Un grafo no dirigido se llama biconectado si hay dos caminos disjuntos de vértice entre dos vértices cualesquiera. En un gráfico biconectado, hay un ciclo simple a través de dos vértices cualesquiera.
Por convención, dos Nodes conectados por un borde forman un gráfico biconectado, pero esto no verifica las propiedades anteriores. Para un gráfico con más de dos vértices, las propiedades anteriores deben estar presentes para que sea Biconectado.
O en otras palabras:
Se dice que un grafo es biconexo si:
- Está conectado, es decir, es posible llegar a cada vértice desde cualquier otro vértice, por un camino simple.
- Incluso después de eliminar cualquier vértice, el gráfico permanece conectado.
Los siguientes son algunos ejemplos:
Ejemplo 1
Ejemplo-2
Ejemplo-3
Ejemplo-4
Ejemplo-5
Ver esto para más ejemplos.
¿Cómo encontrar si un gráfico dado es biconexo o no?
Un grafo conexo es biconexo si es conexo y no tiene ningún punto de articulación . Principalmente necesitamos verificar dos cosas en un gráfico.
- La gráfica está conectada.
- No hay punto de articulación en el gráfico.
Partimos de cualquier vértice y hacemos un recorrido DFS. En DFS transversal, verificamos si hay algún punto de articulación. Si no encontramos ningún punto de articulación, entonces el grafo es Biconexo. Finalmente, debemos verificar si todos los vértices eran accesibles en DFS o no. Si no se pudieran alcanzar todos los vértices, entonces el gráfico ni siquiera es conexo.
A continuación se muestra la implementación del enfoque anterior.
C++
// A C++ program to find if a given undirected graph is
// biconnected
#include<iostream>
#include <list>
#define NIL -1
using namespace std;
// A class that represents an undirected graph
class Graph
{
int V; // No. of vertices
list<int> *adj; // A dynamic array of adjacency lists
bool isBCUtil(int v, bool visited[], int disc[], int low[],
int parent[]);
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // to add an edge to graph
bool isBC(); // returns true if graph is Biconnected
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v); // Note: the graph is undirected
}
// A recursive function that returns true if there is an articulation
// point in given graph, otherwise returns false.
// This function is almost same as isAPUtil() here ( http://goo.gl/Me9Fw )
// u --> The vertex to be visited next
// visited[] --> keeps track of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
bool Graph::isBCUtil(int u, bool visited[], int disc[],int low[],int parent[])
{
// A static variable is used for simplicity, we can avoid use of static
// variable by passing a pointer.
static int time = 0;
// Count of children in DFS Tree
int children = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices adjacent to this
list<int>::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
{
int v = *i; // v is current adjacent of u
// If v is not visited yet, then make it a child of u
// in DFS tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
// check if subgraph rooted with v has an articulation point
if (isBCUtil(v, visited, disc, low, parent))
return true;
// Check if the subtree rooted with v has a connection to
// one of the ancestors of u
low[u] = min(low[u], low[v]);
// u is an articulation point in following cases
// (1) u is root of DFS tree and has two or more children.
if (parent[u] == NIL && children > 1)
return true;
// (2) If u is not root and low value of one of its child is
// more than discovery value of u.
