mirror of
https://github.com/TheAlgorithms/Python.git
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716beb32ed
* Improved prime_numbers.py * update prime_numbers.py * Increase the timeit number to 1_000_000 Co-authored-by: Christian Clauss <cclauss@me.com>
122 lines
3.1 KiB
Python
122 lines
3.1 KiB
Python
import math
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from typing import Generator
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def slow_primes(max: 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 + 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: 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 + 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: 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 + 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 > 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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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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# Let's benchmark them side-by-side...
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from timeit import timeit
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print(
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timeit(
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"slow_primes(1_000_000_000_000)",
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setup="from __main__ import slow_primes",
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number=1_000_000,
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)
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)
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print(
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timeit(
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"primes(1_000_000_000_000)",
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setup="from __main__ import primes",
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number=1_000_000,
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)
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)
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print(
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timeit(
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"fast_primes(1_000_000_000_000)",
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setup="from __main__ import fast_primes",
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number=1_000_000,
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)
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)
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