使用递归和回溯找到所有可能的多米诺骨牌链

Question

我正在做一个挑战，我需要找到线性多米诺骨牌链的所有可能性。我了解递归的原理，但不知道如何将其转化为代码。如果有人可以通过简单的步骤解释问题（解决方案），我可以按照这些步骤进行编码。

Example:

Tiles : [3/4] [5/6] [1/4] [1/6]

Possible chain : [3/4]-[4/1]-[1/6]-[6/5]

允许翻转方块。（切换数字）

Answer 1

过程非常简单：从一组多米诺骨牌 D 和一个空链 C 开始。

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代码：

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 + "]";
    }
}

算法：

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;
}

你的例子：

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);
}

使用递归和回溯找到所有可能的多米诺骨牌链

Find all possible dominoes chains with recursion and backtracking

java

algorithm

recursion

backtracking