Develop Reference

Overflowing stack with recursion that should terminate?

I am attempting to make a random maze generator using Java and the recursive backtracking algorithm. I am getting stack overflow when I try to run this code. I know some about stack, I don't think this is infinite recursion. My guess is that I have a big logic error. Do I have to allocate more memory?
The stack trace:
Exception in thread "main" java.lang.StackOverflowError
at java.base/java.util.Vector.elementAt(Vector.java:499)
at java.base/java.util.Stack.peek(Stack.java:103)
at java.base/java.util.Stack.pop(Stack.java:84)
at mazeMaker.Maze.generateMaze(Maze.java:115)
at mazeMaker.Maze.generateMaze(Maze.java:115)
...
at mazeMaker.Maze.generateMaze(Maze.java:115)
at mazeMaker.Maze.generateMaze(Maze.java:115)
Main.java
package mazeMaker;
public class Main
{
public static void main(String[] args)
{
Maze mainMaze = new Maze(20, 30);
}
}
Maze.java
package mazeMaker;
import java.util.Random;
import java.util.Stack;
public class Maze
{
public int xSize = 0;
public int ySize = 0;
public int totalDimensions = 0;
Random randomGenerator = new Random();
public Cell[][] cellData;
public Stack<Cell> cellStack = new Stack<Cell>();
Cell tempCell; // Temporary variable used for maze generation
public Maze(int xSize, int ySize)
{
cellData = new Cell[xSize][ySize];
this.xSize = xSize;
this.ySize = ySize;
this.totalDimensions = this.xSize * this.ySize;
// Initialize array objects
for (int i = 0; i < this.xSize; i++)
{
for (int j = 0; j < this.ySize; j++)
{
cellData[i][j] = new Cell();
}
}
// Assign x and y positions
for (int i = 0; i < this.xSize; i++)
{
for (int j = 0; j < this.ySize; j++)
{
cellData[i][j].xPos = i;
cellData[i][j].yPos = j;
}
}
initBoundries();
generateMaze();
}
private void initBoundries()
{
// Initialize the border cells as visited so we don't go out of bounds
int m = this.xSize;
int n = this.ySize;
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
if (i == 0 || j == 0 || i == n - 1 || j == n - 1)
cellData[i][j].hasBeenVisited = true;
}
}
}
private void generateMaze(int x, int y)
{
// Set current cell as visited
cellData[x][y].hasBeenVisited = true;
// While there are unvisited neighbors
while (!cellData[x][y+1].hasBeenVisited || !cellData[x+1][y].hasBeenVisited || !cellData[x][y-1].hasBeenVisited || !cellData[x-1][y].hasBeenVisited)
{
// Select a random neighbor
while (true)
{
int r = randomGenerator.nextInt(4);
if (r == 0 && !cellData[x][y+1].hasBeenVisited)
{
cellStack.push(cellData[x][y]);
cellData[x][y].hasNorthWall = false;
cellData[x][y+1].hasSouthWall = false;
generateMaze(x, y + 1);
break;
}
else if (r == 1 && !cellData[x+1][y].hasBeenVisited)
{
cellStack.push(cellData[x][y]);
cellData[x][y].hasEastWall = false;
cellData[x+1][y].hasWestWall = false;
generateMaze(x+1, y);
break;
}
else if (r == 2 && !cellData[x][y-1].hasBeenVisited)
{
cellStack.push(cellData[x][y]);
cellData[x][y].hasSouthWall = false;
cellData[x][y-1].hasNorthWall = false;
generateMaze(x, y-1);
break;
}
else if (r == 3 && !cellData[x-1][y].hasBeenVisited)
{
cellStack.push(cellData[x][y]);
cellData[x][y].hasWestWall = false;
cellData[x-1][y].hasEastWall = false;
generateMaze(x-1, y);
break;
}
}
}
// There are no unvisited neighbors
tempCell = cellStack.pop();
generateMaze(tempCell.xPos, tempCell.yPos);
}
// Begin generating maze at top left corner
private void generateMaze()
{
generateMaze(1,1);
}
}
Cell.java
package mazeMaker;
public class Cell
{
public boolean isCurrentCell;
public boolean hasBeenVisited;
public boolean hasNorthWall;
public boolean hasSouthWall;
public boolean hasEastWall;
public boolean hasWestWall;
public int xPos;
public int yPos;
}

