Tarjan’s Algorithm to find Strongly Connected Components

23 Jun, 2022
A directed graph is strongly connected if there is a path between all pairs of vertices. A strongly connected component (SCC) of a directed graph is a maximal strongly connected subgraph. For example, there are 3 SCCs in the following graph.
We have discussed Kosaraju’s algorithm for strongly connected components. The previously discussed algorithm requires two DFS traversals of a Graph. In this post, Tarjan’s algorithm is discussed that requires only one DFS traversal.
Tarjan Algorithm is based on the following facts: 
DFS search produces a DFS tree/forest Strongly Connected Components form subtrees of the DFS tree. If we can find the head of such subtrees, we can print/store all the nodes in that subtree (including the head) and that will be one SCC. There is no back edge from one SCC to another (There can be cross edges, but cross edges will not be used while processing the graph).
DFS search produces a DFS tree/forest 
Strongly Connected Components form subtrees of the DFS tree. 
If we can find the head of such subtrees, we can print/store all the nodes in that subtree (including the head) and that will be one SCC. 
There is no back edge from one SCC to another (There can be cross edges, but cross edges will not be used while processing the graph).
To find the head of an SCC, we calculate disc and low array (as done for articulation point, bridge, and biconnected component). As discussed in the previous posts, low[u] indicates the earliest visited vertex (the vertex with minimum discovery time) that can be reached from a subtree rooted with u. A node u is head if disc[u] = low[u].
Below image is an illustration of the approach:  
Strongly Connected Component relates to directed graph only, but Disc and Low values relate to both directed and undirected graph, so in the above pic we have taken an undirected graph.
In the above Figure, we have shown a graph and one of the DFS tree (There could be different DFS trees on the same graph depending on the order in which edges are traversed). 
In a DFS tree, continuous arrows are tree edges, and dashed arrows are back edges (DFS Tree Edges )
Disc and Low values are shown in the Figure for every node as (Disc/Low). 
Disc: This is the time when a node is visited 1st time while DFS traversal. For nodes A, B, C, .., and J in the DFS tree, Disc values are 1, 2, 3, .., 10. 
Low: In the DFS tree, Tree edges take us forward, from the ancestor node to one of its descendants. For example, from node C, tree edges can take us to node G, node I, etc. Back edges take us backward, from a descendant node to one of its ancestors. 
For example, from node G, Back edges take us to E or C. If we look at both Tree and Back edge together, then we can see that if we start traversal from one node, we may go down the tree via Tree edges and then go up via back edges. For example, from node E, we can go down to G and then go up to C. Similarly from E, we can go down to I or J and then go up to F. “Low” value of a node tells the topmost reachable ancestor (with minimum possible Disc value) via the subtree of that node. So for any node, a Low value is equal to its Disc value anyway (A node is the ancestor of itself). Then we look into its subtree and see if there is any node that can take us to any of its ancestors. If there are multiple back edges in the subtree that take us to different ancestors, then we take the one with the minimum Disc value (i.e. the topmost one). If we look at node F, it has two subtrees. Subtree with node G takes us to E and C. The other subtree takes us back to F only. Here topmost ancestor is C where F can reach and so the Low value of F is 3 (The Disc value of C). 
Based on the above discussion, it should be clear that the Low values of B, C, and D are 1 (As A is the topmost node where B, C, and D can reach). In the same way, Low values of E, F, and G are 3, and Low values of H, I, and J are 6.For any node u, when DFS starts, Low will be set to its Disc 1st. 
Then later on DFS will be performed on each of its children v one by one, Low value of u can change it in two cases: 
Case1 (Tree Edge): If node v is not visited already, then after the DFS of v is complete, a minimum of low[u] and low[v] will be updated to low[u]. low[u] = min(low[u], low[v]); 
Case 2 (Back Edge): When child v is already visited, then a minimum of low[u] and Disc[v] will be updated to low[u]. low[u] = min(low[u], disc[v]); 
In case two, can we take low[v] instead of the disc[v] ?? . Answer is NO. If you can think why the answer is NO, you probably understood the Low and Disc concept.
