bancaldo™

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%?

python:

import time

ts = time.time()

_MEMO = {3:3, 5:5, 7:8}

def is_prime(num):
    if num <= 1: return False
    elif num == 2: return True
    elif num % 2 == 0: return False
    else:
        d = 3
        r = int(num**0.5)
        while d <= r:
            if num % d == 0: return False
            d += 2
        return True

def get_spiral_primes(side):
    count = 0
    for i in xrange(1,4):
        if is_prime(side**2 - i*side + i):
            count += 1
    return count

def spiral(side):
    _MEMO[side] = _MEMO[side-2] + get_spiral_primes(side)
    return _MEMO[side]

for i in xrange(7, 1000000, 2):
    if spiral(i)/float(i * 2 - 1) < 0.1:
        break

print "PE58: %s\nelapsed time: %ssec" %(i, time.time() - ts)