Rat in a Maze | Backtracking-2

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 matrx.

{1, 0, 0, 0}
{1, 1, 0, 0}
{0, 1, 0, 0}
{0, 1, 1, 1}
All enteries 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 funtion, 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++

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/* 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
        && maze[x][y] == 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 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;
}

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Java

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/* 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) {
            // 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

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Python3

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# Python3 program to solve Rat in a Maze 
# problem using backracking 
  
# Maze size
N = 4
  
# A utility function to print solution matrix sol
def 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 Maze
def 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 problem
def 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:
        # 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
  
# 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 Shiv Shankar

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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 upperbound 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 Maze

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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