if (parent[u] != NIL && low[v] >= disc[u])
return true;
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = min(low[u], disc[v]);
}
return false;
}
// The main function that returns true if graph is Biconnected,
// otherwise false. It uses recursive function isBCUtil()
bool Graph::isBC()
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
int *disc = new int[V];
int *low = new int[V];
int *parent = new int[V];
// Initialize parent and visited, and ap(articulation point)
// arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
}
// Call the recursive helper function to find if there is an articulation
// point in given graph. We do DFS traversal starting from vertex 0
if (isBCUtil(0, visited, disc, low, parent) == true)
return false;
// Now check whether the given graph is connected or not. An undirected
// graph is connected if all vertices are reachable from any starting
// point (we have taken 0 as starting point)
for (int i = 0; i < V; i++)
if (visited[i] == false)
return false;
return true;
}
// Driver program to test above function
int main()
{
// Create graphs given in above diagrams
Graph g1(2);
g1.addEdge(0, 1);
g1.isBC()? cout << "Yes\n" : cout << "No\n";
Graph g2(5);
g2.addEdge(1, 0);
g2.addEdge(0, 2);
g2.addEdge(2, 1);
g2.addEdge(0, 3);
g2.addEdge(3, 4);
g2.addEdge(2, 4);
g2.isBC()? cout << "Yes\n" : cout << "No\n";
Graph g3(3);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.isBC()? cout << "Yes\n" : cout << "No\n";
Graph g4(5);
g4.addEdge(1, 0);
g4.addEdge(0, 2);
g4.addEdge(2, 1);
g4.addEdge(0, 3);
g4.addEdge(3, 4);
g4.isBC()? cout << "Yes\n" : cout << "No\n";
Graph g5(3);
g5.addEdge(0, 1);
g5.addEdge(1, 2);
g5.addEdge(2, 0);
g5.isBC()? cout << "Yes\n" : cout << "No\n";
return 0;
}
Java
// A Java program to find if a given undirected graph is
// biconnected
import java.io.*;
import java.util.*;
import java.util.LinkedList;
// This class represents a directed graph using adjacency
// list representation
class Graph
{
private int V; // No. of vertices
// Array of lists for Adjacency List Representation
private LinkedList<Integer> adj[];
int time = 0;
static final int NIL = -1;
// Constructor
Graph(int v)
{
V = v;
adj = new LinkedList[v];
for (int i=0; i<v; ++i)
adj[i] = new LinkedList();
}
//Function to add an edge into the graph
void addEdge(int v, int w)
{
adj[v].add(w); //Note that the graph is undirected.
adj[w].add(v);
}
// A recursive function that returns true if there is an articulation
// point in given graph, otherwise returns false.
// This function is almost same as isAPUtil() @ http://goo.gl/Me9Fw
// u --> The vertex to be visited next
// visited[] --> keeps track of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
boolean isBCUtil(int u, boolean visited[], int disc[],int low[],
int parent[])
{
// Count of children in DFS Tree
int children = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices adjacent to this
Iterator<Integer> i = adj[u].iterator();
while (i.hasNext())
{
int v = i.next(); // v is current adjacent of u
// If v is not visited yet, then make it a child of u
// in DFS tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
// check if subgraph rooted with v has an articulation point
if (isBCUtil(v, visited, disc, low, parent))
return true;
// Check if the subtree rooted with v has a connection to
// one of the ancestors of u
low[u] = Math.min(low[u], low[v]);
// u is an articulation point in following cases
// (1) u is root of DFS tree and has two or more children.
if (parent[u] == NIL && children > 1)
return true;
// (2) If u is not root and low value of one of its
// child is more than discovery value of u.
if (parent[u] != NIL && low[v] >= disc[u])
return true;
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = Math.min(low[u], disc[v]);
}
return false;
}
// The main function that returns true if graph is Biconnected,
// otherwise false. It uses recursive function isBCUtil()
boolean isBC()
{
// Mark all the vertices as not visited
boolean visited[] = new boolean[V];
int disc[] = new int[V];
int low[] = new int[V];
int parent[] = new int[V];
// Initialize parent and visited, and ap(articulation point)
// arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
}
// Call the recursive helper function to find if there is an
// articulation/ point in given graph. We do DFS traversal
// starting from vertex 0
if (isBCUtil(0, visited, disc, low, parent) == true)
return false;
// Now check whether the given graph is connected or not.