The method generateMaze can never terminate not even by chance for some simple reason:
For terminating the generateMaze method would need to finish it's execution - it has to return.
There are no return statements in this method, therefore it has to pass the while loops and then continue until the execution reaches and finishes the last statement of the method.
However the last statement is generateMaze(tempCell.xPos, tempCell.yPos); which starts a new recursion, therefore your code can never ever terminate!

I tried to run your project on my own environment but unfortunately, I was not able to reproduce your issue.
However, I was facing an IndexOutOfBound exception in the method generateMaze. While I was solving this, I figured out that there was an issue in the initBoudaries method.
Indeed, when you set the boolean hasBeenVisited to true, you do not use the right variable in the IF clause. Here is the version I tried instead :
private void initBoundries()
{
// Initialize the border cells as visited so we don't go out of bounds
for (int i = 0; i < this.xSize; i++)
{
for (int j = 0; j < ySize; j++)
{
if (i == 0 || j == 0 || i == xSize - 1 || j == ySize - 1)
cellData[i][j].hasBeenVisited = true;
}
}
}
Now about the emptyStackException, I think that if this stack is empty, this means that there is no more cell to handle (as you mentioned in your comment) and the program must end. If I am right, just make sure to test if your stack is empty before call the method pop() on it like this :
// There are no unvisited neighbors
if (!cellStack.isEmpty()) {
tempCell = cellStack.pop();
generateMaze(tempCell.xPos, tempCell.yPos);
}
Hope it will help.

Java 2D Array Specific Move

I have made a class where a 6x10 2D array is generated to act as a board.
A random starting location is then generated in the constructor.I only want adjacent moves to be possible.
For example, if the random location has been generated as (2,3) then for example the user enters (1,2) it would be a valid move, but (6,1) would be an invalid move.
Then if the user enters say (1,2), they can then go to any adjacent cell from (1,2).
I have included the class below, and the adjacent method I tried to make to test it, but I'm a bit confused on how I am approaching this.
import java.util.Arrays;
import java.util.Random;
public class Test {
public static final int ROWS = 6;
public static final int COLUMNS = 10;
public int[][] board;
public static void main(String[] args)
{
Test t = new Test();
t.getBoard();
t.makeMove(6,1); //I want this to be an invalid move.
t.getBoard();
t.makeMove(1,2); // this should be a valid move
t.getBoard();
}
public Test()
{
board = new int[ROWS][COLUMNS];
createRandomLocation();
}
public void createRandomLocation()
{
Random rand = new Random();
int x = rand.nextInt(6);
int y = rand.nextInt(10);
board[x][y] = 1;
}
public void makeMove(int x,int y){
if (Math.abs(x-cur_x)==0 || Math.abs(y-cur_y)==0) {
board[x][y] = 1;
}
public String getBoard() {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[i].length; j++) {
System.out.print(board[i][j] + " ");
}
System.out.println();
}
System.out.println();
return Arrays.deepToString(board);
}
}
Adjacent:
/*public boolean isMoveAllowed(int [][] array,int x, int y){
boolean adjacent = false;
int trueCount = 0;
if(array[x-1][y-1] == 0) trueCount++; //topleft
if(array[x-1][y] == 0) trueCount++; //top
if(array[x-1][y+1] == 0) trueCount++;//topright
if(array[x][y+1] == 0) trueCount++;//right
if(array[x][y-1] == 0) trueCount++;//left
if(array[x+1][y-1] == 0) trueCount++;//bottomleft
if(array[x+1][y] == 0) trueCount++;//bottom
if(array[x+1][y+1] == 0) trueCount++; //bottomright
if (trueCount == 8)
{
adjacent = true;
}
return adjacent;
}*/

