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

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

https://docs.codewars.com/references/markdown/extensions/#math-typesetting
This may be of use to you.

No random tests (at least in python).

https://docs.codewars.com/authoring/guidelines/submission-tests/#random-tests

Duplicate.

Is latex an option in the markdown?