Strongly Connected Components

Strongly Connected Components (SCCs) are a fundamental concept in graph theory and algorithms. In a directed graph, a Strongly Connected Component is a subset of vertices where every vertex in the subset is reachable from every other vertex in the same subset by traversing the directed edges. Finding the SCCs of a graph can provide important insights into the structure and connectivity of the graph, with applications in various fields such as social network analysis, web crawling, and network routing. This tutorial will explore the definition, properties, and efficient algorithms for identifying Strongly Connected Components in graph data structures

What is Strongly Connected Components (SCCs)?

A strongly connected component of a directed graph is a maximal subgraph where every pair of vertices is mutually reachable. This means that for any two nodes A and B in this subgraph, there is a path from A to B and a path from B to A.

For example: The below graph has two strongly connected components {1,2,3,4} and {5,6,7} since there is path from each vertex to every other vertex in the same strongly connected component. 

Strongly Connected Component

Why Strongly Connected Components (SCCs) are Important?

Understanding SCCs is crucial for various applications such as:

• Network Analysis: Identifying clusters of tightly interconnected nodes.
• Optimizing Web Crawlers: Determining parts of the web graph that are closely linked.
• Dependency Resolution: In software, understanding which modules are interdependent.

Difference Between Connected and Strongly Connected Components (SCCs)

Connectivity in a undirected graph refers to whether two vertices are reachable from each other or not. Two vertices are said to be connected if there is path between them. Meanwhile Strongly Connected is applicable only to directed graphs. A subgraph of a directed graph is considered to be an Strongly Connected Components (SCC) if and only if for every pair of vertices A and B, there exists a path from A to B and a path from B to A. Let’s see why the standard dfs method to find connnected components in a graph cannot be used to determine strongly connected components.

Consider the following graph:

Now, let’s start a dfs call from vertex 1 to visit other vertices.

Why conventional DFS method cannot be used to find the strongly connected components?

All the vertices can be reached from vertex 1. But vertices 1 and 5,6,7 can not be in the same strongly connected component because there is no directed path from vertex 5,6 or 7 to vertex 1. The graph has two strongly connected components {1,2,3,4} and {5,6,7}. So the conventional dfs method cannot be used to find the strongly connected components.

Connecting Two Strongly Connected Component by a Unidirectional Edge

Two different connected components becomes a single component if a edge is added between a vertex from one component to a vertex of other component. But this is not the case in strongly connected components. Two strongly connected components doesn’t become a single strongly connected component if there is only a unidirectional edge from one SCC to other SCC.

Brute Force Approach for Finding Strongly Connected Components

The simple method will be for each vertex i (which is not a part of any strongly component) find the vertices which will be the part of strongly connected component containing vertex i. Two vertex i and j will be in the same strongly connected component if they there is a directed path from vertex i to vertex j and vice-versa.

Let’s understand the approach with the help of following example:

• Starting with vertex 1. There is path from vertex 1 to vertex 2 and vice-versa. Similarly there is a path from vertex 1 to vertex 3 and vice versa. So, vertex 2 and 3 will be in the same Strongly Connected Component as vertex 1. Although there is directed path form vertex 1 to vertex 4 and vertex 5. But there is no directed path from vertex 4,5 to vertex 1 so vertex 4 and 5 won’t be in the same Strongly Connected Component as vertex 1. Thus Vertex 1,2 and 3 forms a Strongly Connected Component.
• For Vertex 2 and 3, there Strongly Connected Component has already been determined.
• For Vertex 4, there is a path from vertex 4 to vertex 5 but there is no path from vertex 5 to vertex 4. So vertex 4 and 5 won’t be in the Same Strongly Connected Component. So both Vertex 4 and Vertex 5 will be part of Single Strongly Connected Component.
• Hence there will be 3 Strongly Connected Component {1,2,3}, {4} and {5}.

Below is the implementation of above approach:

C++

#include <bits/stdc++.h>
using namespace std;

class GFG {
public:
    // dfs Function to reach destination
    bool dfs(int curr, int des, vector<vector<int> >& adj,
             vector<int>& vis)
    {

        // If curr node is destination return true
        if (curr == des) {
            return true;
        }
        vis[curr] = 1;
        for (auto x : adj[curr]) {
            if (!vis[x]) {
                if (dfs(x, des, adj, vis)) {
                    return true;
                }
            }
        }
        return false;
    }

    // To tell whether there is path from source to
    // destination
    bool isPath(int src, int des, vector<vector<int> >& adj)
    {
        vector<int> vis(adj.size() + 1, 0);
        return dfs(src, des, adj, vis);
    }

    // Function to return all the strongly connected
    // component of a graph.
    vector<vector<int> > findSCC(int n,
                                 vector<vector<int> >& a)
    {
        // Stores all the strongly connected components.
        vector<vector<int> > ans;

