Prime factorization program

Question

Compute the prime factors of a given natural number. A prime number is only evenly divisible by itself and 1. Note that 1 is not a prime number.

My solution will catch every prime up to and including 896803. This is the fastest solution I could come up with. I know that hard code can only go so far.

def prime_factors(n):
 factors = []
 found_last_factor = False
 if n == 1: return factors
 if is_prime(n): return [n]
 while found_last_factor == False: 
 for f in range(2, int(n/2+1)):
 if n % f == 0 and is_prime(f):
 n /= f
 factors.append(f)
 if is_prime(n): 
 factors.append(n)
 found_last_factor = True
 break
 return sorted(factors)
def is_prime(f: int) -> bool:
 if f is 1: return False
 for n in (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 
 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 
 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 
 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 
 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 
 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 
 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 
 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 
 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 
 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 
 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 
 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947):
 if f == n: return True
 elif f % n == 0: return False
 elif f / n < n: return True

• Janne Karila: A test case that can be timed would be good, when you care about the fastest code.
• 200_success: Please see What to do when someone answers . I have rolled back Rev 4 → 3.

Answer 1

 while found_last_factor == False:

Style is subjective, but I know I'm not the only develop who considers comparisons to Boolean literals to be bad style. while not found_last_factor: is an alternative which I consider preferable.

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There are various inefficiencies in this code:

 while found_last_factor == False:
 for f in range(2, int(n/2+1)):
 if n % f == 0 and is_prime(f):
 n /= f
 factors.append(f)
 if is_prime(n): 
 factors.append(n)
 found_last_factor = True
 break
 return sorted(factors)

1. If it were modified to handle repeated prime factors in an inner loop rather than in the outmost loop then
   1. The test is_prime(f) would be unnecessary, because n % f == 0 would already imply the primality of f.
   2. Primes would be discovered in order, removing the need to call sorted.
   3. A lot of unnecessary iterations of f would be saved. You rebutted AJNeufeld's observation on this subject, so consider an alternative perspective: add a line print(f) at the top of the for f loop, and then call prime_factors(1024).
2. The call to is_prime(n) tests primality by trial division, and is called from a loop which is factoring n by trial division. That's duplication of both code and run time.

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/= does floating point division. Use //= for integer division. Similarly int(n/2+1) could be n//2 + 1.

• user85459: I'm not sure how to fix your second point. The value of n changes. It wouldn't make sense to loop up to it if I could just check it right there with is_prime(n),
• Peter Taylor: You can take some of the reasoning used in is_prime and pull it into prime_factors: namely that when f * f > n either n is prime or n is 1. That's a sufficient condition to break.

Answer 2

is_prime() can terminate without returning a value (returning None) if the input is large, but not divisible by a prime under 1000. You should handle that explicitly (perhaps raise an exception).

Bug? prime_factors(45) will return [3,5], but 3*5 != 45. You might want to handle factors like \$p^n\$, where p is a prime. Update Ok, not a bug. (I didn't run the code with an official interpreter; just my eyeball interpreter, which is a bit buggy.) The while loop around the for loop will eventually catch that the repeated primes. Still, you're doing a lot of extra work when you could just keep factoring out the prime factor. If that is done, then you don't need the outer while loop.

Your loop for f in range(2, int(n/2+1)): does not update the end point as n gets smaller with successive factors being found. In the case of 45, it loops up to 22, despite finding the last factor of 5 seventeen iterations earlier.

• user85459: prime_factors(45) gives me [3, 3, 5]. I ran the code in the debugger and it did not do any extra loops.
• Peter Taylor: Actually, given 45 it returns [3, 3.0, 5]
• user85459: @PeterTaylor I fixed that