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5c92b7390e
* prime_numbers.py: Tighten up the benchmarks
* updating DIRECTORY.md
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
110 lines
3.0 KiB
Python
110 lines
3.0 KiB
Python
import math
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from collections.abc import Generator
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def slow_primes(max_n: int) -> Generator[int, None, None]:
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"""
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Return a list of all primes numbers up to max.
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>>> list(slow_primes(0))
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[]
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>>> list(slow_primes(-1))
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[]
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>>> list(slow_primes(-10))
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[]
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>>> list(slow_primes(25))
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[2, 3, 5, 7, 11, 13, 17, 19, 23]
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>>> list(slow_primes(11))
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[2, 3, 5, 7, 11]
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>>> list(slow_primes(33))
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[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
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>>> list(slow_primes(10000))[-1]
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9973
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"""
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numbers: Generator = (i for i in range(1, (max_n + 1)))
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for i in (n for n in numbers if n > 1):
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for j in range(2, i):
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if (i % j) == 0:
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break
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else:
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yield i
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def primes(max_n: int) -> Generator[int, None, None]:
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"""
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Return a list of all primes numbers up to max.
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>>> list(primes(0))
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[]
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>>> list(primes(-1))
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[]
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>>> list(primes(-10))
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[]
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>>> list(primes(25))
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[2, 3, 5, 7, 11, 13, 17, 19, 23]
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>>> list(primes(11))
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[2, 3, 5, 7, 11]
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>>> list(primes(33))
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[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
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>>> list(primes(10000))[-1]
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9973
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"""
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numbers: Generator = (i for i in range(1, (max_n + 1)))
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for i in (n for n in numbers if n > 1):
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# only need to check for factors up to sqrt(i)
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bound = int(math.sqrt(i)) + 1
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for j in range(2, bound):
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if (i % j) == 0:
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break
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else:
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yield i
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def fast_primes(max_n: int) -> Generator[int, None, None]:
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"""
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Return a list of all primes numbers up to max.
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>>> list(fast_primes(0))
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[]
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>>> list(fast_primes(-1))
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[]
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>>> list(fast_primes(-10))
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[]
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>>> list(fast_primes(25))
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[2, 3, 5, 7, 11, 13, 17, 19, 23]
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>>> list(fast_primes(11))
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[2, 3, 5, 7, 11]
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>>> list(fast_primes(33))
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[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
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>>> list(fast_primes(10000))[-1]
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9973
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"""
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numbers: Generator = (i for i in range(1, (max_n + 1), 2))
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# It's useless to test even numbers as they will not be prime
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if max_n > 2:
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yield 2 # Because 2 will not be tested, it's necessary to yield it now
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for i in (n for n in numbers if n > 1):
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bound = int(math.sqrt(i)) + 1
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for j in range(3, bound, 2):
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# As we removed the even numbers, we don't need them now
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if (i % j) == 0:
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break
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else:
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yield i
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def benchmark():
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"""
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Let's benchmark our functions side-by-side...
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"""
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from timeit import timeit
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setup = "from __main__ import slow_primes, primes, fast_primes"
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print(timeit("slow_primes(1_000_000_000_000)", setup=setup, number=1_000_000))
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print(timeit("primes(1_000_000_000_000)", setup=setup, number=1_000_000))
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print(timeit("fast_primes(1_000_000_000_000)", setup=setup, number=1_000_000))
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if __name__ == "__main__":
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number = int(input("Calculate primes up to:\n>> ").strip())
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for ret in primes(number):
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print(ret)
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benchmark()
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