Merge branch 'master' of https://github.com/dmsm/DibTeX

Assigment 08.tex → Assigment08.tex

@@ -9,94 +9,75 @@
 \usepackage{enumerate}
 \usepackage{tikz}
 \usepackage{mathtools}
-\DeclarePairedDelimiter\ceil{\lceil}{\\rceil}
-\DeclarePairedDelimiter\\floor{\\lfloor}{\\rfloor}
+\DeclarePairedDelimiter\ceil{\lceil}{\rceil}
+\DeclarePairedDelimiter\floor{\lfloor}{\rfloor}
 % include this if you want to import graphics files with /includegraphics
-\\name{Assigment 08}
+\name{Assigment 08}
 \duedate{2014-11-03}
-\\begin{document}
-\begin{problem}[Shahriari 8.1.1][20]\\
+\begin{document}
+\begin{problem}[Shahriari 8.1.1][20]
 \\
 A coffee company was willing to pay \$1 for each person interviewed about his or her lieks and dislikes on types of cofee. Of the persons interviewed, 270 liked ground coffee, 200 liked instant coffee, 70 liked both, and 50 did not like either choice. What is the total amount of money the company had to pay?
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.1.3][20]\\
+\begin{problem}[Shahriari 8.1.3][20]
 \\
 An advertising agency finds that of its 170 clients, 115 use television, 100 use radio, 130 use magaines, 75 use television and radio, 95 use radio and magazines, 85 use television and magazines and 70 use all three. How many clients use only magazines? How many client use none of these media?
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.1.4][20]\\
+\begin{problem}[Shahriari 8.1.4][20]
 \\
 %powers (9883)
 How many integers between 1 and 10,000 are neither perfect squares nor perfect cubes.
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.1.6][20]\\
+\begin{problem}[Shahriari 8.1.6][20]
 \\
 How many permutations of the letters \textbf{SCRIPPS} have no two consecutive letters the same?
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.1.8][20]\\
+\begin{problem}[Shahriari 8.1.8][20]
 \\
 Lewis Carrol speaks of a battle among 100 combatants in which 80 lost an arm, 85 a leg, 70 an eye and 75 an ear. Some number $p$ of people lost all four. Find the lower and upper bound for $p$.
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.1.10][20]\\
+\begin{problem}[Shahriari 8.1.10][20]
 \\
 How many five-card hands contain a jack, a queen and king?
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.1.14][20]\\
+\begin{problem}[Shahriari 8.1.14][20]
 \\
 How many permutations of the 26 letters are there that contain none of the sequences MATH, RUNS, FROM or JOE?
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.1.15][20]\\
+\begin{problem}[Shahriari 8.1.15][20]
 \\
 Find the number of primes less than 100 without actually finding all the primes
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.2.1][20]\\
+\begin{problem}[Shahriari 8.2.1][20]
 
 Determine the number of 10-combintions of the multiset
 $$S = \{ \infty \cdot a, 3 \cdot b, 5 \cdot c, 7 \cdot d \}$$
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.2.3][20]\\
+\begin{problem}[Shahriari 8.2.3][20]
 \\
 A bakery sells seven kinds of doughnuts. How many ways are there to choose one dozen donuts if no more than three donughts of any kind are used?
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.3.1][20]\\
-
-\item Determine the number of permutations of $\{1,2,\cdots, 8 \}$ in which no even integer is in its natural position.
-\item Determine the number of permutations of $\{1,2,\cdots, 8 \}$ in which exactly 4 integers are in their natural position.
-\end{enumerate}
-
-\end{problem}
-\begin{problem}[Shahriari 8.3.2][20]\\
-
-\item Eight girls are seated around a carousel. In how many ways can they change seats so that each has a different girl in front of her?
-\item Eight boys are seated around a carousel but facing inward, so that each boy faces one another. In how many ways can they change seats so that each faces a different boy?
-\end{enumerate}
-
-\end{problem}
-\begin{problem}[Shahriari 8.3.6][20]\\
-\\
-For each of the $6 \times 6$ boards with forbidden positions in Figure 8.3 in Shahriari's book on page 147, find the number of ways to place six non-attacking rooks in the non-forbidden positions.
-
-\end{problem}
-\begin{problem}[Shahriari 8.3.10][20]\\
+\begin{problem}[Shahriari 8.3.10][20]
 \\
 Fereydoon, Faranak, Roudabeh, Tahminen and Rostem are students in the class. Each is to be assigned a differnt dialogue form the following list: Crito, Euthyphro, Protagoras, Meno, Parmenides. Fereydoon is not interested in the Crito or the Protagoras; Faranak is not interested in the Meno; Roudabeh is not interested in the Euthyphro or the Parmenides; Tahmineh is not interested in the Euthyphro and Rostam is not interested in the Crito or the Protagoras. In how many ways can we assign the five dialogues to the five students given the above constraints?
-
+\
 \end{problem}
-\begin{problem}[Shahriari 8.X.1][20]\\
+\begin{problem}[Shahriari 8.X.1][20]
 \\
  Twelve citizens vote for two candidates in a college election. It happens that the vote
 is tied 6 to 6. However, for dramatic effect, the ballots are shuffled and read out one at a time
 and the cumulative score announced. What is the probability that at some point there is a
 gap of 3 votes between the two candidates? Is it closer to 47\% or 48\%?
-
+\
 \end{problem}
 \end{document}