// An undirected graph is connected if all vertices are
// reachable from any starting point (we have taken 0 as
// starting point)
for (int i = 0; i < V; i++)
if (visited[i] == false)
return false;
return true;
}
// Driver method
public static void main(String args[])
{
// Create graphs given in above diagrams
Graph g1 =new Graph(2);
g1.addEdge(0, 1);
if (g1.isBC())
System.out.println("Yes");
else
System.out.println("No");
Graph g2 =new Graph(5);
g2.addEdge(1, 0);
g2.addEdge(0, 2);
g2.addEdge(2, 1);
g2.addEdge(0, 3);
g2.addEdge(3, 4);
g2.addEdge(2, 4);
if (g2.isBC())
System.out.println("Yes");
else
System.out.println("No");
Graph g3 = new Graph(3);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
if (g3.isBC())
System.out.println("Yes");
else
System.out.println("No");
Graph g4 = new Graph(5);
g4.addEdge(1, 0);
g4.addEdge(0, 2);
g4.addEdge(2, 1);
g4.addEdge(0, 3);
g4.addEdge(3, 4);
if (g4.isBC())
System.out.println("Yes");
else
System.out.println("No");
Graph g5= new Graph(3);
g5.addEdge(0, 1);
g5.addEdge(1, 2);
g5.addEdge(2, 0);
if (g5.isBC())
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Aakash Hasija
Python3
# A Python program to find if a given undirected graph is
# biconnected
from collections import defaultdict
#This class represents an undirected graph using adjacency list representation
class Graph:
def __init__(self,vertices):
self.V= vertices #No. of vertices
self.graph = defaultdict(list) # default dictionary to store graph
self.Time = 0
# function to add an edge to graph
def addEdge(self,u,v):
self.graph[u].append(v)
self.graph[v].append(u)
'''A recursive function that returns true if there is an articulation
point in given graph, otherwise returns false.
This function is almost same as isAPUtil()
u --> The vertex to be visited next
visited[] --> keeps track of visited vertices
disc[] --> Stores discovery times of visited vertices
parent[] --> Stores parent vertices in DFS tree'''
def isBCUtil(self,u, visited, parent, low, disc):
#Count of children in current node
children =0
# Mark the current node as visited and print it
visited[u]= True
# Initialize discovery time and low value
disc[u] = self.Time
low[u] = self.Time
self.Time += 1
#Recur for all the vertices adjacent to this vertex
for v in self.graph[u]:
# If v is not visited yet, then make it a child of u
# in DFS tree and recur for it
if visited[v] == False :
parent[v] = u
children += 1
if self.isBCUtil(v, visited, parent, low, disc):
return True
# Check if the subtree rooted with v has a connection to
# one of the ancestors of u
low[u] = min(low[u], low[v])
# u is an articulation point in following cases
# (1) u is root of DFS tree and has two or more children.
if parent[u] == -1 and children > 1:
return True
#(2) If u is not root and low value of one of its child is more
# than discovery value of u.
if parent[u] != -1 and low[v] >= disc[u]:
return True
elif v != parent[u]: # Update low value of u for parent function calls.
low[u] = min(low[u], disc[v])
return False
# The main function that returns true if graph is Biconnected,
# otherwise false. It uses recursive function isBCUtil()
def isBC(self):
# Mark all the vertices as not visited and Initialize parent and visited,
# and ap(articulation point) arrays
visited = [False] * (self.V)
disc = [float("Inf")] * (self.V)
low = [float("Inf")] * (self.V)
parent = [-1] * (self.V)
# Call the recursive helper function to find if there is an
# articulation points in given graph. We do DFS traversal starting
# from vertex 0
if self.isBCUtil(0, visited, parent, low, disc):
return False
'''Now check whether the given graph is connected or not.
An undirected graph is connected if all vertices are
reachable from any starting point (we have taken 0 as
starting point)'''
if any(i == False for i in visited):
return False
return True
# Create a graph given in the above diagram
g1 = Graph(2)
g1.addEdge(0, 1)
print ("Yes" if g1.isBC() else "No")
g2 = Graph(5)
g2.addEdge(1, 0)
g2.addEdge(0, 2)
g2.addEdge(2, 1)
g2.addEdge(0, 3)
g2.addEdge(3, 4)
g2.addEdge(2, 4)
print ("Yes" if g2.isBC() else "No")
g3 = Graph(3)
g3.addEdge(0, 1)
g3.addEdge(1, 2)
print ("Yes" if g3.isBC() else "No")
g4 = Graph (5)
g4.addEdge(1, 0)
g4.addEdge(0, 2)
g4.addEdge(2, 1)
g4.addEdge(0, 3)
g4.addEdge(3, 4)
print ("Yes" if g4.isBC() else "No")
g5 = Graph(3)
g5.addEdge(0, 1)
g5.addEdge(1, 2)
g5.addEdge(2, 0)
print ("Yes" if g5.isBC() else "No")
#This code is contributed by Neelam Yadav
C#
// A C# program to find if a given undirected
// graph is biconnected
using System;
using System.Collections.Generic;
// This class represents a directed graph
// using adjacency list representation
class Graph{
// No. of vertices
public int V;
// Array of lists for Adjacency
// List Representation
public List<int> []adj;
int time = 0;
static readonly int NIL = -1;
// Constructor
Graph(int v)
{
V = v;
adj = new List<int>[v];
for(int i = 0; i < v; ++i)
adj[i] = new List<int>();
}
// Function to add an edge into the graph
void addEdge(int v, int w)
{
// Note that the graph is undirected.
