# 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

## 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 pi such that pi  ≤  √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))
```

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)))

# 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.

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