Check if a graph is strongly connected | Set 1 (Kosaraju using DFS) - GeeksforGeeks

18 Jan, 2022
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 pair of vertices. For example, following is a strongly connected graph. 
It is easy for undirected graph, we can just do a BFS and DFS starting from any vertex. If BFS or DFS visits all vertices, then the given undirected graph is connected. This approach won’t work for a directed graph. For example, consider the following graph which is not strongly connected. If we start DFS (or BFS) from vertex 0, we can reach all vertices, but if we start from any other vertex, we cannot reach all vertices. 
How to do for directed graph? 
A simple idea is to use a all pair shortest path algorithm like Floyd Warshall or find Transitive Closure of graph. Time complexity of this method would be O(v3).We can also do DFS V times starting from every vertex. If any DFS, doesn’t visit all vertices, then graph is not strongly connected. This algorithm takes O(V*(V+E)) time which can be same as transitive closure for a dense graph.A better idea can be Strongly Connected Components (SCC) algorithm. We can find all SCCs in O(V+E) time. If number of SCCs is one, then graph is strongly connected. The algorithm for SCC does extra work as it finds all SCCs. Following is Kosaraju’s DFS based simple algorithm that does two DFS traversals of graph: 1) Initialize all vertices as not visited.2) Do a DFS traversal of graph starting from any arbitrary vertex v. If DFS traversal doesn’t visit all vertices, then return false.3) Reverse all arcs (or find transpose or reverse of graph) 4) Mark all vertices as not-visited in reversed graph.5) Do a DFS traversal of reversed graph starting from same vertex v (Same as step 2). If DFS traversal doesn’t visit all vertices, then return false. Otherwise return true.The idea is, if every node can be reached from a vertex v, and every node can reach v, then the graph is strongly connected. In step 2, we check if all vertices are reachable from v. In step 4, 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). Following is the implementation of above algorithm. 
C++
Java
Python3
Javascript
// C++ program to check if a given directed graph is strongly// connected or not#include <iostream>#include <list>#include <stack>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 DFSUtil(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 vvoid Graph::DFSUtil(int v, bool visited[]){    // Mark the current node as visited and print it    visited[v] = true;     // Recur for all the vertices adjacent to this vertex    list<int>::iterator i;    for (i = adj[v].begin(); i != adj[v].end(); ++i)        if (!visited[*i])            DFSUtil(*i, visited);} // Function that returns reverse (or transpose) of this graphGraph 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 connectedbool Graph::isSC(){    // St1p 1: Mark all the vertices as not visited (For first DFS)    bool visited[V];    for (int i = 0; i < V; i++)        visited[i] = false;     // Step 2: Do DFS traversal starting from first vertex.    DFSUtil(0, visited);      // If DFS 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 DFS)    for(int i = 0; i < V; i++)        visited[i] = false;     // Step 5: Do DFS for reversed graph starting from first vertex.    // Starting Vertex must be same starting point of first DFS    gr.DFSUtil(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 functionsint 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 program to check if a given directed graph is strongly// connected or notimport java.io.*;import java.util.*;import java.util.LinkedList; // This class represents a directed graph using adjacency// list representationclass Graph{    private int V;   // No. of vertices    private LinkedList<Integer> adj[]; //Adjacency List     //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); }     // A recursive function to print DFS starting from v    void DFSUtil(int v,Boolean visited[])    {        // Mark the current node as visited and print it        visited[v] = true;         int n;         // Recur for all the vertices adjacent to this vertex        Iterator<Integer> i = adj[v].iterator();        while (i.hasNext())        {            n = i.next();            if (!visited[n])                DFSUtil(n,visited);        }    }     // Function that returns 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            Iterator<Integer> i = adj[v].listIterator();            while (i.hasNext())                g.adj[i.next()].add(v);        }        return g;    }     // The main function that returns true if graph is strongly    // connected    Boolean isSC()    {        // Step 1: Mark all the vertices as not visited        // (For first DFS)        Boolean visited[] = new Boolean[V];        for (int i = 0; i < V; i++)            visited[i] = false;         // Step 2: Do DFS traversal starting from first vertex.        DFSUtil(0, visited);         // If DFS 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 DFS)        for (int i = 0; i < V; i++)            visited[i] = false;         // Step 5: Do DFS for reversed graph starting from        // first vertex. Starting Vertex must be same starting        // point of first DFS        gr.DFSUtil(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;    }     public static void main(String args[])    {        // Create graphs given in the above diagrams        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 Aakash Hasija