adj[v].Add(w);
adj[w].Add(v);
}
// A recursive function that returns true
// if there is an articulation point in
// given graph, otherwise returns false.
// This function is almost same as isAPUtil()
// @ http://goo.gl/Me9Fw
// u --> The vertex to be visited next
// visited[] --> keeps track of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
bool isBCUtil(int u, bool []visited,
int []disc,int []low,
int []parent)
{
// Count of children in DFS Tree
int children = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices adjacent to this
foreach(int i in adj[u])
{
// v is current adjacent of u
int v = i;
// If v is not visited yet, then
// make it a child of u in DFS
// tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
// Check if subgraph rooted with v
// has an articulation point
if (isBCUtil(v, visited, disc,
low, parent))
return true;
// Check if the subtree rooted with
// v has a connection to one of
// the ancestors of u
low[u] = Math.Min(low[u], low[v]);
// u is an articulation point in
// following cases
// (1) u is root of DFS tree and
// has two or more children.
if (parent[u] == NIL && children > 1)
return true;
// (2) If u is not root and low
// value of one of its child is
// more than discovery value of u.
if (parent[u] != NIL && low[v] >= disc[u])
return true;
}
// Update low value of u for
// parent function calls.
else if (v != parent[u])
low[u] = Math.Min(low[u], disc[v]);
}
return false;
}
// The main function that returns true
// if graph is Biconnected, otherwise
// false. It uses recursive function
// isBCUtil()
bool isBC()
{
// Mark all the vertices as not visited
bool []visited = new bool[V];
int []disc = new int[V];
int []low = new int[V];
int []parent = new int[V];
// Initialize parent and visited,
// and ap(articulation point)
// arrays
for(int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
}
// Call the recursive helper function to
// find if there is an articulation/ point
// in given graph. We do DFS traversal
// starting from vertex 0
if (isBCUtil(0, visited, disc,
low, parent) == true)
return false;
// Now check whether the given graph
// is connected or not. An undirected
// graph is connected if all vertices are
// reachable from any starting point
// (we have taken 0 as starting point)
for(int i = 0; i < V; i++)
if (visited[i] == false)
return false;
return true;
}
// Driver code
public static void Main(String []args)
{
// Create graphs given in above diagrams
Graph g1 = new Graph(2);
g1.addEdge(0, 1);
if (g1.isBC())
Console.WriteLine("Yes");
else
Console.WriteLine("No");
Graph g2 = new Graph(5);
g2.addEdge(1, 0);
g2.addEdge(0, 2);
g2.addEdge(2, 1);
g2.addEdge(0, 3);
g2.addEdge(3, 4);
g2.addEdge(2, 4);
if (g2.isBC())
Console.WriteLine("Yes");
else
Console.WriteLine("No");
Graph g3 = new Graph(3);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
if (g3.isBC())
Console.WriteLine("Yes");
else
Console.WriteLine("No");
Graph g4 = new Graph(5);
g4.addEdge(1, 0);
g4.addEdge(0, 2);
g4.addEdge(2, 1);
g4.addEdge(0, 3);
g4.addEdge(3, 4);
if (g4.isBC())
Console.WriteLine("Yes");
else
Console.WriteLine("No");
Graph g5 = new Graph(3);
g5.addEdge(0, 1);
g5.addEdge(1, 2);
g5.addEdge(2, 0);
if (g5.isBC())
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// A Javascript program to find if a given undirected graph is
// biconnected
// This class represents a directed graph using adjacency
// list representation
class Graph
{
// Constructor
constructor(v)
{
this.V = v;
this.adj = new Array(v);
this.NIL = -1;
this.time = 0;
for (let i=0; i<v; ++i)
this.adj[i] = [];
}
//Function to add an edge into the graph
addEdge(v,w)
{
this.adj[v].push(w); //Note that the graph is undirected.
