Connectivity Graph Algorithms

JAVA Programming – Biconnected graph

JAVA Programming - Biconnected graph,An undirected graph is called Biconnected if there are two vertex-disjoint paths between any two vertices.

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.

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.

Java programming:

// 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");
    }
}

Output:

Yes
Yes
No
No
Yes
READ  Eulerian path and circuit for undirected graph

About the author

Venkatesan Prabu

Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.

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