Java 谁能帮我弄一下这个骑士';什么是旅游代码?
代码对我来说似乎很好,但输出太短,当它应该是“是”时,答案是“否”(从a开始,0,骑士应该能够遍历整个棋盘) 顺便说一下,我的Java 谁能帮我弄一下这个骑士';什么是旅游代码?,java,recursion,Java,Recursion,代码对我来说似乎很好,但输出太短,当它应该是“是”时,答案是“否”(从a开始,0,骑士应该能够遍历整个棋盘) 顺便说一下,我的positionsListedarray是[9][9]的原因是我希望值为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 wit
positionsListed
array是[9][9]的原因是我希望值为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;
}
您实现的算法如下所示:
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;
}
}
按如下顺序从正方形排列可能的骑士招式:
在每次移动中,选择仍然允许的编号最低的移动
(A) 如果覆盖所有方块,则返回true
(B) 如果无法继续移动,请返回false
您的代码有两个问题。第一,已经指出,你错过了支票(A)。第二个更严重的问题是,该算法不起作用。事实上,你最终会得到以下结果:
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;
}
}
在这张图片中,黑骑士代表开始和结束的方块,而白骑士代表所有其他被覆盖的方块。如果你遵循你的算法,你最终会在一个正方形中结束,在那里你无法到达任何其他没有被覆盖的正方形。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
如果这确实是您想要实现的算法,那么就没有理由使用递归,因为
for
循环也可以工作。当我们当前所在的方格中存在有效的骑士移动时,您的函数recursive
返回true。这实际上不是一个递归算法,有两个原因:
recursive
函数不是幂等函数-它有一个副作用,即填充positionsVisited
数组的一个平方李>
recursive
函数调用全局变量-positionsVisited
(我说的是“全局变量”,而不是“私有字段”,因为您编写的基本上是过程代码)李>
相反,函数recursive
应该告诉您一个更一般的信息:给定棋盘上的一个特定方块,以及我们不允许访问的一组特定方块,是否有骑士之旅?(当然,使用正方形a1和访问位置的空数组调用该函数将告诉您是否存在骑士之旅。)该函数还可以返回一个字符串,该字符串将告诉您骑士之旅是什么
递归函数的结构应如下所示:
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,换言之:
递归检查骑士离开当前方格的所有方格,传入新方格和通过添加当前方格而形成的新访问方格数组。如果有来自其中一个方块的骑士之旅,请将当前方块附加到其开头,以获得来自当前方块的骑士之旅
而不是你写的是:
递归地建立一个骑士之旅,从一个正方形开始,在每一步(相当随意地)选择一个合法的骑士移动。如果我们到了一个不能再做骑士招式的位置,返回false
你能理解为什么第一个递归算法(当然更复杂)有效而你的算法无效吗 除了“所有已访问”检查的问题外,我还看到至少一个问题,在类字段中。当算法通过递归的一个分支时,它将一些方块标记为已访问,并且由于该信息是类字段,所以当它失败当前分支并启动另一个分支时,它会看到以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有自己的实际副本,会怎么样
最终版本
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;
}
}
公共类骑士俱乐部{
private final static Map letters=new HashMap();
静止的{
字母。放入(0,'a');
字母。将(1,'b');
字母。将(2,'c');
字母。放入(3,'d');
字母。将(4,'e');
字母。放入(5,'f');
}
公共静态void main(字符串[]args){
if(递归(0,0,0,新布尔[6][6],1,”)){
System.out.println(“是”);
}否则{
系统输出打印项次(“否”);
}
}
私有静态布尔值allVisited(布尔值[][]位置Visited){
对于(int i=0;i5 | |位置X<0 | |位置Y>5 | |位置Y<0){
返回false;
}
如果(位置查看[positionX][positionY]){
返回false;
}
//行动
positionsVisited[positionX][positionY]=真;
字符c=字母.get(位置x);
currentMoves++=“c+(位置Y+1)+”(移动类型:“+(移动类型+1)+”)\r\n;
if(递归(位置X+2、位置Y+1、1、cloneArray(位置可见)、移动计数器+1、当前移动)){
返回true;
}else if(递归(位置X+1、位置Y+2、2、cloneArray(位置可见)、移动计数器+1、当前移动)){
返回true;
}else if(递归)