包括非素数在内的素数检查器

Question

I am trying to solve Project Euler number 7.

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?

First thing that came into my mind was using length of list. This was very ineffective solution as it took over a minute. This is the used code.

def ch7():
    primes = []
    x = 2
    while len(primes) != 10001:
        for i in range(2, x):
            if x % i == 0:
                 break
        else:
            primes.append(x)
        x += 1
    
    print(primes[-1])

ch7()

 # Output is: 104743. 

This works well but I wanted to reach faster solution. Therefore I did a bit of research and found out that in order to know if a number is a prime, we need to test whether it is divisible by any number up to its square root eg in order to know if 100 is a prime we dont need to divide it by every number up to 100, but only up to 10.

When I implemented this finding weird thing happened. The algorithm included some non primes. To be exact 66 of them. This is the adjusted code:

import math

primes = []
def ch7():
    x = 2
    while len(primes) != 10001:
        for i in range(2, math.ceil(math.sqrt(x))):
            if x % i == 0:
                break
        else:
            primes.append(x)
        x += 1
    
    print(primes[-1])

ch7()

# Output is 104009

This solution takes under a second but it includes some non primes. I used math.ceil() in order to get int instead of float but I figured it should not be a problem since it still tests by every int up to square root of x.

Thank you for any suggestions.

Answer 1

Your solution generates a list of primes, but doens't use that list for anything but extracting the last element. We can toss that list , and cut the time of the code in half by treating 2 as a special case, and only testing odd numbers:

def ch7(limit=10001):  # assume limit is >= 1
    prime = 2
    number = 3
    count = 1

    while count < limit:
        for divisor in range(3, int(number ** 0.5) + 1, 2):
            if number % divisor == 0:
                break
        else:  # no break
            prime = number
            count += 1

        number += 2

    return prime

print(ch7())

But if you're going to collect a list of primes, you can use that list to get even more speed out of the program (about 10% for the test limits in use) by using those primes as divisors instead of odd numbers:

def ch7(limit=10001):  # assume limit is >= 1
    primes = [2]
    number = 3

    while len(primes) < limit:
        for prime in primes:
            if prime * prime > number:  # look no further
                primes.append(number)
                break

            if number % prime == 0:  # composite
                break
        else:  # may never be needed but prime gaps can be arbitrarily large
            primes.append(number)

        number += 2

    return primes[-1]


print(ch7())

BTW, your second solution, even with the + 1 fix you mention in the comments, comes up with one prime beyond the correct answer. This is due to the way your code (mis)handles the prime 2.