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 Following is definite Kosaraju’s calculation.
How accomplishes this work?
So how would we discover this grouping of picking vertices as beginning stages of DFS?
Last Updated: Mar 27, 2024

Strongly Connected Components in Directed Graphs

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We can discover all emphatically associated segments in O(V+E) time utilising Kosaraju‘s calculation. Let us take a look at the article to understand the directed graphs with strongly connected components.

 Following is definite Kosaraju’s calculation.

  • Create an unfilled stack ‘S’ and do DFS crossing of a diagram. In DFS crossing, subsequent to calling recursive DFS for nearby vertices of a vertex, push the vertex to stack. In the above chart, in the event that we start DFS from vertex 0, we get vertices in a stack as 1, 2, 4, 3, 0.
  • Reverse bearings, everything being equal, to acquire the rendered diagram.
  • One by one pop a vertex from S while S isn’t vacant. Leave the popped vertex alone ‘v’. Accept v as source and do DFS (call DFSUtil (v)). The DFS beginning from v prints unequivocally associated part of v.

In the above model, we measure vertices all together 0, 3, 4, 2, 1 (one by one flew from stack).

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How accomplishes this work?

The above calculation is DFS-based. It does DFS multiple times. DFS of a diagram creates a solitary tree if all vertices are reachable from the DFS beginning stage. In any case, DFS produces a woodland. So DFS of a chart with just a single SCC consistently delivers a tree. The significant highlight note is DFS may create a tree or a woodland when there are more than one SCCs relying on the picked beginning stage.

For instance, in the above outline, on the off chance that we start DFS from vertices 0 or 1 or 2, we get a tree as yield. Furthermore, on the off chance that we start from 3 or 4, we get a woodland. To discover and print all SCCs, we would need to begin DFS from vertex 4 (which is a sink vertex), at that point move to 3 which is a sink in the staying (set barring 4) lastly any of the rest of the vertices (0, 1, 2).

So how would we discover this grouping of picking vertices as beginning stages of DFS?

 Shockingly, there is no immediate path for getting this succession. In any case, on the off chance that we do a DFS of chart and store vertices as per their completion times, we ensure that the completion season of a vertex that interfaces with different SCCs (other that its own SCC), will consistently be more noteworthy than finish season of vertices in the other SCC (See this for evidence).

For instance, in DFS of the above model chart, finish season of 0 is consistently more prominent than 3 and 4 (independent of the arrangement of vertices considered for DFS). What’s more, finish season of 3 is consistently more noteworthy than 4. DFS doesn’t ensure about different vertices, for instance, finish seasons of 1 and 2 might be more modest or more prominent than 3 and 4 relying on the succession of vertices considered for DFS. So to utilize this property, we do DFS crossing of the complete chart and push each completed vertex to a stack. In stack, 3 consistently shows up after 4, and 0 show up after both 3 and 4.

In the subsequent stage, we invert the chart. Think about the chart of SCCs. In the switched chart, the edges that associate two segments are turned around. So the SCC {0, 1, 2} will become sink and the SCC {4} will become a source. As examined above, in the stack, we generally have 0 preceding 3 and 4. So on the off chance that we do a DFS of the turned around chart utilising grouping of vertices in stack, we measure vertices from sink to source (in switched diagram). That is the thing that we needed to accomplish and that is completely expected to print SCCs individually.

// Kosaraju’s algorithm to find strongly connected components of graph in C++

#include <iostream>

#include <list>

#include <stack>

using namespace std;

class Graph {
int V;
list *adj;
void fillorder(int s , bool visited[], stack &stack);
void DFS(int s, bool visitedV[]);

Graph(int V);
void addEdge(int s, int d);
void printSCC();
Graph transpose();

Graph::Graph(int V) {
this->V = V;
adj = new list[V];

// DFS
void Graph::DFS(int s, bool visitedV[]) {
visitedV[s] = true;
cout << s << ” “;

list::iterator i;
for (i = adj[s].begin(); i != adj[s].end(); ++i)
if (!visitedV[i]) DFS(i, visitedV);

// Doing Transpose
Graph Graph::transpose() {
Graph g(V);
for (int s = 0; s < V; s++) { list::iterator i;
for (i = adj[s].begin(); i != adj[s].end(); ++i) {
return g;

// Adding edge into the graph
void Graph::addEdge(int s, int d) {

void graph :: fillorder(int s, bool visited[], stack&stack){
visitedV[s] = true;

list::iterator i;
for (i = adj[s].begin(); i != adj[s].end(); ++i)
if (!visitedV[i]) fillOrder(i, visitedV, Stack);


// Print strongly connected component of graph
void Graph::printSCC() {
stack Stack;

bool *visitedV = new bool[V];
for (int i = 0; i < V; i++)
visitedV[i] = false;

for (int i = 0; i < V; i++)
if (visitedV[i] == false)
fillOrder(i, visitedV, Stack);

Graph gr = transpose();

for (int i = 0; i < V; i++)
visitedV[i] = false;

while (Stack.empty() == false) {
int s =;

if (visitedV[s] == false) {
  gr.DFS(s, visitedV);
  cout << endl;


int main() {
Graph g(8);
g.addEdge(0, 1);
g.addEdge(1, 2);
g.addEdge(2, 3);
g.addEdge(2, 4);
g.addEdge(3, 0);
g.addEdge(4, 5);
g.addEdge(5, 6);
g.addEdge(6, 4);
g.addEdge(6, 7);

cout << “Strongly Connected Components of graph are:\n”;

Time Complexity: The above calculation calls DFS, discovers converse of the diagram and again calls DFS. DFS takes O(V+E) for a chart spoke to utilising nearness list. Turning around a chart likewise takes O(V+E) time. For turning around the diagram, we straightforward navigate all contiguousness records.

The above calculation is an asymptotically best calculation, however, there are different calculations like Tarjan’s calculation and way based which have same time multifaceted nature yet discover SCCs utilising single DFS. The Tarjan’s calculation is talked about in the accompanying post.


SCC calculations can be utilised as an initial phase in many chart calculations that work just on an unequivocally associated diagram. In informal communities, a gathering of individuals is commonly emphatically associated (For instance, understudies of a class or some other regular spot).

Numerous individuals in these gatherings by and large like some basic pages or play regular games. The SCC calculations can be utilised to discover such gatherings and propose the ordinarily enjoyed pages or games to the individuals in the gathering who have not yet loved normally loved a page or played a game.

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