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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
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    \$\begingroup\$ A test case that can be timed would be good, when you care about the fastest code. \$\endgroup\$ Oct 13, 2018 at 20:49
  • \$\begingroup\$ Please see What to do when someone answers. I have rolled back Rev 4 → 3. \$\endgroup\$ Oct 14, 2018 at 19:36

2 Answers 2

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    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.


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.

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

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  • \$\begingroup\$ 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), \$\endgroup\$
    – user85459
    Oct 14, 2018 at 18:30
  • \$\begingroup\$ 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. \$\endgroup\$ Oct 14, 2018 at 19:58
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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.

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  • \$\begingroup\$ prime_factors(45) gives me [3, 3, 5]. I ran the code in the debugger and it did not do any extra loops. \$\endgroup\$
    – user85459
    Oct 13, 2018 at 21:26
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    \$\begingroup\$ Actually, given 45 it returns [3, 3.0, 5] \$\endgroup\$ Oct 13, 2018 at 22:03
  • \$\begingroup\$ @PeterTaylor I fixed that \$\endgroup\$
    – user85459
    Oct 13, 2018 at 22:24

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