**Problem 58:**

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

3736 35 34 33 3231381716 15 141330 39 1854312 29 40 19 6 1 2 11 28 41 2078 9 10 27 42 21 22 23 24 25 264344 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%?

**Idea:**

Even though I already have a spiral generating method used in Problem 28, I don’t need to generate anything: the diagonals are easily recognized by a pattern of numbers equally spaced apart per layer. Determine which numbers are on a diagonal, check to see if it’s prime, and keep track of the total numbers on the diagonal. While the ratio of primes along the diagonals to the total number of numbers along the diagonal is greater than 10%, keep adding a new layer.

int answer = -1; boolean LESS_THAN_10_PERCENT = false; double numberOfPrimes = 0; int totalNumbers = 1; int increment = 0; int start = 1; while(!LESS_THAN_10_PERCENT) { increment += 2; for(int i = 0; i < 4; i++) { start += increment; totalNumbers++; if(EulerUtils.isProbablePrime(start)) { numberOfPrimes++; } } double percentOfPrimes = numberOfPrimes/totalNumbers; if(percentOfPrimes < 0.1) { answer = increment+1; System.out.println("Calculated spiral of width: "+(increment+1)+ " and ratio of: "+percentOfPrimes); LESS_THAN_10_PERCENT = true; } }