Same Low and Disc values help to solve other graph problems like articulation point, bridge, and biconnected component.
To track the subtree rooted at the head, we can use a stack (keep pushing node while visiting). When a head node is found, pop all nodes from the stack till you get head out of stack.
To make sure, we don’t consider cross edges, when we reach a node that is already visited, we should process the visited node only if it is present in the stack, else ignore the node.  
Following is the implementation of Tarjan’s algorithm to print all SCCs. 
C++
Java
Python3
Javascript
// A C++ program to find strongly connected components in a given// directed graph using Tarjan's algorithm (single DFS)#include<iostream>#include <list>#include <stack>#define NIL -1using namespace std; // A class that represents an directed graphclass Graph{    int V; // No. of vertices    list<int> *adj; // A dynamic array of adjacency lists     // A Recursive DFS based function used by SCC()    void SCCUtil(int u, int disc[], int low[],                stack<int> *st, bool stackMember[]);public:    Graph(int V); // Constructor    void addEdge(int v, int w); // function to add an edge to graph    void SCC(); // prints strongly connected components}; Graph::Graph(int V){    this->V = V;    adj = new list<int>[V];} void Graph::addEdge(int v, int w){    adj[v].push_back(w);} // A recursive function that finds and prints strongly connected// components using DFS traversal// u --> The vertex to be visited next// disc[] --> Stores discovery times of visited vertices// low[] -- >> earliest visited vertex (the vertex with minimum//             discovery time) that can be reached from subtree//             rooted with current vertex// *st -- >> To store all the connected ancestors (could be part//         of SCC)// stackMember[] --> bit/index array for faster check whether//                 a node is in stackvoid Graph::SCCUtil(int u, int disc[], int low[], stack<int> *st,                    bool stackMember[]){    // A static variable is used for simplicity, we can avoid use    // of static variable by passing a pointer.    static int time = 0;     // Initialize discovery time and low value    disc[u] = low[u] = ++time;    st->push(u);    stackMember[u] = true;     // Go through all vertices adjacent to this    list<int>::iterator i;    for (i = adj[u].begin(); i != adj[u].end(); ++i)    {        int v = *i; // v is current adjacent of 'u'         // If v is not visited yet, then recur for it        if (disc[v] == -1)        {            SCCUtil(v, disc, low, st, stackMember);             // Check if the subtree rooted with 'v' has a            // connection to one of the ancestors of 'u'            // Case 1 (per above discussion on Disc and Low value)            low[u] = min(low[u], low[v]);        }         // Update low value of 'u' only of 'v' is still in stack        // (i.e. it's a back edge, not cross edge).        // Case 2 (per above discussion on Disc and Low value)        else if (stackMember[v] == true)            low[u] = min(low[u], disc[v]);    }     // head node found, pop the stack and print an SCC    int w = 0; // To store stack extracted vertices    if (low[u] == disc[u])    {        while (st->top() != u)        {            w = (int) st->top();            cout << w << " ";            stackMember[w] = false;            st->pop();        }        w = (int) st->top();        cout << w << "\n";        stackMember[w] = false;        st->pop();    }} // The function to do DFS traversal. It uses SCCUtil()void Graph::SCC(){    int *disc = new int[V];    int *low = new int[V];    bool *stackMember = new bool[V];    stack<int> *st = new stack<int>();     // Initialize disc and low, and stackMember