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58_SpiralPrimes.py
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58_SpiralPrimes.py
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'''
Created on Dec 1, 2023
Starting with
and spiralling anticlockwise in the following way, a square spiral with side length
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
out of the numbers lying along both diagonals are prime; that is, a ratio of 8/3 ~ 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%?
@author: mstackpo
'''
from datetime import datetime
from euler import is_prime
if __name__ == '__main__':
start = datetime.now()
diagonal = [1]
n = 0
prime_count = 0
diagonal_count = 1
ratio = prime_count / diagonal_count
while ratio >= 0.1 or n == 0:
n += 1
new_prime_candidates = [(2*n+1)**2 - i * 2 * n for i in range(1, 4)]
# diagonal += new_diagonal_elements
prime_count += sum(is_prime(d) for d in new_prime_candidates)
diagonal_count += 4
ratio = prime_count / diagonal_count
print(f'{ratio}')
print(f'{2 * n + 1}')
end = datetime.now()
print(f'\nruntime = {end - start}')