mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-21 02:30:15 +00:00
bc8df6de31
* [pre-commit.ci] pre-commit autoupdate updates: - [github.com/astral-sh/ruff-pre-commit: v0.2.2 → v0.3.2](https://github.com/astral-sh/ruff-pre-commit/compare/v0.2.2...v0.3.2) - [github.com/pre-commit/mirrors-mypy: v1.8.0 → v1.9.0](https://github.com/pre-commit/mirrors-mypy/compare/v1.8.0...v1.9.0) * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci --------- Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
106 lines
2.7 KiB
Python
106 lines
2.7 KiB
Python
"""
|
|
Project Euler Problem 58:https://projecteuler.net/problem=58
|
|
|
|
|
|
Starting with 1 and spiralling anticlockwise in the following way,
|
|
a square spiral with side length 7 is formed.
|
|
|
|
37 36 35 34 33 32 31
|
|
38 17 16 15 14 13 30
|
|
39 18 5 4 3 12 29
|
|
40 19 6 1 2 11 28
|
|
41 20 7 8 9 10 27
|
|
42 21 22 23 24 25 26
|
|
43 44 45 46 47 48 49
|
|
|
|
It is interesting to note that the odd squares lie along the bottom right
|
|
diagonal ,but what is more interesting is that 8 out of the 13 numbers
|
|
lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
|
|
|
|
If one complete new layer is wrapped around the spiral above,
|
|
a square spiral with side length 9 will be formed.
|
|
If this process is continued,
|
|
what is the side length of the square spiral for which
|
|
the ratio of primes along both diagonals first falls below 10%?
|
|
|
|
Solution: We have to find an odd length side for which square falls below
|
|
10%. With every layer we add 4 elements are being added to the diagonals
|
|
,lets say we have a square spiral of odd length with side length j,
|
|
then if we move from j to j+2, we are adding j*j+j+1,j*j+2*(j+1),j*j+3*(j+1)
|
|
j*j+4*(j+1). Out of these 4 only the first three can become prime
|
|
because last one reduces to (j+2)*(j+2).
|
|
So we check individually each one of these before incrementing our
|
|
count of current primes.
|
|
|
|
"""
|
|
|
|
import math
|
|
|
|
|
|
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.
|
|
|
|
>>> is_prime(0)
|
|
False
|
|
>>> is_prime(1)
|
|
False
|
|
>>> is_prime(2)
|
|
True
|
|
>>> is_prime(3)
|
|
True
|
|
>>> is_prime(27)
|
|
False
|
|
>>> is_prime(87)
|
|
False
|
|
>>> is_prime(563)
|
|
True
|
|
>>> is_prime(2999)
|
|
True
|
|
>>> is_prime(67483)
|
|
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(math.sqrt(number) + 1), 6):
|
|
if number % i == 0 or number % (i + 2) == 0:
|
|
return False
|
|
return True
|
|
|
|
|
|
def solution(ratio: float = 0.1) -> int:
|
|
"""
|
|
Returns the side length of the square spiral of odd length greater
|
|
than 1 for which the ratio of primes along both diagonals
|
|
first falls below the given ratio.
|
|
>>> solution(.5)
|
|
11
|
|
>>> solution(.2)
|
|
309
|
|
>>> solution(.111)
|
|
11317
|
|
"""
|
|
|
|
j = 3
|
|
primes = 3
|
|
|
|
while primes / (2 * j - 1) >= ratio:
|
|
for i in range(j * j + j + 1, (j + 2) * (j + 2), j + 1):
|
|
primes += is_prime(i)
|
|
j += 2
|
|
return j
|
|
|
|
|
|
if __name__ == "__main__":
|
|
import doctest
|
|
|
|
doctest.testmod()
|