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43 lines
1.0 KiB
Python
43 lines
1.0 KiB
Python
from itertools import compress, repeat
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from math import ceil, sqrt
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def odd_sieve(num: int) -> list[int]:
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"""
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Returns the prime numbers < `num`. The prime numbers are calculated using an
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odd sieve implementation of the Sieve of Eratosthenes algorithm
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(see for reference https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes).
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>>> odd_sieve(2)
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[]
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>>> odd_sieve(3)
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[2]
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>>> odd_sieve(10)
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[2, 3, 5, 7]
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>>> odd_sieve(20)
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[2, 3, 5, 7, 11, 13, 17, 19]
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"""
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if num <= 2:
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return []
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if num == 3:
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return [2]
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# Odd sieve for numbers in range [3, num - 1]
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sieve = bytearray(b"\x01") * ((num >> 1) - 1)
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for i in range(3, int(sqrt(num)) + 1, 2):
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if sieve[(i >> 1) - 1]:
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i_squared = i**2
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sieve[(i_squared >> 1) - 1 :: i] = repeat(
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0, ceil((num - i_squared) / (i << 1))
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)
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return [2, *list(compress(range(3, num, 2), sieve))]
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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