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2104fa7aeb
* fixes #5434 * 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>
85 lines
1.8 KiB
Python
85 lines
1.8 KiB
Python
"""
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Project Euler Problem 7: https://projecteuler.net/problem=7
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10001st prime
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By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we
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can see that the 6th prime is 13.
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What is the 10001st prime number?
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References:
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- https://en.wikipedia.org/wiki/Prime_number
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"""
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from math import sqrt
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def is_prime(number: int) -> bool:
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"""Checks to see if a number is a prime in O(sqrt(n)).
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A number is prime if it has exactly two factors: 1 and itself.
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Returns boolean representing primality of given number (i.e., if the
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result is true, then the number is indeed prime else it is not).
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>>> is_prime(2)
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True
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>>> is_prime(3)
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True
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>>> is_prime(27)
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False
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>>> is_prime(2999)
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True
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>>> is_prime(0)
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False
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>>> is_prime(1)
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False
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"""
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if 1 < number < 4:
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# 2 and 3 are primes
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return True
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elif number < 2 or number % 2 == 0 or number % 3 == 0:
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# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
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return False
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# All primes number are in format of 6k +/- 1
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for i in range(5, int(sqrt(number) + 1), 6):
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if number % i == 0 or number % (i + 2) == 0:
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return False
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return True
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def solution(nth: int = 10001) -> int:
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"""
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Returns the n-th prime number.
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>>> solution(6)
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13
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>>> solution(1)
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2
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>>> solution(3)
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5
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>>> solution(20)
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71
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>>> solution(50)
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229
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>>> solution(100)
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541
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"""
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count = 0
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number = 1
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while count != nth and number < 3:
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number += 1
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if is_prime(number):
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count += 1
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while count != nth:
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number += 2
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if is_prime(number):
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count += 1
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return number
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if __name__ == "__main__":
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print(f"{solution() = }")
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