Question

So I’m doing Project Euler because dear gods do I need to practice writing code, and also my math skills are rusty as something very rusty. Thusly; Project Euler. I’m sure most here have already seen or heard of the problem, but I’ll put it here just for completeness:

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

For this, I’ve written two functions:

from math import sqrt

def isprime(n):
    if n == 1:
        return False
    elif n == 2:
        return True
    elif n % 2 == 0:
        return False
    for x in range(3, round(sqrt(n))+1, 2):
        if n % x == 0:
            return False
    else:
        return True

This just checks any fed number for primality. It’s working as intended (as far as I know), but now that I’ve said that I grow unsure. Anyway, it checks for special cases first: 1 (never prime), 2 (prime) or if it’s divisible by 2 (not prime). If none of the special cases happen, it runs a general primality test.

This is my factorization code:

def factorization(n):
   factor = 2
   x = 3
   while True:
       if n % x == 0:
           if isprime(x):
               factor = x
               n = n // x
               if n == 1:
                   return factor
           else:
               return factor
       x += 2

And this is definitely not working as intended. It is, sadly, working for the particular value of the Project Euler problem, but it doesn’t work for, say, 100. I’m unsure what I need to do to fix this: what happens is that if it’s a number like 100, it will correctly find the first 5 (2*2*5), but after that will loop around and set x = 7, which will make the entire thing loop infinitely because the answer is 2*2*5*5. Would recursion help here? I tried it, but it didn’t get any prettier (it would still go into an endless loop for some numbers). I’m unsure how to solve this now.

Answer 1

You’re on a good track, but you need to take account of the possibility of repeating factors. You can do that with something like this:

factors = []

while num % 2 == 0:
  factors.append(2)
  num /= 2

The idea here being that you’re going to continue adding 2’s to the factor list until the number you’re testing becomes odd. You can use similar logic for other factors as well to enhance your factorization method.