python 代码未运行。我正在计算一个数字的素数（项目欧拉问题 - 3）

Question

I am trying to calculate prime factors of a number in python python 3.8.2 32bit (when i pass 35 in code its output should be 7,and so on).
But for some reason the program does not return any answer (when i click run the cmd does not output anything).But when i run this javascript, the exact same code works there.
(Previously i had an array(list) to which i would append prime factors and at last i would pop the last element which worked for smaller numbers, but for really large numbers i would get a Memory error,so i converted it to use only one variable which will be updated for every while loop).
What is going on here??

My code is:

import math
# Computes only prime factors of n
def compute(n):
    arr = 0
    # Checks if n is divisible by 2, and if it is divisible,returns 2 because there will be no any other 
    # prime factor.
    if n % 2 == 0:
        return 2
    # Now that 2 is eliminated we only check for odd numbers upto (square root of n)+1 
    for i in range(1, round(math.sqrt(n)) + 1, 2):
        while n % i == 0:
            arr = n/i
            n /= i

    return str(arr)

print(compute(81))

I am a newbie in python so plz tell me if i have made any silly mistakes. Ty.

Answer 1

For numbers that are not divisible by 2, your code runs into an infinite loop at while n%i == 0

For numbers that are divisible by 2, your function returns 2. The execution of the return statement exits the function.

Even if you change the while in while n%i == 0 to if n% == 0 the function will not work.

You will have to restructure your code.

An easy fix would be to check for all factors of a number till n/2 + 1 and return the factors (in a list) that are prime (which can be checked using a separate isprime function.

def isprime(n):
    for x in range(2,n//2+1):
        if n%x==0:
            return False
    return True

def compute(n):
    arr = [] 
    for i in range(2, n//2+1):
       if n % i == 0:
            if isprime(i):
                arr.append(i)

    return arr

Answer 2

If you want all prime factors (which i guess) you shouldn't return values before you have all prime factors in a list for example.

And with this programm you only check once for a 2. But 4 has 2*2. Put it in a loop.

Answer 3

So this code saves all prime factors in a list:

from math import sqrt, ceil

def get_prime_factors(n):
    prime_factors = []

    while n > 1:
        for i in range (2, ceil(sqrt(n))+1):
            if n % i == 0:
                possible_prime_number = i
                while True:
                    for j in range(2, possible_prime_number//2-1):
                        if possible_prime_number % j == 0:
                            possible_prime_number //= j
                            break
                    prime_factors.append(possible_prime_number)
                    n //= possible_prime_number
                    break

    return prime_factors

if you only want the highest one replace return prime_factors with return max(prime_factors)

Answer 4

OK. I think counting upto 600 billion is ridiculous(considering that i am on a 32 bit system). So here is the final code that i am setteling on which gives answer in acceptable time upto 99 million, without using any array and only 1 for loop.

from math import ceil
# Computes only prime factors of n
def compute(n):
    arr = 0
    # Checks if n is divisible by 2, and if it is divisible,sets arr =2 
    if n % 2 == 0:
        arr = 2
    # Now that 2 is eliminated we only check for odd numbers upto (square root of n)+1 
    for i in range(1, ceil(n/2) + 1, 2):
        if n % i == 0:
            arr = i
            n //= i

    return str(arr)

print(compute(999999999))

PS:- If you see any possible improvement in my code plz tell me.