list of winning numbers and then a list of numbers you have. -You organize the information into a table (your puzzle input).

    ...

    Advent of code 2023 - Day 3: Gear Ratios

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 3 - see -https:adventofcode.com/2023/day/3

    Part 1 -#

    Lets first read the task:

    The engine schematic (your puzzle input) consists of a visual representation of -the engine. There are lots of numbers and symbols you don’t really understand, -but apparently any number adjacent to a symbol, even diagonally, is a “part -number” and should be included in your sum. (Periods (.) do not count as a -symbol.)

    ...

    Advent of code 2023 - Day 2: Cube Conundrum

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 2 - see -https:adventofcode.com/2023/day/2

    Part 1 -#

    Lets first read the task:

    As you walk, the Elf shows you a small bag and some cubes which are either red, -green, or blue. Each time you play this game, he will hide a secret number of -cubes of each color in the bag, and your goal is to figure out information about -the number of cubes.

    ...

    Advent of code 2023 - Day 1: Trebuchet?!

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 1 - see -https:adventofcode.com/2023/day/1

    Part 1 -#

    Our first task is the following:

    The newly-improved calibration document consists of lines of text; each line -originally contained a specific calibration value that the Elves now need to -recover. On each line, the calibration value can be found by combining the first -digit and the last digit (in that order) to form a single two-digit number.

    ...
    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/posts/index.xml b/aseltmann.github.io-hugo/public/posts/index.xml deleted file mode 100644 index cb446fc..0000000 --- a/aseltmann.github.io-hugo/public/posts/index.xml +++ /dev/null @@ -1,225 +0,0 @@ -Posts on Blog & Noteshttp://localhost:1313/posts/Recent content in Posts on Blog & NotesHugoenSun, 22 Dec 2024 01:46:31 +0100Advent of code 2024 - Day 2: Red-Nosed Reportshttp://localhost:1313/posts/aoc2024-day02/Tue, 10 Dec 2024 01:43:00 +0100http://localhost:1313/posts/aoc2024-day02/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1-safe-reports">Part 1: Safe reports</a></li> -<li> - <a href="http://localhost:1313/#part-2-problem-dampener">Part 2: Problem Dampener</a></li> -</ul> -</div> -<!--endtoc--> -<p>I try to solve this year&rsquo;s <strong>Advent of Code 2024</strong> riddles in R. This is Day 2 - see - <a href="https:adventofcode.com/2024/day/2">https:adventofcode.com/2024/day/2</a></p> -<h2 id="part-1-safe-reports"> - Part 1: Safe reports - <a class="anchor" href="#part-1-safe-reports">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The unusual data (your puzzle input) consists of many <strong>reports</strong>, -one report per line. Each report is a list of numbers called <strong>levels</strong> that are -separated by spaces. For example:</p>Advent of code 2024 - Day 1: Historian Hysteriahttp://localhost:1313/posts/aoc2024-day01/Sat, 07 Dec 2024 00:22:00 +0100http://localhost:1313/posts/aoc2024-day01/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1-locations">Part 1: Locations</a></li> -<li> - <a href="http://localhost:1313/#part-2-similarities">Part 2: Similarities</a></li> -</ul> -</div> -<!--endtoc--> -<p>I try to solve this year&rsquo;s <strong>Advent of Code 2024</strong> riddles in R. This is Day 1 - see - <a href="https:adventofcode.com/2024/day/1">https:adventofcode.com/2024/day/1</a></p> -<h2 id="part-1-locations"> - Part 1: Locations - <a class="anchor" href="#part-1-locations">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>Throughout the Chief&rsquo;s office, the historically significant locations are listed -not by name but by a unique number called the <strong>location ID</strong>. To make sure they -don&rsquo;t miss anything, The Historians split into two groups, each searching the -office and trying to create their own complete list of location IDs.</p>Advent of code 2023 - Day 7: Camel Cardshttp://localhost:1313/posts/aoc2023-day07/Thu, 04 Jan 2024 16:32:00 +0100http://localhost:1313/posts/aoc2023-day07/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 7 - see - <a href="https:adventofcode.com/2023/day/7">https:adventofcode.com/2023/day/7</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>In Camel Cards, you get a list of <strong>hands</strong>, and your goal is to order them based -on the <strong>strength</strong> of each hand. A hand consists of <strong>five cards</strong> labeled one of -<code>A</code>, <code>K</code>, <code>Q</code>, <code>J</code>, <code>T</code>, <code>9</code>, <code>8</code>, <code>7</code>, <code>6</code>, <code>5</code>, <code>4</code>, <code>3</code>, or <code>2</code>. The relative -strength of each card follows this order, where <code>A</code> is the highest and <code>2</code> is -the lowest.</p>Advent of code 2023 - Day 6: Wait For Ithttp://localhost:1313/posts/aoc2023-day06/Thu, 04 Jan 2024 00:44:00 +0100http://localhost:1313/posts/aoc2023-day06/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 6 - see - <a href="https:adventofcode.com/2023/day/6">https:adventofcode.com/2023/day/6</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>You manage to sign up as a competitor in the boat races just in time. The -organizer explains that it&rsquo;s not really a traditional race - instead, you will -get a fixed amount of time during which your boat has to travel as far as it -can, and you win if your boat goes the farthest.</p>Advent of code 2023 - Day 5: If You Give A Seed A Fertilizerhttp://localhost:1313/posts/aoc2023-day05/Sat, 30 Dec 2023 01:10:00 +0100http://localhost:1313/posts/aoc2023-day05/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 5 - see - <a href="https:adventofcode.com/2023/day/5">https:adventofcode.com/2023/day/5</a></p> -<p>Update <span class="timestamp-wrapper"><span class="timestamp">[2023-12-31 So]</span></span>:</p> -<ul> -<li>substitute <code>get_generator()</code> (own implementation) with <code>range()</code> (Python inbuilt)</li> -<li>improve grid search so that it goes through all location ranges, still -starting with the lowest range</li> -</ul> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The almanac (your puzzle input) lists all of the seeds that need to be planted. -It also lists what type of soil to use with each kind of seed, what type of -fertilizer to use with each kind of soil, what type of water to use with each -kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is -identified with a number, but numbers are reused by each category - that is, -soil <code>123</code> and fertilizer <code>123</code> aren&rsquo;t necessarily related to each other.</p>Advent of code 2023 - Day 4: Scratchcardshttp://localhost:1313/posts/aoc2023-day04/Tue, 19 Dec 2023 00:25:00 +0100http://localhost:1313/posts/aoc2023-day04/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 4 - see - <a href="https:adventofcode.com/2023/day/4">https:adventofcode.com/2023/day/4</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The Elf leads you over to the pile of colorful cards. There, you discover dozens -of scratchcards, all with their opaque covering already scratched off. Picking -one up, it looks like each card has two lists of numbers separated by a vertical -bar (<code>|</code>): a list of <strong>winning numbers</strong> and then a list of <strong>numbers you have</strong>. -You organize the information into a table (your puzzle input).