arrays    for (int i = 0; i < V; i++)    {        disc[i] = NIL;        low[i] = NIL;        stackMember[i] = false;    }     // Call the recursive helper function to find strongly    // connected components in DFS tree with vertex 'i'    for (int i = 0; i < V; i++)        if (disc[i] == NIL)            SCCUtil(i, disc, low, st, stackMember);} // Driver program to test above functionint main(){    cout << "\nSCCs in first graph \n";    Graph g1(5);    g1.addEdge(1, 0);    g1.addEdge(0, 2);    g1.addEdge(2, 1);    g1.addEdge(0, 3);    g1.addEdge(3, 4);    g1.SCC();     cout << "\nSCCs in second graph \n";    Graph g2(4);    g2.addEdge(0, 1);    g2.addEdge(1, 2);    g2.addEdge(2, 3);    g2.SCC();     cout << "\nSCCs in third graph \n";    Graph g3(7);    g3.addEdge(0, 1);    g3.addEdge(1, 2);    g3.addEdge(2, 0);    g3.addEdge(1, 3);    g3.addEdge(1, 4);    g3.addEdge(1, 6);    g3.addEdge(3, 5);    g3.addEdge(4, 5);    g3.SCC();     cout << "\nSCCs in fourth graph \n";    Graph g4(11);    g4.addEdge(0,1);g4.addEdge(0,3);    g4.addEdge(1,2);g4.addEdge(1,4);    g4.addEdge(2,0);g4.addEdge(2,6);    g4.addEdge(3,2);    g4.addEdge(4,5);g4.addEdge(4,6);    g4.addEdge(5,6);g4.addEdge(5,7);g4.addEdge(5,8);g4.addEdge(5,9);    g4.addEdge(6,4);    g4.addEdge(7,9);    g4.addEdge(8,9);    g4.addEdge(9,8);    g4.SCC();     cout << "\nSCCs in fifth graph \n";    Graph g5(5);    g5.addEdge(0,1);    g5.addEdge(1,2);    g5.addEdge(2,3);    g5.addEdge(2,4);    g5.addEdge(3,0);    g5.addEdge(4,2);    g5.SCC();     return 0;}
// Java program to find strongly connected// components in a given directed graph// using Tarjan's algorithm (single DFS)import java.io.*;import java.util.*; // This class represents a directed graph// using adjacency list representationclass Graph{ // No. of vertices   private int V; //Adjacency Listsprivate LinkedList<Integer> adj[];private int Time; // Constructor@SuppressWarnings("unchecked")Graph(int v){    V = v;    adj = new LinkedList[v];         for(int i = 0; i < v; ++i)        adj[i] = new LinkedList();             Time = 0;} // Function to add an edge into the graphvoid addEdge(int v,int w){    adj[v].add(w);} // A recursive function that finds and prints strongly// connected components using DFS traversal// u --> The vertex to be visited next// disc[] --> Stores discovery times of visited vertices// low[] -- >> earliest visited vertex (the vertex with//             minimum discovery time) that can be reached//             from subtree rooted with current vertex// st -- >> To store all the connected ancestors (could be part//         of SCC)// stackMember[] --> bit/index array for faster check//                   whether a node is in stackvoid SCCUtil(int u, int low[], int disc[],             boolean stackMember[],             Stack<Integer> st){         // Initialize discovery time and low value    disc[u] = Time;    low[u] = Time;    Time += 1;    stackMember[u] = true;    st.push(u);     int n;         // Go through all vertices adjacent to this    Iterator<Integer> i = adj[u].iterator();         while (i.hasNext())    {        n = i.next();                 if (disc[n] == -1)        {            SCCUtil(n, low, disc, stackMember, st);                         // Check if the subtree rooted with v            // has a connection to one of the            // ancestors of u            // Case 1 (per above discussion on            // Disc and Low value)            low[u] = Math.min(low[u], low[n]);        }        else if (stackMember[n] == true)        {                         // Update low value of 'u' only if 'v' is            // still in stack (i.e. it's a back edge,            // not cross edge).            // Case 2 (per above discussion on Disc            // and Low value)            low[u] = Math.min(low[u], disc[n]);        }    }     // head node found, pop the stack and print an SCC    // To store stack extracted vertices    int w = -1;    if (low[u] == disc[u])    {        while (w != u)        {            w = (int)st.pop();            System.out.print(w + " ");            stackMember[w] = false;        }        System.out.println();    }} // The function to do DFS traversal.// It uses SCCUtil()void SCC(){         // Mark all the vertices as not visited    // and Initialize parent and visited,    // and ap(articulation point) arrays    int disc[] = new int[V];    int low[] = new int[V];    for(int i = 0;i < V; i++)    {        disc[i] = -1;        low[i] = -1;    }         boolean stackMember[] = new boolean[V];    Stack<Integer> st = new Stack<Integer>();         // Call the recursive helper function    // to find articulation points    // in DFS tree rooted with vertex 'i'    for(int i = 0; i < V; i++)    {        if (disc[i] == -1)            SCCUtil(i, low, disc,                    stackMember, st);    }} // Driver codepublic static void main(String args[]){         // Create a graph given in the above diagram    Graph g1 = new Graph(5);     g1.addEdge(1, 0);    g1.addEdge(0, 2);    g1.addEdge(2, 1);    g1.addEdge(0, 3);    g1.addEdge(3, 4);    System.out.println("SSC in first graph ");    g1.SCC();     Graph g2 = new Graph(4);    g2.addEdge(0, 1);    g2.addEdge(1, 2);    g2.addEdge(2, 3);    System.out.println("\nSSC in second graph ");    g2.SCC();         Graph g3 = new Graph(7);    g3.addEdge(0, 1);    g3.addEdge(1, 2);    g3.addEdge(2, 0);    g3.addEdge(1, 3);    g3.addEdge(1, 4);    g3.addEdge(1, 6);    g3.addEdge(3, 5);    g3.addEdge(4, 5);    System.out.println("\nSSC in third graph ");    g3.SCC();         Graph g4 = new Graph(11);    g4.addEdge(0, 1);    g4.addEdge(0, 3);    g4.addEdge(1, 2);    g4.addEdge(1, 4);    g4.addEdge(2, 0);    g4.addEdge(2, 6);    g4.addEdge(3, 2);    g4.addEdge(4, 5);    g4.addEdge(4, 6);    g4.addEdge(5, 6);    g4.addEdge(5, 7);    g4.addEdge(5, 8);    g4.addEdge(5, 9);    g4.addEdge(6, 4);    g4.addEdge(7, 9);    g4.addEdge(8, 9);    g4.addEdge(9, 8);    System.out.println("\nSSC in fourth graph ");    g4.SCC();         Graph g5 = new Graph (5);    g5.addEdge(0, 1);    g5.addEdge(1, 2);    g5.addEdge(2, 3);    g5.addEdge(2, 4);    g5.addEdge(3, 0);    g5.addEdge(4, 2);    System.out.println("\nSSC in fifth graph ");    g5.SCC();}} // This code is contributed by// Prateek Gupta (@prateekgupta10)
# Python program to find strongly connected components in a given# directed graph using Tarjan's algorithm (single DFS)#Complexity : O(V+E)  from collections import defaultdict  #This class represents an directed graph# using adjacency list representationclass Graph:      def __init__(self,vertices):        #No. of vertices        self.V= vertices                 # default dictionary to store graph        self.graph = defaultdict(list)                 self.Time = 0      # function to add an edge to graph    def addEdge(self,u,v):        self.graph[u].append(v)               '''A recursive function that find finds and prints strongly connected    components using DFS traversal    u --> The vertex to be visited next    disc[] --> Stores discovery times of visited vertices    low[] -- >> earliest visited vertex (the vertex with minimum                discovery time) that can be reached from subtree                rooted with current vertex     st -- >> To store all the connected ancestors (could be part           of SCC)     stackMember[] --> bit/index array for faster check whether                  a node is in stack    '''    def SCCUtil(self,u, low, disc, stackMember, st):         # Initialize discovery time