Biconnected graph

An undirected graph is called Biconnected if there are two vertex-disjoint paths between any two vertices. In a Biconnected Graph, there is a simple cycle through any two vertices.
By convention, two nodes connected by an edge form a biconnected graph, but this does not verify the above properties. For a graph with more than two vertices, the above properties must be there for it to be Biconnected.

Following are some examples.

Biconnected1
Biconnected1
Biconnected
Biconnected4
Biconnected5

See this for more examples.









How to find if a given graph is Biconnected or not?

A connected graph is Biconnected if it is connected and doesn’t have any Articulation Point. We mainly need to check two things in a graph.
1) The graph is connected.
2) There is not articulation point in graph.

We start from any vertex and do DFS traversal. In DFS traversal, we check if there is any articulation point. If we don’t find any articulation point, then the graph is Biconnected. Finally, we need to check whether all vertices were reachable in DFS or not. If all vertices were not reachable, then the graph is not even connected.
Following is C++ implementation of above approach.

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 tract 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 aadjacent 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 chilren.
            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 starring 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 tract 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 aadjacent 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 chilren.
                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
        // starring 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

Python

# 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 tract 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 chilren.
				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


Output:
Yes
Yes
No
No
Yes

Time Complexity: The above function is a simple DFS with additional arrays. So time complexity is same as DFS which is O(V+E) for adjacency list representation of graph.

References:
http://www.cs.purdue.edu/homes/ayg/CS251/slides/chap9d.pdf

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