Your problem description has the answer baked into it already. You want any move from (a,b) to (c,d) to be legal if the distance between a and c, and b and d, is zero or one. So if you see Math.abs(a-c)>1, that's an illegal move. So: have the current position stored in some variables, and compare them to the desired new location:
public static void main(String[] args)
{
Board b = new Board(6, 10);
try {
b.tryMove(6,1);
} catch(IllegalMoveException e) {
// do whatever you need to do to inform the user that move is illegal
}
}
With the Board class responsible for tracking coordinates:
class Board {
protected int cur_x, cur_y, rows, cols;
public Board(int rows, int cols) {
this.rows = rows;
this.cols = cols;
this.setRandomPosition();
}
public void setRandomPosition() {
cur_x = (int) Math.round(Math.random() * cols);
cur_y = (int) Math.round(Math.random() * rows);
}
public void tryMove(int x, int y) throws IllegalMoveException {
if (Math.abs(x-cur_x)>1 || Math.abs(y-cur_y)>1) {
throw new IllegalMoveException(...);
}
// bounds check omitted here, but: ensure that
// 0<=x<cols and 0<=y<rows, otherwise throw an
// IllegalMoveException as well.
cur_x = x;
cur_y = y;
}
// with getters for the current x and y, etc.
}

It would be much easier to test for a true case rather than a false case like you currently have, the isMoveAllowed method should look something like this:
public boolean isMoveAllowed(int[][] array, int x, int y) {
return ((array[x + 1][y] == 1) ||
(array[x - 1][y] == 1) ||
(array[x][y + 1] == 1) ||
(array[x][y - 1] == 1));
}
This will return true if the move is adjacent to the current player position

Yahtzee two pair can't get any value

I am in a beginner's java class. I have had this Yahtzee program now for weeks and I still cannot get this figured out.
Having problem getting a points from my checkTwopairs(). checkTwopairs() can see if there is a two pair. But I have a hard problem getting the points out of it. Any good way to do it?
public int checknumber(int nr) {
int sum = 0;
for (Dice d : diceList) {
if (d.getValue() == nr ){
sum++;
}
}
return sum;
}
public void checkTwopairs() {
for (int i = 1; i <= 6; i++) {
int a = checknumber(i);
if (a == 2) {
if (twopair == true && a == 2) {
} else {
twopair = true;
}
}
}
twopair = false;
}

I'm guessing you're looking for 2 pairs, meaning you get dice with {5,5,2,2,1} where you have 2 pairs -- a pair of 5's, and a pair of 2's.
I think you may be looking for something like this to modify your code:
public void checkTwopairs() {
boolean firstPair = false; // New local variable.
for (int i = 1; i <= 6; i++) {
int a = checknumber(i);
if (a == 2) {
//if (twopair == true && a == 2) <--- This second clause is unnecessary as we can only get here if a == 2
if (firstPair == true) {
twopair = true;
return; //This returns the method, so twopair cannot be set to false if two pairs are found
} else {
firstPair = true;
}
}
}
//Checked all dice, less than two pairs were found.
twopair = false;
}

Here is a one-method solution to replace both of those methods:
public void checkTwoPairs() {
int numPairs = 0;
//two pairs haven't been found yet
boolean twoPair = false;
//for each possible number
for(int i = 1; i <= 6; i++) {
int sum = 0;
//for each dice object
for(Dice d : diceList) {
//if i is the dice value
if(d.getValue() == i) {
//increment the number of matches
sum++;
}
}
//sum % 2 - gives you the number of pairs for that number checked
numPairs += (sum%2);
//if you have 2 pairs
if(numPairs == 2) {
//set twoPair = true ... and break out of the loop
twoPair = true;
break;
}
//if you don't have 2 pairs yet, go back through the loop
}
}

Array programming - check winner in a Tic Tac Toe game for an nxn board with n players