        // Stores whether a vertex is a part of any Strongly
        // Connected Component
        vector<int> is_scc(n + 1, 0);

        vector<vector<int> > adj(n + 1);

        for (int i = 0; i < a.size(); i++) {
            adj[a[i][0]].push_back(a[i][1]);
        }

        // Traversing all the vertices
        for (int i = 1; i <= n; i++) {

            if (!is_scc[i]) {

                // If a vertex i is not a part of any SCC
                // insert it into a new SCC list and check
                // for other vertices whether they can be
                // thr part of thidl ist.
                vector<int> scc;
                scc.push_back(i);

                for (int j = i + 1; j <= n; j++) {

                    // If there is a path from vertex i to
                    // vertex j and vice versa put vertex j
                    // into the current SCC list.
                    if (!is_scc[j] && isPath(i, j, adj)
                        && isPath(j, i, adj)) {
                        is_scc[j] = 1;
                        scc.push_back(j);
                    }
                }

                // Insert the SCC containing vertex i into
                // the final list.
                ans.push_back(scc);
            }
        }
        return ans;
    }
};

// Driver Code Starts

int main()
{

    GFG obj;
    int V = 5;
    vector<vector<int> > edges{
        { 1, 3 }, { 1, 4 }, { 2, 1 }, { 3, 2 }, { 4, 5 }
    };
    vector<vector<int> > ans = obj.findSCC(V, edges);
    cout << "Strongly Connected Components are:\n";
    for (auto x : ans) {
        for (auto y : x) {
            cout << y << " ";
        }
        cout << "\n";
    }
}

Java

import java.util.ArrayList;
import java.util.List;

class GFG {

    // dfs Function to reach destination
    boolean dfs(int curr, int des, List<List<Integer>> adj,
                List<Integer> vis) {

        // If curr node is destination return true
        if (curr == des) {
            return true;
        }
        vis.set(curr, 1);
        for (int x : adj.get(curr)) {
            if (vis.get(x) == 0) {
                if (dfs(x, des, adj, vis)) {
                    return true;
                }
            }
        }
        return false;
    }

    // To tell whether there is path from source to
    // destination
    boolean isPath(int src, int des, List<List<Integer>> adj) {
        List<Integer> vis = new ArrayList<>(adj.size() + 1);
        for (int i = 0; i <= adj.size(); i++) {
            vis.add(0);
        }
        return dfs(src, des, adj, vis);
    }

    // Function to return all the strongly connected
    // component of a graph.
    List<List<Integer>> findSCC(int n, List<List<Integer>> a) {
        // Stores all the strongly connected components.
        List<List<Integer>> ans = new ArrayList<>();

        // Stores whether a vertex is a part of any Strongly
        // Connected Component
        List<Integer> is_scc = new ArrayList<>(n + 1);
        for (int i = 0; i <= n; i++) {
            is_scc.add(0);
        }

        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            adj.add(new ArrayList<>());
        }

        for (List<Integer> edge : a) {
            adj.get(edge.get(0)).add(edge.get(1));
        }

        // Traversing all the vertices
        for (int i = 1; i <= n; i++) {

            if (is_scc.get(i) == 0) {

                // If a vertex i is not a part of any SCC
                // insert it into a new SCC list and check
                // for other vertices whether they can be
                // the part of this list.
                List<Integer> scc = new ArrayList<>();
                scc.add(i);

                for (int j = i + 1; j <= n; j++) {

                    // If there is a path from vertex i to
                    // vertex j and vice versa, put vertex j
                    // into the current SCC list.
                    if (is_scc.get(j) == 0 && isPath(i, j, adj)
                            && isPath(j, i, adj)) {
                        is_scc.set(j, 1);
                        scc.add(j);
                    }
                }

                // Insert the SCC containing vertex i into
                // the final list.
                ans.add(scc);
            }
        }
        return ans;
    }
}

public class Main {

    public static void main(String[] args) {

        GFG obj = new GFG();
        int V = 5;
        List<List<Integer>> edges = new ArrayList<>();
        edges.add(new ArrayList<>(List.of(1, 3)));
        edges.add(new ArrayList<>(List.of(1, 4)));
        edges.add(new ArrayList<>(List.of(2, 1)));
        edges.add(new ArrayList<>(List.of(3, 2)));
        edges.add(new ArrayList<>(List.of(4, 5)));
        List<List<Integer>> ans = obj.findSCC(V, edges);
        System.out.println("Strongly Connected Components are:");
        for (List<Integer> x : ans) {
            for (int y : x) {
                System.out.print(y + " ");
            }
            System.out.println();
        }
    }
}

