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.
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 if '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] < disc[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 |
C#
// A C# program to find biconnected components in a given // undirected graph using System; using System.Collections.Generic; // This class represents a directed graph using adjacency // list representation public class Graph { private int V, E; // No. of vertices & Edges respectively private List< int > []adj; // Adjacency List // Count is number of biconnected components. time is // used to find discovery times int count = 0, time = 0; class Edge { public int u; public int v; public Edge( int u, int v) { this .u = u; this .v = v; } }; // Constructor public Graph( int v) { V = v; E = 0; 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) { 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, List<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 foreach ( int it in adj[u]) { int v = it; // 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[st.Count-1].u != u || st[st.Count-1].v != v) { Console.Write(st[st.Count - 1].u + "--" + st[st.Count - 1].v + " " ); st.RemoveAt(st.Count - 1); } Console.WriteLine(st[st.Count - 1].u + "--" + st[st.Count - 1].v + " " ); st.RemoveAt(st.Count - 1); count++; } } // Update low value of 'u' only if '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] < disc[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]; List<Edge> st = new List<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.Count > 0) { j = 1; Console.Write(st[st.Count - 1].u + "--" + st[st.Count - 1].v + " " ); st.RemoveAt(st.Count - 1); } if (j == 1) { Console.WriteLine(); count++; } } } // Driver code 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(); Console.WriteLine( "Above are " + g.count + " biconnected components in graph" ); } } // This code is contributed by PrinciRaj1992 |
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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