Question

I’m trying to solve a maze using recursion. It’s declared Cell [][] maze.

public class Cell {
    private Wall left;
    private Wall right;
    private Wall up;
    private Wall down;
    private boolean end;

    // Setters and getters not shown
}

If there is no Wall for some side of the cell then it has value null, else it refers to a Wall object. Wall references are consistent: Both cells adjacent to single wall refer to it with the appropriate fields. If a wall is missing, then both adjacent cells have corresponding null entries. Here is the search:

public boolean solveMaze(Cell[][] maze, int i, int j) {

    if (maze[i][j].isEnd()){
        System.out.println(maze[i][j].toString());
        return true;
    }
    if (maze[i][j].getDown() == null) {
        return solveMaze(maze, i, j + 1); 
    }
    if (maze[i][j].getUp() == null) {
        return solveMaze(maze, i, j - 1) ;
    }
    if (maze[i][j].getLeft() == null) {
        return solveMaze(maze, i - 1, j);
    }
    if (maze[i][j].getRight() == null) {
        return solveMaze(maze, i + 1, j) ;  
    }
    return false;
}

I’m getting a Stack Overflow error. What is wrong with my recursion stop condition?

Update:

With your highly appreciated help I solved this problem: This is correct solution which works flawless:

public boolean solveMaze(Cell[][] maze, int i, int j){

    if (maze[i][j].isEnd()){
        System.out.println("Maze Exit :["+i+","+j+"]" );
        return true;
    }

    if (maze[i][j].isVisited()){
        return false;
    }

    maze[i][j].setVisited(true);

    if ((maze[i][j].getButtom() == null) ){
        if (solveMaze(maze,i,j+1)==true)
            return true;
    }

    if ((maze[i][j].getUp() == null) ){
    if ( solveMaze(maze,i,j-1) ==true )
        return true;
    }

    if ((maze[i][j].getLeft() == null)){
        if (solveMaze(maze,i-1,j))
            return true;
    }

    if ((maze[i][j].getRight() == null)){
        if (solveMaze(maze,i+1,j)) 
            return true;
    }

    maze[i][j].setVisited(false);
    return false;
} 

may be it will be helpful for any body in the future.

Answer 1

If the maze has a cycle, the solver can run around this cycle forever, which will cause the stack overflow you’re seeing. You need a way of determining when you’re seeing a maze square that’s already been seen. In this case you should backtrack immediately.

This can be done either with a boolean flag visited in each cell initially set to false and then set true for each square you search, or you can maintain a separate Set of (i,j) pairs that have been searched, which is initially empty.

NB: Your use of i and j is unconventional. If someone else wrote the maze reading code with the conventional usage, this could be causing a problem. In math, i is usually used for the row number and j for the column. With this convention your wall tests do not agree with your increments and decrements. Missing the bottom wall would require you to increment i for example.