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 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>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA