Rat in a Maze | Backtracking-2 - GeeksforGeeks

22 Apr, 2022
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.
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++
C
Java
Python3
C#
Javascript
// 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 mazebool 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 problembool 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 functionint 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)
// 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 mazebool 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 problembool 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 functionint 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 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 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;    }     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 program to solve Rat in a Maze# problem using backtracking # Maze sizeN = 4 # A utility function to print solution matrix soldef printSolution( sol ):         for i in sol:        for j in i:            print(str(j) + " ", end ="")        print("") # A utility function to check if x, y is valid# index for N * N Mazedef isSafe( maze, x, y ):         if x >= 0 and x < N and y >= 0 and y < N and 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 feasable solutions. """def solveMaze( maze ):         # Creating a 4 * 4 2-D list    sol = [ [ 0 for j in range(4) ] for i in range(4) ]         if solveMazeUtil(maze, 0, 0, sol) == False:        print("Solution doesn't exist");        return False         printSolution(sol)    return True     # A recursive utility function to solve Maze problemdef solveMazeUtil(maze, x, y, sol):         # if (x, y is goal) return True    if x == N - 1 and y == N - 1 and 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 moving in y direction doesn't give solution then        # Move back in x direction        if solveMazeUtil(maze, x - 1, y, sol) == True:            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) == 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 # Driver program to test above functionif __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 Shiv Shankar
// C# program to solve Rat in a Maze// problem using backtrackingusing System; class RatMaze{ // Size of the mazestatic 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 mazebool 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// problembool 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 Codepublic 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
<script>/* Javascript program to solve Rat in a Maze problem using backtracking */ // Size of the mazelet 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.
Space Complexity: 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 MazePlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Kakashi11
Debomit Dey
andrew1234
GauravRajput1
khushboogoyal499
sumitu0812
avanitrachhadiya2155
sweetyty
rajatjain0807
Amazon
Drishti-Soft
Expedia
Flipkart
Grofers
MakeMyTrip
Numerify
Paytm
Visa
Yatra.com
Zycus
Backtracking
Matrix
Paytm
Flipkart
Amazon
MakeMyTrip
Visa
Yatra.com
Drishti-Soft
Zycus
Expedia
Grofers
Matrix
Backtracking
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