Check if a given Graph is 2-edge connected or not

Given an undirected graph G, with V vertices and E edges, the task is to check whether the graph is 2-edge connected or not.

A graph is said to be 2-edge connected if, on removing any edge of the graph, it still remains connected, i.e. it contains no Bridges.

Examples:

Input: V = 8, E = 10

Output: Yes
Explanation:
Given any vertex in the graph, we can reach any other vertex in the graph. Moreover, removing any edge from the graph does not affect its connectivity. So, the graph is said to be 2-edge connected.

Input: V = 8, E = 9

Output: No
Explanation:
On removal of the edge between vertex 3 and vertex 4, the graph is not connected anymore. So, the graph is not 2-edge connected.



Naive Approach: The naive approach is to check that on removing any edge X, if the remaining graph G – X is connected or not. If the graph remains connected on removing every edge one by one then it is a 2-edge connected graph. To implement the above idea, remove an edge and perform Depth First Search(DFS) or Breadth-First Search(BFS) from any vertex and check if all vertices are covered or not. Repeat this process for all E edges. If all vertices cannot be traversed for any edge, print No. Otherwise, print Yes.
Time Complexity: O(E * ( V + E))
Auxiliary Space: O(1)

Efficient Approach: Since the given graph is undirected, the problem can be solved only by counting the number of edges connected to the nodes. If for any of the nodes, the number of edges connected to it is 1 it means on removing this edge the node becomes disconnected and it can’t be reached from any other node therefore the graph is not 2-edge connected. Below are the steps:

  1. Create an array noOfEdges[] of size V which will store the number of edges connected to a node.
  2. For every edge (u, v) increment the number of edges for node u and v.
  3. Now iterate over the array noOfEdges[] and check if any of the edge has only 1 edge connected to it. If yes then the graph is not 2-edge connected.
  4. Otherwise the graph is 2-edge connected.

Below is the implementation of the above approach:

C++14

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// C++14 program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Definition of a graph
class Graph {
  
    // No. of vertices
    int V;
  
    // To create adjacency list
    list<int>* adj;
  
public:
    // Constructor
    Graph(int V);
  
    // Function to add an edge to graph
    void addEdge(int v, int w);
  
    // Function to check 2-edge
    // 2-edge connectivity
    void twoEdge(int v);
};
  
// Initialize the graph
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}
  
// Adding edges to adjacency list
void Graph::addEdge(int v, int w)
{
    adj[v - 1].push_back(w - 1);
    adj[w - 1].push_back(v - 1);
}
  
// Function to find if the graph is
// 2 edge connected or not
void Graph::twoEdge(int v)
{
    // To store number of edges for
    // each node
    int noOfEdges[v];
  
    for (int i = 0; i < v; i++) {
        noOfEdges[i] = adj[i].size();
    }
  
    bool flag = true;
  
    // Check the number of edges
    // connected to each node
    for (int i = 0; i < v; i++) {
  
        if (noOfEdges[i] < 2) {
            flag = false;
            break;
        }
    }
  
    // Print the result
    if (flag)
        cout << "Yes";
    else
        cout << "No";
}
  
// Driver Code
int main()
{
    // Number of nodes and edges
    int V = 8;
    int E = 10;
  
    // Given Edges
    int edges[E][2] = { { 1, 2 }, { 1, 8 }, { 1, 6 }, 
                        { 2, 3 }, { 2, 4 }, { 3, 7 }, 
                        { 3, 4 }, { 7, 5 }, { 7, 6 }, 
                        { 7, 8 } };
  
    // Initialize the graph
    Graph g(V);
  
    // Adding the edges to graph
    for (int i = 0; i < E; i++) {
        g.addEdge(edges[i][0], edges[i][1]);
    }
  
    // Function call
    g.twoEdge(V);
  
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
  
class Graph{
      
// No. of vertices     
private int V; 
  
// Array of lists for Adjacency
// List Representation 
private LinkedList<Integer> adj[]; 
  
// Constructor 
@SuppressWarnings("unchecked")
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 - 1].add(w - 1); // Add w to v's list. 
    adj[w - 1].add(v - 1);
  
// Function to find if the graph is
// 2 edge connected or not
void twoEdge(int v)
{
      
    // To store number of edges for
    // each node
    int[] noOfEdges = new int[v];
    for(int i = 0; i < v; i++) 
    {
        noOfEdges[i] = adj[i].size();
    }
  
    boolean flag = true;
      
    // Check the number of edges
    // connected to each node
    for(int i = 0; i < v; i++)
    {
        if (noOfEdges[i] < 2)
        {
            flag = false;
            break;
        }
    }
      
    // Print the result
    if (flag)
        System.out.print("Yes");
    else
        System.out.print("No");
}
  
// Driver code
public static void main (String[] args) 
{
      
    // Number of nodes and edges
    int V = 8;
    int E = 10;
      
    // Given Edges
    int edges[][] = { { 1, 2 }, { 1, 8 },
                      { 1, 6 }, { 2, 3 },
                      { 2, 4 }, { 3, 7 }, 
                      { 3, 4 }, { 7, 5 }, 
                      { 7, 6 }, { 7, 8 } };
      
    Graph g = new Graph(V); 
      
    // Adding the edges to graph
    for(int i = 0; i < E; i++) 
    {
        g.addEdge(edges[i][0], edges[i][1]);
    }
      
    // Function call
    g.twoEdge(V);
}
}
  
// This code is contributed by offbeat

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Output:

No

Time Complexity: O(V + E)
Auxiliary Space: O(V)

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