and low value        disc[u] = self.Time        low[u] = self.Time        self.Time += 1        stackMember[u] = True        st.append(u)         # Go through all vertices adjacent to this        for v in self.graph[u]:                         # If v is not visited yet, then recur for it            if disc[v] == -1 :                             self.SCCUtil(v, low, disc, stackMember, st)                 # Check if the subtree rooted with v has a connection to                # one of the ancestors of u                # Case 1 (per above discussion on Disc and Low value)                low[u] = min(low[u], low[v])                                     elif stackMember[v] == True:                 '''Update low value of 'u' only if 'v' is still in stack                (i.e. it's a back edge, not cross edge).                Case 2 (per above discussion on Disc and Low value) '''                low[u] = min(low[u], disc[v])         # head node found, pop the stack and print an SCC        w = -1 #To store stack extracted vertices        if low[u] == disc[u]:            while w != u:                w = st.pop()                print (w, end=" ")                stackMember[w] = False                             print()                       #The function to do DFS traversal.    # It uses recursive SCCUtil()    def SCC(self):          # Mark all the vertices as not visited        # and Initialize parent and visited,        # and ap(articulation point) arrays        disc = [-1] * (self.V)        low = [-1] * (self.V)        stackMember = [False] * (self.V)        st =[]                  # Call the recursive helper function        # to find articulation points        # in DFS tree rooted with vertex 'i'        for i in range(self.V):            if disc[i] == -1:                self.SCCUtil(i, low, disc, stackMember, st)        # Create a graph given in the above diagramg1 = Graph(5)g1.addEdge(1, 0)g1.addEdge(0, 2)g1.addEdge(2, 1)g1.addEdge(0, 3)g1.addEdge(3, 4)print ("SSC in first graph ")g1.SCC() g2 = Graph(4)g2.addEdge(0, 1)g2.addEdge(1, 2)g2.addEdge(2, 3)print ("nSSC in second graph ")g2.SCC()   g3 = Graph(7)g3.addEdge(0, 1)g3.addEdge(1, 2)g3.addEdge(2, 0)g3.addEdge(1, 3)g3.addEdge(1, 4)g3.addEdge(1, 6)g3.addEdge(3, 5)g3.addEdge(4, 5)print ("nSSC in third graph ")g3.SCC() g4 = Graph(11)g4.addEdge(0, 1)g4.addEdge(0, 3)g4.addEdge(1, 2)g4.addEdge(1, 4)g4.addEdge(2, 0)g4.addEdge(2, 6)g4.addEdge(3, 2)g4.addEdge(4, 5)g4.addEdge(4, 6)g4.addEdge(5, 6)g4.addEdge(5, 7)g4.addEdge(5, 8)g4.addEdge(5, 9)g4.addEdge(6, 4)g4.addEdge(7, 9)g4.addEdge(8, 9)g4.addEdge(9, 8)print ("nSSC in fourth graph ")g4.SCC();  g5 = Graph (5)g5.addEdge(0, 1)g5.addEdge(1, 2)g5.addEdge(2, 3)g5.addEdge(2, 4)g5.addEdge(3, 0)g5.addEdge(4, 2)print ("nSSC in fifth graph ")g5.SCC(); #This code is contributed by Neelam Yadav
<script> // Javascript program to find strongly connected// components in a given directed graph// using Tarjan's algorithm (single DFS) // This class represents a directed graph// using adjacency list representationclass Graph{     // Constructorconstructor(v){    this.V = v;    this.adj = new Array(v);       for(let i = 0; i < v; ++i)        this.adj[i] = [];           this.Time = 0;} // Function to add an edge into the graphaddEdge(v, w){    this.adj[v].push(w);} // A recursive function that finds and prints strongly// connected components using DFS traversal// u --> The vertex to be visited next// disc[] --> Stores discovery times of visited vertices// low[] -- >> earliest visited vertex (the vertex with//             minimum discovery time) that can be reached//             from subtree rooted with current vertex// st -- >> To store all the connected ancestors (could be//          part of SCC)// stackMember[] --> bit/index