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* Add more ruff rules * Add more ruff rules * pre-commit: Update ruff v0.0.269 -> v0.0.270 * Apply suggestions from code review * Fix doctest * Fix doctest (ignore whitespace) * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci --------- Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com> Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
66 lines
1.7 KiB
Python
66 lines
1.7 KiB
Python
"""
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Sieve of Eratosthones
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The sieve of Eratosthenes is an algorithm used to find prime numbers, less than or
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equal to a given value.
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Illustration:
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https://upload.wikimedia.org/wikipedia/commons/b/b9/Sieve_of_Eratosthenes_animation.gif
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Reference: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
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doctest provider: Bruno Simas Hadlich (https://github.com/brunohadlich)
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Also thanks to Dmitry (https://github.com/LizardWizzard) for finding the problem
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"""
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from __future__ import annotations
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import math
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def prime_sieve(num: int) -> list[int]:
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"""
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Returns a list with all prime numbers up to n.
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>>> prime_sieve(50)
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[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
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>>> prime_sieve(25)
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[2, 3, 5, 7, 11, 13, 17, 19, 23]
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>>> prime_sieve(10)
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[2, 3, 5, 7]
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>>> prime_sieve(9)
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[2, 3, 5, 7]
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>>> prime_sieve(2)
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[2]
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>>> prime_sieve(1)
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[]
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"""
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if num <= 0:
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msg = f"{num}: Invalid input, please enter a positive integer."
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raise ValueError(msg)
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sieve = [True] * (num + 1)
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prime = []
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start = 2
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end = int(math.sqrt(num))
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while start <= end:
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# If start is a prime
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if sieve[start] is True:
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prime.append(start)
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# Set multiples of start be False
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for i in range(start * start, num + 1, start):
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if sieve[i] is True:
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sieve[i] = False
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start += 1
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for j in range(end + 1, num + 1):
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if sieve[j] is True:
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prime.append(j)
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return prime
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if __name__ == "__main__":
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print(prime_sieve(int(input("Enter a positive integer: ").strip())))
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