Check if a given directed graph is strongly connected | Set 2 (Kosaraju using BFS)
Given a directed graph, find out whether the graph is strongly connected or not. A directed graph is strongly connected if there is a path between any two pairs of vertices. There are different methods to check the connectivity of directed graph but one of the optimized method is Kosaraju’s DFS based simple algorithm.
Kosaraju’s BFS based simple algorithm also work on the same principle as DFS based algorithm does.
Following is Kosaraju’s BFS based simple algorithm that does two BFS traversals of graph: 1) Initialize all vertices as not visited. 2) Do a BFS traversal of graph starting from any arbitrary vertex v. If BFS traversal doesn’t visit all vertices, then return false. 3) Reverse all edges (or find transpose or reverse of graph) 4) Mark all vertices as not visited in reversed graph. 5) Again do a BFS traversal of reversed graph starting from same vertex v (Same as step 2). If BFS traversal doesn’t visit all vertices, then return false. Otherwise, return true.
The idea is again simple if every node can be reached from a vertex v, and every node can reach same vertex v, then the graph is strongly connected. In step 2, we check if all vertices are reachable from v. In step 5, we check if all vertices can reach v (In reversed graph, if all vertices are reachable from v, then all vertices can reach v in original graph).
Explanation with some examples:
Example 1 :
Given a directed to check if it is strongly connected or not.
- step 1: Starting with vertex 2 BFS obtained is 2 3 4 0 1
- step 2: After reversing the given graph we got listed graph.
- step 3: Again after starting with vertex 2 the BFS is 2 1 4 0 3
- step 4: No vertex in both case (step 1 & step 3) remains unvisited.
- step 5: So, given graph is strongly connected.
Example 2 :
Given a directed to check if it is strongly connected or not.
- step 1: Starting with vertex 2 BFS obtained is 2 3 4
- step 2: After reversing the given graph we got listed graph.
- step 3: Again after starting with vertex 2 the BFS is 2 1 0
- step 4: vertex 0, 1 in original graph and 3, 4 in reverse graph remains unvisited.
- step 5: So, given graph is not strongly connected.
Following is the implementation of above algorithm.
C++
// C++ program to check if a given directed graph // is strongly connected or not with BFS use #include <bits/stdc++.h> using namespace std; class Graph { int V; // No. of vertices list< int > *adj; // An array of adjacency lists // A recursive function to print DFS starting from v void BFSUtil( int v, bool visited[]); public : // Constructor and Destructor Graph( int V) { this ->V = V; adj = new list< int >[V];} ~Graph() { delete [] adj; } // Method to add an edge void addEdge( int v, int w); // The main function that returns true if the // graph is strongly connected, otherwise false bool isSC(); // Function that returns reverse (or transpose) // of this graph Graph getTranspose(); }; // A recursive function to print DFS starting from v void Graph::BFSUtil( int v, bool visited[]) { // Create a queue for BFS list< int > queue; // Mark the current node as visited and enqueue it visited[v] = true ; queue.push_back(v); // 'i' will be used to get all adjacent vertices // of a vertex list< int >::iterator i; while (!queue.empty()) { // Dequeue a vertex from queue v = queue.front(); queue.pop_front(); // Get all adjacent vertices of the dequeued vertex s // If a adjacent has not been visited, then mark it // visited and enqueue it for (i = adj[v].begin(); i != adj[v].end(); ++i) { if (!visited[*i]) { visited[*i] = true ; queue.push_back(*i); } } } } // Function that returns reverse (or transpose) of this graph Graph Graph::getTranspose() { Graph g(V); for ( int v = 0; v < V; v++) { // Recur for all the vertices adjacent to this vertex list< int >::iterator i; for (i = adj[v].begin(); i != adj[v].end(); ++i) g.adj[*i].push_back(v); } return g; } void Graph::addEdge( int v, int w) { adj[v].push_back(w); // Add w to v’s list. } // The main function that returns true if graph // is strongly connected bool Graph::isSC() { // Step 1: Mark all the vertices as not // visited (For first BFS) bool visited[V]; for ( int i = 0; i < V; i++) visited[i] = false ; // Step 2: Do BFS traversal starting // from first vertex. BFSUtil(0, visited); // If BFS traversal doesn’t visit all // vertices, then return false. for ( int i = 0; i < V; i++) if (visited[i] == false ) return false ; // Step 3: Create a reversed graph Graph gr = getTranspose(); // Step 4: Mark all the vertices as not // visited (For second BFS) for ( int i = 0; i < V; i++) visited[i] = false ; // Step 5: Do BFS for reversed graph // starting from first vertex. // Starting Vertex must be same starting // point of first DFS gr.BFSUtil(0, visited); // If all vertices are not visited in // second DFS, then return false for ( int i = 0; i < V; i++) if (visited[i] == false ) return false ; return true ; } // Driver program to test above functions int main() { // Create graphs given in the above diagrams Graph g1(5); g1.addEdge(0, 1); g1.addEdge(1, 2); g1.addEdge(2, 3); g1.addEdge(3, 0); g1.addEdge(2, 4); g1.addEdge(4, 2); g1.isSC()? cout << "Yes\n" : cout << "No\n" ; Graph g2(4); g2.addEdge(0, 1); g2.addEdge(1, 2); g2.addEdge(2, 3); g2.isSC()? cout << "Yes\n" : cout << "No\n" ; return 0; } |
Java
// Java program to check if a given directed graph // is strongly connected or not with BFS use import java.util.*; class Graph { int V; // No. of vertices // An array of adjacency lists ArrayList<ArrayList<Integer> > adj; // Constructor and Destructor Graph( int V) { this .V = V; adj = new ArrayList<>(); for ( int i = 0 ; i < V; i++) { adj.add( new ArrayList<>()); } } // Method to add an edge void addEdge( int v, int w) { adj.get(v).add(w); // Add w to v’s list. } // A recursive function to print DFS starting from v void BFSUtil( int v, boolean visited[]) { // Create a queue for BFS Queue<Integer> queue = new ArrayDeque<>(); // Mark the current node as visited and enqueue it visited[v] = true ; queue.add(v); while (!queue.isEmpty()) { // Dequeue a vertex from queue v = queue.peek(); queue.remove(); // Get all adjacent vertices of the dequeued // vertex s If a adjacent has not been visited, // then mark it visited and enqueue it for (Integer i : adj.get(v)) { if (!visited[i]) { visited[i] = true ; queue.add(i); } } } } // The main function that returns true if the // graph is strongly connected, otherwise false boolean isSC() { // Step 1: Mark all the vertices as not // visited (For first BFS) boolean [] visited = new boolean [V]; for ( int i = 0 ; i < V; i++) visited[i] = false ; // Step 2: Do BFS traversal starting // from first vertex. BFSUtil( 0 , visited); // If BFS traversal doesn’t visit all // vertices, then return false. for ( int i = 0 ; i < V; i++) if (visited[i] == false ) return false ; // Step 3: Create a reversed graph Graph gr = getTranspose(); // Step 4: Mark all the vertices as not // visited (For second BFS) for ( int i = 0 ; i < V; i++) visited[i] = false ; // Step 5: Do BFS for reversed graph // starting from first vertex. // Starting Vertex must be same starting // point of first DFS gr.BFSUtil( 0 , visited); // If all vertices are not visited in // second DFS, then return false for ( int i = 0 ; i < V; i++) if (visited[i] == false ) return false ; return true ; } // Function that returns reverse (or transpose) // of this graph Graph getTranspose() { Graph g = new Graph(V); for ( int v = 0 ; v < V; v++) { // Recur for all the vertices adjacent to this // vertex for (Integer i : adj.get(v)) g.adj.get(i).add(v); } return g; } } public class GFG { // Driver program to test above functions public static void main(String[] args) { Graph g1 = new Graph( 5 ); g1.addEdge( 0 , 1 ); g1.addEdge( 1 , 2 ); g1.addEdge( 2 , 3 ); g1.addEdge( 3 , 0 ); g1.addEdge( 2 , 4 ); g1.addEdge( 4 , 2 ); if (g1.isSC()) System.out.println( "Yes" ); else System.out.println( "No" ); Graph g2 = new Graph( 4 ); g2.addEdge( 0 , 1 ); g2.addEdge( 1 , 2 ); g2.addEdge( 2 , 3 ); if (g2.isSC()) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by karandeep1234. |
Python3
# Python3 program to check if a given directed graph # is strongly connected or not with BFS use from collections import deque # A recursive function to print DFS starting from v def BFSUtil(adj, v, visited): # Create a queue for BFS queue = deque() # Mark the current node as visited # and enqueue it visited[v] = True queue.append(v) # 'i' will be used to get all adjacent # vertices of a vertex while ( len (queue) > 0 ): # Dequeue a vertex from queue v = queue.popleft() #print(v) #queue.pop_front() # Get all adjacent vertices of the # dequeued vertex s. If a adjacent # has not been visited, then mark it # visited and enqueue it for i in adj[v]: if (visited[i] = = False ): visited[i] = True queue.append(i) return visited # Function that returns reverse # (or transpose) of this graph def getTranspose(adj, V): g = [[] for i in range (V)] for v in range (V): # Recur for all the vertices adjacent to # this vertex # list<int>::iterator i for i in adj[v]: g[i].append(v) return g def addEdge(adj, v, w): # Add w to v’s list. adj[v].append(w) return adj # The main function that returns True if graph # is strongly connected def isSC(adj, V): # Step 1: Mark all the vertices as not # visited (For first BFS) visited = [ False ] * V # Step 2: Do BFS traversal starting # from first vertex. visited = BFSUtil(adj, 0 , visited) # print(visited) # If BFS traversal doesn’t visit all # vertices, then return false. for i in range (V): if (visited[i] = = False ): return False # Step 3: Create a reversed graph adj = getTranspose(adj, V) # Step 4: Mark all the vertices as not # visited (For second BFS) for i in range (V): visited[i] = False # Step 5: Do BFS for reversed graph # starting from first vertex. # Starting Vertex must be same starting # point of first DFS visited = BFSUtil(adj, 0 , visited) # If all vertices are not visited in # second DFS, then return false for i in range (V): if (visited[i] = = False ): return False return True # Driver code if __name__ = = '__main__' : # Create graphs given in the above diagrams g1 = [[] for i in range ( 5 )] g1 = addEdge(g1, 0 , 1 ) g1 = addEdge(g1, 1 , 2 ) g1 = addEdge(g1, 2 , 3 ) g1 = addEdge(g1, 3 , 0 ) g1 = addEdge(g1, 2 , 4 ) g1 = addEdge(g1, 4 , 2 ) #print(g1) print ( "Yes" if isSC(g1, 5 ) else "No" ) g2 = [[] for i in range ( 4 )] g2 = addEdge(g2, 0 , 1 ) g2 = addEdge(g2, 1 , 2 ) g2 = addEdge(g2, 2 , 3 ) print ( "Yes" if isSC(g2, 4 ) else "No" ) # This code is contributed by mohit kumar 29 |
C#
using System; using System.Collections.Generic; using System.Linq; class Graph { private int V; private List< int >[] adj; public Graph( int v) { V = v; adj = new List< int >[v]; for ( int i = 0; i < v; i++) { adj[i] = new List< int >(); } } public void addEdge( int v, int w) { adj[v].Add(w); } public bool isSC() { bool [] visited = new bool [V]; DFSUtil(0, visited); for ( int i = 0; i < V; i++) { if (visited[i] == false ) { return false ; } } Graph gt = getTranspose(); for ( int i = 0; i < V; i++) { visited[i] = false ; } gt.DFSUtil(0, visited); for ( int i = 0; i < V; i++) { if (visited[i] == false ) { return false ; } } return true ; } private void DFSUtil( int v, bool [] visited) { visited[v] = true ; for ( int i = 0; i < adj[v].Count; i++) { int n = adj[v][i]; if (!visited[n]) { DFSUtil(n, visited); } } } private Graph getTranspose() { Graph g = new Graph(V); for ( int v = 0; v < V; v++) { for ( int i = 0; i < adj[v].Count; i++) { g.adj[adj[v][i]].Add(v); } } return g; } public static void Main( string [] args) { Graph g1 = new Graph(5); g1.addEdge(0, 1); g1.addEdge(1, 2); g1.addEdge(2, 3); g1.addEdge(3, 0); g1.addEdge(2, 4); g1.addEdge(4, 2); if (g1.isSC()) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); Graph g2 = new Graph(4); g2.addEdge(0, 1); g2.addEdge(1, 2); g2.addEdge(2, 3); if (g2.isSC()) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } |
Javascript
<script> // javascript program to check if a given directed graph // is strongly connected or not with BFS use class Graph { constructor(V){ this .V = V; // No. of vertices this .adj = new Array(V); // An array of adjacency lists for (let i = 0; i < V; i++){ this .adj[i] = new Array(); } } // A recursive function to print DFS starting from v BFSUtil(v, visited) { // Create a queue for BFS let queue = []; // Mark the current node as visited and enqueue it visited[v] = true ; queue.push(v); while (queue.length > 0) { // Dequeue a vertex from queue v = queue[0]; queue.shift(); // Get all adjacent vertices of the dequeued vertex s // If a adjacent has not been visited, then mark it // visited and enqueue it for (let i = 0; i < this .adj[v].length; i++) { let x = this .adj[v][i]; if (!visited[x]) { visited[x] = true ; queue.push(x); } } } } // Function that returns reverse (or transpose) of this graph getTranspose() { let g = new Graph( this .V); for (let v = 0; v < this .V; v++) { // Recur for all the vertices adjacent to this vertex for (let i = 0; i < this .adj[v].length; i++){ let x = this .adj[v][i]; g.adj[x].push(v); } } return g; } addEdge(v, w) { this .adj[v].push(w); // Add w to v’s list. } // The main function that returns true if graph // is strongly connected isSC() { // Step 1: Mark all the vertices as not // visited (For first BFS) let visited = new Array( this .V); for (let i = 0; i < this .V; i++) visited[i] = false ; // Step 2: Do BFS traversal starting // from first vertex. this .BFSUtil(0, visited); // If BFS traversal doesn’t visit all // vertices, then return false. for (let i = 0; i < this .V; i++) if (visited[i] == false ) return false ; // Step 3: Create a reversed graph let gr = this .getTranspose(); // Step 4: Mark all the vertices as not // visited (For second BFS) for (let i = 0; i < this .V; i++) visited[i] = false ; // Step 5: Do BFS for reversed graph // starting from first vertex. // Starting Vertex must be same starting // point of first DFS gr.BFSUtil(0, visited); // If all vertices are not visited in // second DFS, then return false for (let i = 0; i < this .V; i++) if (visited[i] == false ) return false ; return true ; } }; // Driver program to test above functions // Create graphs given in the above diagrams let g1 = new Graph(5); g1.addEdge(0, 1); g1.addEdge(1, 2); g1.addEdge(2, 3); g1.addEdge(3, 0); g1.addEdge(2, 4); g1.addEdge(4, 2); if (g1.isSC()){ document.write( "Yes" ); } else { document.write( "No" ); } let g2 = new Graph(4); g2.addEdge(0, 1); g2.addEdge(1, 2); g2.addEdge(2, 3); if (g2.isSC()){ document.write( "Yes" ); } else { document.write( "No" ); } // the code is contributed by Nidhi goel. </script> |
Yes No
Time Complexity: Time complexity of above implementation is same as Breadth First Search which is O(V+E) if the graph is represented using adjacency matrix representation.
Can we improve further?
The above approach requires two traversals of graph. We can find whether a graph is strongly connected or not in one traversal using Tarjan’s Algorithm to find Strongly Connected Components.