See you've already accepted an answer, but I'll add this in anyway...
Recursion can be a really elegant way to solve some problems, but it can take a bit to get your head around. So this is not an exact answer to why your code doesn't work, but more of a higher level of how to use recursion in problems like this.
Recursion problems generally have two parts to the data: some overall puzzle state, and some state associated with the current attempt. The whole recursion thing works because each time you call the recursive function you push some new state onto the call stack, and when the function exits it is removed for you, leaving you ready to try the next option. You can also manipulate the overall puzzle state within the recursive function, but generally when you're starting I'd suggest that any changes you make to the puzzle state in your function should be reverted when it exits.
So in your case, the maze itself is puzzle state, the current path is a temporary change to the overall puzzle state, and the current position is transient state associated with the current call stack.
So the overall solution starts to take the form:
// global state
private static int[][] maze;
private static boolean solve(int r, int c) {
// return true if I'm at the exit, false otherwise
}
and the main function simply provides the starting coords:
public static void main(String[] args) {
if (solve(1, 0)) {
print();
} else {
System.out.println("no solution found");
}
}
So the next step is the body of the "solve" function (I've set the exit position to 3 in the maze data - see the full solution at the end), which becomes:
private static boolean solve(int r, int c) {
if (maze[r][c] == 3) {
// we've found the exit
return true;
}
// push the current position onto the path
maze[r][c] == 2;
// try up / down / left / right - if any of these return true then we're done
if (available(r - 1, c) && solve(r - 1, c)) {
return true;
}
if (available(r + 1, c) && solve(r + 1, c)) {
return true;
}
if (available(r, c - 1) && solve(r, c - 1)) {
return true;
}
if (available(r, c + 1) && solve(r, c + 1)) {
return true;
}
// no result found from the current position so return false
// ... but have to revert the temporary state before doing so
maze[r][c] = 0;
return false;
}
This first checks the simple case of if we're at the exit, and returns true if that is the case. If not, we push the current cell on to the path, and look for an available neighbour. If we find one we try each in turn, and this is the core of the recursion... If none of the available neighbours work then we've failed, and have to backtrack.
Finally, if we're backtracking, we have to remove the current cell from the path.
And that's about it. The 'available' function simply checks if the potential cell is in bounds and not a wall or already on the current path:
private static boolean available(int r, int c) {
return r >= 0 && r < maze.length
&& c >= 0 && c < maze[r].length
&& (maze[r][c] == 0 || maze[r][c] == 3);
}
Full code:
public class Maze2 {
private static int[][] maze =
{{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
{0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1},
{1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1},
{1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
{1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1},
{1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1},
{1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1},
{1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
{1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1},
{1,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1},
{1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1},
{1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1},
{1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1},
{1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1},
{1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1},
{1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1},
{1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1},
{1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1},
{1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1},
{1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
{1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1},
{1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,1},
{1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1},
{1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
{1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1},
{1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
{1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1},
{1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,1},
{1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1},
{1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1},
{1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1},
{1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
{1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1},
{1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1},
{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,3,1}};
public static void main(String[] args) {
if (solve(1, 0)) {
print();
} else {
System.out.println("no solution found");
}
}
private static boolean solve(int r, int c) {
// if we're at the goal then we've solved it
if (maze[r][c] == 3) {
return true;
}
// mark the current cell as on the path
maze[r][c] = 2;
// try all available neighbours - if any of these return true then we're solved
if (available(r - 1, c) && solve(r - 1, c)) {
return true;
}
if (available(r + 1, c) && solve(r + 1, c)) {
return true;
}
if (available(r, c - 1) && solve(r, c - 1)) {
return true;
}
if (available(r, c + 1) && solve(r, c + 1)) {
return true;
}
// nothing found so remove the current cell from the path and backtrack
maze[r][c] = 0;
return false;
}
// cell is available if it is in the maze and either a clear space or the
// goal - it is not available if it is a wall or already on the current path
private static boolean available(int r, int c) {
return r >= 0 && r < maze.length
&& c >= 0 && c < maze[r].length
&& (maze[r][c] == 0 || maze[r][c] == 3);
}
// use symbols to make reading the output easier...
private static final char[] SYMBOLS = {' ', '#', '.', '*' };
private static void print(){
for(int row = 0; row < maze.length; ++row) {
for(int col = 0; col < maze[row].length; ++col) {
System.out.print(SYMBOLS[maze[row][col]]);
}
System.out.println();
}
}
}
Finally, if you want to print out all possible solutions instead of just the first one found, then just change the top of the solved function to:
// if we're at the goal then print it but return false to continue searching
if (maze[r][c] == 3) {
print();
return false;
}
Happy recursing!!!