Spiral Primes

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 2643 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%?

This problem actually belongs to a class of problems that require some use of programming to be effectively solved. We can’t solve this efficiently just by pure manual calculation. However, I can provide an algorithm in pseudo-code for how to solve this, and one brief example on how the algorithm works.

Algorithm:

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1. Initialize prime count (PC) = 3, total count (TC) = 5, layer = 1.
Note: The initial values are given based on the problem statement from the second layer (3×3)

2. While PC / TC >= 10 %:
For i from 1 to 4:
– Compute the current number (CN) in the diagonal, CN = (2*layer + 1)^2 – i * (2*layer).
– If CN is a prime number, increment PC by 1.
– Increment layer by 1.
– Update TC = TC + 4.

3. The side length of the square spiral for which the ratio of primes along both diagonals falls below 10% is 2*layer + 1
“`

Brief Explanation:
To generate the numbers along the diagonals of the spiral, consider the following: The largest number on layer n is (2n+1)^2, moving anticlockwise we find the other corners at (2n+1)^2 – 2n, (2n+1)^2 – 4n and (2n+1)^2 – 6n.

Remember to check if the numbers generated are primes and stop the program once the ratio of primes falls below 10%.

Now you just need to implement this in any programming language of your choice to get the answer.

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