Biconnected Components

A biconnected component is a maximal biconnected subgraph.

Biconnected Graph is already discussed here. In this article, we will see how to find biconnected component in a graph using algorithm by John Hopcroft and Robert Tarjan.

Biconnected Components

In above graph, following are the biconnected components:

  • 4–2 3–4 3–1 2–3 1–2
  • 8–9
  • 8–5 7–8 5–7
  • 6–0 5–6 1–5 0–1
  • 10–11

Algorithm is based on Disc and Low Values discussed in Strongly Connected Components Article.



Idea is to store visited edges in a stack while DFS on a graph and keep looking for Articulation Points (highlighted in above figure). As soon as an Articulation Point u is found, all edges visited while DFS from node u onwards will form one biconnected component. When DFS completes for one connected component, all edges present in stack will form a biconnected component.
If there is no Articulation Point in graph, then graph is biconnected and so there will be one biconnected component which is the graph itself.

C++

// A C++ program to find biconnected components in a given undirected graph
#include <iostream>
#include <list>
#include <stack>
#define NIL -1
using namespace std;
int count = 0;
class Edge {
public:
    int u;
    int v;
    Edge(int u, int v);
};
Edge::Edge(int u, int v)
{
    this->u = u;
    this->v = v;
}
  
// A class that represents an directed graph
class Graph {
    int V; // No. of vertices
    int E; // No. of edges
    list<int>* adj; // A dynamic array of adjacency lists
  
    // A Recursive DFS based function used by BCC()
    void BCCUtil(int u, int disc[], int low[],
                 list<Edge>* st, int parent[]);
  
public:
    Graph(int V); // Constructor
    void addEdge(int v, int w); // function to add an edge to graph
    void BCC(); // prints strongly connected components
};
  
Graph::Graph(int V)
{
    this->V = V;
    this->E = 0;
    adj = new list<int>[V];
}
  
void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w);
    E++;
}
  
// A recursive function that finds and prints strongly connected
// components using DFS traversal
// u --> The vertex to be visited next
// disc[] --> Stores discovery times of visited vertices
// low[] -- >> earliest visited vertex (the vertex with minimum
// discovery time) that can be reached from subtree
// rooted with current vertex
// *st -- >> To store visited edges
void Graph::BCCUtil(int u, int disc[], int low[], list<Edge>* st,
                    int parent[])
{
    // A static variable is used for simplicity, we can avoid use
    // of static variable by passing a pointer.
    static int time = 0;
  
    // Initialize discovery time and low value
    disc[u] = low[u] = ++time;
    int children = 0;
  
    // 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 recur for it
        if (disc[v] == -1) {
            children++;
            parent[v] = u;
            // store the edge in stack
            st->push_back(Edge(u, v));
            BCCUtil(v, disc, low, st, parent);
  
            // Check if the subtree rooted with 'v' has a
            // connection to one of the ancestors of 'u'
            // Case 1 -- per Strongly Connected Components Article
            low[u] = min(low[u], low[v]);
  
            // If u is an articulation point,
            // pop all edges from stack till u -- v
            if ((disc[u] == 1 && children > 1) || (disc[u] > 1 && low[v] >= disc[u])) {
                while (st->back().u != u || st->back().v != v) {
                    cout << st->back().u << "--" << st->back().v << " ";
                    st->pop_back();
                }
                cout << st->back().u << "--" << st->back().v;
                st->pop_back();
                cout << endl;
                count++;
            }
        }
  
        // Update low value of 'u' only of 'v' is still in stack
        // (i.e. it's a back edge, not cross edge).
        // Case 2 -- per Strongly Connected Components Article
        else if (v != parent[u]) {
            low[u] = min(low[u], disc[v]);
            if (disc[v] < disc[u]) {
                st->push_back(Edge(u, v));
            }
        }
    }
}
  
// The function to do DFS traversal. It uses BCCUtil()
void Graph::BCC()
{
    int* disc = new int[V];
    int* low = new int[V];
    int* parent = new int[V];
    list<Edge>* st = new list<Edge>[E];
  
    // Initialize disc and low, and parent arrays
    for (int i = 0; i < V; i++) {
        disc[i] = NIL;
        low[i] = NIL;
        parent[i] = NIL;
    }
  
    for (int i = 0; i < V; i++) {
        if (disc[i] == NIL)
            BCCUtil(i, disc, low, st, parent);
  
        int j = 0;
        // If stack is not empty, pop all edges from stack
        while (st->size() > 0) {
            j = 1;
            cout << st->back().u << "--" << st->back().v << " ";
            st->pop_back();
        }
        if (j == 1) {
            cout << endl;
            count++;
        }
    }
}
  
// Driver program to test above function
int main()
{
    Graph g(12);
    g.addEdge(0, 1);
    g.addEdge(1, 0);
    g.addEdge(1, 2);
    g.addEdge(2, 1);
    g.addEdge(1, 3);
    g.addEdge(3, 1);
    g.addEdge(2, 3);
    g.addEdge(3, 2);
    g.addEdge(2, 4);
    g.addEdge(4, 2);
    g.addEdge(3, 4);
    g.addEdge(4, 3);
    g.addEdge(1, 5);
    g.addEdge(5, 1);
    g.addEdge(0, 6);
    g.addEdge(6, 0);
    g.addEdge(5, 6);
    g.addEdge(6, 5);
    g.addEdge(5, 7);
    g.addEdge(7, 5);
    g.addEdge(5, 8);
    g.addEdge(8, 5);
    g.addEdge(7, 8);
    g.addEdge(8, 7);
    g.addEdge(8, 9);
    g.addEdge(9, 8);
    g.addEdge(10, 11);
    g.addEdge(11, 10);
    g.BCC();
    cout << "Above are " << count << " biconnected components in graph";
    return 0;
}

Java

// A Java program to find biconnected components in a given
// undirected graph
import java.io.*;
import java.util.*;
  
// This class represents a directed graph using adjacency
// list representation
class Graph {
    private int V, E; // No. of vertices & Edges respectively
    private LinkedList<Integer> adj[]; // Adjacency List
  
    // Count is number of biconnected components. time is
    // used to find discovery times
    static int count = 0, time = 0;
  
    class Edge {
        int u;
        int v;
        Edge(int u, int v)
        {
            this.u = u;
            this.v = v;
        }
    };
  
    // Constructor
    Graph(int v)
    {
        V = v;
        E = 0;
        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);
        E++;
    }
  
    // A recursive function that finds and prints strongly connected
    // components using DFS traversal
    // u --> The vertex to be visited next
    // disc[] --> Stores discovery times of visited vertices
    // low[] -- >> earliest visited vertex (the vertex with minimum
    // discovery time) that can be reached from subtree
    // rooted with current vertex
    // *st -- >> To store visited edges
    void BCCUtil(int u, int disc[], int low[], LinkedList<Edge> st,
                 int parent[])
    {
  
        // Initialize discovery time and low value
        disc[u] = low[u] = ++time;
        int children = 0;
  
        // Go through all vertices adjacent to this
        Iterator<Integer> it = adj[u].iterator();
        while (it.hasNext()) {
            int v = it.next(); // v is current adjacent of 'u'
  
            // If v is not visited yet, then recur for it
            if (disc[v] == -1) {
                children++;
                parent[v] = u;
  
                // store the edge in stack
                st.add(new Edge(u, v));
                BCCUtil(v, disc, low, st, parent);
  
                // Check if the subtree rooted with 'v' has a
                // connection to one of the ancestors of 'u'
                // Case 1 -- per Strongly Connected Components Article
                if (low[u] > low[v])
                    low[u] = low[v];
  
                // If u is an articulation point,
                // pop all edges from stack till u -- v
                if ((disc[u] == 1 && children > 1) || (disc[u] > 1 && low[v] >= disc[u])) {
                    while (st.getLast().u != u || st.getLast().v != v) {
                        System.out.print(st.getLast().u + "--" + st.getLast().v + " ");
                        st.removeLast();
                    }
                    System.out.println(st.getLast().u + "--" + st.getLast().v + " ");
                    st.removeLast();
  
                    count++;
                }
            }
  
            // Update low value of 'u' only of 'v' is still in stack
            // (i.e. it's a back edge, not cross edge).
            // Case 2 -- per Strongly Connected Components Article
            else if (v != parent[u] && disc[v] < low[u]) {
                if (low[u] > disc[v])
                    low[u] = disc[v];
                st.add(new Edge(u, v));
            }
        }
    }
  
    // The function to do DFS traversal. It uses BCCUtil()
    void BCC()
    {
        int disc[] = new int[V];
        int low[] = new int[V];
        int parent[] = new int[V];
        LinkedList<Edge> st = new LinkedList<Edge>();
  
        // Initialize disc and low, and parent arrays
        for (int i = 0; i < V; i++) {
            disc[i] = -1;
            low[i] = -1;
            parent[i] = -1;
        }
  
        for (int i = 0; i < V; i++) {
            if (disc[i] == -1)
                BCCUtil(i, disc, low, st, parent);
  
            int j = 0;
  
            // If stack is not empty, pop all edges from stack
            while (st.size() > 0) {
                j = 1;
                System.out.print(st.getLast().u + "--" + st.getLast().v + " ");
                st.removeLast();
            }
            if (j == 1) {
                System.out.println();
                count++;
            }
        }
    }
  
    public static void main(String args[])
    {
        Graph g = new Graph(12);
        g.addEdge(0, 1);
        g.addEdge(1, 0);
        g.addEdge(1, 2);
        g.addEdge(2, 1);
        g.addEdge(1, 3);
        g.addEdge(3, 1);
        g.addEdge(2, 3);
        g.addEdge(3, 2);
        g.addEdge(2, 4);
        g.addEdge(4, 2);
        g.addEdge(3, 4);
        g.addEdge(4, 3);
        g.addEdge(1, 5);
        g.addEdge(5, 1);
        g.addEdge(0, 6);
        g.addEdge(6, 0);
        g.addEdge(5, 6);
        g.addEdge(6, 5);
        g.addEdge(5, 7);
        g.addEdge(7, 5);
        g.addEdge(5, 8);
        g.addEdge(8, 5);
        g.addEdge(7, 8);
        g.addEdge(8, 7);
        g.addEdge(8, 9);
        g.addEdge(9, 8);
        g.addEdge(10, 11);
        g.addEdge(11, 10);
  
        g.BCC();
  
        System.out.println("Above are " + g.count + " biconnected components in graph");
    }
}
// This code is contributed by Aakash Hasija

Python

# Python program to find biconnected components in a given
# undirected graph
# Complexity : O(V + E)
  
   
from collections import defaultdict
   
# This class represents an directed graph 
# using adjacency list representation
class Graph:
   
    def __init__(self, vertices):
        # No. of vertices
        self.V = vertices 
          
        # default dictionary to store graph
        self.graph = defaultdict(list)
          
        # time is used to find discovery times
        self.Time = 0 
          
        # Count is number of biconnected components
        self.count = 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 finds and prints strongly connected
    components using DFS traversal
    u --> The vertex to be visited next
    disc[] --> Stores discovery times of visited vertices
    low[] -- >> earliest visited vertex (the vertex with minimum
               discovery time) that can be reached from subtree
               rooted with current vertex
    st -- >> To store visited edges'''
    def BCCUtil(self, u, parent, low, disc, st):
  
        # Count of children in current node 
        children = 0
  
        # 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 disc[v] == -1 :
                parent[v] = u
                children += 1
                st.append((u, v)) # store the edge in stack
                self.BCCUtil(v, parent, low, disc, st)
  
                # Check if the subtree rooted with v has a connection to
                # one of the ancestors of u
                # Case 1 -- per Strongly Connected Components Article
                low[u] = min(low[u], low[v])
  
                # If u is an articulation point, pop 
                # all edges from stack till (u, v)
                if parent[u] == -1 and children > 1 or parent[u] != -1 and low[v] >= disc[u]:
                    self.count += 1 # increment count
                    w = -1
                    while w != (u, v):
                        w = st.pop()
                        print w,
                    print""
              
            elif v != parent[u] and low[u] > disc[v]:
                '''Update low value of 'u' only of 'v' is still in stack
                (i.e. it's a back edge, not cross edge).
                Case 2 
                -- per Strongly Connected Components Article'''
  
                low[u] = min(low [u], disc[v])
      
                st.append((u, v))
  
  
    # The function to do DFS traversal. 
    # It uses recursive BCCUtil()
    def BCC(self):
          
        # Initialize disc and low, and parent arrays
        disc = [-1] * (self.V)
        low = [-1] * (self.V)
        parent = [-1] * (self.V)
        st = []
  
        # Call the recursive helper function to 
        # find articulation points
        # in DFS tree rooted with vertex 'i'
        for i in range(self.V):
            if disc[i] == -1:
                self.BCCUtil(i, parent, low, disc, st)
  
            # If stack is not empty, pop all edges from stack
            if st:
                self.count = self.count + 1
  
                while st:
                    w = st.pop()
                    print w,
                print ""
  
# Create a graph given in the above diagram
  
g = Graph(12)
g.addEdge(0, 1)
g.addEdge(1, 2)
g.addEdge(1, 3)
g.addEdge(2, 3)
g.addEdge(2, 4)
g.addEdge(3, 4)
g.addEdge(1, 5)
g.addEdge(0, 6)
g.addEdge(5, 6)
g.addEdge(5, 7)
g.addEdge(5, 8)
g.addEdge(7, 8)
g.addEdge(8, 9)
g.addEdge(10, 11)
  
g.BCC();
print ("Above are % d biconnected components in graph" %(g.count));
  
# This code is contributed by Neelam Yadav

Output:

4--2 3--4 3--1 2--3 1--2
8--9
8--5 7--8 5--7
6--0 5--6 1--5 0--1 
10--11
Above are 5 biconnected components in graph

This article is contributed by Anurag Singh. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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Improved By : VismayMalondkar

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