// This code is contributed by shivamgupta310570

Python

class GFG:
    # dfs Function to reach destination
    def dfs(self, curr, des, adj, vis):
        # If current node is the destination, return True
        if curr == des:
            return True
        vis[curr] = 1
        for x in adj[curr]:
            if not vis[x]:
                if self.dfs(x, des, adj, vis):
                    return True
        return False
    
    # To tell whether there is a path from source to destination
    def isPath(self, src, des, adj):
        vis = [0] * (len(adj) + 1)
        return self.dfs(src, des, adj, vis)
    
    # Function to return all the strongly connected components of a graph.
    def findSCC(self, n, a):
        # Stores all the strongly connected components.
        ans = []
        
        # Stores whether a vertex is a part of any Strongly Connected Component
        is_scc = [0] * (n + 1)
        
        adj = [[] for _ in range(n + 1)]
        
        for i in range(len(a)):
            adj[a[i][0]].append(a[i][1])
        
        # Traversing all the vertices
        for i in range(1, n + 1):
            if not is_scc[i]:
                # If a vertex i is not a part of any SCC, insert it into a new SCC list
                # and check for other vertices whether they can be part of this list.
                scc = [i]
                for j in range(i + 1, n + 1):
                    # If there is a path from vertex i to vertex j and vice versa,
                    # put vertex j into the current SCC list.
                    if not is_scc[j] and self.isPath(i, j, adj) and self.isPath(j, i, adj):
                        is_scc[j] = 1
                        scc.append(j)
                # Insert the SCC containing vertex i into the final list.
                ans.append(scc)
        return ans

# Driver Code Starts
if __name__ == "__main__":
    obj = GFG()
    V = 5
    edges = [
        [1, 3], [1, 4], [2, 1], [3, 2], [4, 5]
    ]
    ans = obj.findSCC(V, edges)
    print("Strongly Connected Components are:")
    for x in ans:
        for y in x:
            print(y, end=" ")
        print()
        
# This code is contributed by shivamgupta310570

C#

using System;
using System.Collections.Generic;

class GFG
{
    // dfs Function to reach destination
    public bool Dfs(int curr, int des, List<List<int>> adj, List<int> vis)
    {
        // If curr node is the destination, return true
        if (curr == des)
        {
            return true;
        }
        vis[curr] = 1;
        foreach (var x in adj[curr])
        {
            if (vis[x] == 0)
            {
                if (Dfs(x, des, adj, vis))
                {
                    return true;
                }
            }
        }
        return false;
    }

    // To tell whether there is a path from source to destination
    public bool IsPath(int src, int des, List<List<int>> adj)
    {
        var vis = new List<int>(adj.Count + 1);
        for (int i = 0; i < adj.Count + 1; i++)
        {
            vis.Add(0);
        }
        return Dfs(src, des, adj, vis);
    }

    // Function to return all the strongly connected components of a graph
    public List<List<int>> FindSCC(int n, List<List<int>> a)
    {
        // Stores all the strongly connected components
        var ans = new List<List<int>>();

        // Stores whether a vertex is a part of any Strongly Connected Component
        var isScc = new List<int>(n + 1);
        for (int i = 0; i < n + 1; i++)
        {
            isScc.Add(0);
        }

        var adj = new List<List<int>>(n + 1);
        for (int i = 0; i < n + 1; i++)
        {
            adj.Add(new List<int>());
        }

        for (int i = 0; i < a.Count; i++)
        {
            adj[a[i][0]].Add(a[i][1]);
        }

        // Traversing all the vertices
        for (int i = 1; i <= n; i++)
        {
            if (isScc[i] == 0)
            {
                // If a vertex i is not a part of any SCC
                // insert it into a new SCC list and check
                // for other vertices whether they can be
                // the part of this list.
                var scc = new List<int>();
                scc.Add(i);

                for (int j = i + 1; j <= n; j++)
                {
                    // If there is a path from vertex i to
                    // vertex j and vice versa, put vertex j
                    // into the current SCC list.
                    if (isScc[j] == 0 && IsPath(i, j, adj) && IsPath(j, i, adj))
                    {
                        isScc[j] = 1;
                        scc.Add(j);
                    }
                }

                // Insert the SCC containing vertex i into
                // the final list.
                ans.Add(scc);
            }
        }
        return ans;
    }
}

// Driver Code Starts
class Program
{
    static void Main(string[] args)
    {
        GFG obj = new GFG();
        int V = 5;
        List<List<int>> edges = new List<List<int>>
        {
            new List<int> { 1, 3 }, new List<int> { 1, 4 }, new List<int> { 2, 1 },
            new List<int> { 3, 2 }, new List<int> { 4, 5 }
        };
        List<List<int>> ans = obj.FindSCC(V, edges);
        Console.WriteLine("Strongly Connected Components are:");
        foreach (var x in ans)
        {
            foreach (var y in x)
            {
                Console.Write(y + " ");
            }
            Console.WriteLine();
        }
    }
}


// This code is contributed by shivamgupta310570

JavaScript

class GFG {
    // Function to reach the destination using DFS
    dfs(curr, des, adj, vis) {
        // If the current node is the destination, return true
        if (curr === des) {
            return true;
        }
        vis[curr] = 1;
        for (let x of adj[curr]) {
            if (!vis[x]) {
                if (this.dfs(x, des, adj, vis)) {
                    return true;
                }
            }
        }
        return false;
    }

    // Check whether there is a path from source to destination
    isPath(src, des, adj) {
        const vis = new Array(adj.length + 1).fill(0);
        return this.dfs(src, des, adj, vis);
    }

    // Function to find all strongly connected components of a graph
    findSCC(n, a) {
        // Stores all strongly connected components
        const ans = [];

        // Stores whether a vertex is part of any Strongly Connected Component
        const is_scc = new Array(n + 1).fill(0);
        const adj = new Array(n + 1).fill().map(() => []);

        for (let i = 0; i < a.length; i++) {
            adj[a[i][0]].push(a[i][1]);
        }

        // Traversing all the vertices
        for (let i = 1; i <= n; i++) {
            if (!is_scc[i]) {
                // If a vertex i is not part of any SCC,
                // insert it into a new SCC list and check
                // for other vertices that can be part of this list.
                const scc = [i];
                for (let j = i + 1; j <= n; j++) {
                    // If there is a path from vertex i to
                    // vertex j and vice versa, put vertex j
                    // into the current SCC list.
                    if (!is_scc[j] && this.isPath(i, j, adj) && this.isPath(j, i, adj)) {
                        is_scc[j] = 1;
                        scc.push(j);
                    }
                }
                // Insert the SCC containing vertex i into the final list.
                ans.push(scc);
            }
        }
        return ans;
    }
}

// Driver Code Starts
const obj = new GFG();
const V = 5;
const edges = [
    [1, 3], [1, 4], [2, 1], [3, 2], [4, 5]
];
const ans = obj.findSCC(V, edges);
console.log("Strongly Connected Components are:");
for (let x of ans) {
    console.log(x.join(" "));
}

// This code is contributed by shivamgupta310570

Output

Strongly Connected Components are:
1 2 3 
4 
5 

Time complexity: O(n * (n + m)), because for each pair of vertices we are checking whether a path exists between them.
Auxiliary Space: O(N)

Efficient Approach for Finding Strongly Connected Components (SCCs)

To find SCCs in a graph, we can use algorithms like Kosaraju’s Algorithm or Tarjan’s Algorithm. Let’s explore these algorithms step-by-step.

1. Kosaraju’s Algorithm:

Kosaraju’s Algorithm involves two main phases:

1. Performing Depth-First Search (DFS) on the Original Graph:
   • We first do a DFS on the original graph and record the finish times of nodes (i.e., the time at which the DFS finishes exploring a node completely).
2. Performing DFS on the Transposed Graph:
   • We then reverse the direction of all edges in the graph to create the transposed graph.
   • Next, we perform a DFS on the transposed graph, considering nodes in decreasing order of their finish times recorded in the first phase.
   • Each DFS traversal in this phase will give us one SCC.

Here’s a simplified version of Kosaraju’s Algorithm:

1. DFS on Original Graph: Record finish times.
2. Transpose the Graph: Reverse all edges.
3. DFS on Transposed Graph: Process nodes in order of decreasing finish times to find SCCs.

2. Tarjan’s Algorithm:

Tarjan’s Algorithm is more efficient because it finds SCCs in a single DFS pass using a stack and some additional bookkeeping:

1. DFS Traversal: During the DFS, maintain an index for each node and the smallest index (low-link value) that can be reached from the node.
2. Stack: Keep track of nodes currently in the recursion stack (part of the current SCC being explored).
3. Identifying SCCs: When a node’s low-link value equals its index, it means we have found an SCC. Pop all nodes from the stack until we reach the current node.

Here’s a simplified outline of Tarjan’s Algorithm:

1. Initialize index to 0.
2. For each unvisited node, perform DFS.
   • Set the node’s index and low-link value.
   • Push the node onto the stack.
   • For each adjacent node, either perform DFS if it’s not visited or update the low-link value if it’s in the stack.
   • If the node’s low-link value equals its index, pop nodes from the stack to form an SCC.

Conclusion

Understanding and finding strongly connected components in a directed graph is essential for many applications in computer science. Kosaraju’s and Tarjan’s algorithms are efficient ways to identify SCCs, each with their own approach and advantages. By mastering these concepts, you can better analyze and optimize the structure and behavior of complex networks.