// CSE 143, Summer 2012
// Uses recursive backtracking to solve the problem of whether
// a group of dominoes can be lined up to produce a chain that starts and ends
// with a particular number.

import java.util.*;   // for ArrayList

public class SolveDominoes {
    public static void main(String[] args) {
        // [(1|4), (2|6), (4|5), (1|5), (3|5)]
        List<Domino> dominoes = new ArrayList<Domino>();
        dominoes.add(new Domino(1, 4));
        dominoes.add(new Domino(2, 6));
        dominoes.add(new Domino(4, 5));
        dominoes.add(new Domino(1, 5));
        dominoes.add(new Domino(3, 5));
        System.out.println(hasChain(dominoes, 5, 5));   // true
        System.out.println(hasChain(dominoes, 1, 5));   // true
        System.out.println(hasChain(dominoes, 1, 3));   // true
        System.out.println(hasChain(dominoes, 1, 6));   // false
        System.out.println(hasChain(dominoes, 1, 2));   // false
    }
    
	// Returns true if a chain of dominoes can be made from the given list
	// that starts and ends with the given numbers. Otherwise returns false.
	// Assumes that if start == end, it's trivial, so we should return true.
	// Precondition: 0 <= start,end <= 6
	// Precondition: dominoes != null
    public static boolean hasChain(List<Domino> dominoes, int start, int end) {
   		
			if (start == end) {
				return true;         // base case
			} else {
				for (int i = 0; i < dominoes.size(); i++) {
				
					//choose: pick a domino to start the chain
					Domino d = dominoes.remove(i);
					
					//explore
					if (d.first() == start) {
						if (hasChain(dominoes, d.second(), end)) {
							return true;
						}
					} else if (d.second() == start) {
						if (hasChain(dominoes, d.first(), end)) {
							return true;
						}
					}
					
					//un-choose
					dominoes.add(i, d);
				}
				return false;    // tried all dominoes and none worked
			}
    }
}