La ruta hamiltoniana en un gráfico no dirigido es una ruta que visita cada vértice exactamente una vez. Un ciclo hamiltoniano (o circuito hamiltoniano) es un camino hamiltoniano tal que hay un borde (en el gráfico) desde el último vértice hasta el primer vértice del camino hamiltoniano. Determinar si un gráfico dado contiene ciclo hamiltoniano o no. Si contiene, imprime la ruta. A continuación se muestran la entrada y la salida de la función requerida.
Entrada:
un gráfico de array 2D [V] [V] donde V es el número de vértices en el gráfico y el gráfico [V] [V] es la representación de array de adyacencia del gráfico. Un valor graph[i][j] es 1 si hay un borde directo de i a j, de lo contrario, graph[i][j] es 0.
Salida:
Una ruta de array [V] que debe contener la ruta hamiltoniana. path[i] debe representar el i-ésimo vértice en el camino hamiltoniano. El código también debería devolver falso si no hay un ciclo hamiltoniano en el gráfico.
Por ejemplo, un ciclo hamiltoniano en el siguiente gráfico es {0, 1, 2, 4, 3, 0}.
(0)--(1)--(2) | / \ | | / \ | | / \ | (3)-------(4)
Y el siguiente gráfico no contiene ningún ciclo hamiltoniano.
(0)--(1)--(2) | / \ | | / \ | | / \ | (3) (4)
Algoritmo ingenuo
Genere todas las configuraciones posibles de vértices e imprima una configuración que satisfaga las restricciones dadas. ¡Habrá n! (n factorial) configuraciones.
while there are untried conflagrations
{
generate the next configuration
if ( there are edges between two consecutive vertices of this
configuration and there is an edge from the last vertex to
the first ).
{
print this configuration;
break;
}
}
Algoritmo
de retroceso Cree una array de ruta vacía y agréguele el vértice 0. Agregue otros vértices, comenzando desde el vértice 1. Antes de agregar un vértice, verifique si es adyacente al vértice agregado anteriormente y si aún no lo está. Si encontramos tal vértice, agregamos el vértice como parte de la solución. Si no encontramos un vértice, devolvemos falso.
Implementación de la solución Backtracking
Las siguientes son implementaciones de la solución Backtracking.
C++
/* C++ program for solution of Hamiltonian
Cycle problem using backtracking */
#include <bits/stdc++.h>
using namespace std;
// Number of vertices in the graph
#define V 5
void printSolution(int path[]);
/* A utility function to check if
the vertex v can be added at index 'pos'
in the Hamiltonian Cycle constructed
so far (stored in 'path[]') */
bool isSafe(int v, bool graph[V][V],
int path[], int pos)
{
/* Check if this vertex is an adjacent
vertex of the previously added vertex. */
if (graph [path[pos - 1]][ v ] == 0)
return false;
/* Check if the vertex has already been included.
This step can be optimized by creating
an array of size V */
for (int i = 0; i < pos; i++)
if (path[i] == v)
return false;
return true;
}
/* A recursive utility function
to solve hamiltonian cycle problem */
bool hamCycleUtil(bool graph[V][V],
int path[], int pos)
{
/* base case: If all vertices are
included in Hamiltonian Cycle */
if (pos == V)
{
// And if there is an edge from the
// last included vertex to the first vertex
if (graph[path[pos - 1]][path[0]] == 1)
return true;
else
return false;
}
// Try different vertices as a next candidate
// in Hamiltonian Cycle. We don't try for 0 as
// we included 0 as starting point in hamCycle()
for (int v = 1; v < V; v++)
{
/* Check if this vertex can be added
// to Hamiltonian Cycle */
if (isSafe(v, graph, path, pos))
{
path[pos] = v;
/* recur to construct rest of the path */
if (hamCycleUtil (graph, path, pos + 1) == true)
return true;
/* If adding vertex v doesn't lead to a solution,
then remove it */
path[pos] = -1;
}
}
/* If no vertex can be added to
Hamiltonian Cycle constructed so far,
then return false */
return false;
}
/* This function solves the Hamiltonian Cycle problem
using Backtracking. It mainly uses hamCycleUtil() to
solve the problem. It returns false if there is no
Hamiltonian Cycle possible, otherwise return true
and prints the path. Please note that there may be
more than one solutions, this function prints one
of the feasible solutions. */
bool hamCycle(bool graph[V][V])
{
int *path = new int[V];
for (int i = 0; i < V; i++)
path[i] = -1;
/* Let us put vertex 0 as the first vertex in the path.
