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 any

Answer 1

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}

Answer 2

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
...

Answer 3

You should probably do some prime detection which you could look here, Fast algorithm for finding prime numbers?

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

Answer 4

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))

Answer 5

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.

Answer 6

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.