Python/project_euler/problem_007/sol1.py
Nikos Giachoudis 2104fa7aeb
Unify O(sqrt(N)) is_prime functions under project_euler ()
* fixes 

* fixes broken solution

* removes assert

* removes assert

* Apply suggestions from code review

Co-authored-by: John Law <johnlaw.po@gmail.com>

* Update project_euler/problem_003/sol1.py

Co-authored-by: John Law <johnlaw.po@gmail.com>
2022-09-14 09:40:04 +01:00

85 lines
1.8 KiB
Python

"""
Project Euler Problem 7: https://projecteuler.net/problem=7
10001st prime
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we
can see that the 6th prime is 13.
What is the 10001st prime number?
References:
- https://en.wikipedia.org/wiki/Prime_number
"""
from math import sqrt
def is_prime(number: int) -> bool:
"""Checks to see if a number is a prime in O(sqrt(n)).
A number is prime if it has exactly two factors: 1 and itself.
Returns boolean representing primality of given number (i.e., if the
result is true, then the number is indeed prime else it is not).
>>> is_prime(2)
True
>>> is_prime(3)
True
>>> is_prime(27)
False
>>> is_prime(2999)
True
>>> is_prime(0)
False
>>> is_prime(1)
False
"""
if 1 < number < 4:
# 2 and 3 are primes
return True
elif number < 2 or number % 2 == 0 or number % 3 == 0:
# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
return False
# All primes number are in format of 6k +/- 1
for i in range(5, int(sqrt(number) + 1), 6):
if number % i == 0 or number % (i + 2) == 0:
return False
return True
def solution(nth: int = 10001) -> int:
"""
Returns the n-th prime number.
>>> solution(6)
13
>>> solution(1)
2
>>> solution(3)
5
>>> solution(20)
71
>>> solution(50)
229
>>> solution(100)
541
"""
count = 0
number = 1
while count != nth and number < 3:
number += 1
if is_prime(number):
count += 1
while count != nth:
number += 2
if is_prime(number):
count += 1
return number
if __name__ == "__main__":
print(f"{solution() = }")