1. Introduction

The Chess Knight Problem is about finding the shortest path for a knight on a chessboard to move from one cell to another. Given the nature of the knight’s movement (L-shape), it’s crucial to determine the shortest sequence of moves it needs to reach a specified destination.

2. Program Steps

1. Initialize a queue to process each cell.

2. Mark the source cell as visited.

3. While the queue is not empty, dequeue the front cell and enqueue its unvisited neighbors.

4. Return the distance for the destination cell once reached.

3. Code Program

import java.util.LinkedList;
import java.util.Queue;

class Cell {
    int x, y, dist;

    Cell(int x, int y, int dist) {
        this.x = x;
        this.y = y;
        this.dist = dist;
    }
}

public class ChessKnightProblem {

    private static final int[] row = { -2, -1, 1, 2, -2, -1, 1, 2 };
    private static final int[] col = { -1, -2, -2, -1, 1, 2, 2, 1 };

    private static boolean isValid(int x, int y, boolean[][] visited) {
        return x >= 0 && y >= 0 && x < 8 && y < 8 && !visited[x][y];
    }

    public static int BFS(Cell src, Cell dest) {
        boolean[][] visited = new boolean[8][8];

        Queue<Cell> q = new LinkedList<>();
        q.add(src);

        while (!q.isEmpty()) {
            Cell cell = q.poll();

            int x = cell.x;
            int y = cell.y;
            int dist = cell.dist;

            if (x == dest.x && y == dest.y)
                return dist;

            for (int i = 0; i < 8; i++) {
                int newX = x + row[i];
                int newY = y + col[i];

                if (isValid(newX, newY, visited)) {
                    visited[newX][newY] = true;
                    q.add(new Cell(newX, newY, dist + 1));
                }
            }
        }

        return Integer.MAX_VALUE;
    }

    public static void main(String[] args) {
        Cell source = new Cell(0, 0, 0);
        Cell dest = new Cell(7, 7, 0);

        System.out.println("Shortest path is: " + BFS(source, dest));
    }
}

Output:

Shortest path is: 6

4. Step By Step Explanation

1. The Cell class represents a cell on the chessboard with its x and y coordinates and the distance from the source.

2. row and col arrays define the possible movements of the knight in all 8 possible directions.

3. isValid checks whether a move is valid, i.e., within the boundaries of the board and not previously visited.

4. BFS (Breadth-First Search) is used to find the shortest path. We start from the source cell and explore its neighbors, then the neighbors of its neighbors, and so on.

5. The function returns the dist value when the destination cell is dequeued, which represents the minimum number of moves required by the knight to reach the destination.