Python/project_euler/problem_058/sol1.py
Paulo S. G. Ferraz 1400cb86ff
Remove duplicate is_prime related functions (#5892)
* Fixes (#5434)

* Update ciphers.rabin_miller.py
         maths.miller_rabin.py

* Fixing ERROR maths/miller_rabin.py - ModuleNotFoundError and changing project_euler's isPrime to is_prime function names

* Update sol1.py

* fix: try to change to list

* fix pre-commit

* fix capital letters

* Update miller_rabin.py

* Update rabin_miller.py

Co-authored-by: John Law <johnlaw.po@gmail.com>
2022-04-09 01:40:45 +08:00

88 lines
2.3 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.
"""
from math import isqrt
def is_prime(number: int) -> int:
"""
Returns whether the given number is prime or not
>>> is_prime(1)
0
>>> is_prime(17)
1
>>> is_prime(10000)
0
"""
if number == 1:
return 0
if number % 2 == 0 and number > 2:
return 0
for i in range(3, isqrt(number) + 1, 2):
if number % i == 0:
return 0
return 1
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()