5 Ways Validate Doubles Domino Board Java

Validating a Doubles Domino Board in Java requires checking if the dominoes are arranged in a way that each domino shares a common number with its adjacent domino, and the numbers on the dominoes are from 0 to 6. Here, we'll explore five different approaches to validate a Doubles Domino Board in Java.

Naturally worded primary topic section with semantic relevance

A Doubles Domino Board consists of a set of dominoes where each domino has two halves, each half representing a number from 0 to 6. The goal is to arrange these dominoes in a line such that each domino shares a common number with its adjacent domino. We’ll start by defining a Domino class to represent each domino, and then implement different validation methods.

public class Domino {
    private int left;
    private int right;

    public Domino(int left, int right) {
        this.left = left;
        this.right = right;
    }

    // Getters and setters
    public int getLeft() {
        return left;
    }

    public void setLeft(int left) {
        this.left = left;
    }

    public int getRight() {
        return right;
    }

    public void setRight(int right) {
        this.right = right;
    }
}

Specific subtopic with natural language phrasing

One approach to validate a Doubles Domino Board is to use a simple iterative method. We’ll iterate through the dominoes and check if each domino shares a common number with its adjacent domino.

public class DoublesDominoBoardValidator {
    public boolean validate(Domino[] dominoes) {
        for (int i = 0; i < dominoes.length - 1; i++) {
            if (dominoes[i].getRight()!= dominoes[i + 1].getLeft()) {
                return false;
            }
        }
        return true;
    }
}

Another approach with technical accuracy

Another approach is to use a recursive method. We’ll recursively check each domino and its adjacent domino to ensure they share a common number.

public class RecursiveDoublesDominoBoardValidator {
    public boolean validate(Domino[] dominoes) {
        return validateRecursive(dominoes, 0);
    }

    private boolean validateRecursive(Domino[] dominoes, int index) {
        if (index >= dominoes.length - 1) {
            return true;
        }
        if (dominoes[index].getRight()!= dominoes[index + 1].getLeft()) {
            return false;
        }
        return validateRecursive(dominoes, index + 1);
    }
}

Using a graph-based approach

We can also use a graph-based approach to validate a Doubles Domino Board. We’ll create a graph where each domino is a node, and two nodes are connected if the corresponding dominoes share a common number.

import java.util.*;

public class GraphDoublesDominoBoardValidator {
    public boolean validate(Domino[] dominoes) {
        Map<Domino, List<Domino>> graph = new HashMap<>();
        for (Domino domino : dominoes) {
            graph.put(domino, new ArrayList<>());
        }
        for (int i = 0; i < dominoes.length - 1; i++) {
            if (dominoes[i].getRight() == dominoes[i + 1].getLeft()) {
                graph.get(dominoes[i]).add(dominoes[i + 1]);
            }
        }
        // Check if the graph is connected
        return isGraphConnected(graph);
    }

    private boolean isGraphConnected(Map<Domino, List<Domino>> graph) {
        // Perform a depth-first search to check if the graph is connected
        Set<Domino> visited = new HashSet<>();
        dfs(graph, graph.keySet().iterator().next(), visited);
        return visited.size() == graph.size();
    }

    private void dfs(Map<Domino, List<Domino>> graph, Domino node, Set<Domino> visited) {
        visited.add(node);
        for (Domino neighbor : graph.get(node)) {
            if (!visited.contains(neighbor)) {
                dfs(graph, neighbor, visited);
            }
        }
    }
}

Dynamic programming approach

We can also use dynamic programming to validate a Doubles Domino Board. We’ll create a 2D table where each cell represents whether it’s possible to arrange the dominoes up to that point.

public class DynamicProgrammingDoublesDominoBoardValidator {
    public boolean validate(Domino[] dominoes) {
        boolean[][] dp = new boolean[dominoes.length][7];
        dp[0][dominoes[0].getLeft()] = true;
        dp[0][dominoes[0].getRight()] = true;
        for (int i = 1; i < dominoes.length; i++) {
            for (int j = 0; j < 7; j++) {
                if (dp[i - 1][j] && (j == dominoes[i].getLeft() || j == dominoes[i].getRight())) {
                    dp[i][dominoes[i].getLeft()] = true;
                    dp[i][dominoes[i].getRight()] = true;
                }
            }
        }
        return dp[dominoes.length - 1][dominoes[dominoes.length - 1].getLeft()];
    }
}

Using a backtracking approach

Finally, we can use a backtracking approach to validate a Doubles Domino Board. We’ll try to arrange the dominoes one by one, and if we reach a point where it’s not possible to arrange the next domino, we’ll backtrack and try a different arrangement.

public class BacktrackingDoublesDominoBoardValidator {
    public boolean validate(Domino[] dominoes) {
        return backtrack(dominoes, 0, new Domino[dominoes.length]);
    }

    private boolean backtrack(Domino[] dominoes, int index, Domino[] arrangement) {
        if (index == dominoes.length) {
            return true;
        }
        for (int i = 0; i < dominoes.length; i++) {
            if (arrangement[index] == null) {
                arrangement[index] = dominoes[i];
                if (index > 0 && arrangement[index - 1].getRight()!= arrangement[index].getLeft()) {
                    continue;
                }
                if (backtrack(dominoes, index + 1, arrangement)) {
                    return true;
                }
                arrangement[index] = null;
            }
        }
        return false;
    }
}