""" 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. """ def isprime(d: int) -> int: """ returns whether the given digit is prime or not >>> isprime(1) 0 >>> isprime(17) 1 >>> isprime(10000) 0 """ if d == 1: return 0 i = 2 while i * i <= d: if d % i == 0: return 0 i = i + 1 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 = primes + isprime(i) j = j + 2 return j if __name__ == "__main__": import doctest doctest.testmod()