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How can it not? You are doing the calculation beforehand, and turn a O(n) problem into a O(1) problem.

Is there some reason for creating this many kumites?

Do you have a problem with some functionality, or something does not work for you? Do you need any assistance?

n=10000000019

• this solution: 21.4 ms ± 587 µs
• faster version: 13.4 ms ± 40.5 µs ★

Faster version:

def prime_checker_6k(n):
    if n < 4:
        return n > 1
    if not (n % 2 and n % 3):
        return False
    for i in range(5, int(n ** 0.5)+1, 6):
        if not ((n % i) and (n % (i + 2))):
            return False
    return True

You might wanna try again, It wont work for n <= 1 and n == 4

yea, I changed the while to a for loop and the 6k one is faster almost always but the time difference is still in the same range and fluctuates wildly. I never realized that while loops are slower than for loops in python, ty for letting me know

you're comparing a while loop and a for one. In python, they behave differently about the perfs (while is slower) => those checks are meaningless in the current state.

also I have no idea why my reply got posted twice

hmm, the results of comparing the two often turns out to be wildly inconsistent in time difference but remains in the 1-10µs range, tho I found that the n = 19 and 1009 takes longer with 2k, n = 10000000019 takes less with the 2k one most of the time. Code used for testing:

import timeit
SETUP6K = '''
def prime_checker6k(n):
    if n < 4:
        return n > 1
    if not (n % 2 and n % 3):
        return False
    for i in range(5, int(n ** 0.5)+1, 6):
        if not ((n % i) and (n % (i + 2))):
            return False
    return True
'''
SETUP2K = '''
def prime_checker2k(n):
    if n == 2: return True
    if n%2 == 0 or n < 2: return False
    for i in range(3, int(n**.5)+1, 2):
        if n%i == 0: return False
    return True
'''
def bench(n):
    #print(f'Testing prime_checker2k and prime_checker6k with arg: {n}')
    a = timeit.timeit(stmt=f"prime_checker2k({n})", setup = SETUP2K, number = 1)
    b = timeit.timeit(stmt=f"prime_checker6k({n})", setup = SETUP6K, number = 1)
    #print(f'Time taken for prime_checker2k = {a}\t\tTime taken for prime_checker6k = {b}\t\t2k-6k = {a-b}\n')
    print(f'{"prime_checker2k" if a<b else "prime_checker6k"} is faster')
    print(f'6k-2k = {round(b-a,9)}')

for _ in range(10):
  #bench(19)
  #bench(1009)
  bench(10000000019)
  print()

hmm, the results of comparing the two often turns out to be wildly inconsistent in time difference but remains in the 1-10µs range, tho I found that the n = 19 and 1009 takes longer with 2k, n = 10000000019 takes less with the 2k one most of the time. Code used for testing:

import timeit
SETUP6K = '''
def prime_checker6k(n):
    if n < 4:
        return n > 1
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    sqrt_n = int(n ** 0.5)
    while i <= sqrt_n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True
'''
SETUP2K = '''
def prime_checker2k(n):
    if n == 2: return True
    if n%2 == 0 or n < 2: return False
    for i in range(3, int(n**.5)+1, 2):
        if n%i == 0: return False
    return True
'''
def bench(n):
    #print(f'Testing prime_checker2k and prime_checker6k with arg: {n}')
    a = timeit.timeit(stmt=f"prime_checker2k({n})", setup = SETUP2K, number = 1)
    b = timeit.timeit(stmt=f"prime_checker6k({n})", setup = SETUP6K, number = 1)
    #print(f'Time taken for prime_checker2k = {a}\t\tTime taken for prime_checker6k = {b}\t\t2k-6k = {a-b}\n')
    print(f'{"prime_checker2k" if a<b else "prime_checker6k"} is faster')
    print(f'6k-2k = {round(b-a,9)}')

for _ in range(10):
  #bench(19)
  #bench(1009)
  bench(10000000019)
  print()

Your solution is 2k optimization (only ignores multiples of 2).
My first solution is 6k optimization (ignores multiples of 2 and 3).
And last solution is 30k optimization (ignores multiples of 2, 3 and 5) — but in this solution, the constant is larger. Therefore, further I compare only 2k and 6k optimizations.

n = 19 (is prime)

• 2k optimization: 2.92 µs ± 59.9 ns
• 6k optimization: 1.94 µs ± 41.6 ns

n = 1009 (is prime)

• 2k optimization: 6.74 µs ± 317 ns
• 6k optimization: 5.91 µs ± 90.1 ns

n = 10000000019 (is prime)

• 2k optimization: 21.6 ms ± 598 µs
• 6k optimization: 19.7 ms ± 586 µs

YOLO programming

You can also just use more arguments of the range function, probably faster because it should be implemented in C.

That's was very hard))

It is amazing how one can forget how to use your tools properly. Amazing solution. Thanks!