Fast prime factorization module

Python Problem Overview

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

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 optimization 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

    try: return totients[n]
    except KeyError: pass

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

    totients[n] = tot
    return 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)

Python Solutions

Solution 1 - Python

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

pip install sympy

Use the function sympy.ntheory.factorint

> Given a positive integer n, factorint(n) returns a dict containing the prime factors of n as keys and their respective multiplicities as values. For example:

Example:

>>> 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}

Solution 2 - Python

There is no need to calculate smallprimes using primesbelow, use smallprimeset for that.

smallprimes = (2,) + tuple(n for n in xrange(3,1000,2) if n in smallprimeset)

Divide your primefactors into two functions for handling smallprimes and other for pollard_brent, this can save a couple of iterations as all the powers of smallprimes will be divided from n.

def primefactors(n, sort=False):
    factors = []
   
    limit = int(n ** .5) + 1
    for checker in smallprimes:
        print smallprimes[-1]
        if checker > limit: break
        while n % checker == 0:
            factors.append(checker)
            n //= checker


    if n < 2: return factors
    else : 
        factors.extend(bigfactors(n,sort))
        return factors

def bigfactors(n, sort = False):
    factors = []
    while n > 1:
        if isprime(n):
            factors.append(n)
            break
        factor = pollard_brent(n) 
        factors.extend(bigfactors(factor,sort)) # recurse to factor the not necessarily prime factor returned by pollard-brent
        n //= factor

    if sort: factors.sort()    
    return factors

By considering verified results of Pomerance, Selfridge and Wagstaff and Jaeschke, you can reduce the repetitions in isprime which uses Miller-Rabin primality test. From Wiki.

if n < 1,373,653, it is enough to test a = 2 and 3;
if n < 9,080,191, it is enough to test a = 31 and 73;
if n < 4,759,123,141, it is enough to test a = 2, 7, and 61;
if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7, and 11;
if n < 3,474,749,660,383, it is enough to test a = 2, 3, 5, 7, 11, and 13;
if n < 341,550,071,728,321, it is enough to test a = 2, 3, 5, 7, 11, 13, and 17.

Edit 1: Corrected return call of if-else to append bigfactors to factors in primefactors.

Solution 3 - Python

Even on the current one, there are several spots to be noticed.

Don't do checker*checker every loop, use s=ceil(sqrt(num)) and checher < s
checher should plus 2 each time, ignore all even numbers except 2
Use divmod instead of % and //

Solution 4 - Python

You should probably do some prime detection which you could look here, https://stackoverflow.com/questions/3808171/fast-algorithm-for-finding-prime-numbers

You should read that entire blog though, there is a few algorithms that he lists for testing primality.

Solution 5 - Python

There's a python library with a collection of primality tests (including incorrect ones for what not to do). It's called pyprimes. Figured it's worth mentioning for posterity's purpose. I don't think it includes the algorithms you mentioned.

Solution 6 - Python

You could factorize up to a limit then use brent to get higher factors

from fractions import gcd
from random import randint

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

def factorize(n1):
    if n1==0: return []
    if n1==1: return [1]
    n=n1
    b=[]
    p=0
    mx=1000000
    while n % 2 ==0 : b.append(2);n//=2
    while n % 3 ==0 : b.append(3);n//=3
    i=5
    inc=2
    while i <=mx:
       while n % i ==0 : b.append(i); n//=i
       i+=inc
       inc=6-inc
    while n>mx:
      p1=n
      while p1!=p:
          p=p1
          p1=brent(p)
      b.append(p1);n//=p1 
    if n!=1:b.append(n)   
    return sorted(b)

from functools import reduce
#n= 2**1427 * 31 #
n= 67898771  * 492574361 * 10000223 *305175781* 722222227*880949 *908909
li=factorize(n)
print (li)
print (n - reduce(lambda x,y :x*y ,li))

Solution 7 - Python

I just ran into a bug in this code when factoring the number 2**1427 * 31.

  File "buckets.py", line 48, in prettyprime
    factors = primefactors.primefactors(n, sort=True)
  File "/private/tmp/primefactors.py", line 83, in primefactors
    limit = int(n ** .5) + 1
OverflowError: long int too large to convert to float

This code snippet:

limit = int(n ** .5) + 1
for checker in smallprimes:
    if checker > limit: break
    while n % checker == 0:
        factors.append(checker)
        n //= checker
        limit = int(n ** .5) + 1
        if checker > limit: break

should be rewritten into

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

which will likely perform faster on realistic inputs anyway. Square root is slow — basically the equivalent of many multiplications —, smallprimes only has a few dozen members, and this way we remove the computation of n ** .5 from the tight inner loop, which is certainly helpful when factoring numbers like 2**1427. There's simply no reason to compute sqrt(2**1427), sqrt(2**1426), sqrt(2**1425), etc. etc., when all we care about is "does the [square of the] checker exceed n".

The rewritten code doesn't throw exceptions when presented with big numbers, and is roughly twice as fast according to timeit (on sample inputs 2 and 2**718 * 31).

Also notice that isprime(2) returns the wrong result, but this is okay as long as we don't rely on it. IMHO you should rewrite the intro of that function as

if n <= 3:
    return n >= 2
...

Content Type	Original Author	Original Content on Stackoverflow
Question	orlp	View Question on Stackoverflow
Solution 1 - Python	Colonel Panic	View Answer on Stackoverflow
Solution 2 - Python	Rozuur	View Answer on Stackoverflow
Solution 3 - Python	Kabie	View Answer on Stackoverflow
Solution 4 - Python	milkypostman	View Answer on Stackoverflow
Solution 5 - Python	Ehtesh Choudhury	View Answer on Stackoverflow
Solution 6 - Python	Antoni Gual Via	View Answer on Stackoverflow
Solution 7 - Python	Quuxplusone	View Answer on Stackoverflow