## 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.

3736 35 34 33 3231

381716 15 141330

39 1854312 29

40 19 6 1 2 11 28

41 2078 9 10 27

42 21 22 23 24 25 26

4344 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.

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