Hemos discutido el problema de Backtracking y Knight’s tour en el Set 1 . Analicemos Rat in a Maze como otro problema de ejemplo que se puede resolver usando Backtracking.
Un Laberinto se da como una array binaria N*N de bloques donde el bloque de origen es el bloque superior izquierdo, es decir, laberinto[0][0] y el bloque de destino es el bloque inferior derecho, es decir, laberinto[N-1][N-1] . Una rata parte de la fuente y tiene que llegar al destino. La rata solo puede moverse en dos direcciones: hacia adelante y hacia abajo.
C++
// C++ program to solve Rat in a Maze problem using // backtracking #include <bits/stdc++.h> using namespace std; // 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++) cout<<" "<<sol[i][j]<<" "; cout<<endl; } } // 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) { cout<<"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 && maze[x][y] == 1) { sol[x][y] = 1; return true; } // Check if maze[x][y] is valid if (isSafe(maze, x, y) == true) { // Check if the current block is already part of // solution path. if (sol[x][y] == 1) return false; // 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; } // This code is contributed by Aditya Kumar (adityakumar129)
C
// C++ program to solve Rat in a Maze problem using // backtracking #include <stdio.h> #include <stdbool.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 && maze[x][y] == 1) { sol[x][y] = 1; return true; } // Check if maze[x][y] is valid if (isSafe(maze, x, y) == true) { // Check if the current block is already part of // solution path. if (sol[x][y] == 1) return false; // 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; } // This code is contributed by Aditya Kumar (adityakumar129)
Java
/* Java program to solve Rat in a Maze problem using backtracking */ public class RatMaze { // Size of the maze static int N; /* A utility function to print solution matrix sol[N][N] */ void printSolution(int sol[][]) { for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) System.out.print( " " + sol[i][j] + " "); System.out.println(); } } /* A utility function to check if x, y is valid index for N*N maze */ boolean isSafe( int maze[][], int x, int y) { // if (x, y outside maze) return false return (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1); } /* 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.*/ boolean solveMaze(int maze[][]) { int sol[][] = new int[N][N]; if (solveMazeUtil(maze, 0, 0, sol) == false) { System.out.print("Solution doesn't exist"); return false; } printSolution(sol); return true; } /* A recursive utility function to solve Maze problem */ boolean solveMazeUtil(int maze[][], int x, int y, int sol[][]) { // if (x, y is goal) return true if (x == N - 1 && y == N - 1 && maze[x][y] == 1) { sol[x][y] = 1; return true; } // Check if maze[x][y] is valid if (isSafe(maze, x, y) == true) { // Check if the current block is already part of solution path. if (sol[x][y] == 1) return false; // mark x, y as part of solution path sol[x][y] = 1; /* Move forward in x direction */ if (solveMazeUtil(maze, x + 1, y, sol)) return true; /* If moving in x direction doesn't give solution then Move down in y direction */ if (solveMazeUtil(maze, x, y + 1, sol)) return true; /* If none of the above movements works then BACKTRACK: unmark x, y as part of solution path */ sol[x][y] = 0; return false; } return false; } public static void main(String args[]) { RatMaze rat = new RatMaze(); int maze[][] = { { 1, 0, 0, 0 }, { 1, 1, 0, 1 }, { 0, 1, 0, 0 }, { 1, 1, 1, 1 } }; N = maze.length; rat.solveMaze(maze); } } // This code is contributed by Abhishek Shankhadhar
Python3
