We have discussed Backtracking and Knight’s tour problem in Set 1. Let us discuss Rat in a Maze as another example problem that can be solved using Backtracking.
A Maze is given as N*N binary matrix of blocks where source block is the upper left most block i.e., maze[0][0] and destination block is lower rightmost block i.e., maze[N-1][N-1]. A rat starts from source and has to reach the destination. The rat can move only in two directions: forward and down.
In the maze matrix, 0 means the block is a dead end and 1 means the block can be used in the path from source to destination. Note that this is a simple version of the typical Maze problem. For example, a more complex version can be that the rat can move in 4 directions and a more complex version can be with a limited number of moves.
Following is an example maze.
Gray blocks are dead ends (value = 0).

Following is a binary matrix representation of the above maze.
{1, 0, 0, 0}
{1, 1, 0, 1}
{0, 1, 0, 0}
{1, 1, 1, 1}
Following is a maze with highlighted solution path.

Following is the solution matrix (output of program) for the above input matrix.
{1, 0, 0, 0}
{1, 1, 0, 0}
{0, 1, 0, 0}
{0, 1, 1, 1}
All entries in solution path are marked as 1.
Backtracking Algorithm: Backtracking is an algorithmic-technique for solving problems recursively by trying to build a solution incrementally. Solving one piece at a time, and removing those solutions that fail to satisfy the constraints of the problem at any point of time (by time, here, is referred to the time elapsed till reaching any level of the search tree) is the process of backtracking.
Approach: Form a recursive function, which will follow a path and check if the path reaches the destination or not. If the path does not reach the destination then backtrack and try other paths.
Algorithm:
- Create a solution matrix, initially filled with 0’s.
- Create a recursive function, which takes initial matrix, output matrix and position of rat (i, j).
- if the position is out of the matrix or the position is not valid then return.
- Mark the position output[i][j] as 1 and check if the current position is destination or not. If destination is reached print the output matrix and return.
- Recursively call for position (i+1, j) and (i, j+1).
- Unmark position (i, j), i.e output[i][j] = 0.
C++
#include <bits/stdc++.h>
using namespace std;
#define N 4
bool solveMazeUtil( int maze[N][N], int x, int y, int sol[N][N]);
void printSolution( int sol[N][N])
{
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < N; j++)
cout<< " " <<sol[i][j]<< " " ;
cout<<endl;
}
}
bool isSafe( int maze[N][N], int x, int y)
{
if (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)
return true ;
return false ;
}
bool solveMaze( int maze[N][N])
{
int sol[N][N] = { { 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 } };
if (solveMazeUtil(maze, 0, 0, sol) == false ) {
cout<< "Solution doesn't exist" ;
return false ;
}
printSolution(sol);
return true ;
}
bool solveMazeUtil( int maze[N][N], int x, int y, int sol[N][N])
{
if (x == N - 1 && y == N - 1 && maze[x][y] == 1) {
sol[x][y] = 1;
return true ;
}
if (isSafe(maze, x, y) == true ) {
if (sol[x][y] == 1)
return false ;
sol[x][y] = 1;
if (solveMazeUtil(maze, x + 1, y, sol) == true )
return true ;
if (solveMazeUtil(maze, x - 1, y, sol) == true )
return true ;
if (solveMazeUtil(maze, x, y + 1, sol) == true )
return true ;
if (solveMazeUtil(maze, x, y - 1, sol) == true )
return true ;
sol[x][y] = 0;
return false ;
}
return false ;
}
int main()
{
int maze[N][N] = { { 1, 0, 0, 0 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 0 },
{ 1, 1, 1, 1 } };
solveMaze(maze);
return 0;
}
|
C
#include <stdio.h>
#include <stdbool.h>
#define N 4
bool solveMazeUtil( int maze[N][N], int x, int y, int sol[N][N]);
void printSolution( int sol[N][N])
{
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < N; j++)
printf ( " %d " , sol[i][j]);
printf ( "\n" );
}
}
bool isSafe( int maze[N][N], int x, int y)
{
if (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)
return true ;
return false ;
}
bool solveMaze( int maze[N][N])
{
int sol[N][N] = { { 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 0, 0 } };
if (solveMazeUtil(maze, 0, 0, sol) == false ) {
printf ( "Solution doesn't exist" );
return false ;
}
printSolution(sol);
return true ;
}
bool solveMazeUtil( int maze[N][N], int x, int y, int sol[N][N])
{
if (x == N - 1 && y == N - 1 && maze[x][y] == 1) {
sol[x][y] = 1;
return true ;
}
if (isSafe(maze, x, y) == true ) {
if (sol[x][y] == 1)
return false ;
sol[x][y] = 1;
if (solveMazeUtil(maze, x + 1, y, sol) == true )
return true ;
if (solveMazeUtil(maze, x, y + 1, sol) == true )
return true ;
sol[x][y] = 0;
return false ;
}
return false ;
}
int main()
{
int maze[N][N] = { { 1, 0, 0, 0 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 0 },
{ 1, 1, 1, 1 } };
solveMaze(maze);
return 0;
}
|
Java
public class RatMaze {
static int N;
void printSolution( int sol[][])
{
for ( int i = 0 ; i < N; i++) {
for ( int j = 0 ; j < N; j++)
System.out.print(
" " + sol[i][j] + " " );
System.out.println();
}
}
boolean isSafe(
int maze[][], int x, int y)
{
return (x >= 0 && x < N && y >= 0
&& y < N && maze[x][y] == 1 );
}
boolean solveMaze( int maze[][])
{
int sol[][] = new int [N][N];
if (solveMazeUtil(maze, 0 , 0 , sol) == false ) {
System.out.print( "Solution doesn't exist" );
return false ;
}
printSolution(sol);
return true ;
}
boolean solveMazeUtil( int maze[][], int x, int y,
int sol[][])
{
if (x == N - 1 && y == N - 1
&& maze[x][y] == 1 ) {
sol[x][y] = 1 ;
return true ;
}
if (isSafe(maze, x, y) == true ) {
if (sol[x][y] == 1 )
return false ;
sol[x][y] = 1 ;
if (solveMazeUtil(maze, x + 1 , y, sol))
return true ;
if (solveMazeUtil(maze, x, y + 1 , sol))
return true ;
sol[x][y] = 0 ;
return false ;
}
return false ;
}
public static void main(String args[])
{
RatMaze rat = new RatMaze();
int maze[][] = { { 1 , 0 , 0 , 0 },
{ 1 , 1 , 0 , 1 },
{ 0 , 1 , 0 , 0 },
{ 1 , 1 , 1 , 1 } };
N = maze.length;
rat.solveMaze(maze);
}
}
|
Python3
n = 4
def isValid(n, maze, x, y, res):
if x > = 0 and y > = 0 and x < n and y < n and maze[x][y] = = 1 and res[x][y] = = 0 :
return True
return False
def RatMaze(n, maze, move_x, move_y, x, y, res):
if x = = n - 1 and y = = n - 1 :
return True
for i in range ( 4 ):
x_new = x + move_x[i]
y_new = y + move_y[i]
if isValid(n, maze, x_new, y_new, res):
res[x_new][y_new] = 1
if RatMaze(n, maze, move_x, move_y, x_new, y_new, res):
return True
res[x_new][y_new] = 0
return False
def solveMaze(maze):
res = [[ 0 for i in range (n)] for i in range (n)]
res[ 0 ][ 0 ] = 1
move_x = [ - 1 , 1 , 0 , 0 ]
move_y = [ 0 , 0 , - 1 , 1 ]
if RatMaze(n, maze, move_x, move_y, 0 , 0 , res):
for i in range (n):
for j in range (n):
print (res[i][j], end = ' ' )
print ()
else :
print ( 'Solution does not exist' )
if __name__ = = "__main__" :
maze = [[ 1 , 0 , 0 , 0 ],
[ 1 , 1 , 0 , 1 ],
[ 0 , 1 , 0 , 0 ],
[ 1 , 1 , 1 , 1 ]]
solveMaze(maze)
|
C#
using System;
class RatMaze{
static int N;
void printSolution( int [,]sol)
{
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
Console.Write( " " + sol[i, j] + " " );
Console.WriteLine();
}
}
bool isSafe( int [,]maze, int x, int y)
{
return (x >= 0 && x < N && y >= 0 &&
y < N && maze[x, y] == 1);
}
bool solveMaze( int [,]maze)
{
int [,]sol = new int [N, N];
if (solveMazeUtil(maze, 0, 0, sol) == false )
{
Console.Write( "Solution doesn't exist" );
return false ;
}
printSolution(sol);
return true ;
}
bool solveMazeUtil( int [,]maze, int x, int y,
int [,]sol)
{
if (x == N - 1 && y == N - 1 &&
maze[x, y] == 1)
{
sol[x, y] = 1;
return true ;
}
if (isSafe(maze, x, y) == true )
{
if (sol[x, y] == 1)
return false ;
sol[x, y] = 1;
if (solveMazeUtil(maze, x + 1, y, sol))
return true ;
if (solveMazeUtil(maze, x, y + 1, sol))
return true ;
if (solveMazeUtil(maze, x - 1, y, sol))
return true ;
if (solveMazeUtil(maze, x, y - 1, sol))
return true ;
sol[x, y] = 0;
return false ;
}
return false ;
}
public static void Main(String []args)
{
RatMaze rat = new RatMaze();
int [,]maze = { { 1, 0, 0, 0 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 0 },
{ 1, 1, 1, 1 } };
N = maze.GetLength(0);
rat.solveMaze(maze);
}
}
|
Javascript
<script>
let N;
function printSolution(sol)
{
for (let i = 0; i < N; i++) {
for (let j = 0; j < N; j++)
document.write(
" " + sol[i][j] + " " );
document.write( "<br>" );
}
}
function isSafe(maze,x,y)
{
return (x >= 0 && x < N && y >= 0
&& y < N && maze[x][y] == 1);
}
function solveMaze(maze)
{
let sol = new Array(N);
for (let i=0;i<N;i++)
{
sol[i]= new Array(N);
for (let j=0;j<N;j++)
{
sol[i][j]=0;
}
}
if (solveMazeUtil(maze, 0, 0, sol) == false ) {
document.write( "Solution doesn't exist" );
return false ;
}
printSolution(sol);
return true ;
}
function solveMazeUtil(maze,x,y,sol)
{
if (x == N - 1 && y == N - 1
&& maze[x][y] == 1) {
sol[x][y] = 1;
return true ;
}
if (isSafe(maze, x, y) == true ) {
if (sol[x][y] == 1)
return false ;
sol[x][y] = 1;
if (solveMazeUtil(maze, x + 1, y, sol))
return true ;
if (solveMazeUtil(maze, x, y + 1, sol))
return true ;
if (solveMazeUtil(maze, x - 1, y, sol))
return true ;
if (solveMazeUtil(maze, x, y - 1, sol))
return true ;
sol[x][y] = 0;
return false ;
}
return false ;
}
let maze=[[ 1, 0, 0, 0 ],
[ 1, 1, 0, 1 ],
[ 0, 1, 0, 0 ],
[ 1, 1, 1, 1 ] ];
N = maze.length;
solveMaze(maze);
</script>
|
Output:
The 1 values show the path for rat
1 0 0 0
1 1 0 0
0 1 0 0
0 1 1 1
Complexity Analysis:
- Time Complexity: O(2^(n^2)).
The recursion can run upper-bound 2^(n^2) times. - Auxiliary Space: O(n^2).
Output matrix is required so an extra space of size n*n is needed.
Below is an extended version of this problem. Count number of ways to reach destination in a Maze
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.