Find all possible dominoes chains with recursion and backtracking [closed]

Answer 1

The process is very simple: you start with a collection of dominoes D, and an empty chain C.

for each domino in the collection:
    see if it can be added to the chain (either the chain is empty, or the first 
    number is the same as the second number of the last domino in the chain.
    if it can, 
        append the domino to the chain,
        then print this new chain as it is a solution,
        then call recursively with D - {domino} and C + {domino}

    repeat with the flipped domino

Java code:

public class Domino {
    public final int a;
    public final int b;

    public Domino(int a, int b) {
        this.a = a;
        this.b = b;
    }

    public Domino flipped() {
        return new Domino(b, a);
    }

    @Override
    public String toString() {
        return "[" + a + "/" + b + "]";
    }
}

Algorithm:

private static void listChains(List<Domino> chain, List<Domino> list) {
    for (int i = 0; i < list.size(); ++i) {
        Domino dom = list.get(i);
        if (canAppend(dom, chain)) {
            chain.add(dom);
            System.out.println(chain);
            Domino saved = list.remove(i);
            listChains(chain, list);
            list.add(i, saved);
            chain.remove(chain.size()-1);
        }
        dom = dom.flipped();
        if (canAppend(dom, chain)) {
            chain.add(dom);
            System.out.println(chain);
            Domino saved = list.remove(i);
            listChains(chain, list);
            list.add(i, saved);
            chain.remove(chain.size()-1);
        }
    }
}

private static boolean canAppend(Domino dom, List<Domino> to) {
    return to.isEmpty() || to.get(to.size()-1).b == dom.a;
}

Your example:

public static void main(String... args) {
    List<Domino> list = new ArrayList<>();
    // [3/4] [5/6] [1/4] [1/6]
    list.add(new Domino(3, 4));
    list.add(new Domino(5, 6));
    list.add(new Domino(1, 4));
    list.add(new Domino(1, 6));

    List<Domino> chain = new ArrayList<>();
    listChains(chain, list);
}