Problem 3 - Factoring Primes

Problem 3

The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?

This is the first problem I found with an interesting real world application.  Factoring primes is hard but multiplying out two primes is trivial.  That concept, called a one-way function, underpins modern cryptography.  The generation of large primes is a hard problem, but it is much harder to figure out which two primes were multiplied to make a given number than it is to generate those primes in the first place.  That's because modern computers are very, very good at multiplication and addition, but relatively slow at division.  Likewise, the brute force guessing of prime factors is very time consuming because there are a large number of primes.

All cryptography can be broken given enough time.  The trick is to make the math problem hard enough that the attacker gives up and does something else before he can decrypt your secrets.


Here's the code for Problem 3:


def isprime(val): #Checks to see if a number is prime, brute force
for x in range(2, int(val**0.5)+1):
if val % x == 0: return False
return True

def factor(val): #Get the prime factors of a number
if isprime(val): return [val]
i = 2
thesefactors = []
tempval = val
while (i <= tempval):
if tempval%i == 0:
thesefactors += [i]
tempval = tempval/i
i = 2
i = i + 1
if len(thesefactors) <= 1:
return [val]
return thesefactors

if __name__ == '__main__':
import sys
factors = factor(int(sys.argv[1]))
print (factors)

This is run using the following command (with the correct number argument for the question asked):

python 600851475143

The correct answer is 6857.


Again, this is a brute force approach.  I have included a refinement called Newton's Improvement where you don't need to check if numbers larger than the square root of that number are factors.  This helps improve both subroutines.  These subroutines were both used in the solutions of later problems.

for x in range(2, int(val**0.5)+1):


The question asked was too easy to make a good benchmark, so I created my own composite using the solution of a later problem involving counting primes.  The two primes I multiplied were 1299379 and 1299709.  The multiplication takes the blink of an eye, but the factoring takes many, many blinks.

As before, pypy is about 7-10x faster than Python2, with Jython being much slower and Python3 somewhere in the middle.

Problem #3 Benchmarks

time python 1688814580711

[1299379, 1299709]

real 0m0.329s

user 0m0.316s

sys 0m0.012s

time python -O 1688814580711

[1299379, 1299709]

real 0m0.380s

user 0m0.372s

sys 0m0.004s

time python3 1688814580711

[1299379, 1299709]

real 0m0.715s

user 0m0.704s

sys 0m0.008s

time python3 -O 1688814580711

[1299379, 1299709]

real 0m0.772s

user 0m0.764s

sys 0m0.004s

time pypy 1688814580711

[1299379, 1299709]

real 0m0.083s

user 0m0.080s

sys 0m0.000s

time pypy -O 1688814580711

[1299379, 1299709]

real 0m0.083s

user 0m0.076s

sys 0m0.004s


The Problem 3 solution is a program with both real world uses and utility as a library for later problems.  The problem of locating and factoring primes is the holy grail of cryptography and this simple implementation gives some insight into prime numbers and the problem of factoring primes.