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Python Program for Rat in a Maze | Backtracking-2

  • Last Updated : 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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Python3




# Python3 program to solve Rat in a Maze
# problem using backtracking
 
# 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:
        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
Output:
1 0 0 0 
1 1 0 0 
0 1 0 0 
0 1 1 1

Please refer complete article on Rat in a Maze | Backtracking-2 for more details!




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