euler_p58.py

#Euler - problem 58
"""
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: diagonal elements are 1, 3, 5, 7, 9, 13, 17, 21, 25, 31, 37, ...
# therefore, increment the progression by 2 after every four numbers

def isPrime(num):
    if num in [2, 3, 5, 7, 11, 13, 17, 19, 23]:
        return True
    elif num%2 == 0: return False
    elif num%3 == 0: return False
    else:
        i = 5
        k = 2
        while i**2 <= num:
            if num%i == 0: return False
            i += k
            k = 6 - k
        return True

ratio = 0
nprimes, primeratio = 0, 1
spiralsize = 1
while primeratio > 0.1:
    spiralsize += 2
    ratio += 2
    nprimes += sum([isPrime((spiralsize - 2)**2 + ratio*i) for i in range(1, 4)])
    primeratio = nprimes/(2*spiralsize - 1)

print(str(spiralsize) + ' : ' + str(primeratio))