Pollard algorithm now traverses the entire divisor group modulo n, wh…

…ich will lead to the correct result in all cases, new demo in readme

README.md

@@ -2,33 +2,79 @@
 This is my implementation of RSA cracking algorithm based on Pollard factorization. It will work if the prime number `n` as part of the public key has not a very big maximum prime factor. Of course, all modern RSA implementations don't have this security hole so this is more of an educational project that doesn't have much real-world applications.
 
 # Usage
-In order to start the script, simply run `python pollard.py`
+In order to start the script, simply run `python cracker.py`
 
 You will need 4 inputs:
-1. `b` - the maximum prime factor of `n`.
-2. `n` - prime number in the public key so that `m = c ^ d mod n` where `m` is the message, `c` is the encrypted message, `d` is the private key.
+1. `n` - prime number in the public key so that `m = c ^ d mod n` where `m` is the message, `c` is the encrypted message, `d` is the private key.
+2. `e` - the second public key component (so that `gcd(e, phi(n)) = 1`).
 3. `c` - the encrypted message.
-4. `e` - the second public key component (so that `gcd(e, phi(n)) = 1`).
+4. `b` - the maximum prime factor of `n`. This field is optional. However, if you leave it blank, the brute-force algorithm will be used for factorizing `n`.
 
 # Demo
 ```
-Enter b (the maximum prime number factor): 200000
 Enter n (as part of the public key): 186444745729857899758373984272541398503249351266417000699738642133172271283265124803102459
-Enter c (the encrypted message to decrypt): 159178142916077677757648147687519523540045276157456113470673097514775229976995968698190914
 Enter e (as part of the public key): 65537
-Factorizing n to find its prime divisors...
+Enter c (the encrypted message to decrypt): 159178142916077677757648147687519523540045276157456113470673097514775229976995968698190914
+Enter b (the maximum prime factor in n, optional): 200000
+Factorizing n with the Pollard algorithm...
+k has 86871 digits
+Trying to solve with a = 2
 (a ^ k mod n) - 1 = 95072790833932492612013226959843266212858324985456485206430737027466196168851937286400160
 p = 2962244945158622279601750283735698362054297
 q = 62940354083336669282335053470277744418281466547
 Factorizing complete.
 phi(n) = 186444745729857899758373984272541398503249288323100672417910737518517050721785008159581616
 Solving e * d = 1 mod phi(n) with the euklidian algorithm...
-a = -41924961744066952541910026492278294547238731739590378098215520069737488388039514552813743
-b = 14737
 d (private key) = 144519783985790947216463957780263103956010556583510294319695217448779562333745493606767873
 m (decrypted message) = 2020
 ```
 
+# Demo with small numbers
+## Brute-force
+```
+Enter n (as part of the public key): 493
+Enter e (as part of the public key): 45
+Enter c (the encrypted message to decrypt): 56
+Enter b (the maximum prime factor in n, optional):
+Factorizing n with the BRUTEFORCE algorithm...
+p = 17
+q = 29
+Factorizing complete.
+phi(n) = 448
+Solving e * d = 1 mod phi(n) with the euklidian algorithm...
+d (private key) = 229
+m (decrypted message) = 490
+```
+## Pollard
+```
+Enter n (as part of the public key): 493
+Enter e (as part of the public key): 45
+Enter c (the encrypted message to decrypt): 56
+Enter b (the maximum prime factor in n, optional): 493
+Factorizing n with the Pollard algorithm...
+k has 216 digits
+Trying to solve with a = 2
+(a ^ k mod n) - 1 = 0
+Trying to solve with a = 3
+(a ^ k mod n) - 1 = 0
+Trying to solve with a = 4
+(a ^ k mod n) - 1 = 0
+
+...
+
+Trying to solve with a = 509
+(a ^ k mod n) - 1 = 0
+Trying to solve with a = 510
+(a ^ k mod n) - 1 = 203
+p = 29
+q = 17
+Factorizing complete.
+phi(n) = 448
+Solving e * d = 1 mod phi(n) with the euklidian algorithm...
+d (private key) = 229
+m (decrypted message) = 490
+```
+
 # Credits
 * Mirko Hoehn for helping me with the mathematics behind RSA.
 

cracker.py

@@ -14,7 +14,7 @@
     n = int(input("Enter n (as part of the public key): "))
     e = int(input("Enter e (as part of the public key): "))
     c = int(input("Enter c (the encrypted message to decrypt): "))
-    b = int(input("Enter b (the maximum prime factor in n, optional): ") or n)
+    b = int(input("Enter b (the maximum prime factor in n, optional): ") or 1)
 
     # Demo:
     # n = 186444745729857899758373984272541398503249351266417000699738642133172271283265124803102459
@@ -24,11 +24,13 @@
 
     p = None
 
-    if b >= n or b < 1:
+    if b > n or b <= 1:
         print("Factorizing n with the BRUTEFORCE algorithm...")
         p = find_factor(n)
     else:
         print("Factorizing n with the Pollard algorithm...")
+        if b > n - 1:
+            b = n - 1
         p = pollard(b, n)
 
     print("p = " + str(p))

euklidian.py

@@ -74,5 +74,11 @@ def inverse_modulo(a, n):
 
     if b < 0:
         b += n
-
-    return b
+
+    return b
+
+def next_divisor_of(d, n):
+    d = d + 1
+    while d < n and gcd(d, n) != 1:
+        d = d + 1
+    return d

pollard.py

@@ -7,7 +7,7 @@
 
 from prime import nextPrime
 from powmod import aPowbModn
-from euklidian import gcd
+from euklidian import gcd, next_divisor_of
 
 def calculateK(b):
 
@@ -28,7 +28,7 @@ def calculateK(b):
             maximalePrimFaktor = currentPrimFaktor
             currentExponent += 1
 
-        print("Found factor in k: "+str(currentPrime)+" ^ "+str(currentExponent)+" = "+str(maximalePrimFaktor))
+        # print("Found factor in k: "+str(currentPrime)+" ^ "+str(currentExponent)+" = "+str(maximalePrimFaktor))
 
         k *= maximalePrimFaktor
 
@@ -40,25 +40,22 @@ def pollard(b, n):
 
     k = calculateK(b)
 
-    factor = 1
-
-    while n % 2 == 0:
-        factor = factor * 2
-        n = n // 2
+    # print("k = " + str(k))
+    print("k has " + str(len(str(k))) + " digits")
 
+    p = 1
     # a is an arbitraty element in the multiplicative group of divisors of 1 modulo n
-    # as gcd(n-1, n) = 1 for all n, we can pick a = n - 1
-    # or we can pick 2 if n is odd
-    a = 2 # n - 1
+    a = next_divisor_of(1, n)
 
-    # print("k = " + str(k))
-    print("k has " + str(len(str(k))) + " digits")
-    print("a = " + str(a))
+    while p == 1 or p >= n:
+        print("Trying to solve with a = " + str(a))
+
+        aPowKModNminus1 = aPowbModn(a, k, n) - 1
 
-    aPowKModNminus1 = aPowbModn(a, k, n) - 1
+        print("(a ^ k mod n) - 1 = " + str(aPowKModNminus1))
 
-    print("(a ^ k mod n) - 1 = " + str(aPowKModNminus1))
+        p = gcd(aPowKModNminus1, n)
 
-    p = gcd(aPowKModNminus1, n)
+        a = next_divisor_of(a, n)
 
-    return factor * p
+    return p