# Python program to check if a given directed graph is strongly# connected or not from collections import defaultdict # This class represents a directed graph using adjacency list representation  class Graph:     def __init__(self, vertices):        self.V = vertices  # No. of vertices        self.graph = defaultdict(list)  # default dictionary to store graph     # function to add an edge to graph    def addEdge(self, u, v):        self.graph[u].append(v)      # A function used by isSC() to perform DFS    def DFSUtil(self, v, visited):         # Mark the current node as visited        visited[v] = True         # Recur for all the vertices adjacent to this vertex        for i in self.graph[v]:            if visited[i] == False:                self.DFSUtil(i, visited)     # Function that returns reverse (or transpose) of this graph     def getTranspose(self):         g = Graph(self.V)         # Recur for all the vertices adjacent to this vertex        for i in self.graph:            for j in self.graph[i]:                g.addEdge(j, i)         return g     # The main function that returns true if graph is strongly connected    def isSC(self):          # Step 1: Mark all the vertices as not visited (For first DFS)        visited =[False]*(self.V)                 # Step 2: Do DFS traversal starting from first vertex.        self.DFSUtil(0,visited)         # If DFS traversal doesnt visit all vertices, then return false        if any(i == False for i in visited):            return False         # Step 3: Create a reversed graph        gr = self.getTranspose()                 # Step 4: Mark all the vertices as not visited (For second DFS)        visited =[False]*(self.V)         # Step 5: Do DFS for reversed graph starting from first vertex.        # Starting Vertex must be same starting point of first DFS        gr.DFSUtil(0,visited)         # If all vertices are not visited in second DFS, then        # return false        if any(i == False for i in visited):            return False         return True # Create a graph given in the above diagramg1 = 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)print ("Yes" if g1.isSC() else "No") g2 = Graph(4)g2.addEdge(0, 1)g2.addEdge(1, 2)g2.addEdge(2, 3)print ("Yes" if g2.isSC() else "No") # This code is contributed by Neelam Yadav
<script>// Javascript program to check if a given directed graph is strongly// connected or not // This class represents a directed graph using adjacency// list representationclass Graph{    // Constructor    constructor(v)    {        this.V = v;        this.adj = new Array(v);        for (let i = 0; i < v; ++i)            this.adj[i] = [];    }         // Function to add an edge into the graph    addEdge(v,w)    {        this.adj[v].push(w);    }         // A recursive function to print DFS starting from v    DFSUtil(v,visited)    {        // Mark the current node as visited and print it        visited[v] = true;          let n;          // Recur for all the vertices adjacent to this vertex                for(let i of this.adj[v].values())        {            n = i;            if (!visited[n])                this.DFSUtil(n,visited);        }    }         // Function that returns 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 of this.adj[v].values())                g.adj[i].push(v);        }        return g;    }         // The main function that returns true if graph is strongly    // connected    isSC()    {             // Step 1: Mark all the vertices as not visited        // (For first DFS)        let visited = new Array(this.V);        for (let i = 0; i < this.V; i++)            visited[i] = false;          // Step 2: Do DFS traversal starting from first vertex.        this.DFSUtil(0, visited);          // If DFS 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 DFS)        for (let i = 0; i < this.V; i++)            visited[i] = false;          // Step 5: Do DFS for reversed graph starting from        // first vertex. Starting Vertex must be same starting        // point of first DFS        gr.DFSUtil(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;    }} // Create graphs given in the above diagramslet 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<br>");else    document.write("No<br>"); 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");  // This code is contributed by avanitrachhadiya2155</script>
Output:
Yes
No
Time Complexity: Time complexity of above implementation is same as Depth First Search which is O(V+E) if the graph is represented using adjacency list 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.Exercise: Can we use BFS instead of DFS in above algorithm? See this.References: http://www.ieor.berkeley.edu/~hochbaum/files/ieor266-2012.pdfPlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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