Rat in a Maze with multiple steps or jump allowed

This is the variation of Rat in Maze

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 destination. The rat can move only in two directions: forward and down.
In the maze matrix, 0 means the block is dead end and non-zero number means the block can be used in the path from source to destination. The non-zero value of mat[i][j] indicates number of maximum jumps rat can make from cell mat[i][j].

In this variation, Rat is allowed to jump multiple steps at a time instead of 1.

Examples
Examples:

Input : { {2, 1, 0, 0},
         {3, 0, 0, 1},
         {0, 1, 0, 1},
          {0, 0, 0, 1}
        }
Output : { {1, 0, 0, 0},
           {1, 0, 0, 1},
           {0, 0, 0, 1},
           {0, 0, 0, 1}
         }

Explanation 
Rat started with M[0][0] and can jump upto 2 steps right/down. 
Let's try in horizontal direction - 
M[0][1] won't lead to solution and M[0][2] is 0 which is dead end. 
So, backtrack and try in down direction. 
Rat jump down to M[1][0] which eventually leads to solution.  

Input : { 
      {2, 1, 0, 0},
      {2, 0, 0, 1},
      {0, 1, 0, 1},
      {0, 0, 0, 1}
    }
Output : Solution doesn't exist

Naive Algorithm
The Naive Algorithm is to generate all paths from source to destination and one by one check if the generated path satisfies the constraints.

while there are untried paths
{
   generate the next path
   if this path has all blocks as non-zero
   {
      print this path;
   }
}

Backtracking Algorithm

If destination is reached
    print the solution matrix
Else
   a) Mark current cell in solution matrix as 1. 
   b) Move forward/jump (for each valid steps) in horizontal direction 
      and recursively check if this move leads to a solution. 
   c) If the move chosen in the above step doesn't lead to a solution
       then move down and check if this move leads to a solution. 
   d) If none of the above solutions work then unmark this cell as 0 
       (BACKTRACK) and return false.

Implementation of Backtracking solution

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] != 0)
        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 */
        for (int i = 1; i <= maze[x][y] && i < N; i++) {

            /* Move forward in x direction */
            if (solveMazeUtil(maze, x + i, 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 + i, 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] = { { 2, 1, 0, 0 },
                       { 3, 0, 0, 1 },
                       { 0, 1, 0, 1 },
                       { 0, 0, 0, 1 } };

    solveMaze(maze);
    return 0;
}

Output:

My Personal Notes

Ajay Verma

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Practice Tags :

Matrix

Backtracking

C/C++

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