</p>Advent of code 2023 - Day 3: Gear Ratioshttp://localhost:1313/posts/aoc2023-day03/Mon, 18 Dec 2023 01:01:00 +0100http://localhost:1313/posts/aoc2023-day03/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 3 - see - <a href="https:adventofcode.com/2023/day/3">https:adventofcode.com/2023/day/3</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The engine schematic (your puzzle input) consists of a visual representation of -the engine. There are lots of numbers and symbols you don&rsquo;t really understand, -but apparently <strong>any number adjacent to a symbol</strong>, even diagonally, is a &ldquo;part -number&rdquo; and should be included in your sum. (Periods (<code>.</code>) do not count as a -symbol.)</p>Advent of code 2023 - Day 2: Cube Conundrumhttp://localhost:1313/posts/aoc2023-day02/Mon, 11 Dec 2023 23:38:00 +0100http://localhost:1313/posts/aoc2023-day02/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 2 - see - <a href="https:adventofcode.com/2023/day/2">https:adventofcode.com/2023/day/2</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>As you walk, the Elf shows you a small bag and some cubes which are either red, -green, or blue. Each time you play this game, he will hide a secret number of -cubes of each color in the bag, and your goal is to figure out information about -the number of cubes.</p>Advent of code 2023 - Day 1: Trebuchet?!http://localhost:1313/posts/aoc2023-day01/Sun, 10 Dec 2023 03:05:00 +0100http://localhost:1313/posts/aoc2023-day01/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 1 - see - <a href="https:adventofcode.com/2023/day/1">https:adventofcode.com/2023/day/1</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Our first task is the following:</summary> -<div class="details"> -<blockquote> -<p>The newly-improved calibration document consists of lines of text; each line -originally contained a specific calibration value that the Elves now need to -recover. On each line, the calibration value can be found by combining the first -digit and the last digit (in that order) to form a single two-digit number.</p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/posts/page/1/index.html b/aseltmann.github.io-hugo/public/posts/page/1/index.html deleted file mode 100644 index 282c94f..0000000 --- a/aseltmann.github.io-hugo/public/posts/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/posts/ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/sitemap.xml b/aseltmann.github.io-hugo/public/sitemap.xml deleted file mode 100644 index 7084044..0000000 --- a/aseltmann.github.io-hugo/public/sitemap.xml +++ /dev/null @@ -1 +0,0 @@ -http://localhost:1313/en/sitemap.xml2024-12-22T01:46:31+01:00http://localhost:1313/de/sitemap.xml2024-12-22T01:46:51+01:00 \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/svg/calendar.svg b/aseltmann.github.io-hugo/public/svg/calendar.svg deleted file mode 100644 index f848112..0000000 --- a/aseltmann.github.io-hugo/public/svg/calendar.svg +++ /dev/null @@ -1 +0,0 @@ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/svg/edit.svg b/aseltmann.github.io-hugo/public/svg/edit.svg deleted file mode 100644 index 5b54e69..0000000 --- a/aseltmann.github.io-hugo/public/svg/edit.svg +++ /dev/null @@ -1 +0,0 @@ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/svg/menu.svg b/aseltmann.github.io-hugo/public/svg/menu.svg deleted file mode 100644 index 770b192..0000000 --- a/aseltmann.github.io-hugo/public/svg/menu.svg +++ /dev/null @@ -1 +0,0 @@ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/svg/toc.svg b/aseltmann.github.io-hugo/public/svg/toc.svg deleted file mode 100644 index 1889904..0000000 --- a/aseltmann.github.io-hugo/public/svg/toc.svg +++ /dev/null @@ -1 +0,0 @@ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/svg/translate.svg b/aseltmann.github.io-hugo/public/svg/translate.svg deleted file mode 100644 index a1bbe16..0000000 --- a/aseltmann.github.io-hugo/public/svg/translate.svg +++ /dev/null @@ -1 +0,0 @@ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/sw.js b/aseltmann.github.io-hugo/public/sw.js deleted file mode 100644 index ff3a148..0000000 --- a/aseltmann.github.io-hugo/public/sw.js +++ /dev/null @@ -1,51 +0,0 @@ -const cacheName = self.location.pathname -const pages = [ - -]; - -self.addEventListener("install", function (event) { - self.skipWaiting(); - - caches.open(cacheName).then((cache) => { - return cache.addAll(pages); - }); -}); - -self.addEventListener("fetch", (event) => { - const request = event.request; - if (request.method !== "GET") { - return; - } - - /** - * @param {Response} response - * @returns {Promise} - */ - function saveToCache(response) { - if (cacheable(response)) { - return caches - .open(cacheName) - .then((cache) => cache.put(request, response.clone())) - .then(() => response); - } else { - return response; - } - } - - /** - * @param {Error} error - */ - function serveFromCache(error) { - return caches.open(cacheName).then((cache) => cache.match(request.url)); - } - - /** - * @param {Response} response - * @returns {Boolean} - */ - function cacheable(response) { - return response.type === "basic" && response.ok && !response.headers.has("Content-Disposition") - } - - event.respondWith(fetch(request).then(saveToCache).catch(serveFromCache)); -}); diff --git a/aseltmann.github.io-hugo/public/sw.min.6f6f90fcb8eb1c49ec389838e6b801d0de19430b8e516902f8d75c3c8bd98739.js b/aseltmann.github.io-hugo/public/sw.min.6f6f90fcb8eb1c49ec389838e6b801d0de19430b8e516902f8d75c3c8bd98739.js deleted file mode 100644 index 20eebad..0000000 --- a/aseltmann.github.io-hugo/public/sw.min.6f6f90fcb8eb1c49ec389838e6b801d0de19430b8e516902f8d75c3c8bd98739.js +++ /dev/null @@ -1 +0,0 @@ -navigator.serviceWorker&&navigator.serviceWorker.register("/sw.js",{scope:"/"}) \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/R/index.html b/aseltmann.github.io-hugo/public/tags/R/index.html deleted file mode 100644 index d07aff4..0000000 --- a/aseltmann.github.io-hugo/public/tags/R/index.html +++ /dev/null @@ -1,26 +0,0 @@ -R | Blog & Notes - -