sample.tex

@@ -11,7 +11,6 @@
 \usetikzlibrary{graphs,graphs.standard}
 \usepackage{color}
 \usepackage{multicol}
-\usepackage{enumerate}
 \usepackage{tikz}
 \usepackage{mathtools}
 \DeclarePairedDelimiter\ceil{\lceil}{\rceil}
@@ -216,75 +215,6 @@
 \end{align*}
 \end{solution}
 
-\begin{problem}[Shahriari 8.3.1][20]
-\begin{enumerate}[(a)]
-\item Determine the number of permutations of $\{1,2,\cdots, 8 \}$ in which no even integer is in its natural position.
-\item Determine the number of permutations of $\{1,2,\cdots, 8 \}$ in which exactly 4 integers are in their natural position.
-\end{enumerate}
-\end{problem}
-
-\begin{solution}
-\begin{enumerate}[(a)]
-\item We apply theorem 8.25 to the following chessboard. Counting the ways of placing 1,2,3,4 rooks on the forbidden squares is trival, as you are just choosing that many from the 4 available spots.\\
-\begin{tikzpicture}
-\draw[step=1cm,gray,very thin] (-2,-2) grid (6,6);
-\fill[blue!40!white] (-0,4) rectangle (-1,5);
-\fill[blue!40!white] (2,2) rectangle (1,3);
-\fill[blue!40!white] (4,0) rectangle (3,1);
-\fill[blue!40!white] (6,-2) rectangle (5,-1);
-\end{tikzpicture}
-$$\boxed{8!-\binom{4}{1}7!+\binom{4}{2}6!-\binom{4}{3}5!+\binom{4}{4}4!}$$
-\item First we must choose the 4 of the 8 to be ones on the diagonal. Then we must place the remaining 4 in such a way that the other 4 diagonals are now fobidden. 
-$$\binom{8}{4}\times[4!-\binom{4}{1}3!+\binom{4}{2}2!-\binom{4}{3}1!+\binom{4}{4}0!]$$
-\end{enumerate}
-\end{solution}
-
-\begin{problem}[Shahriari 8.3.2][20]
-\begin{enumerate}[(a)]
-\item Eight girls are seated around a carousel. In how many ways can they change seats so that each has a different girl in front of her?
-\item Eight boys are seated around a carousel but facing inward, so that each boy faces one another. In how many ways can they change seats so that each faces a different boy?
-\end{enumerate}
-\end{problem}
-
-\begin{solution}
-\begin{enumerate}[(a)]
-\item Name the 8 girls $1,2,\cdots,8$. 
-$$A_i:=\text{ the circular permutations with $i+1$ in front of $i$}$$
-Noting that $A_8:=$ the circular perms with 1 in front of 8.
-\begin{align*}
-\bigcap_{k=1}^8 A_i^c = \sum_{k=1}^8 \binom{8}{k} (k-1)!
-\end{align*}
-\item 
-\begin{align*}
-\bigcap_{k=1}^8 A_i^c =7!- 2\times \sum_{k=1}^4 \binom{4}{k} (7-k)!
-\end{align*}
-\end{enumerate}
-\end{solution}
-
-\begin{problem}[Shahriari 8.3.6][20]\\
-For each of the $6 \times 6$ boards with forbidden positions in Figure 8.3 in Shahriari's book on page 147, find the number of ways to place six non-attacking rooks in the non-forbidden positions.
-\end{problem}
-
-\begin{solution}
-\begin{enumerate}[(a)]
-\item $r_1 = 6$, $r_2 = 12$ because of the three partitions, choose 2 for the 2 objects, then there are 4 combinations of the two within the chosen partitions (thus $4\times \binom{3}{2}$) , $r_3 = 8$ because one rook must go in each partition, and for each, there is two options (thus $2^3$), $r_i=0$ st $i>3$ because need to place atleast 2 in one partition, which is not allowed.\\
-Thus by Theorem 8.25:
-$$\boxed{6!-6\times5! +12\times 4! -8\times 3!}$$
-
-\item $r_1 = 12$, $r_2 = 2\times \binom{3}{1} + 4^2\times \binom{3}{2} =54$ because either the two are in the same quadrant, or in different quadrants. If they are in the same, first choose the quadrant, and then there are two orthagonal orientations within that quadrant. If they aren't, choose the 2 quadrants for the two rooks, and then there are 4 spaces in each quadrant .\\ $r_3 = 6\times2\times4 + 4^3=112$ since all three can't be in same quadrant, one must be alone and two must the together or they must each be in a different quadrant. There are 6 ways to choose a quadrant for the two and the one. For the quadrant with two, there are two orthagonal orientations within that quadrant and for the quadrant with 1, there are 4 possible locations. If they are each alone, there are four options for the 3 quadrants. \\