# Python3 program to solve Rat in a Maze # problem using backtracking # Maze size n = 4 # A utility function to check if x, y is valid # index for N * N Maze def isValid(n, maze, x, y, res): if x >= 0 and y >= 0 and x < n and y < n and maze[x][y] == 1 and res[x][y] == 0: return True return False # A recursive utility function to solve Maze problem def RatMaze(n, maze, move_x, move_y, x, y, res): # if (x, y is goal) return True if x == n-1 and y == n-1: return True for i in range(4): # Generate new value of x x_new = x + move_x[i] # Generate new value of y y_new = y + move_y[i] # Check if maze[x][y] is valid if isValid(n, maze, x_new, y_new, res): # mark x, y as part of solution path res[x_new][y_new] = 1 if RatMaze(n, maze, move_x, move_y, x_new, y_new, res): return True res[x_new][y_new] = 0 return False def solveMaze(maze): # Creating a 4 * 4 2-D list res = [[0 for i in range(n)] for i in range(n)] res[0][0] = 1 # x matrix for each direction move_x = [-1, 1, 0, 0] # y matrix for each direction move_y = [0, 0, -1, 1] if RatMaze(n, maze, move_x, move_y, 0, 0, res): for i in range(n): for j in range(n): print(res[i][j], end=' ') print() else: print('Solution does not exist') # 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 Anvesh Govind Saxena
C#
// C# program to solve Rat in a Maze // problem using backtracking using System; class RatMaze{ // Size of the maze static int N; // A utility function to print // solution matrix sol[N,N] void printSolution(int [,]sol) { for(int i = 0; i < N; i++) { for(int j = 0; j < N; j++) Console.Write(" " + sol[i, j] + " "); Console.WriteLine(); } } // A utility function to check if x, y // is valid index for N*N maze bool isSafe(int [,]maze, int x, int y) { // If (x, y outside maze) return false return (x >= 0 && x < N && y >= 0 && y < N && maze[x, y] == 1); } // 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) { int [,]sol = new int[N, N]; 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 bool solveMazeUtil(int [,]maze, int x, int y, int [,]sol) { // If (x, y is goal) return true if (x == N - 1 && y == N - 1 && maze[x, y] == 1) { sol[x, y] = 1; return true; } // Check if maze[x,y] is valid if (isSafe(maze, x, y) == true) { // Check if the current block is already part of solution path. if (sol[x, y] == 1) return false; // Mark x, y as part of solution path sol[x, y] = 1; // Move forward in x direction if (solveMazeUtil(maze, x + 1, y, sol)) return true; // If moving in x direction doesn't give // solution then Move down in y direction if (solveMazeUtil(maze, x, y + 1, sol)) return true; // If moving in y direction doesm't give // solution then Move backward in x direction if (solveMazeUtil(maze, x - 1, y, sol)) return true; // If moving in backwards in x direction doesn't give // solution then Move upwards in y direction if (solveMazeUtil(maze, x, y - 1, sol)) return true; // If none of the above movements works 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) { RatMaze rat = new RatMaze(); int [,]maze = { { 1, 0, 0, 0 }, { 1, 1, 0, 1 }, { 0, 1, 0, 0 }, { 1, 1, 1, 1 } }; N = maze.GetLength(0); rat.solveMaze(maze); } } // This code is contributed by gauravrajput1
Javascript
<script> /* Javascript program to solve Rat in a Maze problem using backtracking */ // Size of the maze let N; /* 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 return (x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1); } /* 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.*/ function solveMaze(maze) { let sol = new Array(N); for(let i=0;i<N;i++) { sol[i]=new Array(N); for(let j=0;j<N;j++) { sol[i][j]=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 && maze[x][y] == 1) { sol[x][y] = 1; return true; } // Check if maze[x][y] is valid if (isSafe(maze, x, y) == true) { // Check if the current block is already part of solution path. if (sol[x][y] == 1) return false; // mark x, y as part of solution path sol[x][y] = 1; /* Move forward in x direction */ if (solveMazeUtil(maze, x + 1, y, sol)) return true; /* If moving in x direction doesn't give solution then Move down in y direction */ if (solveMazeUtil(maze, x, y + 1, sol)) return true; /* If moving in y direction doesn't give solution then Move backwards in x direction */ if (solveMazeUtil(maze, x - 1, y, sol)) return true; /* If moving backwards in x direction doesn't give solution then Move upwards in y direction */ if (solveMazeUtil(maze, x, y - 1, sol)) return true; /* If none of the above movements works then BACKTRACK: unmark x, y as part of solution path */ sol[x][y] = 0; return false; } return false; } let maze=[[ 1, 0, 0, 0 ], [ 1, 1, 0, 1 ], [ 0, 1, 0, 0 ], [ 1, 1, 1, 1 ] ]; N = maze.length; solveMaze(maze); // This code is contributed by avanitrachhadiya2155 </script>
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA