Prime factorization algorithm

A prime factorization algorithm is any algorithm (a step-by-step process) by which an integer (whole number) is "decomposed" into a product of factors that are prime numbers. The fundamental theorem of arithmetic guarantees that this decomposition is unique. This article gives a simple example of an algorithm, which works well for numbers whose prime factors are small; faster algorithms for numbers with larger prime factors are discussed in the article on integer factorization. A 'fast' algorithm (which can factorise large numbers in a reasonably small time) is much sought after.

Contents

1 A simple factorization algorithm 1.1 Description 1.2 Example 1.2.1 Code 1.3 Time complexity 2 External link

A simple factorization algorithm

Description

We can describe a recursive algorithm to perform such factorizations: given a number n

• if n is prime, this is the factorization, so stop here.
• if n is composite, divide n by the first prime p₁. If it divides cleanly, recurse with the value n/p₁. Add p₁ to the list of factors obtained for n/p₁ to get a factorization for n. If it does not divide cleanly, divide n by the next prime p₂, and so on.

Note that we need to test only primes p_i such that p_i  ≤  √n.

Example

Suppose we wish to factorize the number 9438.

9438/2 = 4719 with a remainder of 0, so 2 is a factor of 9438. We repeat the algorithm with 4719.

4719/2 = 2359 with a remainder of 1, so 2 is NOT a factor of 4719. We try the next prime, 3.

4719/3 = 1573 with a remainder of 0, so 3 is a factor of 4719. We repeat the algorithm with 1573.

1573/3 = 524 with a remainder of 1, so 3 is NOT a factor of 1573. We try the next prime, 5.

1573/5 = 314 with a remainder of 3, so 5 is NOT a factor of 1573. We try the next prime, 7.

1573/7 = 224 with a remainder of 5, so 7 is NOT a factor of 1573. We try the next prime, 11.

1573/11 = 143 with a remainder of 0, so 11 is a factor of 1573. We repeat the algorithm with 143.

143/11 = 13 with a remainder of 0, so 11 is a factor of 143. We repeat the algorithm with 13.

13/11 = 1 with a remainder of 2, so 11 is NOT a factor of 13. We try the next prime, 13.

13/13 = 1 with a remainder of 0, so 13 is a factor of 13. We stop when we reached 1.

Thus working from top to bottom, we have 9438 = 2 * 3 * 11 * 11 * 13

Code

Here is some code in Python for finding the factors of numbers less than 2147483647:

from math import sqrt
def factorize(n):
    def isPrime(n):
        return not [x for x in xrange(2,int(sqrt(n))+1)
                    if n%x == 0]
    primes = []
    candidates = xrange(2,n+1)
    candidate = 2
    while not primes and candidate in candidates:
        if n%candidate == 0 and isPrime(candidate):
            primes = primes + [candidate] + factorize(n/candidate)
        candidate += 1            
    return primes
print factorize(int(sys.argv[1]))

output:

python factorize.py 9438
[2, 3, 11, 11, 13]

Here is more complex code in Python for finding the factors of any arbitrarily large number:

import sys

ListOfPrimes=[2,3,5,7,11,13,17,19]
maxindex=len(ListOfPrimes)
maxprimeinlist=ListOfPrimes[-1]

# Put Primes in a dictionary
DictPrime={}
DictPrime.fromkeys(ListOfPrimes,True)

def intsqrt(n):
  """ Return the integer square root of a long number """
  def intsqrt_core(digitpair,remainder,results):
    # function intsqrt_core returns (results,remainder)
    if digitpair<100:
      currvalue=remainder*100 + digitpair
      for d in range(9,-1,-1):
        x=(2*10*results + d)*d
        if x <= currvalue:
          remainder= currvalue - x
          results=results*10 + d
          return(results,remainder)
    else:
      (results,remainder)=intsqrt_core(digitpair//100,remainder,results)
      (results,remainder)=intsqrt_core(digitpair%100,remainder,results)
      return(results,remainder)
  (results,remainder)=intsqrt_core(n,0,0)
  return results

def isPrime(n):
  """ Return True if n is a prime """
  if DictPrime.has_key(n):
    return True
  high=intsqrt(n)
  for x in ListOfPrimes:
    if x <= high and n%x == 0:
      return False
    if x >= high:
      return True
  x=maxprimeinlist + 2
  while x<=high:
    if n%x == 0:
      return False
    x += 2
  return True

def factorize(n):
  """ Factorize a integer number """
  primes = []
  index=0
  candidate = ListOfPrimes[index]
  while not primes and candidate <= n:
    if n%candidate == 0 and (index < maxindex or isPrime(candidate)):
      primes = primes + [candidate] + factorize(n//candidate)
    index += 1            
    if index < maxindex:
      candidate = ListOfPrimes[index]
    else:
      candidate += 2
  return primes

def condense(L):
  """ Condense result in list to prime^nth_power format """
  prime,count,list=0,0,[]
  for x in L:
    if x == prime:
      count += 1
    else:
      if prime != 0:
        list = list + [str(prime) + '^' + str(count)]
      prime,count=x,1
  list = list + [str(prime) + '^' + str(count)]
  return list

if __name__ == '__main__':
  print condense(factorize(long(sys.argv[1])))

# Sample output
#
# python factorize.py 173248246132375748867198458668657948626531982421875
# ['3^24', '5^14', '7^33', '13^1']

Time complexity

The algorithm described above works fine for small n, but becomes impractical as n gets larger. For example, for an 18-digit (or 60 bit) number, all primes below about 1,000,000,000 may need to be tested, which is taxing even for a computer. Adding two decimal digits to the original number will multiply the computation time by 10.

The difficulty (large time complexity) of factorization makes it a suitable basis for modern cryptography.

See also: Euler's Theorem, Integer factorization, Trial division

External link

• Prime factorization at Mathworld (http://mathworld.wolfram.com/PrimeFactorizationAlgorithms.html)
• Factorization solver implementing the algorithm described here, work shown (http://www.algebra.com/algebra/homework/divisibility/)fr:Factorisation en nombres premiers

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