问题描述
我尝试使用深度搜索来实现8个皇后，以对任何初始状态都适用于空板(在木板上没有皇后)，但是如果有解决方案，如果没有解决方案，我需要它在初始状态下工作此初始状态的解决方案，它将打印没有解决方案
i am try to implement 8 queen using depth search for any initial state it work fine for empty board(no queen on the board) ,but i need it to work for initial state if there is a solution,if there is no solution for this initial state it will print there is no solution
这是我的代码:
public class depth {
public static void main(String[] args) {
//we create a board
int[][] board = new int[8][8];
board [0][0]=1;
board [1][1]=1;
board [2][2]=1;
board [3][3]=1;
board [4][4]=1;
board [5][5]=1;
board [6][6]=1;
board [7][7]=1;
eightQueen(8, board, 0, 0, false);
System.out.println("the solution as pair");
for(int i=0;i<board.length;i++){
for(int j=0;j<board.length;j++)
if(board[i][j]!=0)
System.out.println(" ("+i+" ,"+j +")");
}
System.out.println("the number of node stored in memory "+count1);
}
public static int count1=0;
public static void eightQueen(int N, int[][] board, int i, int j, boolean found) {
long startTime = System.nanoTime();//time start
if (!found) {
if (IsValid(board, i, j)) {//check if the position is valid
board[i][j] = 1;
System.out.println("[Queen added at (" + i + "," + j + ")");
count1++;
PrintBoard(board);
if (i == N - 1) {//check if its the last queen
found = true;
PrintBoard(board);
double endTime = System.nanoTime();//end the method time
double duration = (endTime - startTime)*Math.pow(10.0, -9.0);
System.out.print("total Time"+"= "+duration+"\n");
}
//call the next step
eightQueen(N, board, i + 1, 0, found);
} else {
//if the position is not valid & if reach the last row we backtracking
while (j >= N - 1) {
int[] a = Backmethod(board, i, j);
i = a[0];
j = a[1];
System.out.println("back at (" + i + "," + j + ")");
PrintBoard(board);
}
//we do the next call
eightQueen(N, board, i, j + 1, false);
}
}
}
public static int[] Backmethod(int[][] board, int i, int j) {
int[] a = new int[2];
for (int x = i; x >= 0; x--) {
for (int y = j; y >= 0; y--) {
//search for the last queen
if (board[x][y] != 0) {
//deletes the last queen and returns the position
board[x][y] = 0;
a[0] = x;
a[1] = y;
return a;
}
}
}
return a;
}
public static boolean IsValid(int[][] board, int i, int j) {
int x;
//check the queens in column
for (x = 0; x < board.length; x++) {
if (board[i][x] != 0) {
return false;
}
}
//check the queens in row
for (x = 0; x < board.length; x++) {
if (board[x][j] != 0) {
return false;
}
}
//check the queens in the diagonals
if (!SafeDiag(board, i, j)) {
return false;
}
return true;
}
public static boolean SafeDiag(int[][] board, int i, int j) {
int xx = i;
int yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy++;
xx++;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy--;
xx--;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy--;
xx++;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy++;
xx--;
}
return true;
}
public static void PrintBoard(int[][] board) {
System.out.print(" ");
for (int j = 0; j < board.length; j++) {
System.out.print(j);
}
System.out.print("\n");
for (int i = 0; i < board.length; i++) {
System.out.print(i);
for (int j = 0; j < board.length; j++) {
if (board[i][j] == 0) {
System.out.print(" ");
} else {
System.out.print("Q");
}
}
System.out.print("\n");
}
}
}
例如，对于此初始状态，它会给我以下错误:
for example for this initial state it give me the following error:
Exception in thread "main" java.lang.StackOverflowError
我被困住了，我认为错误是该方法如何解决这个问题的无限呼吁.
i am stuck, i think the error is infinite call for the method how to solve this problem.
任何想法都会有所帮助，谢谢.
any idea will be helpful,thanks in advance.
注意:宽泛是二维数组，当我放入(1)时，它表示此时是女王.
note:the broad is two dimensional array,when i put (1) it means there queen at this point.
note2:我们将初始状态设置为以下状态:
note2:we i put the initial state as the following it work:
board [0][0]=1;
board [1][1]=1;
board [2][2]=1;
board [3][3]=1;
board [4][4]=1;
board [5][5]=1;
board [6][6]=1;
board [7][1]=1;
推荐答案
要添加到@StephenC的答案中:
To add to @StephenC's answer:
这是一段复杂的代码，特别是如果您没有Java编程经验的话.
This is a heck of a complicated piece of code, especially if you're not experienced in programming Java.
我执行了您的代码，并一遍又一遍地(一遍又一遍)输出它
I executed your code, and it outputs this over and over and over and over (and over)
back at (0,0)
01234567
0
1 Q
2 Q
3 Q
4 Q
5 Q
6 Q
7 Q
back at (0,0)
然后崩溃
Exception in thread "main" java.lang.StackOverflowError
at java.nio.Buffer.<init>(Unknown Source)
...
at java.io.PrintStream.print(Unknown Source)
at java.io.PrintStream.println(Unknown Source)
at Depth.eightQueen(Depth.java:56)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
...
我的第一个直觉总是添加一些System.out.println(...)来找出问题所在，但这在这里不起作用.
My first instinct is always to add some System.out.println(...)s to figure out where stuff is going wrong, but that won't work here.
我唯一看到的两个选择是
The only two options I see are to
- 熟悉调试器并使用它来逐步分析其为何永不停止循环
- 把它弄碎！您如何希望在不将其分解为可消化的大块的情况下处理此类大问题?
更不用说 8皇后的概念很复杂.
Not to mention that the concept of 8-queens is complicated to begin with.
进一步思考:
System.out.println()在当前实现中没有用，因为存在无限的输出.在这里，调试器是更好的解决方案，但是另一种选择是以某种方式限制您的输出.例如，在顶部创建一个计数器
System.out.println()s are not useful as currently implemented, because there's infinite output. A debugger is the better solution here, but another option is to somehow limit your output. For example, create a counter at the top
private static final int iITERATIONS = 0;
，而不是
System.out.println("[ANUMBERFORTRACING]: ... USEFUL INFORMATION ...")
使用
conditionalSDO((iITERATIONS < 5), "[ANUMBERFORTRACING]: ... USEFUL INFORMATION");
这是函数:
private static final void conditionalSDO(boolean b_condition, String s_message) {
if(b_condition) {
System.out.println(s_message);
}
}
另一种选择是不限制输出，而是将其写入文件.
Another alternative is to not limit the output, but to write it to a file.
我希望这些信息对您有所帮助.
I hope this information helps you.
(注意:我编辑了OP的代码以使其可编译.)
(Note: I edited the OP's code to be compilable.)
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