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} }ExplanationRat 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:**

1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1

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