# Fast prime factorization module

### Question

I am looking for an implementation or clear algorithm for getting the prime factorization of N in either python, pseudocode or anything else well-readable. There are a few demands/facts:

• N is between 1 and ~20 digits
• No pre-calculated lookup table, memoization is fine though.
• Need not to be mathematically proven (e.g. could rely on the Goldbach conjecture if needed)
• Need not to be precise, is allowed to be probabilistic/deterministic if needed

I need a fast prime factorization algorithm, not only for itself, but for usage in many other algorithms like calculating the Euler phi(n).

I have tried other algorithms from Wikipedia and such but either I couldn't understand them (ECM) or I couldn't create a working implementation from the algorithm (Pollard-Brent).

I am really interested in the Pollard-Brent algorithm, so any more information/implementations on it would be really nice.

Thanks!

EDIT

After messing around a little I have created a pretty fast prime/factorization module. It combines an optimized trial division algorithm, the Pollard-Brent algorithm, a miller-rabin primality test and the fastest primesieve I found on the internet. gcd is a regular Euclid's GCD implementation (binary Euclid's GCD is much slower then the regular one).

# Bounty

Oh joy, a bounty can be acquired! But how can I win it?

• Find an optimalization or bug in my module.
• Provide alternative/better algorithms/implementations.

The answer which is the most complete/constructive gets the bounty.

And finally the module itself:

``````import random

def primesbelow(N):
# http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
#""" Input N>=6, Returns a list of primes, 2 <= p < N """
correction = N % 6 > 1
N = {0:N, 1:N-1, 2:N+4, 3:N+3, 4:N+2, 5:N+1}[N%6]
sieve = [True] * (N // 3)
sieve[0] = False
for i in range(int(N ** .5) // 3 + 1):
if sieve[i]:
k = (3 * i + 1) | 1
sieve[k*k // 3::2*k] = [False] * ((N//6 - (k*k)//6 - 1)//k + 1)
sieve[(k*k + 4*k - 2*k*(i%2)) // 3::2*k] = [False] * ((N // 6 - (k*k + 4*k - 2*k*(i%2))//6 - 1) // k + 1)
return [2, 3] + [(3 * i + 1) | 1 for i in range(1, N//3 - correction) if sieve[i]]

smallprimeset = set(primesbelow(100000))
_smallprimeset = 100000
def isprime(n, precision=7):
# http://en.wikipedia.org/wiki/Miller-Rabin_primality_test#Algorithm_and_running_time
if n < 1:
raise ValueError("Out of bounds, first argument must be > 0")
elif n <= 3:
return n >= 2
elif n % 2 == 0:
return False
elif n < _smallprimeset:
return n in smallprimeset

d = n - 1
s = 0
while d % 2 == 0:
d //= 2
s += 1

for repeat in range(precision):
a = random.randrange(2, n - 2)
x = pow(a, d, n)

if x == 1 or x == n - 1: continue

for r in range(s - 1):
x = pow(x, 2, n)
if x == 1: return False
if x == n - 1: break
else: return False

return True

# https://comeoncodeon.wordpress.com/2010/09/18/pollard-rho-brent-integer-factorization/
def pollard_brent(n):
if n % 2 == 0: return 2
if n % 3 == 0: return 3

y, c, m = random.randint(1, n-1), random.randint(1, n-1), random.randint(1, n-1)
g, r, q = 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = (pow(y, 2, n) + c) % n

k = 0
while k < r and g==1:
ys = y
for i in range(min(m, r-k)):
y = (pow(y, 2, n) + c) % n
q = q * abs(x-y) % n
g = gcd(q, n)
k += m
r *= 2
if g == n:
while True:
ys = (pow(ys, 2, n) + c) % n
g = gcd(abs(x - ys), n)
if g > 1:
break

return g

smallprimes = primesbelow(1000) # might seem low, but 1000*1000 = 1000000, so this will fully factor every composite < 1000000
def primefactors(n, sort=False):
factors = []

for checker in smallprimes:
while n % checker == 0:
factors.append(checker)
n //= checker
if checker > n: break

if n < 2: return factors

while n > 1:
if isprime(n):
factors.append(n)
break
factor = pollard_brent(n) # trial division did not fully factor, switch to pollard-brent
factors.extend(primefactors(factor)) # recurse to factor the not necessarily prime factor returned by pollard-brent
n //= factor

if sort: factors.sort()

return factors

def factorization(n):
factors = {}
for p1 in primefactors(n):
try:
factors[p1] += 1
except KeyError:
factors[p1] = 1
return factors

totients = {}
def totient(n):
if n == 0: return 1

except KeyError: pass

tot = 1
for p, exp in factorization(n).items():
tot *= (p - 1)  *  p ** (exp - 1)

totients[n] = tot

def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a

def lcm(a, b):
return abs((a // gcd(a, b)) * b)
``````
1
67
8/13/2015 10:55:36 AM

13
1/10/2011 4:27:10 AM

If you don't want to reinvent the wheel, use the library sympy

``````pip install sympy
``````

Use the function `sympy.ntheory.factorint`

``````>>> from sympy.ntheory import factorint
>>> factorint(10**20+1)
{73: 1, 5964848081: 1, 1676321: 1, 137: 1}
``````

You can factor some very large numbers:

``````>>> factorint(10**100+1)
{401: 1, 5964848081: 1, 1676321: 1, 1601: 1, 1201: 1, 137: 1, 73: 1, 129694419029057750551385771184564274499075700947656757821537291527196801: 1}
``````