tarjan's strongly connected components algorithm

Comment

0

# 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 representation
class 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 diagram
g1 = 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

Popularity 7/10 Helpfulness 1/10 Language python

Source: www.geeksforgeeks.org

Tags: algorithm connected-components python

Share

Link to this answer
Share Copy Link

Contributed on Jul 17 2022

Mustafa Marzouk

0 Answers Avg Quality 2/10

tarjan's strongly connected components algorithm

Comment

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 representation
class Graph{
 
// No. of vertices   
private int V;
 
//Adjacency Lists
private 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 graph
void 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 stack
void 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 code
public 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)

Popularity 9/10 Helpfulness 2/10 Language java

Source: www.geeksforgeeks.org

Tags: algorithm algo

Share

Link to this answer
Share Copy Link

Contributed on Jul 17 2022

Mustafa Marzouk

0 Answers Avg Quality 2/10

tarjan's strongly connected components algorithm

Comment

0

// 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 -1
using namespace std;
 
// A class that represents an directed graph
class 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 stack
void 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 function
int 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;
}

Popularity 7/10 Helpfulness 1/10 Language cpp

Source: www.geeksforgeeks.org

Tags: algorithm c++ connected-components co

Share

Link to this answer
Share Copy Link

Contributed on Jul 17 2022

Mustafa Marzouk

0 Answers Avg Quality 2/10

tarjan's strongly connected components algorithm

Comment

0

<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 representation
class Graph{
     
// Constructor
constructor(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 graph
addEdge(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 stack
SCCUtil(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 diagram
let 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>

Popularity 7/10 Helpfulness 1/10 Language javascript

Source: www.geeksforgeeks.org

Tags: algorithm connected-components connected-com

Share

Link to this answer
Share Copy Link

Contributed on Jul 17 2022

Mustafa Marzouk

0 Answers Avg Quality 2/10

tarjan's strongly connected components algorithm

Contents

More Related Answers

tarjan's strongly connected components algorithm

Closely Related Answers

tarjan's strongly connected components algorithm

tarjan's strongly connected components algorithm

tarjan's strongly connected components algorithm

Grepper

Documentation

Social

Legal

Contact

Oops, You will need to install Grepper and log-in to perform this action.