Problem 58 : Spiral primes

Problem 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

def is_prime(n):
    for i in range(3, int(n**0.5)+1, 2):
        if n % i == 0:
            return False
    return True


i = 0

gap = 1
ratio = 1
primes = []
all_numbers = [1]

while ratio > 0.1:
    for j in range(4):
        i += gap
        present_number = 2*i + 1
        all_numbers.append(present_number)
        if is_prime(present_number):
            primes.append(2*i + 1)
    ratio = float(len(primes))/len(all_numbers)
    gap += 1

print (int((2*i+1)**0.5))

Output

26241