array for faster check//                   whether a node is in stackSCCUtil(u, low, disc, stackMember, st){         // Initialize discovery time and low value    disc[u] = this.Time;    low[u] = this.Time;    this.Time += 1;    stackMember[u] = true;    st.push(u);       let n;           // Go through all vertices adjacent to this    for(let i of this.adj[u])    {        n = i;                   if (disc[n] == -1)        {            this.SCCUtil(n, low, disc, stackMember, st);                           // Check if the subtree rooted with v            // has a connection to one of the            // ancestors of u            // Case 1 (per above discussion on            // Disc and Low value)            low[u] = Math.min(low[u], low[n]);        }        else if (stackMember[n] == true)        {                           // Update low value of 'u' only if 'v' is            // still in stack (i.e. it's a back edge,            // not cross edge).            // Case 2 (per above discussion on Disc            // and Low value)            low[u] = Math.min(low[u], disc[n]);        }    }       // Head node found, pop the stack and print an SCC    // To store stack extracted vertices    let w = -1;    if (low[u] == disc[u])    {        while (w != u)        {            w = st.pop();            document.write(w + " ");            stackMember[w] = false;        }        document.write("<br>");    }    }     // The function to do DFS traversal.// It uses SCCUtil()SCC(){         // Mark all the vertices as not visited    // and Initialize parent and visited,    // and ap(articulation point) arrays    let disc = new Array(this.V);    let low = new Array(this.V);    for(let i = 0;i < this.V; i++)    {        disc[i] = -1;        low[i] = -1;    }           let stackMember = new Array(this.V);    let st = [];           // Call the recursive helper function    // to find articulation points    // in DFS tree rooted with vertex 'i'    for(let i = 0; i < this.V; i++)    {        if (disc[i] == -1)            this.SCCUtil(i, low, disc,                         stackMember, st);    }}} // Driver code // Create a graph given in the above diagramlet g1 = new Graph(5); g1.addEdge(1, 0);g1.addEdge(0, 2);g1.addEdge(2, 1);g1.addEdge(0, 3);g1.addEdge(3, 4);document.write("SSC in first graph <br>");g1.SCC(); let g2 = new Graph(4);g2.addEdge(0, 1);g2.addEdge(1, 2);g2.addEdge(2, 3);document.write("\nSSC in second graph<br> ");g2.SCC(); let g3 = new Graph(7);g3.addEdge(0, 1);g3.addEdge(1, 2);g3.addEdge(2, 0);g3.addEdge(1, 3);g3.addEdge(1, 4);g3.addEdge(1, 6);g3.addEdge(3, 5);g3.addEdge(4, 5);document.write("\nSSC in third graph <br>");g3.SCC(); let g4 = new Graph(11);g4.addEdge(0, 1);g4.addEdge(0, 3);g4.addEdge(1, 2);g4.addEdge(1, 4);g4.addEdge(2, 0);g4.addEdge(2, 6);g4.addEdge(3, 2);g4.addEdge(4, 5);g4.addEdge(4, 6);g4.addEdge(5, 6);g4.addEdge(5, 7);g4.addEdge(5, 8);g4.addEdge(5, 9);g4.addEdge(6, 4);g4.addEdge(7, 9);g4.addEdge(8, 9);g4.addEdge(9, 8);document.write("\nSSC in fourth graph<br> ");g4.SCC(); let g5 = new Graph (5);g5.addEdge(0, 1);g5.addEdge(1, 2);g5.addEdge(2, 3);g5.addEdge(2, 4);g5.addEdge(3, 0);g5.addEdge(4, 2);document.write("\nSSC in fifth graph <br>");g5.SCC(); // This code is contributed by avanitrachhadiya2155 </script>
SCCs in first graph 
4
3
1 2 0

SCCs in second graph 
3
2
1
0

SCCs in third graph 
5
3
4
6
2 1 0

SCCs in fourth graph 
8 9
7
5 4 6
3 2 1 0
10

SCCs in fifth graph 
4 3 2 1 0

Time Complexity: The above algorithm mainly calls DFS, DFS takes O(V+E) for a graph represented using an adjacency list. 
frankfeng579
PrateekGupta10
avanitrachhadiya2155
amartyaghoshgfg
raj2002
hardikkoriintern
graph-connectivity
Graph
Graph
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