'''
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%?
'''

import time
import math
import numpy as np
from sympy import isprime

def sieve(n):
    assert n > 1

    ns = [True] * n

    for i in range(2, math.ceil(np.sqrt(n))):
        if ns[i]:
            j = pow(i, 2)

            while j < n:
                ns[j] = False
                j = j + i

    return [i for i,val in enumerate(ns) if val][2:]

def corners(n):
    if n == 1:
        return [1]
    
    layer = list(range((n*2-3)**2+1, (n*2-1)**2+1))
    cList = []
    for i in range(4):
        cList.append(layer[-1 - (2*n-2)*i])
    
    return cList

def main():
    print("Hello, this is Patrick")
    t0 = time.time()

    # primes = set(sieve(1000000000))
    numberPrimes = 0
    n = 2

    while True:
        cs = corners(n)
        for c in cs:
            if isprime(c):
                numberPrimes += 1
        if numberPrimes / ((n-1)*4 + 1) < 0.1:
            print(n*2-1, time.time() - t0)
            break
        n += 1

if __name__ == "__main__":
    main()