## Problem:

Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

Triangle P3,n=n(n+1)/2 1, 3, 6, 10, 15, ...
Square P4,n=n2 1, 4, 9, 16, 25, ...
Pentagonal P5,n=n(3n[−]1)/2 1, 5, 12, 22, 35, ...
Hexagonal P6,n=n(2n[−]1) 1, 6, 15, 28, 45, ...
Heptagonal P7,n=n(5n[−]3)/2 1, 7, 18, 34, 55, ...
Octagonal P8,n=n(3n[−]2) 1, 8, 21, 40, 65, ...

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.

1. The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
2. Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=2882), is represented by a different number in the set.
3. This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.

70600674

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 11 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p011 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p011().run());	}			public String run() {		int max = -1;		max = Math.max(maxProduct(1, 0), max);		max = Math.max(maxProduct(0, 1), max);		max = Math.max(maxProduct(1, 1), max);		max = Math.max(maxProduct(1, -1), max);		return Integer.toString(max);	}			private static int maxProduct(int dx, int dy) {		int max = -1;		for (int y = 0; y < SQUARE.length; y++) {			for (int x = 0; x < SQUARE[y].length; x++)				max = Math.max(product(x, y, dx, dy, 4), max);		}		return max;	}			private static int product(int x, int y, int dx, int dy, int n) {		// First endpoint is assumed to be in bounds. Check if second endpoint is in bounds.		if (!isInBounds(x + (n - 1) * dx, y + (n - 1) * dy))			return -1;				int prod = 1;		for (int i = 0; i < n; i++, x += dx, y += dy)			prod *= SQUARE[y][x];		return prod;	}			private static boolean isInBounds(int x, int y) {		return 0 <= y && y < SQUARE.length && 0 <= x && x < SQUARE[y].length;	}			private static int[][] SQUARE = {		{ 8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91, 8},		{49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00},		{81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65},		{52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91},		{22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},		{24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50},		{32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},		{67,26,20,68,02,62,12,20,95,63,94,39,63, 8,40,91,66,49,94,21},		{24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},		{21,36,23, 9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},		{78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14, 9,53,56,92},		{16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57},		{86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58},		{19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40},		{04,52, 8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},		{88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},		{04,42,16,73,38,25,39,11,24,94,72,18, 8,46,29,32,40,62,76,36},		{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},		{20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54},		{01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48}	};	}