Spiral primes

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%?

Idea

Naive solution.

In [1]:
import sys, os; sys.path.append(os.path.abspath('..'))
from timer import timethis
from math import sqrt
In [2]:
def generate_corners():
    layer = 1
    corner_start = 3
    while True:
        yield range(corner_start, corner_start+8*layer, 2*layer)
        corner_start += 6 * layer + (layer+1)*2
        layer += 1
In [3]:
g = generate_corners()
for _ in range(3):
    print(list(next(g)))

[3, 5, 7, 9]
[13, 17, 21, 25]
[31, 37, 43, 49]
In [4]:
def is_prime(n):
    return all([n % d != 0 for d in range(3, int(sqrt(n))+1, 2)])
In [5]:
is_prime(7)
Out[5]:
True
In [6]:
is_prime(15)
Out[6]:
False
In [7]:
@timethis
def solve():
    all_corners = 1
    prime_corners = 0
    for i, corners in enumerate(generate_corners(), 1):
        all_corners += 4
        prime_corners += sum(map(is_prime, corners))
        if prime_corners / all_corners < 0.1:
            return 2*i+1
In [8]:
solve()

Run for 32.688 seconds
Out[8]:
26241