java - Knight's Tour 代码陷入无限循环，没有达成解决方案

Question

我对 Knight's Tour 的递归回溯方法遇到了无限循环。起初，我认为这个问题通常可能会花费这么多时间，但有些解决方案可以立即解决。请告诉我的代码有什么问题。

package io.github.thegeekybaniya.InterviewPrep.TopTopics.Backtracking;

import java.util.Arrays;

public class KnightsTour {
    private static int counter=0;

    public static void main(String[] args) {

        knightsTour(8);
    }

    private static void knightsTour(int i) {
        int[][] board = new int[i][i];
        for (int[] arr :
                board) {
            Arrays.fill(arr, -1);

        }
        board[0][0] = 0;
        knightsTour(board,0,1);

    }

    private static boolean knightsTour(int[][] board, int cellno, int stepno) {
        if (stepno == board.length * board.length) {
            printBoard(board);
            return true;
        }

        int[][] dirs = {
                {1, 2}, {1, -2}, {-1, 2}, {-1, -2}, {2, 1}, {2, -1}, {-2, 1}, {-2, -1}
        };
        int row = cellno / board.length, col = cellno % board.length;
        for (int i = 0; i < dirs.length; i++) {
            int r = dirs[i][0] + row;
            int c = dirs[i][1] + col;
            if (isSafe(board, r, c)&&board[r][c]==-1) {
                int ncell = r * board.length + c;
                board[r][c] = stepno;
                if (knightsTour(board, ncell, stepno + 1)) {
                    return true;
                } else {
                    board[r][c] = -1;
                }
            }
        }


        return false;
    }

    private static boolean isSafe(int[][] board, int r, int c) {

        return r >= 0 && c >= 0 && r < board.length && c < board.length;
    }

    private static void printBoard(int[][] board) {
        System.out.println(++counter);
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board.length; j++) {
                System.out.print(board[i][j]+" ");

            }
            System.out.println();
        }
    }

}

Answer 1

您的代码中没有错误，只是由于搜索空间巨大，蛮力方法速度较慢。您可以通过实现 Warnsdorf's Rule 来加快搜索速度.这是选择下一步的启发式方法，在这种情况下，您总是会尝试为之后的下一步移动带来最少可用移动的移动。它可以通过几个简单的循环来完成:

int row = cellno / board.length, col = cellno % board.length;

// find move with fewest moves available for the next move:
int minMovesAvailable = 8;
int minMovesDir = 0;
for (int i = 0; i < dirs.length; i++) {
    int r = dirs[i][0] + row;
    int c = dirs[i][1] + col;
    if (isSafe(board, r, c)&&board[r][c]==-1)
    {
        board[r][c] = stepno;
        int movesAvailable = 0;
        for (int j = 0; j < dirs.length; j++) {
            int r2 = dirs[j][0] + r;
            int c2 = dirs[j][1] + c;
            if (isSafe(board, r2, c2)&&board[r2][c2]==-1)
            {
                movesAvailable++;
            }
        }
        board[r][c] = -1;
        if(movesAvailable < minMovesAvailable)
        {
            minMovesAvailable = movesAvailable;
            minMovesDir = i;
        }
    }
}

// now recurse this move first:
// int r = dirs[minMovesDir][0] + row;
// int c = dirs[minMovesDir][1] + col;