# Check if removing a given edge disconnects a graph

Given an undirected graph and an edge, the task is to find if the given edge is a bridge in graph, i.e., removing the edge disconnects the graph.

Following are some example graphs with bridges highlighted with red color.

One solution is to find all bridges in given graph and then check if given edge is a bridge or not.

A simpler solution is to remove the edge, check if graph remains connect after removal or not, finally add the edge back. We can always find if an undirected is connected or not by finding all reachable vertices from any vertex. If count of reachable vertices is equal to number of vertices in graph, then the graph is connected else not. We can find all reachable vertices either using BFS or DFS. Below are complete steps.

1.  Remove the given edge
2. Find all reachable vertices from any vertex. We have chosen first vertex in below implementation.
3. If count of reachable nodes is V, then return false [given is not Bridge]. Else return yes.

Implementation:

## C++

 // C++ program to check if removing an // edge disconnects a graph or not. #include using namespace std;   // Graph class represents a directed graph // using adjacency list representation class Graph {     int V;    // No. of vertices     list *adj;     void DFS(int v, bool visited[]); public:     Graph(int V);   // Constructor       // function to add an edge to graph     void addEdge(int v, int w);       // Returns true if graph is connected     bool isConnected();       bool isBridge(int u, int v); };   Graph::Graph(int V) {     this->V = V;     adj = new list[V]; }   void Graph::addEdge(int u, int v) {     adj[u].push_back(v); // Add w to vâ€™s list.     adj[v].push_back(u); // Add w to vâ€™s list. }   void Graph::DFS(int v, bool visited[]) {     // Mark the current node as visited and print it     visited[v] = true;       // Recur for all the vertices adjacent to     // this vertex     list::iterator i;     for (i = adj[v].begin(); i != adj[v].end(); ++i)         if (!visited[*i])             DFS(*i, visited); }   // Returns true if given graph is connected, else false bool Graph::isConnected() {     bool visited[V];     memset(visited, false, sizeof(visited));       // Find all reachable vertices from first vertex     DFS(0, visited);       // If set of reachable vertices includes all,     // return true.     for (int i=1; i

## Python3

 # Python3 program to check if removing # an edge disconnects a graph or not.   # Graph class represents a directed graph # using adjacency list representation class Graph:       def __init__(self, V):         self.V = V         self.adj = [[] for i in range(V)]           def addEdge(self, u, v):         self.adj[u].append(v) # Add w to vâ€™s list.         self.adj[v].append(u) # Add w to vâ€™s list.           def DFS(self, v, visited):                   # Mark the current node as         # visited and print it         visited[v] = True               # Recur for all the vertices         # adjacent to this vertex         i = 0         while i != len(self.adj[v]):             if (not visited[self.adj[v][i]]):                 self.DFS(self.adj[v][i], visited)             i += 1           # Returns true if given graph is     # connected, else false     def isConnected(self):         visited = [False] * self.V               # Find all reachable vertices         # from first vertex         self.DFS(0, visited)               # If set of reachable vertices         # includes all, return true.         for i in range(1, self.V):             if (visited[i] == False):                 return False               return True       # This function assumes that edge      # (u, v) exists in graph or not,     def isBridge(self, u, v):                   # Remove edge from undirected graph         indU = self.adj[v].index(u)         indV = self.adj[u].index(v)         del self.adj[u][indV]         del self.adj[v][indU]               res = self.isConnected()               # Adding the edge back         self.addEdge(u, v)               # Return true if graph becomes         # disconnected after removing         # the edge.         return (res == False)   # Driver code if __name__ == '__main__':       # Create a graph given in the     # above diagram     g = Graph(4)     g.addEdge(0, 1)     g.addEdge(1, 2)     g.addEdge(2, 3)       if g.isBridge(1, 2):         print("Yes")     else:         print("No")   # This code is contributed by PranchalK

## C#

 using System; using System.Collections.Generic;   class Graph {   int V; // No. of vertices   List> adj;     private void DFS(int v, bool[] visited) {     // Mark the current node as visited and print it     visited[v] = true;     // Recur for all the vertices adjacent to     // this vertex     foreach (int i in adj[v]) {       if (!visited[i]) {         DFS(i, visited);       }     }   }     public Graph(int V) {     this.V = V;     adj = new List>();     for (int i = 0; i < V; i++) {       adj.Add(new List());     }   }     public void addEdge(int u, int v) {     adj[u].Add(v); // Add v to uâ€™s list.     adj[v].Add(u); // Add u to vâ€™s list.   }     // Returns true if given graph is connected, else false   public bool isConnected() {     bool[] visited = new bool[V];       // Find all reachable vertices from first vertex     DFS(0, visited);       // If set of reachable vertices includes all,     // return true.     for (int i = 1; i < V; i++)       if (!visited[i])         return false;       return true;   }     // This function assumes that edge (u, v)   // exists in graph or not,   public bool isBridge(int u, int v) {     // Remove edge from undirected graph     adj[u].Remove(v);     adj[v].Remove(u);       bool res = isConnected();       // Adding the edge back     addEdge(u, v);       // Return true if graph becomes disconnected     // after removing the edge.     return (res == false);   }     // Driver code   public static void Main(string[] args) {     Graph g = new Graph(4);     g.addEdge(0, 1);     g.addEdge(1, 2);     g.addEdge(2, 3);       if (g.isBridge(1, 2)) {       Console.WriteLine("Yes");     }     else {       Console.WriteLine("No");     }   } }

## Javascript

Output

Yes

Time Complexity : O(V + E)

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