- $r_4 = 3\times 4 + 3 \times 2 \times 4 \times 4 = 108$. Since 3 cannot go in the same quadrant, either 2 must go in one quadrant and two in another, or 2 must go in one quadrant. If it is the first case, choose the 2 quadrants, then for each quadrant, there is two orthagonal orientations. If the second case, choose the quadrant with two, determine the orthagonal orientation within that quadrant, then for the other 2 there are 4 possible orientations.\\
-  $r_5 = \binom{3}{1}\times 4\times 2^2 =48$. Since 3 cannot go in the same quadrant, 2 quadrants must both have 2 and the last 1. Choose the quadrant to have one. For that quadrant, the rook has 4 options. For the other two quadrants, there are two orthagonal options each.\\ $r_6 = 2^3$, because there must be 2 rooks in each quadrant. For each quadrant, there are two possible orientations\\
-$r_i=0$ st $i>6$ can't have more than 2 in a given quadrant.\\
-Thus by Theorem 8.25:
-$$\boxed{6!-12\times5! +54\times 4! -112\times 3! + 108\times 2! -48\times 1! + 8}$$
-\item $r_1 = 8$, $r_2 = 3\times 5 + 1 + 6 =22$ because there are 3 cases, either they are both in larger quadrant, or the both the smaller, or 1 in each. If they are in the larger, there is 4 ways by brute force. If they are in the smaller, there is only 1 way. If they are in different, there are $3 \times 5$ ways to order them.  .\\ $r_3 = 1+6\times 3 + 5 = 24$  because there are 3 cases, either all 3 are in larger quadrant, or 2 are in the larger and one in the smaller, 1 in the larger and 2 in the smaller (all 3 can't be in the smaller). If they are in the larger, there is 1 way by brute force. If 2 are in the larger and one in the smaller, there is 6 ways (by brute force in $r_2$) for the two in the larger, and 3 for the smaller. If 1 in the larger and 2 in the smaller, there are 5 ways for the one in the larger and 1 way for the smaller  .\\
- $r_4 = 1\times 3 + 6 = 9$  because there are 2 cases, either 3 are in the larger and one in the smaller, 2 in the larger and 2 in the smaller (more than 2 can't be in the smaller). If they are in the larger, there is 1 way by brute force. If 3 are in the larger and one in the smaller, there is 3 ways (by brute force in $r_3$ for the two in the larger, and 3 for the smaller. If 2 in the larger and 2 in the smaller, there are 6 ways for the two in the larger by brute froce from $r_2$ and 1 way for the smaller  .\\
-  $r_5= 1$  because 3 are in the larger and two in the smaller only has 1 way\\
-$r_i=0$ st $i>5$ can't have more than 2 in a given quadrant.\\
-Thus by Theorem 8.25:
-$$\boxed{6!-8\times5! +22\times 4! -24\times 3! + 9\times 2! -1\times 1!}$$
-\end{enumerate}
-\end{solution}
 
 \begin{problem}[Shahriari 8.3.10][20]\\
 Fereydoon, Faranak, Roudabeh, Tahminen and Rostem are students in the class. Each is to be assigned a differnt dialogue form the following list: Crito, Euthyphro, Protagoras, Meno, Parmenides. Fereydoon is not interested in the Crito or the Protagoras; Faranak is not interested in the Meno; Roudabeh is not interested in the Euthyphro or the Parmenides; Tahmineh is not interested in the Euthyphro and Rostam is not interested in the Crito or the Protagoras. In how many ways can we assign the five dialogues to the five students given the above constraints?

