Rat in a Maze with multiple steps or jump allowed

01 Jul, 2022
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:  
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++
Java
Python3
C#
Javascript
/* C/C++ program to solve Rat in a Maze problemusing 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 matrixsol[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 validindex 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 usingBacktracking. It mainly uses solveMazeUtil() tosolve the problem. It returns false if no pathis possible, otherwise return true and printsthe path in the form of 1s. Please note thatthere 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 functionint 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;}
// Java program to solve Rat in a Maze problem// using backtrackingclass GFG{     // Maze size    static int N = 4;     /* A utility function to print solution matrix    sol[N][N] */    static void printSolution(int sol[][])    {        for (int i = 0; i < N; i++)        {            for (int j = 0; j < N; j++)            {                System.out.printf(" %d ", sol[i][j]);            }            System.out.printf("\n");        }    }     /* A utility function to check if x, y is valid    index for N*N maze */    static boolean isSafe(int maze[][], 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.*/    static boolean solveMaze(int maze[][])    {        int sol[][] = {{0, 0, 0, 0},                       {0, 0, 0, 0},                       {0, 0, 0, 0},                       {0, 0, 0, 0}};         if (solveMazeUtil(maze, 0, 0, sol) == false)        {            System.out.printf("Solution doesn't exist");            return false;        }         printSolution(sol);        return true;    }     /* A recursive utility function to solve Maze problem */    static boolean solveMazeUtil(int maze[][], int x,                                 int y, int sol[][])    {        // 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 Code    public static void main(String[] args)    {        int maze[][] = {{2, 1, 0, 0},                        {3, 0, 0, 1},                        {0, 1, 0, 1},                        {0, 0, 0, 1}};         solveMaze(maze);    }} // This code is contributed by Princi Singh
""" Python3 program to solve Rat in aMaze problem using backtracking """ # Maze sizeN = 4 """ A utility function to print solution matrixsol """def printSolution(sol):    for i in range(N):        for j in range(N):            print(sol[i][j], end = " ")        print()         """ A utility function to check ifx, y is valid index for N*N maze """def isSafe(maze, x, y):         # if (x, y outside maze) return false    if (x >= 0 and x < N and y >= 0 and         y < N and maze[x][y] != 0):        return True    return False """ This function solves the Maze problem usingBacktracking. It mainly uses solveMazeUtil() tosolve the problem. It returns false if no pathis possible, otherwise return True and printsthe path in the form of 1s. Please note thatthere may be more than one solutions,this function prints one of the feasible solutions."""def solveMaze(maze):    sol = [[0, 0, 0, 0],           [0, 0, 0, 0],           [0, 0, 0, 0],           [0, 0, 0, 0]]    if (solveMazeUtil(maze, 0, 0, sol) == False):        print("Solution doesn't exist")        return False    printSolution(sol)    return True     """ A recursive utility functionto 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 """        for i in range(1, N):            if (i <= maze[x][y]):                                 """ 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 Codemaze = [[2, 1, 0, 0],        [3, 0, 0, 1],        [0, 1, 0, 1],        [0, 0, 0, 1]]solveMaze(maze) # This code is contributed by SHUBHAMSINGH10
// C# program to solve Rat in a Maze problem// using backtrackingusing System;     class GFG{     // Maze size    static int N = 4;     /* A utility function to print    solution matrix sol[N, N] */    static void printSolution(int [,]sol)    {        for (int i = 0; i < N; i++)        {            for (int j = 0; j < N; j++)            {                Console.Write(" {0} ", sol[i, j]);            }            Console.Write("\n");        }    }     /* A utility function to check if    x, y is valid index for N*N maze */    static Boolean isSafe(int [,]maze,                          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.*/    static Boolean solveMaze(int [,]maze)    {        int [,]sol = {{0, 0, 0, 0},                      {0, 0, 0, 0},                      {0, 0, 0, 0},                      {0, 0, 0, 0}};         if (solveMazeUtil(maze, 0, 0, sol) == false)        {            Console.Write("Solution doesn't exist");            return false;        }         printSolution(sol);        return true;    }     /* A recursive utility function to solve Maze problem */    static Boolean solveMazeUtil(int [,]maze, int x,                                 int y, int [,]sol)    {        // 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 Code    public static void Main(String[] args)    {        int [,]maze = {{2, 1, 0, 0},                       {3, 0, 0, 1},                       {0, 1, 0, 1},                       {0, 0, 0, 1}};         solveMaze(maze);    }} // This code is contributed by 29AjayKumar
<script> // JavaScript program to solve Rat in a Maze problem// using backtracking     // Maze size    let N = 4;       /* A utility function to print solution matrix    sol[N][N] */    function printSolution(sol)    {        for (let i = 0; i < N; i++)        {            for (let j = 0; j < N; j++)            {                 document.write(sol[i][j] + " ");            }             document.write("<br/>");        }    }       /* A utility function to check if x, y is valid    index for N*N maze */    function isSafe(maze, x, 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 print    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.*/    function solveMaze(maze)    {        let sol = [[0, 0, 0, 0],                       [0, 0, 0, 0],                       [0, 0, 0, 0],                       [0, 0, 0, 0]];           if (solveMazeUtil(maze, 0, 0, sol) == false)        {             document.write("Solution doesn't exist");            return false;        }           printSolution(sol);        return true;    }       /* A recursive utility function to solve Maze problem */    function solveMazeUtil(maze,  x,                                 y, sol)    {        // 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 (let 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 code         let maze = [[2, 1, 0, 0],                        [3, 0, 0, 1],                        [0, 1, 0, 1],                        [0, 0, 0, 1]];           solveMaze(maze); // This code is contributed by splevel62.</script>
1  0  0  0 
1  0  0  1 
0  0  0  1 
0  0  0  1
 
princi singh
29AjayKumar
SHUBHAMSINGH10
splevel62
khushboogoyal499
surinderdawra388
Backtracking
Matrix
Matrix
Backtracking
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