## Problem 58: Spiral primes

*Difficulty: Easy

Problem:

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%?

How to solve:

This problem is kind of the same with Problem 28
We got 4 diagonals in total, let name cur1,cur2,cur3,cur4 , they start with 5,3,9,7, respectively.
Next number of cur1 = 5 + (4 + 8*i)
Next number of cur2 = 3 + (2 + 8*i)
Next number of cur3 = 9 + (8+ 8*i)
Next number of cur4 = 7 + (6+8*i)
i start from 0, the size of spiral start from 3

Implementation: Python 2

Before we can start coding, we must have a function to check primality, here’s mine:

Then we initialize everything as described above:

Now, we can iterate all spirals with size 3,5,7,…… until we got the ratio under 10%

Solution took 6.35 seconds , which is so damn long in my opinion. I don’t know why, maybe because Python 2 is still too slow.