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Rat in Maze

The document describes a problem to find a path for a rat to navigate through a maze from a starting to ending point. The maze is represented as a binary N×N matrix where 1's indicate passable spaces and 0's indicate dead ends. The rat can only move forward or downward through the maze. The document provides a C program that uses backtracking to solve the rat maze problem by recursively exploring all paths from the start to find one that reaches the end.

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276 views2 pages

Rat in Maze

The document describes a problem to find a path for a rat to navigate through a maze from a starting to ending point. The maze is represented as a binary N×N matrix where 1's indicate passable spaces and 0's indicate dead ends. The rat can only move forward or downward through the maze. The document provides a C program that uses backtracking to solve the rat maze problem by recursively exploring all paths from the start to find one that reaches the end.

Uploaded by

Manikandan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as TXT, PDF, TXT or read online on Scribd
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/*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 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 limited number of
moves.

Following is an example maze.*/

/* 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 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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