    R

    Advent of code 2024 - Day 1: Historian Hysteria

    I try to solve this year’s Advent of Code 2024 riddles in R. This is Day 1 - see -https:adventofcode.com/2024/day/1

    Part 1: Locations -#

    Lets first read the task:

    Throughout the Chief’s office, the historically significant locations are listed -not by name but by a unique number called the location ID. To make sure they -don’t miss anything, The Historians split into two groups, each searching the -office and trying to create their own complete list of location IDs.

    ...

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/R/index.xml b/aseltmann.github.io-hugo/public/tags/R/index.xml deleted file mode 100644 index 584ec41..0000000 --- a/aseltmann.github.io-hugo/public/tags/R/index.xml +++ /dev/null @@ -1,46 +0,0 @@ -R on Blog & Noteshttp://localhost:1313/tags/R/Recent content in R on Blog & NotesHugoenSun, 22 Dec 2024 01:46:31 +0100Advent of code 2024 - Day 2: Red-Nosed Reportshttp://localhost:1313/posts/aoc2024-day02/Tue, 10 Dec 2024 01:43:00 +0100http://localhost:1313/posts/aoc2024-day02/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1-safe-reports">Part 1: Safe reports</a></li> -<li> - <a href="http://localhost:1313/#part-2-problem-dampener">Part 2: Problem Dampener</a></li> -</ul> -</div> -<!--endtoc--> -<p>I try to solve this year&rsquo;s <strong>Advent of Code 2024</strong> riddles in R. This is Day 2 - see - <a href="https:adventofcode.com/2024/day/2">https:adventofcode.com/2024/day/2</a></p> -<h2 id="part-1-safe-reports"> - Part 1: Safe reports - <a class="anchor" href="#part-1-safe-reports">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The unusual data (your puzzle input) consists of many <strong>reports</strong>, -one report per line. Each report is a list of numbers called <strong>levels</strong> that are -separated by spaces. For example:</p>Advent of code 2024 - Day 1: Historian Hysteriahttp://localhost:1313/posts/aoc2024-day01/Sat, 07 Dec 2024 00:22:00 +0100http://localhost:1313/posts/aoc2024-day01/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1-locations">Part 1: Locations</a></li> -<li> - <a href="http://localhost:1313/#part-2-similarities">Part 2: Similarities</a></li> -</ul> -</div> -<!--endtoc--> -<p>I try to solve this year&rsquo;s <strong>Advent of Code 2024</strong> riddles in R. This is Day 1 - see - <a href="https:adventofcode.com/2024/day/1">https:adventofcode.com/2024/day/1</a></p> -<h2 id="part-1-locations"> - Part 1: Locations - <a class="anchor" href="#part-1-locations">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>Throughout the Chief&rsquo;s office, the historically significant locations are listed -not by name but by a unique number called the <strong>location ID</strong>. To make sure they -don&rsquo;t miss anything, The Historians split into two groups, each searching the -office and trying to create their own complete list of location IDs.</p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/R/page/1/index.html b/aseltmann.github.io-hugo/public/tags/R/page/1/index.html deleted file mode 100644 index a67d072..0000000 --- a/aseltmann.github.io-hugo/public/tags/R/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/tags/R/ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/index.html b/aseltmann.github.io-hugo/public/tags/index.html deleted file mode 100644 index 5a18aba..0000000 --- a/aseltmann.github.io-hugo/public/tags/index.html +++ /dev/null @@ -1,17 +0,0 @@ -Tags | Blog & Notes - -

    Tags

    R

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/index.xml b/aseltmann.github.io-hugo/public/tags/index.xml deleted file mode 100644 index fdf560d..0000000 --- a/aseltmann.github.io-hugo/public/tags/index.xml +++ /dev/null @@ -1 +0,0 @@ -Tags on Blog & Noteshttp://localhost:1313/tags/Recent content in Tags on Blog & NotesHugoenSun, 22 Dec 2024 01:46:31 +0100Rhttp://localhost:1313/tags/R/Tue, 10 Dec 2024 01:43:00 +0100http://localhost:1313/tags/R/Pythonhttp://localhost:1313/tags/python/Thu, 04 Jan 2024 16:32:00 +0100http://localhost:1313/tags/python/ \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/page/1/index.html b/aseltmann.github.io-hugo/public/tags/page/1/index.html deleted file mode 100644 index 3132ec0..0000000 --- a/aseltmann.github.io-hugo/public/tags/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/tags/ - \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/python/index.html b/aseltmann.github.io-hugo/public/tags/python/index.html deleted file mode 100644 index 2b73347..0000000 --- a/aseltmann.github.io-hugo/public/tags/python/index.html +++ /dev/null @@ -1,70 +0,0 @@ -Python | Blog & Notes - -

    Python

    Advent of code 2023 - Day 7: Camel Cards

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 7 - see -https:adventofcode.com/2023/day/7

    Part 1 -#

    Lets first read the task:

    In Camel Cards, you get a list of hands, and your goal is to order them based -on the strength of each hand. A hand consists of five cards labeled one of -A, K, Q, J, T, 9, 8, 7, 6, 5, 4, 3, or 2. The relative -strength of each card follows this order, where A is the highest and 2 is -the lowest.

    ...

    Advent of code 2023 - Day 6: Wait For It

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 6 - see -https:adventofcode.com/2023/day/6

    Part 1 -#

    Lets first read the task:

    You manage to sign up as a competitor in the boat races just in time. The -organizer explains that it’s not really a traditional race - instead, you will -get a fixed amount of time during which your boat has to travel as far as it -can, and you win if your boat goes the farthest.

    ...

    Advent of code 2023 - Day 5: If You Give A Seed A Fertilizer

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 5 - see -https:adventofcode.com/2023/day/5

    Update [2023-12-31 So]:

    • substitute get_generator() (own implementation) with range() (Python inbuilt)
    • improve grid search so that it goes through all location ranges, still -starting with the lowest range

    Part 1 -#

    Lets first read the task:

    The almanac (your puzzle input) lists all of the seeds that need to be planted. -It also lists what type of soil to use with each kind of seed, what type of -fertilizer to use with each kind of soil, what type of water to use with each -kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is -identified with a number, but numbers are reused by each category - that is, -soil 123 and fertilizer 123 aren’t necessarily related to each other.

    ...

    Advent of code 2023 - Day 4: Scratchcards

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 4 - see -https:adventofcode.com/2023/day/4

    Part 1 -#

    Lets first read the task:

    The Elf leads you over to the pile of colorful cards. There, you discover dozens -of scratchcards, all with their opaque covering already scratched off. Picking -one up, it looks like each card has two lists of numbers separated by a vertical -bar (|): a list of winning numbers and then a list of numbers you have. -You organize the information into a table (your puzzle input).

    ...

    Advent of code 2023 - Day 3: Gear Ratios

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 3 - see -https:adventofcode.com/2023/day/3

    Part 1 -#

    Lets first read the task:

    The engine schematic (your puzzle input) consists of a visual representation of -the engine. There are lots of numbers and symbols you don’t really understand, -but apparently any number adjacent to a symbol, even diagonally, is a “part -number” and should be included in your sum. (Periods (.) do not count as a -symbol.)

    ...

    Advent of code 2023 - Day 2: Cube Conundrum

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 2 - see -https:adventofcode.com/2023/day/2

    Part 1 -#

    Lets first read the task:

    As you walk, the Elf shows you a small bag and some cubes which are either red, -green, or blue. Each time you play this game, he will hide a secret number of -cubes of each color in the bag, and your goal is to figure out information about -the number of cubes.

    ...

    Advent of code 2023 - Day 1: Trebuchet?!

    Table of Contents

    This year I try to record my attempt at solving the Advent of Code 2023 -riddles. This is Day 1 - see -https:adventofcode.com/2023/day/1

    Part 1 -#

    Our first task is the following:

    The newly-improved calibration document consists of lines of text; each line -originally contained a specific calibration value that the Elves now need to -recover. On each line, the calibration value can be found by combining the first -digit and the last digit (in that order) to form a single two-digit number.

    ...

    \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/python/index.xml b/aseltmann.github.io-hugo/public/tags/python/index.xml deleted file mode 100644 index 16f9320..0000000 --- a/aseltmann.github.io-hugo/public/tags/python/index.xml +++ /dev/null @@ -1,180 +0,0 @@ -Python on Blog & Noteshttp://localhost:1313/tags/python/Recent content in Python on Blog & NotesHugoenSun, 22 Dec 2024 01:44:40 +0100Advent of code 2023 - Day 7: Camel Cardshttp://localhost:1313/posts/aoc2023-day07/Thu, 04 Jan 2024 16:32:00 +0100http://localhost:1313/posts/aoc2023-day07/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 7 - see - <a href="https:adventofcode.com/2023/day/7">https:adventofcode.com/2023/day/7</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>In Camel Cards, you get a list of <strong>hands</strong>, and your goal is to order them based -on the <strong>strength</strong> of each hand. A hand consists of <strong>five cards</strong> labeled one of -<code>A</code>, <code>K</code>, <code>Q</code>, <code>J</code>, <code>T</code>, <code>9</code>, <code>8</code>, <code>7</code>, <code>6</code>, <code>5</code>, <code>4</code>, <code>3</code>, or <code>2</code>. The relative -strength of each card follows this order, where <code>A</code> is the highest and <code>2</code> is -the lowest.</p>Advent of code 2023 - Day 6: Wait For Ithttp://localhost:1313/posts/aoc2023-day06/Thu, 04 Jan 2024 00:44:00 +0100http://localhost:1313/posts/aoc2023-day06/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 6 - see - <a href="https:adventofcode.com/2023/day/6">https:adventofcode.com/2023/day/6</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>You manage to sign up as a competitor in the boat races just in time. The -organizer explains that it&rsquo;s not really a traditional race - instead, you will -get a fixed amount of time during which your boat has to travel as far as it -can, and you win if your boat goes the farthest.</p>Advent of code 2023 - Day 5: If You Give A Seed A Fertilizerhttp://localhost:1313/posts/aoc2023-day05/Sat, 30 Dec 2023 01:10:00 +0100http://localhost:1313/posts/aoc2023-day05/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 5 - see - <a href="https:adventofcode.com/2023/day/5">https:adventofcode.com/2023/day/5</a></p> -<p>Update <span class="timestamp-wrapper"><span class="timestamp">[2023-12-31 So]</span></span>:</p> -<ul> -<li>substitute <code>get_generator()</code> (own implementation) with <code>range()</code> (Python inbuilt)</li> -<li>improve grid search so that it goes through all location ranges, still -starting with the lowest range</li> -</ul> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The almanac (your puzzle input) lists all of the seeds that need to be planted. -It also lists what type of soil to use with each kind of seed, what type of -fertilizer to use with each kind of soil, what type of water to use with each -kind of fertilizer, and so on. Every type of seed, soil, fertilizer and so on is -identified with a number, but numbers are reused by each category - that is, -soil <code>123</code> and fertilizer <code>123</code> aren&rsquo;t necessarily related to each other.</p>Advent of code 2023 - Day 4: Scratchcardshttp://localhost:1313/posts/aoc2023-day04/Tue, 19 Dec 2023 00:25:00 +0100http://localhost:1313/posts/aoc2023-day04/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 4 - see - <a href="https:adventofcode.com/2023/day/4">https:adventofcode.com/2023/day/4</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The Elf leads you over to the pile of colorful cards. There, you discover dozens -of scratchcards, all with their opaque covering already scratched off. Picking -one up, it looks like each card has two lists of numbers separated by a vertical -bar (<code>|</code>): a list of <strong>winning numbers</strong> and then a list of <strong>numbers you have</strong>. -You organize the information into a table (your puzzle input).</p>Advent of code 2023 - Day 3: Gear Ratioshttp://localhost:1313/posts/aoc2023-day03/Mon, 18 Dec 2023 01:01:00 +0100http://localhost:1313/posts/aoc2023-day03/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 3 - see - <a href="https:adventofcode.com/2023/day/3">https:adventofcode.com/2023/day/3</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>The engine schematic (your puzzle input) consists of a visual representation of -the engine. There are lots of numbers and symbols you don&rsquo;t really understand, -but apparently <strong>any number adjacent to a symbol</strong>, even diagonally, is a &ldquo;part -number&rdquo; and should be included in your sum. (Periods (<code>.</code>) do not count as a -symbol.)</p>Advent of code 2023 - Day 2: Cube Conundrumhttp://localhost:1313/posts/aoc2023-day02/Mon, 11 Dec 2023 23:38:00 +0100http://localhost:1313/posts/aoc2023-day02/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 2 - see - <a href="https:adventofcode.com/2023/day/2">https:adventofcode.com/2023/day/2</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Lets first read the task:</summary> -<div class="details"> -<blockquote> -<p>As you walk, the Elf shows you a small bag and some cubes which are either red, -green, or blue. Each time you play this game, he will hide a secret number of -cubes of each color in the bag, and your goal is to figure out information about -the number of cubes.</p>Advent of code 2023 - Day 1: Trebuchet?!http://localhost:1313/posts/aoc2023-day01/Sun, 10 Dec 2023 03:05:00 +0100http://localhost:1313/posts/aoc2023-day01/<div class="ox-hugo-toc toc"> -<div class="heading">Table of Contents</div> -<ul> -<li> - <a href="http://localhost:1313/#part-1">Part 1</a></li> -<li> - <a href="http://localhost:1313/#part-2">Part 2</a></li> -</ul> -</div> -<!--endtoc--> -<p>This year I try to record my attempt at solving the <strong>Advent of Code 2023</strong> -riddles. This is Day 1 - see - <a href="https:adventofcode.com/2023/day/1">https:adventofcode.com/2023/day/1</a></p> -<h2 id="part-1"> - Part 1 - <a class="anchor" href="#part-1">#</a> -</h2> -<details> -<summary>Our first task is the following:</summary> -<div class="details"> -<blockquote> -<p>The newly-improved calibration document consists of lines of text; each line -originally contained a specific calibration value that the Elves now need to -recover. On each line, the calibration value can be found by combining the first -digit and the last digit (in that order) to form a single two-digit number.</p> \ No newline at end of file diff --git a/aseltmann.github.io-hugo/public/tags/python/page/1/index.html b/aseltmann.github.io-hugo/public/tags/python/page/1/index.html deleted file mode 100644 index 8324ab1..0000000 --- a/aseltmann.github.io-hugo/public/tags/python/page/1/index.html +++ /dev/null @@ -1,2 +0,0 @@ -http://localhost:1313/tags/python/ - \ No newline at end of file