If there is a Hamiltonian Cycle, then the path can be
started from any point of the cycle as the graph is undirected */
path[0] = 0;
if (hamCycleUtil(graph, path, 1) == false )
{
cout << "\nSolution does not exist";
return false;
}
printSolution(path);
return true;
}
/* A utility function to print solution */
void printSolution(int path[])
{
cout << "Solution Exists:"
" Following is one Hamiltonian Cycle \n";
for (int i = 0; i < V; i++)
cout << path[i] << " ";
// Let us print the first vertex again
// to show the complete cycle
cout << path[0] << " ";
cout << endl;
}
// Driver Code
int main()
{
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3)-------(4) */
bool graph1[V][V] = {{0, 1, 0, 1, 0},
{1, 0, 1, 1, 1},
{0, 1, 0, 0, 1},
{1, 1, 0, 0, 1},
{0, 1, 1, 1, 0}};
// Print the solution
hamCycle(graph1);
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3) (4) */
bool graph2[V][V] = {{0, 1, 0, 1, 0},
{1, 0, 1, 1, 1},
{0, 1, 0, 0, 1},
{1, 1, 0, 0, 0},
{0, 1, 1, 0, 0}};
// Print the solution
hamCycle(graph2);
return 0;
}
// This is code is contributed by rathbhupendra
C
/* C program for solution of Hamiltonian Cycle problem
using backtracking */
#include<stdio.h>
// Number of vertices in the graph
#define V 5
void printSolution(int path[]);
/* A utility function to check if the vertex v can be added at
index 'pos' in the Hamiltonian Cycle constructed so far (stored
in 'path[]') */
bool isSafe(int v, bool graph[V][V], int path[], int pos)
{
/* Check if this vertex is an adjacent vertex of the previously
added vertex. */
if (graph [ path[pos-1] ][ v ] == 0)
return false;
/* Check if the vertex has already been included.
This step can be optimized by creating an array of size V */
for (int i = 0; i < pos; i++)
if (path[i] == v)
return false;
return true;
}
/* A recursive utility function to solve hamiltonian cycle problem */
bool hamCycleUtil(bool graph[V][V], int path[], int pos)
{
/* base case: If all vertices are included in Hamiltonian Cycle */
if (pos == V)
{
// And if there is an edge from the last included vertex to the
// first vertex
if ( graph[ path[pos-1] ][ path[0] ] == 1 )
return true;
else
return false;
}
// Try different vertices as a next candidate in Hamiltonian Cycle.
// We don't try for 0 as we included 0 as starting point in hamCycle()
for (int v = 1; v < V; v++)
{
/* Check if this vertex can be added to Hamiltonian Cycle */
if (isSafe(v, graph, path, pos))
{
path[pos] = v;
/* recur to construct rest of the path */
if (hamCycleUtil (graph, path, pos+1) == true)
return true;
/* If adding vertex v doesn't lead to a solution,
then remove it */
path[pos] = -1;
}
}
/* If no vertex can be added to Hamiltonian Cycle constructed so far,
then return false */
return false;
}
/* This function solves the Hamiltonian Cycle problem using Backtracking.
It mainly uses hamCycleUtil() to solve the problem. It returns false
if there is no Hamiltonian Cycle possible, otherwise return true and
prints the path. Please note that there may be more than one solutions,
this function prints one of the feasible solutions. */
bool hamCycle(bool graph[V][V])
{
int *path = new int[V];
for (int i = 0; i < V; i++)
path[i] = -1;
/* Let us put vertex 0 as the first vertex in the path. If there is
a Hamiltonian Cycle, then the path can be started from any point
of the cycle as the graph is undirected */
path[0] = 0;
if ( hamCycleUtil(graph, path, 1) == false )
{
printf("\nSolution does not exist");
return false;
}
printSolution(path);
return true;
}
/* A utility function to print solution */
void printSolution(int path[])
{
printf ("Solution Exists:"
" Following is one Hamiltonian Cycle \n");
for (int i = 0; i < V; i++)
printf(" %d ", path[i]);
// Let us print the first vertex again to show the complete cycle
printf(" %d ", path[0]);
printf("\n");
}
// driver program to test above function
int main()
{
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3)-------(4) */
bool graph1[V][V] = {{0, 1, 0, 1, 0},
{1, 0, 1, 1, 1},
{0, 1, 0, 0, 1},
{1, 1, 0, 0, 1},
{0, 1, 1, 1, 0},
};
// Print the solution
hamCycle(graph1);
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3) (4) */
bool graph2[V][V] = {{0, 1, 0, 1, 0},
{1, 0, 1, 1, 1},
{0, 1, 0, 0, 1},
{1, 1, 0, 0, 0},
{0, 1, 1, 0, 0},
};
// Print the solution
hamCycle(graph2);
return 0;
}
Java
/* Java program for solution of Hamiltonian Cycle problem
using backtracking */
class HamiltonianCycle
{
final int V = 5;
int path[];
/* A utility function to check if the vertex v can be
added at index 'pos'in the Hamiltonian Cycle
constructed so far (stored in 'path[]') */
boolean isSafe(int v, int graph[][], int path[], int pos)
{
/* Check if this vertex is an adjacent vertex of
the previously added vertex. */
if (graph[path[pos - 1]][v] == 0)
return false;
/* Check if the vertex has already been included.