I am making a tic tac toe game for n number of players on a nxn board, but the winning condition is aways 3 on a row. My so far solution to the problem is: when a move is made the program will check the following square for 3 on a row.
(x-1,y+1) (x,y+1) (x+1,y+1)
(x-1,y) (x,y) (x+1,y)
(x-1,y-1) (x,y-1) (x+1,y-1)
It will check the top (x-1,y+1) (x,y+1) (x+1,y+1) bottom(x-1,y-1) (x,y-1) (x+1,y-1)
sides(x+1,y+1) (x+1,y) (x+1,y-1) , (x-1,y+1) (x-1,y) (x-1,y-1) , the diagonals and the ones going through the middle(x,y).
my code so far is:
public int checkWinning() {
for(int a = 1; a < size-1; a++){
for(int b = 1; b < size-1; b++){
if (board[a][b] == board[a+1][b] && board[a][b] == board[a-1][b]){
return board[a][b];
}else if(board[a][b] == board[a][b+1] && board[a][b] == board[a][b-1]){
return board[a][b];
}else if(board[a][b] == board[a+1][b-1] && board[a][b] == board[a-1][b+1]){
return board[a][b];
}else if(board[a][b] == board[a+1][b+1] && board[a][b] == board[a-1][b-1]){
return board[a][b];
}
}
}
for(int d = 1; d < size-1; d++){
if (board[0][d] == board[0][d-1] && board[0][d] == board[0][d+1]){
return board[0][d];
} else if (board[size-1][d] == board[size-1][d-1] && board[size-1][d] == board[size-1][d+1]){
return board[size-1][d];
}
}
for(int c = 1; c < size-1; c++){
if (board[c][0] == board[c-1][0] && board[c][0] == board[c+1][0]){
return board[c][0];
}else if(board[c][size-1] == board[c-1][size-1] && board[c][size-1] == board[c+1][size-1]){
return board[c][size-1];
}
}
return 0;
}
where the first section is where I check the ones through the middle and diagonals. the second section I check the top an bottom and the top and the thrid section checks the sides.
When it returns 0 is means that there are no winner yet.
#override
public void checkResult() {
int winner = this.board.checkWinning();
if (winner > 0) {
this.ui.showResult("Player "+winner+" wins!");
}
if (this.board.checkFull()) {
this.ui.showResult("This is a DRAW!");
}
}
Board[x][y] -> 2-dimensional array representing the board, The coordinates are counted from top-left (0,0) to bottom-right (size-1, size-1), board[x][y] == 0 signifies free at position (x,y), board[x][y] == i for i > 0 signifies that Player i made a move on (x,y), just so you know it.
my problem is that when i expands the board to a size larger than 3x3 the program somehow overwrites it self or a does not check every thing sides top and bottom every time, and I can't seem too se why.

EDIT:
played with the app for a few minutes... interesting results
java -jar tic-tac-toe.jar 5 20
It was a cats game!!
|1|1|5|5|1|3|5|3|1|5|2|5|1|1|2|
|2|3|2|3|1|5|3|5|3|2|3|1|5|2|2|
|5|4|5|4|1|5|5|4|2|1|4|5|4|2|2|
|3|2|1|5|5|5|2|4|5|3|4|1|2|4|2|
|3|4|1|2|5|4|1|1|4|5|1|3|3|4|1|
|1|5|4|4|3|2|5|1|3|5|1|3|5|3|4|
|2|5|1|4|3|3|3|5|3|1|1|4|3|4|4|
|1|4|5|1|1|5|4|5|2|4|1|1|5|4|3|
|1|3|2|1|4|2|4|3|3|4|5|2|4|3|3|
|5|1|1|3|3|4|4|4|2|2|1|4|3|2|5|
|2|2|3|1|5|5|4|1|3|5|3|2|3|3|2|
|2|4|2|4|4|1|3|1|1|3|1|2|1|2|2|
|2|5|5|1|4|3|4|5|5|4|5|3|3|5|2|
|4|5|2|1|5|3|2|1|3|2|2|2|2|4|4|
|4|1|1|4|5|4|5|4|2|2|3|3|2|2|3|
Played 100 games:
Number wins by Player1: 0
Number wins by Player2: 0
Number wins by Player3: 0
Number wins by Player4: 0
Number wins by Player5: 0
Number of ties: 100
didn't scroll through all 100 games to find the winning board, but I thought this was interesting:
java -jar tic-tac-toe.jar 2 10
Player2 won the game!
|1|1|2|1|2|2| |2|1|2|
|2|2|2|2|2|2|2|2|2|2|
|2|1|2|2|2|1|1|1|1|1|
|1|1|1|1|2|1|2|1|1|1|
|2|2| |1|2|1|1|1|1|2|
|2|2|2|1|1|1| |1|2|2|
|2|2|1|2|2|2|2|2|1|1|
| | |2|2|2|2| |1|1|1|
|1|1|2|2|2|1|1|1|1| |
| | |1|1|1|1|1|2|1| |
Played 100 games:
Number wins by Player1: 0
Number wins by Player2: 1
Number of ties: 99
This does answer your question... but I took it a bit far... decided to implement the solution.
Instead of counting matches... I just check from teh point the last player plays, if all marks in a row column and diagnal match the players, he wins.
package com.clinkworks.example;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class TicTacToe {
private static final String TIE = "TIE";
private static final Map<String, Integer> gamesToWinsMap = new HashMap<String, Integer>();
/**
* accepts input in the following format:
*
* playerCount rowCount columnCount (sets the game with the n players, n columns, and n rows)
* - java -jar tic-tac-toe.jar 2 3 3
* PlayerCount squareSize (defaults to a game with rows and cols the same as squareSize and the player count given)
* - java -jar tic-tac-toe.jar 2 3
* PlayerCount (defaults to a 3 by 3 game)
* - java -jar tic-tac-toe.jar 2
* no input (defaults to a 3 by 3 game with 2 players)
* - java -jar tic-tac-toe.jar
* #param args
*/
public static void main(String[] args) {
int playerCount = 2;
int rows = 3;
int cols = 3;
if(args.length == 3){
playerCount = Integer.valueOf(args[0]);
rows = Integer.valueOf(args[1]);
cols = Integer.valueOf(args[2]);
}
if(args.length == 2){
playerCount = Integer.valueOf(args[0]);
rows = Integer.valueOf(args[1]);
cols = rows;
}
if(args.length == 1){
playerCount = Integer.valueOf(args[0]);
}
for(int i = 1; i <= playerCount; i++){
gamesToWinsMap.put("Player" + i, 0);
}
//lets play 100 games and see the wins and ties
playGames(100, playerCount, rows, cols);
for(int i = 1; i <= playerCount; i++){
System.out.println("Number wins by Player" + i + ": " + gamesToWinsMap.get("Player" + i));
}
System.out.println("Number of ties: " + gamesToWinsMap.get(TIE));
}
public static void playGames(int gamesToPlay, int playerCount, int rows, int cols) {
//play a new game each iteration, in our example, count = 100;
for (int i = 0; i < gamesToPlay; i++) {
playGame(playerCount, rows, cols);
}
}
public static void playGame(int playerCount, int rows, int cols) {
//create a new game board. this initalizes our 2d array and lets the complexity of handling that
// array be deligated to the board object.
Board board = new Board(playerCount, rows, cols);
//we are going to generate a random list of moves. Heres where we are goign to store it
List<Move> moves = new ArrayList<Move>();
//we are creating moves for each space on the board.
for (int row = 0; row < rows; row++) {
for (int col = 0; col < cols; col++) {
moves.add(new Move(row, col));
}
}
//randomize the move list
Collections.shuffle(moves);
//do each move
for (Move move : moves) {
board.play(move);
if(gameOver(board)){
break;
}
}
}
public static boolean gameOver(Board board){
if (board.whoWon() != null) {
System.out.println(board.whoWon() + " won the game!");
System.out.println(board);
Integer winCount = gamesToWinsMap.get(board.whoWon());
winCount = winCount == null ? 1 : winCount + 1;
gamesToWinsMap.put(board.whoWon(), winCount);
return true;
} else if (board.movesLeft() == 0) {
System.out.println("It was a cats game!!");
System.out.println(board);
Integer tieCount = gamesToWinsMap.get(TIE);
tieCount = tieCount == null ? 1 : tieCount + 1;
gamesToWinsMap.put(TIE, tieCount);
return true;
}
return false;
}
public static class Move {
private int row;
private int column;
public Move(int row, int column) {
this.row = row;
this.column = column;
}
public int getRow() {
return row;
}
public int getColumn() {
return column;
}
}
public static class Board {
private final int rowSize;
private final int columnSize;
private final Integer[][] gameBoard;
private int playerCount;
private int currentPlayer;
private String winningPlayer;
public Board() {
gameBoard = new Integer[3][3];
currentPlayer = 1;
winningPlayer = null;
this.rowSize = 3;
this.columnSize = 3;
playerCount = 2;
}
public Board(int players) {
gameBoard = new Integer[3][3];
currentPlayer = 1;
winningPlayer = null;
this.rowSize = 3;
this.columnSize = 3;
playerCount = players;
}
public Board(int rowSize, int columnSize) {
gameBoard = new Integer[rowSize][columnSize];
currentPlayer = 1;
winningPlayer = null;
playerCount = 2;
this.rowSize = rowSize;
this.columnSize = columnSize;
}
public Board(int players, int rowSize, int columnSize) {
gameBoard = new Integer[rowSize][columnSize];
currentPlayer = 1;
winningPlayer = null;
playerCount = players;
this.rowSize = rowSize;
this.columnSize = columnSize;
}
/**
*
* #return the amount of empty spaces remaining on the game board, or if theres a winning player, zero.
*/
public int movesLeft() {
if(whoWon() != null){
return 0;
}
int moveCount = 0;
for (int x = 0; x < getRowSize(); x++) {
for (int y = 0; y < getColumnSize(); y++) {
moveCount += getMoveAt(x, y) == null ? 1 : 0;
}
}
return moveCount;
}
/**
* If someone won, this will return the winning player.
*
* #return the winning player
*/
public String whoWon() {
return winningPlayer;
}
/**
* This move allows the next player to choose where to place their mark.
*
* #param row
* #param column
* #return if the game is over, play will return true, otherwise false.
*/
public boolean play(Move move) {
if (!validMove(move)) {
// always fail early
throw new IllegalStateException("Player " + getCurrentPlayer() + " cannot play at " + move.getRow() + ", " + move.getColumn() + "\n" + toString());
}
doMove(move);
boolean playerWon = isWinningMove(move);
if (playerWon) {
winningPlayer = "Player" + getCurrentPlayer();
return true;
}
shiftPlayer();
boolean outOfMoves = movesLeft() <= 0;
return outOfMoves;
}
public int getRowSize() {
return rowSize;
}
public int getColumnSize() {
return columnSize;
}
public int getCurrentPlayer() {
return currentPlayer;
}
public Integer getMoveAt(int row, int column) {
return gameBoard[row][column];
}
private void doMove(Move move) {
gameBoard[move.getRow()][move.getColumn()] = getCurrentPlayer();
}
private void shiftPlayer() {
if(getCurrentPlayer() == getPlayerCount()){
currentPlayer = 1;
}else{
currentPlayer++;
}
}
private int getPlayerCount() {
return playerCount;
}
private boolean validMove(Move move) {
boolean noMoveAtIndex = false;
boolean indexesAreOk = move.getRow() >= 0 || move.getRow() < getRowSize();
indexesAreOk = indexesAreOk && move.getColumn() >= 0 || move.getColumn() < getColumnSize();
if (indexesAreOk) {
noMoveAtIndex = getMoveAt(move.getRow(), move.getColumn()) == null;
}
return indexesAreOk && noMoveAtIndex;
}
private boolean isWinningMove(Move move) {
// since we check to see if the player won on each move
// we are safe to simply check the last move
return winsDown(move) || winsAcross(move) || winsDiagnally(move);
}
private boolean winsDown(Move move) {
boolean matchesColumn = true;
for (int i = 0; i < getColumnSize(); i++) {
Integer moveOnCol = getMoveAt(move.getRow(), i);
if (moveOnCol == null || getCurrentPlayer() != moveOnCol) {
matchesColumn = false;
break;
}
}
return matchesColumn;
}
private boolean winsAcross(Move move) {
boolean matchesRow = true;
for (int i = 0; i < getRowSize(); i++) {
Integer moveOnRow = getMoveAt(i, move.getColumn());
if (moveOnRow == null || getCurrentPlayer() != moveOnRow) {
matchesRow = false;
break;
}
}
return matchesRow;
}
private boolean winsDiagnally(Move move) {
// diagnals we only care about x and y being teh same...
// only perfect squares can have diagnals
// so we check (0,0)(1,1)(2,2) .. etc
boolean matchesDiagnal = false;
if (isOnDiagnal(move.getRow(), move.getColumn())) {
matchesDiagnal = true;
for (int i = 0; i < getRowSize(); i++) {
Integer moveOnDiagnal = getMoveAt(i, i);
if (moveOnDiagnal == null || moveOnDiagnal != getCurrentPlayer()) {
matchesDiagnal = false;
break;
}
}
}
return matchesDiagnal;
}
private boolean isOnDiagnal(int x, int y) {
if (boardIsAMagicSquare()) {
return x == y;
} else {
return false;
}
}
private boolean boardIsAMagicSquare() {
return getRowSize() == getColumnSize();
}
public String toString() {
StringBuffer stringBuffer = new StringBuffer();
for(int y = 0; y < getColumnSize(); y++) {
for(int x = 0; x < getRowSize(); x++) {
Integer move = getMoveAt(x, y);
String moveToPrint = "";
if (move == null) {
moveToPrint = " ";
} else {
moveToPrint = move.toString();
}
stringBuffer.append("|").append(moveToPrint);
}
stringBuffer.append("|\n");
}
return stringBuffer.toString();
}
}
}

I have to revise my answer. If you want to have three in a row regardless of your board size, your loop code might be sufficient, but you are always checking whether the values of the fields are the same but never make a difference between empty and non-empty fields.
So “empty” can win too, which would effectively hide a possible win of a player. In other words, your code does not work correctly, even for a field size of three. You didn’t test it enough.
If I initialize the board as
int[][] board={
{ 1, 1, 1 },
{ 0, 0, 0 },
{ 0, 0, 0 },
};
your code returns 0 as the second row contains three zeros. I assumed that 0 represents the empty field but the actual value for “empty” doesn’t matter. You have to exclude empty fields from the three-in-a-row check.

You can simplify this a fair amount by breaking the logic up a bit.
First realize that you only need to check for a win around the piece you just placed.
Now we need a way to check whether that move is a winner.
First we need a simple function to check whether a cell matches a given value, returning true if its within bounds and matches.
private boolean cellMatches(int x, int y, int val) {
if (x<0||x>boardWidth)
return false;
if (y<0||y>boardHeight)
return false;
return board[x][y]==val;
}
Now a function that you give a starting position (x and y) and a delta (dx, dy) and it checks up to two cells in that direction returning a count of how many in a row matched value. The for loop may be overkill for two checks but it would easily allow you to expand up to longer lines being used.
private int countMatches(int x, int y, int dx, int dy, int val) {
int count = 0;
for (int step=1;step<=2;step++) {
if (cellMatches(x+dx*step, y+dy*step, val) {
count++;
} else {
return count;
}
}
return count;
}
Now we can use the previous method. When we place a new piece we can just count out in each matching pair of directions. The combined count is the total number in a row. (i.e. two in a row top + 1 bot = a total run length of 4). If any of those run lengths is three then it is a winning move.
private boolean makeMove(int x, int y, int val) {
board[x][y] = val;
int runlength=countMatches(x,y,0,1,val) + countMatches(x,y,0,-1,val);
if (runLength >= 2)
return true;
int runlength=countMatches(x,y,1,0,val) + countMatches(x,y,-1,0,val);
if (runLength >= 2)
return true;
int runlength=countMatches(x,y,1,1,val) + countMatches(x,y,-1,-1,val);
if (runLength >= 2)
return true;
int runlength=countMatches(x,y,1,-1,val) + countMatches(x,y,-1,1,val);
if (runLength >= 2)
return true;
return false;
}
Note that because we need to count the center piece that we placed we just need a run length of two or more.

8 Non-Attacking Queens Algorithm with Recursion

I'm having trouble coding the 8 queens problem. I've coded a class to help me solve it, but for some reason, I'm doing something wrong. I kind of understand what's supposed to happen.
Also, we have to use recursion to solve it but I have no clue how to use the backtracking I've read about, so I just used it in the methods checking if a position is legitimate.
My board is String [] [] board = { { "O", "O"... etc etc with 8 rows and 8 columns.
If I'm getting anything wrong conceptually or making a grave Java mistake, please say so :D
Thanks!
public void solve () {
int Queens = NUM_Queens - 1;
while (Queens > 0) {
for (int col = 0; col < 8; col++) {
int row = -1;
boolean c = false;
while (c = false && row < 8) {
row ++;
c = checkPos (row, col);
}
if (c == true) {
board[row][col] = "Q";
Queens--;
}
else
System.out.println("Error");
}
}
printBoard ();
}
// printing the board
public void printBoard () {
String ret = "";
for (int i = 0; i < 8; i++) {
for (int a = 0; a < 8; a++)
ret += (board[i][a] + ", ");
ret += ("\n");
}
System.out.print (ret);
}
// checking if a position is a legitimate location to put a Queen
public boolean checkPos (int y, int x) {
boolean r = true, d = true, u = true, co = true;
r = checkPosR (y, 0);
co = checkPosC (0, x);
int col = x;
int row = y;
while (row != 0 && col != 0 ) { //setting up to check diagonally downwards
row--;
col--;
}
d = checkPosDD (row, col);
col = x;
row = y;
while (row != 7 && col != 0 ) { //setting up to check diagonally upwards
row++;
col--;
}
d = checkPosDU (row, col);
if (r = true && d = true && u = true && co = true)
return true;
else
return false;
}
// checking the row
public boolean checkPosR (int y, int x) {
if (board[y][x].contentEquals("Q"))
return false;
else if (board[y][x].contentEquals("O") && x == 7)
return true;
else //if (board[y][x].contentEquals("O"))
return checkPosR (y, x+1);
}
// checking the column
public boolean checkPosC (int y, int x) {
if (board[y][x].contentEquals("Q"))
return false;
else if (board[y][x].contentEquals("O") && y == 7)
return true;
else //if (board[y][x].contentEquals("O"))
return checkPosR (y+1, x);
}
// checking the diagonals from top left to bottom right
public boolean checkPosDD (int y, int x) {
if (board[y][x].contentEquals("Q"))
return false;
else if (board[y][x].contentEquals("O") && (x == 7 || y == 7))
return true;
else //if (board[y][x].contentEquals("O"))
return checkPosR (y+1, x+1);
}
// checking the diagonals from bottom left to up right
public boolean checkPosDU (int y, int x) {
if (board[y][x].contentEquals("Q"))
return false;
else if (board[y][x].contentEquals("O") && (x == 7 || y == 0))
return true;
else //if (board[y][x].contentEquals("O"))
return checkPosR (y-1, x+1);
}
}

As this is homework, the solution, but not in code.
Try to write a method that only handles what needs to happen on a single column; this is where you are supposed to use recursion. Do backtracking by checking if a solution exists, if not, undo your last change (i.e. change the queen position) and try again. If you only focus on one part of the problem (one column), this is much easier than thinking about all columns at the same time.
And as Quetzalcoatl points out, you are assigning false to your variable in the first loop. You probably do not want to do that. You should always enable all warnings in your compiler (run javac with -Xlint) and fix them.

You are trying some kind of brute-force, but, as you already mentioned, you have no recursion.
Your programs tries to put a queen on the first possible position. But at the end no solution is found. It follows that your first assumption (the position of your first queen) is invalid. You have to go back to this state. And have to assume that your checkPos(x,y) is false instead of true.
Now some hints:
As mentioned before by NPE int[N] queens is more suitable representation.
sum(queens) have to be 0+1+2+3+4+5+6+7=28, since a position has to be unique.
Instead of checking only the position of the new queen, you may check a whole situation. It is valid if for all (i,j) \in N^2 with queen(i) = j, there exists no (k,l) != (i,j) with abs(k-i) == abs(l-j)