Project Euler > Problem 128 > Hexagonal tile differences (Java Solution)

Problem:

A hexagonal tile with number 1 is surrounded by a ring of six hexagonal tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an anti-clockwise direction.

New rings are added in the same fashion, with the next rings being numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows the first three rings.

By finding the difference between tile n and each its six neighbours we shall define PD(n) to be the number of those differences which are prime.

For example, working clockwise around tile 8 the differences are 12, 29, 11, 6, 1, and 13. So PD(8) = 3.

In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and 10, hence PD(17) = 2.

It can be shown that the maximum value of PD(n) is 3.

If all of the tiles for which PD(n) = 3 are listed in ascending order to form a sequence, the 10th tile would be 271.

Find the 2000th tile in this sequence.

Solution:

669171001

Code:
The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 28
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p028 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p028().run());
	}
	
	
	/* 
	 * From the diagram, let's observe the four corners of an n * n square (where n is odd).
	 * It's not hard to convince yourself that the top right corner always has the value n^2.
	 * Working counterclockwise (backwards), the top left corner has the value n^2 - (n - 1),
	 * the bottom left corner has the value n^2 - 2(n - 1), and the bottom right is n^2 - 3(n - 1).
	 * Putting it all together, this outermost ring contributes 4n^2 - 6(n - 1) to the final sum.
	 * 
	 * Incidentally, the closed form of this sum is (4m^3 + 3m^2 + 8m - 9) / 6, where m = size.
	 */
	private static final int SIZE = 1001;  // Must be odd
	
	public String run() {
		long sum = 1;  // Special case for size 1
		for (int n = 3; n <= SIZE; n += 2)
			sum += 4 * n * n - 6 * (n - 1);
		return Long.toString(sum);
	}
	
}