This step can be optimized by creating an array
of size V */
for (int i = 0; i < pos; i++)
if (path[i] == v)
return false;
return true;
}
/* A recursive utility function to solve hamiltonian
cycle problem */
boolean hamCycleUtil(int graph[][], int path[], int pos)
{
/* base case: If all vertices are included in
Hamiltonian Cycle */
if (pos == V)
{
// And if there is an edge from the last included
// vertex to the first vertex
if (graph[path[pos - 1]][path[0]] == 1)
return true;
else
return false;
}
// Try different vertices as a next candidate in
// Hamiltonian Cycle. We don't try for 0 as we
// included 0 as starting point in hamCycle()
for (int v = 1; v < V; v++)
{
/* Check if this vertex can be added to Hamiltonian
Cycle */
if (isSafe(v, graph, path, pos))
{
path[pos] = v;
/* recur to construct rest of the path */
if (hamCycleUtil(graph, path, pos + 1) == true)
return true;
/* If adding vertex v doesn't lead to a solution,
then remove it */
path[pos] = -1;
}
}
/* If no vertex can be added to Hamiltonian Cycle
constructed so far, then return false */
return false;
}
/* This function solves the Hamiltonian Cycle problem using
Backtracking. It mainly uses hamCycleUtil() to solve the
problem. It returns false if there is no Hamiltonian Cycle
possible, otherwise return true and prints the path.
Please note that there may be more than one solutions,
this function prints one of the feasible solutions. */
int hamCycle(int graph[][])
{
path = new int[V];
for (int i = 0; i < V; i++)
path[i] = -1;
/* Let us put vertex 0 as the first vertex in the path.
If there is a Hamiltonian Cycle, then the path can be
started from any point of the cycle as the graph is
undirected */
path[0] = 0;
if (hamCycleUtil(graph, path, 1) == false)
{
System.out.println("\nSolution does not exist");
return 0;
}
printSolution(path);
return 1;
}
/* A utility function to print solution */
void printSolution(int path[])
{
System.out.println("Solution Exists: Following" +
" is one Hamiltonian Cycle");
for (int i = 0; i < V; i++)
System.out.print(" " + path[i] + " ");
// Let us print the first vertex again to show the
// complete cycle
System.out.println(" " + path[0] + " ");
}
// driver program to test above function
public static void main(String args[])
{
HamiltonianCycle hamiltonian =
new HamiltonianCycle();
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3)-------(4) */
int graph1[][] = {{0, 1, 0, 1, 0},
{1, 0, 1, 1, 1},
{0, 1, 0, 0, 1},
{1, 1, 0, 0, 1},
{0, 1, 1, 1, 0},
};
// Print the solution
hamiltonian.hamCycle(graph1);
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3) (4) */
int graph2[][] = {{0, 1, 0, 1, 0},
{1, 0, 1, 1, 1},
{0, 1, 0, 0, 1},
{1, 1, 0, 0, 0},
{0, 1, 1, 0, 0},
};
// Print the solution
hamiltonian.hamCycle(graph2);
}
}
// This code is contributed by Abhishek Shankhadhar
Python3
# Python program for solution of
# hamiltonian cycle problem
class Graph():
def __init__(self, vertices):
self.graph = [[0 for column in range(vertices)]
for row in range(vertices)]
self.V = vertices
''' Check if this vertex is an adjacent vertex
of the previously added vertex and is not
included in the path earlier '''
def isSafe(self, v, pos, path):
# Check if current vertex and last vertex
# in path are adjacent
if self.graph[ path[pos-1] ][v] == 0:
return False
# Check if current vertex not already in path
for vertex in path:
if vertex == v:
return False
return True
# A recursive utility function to solve
# hamiltonian cycle problem
def hamCycleUtil(self, path, pos):
