Question

我最近推出了一个C程序（https://repl.it/Klpv）来搜索8乘8的骑士之旅。我用JavaScript重写了程序（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。然后停止。最初的C程序中的一条评论说该程序将打印一个可行的解决方案，但我的目标是打印所有（或可定制数量）的解决方案。我尝试过修改程序的'return'命令，以便我可以打印多个解决方案，但我似乎无法做到这一点。任何人都可以帮助我吗？

jquery

这将打印到控制台：

// 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();

编辑：我发现了一个完全符合这一要求的C ++程序。 https://repl.it/L4Pf不再需要帮助了！

Answer 1

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

这样做的一个简单方法是填充一系列路径，直到它足够大而不是返回1.

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

应该变得像

...
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，在返回之前打印并回溯，以便继续搜索解决方案。

修改骑士游览算法以打印所有解决方案

2 个答案: