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How to find a >3 line of consecutive elements in a 2d array

I'm a game where you have to make a line in a row/column of 3 or more (up to 7) consecutive elements in a 2d array. Every time there is one of those lines you add the score and delete those consecutive elements from the row/column (Similar as to candycrash disappearing candies.
If you have something like this ['a' 'b' 'b' 'b' 'a'] the result should be ['a' '' '' '' 'a']
The problem comes to more complex cases like ['b' 'b' ' ' 'b' 'b'], that ends up like this
['' '' '' '' ''] because its also detected as a >=3 elem. consecutive line.
I know its due to the counter that keeps adding to the value it already has, but I couldnt come to any other solution till now
The code I have up to now is this:
**
public int[] checkRows() {
int[] results = new int[BOARD_SIZE];
for (int i = 0; i < BOARD_SIZE; i++) {
int counter = 1;
int counterHead = 1;
int j = 0;
while (j < BOARD_SIZE) {
if (j+1 < BOARD_SIZE) {
if (getBoard()[i][j].getType() == getBoard()[i][j+1].getType() && getBoard()[i][j].getType() != ' ') {
counter++;
}
}
if (counter >= 3)
deleteRows(i, j);
j++;
}
results[i] = counter;
}
return results;
}
**

So we want to see if there are 3 consecutive elements with the same value, but what it looks like to me is we are simply testing if there are three values in a row of the board that are equal. This works if we find a section of consecutive equivalent elements, but as soon as we find a space or non-equal element the logic fails.
Things we need to consider:
If the counter reaches 3 we can break out of your conditional and delete. This case is covered.
If your counter is updated from its original value (counter > 1) and never reaches 3 we know we must have encountered either a space or a non-equivalent value. In this case we should reset the counter and continue our test for the rest of the row.

Your problem is that you are not counting consecutive values, but rather pairs of consecutive values. For 'b', 'b', 'b', there are three b characters, but only 2 pairs of values, 1) the first and second b, and 2) the second and third b. Your counter should start at 1 rather than 0 because every value is by itself a run of 1 "consecutive" values. This, along with resetting the counter when you see a non-matching letter should give you the result you desire.
I haven't tried this as I don't have your data or the structures that this code uses, but try this:
public int[] checkRows() {
int[] results = new int[BOARD_SIZE];
for (int i = 0; i < BOARD_SIZE; i++) {
int counter = 1;
int counterHead = 1;
int j = 1;
while (j < BOARD_SIZE) {
if (j+1 < BOARD_SIZE) {
if (getBoard()[i][j].getType() == getBoard()[i][j+1].getType() && getBoard()[i][j].getType() != ' ') {
counter++;
}
else {
counter = 1;
}
}
if (counter >= 3)
deleteRows(i, j);
j++;
}
results[i] = counter;
}
return results;
}

First of all, I would suggest that for such complex problems, always try to break your problems and then merge those pieces together.
Below is the algorithm for your problem, just for simplicity and ease of explanation I used an Integer array which you can easily replace with a string or character array and change the condition.
A basic function for performing candyCrush Algorithm.
public void candyCrush(){
int[][] result = new int[][]{{1,2,2,2,2},
{1,2,2,2,1},
{1,2,1,1,1},
{2,1,2,1,1}
};
int count = 0;
for (int i = 0; i < result.length; i++) {
for (int j = 0; j < result[i].length; j++) {
if(result[i][j]==-1){
continue;
}
boolean hasRowElements = checkRowForCurrentIndex(result, i, j);
boolean hasColumnElements = checkColumnForCurrentIndex(result, i, j);
if(hasRowElements || hasColumnElements){
count++;
}
}
}
System.out.println(count);
}
private boolean checkRowForCurrentIndex(int[][] result, int row, int col) {
int forTop = row, forBottom = row;
int topStartingLimit = 0, bottomEndingLimit = 0;
while(forTop>= 0){
if(result[row][col] == result[forTop][col] && result[forTop][col]!=-1){
topStartingLimit = forTop;
}else{
break;
}
forTop--;
}
while(forBottom< result.length){
if(result[row][col] == result[forBottom][col] && result[forBottom][col]!=-1){
bottomEndingLimit = forBottom;
}else{
break;
}
forBottom++;
}
if(topStartingLimit==bottomEndingLimit){
return false;
}
if(bottomEndingLimit-topStartingLimit>=2 && bottomEndingLimit-topStartingLimit<= 6) {
deleteRows(result, topStartingLimit, bottomEndingLimit, row, col);
return true;
}
return false;
}
private void deleteRows(int[][] result, int topStartingLimit, int bottomEndingLimit, int row, int col) {
while(topStartingLimit<= bottomEndingLimit){
result[topStartingLimit][col] = -1;
topStartingLimit++;
}
return;
}
private boolean checkColumnForCurrentIndex(int[][] result, int row, int col) {
int forLeft = col, forRight = col;
int leftStartingLimit = 0, rightEndingLimit = 0;
while(forLeft>= 0){
if(result[row][col] == result[row][forLeft] && result[row][forLeft]!=-1){
leftStartingLimit = forLeft;
}else{
break;
}
forLeft--;
}
while(forRight< result[row].length){
if(result[row][col] == result[row][forRight] && result[row][forRight]!=-1){
rightEndingLimit = forRight;
}else{
break;
}
forRight++;
}
if(leftStartingLimit==rightEndingLimit){
return false;
}
if(rightEndingLimit-leftStartingLimit>=2 && rightEndingLimit-leftStartingLimit<= 6) {
deleteCols(result, leftStartingLimit, rightEndingLimit, row, col);
return true;
}
return false;
}
private void deleteCols(int[][] result, int leftStartingLimit, int rightEndingLimit, int row, int col) {
while(leftStartingLimit<= rightEndingLimit){
result[row][leftStartingLimit] = -1;
leftStartingLimit++;
}
return;
}
Definitely, you can optimize this code better by using the same function again for Rows and Cols, but for sake of simplicity. I tried to break everything so that the solution becomes more clear.
Hope this algorithm would work.

Java: How to move a knight on a chessboard until no possible moves are left

I have an assignment to have a knight move around the chessboard until it either completes a full tour or has no where else to go.
im having a problem figuring out how to have it actually stop after there are no more moves. I have the algorithm for the movement down, as well as bounds checking on the chessboard.
setting the loop count to 64 just times out, because the program will try to continually find a spot that doesn't exist if a perfect tour isn't created.
two ideas I have had to fix this is to either increment a variable every time a certain position is checked around a certain spot, and if all possible moves are taken by a previous move, terminate the program. Problem is I have no idea how to actually do this.
A second idea I had is to have the program quit the for loop after 2 seconds(during which the for loop will check each position way more than once) but I feel like my professor would crucify me upside down if I did that
here is my code:
import apcslib.*; //this is only for Format()
public class ktour
{
int[][] kboard = new int[9][9];
int[] vert = new int[9];
int[] horiz = new int[9];
ktour()
{
vert[1] = -2;vert[2] = -1;vert[3] = 1;vert[4] = 2;vert[5] = 2;vert[6] = 1;vert[7] = -1;vert[8] = -2;
horiz[1] = 1;horiz[2] = 2;horiz[3] = 2;horiz[4] = 1;horiz[5] = -1;horiz[6] = -2;horiz[7] = -2;horiz[8] = -1;
path();
}
public void path()
{
int row = 1;
int col = 1;
int loops = 10; //i have this set to 10 for now
int col2 = 1;
int row2 = 1;
int r = (int)(Math.random() * (8) +1); //returns a random from 1 to 9
//System.out.println(r);
kboard[col][row] = 1;
for(int x = 2; x < loops; x++) //this runs the bounds check and places each number for the amount that loops is
{
r = (int)(Math.random() * (8) +1);
col = col2;
row = row2;
col = col + vert[r];
row = row + horiz[r];
while(col <= 0 || col > 8 || row <= 0 || row > 8) //bounds check, will keep running until both row and columb is in the board
{
r = (int)(Math.random() * (8) + 1);
col = col2;
row = row2;
col = col + vert[r];
row = row + horiz[r];
}
if(kboard[col][row] == 0)
{
kboard[col][row] = x;
row2 = row;
col2 = col;
}
else
{
x--; //if the above if is false and a number already occupies the generated spot, x is decremented and the program tries again
}
}
printboard();
}
public void printboard()
{
for(int y = 1; y < 9; y++)
{
System.out.println();
for(int x = 1; x < 9; x++)
{
System.out.print(Format.right(kboard[y][x],3));
}
}
}
}

I was able to fix my lab with the following code. I created a variable called count which I used to check if at any move there were no more moves left. As there are only 8 moves, when the variable reached 9 the code terminated, and printed up to the point it got to.
I had to put multiple if statements excluding r = math.random if count was not 0, meaning I was checking r 1-9, aka every possible move. Therefore, I couldn't use a randomizer, I had to traverse all 8 possible moves.
I also ran into problems when I reached the line where it checks if kboard[col][row] == 0. if you were running through a loop with count greater than 1, it was possible that col or row could be out of bounds, due to lack of a randomizer in the bounds checker. If left without a break, the bounds checker would run forever without a random number generated every time. I fixed this by adding an if statement that allowed the program to proceed if col and row were inside the board. if they were not, x was decremented and count was increased again, signifying a failed attempt.
This way I was able to check all possible moves, disregarding whether or not they were inside the board.
public void path()
{
int row = 1;
int col = 1;
int loops = 64; //i have this set to 10 for now
int col2 = 1;
int row2 = 1;
int count = 0;
boolean end = false;
int r = (int)(Math.random() * (8) +1); //returns a random from 1 to 9
//System.out.println(r);
kboard[col][row] = 1;
for(int x = 2; x < loops; x++) //this runs the bounds check and places each number for the amount that loops is
{
if(count == 0)
r = (int)(Math.random() * (8) +1);
if(count >= 1 && r != 8)
r++;
col = col2;
row = row2;
col = col + vert[r];
row = row + horiz[r];
while(col <= 0 || col > 8 || row <= 0 || row > 8) //bounds check, will keep running until both row and columb is in the board
{
if(count == 0)
r = (int)(Math.random() * (8) + 1);
col = col2;
row = row2;
col = col + vert[r];
row = row + horiz[r];
if(count >= 1)
break;
}
end = false;
if(r == 8 || r == 9)
r = 1;
if(count >= 9)
{
System.out.println("Halting... no where else to go");
loops = 0;
}
if(!(col <= 0 || row <= 0 || row > 8 || col > 8))
{
if(kboard[col][row] == 0)
{
kboard[col][row] = x;
row2 = row;
col2 = col;
count = 0;
}
else
{
count++;
x--; //if the above if is false and a number already occupies the generated spot, x is decremented and the program tries again
}
}
else
{
count++;
x--;
}
}
printboard();
}

Explanation of 8-Queen solution Java code.

I found a 8-Queen solution that I was able to compile and run in Eclipse. I believe this solution follows the backtracking approach and meat of the work is done in the solution and unsafe functions, but I am having a hard time understanding the code paths in there. Can someone please help me understand what this code is doing.
My Source - http://rosettacode.org/wiki/N-queens_problem#Java
I verified the output against the 92 solutions published on other sources. Looks good. So I know the code works.
I have tried to format it and add some basic notes to clear things up -
private static int[] b = new int[8];
private static int s = 0;
public static void main(String[] args) {
// solution from - http://rosettacode.org/wiki/N-queens_problem#Java
new QueenN();
}
public QueenN() {
solution();
}
public void solution() {
int y = 0;
b[0] = -1;
while (y >= 0) {
do {
b[y]++;
} while ((b[y] < 8) && unsafe(y));
if (b[y] < 8) {
if (y < 7) {
b[++y] = -1;
} else {
putboard();
}
} else {
y--;
}
}
}
// check if queen placement clashes with other queens
public static boolean unsafe(int y) {
int x = b[y];
for (int i = 1; i <= y; i++) {
int t = b[y - i];
if (t == x || t == x - i || t == x + i) {
return true;
}
}
return false;
}
// printing solution
public static void putboard() {
System.out.println("\n\nSolution " + (++s));
for (int y = 0; y < 8; y++) {
for (int x = 0; x < 8; x++) {
if (b[y] == x)
System.out.print("|Q");
else
System.out.print("|_");
}
System.out.println("|");
}
}
End.

Let's try to understand the code step by step. First of all, we call the function solution(), which is what leads to the execution of the puzzle answer.
Solution funcion:
public void solution() {
int y = 0;
b[0] = -1;
while (y >= 0) {
do {
b[y]++; //if last cell was unsafe or we reached the end of the board, we go for the next row.
} while ((b[y] < 8) && unsafe(y)); //Checks whether it's the last cell and if it's an unsafe cell (clashing)
if (b[y] < 8) { //We found a safe cell. Hooray!
if (y < 7) { //Did we place the last queen?
b[++y] = -1; //Nope, let's allocate the next one.
} else {
putboard(); //Yup, let's print the result!
}
} else { //If not a single safe cell was found, we reallocate the last queen.
y--;
}
}
}
On the first while, you're going to iterate through each cell in a row (or column, however you prefer it. It's just a rotation matter). On each cell you make the unsafe(y) check, which will return true in case the cell you're placing the queen in is clashing with other queen-occupied cells (as you've already found out in the comment).
Next step, once we've found a safe cell in which to place the actual queen (y), we make a security check: if we have not found a single safe cell for that queen, we have to reallocate last queen.
In case the cell was found and was correct, we check whether it was the last queen (y < 7). If it is so, we proceed to print the result. Otherwise, we just re-start the while loop by placing b[++y] = -1.
Unsafe function:
public static boolean unsafe(int y) {
int x = b[y]; //Let's call the actual cell "x"
for (int i = 1; i <= y; i++) { //For each queen before placed BEFORE the actual one, we gotta check if it's in the same row, column or diagonal.
int t = b[y - i];
if (t == x || t == x - i || t == x + i) {
return true; //Uh oh, clash!
}
}
return false; //Yay, no clashes!
}
This function checks whether the queen we're using clashes with any of the allocated queens before this one. The clashes might happen diagonally, vertically or horizontally: That is why there is a triple OR check before the "return true" statement.
Putboard function:
public static void putboard() {
System.out.println("\n\nSolution " + (++s));
for (int y = 0; y < 8; y++) {
for (int x = 0; x < 8; x++) {
if (b[y] == x)
System.out.print("|Q");
else
System.out.print("|_");
}
System.out.println("|");
}
}
I'm not gonna explain this one that deep, because it is simply a fairly simply line printing function which you can find out how it works by yourself when executing the solution!
Hope this helps.
Cheers!

Why won't the loop stop iterating?

I am a beginner java student writing a gui tic-tac-toe program for my class. (No players, just computer generated).
Everything in my program works as expected, except for one thing; it seems that the placement of my method call for checkWinner is not place correctly, because the assignment for the X's and O's always finish. Why won't the loop end as soon as there is a winner?
It will return the correct winner based on the method call, but the for-loop will continue to iterate and fill in the rest (so sometimes it looks like both the x and o win or one wins twice). I've been going crazy, thinking it might be the placement of my checkWinner method call and if statement. When I set the winner = true; shouldn't that cancel the loop? I have tried putting it between, inside and outside each for-loop with no luck :(
I have marked the area I think is the problem //What is wrong here?// off to the right of that part the code. Thank you for any input!! :)
public void actionPerformed(ActionEvent e)
{
int total = 0, i = 0;
boolean winner = false;
//stop current game if a winner is found
do{
// Generate random # 0-1 for the labels and assign
// X for a 0 value and O for a 1 value
for (int row = 0; row < gameboard.length; row++) //rows
{
for (int col = 0; col < gameboard[row].length; col++) //columns
{
//Generate random number
gameboard[row][col] = (int)(Math.random() * 2);
//Assign proper values
if(gameboard[row][col] == 0)
{
labels[i].setText("X");
gameboard[row][col] = 10; //this will help check for the winner
}
else if(gameboard[row][col] == 1)
{
labels[i].setText("O");
gameboard[row][col] = 100; //this will help check for winner
}
/**Send the array a the method to find a winner
The x's are counted as 10s
The 0s are counted as 100s
if any row, column or diag = 30, X wins
if any row, column or diag = 300, Y wins
else it will be a tie
*/
total = checkWinner(gameboard); **//Is this okay here??//**
if(total == 30 || total == 300) //
winner = true; //Shouldn't this cancel the do-while?
i++; //next label
}
}//end for
}while(!winner);//end while
//DISPLAY WINNER
if(total == 30)
JOptionPane.showMessageDialog(null, "X is the Winner!");
else if(total == 300)
JOptionPane.showMessageDialog(null, "0 is the Winner!");
else
JOptionPane.showMessageDialog(null, "It was a tie!");
}

The easiest way would be to break all loops at once. (Even if some people dont like this)
outerwhile: while(true){
// Generate random # 0-1 for the labels and assign
// X for a 0 value and O for a 1 value
for (int row = 0; row < gameboard.length; row++) //rows
{
for (int col = 0; col < gameboard[row].length; col++) //columns
{
total = checkWinner(gameboard);
if(total == 30 || total == 300)
break outerwhile; //leave outer while, implicit canceling all inner fors.
i++; //next label
}
}//end for
}//end while
This However would not allow for the "tie" option, because the while will basically restart a game, if no winner has been found. To allow tie, you dont need the outer while at all, and can leave both fors at once, when a winner is found:
Boolean winner = false;
outerfor: for (int row = 0; row < gameboard.length; row++) //rows
{
for (int col = 0; col < gameboard[row].length; col++) //columns
{
total = checkWinner(gameboard);
if(total == 30 || total == 300){
winner = true;
break outerfor; //leave outer for, implicit canceling inner for.
}
i++; //next label
}
}//end for
if (winner){
//winner
}else{
//tie.
}

First of all, your code iterates through a board and generates random marks of X and O. This leads to some very odd board states, being always filled row-by-row, and possibly with unbalanced number of X and O marks.
IMHO you should organize your code in opposite manner to fill a board similary to a true game. I mean a series of 9 marks 'XOXOXOXOX' spreaded over the board.
Let Labels labels be a nine-character array, initialized to 9 spaces.
public int doGame( Labels labels)
{
labels = " ";
int itisXmove = true; // player X or O turn
for( int movesLeft = 9; movesLeft > 0; movesLeft --)
{
int position = // 0 .. movesLeft-1
(int) Math.floor(Math.random() * movesLeft);
for( int pos = 0; pos < 9; pos ++) // find position
if( labels[ pos] == " ") // unused pos?
if( position-- == 0) // countdown
{
if( itisXmove) // use the pos
labels[ pos] = "X"; // for current player
else
labels[ pos] = "O";
break;
}
int result = checkWinner( labels); // who wins (non-zero)?
if( result != 0)
return result;
itisXmove = ! itisXmove; // next turn
}
return 0; // a tie
}
then
public void actionPerformed(ActionEvent e)
{
Labels labels;
int result = doGame( labels);
if( result == valueForX)
JOptionPane.showMessageDialog(null, "X is the Winner!");
else if( result == valueForO)
JOptionPane.showMessageDialog(null, "O is the Winner!");
else
JOptionPane.showMessageDialog(null, "It's a tie!");
for( int rowpos = 0; rowpos < 9; rowpos += 3)
{
for( int colpos = 0; colpos < 3; colpos ++)
/* output (char)label[ rowpos + colpos] */;
/* output (char)newline */;
}
}

I think you should change your loop condition and add one more bool.
You have a "tie" condition but currently you only check for winner. The only explanation without the checkWinner code is that you are encountering a tie every time.
So...
boolean tie;
boolean winner;
do {
//your stuff
}
while(!(tie || winner))
Edit: I didn't realize you put the while loop outside your for loop, you will need to break out of your for loops in order for the while condition to be checked.
//stop current game if a winner is found
do{
for (int row = 0; row < gameboard.length; row++) //rows
{
for (int col = 0; col < gameboard[row].length; col++) //columns
{
if(winner || tie)
break;
}//end for
if(winner || tie)
break;
}//end for
}while(!(winner || tie));//end while
//the rest of your stuff here

You're not checking the value of winner until both for loops complete. Add a break right after you set winner = true, and add an
if (winner)
{
break;
}
to the beginning or end of your outer for loop.

Your issue is that your do/while statement is wrapped around the for statements. So the for statements end up running their entire cycle before it ever reaches the while statement. A solution to get around this is checking for a winner in the for statements and breaking:
//stop current game if a winner is found
do {
for (int row = 0; row < gameboard.length; row++) //rows
{
for (int col = 0; col < gameboard[row].length; col++) //columns
{
// ... your other code ...
total = checkWinner(gameboard);
if(total == 30 || total == 300) {
winner = true;
break; // end current for-loop
}
i++; //next label
}
if (winner) break; // we have a winner so we want to kill the for-loop
} //end for
} while(!winner); //end while
So you should be able to just loop through the two for-statements and break upon a winner. Your code also does not seem to handle a tied case, but I am guessing you already know that.

java:generalized 8 queen to work for any initial state using depth first search [duplicate]

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
Here is my code:
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");
}
}
}
for example for this initial state it give me the following error:
Exception in thread "main" java.lang.StackOverflowError
i am stuck, i think the error is infinite call for the method how to solve this problem.
any idea will be helpful,thanks in advance.
note:the broad is two dimensional array,when i put (1) it means there queen at this point.
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;

[EDIT: Added conditional output tip.]
To add to #StephenC's answer:
This is a heck of a complicated piece of code, especially if you're not experienced in programming Java.
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)
And then crashes with this
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)
...
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.
The only two options I see are to
Get familiar with a debugger and use it to step through and analyze why it's never stopping the loop
Break it down man! How can you hope to deal with a massive problem like this without breaking it into digestible chunks???
Not to mention that the concept of 8-queens is complicated to begin with.
One further thought:
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;
and instead of
System.out.println("[ANUMBERFORTRACING]: ... USEFUL INFORMATION ...")
use
conditionalSDO((iITERATIONS < 5), "[ANUMBERFORTRACING]: ... USEFUL INFORMATION");
Here is the function:
private static final void conditionalSDO(boolean b_condition, String s_message) {
if(b_condition) {
System.out.println(s_message);
}
}
Another alternative is to not limit the output, but to write it to a file.
I hope this information helps you.
(Note: I edited the OP's code to be compilable.)

You asked for ideas on how to solve it (as distinct from solutions!) so, here's a couple of hints:
Hint #1:
If you get a StackOverflowError in a recursive program it can mean one of two things:
your problem is too "deep", OR
you've got a bug in your code that is causing it to recurse infinitely.
In this case, the depth of the problem is small (8), so this must be a recursion bug.
Hint #2:
If you examine the stack trace, you will see the method names and line numbers for each of the calls in the stack. This ... and some thought ... should help you figure out the pattern of recursion in your code (as implemented!).
Hint #3:
Use a debugger Luke ...
Hint #4:
If you want other people to read your code, pay more attention to style. Your indentation is messed up in the most important method, and you have committed the (IMO) unforgivable sin of ignoring the Java style rules for identifiers. A method name MUST start with a lowercase letter, and a class name MUST start with an uppercase letter.
(I stopped reading your code very quickly ... on principle.)

Try to alter your method IsValid in the lines where for (x = 0; x < board.length - 1; x++).
public static boolean IsValid(int[][] board, int i, int j) {
int x;
//check the queens in column
for (x = 0; x < board.length - 1; x++) {
if (board[i][x] != 0) {
return false;
}
}
//check the queens in row
for (x = 0; x < board.length - 1; x++) {
if (board[x][j] != 0) {
return false;
}
}
//check the queens in the diagonals
if (!SafeDiag(board, i, j)) {
return false;
}
return true;
}