# base case: if all vertices are
# included in the path
if pos == self.V:
# Last vertex must be adjacent to the
# first vertex in path to make a cycle
if self.graph[ path[pos-1] ][ path[0] ] == 1:
return True
else:
return False
# Try different vertices as a next candidate
# in Hamiltonian Cycle. We don't try for 0 as
# we included 0 as starting point in hamCycle()
for v in range(1,self.V):
if self.isSafe(v, pos, path) == True:
path[pos] = v
if self.hamCycleUtil(path, pos+1) == True:
return True
# Remove current vertex if it doesn't
# lead to a solution
path[pos] = -1
return False
def hamCycle(self):
path = [-1] * self.V
''' Let us put vertex 0 as the first vertex
in the path. If there is a Hamiltonian Cycle,
then the path can be started from any point
of the cycle as the graph is undirected '''
path[0] = 0
if self.hamCycleUtil(path,1) == False:
print ("Solution does not exist\n")
return False
self.printSolution(path)
return True
def printSolution(self, path):
print ("Solution Exists: Following",
"is one Hamiltonian Cycle")
for vertex in path:
print (vertex, end = " ")
print (path[0], "\n")
# Driver Code
''' Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3)-------(4) '''
g1 = Graph(5)
g1.graph = [ [0, 1, 0, 1, 0], [1, 0, 1, 1, 1],
[0, 1, 0, 0, 1,],[1, 1, 0, 0, 1],
[0, 1, 1, 1, 0], ]
# Print the solution
g1.hamCycle();
''' Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3) (4) '''
g2 = Graph(5)
g2.graph = [ [0, 1, 0, 1, 0], [1, 0, 1, 1, 1],
[0, 1, 0, 0, 1,], [1, 1, 0, 0, 0],
[0, 1, 1, 0, 0], ]
# Print the solution
g2.hamCycle();
# This code is contributed by Divyanshu Mehta
C#
// C# program for solution of Hamiltonian
// Cycle problem using backtracking
using System;
public class HamiltonianCycle
{
readonly int V = 5;
int []path;
/* A utility function to check
if the vertex v can be added at
index 'pos'in the Hamiltonian Cycle
constructed so far (stored in 'path[]') */
bool isSafe(int v, int [,]graph,
int []path, int pos)
{
/* Check if this vertex is
an adjacent vertex of the
previously added vertex. */
if (graph[path[pos - 1], v] == 0)
return false;
/* Check if the vertex has already
been included. This step can be
optimized by creating an array
of size V */
for (int i = 0; i < pos; i++)
if (path[i] == v)
return false;
return true;
}
/* A recursive utility function
to solve hamiltonian cycle problem */
bool hamCycleUtil(int [,]graph, int []path, int pos)
{
/* base case: If all vertices
are included in Hamiltonian Cycle */
if (pos == V)
{
// And if there is an edge from the last included
// vertex to the first vertex
if (graph[path[pos - 1],path[0]] == 1)
return true;
else
return false;
}
// Try different vertices as a next candidate in
// Hamiltonian Cycle. We don't try for 0 as we
// included 0 as starting point in hamCycle()
for (int v = 1; v < V; v++)
{
/* Check if this vertex can be
added to Hamiltonian Cycle */
if (isSafe(v, graph, path, pos))
{
path[pos] = v;
/* recur to construct rest of the path */
if (hamCycleUtil(graph, path, pos + 1) == true)
return true;
/* If adding vertex v doesn't
lead to a solution, then remove it */
path[pos] = -1;
}
}
/* If no vertex can be added to Hamiltonian Cycle
constructed so far, then return false */
return false;
}
/* This function solves the Hamiltonian
Cycle problem using Backtracking. It
mainly uses hamCycleUtil() to solve the
problem. It returns false if there
is no Hamiltonian Cycle possible,
otherwise return true and prints the path.
Please note that there may be more than
one solutions, this function prints one
of the feasible solutions. */
int hamCycle(int [,]graph)
{
path = new int[V];
for (int i = 0; i < V; i++)
path[i] = -1;
/* Let us put vertex 0 as the first
vertex in the path. If there is a
Hamiltonian Cycle, then the path can be
started from any point of the cycle
as the graph is undirected */
path[0] = 0;
if (hamCycleUtil(graph, path, 1) == false)
{
Console.WriteLine("\nSolution does not exist");
return 0;
}
printSolution(path);
return 1;
}
/* A utility function to print solution */
void printSolution(int []path)
{
Console.WriteLine("Solution Exists: Following" +
" is one Hamiltonian Cycle");
for (int i = 0; i < V; i++)
Console.Write(" " + path[i] + " ");
// Let us print the first vertex again
// to show the complete cycle
Console.WriteLine(" " + path[0] + " ");
}
// Driver code
public static void Main(String []args)
{
HamiltonianCycle hamiltonian =
new HamiltonianCycle();
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3)-------(4) */
int [,]graph1= {{0, 1, 0, 1, 0},
{1, 0, 1, 1, 1},
{0, 1, 0, 0, 1},
{1, 1, 0, 0, 1},
{0, 1, 1, 1, 0},
};
// Print the solution
hamiltonian.hamCycle(graph1);
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3) (4) */
int [,]graph2 = {{0, 1, 0, 1, 0},
{1, 0, 1, 1, 1},
{0, 1, 0, 0, 1},
{1, 1, 0, 0, 0},
{0, 1, 1, 0, 0},
};
// Print the solution
hamiltonian.hamCycle(graph2);
}
}
// This code contributed by Rajput-Ji
PHP
<?php
// PHP program for solution of
// Hamiltonian Cycle problem
// using backtracking
$V = 5;
/* A utility function to check if
the vertex v can be added at index 'pos'
in the Hamiltonian Cycle constructed so far
(stored in 'path[]') */
function isSafe($v, $graph, &$path, $pos)
{
/* Check if this vertex is
an adjacent vertex of the
previously added vertex. */
if ($graph[$path[$pos - 1]][$v] == 0)
return false;
/* Check if the vertex has already been included.
This step can be optimized by creating an array
of size V */
for ($i = 0; $i < $pos; $i++)
if ($path[$i] == $v)
return false;
return true;
}
/* A recursive utility function
to solve hamiltonian cycle problem */
function hamCycleUtil($graph, &$path, $pos)
{
global $V;
/* base case: If all vertices are included in
Hamiltonian Cycle */
if ($pos == $V)
{
// And if there is an edge from the
// last included vertex to the first vertex
if ($graph[$path[$pos - 1]][$path[0]] == 1)
return true;
else
return false;
}
// Try different vertices as a next candidate in
// Hamiltonian Cycle. We don't try for 0 as we
// included 0 as starting point hamCycle()
for ($v = 1; $v < $V; $v++)
{
/* Check if this vertex can be added
to Hamiltonian Cycle */
if (isSafe($v, $graph, $path, $pos))
{
$path[$pos] = $v;
/* recur to construct rest of the path */
if (hamCycleUtil($graph, $path,
$pos + 1) == true)
return true;
/* If adding vertex v doesn't lead to a solution,
then remove it */
$path[$pos] = -1;
}
}
/* If no vertex can be added to Hamiltonian Cycle
constructed so far, then return false */
return false;
}
/* This function solves the Hamiltonian Cycle problem using
Backtracking. It mainly uses hamCycleUtil() to solve the
problem. It returns false if there is no Hamiltonian Cycle
possible, otherwise return true and prints the path.
Please note that there may be more than one solutions,
this function prints one of the feasible solutions. */
function hamCycle($graph)
{
global $V;
$path = array_fill(0, $V, 0);
for ($i = 0; $i < $V; $i++)
$path[$i] = -1;
/* Let us put vertex 0 as the first vertex in the path.
If there is a Hamiltonian Cycle, then the path can be
started from any point of the cycle as the graph is
undirected */
$path[0] = 0;
if (hamCycleUtil($graph, $path, 1) == false)
{
echo("\nSolution does not exist");
return 0;
}
printSolution($path);
return 1;
}
/* A utility function to print solution */
function printSolution($path)
{
global $V;
echo("Solution Exists: Following is ".
