Statement

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

The key for solving this is finding an efficient way of getting the numbers in the diagonal. We start with the number one, the next four are at distance 2 of the previous. $3 = 1 + 2$, $5 = 3 + 2$, , $7 = 5 + 2$ and , $9 = 7 + 2$.
For the next 4, they are at distance 4. And for the other group of four they are at distance 6 and so on.

Once identified the method it's a matter of iterating and having the percentage of prime numbers that had appeared.

I decided to use an approach using a generator function.

from CommonFunctions import is_prime
from itertools import takewhile
 
def sqr_diag_generator():
    cant_tot = 1
    cant_primes = 0
    num = 1
    to_sum = 2
    while True:
        for i in range(0,4):
            num += to_sum
            if is_prime(num):
                cant_primes += 1
        to_sum += 2
        cant_tot += 4
 
        yield cant_primes * 100 / cant_tot 
 
if __name__ == '__main__':
    result = 3 + sum(2 for i in takewhile(lambda p: p >= 10, sqr_diag_generator()))
    print("The result is:", result)

The Python file is available for download here.