website/aggregator.py

@@ -27,8 +27,8 @@ def print_header(submission_file):
     print(r"\usepackage{enumerate}", file=submission_file)
     print(r"\usepackage{tikz}", file=submission_file)
     print(r"\usepackage{mathtools}", file=submission_file)
-    print(r"\DeclarePairedDelimiter\ceil{\lceil}{\\rceil}", file=submission_file)
-    print(r"\DeclarePairedDelimiter\\floor{\\lfloor}{\\rfloor}", file=submission_file)
+    print(r"\DeclarePairedDelimiter\ceil{\lceil}{\rceil}", file=submission_file)
+    print(r"\DeclarePairedDelimiter\floor{\lfloor}{\rfloor}", file=submission_file)
     print(r"% include this if you want to import graphics files with /includegraphics", file=submission_file)
 
 def aggregate(self, student_id, submissions):

website/migrations/0001_initial.py

@@ -28,6 +28,7 @@ class Migration(migrations.Migration):
             fields=[
                 ('id', models.AutoField(verbose_name='ID', serialize=False, auto_created=True, primary_key=True)),
                 ('name', models.CharField(max_length=255)),
+                ('contents', models.TextField()),
                 ('points', models.IntegerField()),
                 ('solution', models.TextField()),
             ],
@@ -41,7 +42,7 @@ class Migration(migrations.Migration):
                 ('id', models.AutoField(verbose_name='ID', serialize=False, auto_created=True, primary_key=True)),
                 ('contents', models.TextField()),
                 ('score', models.IntegerField(null=True, blank=True)),
-                ('problems', models.ManyToManyField(to='website.Problem')),
+                ('problem', models.ForeignKey(to='website.Problem')),
             ],
             options={
             },

website/views.py

@@ -5,6 +5,7 @@
 from django.contrib.auth import authenticate, login, logout
 from django.http import HttpResponseRedirect, HttpResponse
 import re
+import os
 import aggregator
 from website.models import Assignment, Problem
 import package_problems
@@ -51,7 +52,6 @@ def process_prof_file(file):
     asgt = Assignment(name=name, due_date=due_date)
     asgt.save()
 
-
     problem_locs = [m.end() for m in re.finditer("begin{problem}", data)]
 
     for loc in problem_locs:
@@ -60,21 +60,22 @@ def process_prof_file(file):
         p.name = m.group(1)
         n = re.search(r"\[([^]]*)\]", data[loc+m.end():])
         p.points = n.group(1)
-        o = re.search(r"([^]]*)\\end{problem}", data[loc + m.end() + n.end():])
+        o = re.search(r"([^]]*)\end{problem}", data[loc + m.end() + n.end():])
         p.contents = o.group(1)
         q = re.search(r"begin{solution}([^]]*)\\end{solution}", data[loc + m.end() + n.end() + o.end():])
         p.solution = q.group(1)
         p.save()
         asgt.problems.add(p)
 
     strip_solutions(asgt)
+    os.system("pdflatex %s" % asgt.name.replace(" ", "") + aggregator.TEX_FILE)
 
 def strip_solutions(asgt):
-    problems_file = open(asgt.name + aggregator.TEX_FILE, 'w')
+    problems_file = open(asgt.name.replace(" ", "") + aggregator.TEX_FILE, 'w')
     aggregator.print_header(problems_file)
-    print(r"\\name{" + asgt.name + "}", file=problems_file)
+    print(r"\name{" + asgt.name + "}", file=problems_file)
     print(r"\duedate{" + asgt.due_date + "}", file=problems_file)
-    print(r"\\begin{document}", file=problems_file)
+    print(r"\begin{document}", file=problems_file)
     for p in asgt.problems.all():
         print(r"\begin{problem}[" + p.name + "][" + str(p.points) + r"]", file=problems_file)
         print(p.contents, file=problems_file)