"one Hamiltonian Cycle\n");
for ($i = 0; $i < $V; $i++)
echo(" ".$path[$i]." ");
// Let us print the first vertex again to show the
// complete cycle
echo(" ".$path[0]." \n");
}
// Driver Code
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3)-------(4) */
$graph1 = array(array(0, 1, 0, 1, 0),
array(1, 0, 1, 1, 1),
array(0, 1, 0, 0, 1),
array(1, 1, 0, 0, 1),
array(0, 1, 1, 1, 0),
);
// Print the solution
hamCycle($graph1);
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3) (4) */
$graph2 = array(array(0, 1, 0, 1, 0),
array(1, 0, 1, 1, 1),
array(0, 1, 0, 0, 1),
array(1, 1, 0, 0, 0),
array(0, 1, 1, 0, 0));
// Print the solution
hamCycle($graph2);
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript program for solution of Hamiltonian
// Cycle problem using backtracking
class HamiltonianCycle {
constructor() {
this.V = 5;
this.path = [];
}
/* A utility function to check
if the vertex v can be added at
index 'pos'in the Hamiltonian Cycle
constructed so far (stored in 'path[]') */
isSafe(v, graph, path, pos) {
/* Check if this vertex is
an adjacent vertex of the
previously added vertex. */
if (graph[path[pos - 1]][v] == 0) return false;
/* Check if the vertex has already
been included. This step can be
optimized by creating an array
of size V */
for (var i = 0; i < pos; i++) if (path[i] == v) return false;
return true;
}
/* A recursive utility function
to solve hamiltonian cycle problem */
hamCycleUtil(graph, path, pos) {
/* base case: If all vertices
are included in Hamiltonian Cycle */
if (pos == this.V) {
// And if there is an edge from the last included
// vertex to the first vertex
if (graph[path[pos - 1]][path[0]] == 1) return true;
else return false;
}
// Try different vertices as a next candidate in
// Hamiltonian Cycle. We don't try for 0 as we
// included 0 as starting point in hamCycle()
for (var v = 1; v < this.V; v++) {
/* Check if this vertex can be
added to Hamiltonian Cycle */
if (this.isSafe(v, graph, path, pos)) {
path[pos] = v;
/* recur to construct rest of the path */
if (this.hamCycleUtil(graph, path, pos + 1) == true) return true;
/* If adding vertex v doesn't
lead to a solution, then remove it */
path[pos] = -1;
}
}
/* If no vertex can be added to Hamiltonian Cycle
constructed so far, then return false */
return false;
}
/* This function solves the Hamiltonian
Cycle problem using Backtracking. It
mainly uses hamCycleUtil() to solve the
problem. It returns false if there
is no Hamiltonian Cycle possible,
otherwise return true and prints the path.
Please note that there may be more than
one solutions, this function prints one
of the feasible solutions. */
hamCycle(graph) {
this.path = new Array(this.V).fill(0);
for (var i = 0; i < this.V; i++) this.path[i] = -1;
/* Let us put vertex 0 as the first
vertex in the path. If there is a
Hamiltonian Cycle, then the path can be
started from any point of the cycle
as the graph is undirected */
this.path[0] = 0;
if (this.hamCycleUtil(graph, this.path, 1) == false) {
document.write("<br>Solution does not exist");
return 0;
}
this.printSolution(this.path);
return 1;
}
/* A utility function to print solution */
printSolution(path) {
document.write(
"Solution Exists: Following" + " is one Hamiltonian Cycle <br>"
);
for (var i = 0; i < this.V; i++) document.write(" " + path[i] + " ");
// Let us print the first vertex again
// to show the complete cycle
document.write(" " + path[0] + " <br>");
}
}
// Driver code
var hamiltonian = new HamiltonianCycle();
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3)-------(4) */
var graph1 = [
[0, 1, 0, 1, 0],
[1, 0, 1, 1, 1],
[0, 1, 0, 0, 1],
[1, 1, 0, 0, 1],
[0, 1, 1, 1, 0],
];
// Print the solution
hamiltonian.hamCycle(graph1);
/* Let us create the following graph
(0)--(1)--(2)
| / \ |
| / \ |
| / \ |
(3) (4) */
var graph2 = [
[0, 1, 0, 1, 0],
[1, 0, 1, 1, 1],
[0, 1, 0, 0, 1],
[1, 1, 0, 0, 0],
[0, 1, 1, 0, 0],
];
// Print the solution
hamiltonian.hamCycle(graph2);
// This code is contributed by rdtank.
</script>
Producción:
Solution Exists: Following is one Hamiltonian Cycle 0 1 2 4 3 0 Solution does not exist
Tenga en cuenta que el código anterior siempre imprime un ciclo a partir de 0. El punto de inicio no debería importar, ya que el ciclo se puede iniciar desde cualquier punto. Si desea cambiar el punto de partida, debe realizar dos cambios en el código anterior.
Cambiar «ruta [0] = 0;» a “ruta[0] = s;” donde s es su nuevo punto de partida. También cambie el ciclo “for (int v = 1; v < V; v++)” en hamCycleUtil() a “for (int v = 0; v < V; v++)”.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA