Check if a Given Graph is Strongly Connected ( Kosaraju's BFS )🥷🏻

Introduction🎄

Hi Ninja🥷! In this article, we will learn to solve the problem of checking if a given graph is strongly connected ( Kosaraju's BFS ). A strongly connected component of a graph is a part in which there is a path from every vertex to another vertex. It is only valid in the case of directed graphs.

We hope you will enjoy and have fun learning a hard-level DSA problem on graph data structure "Check if a given graph is strongly connected ( Kosaraju's BFS )".

Problem Description🗒️

Given a directed graph, we have to find out whether the graph is strongly connected or not. A directed graph is strongly connected if there is a path from every vertex to another vertex.

There are many algorithms to solve this problem, but one of the most efficient ones is Kosaraju’s BFS-based algorithm.

Approach 🤖

Following are the steps involved in Kosaraju’s BFS-based algorithm.

Algorithm👽

• We do two BFS traversals of the given graph:
• First, we Initialize all vertices as not visited.
• Then we do a BFS traversal of a graph starting from any vertex, let's say vertex V.
• After BFS traversal, if we cannot visit all vertices, then return false.
• Second, We reverse all the edges present in the graph.
• We Initialize all vertices as not visited in the reversed graph.
• Then we repeat a BFS traversal of the reversed graph starting from the same vertex V.
• After BFS traversal, if we cannot visit all vertices, then return false.
• Otherwise, return true.

Sample Examples😄

Input:

Output:

Explanation:

We say a directed graph is strongly connected if there is a path from every vertex to another vertex.

Here, If we start from vertex 1, we get BFS as 1 2 3 4 5

After reversing the given graph, we get,

Again, starting with vertex 1, we get BFS as 1 0

Vertex 0 in the original graph and 2 3 4 5 in the reverse graph remain unvisited.

So, this graph is not strongly connected.
 

Input:

Output:

Given graph is strongly connected.

Explanation:

If we start from vertex 2, we get BFS as 2 4 3 5 0 6 1

After reversing the given graph, we get,

Again, starting with vertex 2, we get BFS as 2 4 1 6 0 5 3

No vertex in the given and reversed graph remains unvisited.

So, this graph is strongly connected.

C++ Code⌨️

// Program to check if the given directed graph
// is strongly connected or not.
#include <bits/stdc++.h>
using namespace std;


class Graph
{
	int vertices; // To store no. of vertices
	list<int> *adjList;


	// Function to print DFS starting from v
	void BFSUtil(int v, bool visited[]);
	public:


	Graph(int V) { this->vertices = V; adjList = new list<int>[V];}
	~Graph() { delete [] adjList; }


	// Function to add an edge
	void addEdge(int v, int w);


	// Function that returns if the
	// graph is strongly connected or not.
	bool isStronglyConnected();


	// 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[])
	{
	// Creating a queue for BFS traversal
	list<int> queue;
	visited[v] = true;
	queue.push_back(v);


	list<int>::iterator i;


		while (!queue.empty())
		{
			// Dequeue a vertex from queue
			v = queue.front();
			queue.pop_front();
			for (i = adjList[v].begin(); i != adjList[v].end(); ++i)
			{
				if (!visited[*i])
				{
					visited[*i] = true;
					queue.push_back(*i);
				}
			}
		}
	}


	Graph Graph::getTranspose()
	{
		Graph g(vertices);
		for (int v = 0; v < vertices; v++)
		{
			list<int>::iterator i;
			for (i = adjList[v].begin(); i != adjList[v].end(); ++i)
			g.adjList[*i].push_back(v);
		}
		return g;
	}


	void Graph::addEdge(int v, int w)
	{
		adjList[v].push_back(w); 
	}


	// Function that returns true if graph
	// is strongly connected
	bool Graph::isStronglyConnected()
	{
		// Step 1: Mark all the vertices as not
		// visited (For first BFS)
		bool visited[vertices];
		for (int i = 0; i < vertices; i++)
		visited[i] = false;


		// Do BFS traversal starting from the first vertex.
		BFSUtil(0, visited);


		// If BFS traversal doesn’t visit all vertices
		// return false.
		for (int i = 0; i < vertices; i++)
		if (visited[i] == false)
		return false;


		//Create a reversed graph
		Graph gr = getTranspose();


		// Step 4: Mark all the vertices as not
		// visited (For second BFS)
		for(int i = 0; i < vertices; i++)
		visited[i] = false;


		gr.BFSUtil(0, visited);


		// If all vertices are not visited in the second DFS
		//return false
		for (int i = 0; i < vertices; i++)
		if (visited[i] == false)
		return false;


		return true;
	}


	// Driver program to test the above functions
	int main()
	{
		Graph g1(7);
		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.addEdge(4, 5);
		g1.addEdge(5, 6);
		g1.addEdge(6, 4);

		g1.isStronglyConnected()?cout<<"Yes":cout<<"No";
		return 0;
	}

Output: 

Yes

Complexity Analysis

Time Complexity: O(V+E), where V and E are the numbers of vertices and edges, respectively. Time taken by this algorithm if the graph is represented using adjacency matrix representation is the same as BFS traversal.

Space Complexity: O(V), Where V is the number of vertices in the graph. We have to maintain a visited array and list of size V.