58_spiral_primes.py

"""
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%?
"""

# stuff to note
# lower right diagonal - squares of odd integers (3^2, 5^2)
# lower left diagonal - squares of even integers + that even integer + 1 (2^2+3)
# upper left diagonal - squares of even integers + 1 (2^2 + 1)
# upper right diagonal - squares of odd integers + that odd integer + 1 (1^2 + 2)
import time

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

start = time.time()
nprimes = 0
ntotal = 1

row = 1

while True:
    new_entries = (row**2 + row + 1, (row+1)**2 + 1, (row+1)**2 + row + 2)
    nprimes += len(tuple(filter(None, map(is_prime, new_entries))))
    ntotal += 4
    if nprimes/ntotal < 0.1:
        break
    row += 2
    
print(row+2)
print(time.time() - start)