# 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

The key for solving this is finding an efficient way of getting the numbers in the diagonal. We start with the number one, the next four are at distance 2 of the previous. $3 = 1 + 2$, $5 = 3 + 2$, , $7 = 5 + 2$ and , $9 = 7 + 2$.

For the next 4, they are at distance 4. And for the other group of four they are at distance 6 and so on.

Once identified the method it's a matter of iterating and having the percentage of prime numbers that had appeared.

I decided to use an approach using a generator function.

from CommonFunctions import is_prime from itertools import takewhile def sqr_diag_generator(): cant_tot = 1 cant_primes = 0 num = 1 to_sum = 2 while True: for i in range(0,4): num += to_sum if is_prime(num): cant_primes += 1 to_sum += 2 cant_tot += 4 yield cant_primes * 100 / cant_tot if __name__ == '__main__': result = 3 + sum(2 for i in takewhile(lambda p: p >= 10, sqr_diag_generator())) print("The result is:", result)

The Python file is available for download here.