Python/maths/prime_numbers.py
Muhammad Umer Farooq cffdf99c55
Updated prime_numbers.py testcases. (#9851)
* Updated prime_numbers.py testcases.

* revert __main__ code.
2023-10-05 20:44:55 +02:00

110 lines
3.0 KiB
Python

import math
from collections.abc import Generator
def slow_primes(max_n: int) -> Generator[int, None, None]:
"""
Return a list of all primes numbers up to max.
>>> list(slow_primes(0))
[]
>>> list(slow_primes(-1))
[]
>>> list(slow_primes(-10))
[]
>>> list(slow_primes(25))
[2, 3, 5, 7, 11, 13, 17, 19, 23]
>>> list(slow_primes(11))
[2, 3, 5, 7, 11]
>>> list(slow_primes(33))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
>>> list(slow_primes(1000))[-1]
997
"""
numbers: Generator = (i for i in range(1, (max_n + 1)))
for i in (n for n in numbers if n > 1):
for j in range(2, i):
if (i % j) == 0:
break
else:
yield i
def primes(max_n: int) -> Generator[int, None, None]:
"""
Return a list of all primes numbers up to max.
>>> list(primes(0))
[]
>>> list(primes(-1))
[]
>>> list(primes(-10))
[]
>>> list(primes(25))
[2, 3, 5, 7, 11, 13, 17, 19, 23]
>>> list(primes(11))
[2, 3, 5, 7, 11]
>>> list(primes(33))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
>>> list(primes(1000))[-1]
997
"""
numbers: Generator = (i for i in range(1, (max_n + 1)))
for i in (n for n in numbers if n > 1):
# only need to check for factors up to sqrt(i)
bound = int(math.sqrt(i)) + 1
for j in range(2, bound):
if (i % j) == 0:
break
else:
yield i
def fast_primes(max_n: int) -> Generator[int, None, None]:
"""
Return a list of all primes numbers up to max.
>>> list(fast_primes(0))
[]
>>> list(fast_primes(-1))
[]
>>> list(fast_primes(-10))
[]
>>> list(fast_primes(25))
[2, 3, 5, 7, 11, 13, 17, 19, 23]
>>> list(fast_primes(11))
[2, 3, 5, 7, 11]
>>> list(fast_primes(33))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
>>> list(fast_primes(1000))[-1]
997
"""
numbers: Generator = (i for i in range(1, (max_n + 1), 2))
# It's useless to test even numbers as they will not be prime
if max_n > 2:
yield 2 # Because 2 will not be tested, it's necessary to yield it now
for i in (n for n in numbers if n > 1):
bound = int(math.sqrt(i)) + 1
for j in range(3, bound, 2):
# As we removed the even numbers, we don't need them now
if (i % j) == 0:
break
else:
yield i
def benchmark():
"""
Let's benchmark our functions side-by-side...
"""
from timeit import timeit
setup = "from __main__ import slow_primes, primes, fast_primes"
print(timeit("slow_primes(1_000_000_000_000)", setup=setup, number=1_000_000))
print(timeit("primes(1_000_000_000_000)", setup=setup, number=1_000_000))
print(timeit("fast_primes(1_000_000_000_000)", setup=setup, number=1_000_000))
if __name__ == "__main__":
number = int(input("Calculate primes up to:\n>> ").strip())
for ret in primes(number):
print(ret)
benchmark()