Question

代码对我来说似乎很好，但是输出太短了，答案是肯定的，因为它应该是肯定的（从0开始，骑士应该能够游遍整个棋盘）

顺便说一句，我的positionsVisted数组是[9] [9]的原因是因为我希望值为1-8，以匹配输出。

public class KnightChessRiddle {
static // given a chess board and a 0,0 starting point, can a knight pass through
// all the the squares without passing
// a square twice

int[][] positionsVisited = new int[9][9];
static int positionX = 1;
static int positionY = 1;
boolean stop = false;
static boolean continUe = false;
static int moveCounter = -1;

public static void main(String[] args) {
    if (recursive(1, 1, 0)) {
        System.out.println("yes");
    }
    else System.out.println("no");
}

public static boolean recursive(int x, int y, int moveType){


    if (x>8||x<=0||y>8||y<=0) return false;
    if (positionsVisited[x][y]==1) {
        return false;
    }
    positionX = x;
    positionY = y;
    positionsVisited[positionX][positionY]++;

    char c;
    c='a';
    switch (positionX) {
    case 1:
        c='a';
        break;
    case 2:
        c='b';
        break;
    case 3:
        c='c';
        break;
    case 4:
        c='d';
        break;
    case 5:
        c='e';
        break;
    case 6:
        c='f';
        break;
    case 7:
        c='g';
        break;
    case 8:
        c='h';
        break;

    default: 
        break;
    }
    moveCounter++;
    System.out.println("doing move "+moveType+" move count: "+moveCounter);
    System.out.println("Knight is in "+ c +","+positionY);

    try {
        Thread.sleep(100);
    } catch (InterruptedException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    if (recursive(positionX+2, positionY+1, 1)) {
        return true;
    }
    else if (recursive(positionX+1, positionY+2, 2) ) {
        return true;
    }
    else if (recursive(positionX+2, positionY-1, 3) ) {
        return true;
    }
    else if (recursive(positionX+1, positionY-2, 4)) {
        return true;
    }
    else if (recursive(positionX-2, positionY+1, 5)) {
        return true;
    }
    else if (recursive(positionX-1, positionY+2, 6)) {
        return true;
    }
    else if (recursive(positionX-2, positionY-1, 7)) {
        return true;
    }
    else if (recursive(positionX-1, positionY-2, 8)) {
        return true;
    }
    else return false;
}

这是该计划的输出：

doing move 0 move count: 0
Knight is in a,1
doing move 1 move count: 1
Knight is in c,2
doing move 1 move count: 2
Knight is in e,3
doing move 1 move count: 3
Knight is in g,4
doing move 2 move count: 4
Knight is in h,6
doing move 5 move count: 5
Knight is in f,7
doing move 1 move count: 6
Knight is in h,8
doing move 8 move count: 7
Knight is in g,6
doing move 4 move count: 8
Knight is in h,4
doing move 5 move count: 9
Knight is in f,5
doing move 2 move count: 10
Knight is in g,7
doing move 4 move count: 11
Knight is in h,5
doing move 5 move count: 12
Knight is in f,6
doing move 1 move count: 13
Knight is in h,7
doing move 5 move count: 14
Knight is in f,8
doing move 7 move count: 15
Knight is in d,7
doing move 4 move count: 16
Knight is in e,5
doing move 4 move count: 17
Knight is in f,3
doing move 2 move count: 18
Knight is in g,5
doing move 4 move count: 19
Knight is in h,3
doing move 5 move count: 20
Knight is in f,4
doing move 4 move count: 21
Knight is in g,2
doing move 7 move count: 22
Knight is in e,1
doing move 6 move count: 23
Knight is in d,3
doing move 3 move count: 24
Knight is in f,2
doing move 3 move count: 25
Knight is in h,1
doing move 6 move count: 26
Knight is in g,3
doing move 5 move count: 27
Knight is in e,4
doing move 5 move count: 28
Knight is in c,5
doing move 1 move count: 29
Knight is in e,6
doing move 5 move count: 30
Knight is in c,7
doing move 1 move count: 31
Knight is in e,8
doing move 8 move count: 32
Knight is in d,6
doing move 5 move count: 33
Knight is in b,7
doing move 1 move count: 34
Knight is in d,8
doing move 8 move count: 35
Knight is in c,6
doing move 1 move count: 36
Knight is in e,7
doing move 1 move count: 37
Knight is in g,8
no

Answer 1

您实施的算法如下：

按如下方式从正方形命令可能的骑士移动：



在每次移动中，选择仍然允许的编号最小的移动。

（A）如果您覆盖所有方格，请返回true。

（B）如果您无法再移动，请返回false。

您的代码存在两个问题。第一个已经指出的是你错过了支票（A）。第二个更严重的问题是这个算法不起作用。事实上，你最终会得到以下结论：

在这张照片中，黑色骑士代表开始和结束方块，而白色骑士则是所有其他方块。如果你遵循你的算法，你最终会进入一个方格，在那里你无法到达任何其他尚未被覆盖的方格。这并不意味着你不能在棋盘上进行骑士之旅，只是你的算法不起作用。

如果这确实是你想要实现的算法，那么就没有理由使用递归，因为for循环也可以正常工作。当有效骑士从我们当前所在的方格移动时，你的函数recursive会返回true。有两个原因，这实际上不是一个递归算法：

recursive函数不是幂等的 - 它有副作用，即填充positionsVisited数组的正方形。
recursive函数调用全局变量 - positionsVisited（我说＆＃39;全局变量＆＃39;，而不是＆＃39;私有字段＆＃39;，因为你有什么写的基本上是程序代码）。

