Sometimes, your interests converge. Over on Programming Praxis, he had a coding challenge to implement Monte Carlo factorization. The last couple of days, I have been thinking a bit more about factorization, so it seemed like a good idea to give it a whack. I cribbed a version of the Rabin Miller primality tester off the web, and then coded the rest up in python.
#!/usr/bin/env python # # a basic implementation of the Pollard rho factorization # Written by Mark VandeWettering <markv@pixar.com> # import sys import locale import random class FactorError(Exception): def __init__(self, value): self.value = value def __str__(self): return repr(self.value) def miller_rabin_pass(a, n): d = n - 1 s = 0 while d % 2 == 0: d >>= 1 s += 1 a_to_power = pow(a, d, n) if a_to_power == 1: return True for i in xrange(s-1): if a_to_power == n - 1: return True a_to_power = (a_to_power * a_to_power) % n return a_to_power == n - 1 def isprime(n): for repeat in xrange(20): a = 0 while a == 0: a = random.randrange(n) if not miller_rabin_pass(a, n): return False return True def gcd(a, b): while b != 0: a, b = b, a%b return a def findfactor(n): for c in range(1, 50): x = y = random.randint(1, n-1) x = (x * x + c) % n y = (y * y + c) % n y = (y * y + c) % n while True: t = gcd(n, abs(x-y)) if t == 1: x = (x * x + c) % n y = (y * y + c) % n y = (y * y + c) % n elif t == n: break else: return t raise FactorError("couldn't find a factor.") def factor(n): r = [] while True: if isprime(n): r.append(n) break try: f = findfactor(n) r.append(f) n = n / f except FactorError, msg: r.append(n) break r.sort() return r def doit(n): flist = factor(n) print locale.format("%d", n, True), "=" for f in flist: print "\t%s" % locale.format("%d", f, True) locale.setlocale(locale.LC_ALL, "") doit(2**98-1)
This is a fairly challenging factorization, and after a few seconds, it outputs:
316,912,650,057,057,350,374,175,801,343 = 3 43 127 4,363,953,127,297 4,432,676,798,593
Thanks for mentioning my blog. Perhaps you could contribute your solution as a comment.
Phil
Thanks for this program. You clearly put some effort into it. I found so many silly programs on the web that did nothing better than divide by all the odd numbers below the sqrt of the number to be factored.
I was looking some web materials regarding factoring software but this is not I was looking for. I was looking for something related to invoice management (also called factoring). However it’s good to see the old-friend Python programming langugage 🙂
The program doesnt work for me
Did not work first up. Tried some changes to allow for changes from Python 2 to 3 (e.g. print statements need ()). When run, obtained 3 and the large factor, immediately after ran again and obtained 43 and the large factor (no 3) ?? Again, got 129 and the large factor.
Gave up. I have only just started on Python, so some of the statements are a bit obscure. For example, why are there duplicate lines in the function find factor??