Python/maths/prime_numbers.py
Leoriem-code 716beb32ed
Improved prime_numbers.py (#5592)
* Improved prime_numbers.py

* update prime_numbers.py

* Increase the timeit number to 1_000_000

Co-authored-by: Christian Clauss <cclauss@me.com>
2021-10-26 09:21:44 +02:00

122 lines
3.1 KiB
Python

import math
from typing import Generator
def slow_primes(max: 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(10000))[-1]
9973
"""
numbers: Generator = (i for i in range(1, (max + 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: 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(10000))[-1]
9973
"""
numbers: Generator = (i for i in range(1, (max + 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: 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(10000))[-1]
9973
"""
numbers: Generator = (i for i in range(1, (max + 1), 2))
# It's useless to test even numbers as they will not be prime
if max > 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
if __name__ == "__main__":
number = int(input("Calculate primes up to:\n>> ").strip())
for ret in primes(number):
print(ret)
# Let's benchmark them side-by-side...
from timeit import timeit
print(
timeit(
"slow_primes(1_000_000_000_000)",
setup="from __main__ import slow_primes",
number=1_000_000,
)
)
print(
timeit(
"primes(1_000_000_000_000)",
setup="from __main__ import primes",
number=1_000_000,
)
)
print(
timeit(
"fast_primes(1_000_000_000_000)",
setup="from __main__ import fast_primes",
number=1_000_000,
)
)