this.adj[w].push(v);
}
// A recursive function that returns true if there is an articulation
// point in given graph, otherwise returns false.
// This function is almost same as isAPUtil() @ http://goo.gl/Me9Fw
// u --> The vertex to be visited next
// visited[] --> keeps track of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
isBCUtil(u,visited,disc,low,parent)
{
// Count of children in DFS Tree
let children = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++this.time;
// Go through all vertices adjacent to this
for(let i of this.adj[u])
{
let v = i; // v is current adjacent of u
// If v is not visited yet, then make it a child of u
// in DFS tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
// check if subgraph rooted with v has an articulation point
if (this.isBCUtil(v, visited, disc, low, parent))
return true;
// Check if the subtree rooted with v has a connection to
// one of the ancestors of u
low[u] = Math.min(low[u], low[v]);
// u is an articulation point in following cases
// (1) u is root of DFS tree and has two or more children.
if (parent[u] == this.NIL && children > 1)
return true;
// (2) If u is not root and low value of one of its
// child is more than discovery value of u.
if (parent[u] != this.NIL && low[v] >= disc[u])
return true;
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = Math.min(low[u], disc[v]);
}
return false;
}
// The main function that returns true if graph is Biconnected,
// otherwise false. It uses recursive function isBCUtil()
isBC()
{
// Mark all the vertices as not visited
let visited = new Array(this.V);
let disc = new Array(this.V);
let low = new Array(this.V);
let parent = new Array(this.V);
// Initialize parent and visited, and ap(articulation point)
// arrays
for (let i = 0; i < this.V; i++)
{
parent[i] = this.NIL;
visited[i] = false;
}
// Call the recursive helper function to find if there is an
// articulation/ point in given graph. We do DFS traversal
// starting from vertex 0
if (this.isBCUtil(0, visited, disc, low, parent) == true)
return false;
// Now check whether the given graph is connected or not.
// An undirected graph is connected if all vertices are
// reachable from any starting point (we have taken 0 as
// starting point)
for (let i = 0; i < this.V; i++)
if (visited[i] == false)
return false;
return true;
}
}
// Driver method
// Create graphs given in above diagrams
let g1 =new Graph(2);
g1.addEdge(0, 1);
if (g1.isBC())
document.write("Yes<br>");
else
document.write("No<br>");
let g2 =new Graph(5);
g2.addEdge(1, 0);
g2.addEdge(0, 2);
g2.addEdge(2, 1);
g2.addEdge(0, 3);
g2.addEdge(3, 4);
g2.addEdge(2, 4);
if (g2.isBC())
document.write("Yes<br>");
else
document.write("No<br>");
let g3 = new Graph(3);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
if (g3.isBC())
document.write("Yes<br>");
else
document.write("No<br>");
let g4 = new Graph(5);
g4.addEdge(1, 0);
g4.addEdge(0, 2);
g4.addEdge(2, 1);
g4.addEdge(0, 3);
g4.addEdge(3, 4);
if (g4.isBC())
document.write("Yes<br>");
else
document.write("No<br>");
let g5= new Graph(3);
g5.addEdge(0, 1);
g5.addEdge(1, 2);
g5.addEdge(2, 0);
if (g5.isBC())
document.write("Yes<br>");
else
document.write("No<br>");
// This code is contributed by avanitrachhadiya2155
</script>
Producción
Yes Yes No No Yes
Complejidad de tiempo: la función anterior es un DFS simple con arrays adicionales. Entonces, la complejidad del tiempo es la misma que DFS, que es O (V + E) para la representación de la lista de adyacencia del gráfico.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA