修改 Knight's Tour 算法以打印所有解决方案

Question

我最近提取了一个 C 程序 (https://repl.it/Klpv)，它在 8 x 8 板上搜索骑士之旅。我用 JavaScript 重新编写了程序（因为我更了解 JS），然后我修改了程序，以便可以在任何给定的板尺寸上搜索解决方案。

目前，它使用带有回溯的递归算法来打印它找到的第一个解决方案。然后它停止。原始 C 程序中的一个 cmets 表示该程序将打印一种可行的解决方案，但我的目标是打印所有（或可定制数量的）解决方案。我已经尝试稍微修改程序的“返回”命令，以便我可以让它打印多个解决方案，但我似乎无法做到这一点。谁能帮帮我？

// JavaScript Knight's Tour Algorithm

/* A utility function to check if i,j are valid indexes
   for width*height chessboard */
var isSafe = function(x, y, sol) {
  return (x >= 0 && x < width && y >= 0 && y < height && sol[x][y] == -1);
}

/* A utility function to print solution matrix sol[width][height] */
var printSolution = function(sol) {
  var solution = "Knight's Tour for "+width+" by "+height+" board:\n";
  for (i = 0; i < width; i++) {
    for (j = 0; j < height; j++) {
      solution += sol[i][j] + "\t";
    }
    solution += "\n";
  }

  console.log(solution)
}

/* This function solves the Knight Tour problem using
   Backtracking.  This function mainly uses solveKTUtil()
   to solve the problem. It returns false if no complete
   tour is possible, otherwise return true and prints the
   tour.
   Please note that there may be more than one solutions,
   this function prints one of the feasible solutions.  */
var solveKT = function() {
  /* Create two-dimentional array */
  var sol = new Array(width);
  for (i = 0; i < sol.length; i++) {
    sol[i] = new Array(height)
  }

  /* Initialization of solution matrix - set all to -1 */
  for (i = 0; i < width; i++) {
    for (j = 0; j < height; j++) {
      sol[i][j] = -1
    }
  }

  /* xMove[] and yMove[] define next move of Knight.
     xMove[] is for next value of x coordinate
     yMove[] is for next value of y coordinate */
  var xMove = [2, 1, -1, -2, -2, -1, 1, 2];
  var yMove = [1, 2, 2, 1, -1, -2, -2, -1];

  // Since the Knight is initially at the first block
  sol[0][0] = 0;

  /* Start from 0,0 and explore all tours using
     solveKTUtil() */
  if (solveKTUtil(0, 0, 1, sol, xMove, yMove) == 0) {
    console.log("Solution does not exist");
    return 0;
  } else {
    printSolution(sol);
  }
}

/* A recursive utility function to solve Knight Tour
   problem */
var solveKTUtil = function(x, y, movei, sol, xMove, yMove) {
  var k, next_x, next_y;
  if (movei == width * height) {
    return 1;
  }

  /* Try all next moves from the current coordinate x, y */
  for (k = 0; k < 8; k++) {
    next_x = x + xMove[k];
    next_y = y + yMove[k];
    if (isSafe(next_x, next_y, sol)) {
      sol[next_x][next_y] = movei;
      if (solveKTUtil(next_x, next_y, movei + 1, sol, xMove, yMove) == 1) {
        return 1;
      } else {
        sol[next_x][next_y] = -1; // backtracking
      }
    }
  }

  return 0;
}

var width = 5;
var height = 6;
solveKT();

这将打印到控制台：

Knight's Tour for 5 by 6 board:
0   27  22  15  6   11  
23  16  7   12  21  14  
28  1   26  19  10  5   
17  24  3   8   13  20  
2   29  18  25  4   9   

编辑：我找到了一个 C++ 程序可以做到这一点。 https://repl.it/L4Pf不再需要帮助！

Answer 1

这个问题有很多解决方案（根据this question 的回答，超过 10^16 个），所以你必须满足于其中的几个。

一个简单的方法是填充一个路径数组，直到它对你来说足够大，而不是返回 1。

...
if (solveKTUtil(next_x, next_y, movei + 1, sol, xMove, yMove) == 1) {
    return 1;
  }
...

应该变成类似

...
if (solveKTUtil(next_x, next_y, movei + 1, sol, xMove, yMove) == 1) {
    results.push(deepCopy(sol)) //copy the sol
    if(results.length > desired_number_of_paths){
        return 1;
    }
  }
...

Answer 2

您还可以将解决方案修改为：

var solveKTUtil = function(x, y, movei, sol, xMove, yMove) {
    var k, next_x, next_y;

    if (movei === width * height) {
        //Solution found, print it
        printSolution(sol);
        //Backtrack
        sol[x][y] = -1;
    }
    ....

这是：找到解决方案后，将其打印出来，并在返回之前回溯，以便继续搜索解决方案。