Project Euler – Problem 58

Problem 58:
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%?

import java.math.BigInteger;
 
class runner
{
	public static void main (String[] args) throws java.lang.Exception
	{
		long time = System.currentTimeMillis();
 
		int diff_p = 0; int prev = 0;
		int i=1;
 
		int numOfPrime = 0;
		double totalDiag = 0;
		while(true){
			int difference = 2*diff_p;//from 8*diff_p/4
			for(int j=1;j<5;j++){
				BigInteger x = new BigInteger(""+(difference*j+prev));
				if(x.isProbablePrime(100)){
					numOfPrime++;
				}
				totalDiag++;
			}
			double cur = numOfPrime/totalDiag;
			if(numOfPrime> 0 && cur < .1){
				break;
			}
 
			prev = i*i;
			diff_p++;
			i+=2;
		}
 
		System.out.println(i);
		System.out.println("time:"+(System.currentTimeMillis()-time));
	}
}


Note: Got lazy and ended up using BigInteger for primality test.