相反，函数recursive应该告诉你一个更为一般的信息：给定棋盘上的特定方块，以及我们不允许访问的特定方块，是否有骑士＆＃39;参观剩下的广场？（当然，用方形a1调用该函数和一个空的访问位置数组将告诉你是否有一个骑士游览。）该函数还可以返回一个字符串，它会告诉你什么是骑士＆＃ 39; s tour is。

递归函数的结构应该类似于以下内容：

private boolean isKnightsTour(Square currentSquare,
                              int[9][9] visitedSquares,
                              KnightsTour tour)
{
  // Append the current square to the array of visited squares.
  int[9][9] newVisitedSquares = visitedSquares;
  newVisitedSquares[currentSquare.getX()][currentSquare.getY()] = 1;

  // If we have visited all the squares, there is a knight's tour.  
  // Add some code here to check for that.
  if (allSquaresVisited()) {
    tour = new KnightsTour(currentSquare);
    return true;
  }

  // Test all squares a knight's move away.  If you get a knight's tour, 
  // append the current square to the start and return that.  
  KnightsTour newTour;
  if (isKnightsTour(currentSquare.doMove1(), newVisitedSquares, newTour) {
    newTour.appendStart(currentSquare);
    tour = newTour;
    return true;
  }

  // Repeat for the other knight's moves.

  else {
    tour = null;
    return false;
  }
}

或者，用文字：

递归检查骑士从当前方块移开的所有方格，传递新方块和通过添加当前方块形成的新访问方块阵列。如果从其中一个方格中有骑士之旅，请将当前方格附加到其开头，以便从当前方块获得骑士之旅。

而不是你所写的：

通过在广场上开始并且（相当随意地）选择合法骑士在每一步的移动来递归地建立一个骑士之旅。如果我们到达一个我们无法再进行骑士行动的位置，请返回false。

你能看出为什么第一个（不可否认的是更复杂的）递归算法起作用而你的不是吗？

Answer 2

除了“所有被访问”的问题之外，我看到至少还有一个问题，即在类字段中。当算法通过一个递归分支时，它会将一些正方形标记为已访问，并且由于此信息是类字段，当它失败当前分支并启动另一个分支时，它会看到之前尝试的所有无效信息。

如果您尝试将positionsVisited，positionX和positionY作为方法参数传递并将其从类字段中删除，那么每个方法调用都会拥有自己的实际副本，该怎么办？

最终版

public class KnightChessRiddle {

    private final static Map<Integer, Character> letters = new HashMap<>();

    static {
        letters.put(0, 'a');
        letters.put(1, 'b');
        letters.put(2, 'c');
        letters.put(3, 'd');
        letters.put(4, 'e');
        letters.put(5, 'f');
    }

    public static void main(String[] args) {
        if (recursive(0, 0, 0, new boolean[6][6], 1, "")) {
            System.out.println("yes");
        } else {
            System.out.println("no");
        }
    }

    private static boolean allVisited(boolean[][] positionsVisited) {
        for (int i = 0; i < positionsVisited.length; i++) {
            for (int j = 0; j < positionsVisited.length; j++) {
                if (!positionsVisited[i][j]) {
                    return false;
                }
            }
        }
        return true;
    }

    private static boolean recursive(int positionX, int positionY, int moveType,
            boolean[][] positionsVisited, int moveCounter, String currentMoves) {

        // checks
        if (allVisited(positionsVisited)) {
            System.out.println(currentMoves);
            return true;
        }

        if (positionX > 5 || positionX < 0 || positionY > 5 || positionY < 0) {
            return false;
        }

        if (positionsVisited[positionX][positionY]) {
            return false;
        }

        // make move
        positionsVisited[positionX][positionY] = true;

        char c = letters.get(positionX);
        currentMoves += "" + c + (positionY + 1) + " (move type: " + (moveType + 1) + ")\r\n";

        if (recursive(positionX + 2, positionY + 1, 1, cloneArray(positionsVisited), moveCounter + 1, currentMoves)) {
            return true;
        } else if (recursive(positionX + 1, positionY + 2, 2, cloneArray(positionsVisited), moveCounter + 1, currentMoves)) {
            return true;
        } else if (recursive(positionX + 2, positionY - 1, 3, cloneArray(positionsVisited), moveCounter + 1, currentMoves)) {
            return true;
        } else if (recursive(positionX + 1, positionY - 2, 4, cloneArray(positionsVisited), moveCounter + 1, currentMoves)) {
            return true;
        } else if (recursive(positionX - 2, positionY + 1, 5, cloneArray(positionsVisited), moveCounter + 1, currentMoves)) {
            return true;
        } else if (recursive(positionX - 1, positionY + 2, 6, cloneArray(positionsVisited), moveCounter + 1, currentMoves)) {
            return true;
        } else if (recursive(positionX - 2, positionY - 1, 7, cloneArray(positionsVisited), moveCounter + 1, currentMoves)) {
            return true;
        } else if (recursive(positionX - 1, positionY - 2, 8, cloneArray(positionsVisited), moveCounter + 1, currentMoves)) {
            return true;
        } else {
            return false;
        }

    }

    private static boolean[][] cloneArray(boolean[][] src) {
        boolean[][] newPositions = new boolean[src.length][src.length];
        for (int i = 0; i < src.length; i++) {
            System.arraycopy(src[i], 0, newPositions[i], 0, src[0].length);
        }
        return newPositions;
    } 
}

以下是6x6板的工作变化。使用8x8电路板计算会占用我的机器上太多时间。如果随机选择移动选择，也许可以更快。

有人可以帮助我使用这个骑士的旅行代码吗？

2 个答案: