C Program for Rat in a Maze | Backtracking-2 - GeeksforGeeks

02 Aug, 2021
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 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.

C/C++
/* C/C++ program to solve Rat in a Maze problem using   backtracking */#include <stdio.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) {        sol[x][y] = 1;        return true;    }     // Check if maze[x][y] is valid    if (isSafe(maze, x, y) == true) {        // 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;}
1  0  0  0 
1  1  0  0 
0  1  0  0 
0  1  1  1

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