Rat in a Maze

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: 

1. Create a solution matrix, initially filled with 0’s.
2. Create a recursive function, which takes initial matrix, output matrix and position of rat (i, j).
3. if the position is out of the matrix or the position is not valid then return.
4. 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.
5. Recursively call for position (i+1, j) and (i, j+1).
6. Unmark position (i, j), i.e output[i][j] = 0.

C++

// C++ program to solve Rat in a Maze problem using
// backtracking
#include <bits/stdc++.h>
using namespace std;
// Maze size
#define N 4

bool solveMazeUtil(int maze[N][N], int x, int y,int sol[N][N]);

// A utility function to print solution matrix 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;
    }
}

// A utility function to check if x, y is valid index for
// N*N maze
bool isSafe(int maze[N][N], int x, int y)
{
    // if (x, y outside maze) return false
    if (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)
        return true;
    return false;
}

// This function solves the Maze problem using Backtracking.
// It mainly uses solveMazeUtil() to solve the problem. It
// returns false if no path is possible, otherwise return
// true and prints the path in the form of 1s. Please note
// that there may be more than one solutions, this function
// prints one of the feasible solutions.
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;
}

// A recursive utility function to solve Maze problem
bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N])
{
    // if (x, y is goal) return true
    if (x == N - 1 && y == N - 1 && maze[x][y] == 1) {
        sol[x][y] = 1;
        return true;
    }
    // Check if maze[x][y] is valid
    if (isSafe(maze, x, y) == true) {
        // Check if the current block is already part of
        // solution path.
        if (sol[x][y] == 1)
            return false;
        // mark x, y as part of solution path
        sol[x][y] = 1;
        /* Move forward in x direction */
        if (solveMazeUtil(maze, x + 1, y, sol) == true)
            return true;
        // If moving right didn't work
        // move left
          if (solveMazeUtil(maze, x - 1, y, sol) == true)
            return true;
        // If moving in x direction doesn't give solution
        // then Move down in y direction
        if (solveMazeUtil(maze, x, y + 1, sol) == true)
            return true;
        // If moving down didn't work
        // move up
          if (solveMazeUtil(maze, x, y - 1, sol) == true)
            return true;
        // If none of the above movements work then
        // BACKTRACK: unmark x, y as part of solution path
        sol[x][y] = 0;
        return false;
    }
    return false;
}

// driver program to test above function
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;
}

// This code is contributed by Aditya Kumar (adityakumar129)
// Changes made for N>4 by Ch. Abdul Wasay

C

// C++ program to solve Rat in a Maze problem using
// backtracking
#include <stdio.h>
#include <stdbool.h>
// Maze size
#define N 4

bool solveMazeUtil(int maze[N][N], int x, int y,int sol[N][N]);

// A utility function to print solution matrix 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");
    }
}

// A utility function to check if x, y is valid index for
// N*N maze
bool isSafe(int maze[N][N], int x, int y)
{
    // if (x, y outside maze) return false
    if (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)
        return true;
    return false;
}

// This function solves the Maze problem using Backtracking.
// It mainly uses solveMazeUtil() to solve the problem. It
// returns false if no path is possible, otherwise return
// true and prints the path in the form of 1s. Please note
// that there may be more than one solutions, this function
// prints one of the feasible solutions.
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;
}

// A recursive utility function to solve Maze problem
bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N])
{
    // if (x, y is goal) return true
    if (x == N - 1 && y == N - 1 && maze[x][y] == 1) {
        sol[x][y] = 1;
        return true;
    }
    // Check if maze[x][y] is valid
    if (isSafe(maze, x, y) == true) {
        // Check if the current block is already part of
        // solution path.
        if (sol[x][y] == 1)
            return false;
        // mark x, y as part of solution path
        sol[x][y] = 1;
        /* Move forward in x direction */
        if (solveMazeUtil(maze, x + 1, y, sol) == true)
            return true;
        // If moving in x direction doesn't give solution
        // then Move down in y direction
        if (solveMazeUtil(maze, x, y + 1, sol) == true)
            return true;
        // If none of the above movements work then
        // BACKTRACK: unmark x, y as part of solution path
        sol[x][y] = 0;
        return false;
    }
    return false;
}

// driver program to test above function
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;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

/* Java program to solve Rat in
 a Maze problem using backtracking */

public class RatMaze {

    // Size of the maze
    static int N;

    /* A utility function to print
    solution matrix sol[N][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();
        }
    }

    /* A utility function to check
        if x, y is valid index for N*N maze */
    boolean isSafe(
        int maze[][], int x, int y)
    {
        // if (x, y outside maze) return false
        return (x >= 0 && x < N && y >= 0
                && y < N && maze[x][y] == 1);
    }

    /* This function solves the Maze problem using
    Backtracking. It mainly uses solveMazeUtil()
    to solve the problem. It returns false if no
    path is possible, otherwise return true and
    prints the path in the form of 1s. Please note
    that there may be more than one solutions, this
    function prints one of the feasible solutions.*/
    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;
    }

    /* A recursive utility function to solve Maze
    problem */
    boolean solveMazeUtil(int maze[][], int x, int y,
                          int sol[][])
    {
        // if (x, y is goal) return true
        if (x == N - 1 && y == N - 1
            && maze[x][y] == 1) {
            sol[x][y] = 1;
            return true;
        }

        // Check if maze[x][y] is valid
        if (isSafe(maze, x, y) == true) {
              // Check if the current block is already part of solution path.
              if (sol[x][y] == 1)
                  return false;

            // mark x, y as part of solution path
            sol[x][y] = 1;

            /* Move forward in x direction */
            if (solveMazeUtil(maze, x + 1, y, sol))
                return true;

            /* If moving in x direction doesn't give
            solution then Move down in y direction */
            if (solveMazeUtil(maze, x, y + 1, sol))
                return true;

            /* If none of the above movements works then
            BACKTRACK: unmark x, y as part of solution
            path */
            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);
    }
}
// This code is contributed by Abhishek Shankhadhar

Python3

# Python3 program to solve Rat in a Maze
# problem using backtracking

# Maze size
n = 4

# A utility function to check if x, y is valid
# index for N * N Maze


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

# A recursive utility function to solve Maze problem


def RatMaze(n, maze, move_x, move_y, x, y, res):
    # if (x, y is goal) return True
    if x == n-1 and y == n-1:
        return True
    for i in range(4):
        # Generate new value of x
        x_new = x + move_x[i]

        # Generate new value of y
        y_new = y + move_y[i]

        # Check if maze[x][y] is valid
        if isValid(n, maze, x_new, y_new, res):

            # mark x, y as part of solution path
            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):
    # Creating a 4 * 4 2-D list
    res = [[0 for i in range(n)] for i in range(n)]
    res[0][0] = 1

    # x matrix for each direction
    move_x = [-1, 1, 0, 0]

    # y matrix for each direction
    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')


# Driver program to test above function
if __name__ == "__main__":
    # Initialising the maze
    maze = [[1, 0, 0, 0],
             [1, 1, 0, 1],
             [0, 1, 0, 0],
             [1, 1, 1, 1]]

    solveMaze(maze)

# This code is contributed by Anvesh Govind Saxena

C#

// C# program to solve Rat in a Maze
// problem using backtracking
using System;

class RatMaze{

// Size of the maze
static int N;

// A utility function to print
// solution matrix sol[N,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();
    }
}

// A utility function to check if x, y
// is valid index for N*N maze
bool isSafe(int [,]maze, int x, int y)
{

    // If (x, y outside maze) return false
    return (x >= 0 && x < N && y >= 0 &&
            y < N && maze[x, y] == 1);
}

// This function solves the Maze problem using
// Backtracking. It mainly uses solveMazeUtil()
// to solve the problem. It returns false if no
// path is possible, otherwise return true and
// prints the path in the form of 1s. Please note
// that there may be more than one solutions, this
// function prints one of the feasible solutions.
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;
}

// A recursive utility function to solve Maze
// problem
bool solveMazeUtil(int [,]maze, int x, int y,
                   int [,]sol)
{

    // If (x, y is goal) return true
    if (x == N - 1 && y == N - 1 &&
        maze[x, y] == 1)
    {
        sol[x, y] = 1;
        return true;
    }

    // Check if maze[x,y] is valid
    if (isSafe(maze, x, y) == true)
    {
          // Check if the current block is already part of solution path.
          if (sol[x, y] == 1)
              return false;

        // Mark x, y as part of solution path
        sol[x, y] = 1;

        // Move forward in x direction
        if (solveMazeUtil(maze, x + 1, y, sol))
            return true;

        // If moving in x direction doesn't give
        // solution then Move down in y direction
        if (solveMazeUtil(maze, x, y + 1, sol))
            return true;

          // If moving in y direction doesm't give
        // solution then Move backward in x direction
        if (solveMazeUtil(maze, x - 1, y, sol))
            return true;

        // If moving in backwards in x direction doesn't give
        // solution then Move upwards in y direction
        if (solveMazeUtil(maze, x, y - 1, sol))
            return true;

        // If none of the above movements works then
        // BACKTRACK: unmark x, y as part of solution
        // path
        sol[x, y] = 0;
        return false;
    }
    return false;
}

// Driver Code
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);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>
/* Javascript program to solve Rat in
 a Maze problem using backtracking */

// Size of the maze
let N;

/* A utility function to print
    solution matrix sol[N][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>");
        }
}

/* A utility function to check
        if x, y is valid index for N*N maze */
function isSafe(maze,x,y)
{
    // if (x, y outside maze) return false
        return (x >= 0 && x < N && y >= 0
                && y < N && maze[x][y] == 1);
}

/* This function solves the Maze problem using
    Backtracking. It mainly uses solveMazeUtil()
    to solve the problem. It returns false if no
    path is possible, otherwise return true and
    prints the path in the form of 1s. Please note
    that there may be more than one solutions, this
    function prints one of the feasible solutions.*/
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;
}
/* A recursive utility function to solve Maze
    problem */
function solveMazeUtil(maze,x,y,sol)
{
    // if (x, y is goal) return true
        if (x == N - 1 && y == N - 1
            && maze[x][y] == 1) {
            sol[x][y] = 1;
            return true;
        }

        // Check if maze[x][y] is valid
        if (isSafe(maze, x, y) == true) {
              // Check if the current block is already part of solution path.
              if (sol[x][y] == 1)
                  return false;

            // mark x, y as part of solution path
            sol[x][y] = 1;

            /* Move forward in x direction */
            if (solveMazeUtil(maze, x + 1, y, sol))
                return true;

            /* If moving in x direction doesn't give
            solution then Move down in y direction */
            if (solveMazeUtil(maze, x, y + 1, sol))
                return true;

            /* If moving in y direction doesn't give
            solution then Move backwards in x direction */
            if (solveMazeUtil(maze, x - 1, y, sol))
                return true;

            /* If moving backwards in x direction doesn't give
            solution then Move upwards in y direction */
            if (solveMazeUtil(maze, x, y - 1, sol))
                return true;

            /* If none of the above movements works then
            BACKTRACK: unmark x, y as part of solution
            path */
            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);


// This code is contributed by avanitrachhadiya2155
</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.