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.
Or in other words: 

A graph is said to be Biconnected if: 

1. It is connected, i.e. it is possible to reach every vertex from every other vertex, by a simple path. 
2. Even after removing any vertex the graph remains connected.

Following are some examples:

Example-1

Example-2

Example-3

Example-4

Example-5

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 the 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 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>

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.

Auxiliary Space : O(V)