An undirected graph is called Biconnected if there are two vertex-disjoint paths between any two vertices. In a Biconnected Graph, there is a simple cycle through any two vertices.

By convention, two nodes connected by an edge form a biconnected graph, but this does not verify the above properties. For a graph with more than two vertices, the above properties must be there for it to be Biconnected.

Following are some examples.

**See this for more examples.**

**How to find if a given graph is Biconnected or not?**

*A connected graph is Biconnected if it is connected and doesn’t have any Articulation Point*. We mainly need to check two things in a graph.

1) The graph is connected.

2) There is not articulation point in graph.

We start from any vertex and do DFS traversal. In DFS traversal, we check if there is any articulation point. If we don’t find any articulation point, then the graph is Biconnected. Finally, we need to check whether all vertices were reachable in DFS or not. If all vertices were not reachable, then the graph is not even connected.

Following is C++ implementation of above approach.

## C++

// A C++ program to find if a given undirected graph is // biconnected #include<iostream> #include <list> #define NIL -1 using namespace std; // A class that represents an undirected graph class Graph { int V; // No. of vertices list<int> *adj; // A dynamic array of adjacency lists bool isBCUtil(int v, bool visited[], int disc[], int low[], int parent[]); public: Graph(int V); // Constructor void addEdge(int v, int w); // to add an edge to graph bool isBC(); // returns true if graph is Biconnected }; Graph::Graph(int V) { this->V = V; adj = new list<int>[V]; } void Graph::addEdge(int v, int w) { adj[v].push_back(w); adj[w].push_back(v); // Note: the graph is undirected } // A recursive function that returns true if there is an articulation // point in given graph, otherwise returns false. // This function is almost same as isAPUtil() here ( http://goo.gl/Me9Fw ) // u --> The vertex to be visited next // visited[] --> keeps tract of visited vertices // disc[] --> Stores discovery times of visited vertices // parent[] --> Stores parent vertices in DFS tree bool Graph::isBCUtil(int u, bool visited[], int disc[],int low[],int parent[]) { // A static variable is used for simplicity, we can avoid use of static // variable by passing a pointer. static int time = 0; // Count of children in DFS Tree int children = 0; // Mark the current node as visited visited[u] = true; // Initialize discovery time and low value disc[u] = low[u] = ++time; // Go through all vertices aadjacent 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 make it a child of u // in DFS tree and recur for it if (!visited[v]) { children++; parent[v] = u; // check if subgraph rooted with v has an articulation point if (isBCUtil(v, visited, disc, low, parent)) return true; // Check if the subtree rooted with v has a connection to // one of the ancestors of u low[u] = min(low[u], low[v]); // u is an articulation point in following cases // (1) u is root of DFS tree and has two or more chilren. if (parent[u] == NIL && children > 1) return true; // (2) If u is not root and low value of one of its child is // more than discovery value of u. if (parent[u] != NIL && low[v] >= disc[u]) return true; } // Update low value of u for parent function calls. else if (v != parent[u]) low[u] = min(low[u], disc[v]); } return false; } // The main function that returns true if graph is Biconnected, // otherwise false. It uses recursive function isBCUtil() bool Graph::isBC() { // Mark all the vertices as not visited bool *visited = new bool[V]; int *disc = new int[V]; int *low = new int[V]; int *parent = new int[V]; // Initialize parent and visited, and ap(articulation point) // arrays for (int i = 0; i < V; i++) { parent[i] = NIL; visited[i] = false; } // Call the recursive helper function to find if there is an articulation // point in given graph. We do DFS traversal starring from vertex 0 if (isBCUtil(0, visited, disc, low, parent) == true) return false; // Now check whether the given graph is connected or not. An undirected // graph is connected if all vertices are reachable from any starting // point (we have taken 0 as starting point) for (int i = 0; i < V; i++) if (visited[i] == false) return false; return true; } // Driver program to test above function int main() { // Create graphs given in above diagrams Graph g1(2); g1.addEdge(0, 1); g1.isBC()? cout << "Yes\n" : cout << "No\n"; Graph g2(5); g2.addEdge(1, 0); g2.addEdge(0, 2); g2.addEdge(2, 1); g2.addEdge(0, 3); g2.addEdge(3, 4); g2.addEdge(2, 4); g2.isBC()? cout << "Yes\n" : cout << "No\n"; Graph g3(3); g3.addEdge(0, 1); g3.addEdge(1, 2); g3.isBC()? cout << "Yes\n" : cout << "No\n"; Graph g4(5); g4.addEdge(1, 0); g4.addEdge(0, 2); g4.addEdge(2, 1); g4.addEdge(0, 3); g4.addEdge(3, 4); g4.isBC()? cout << "Yes\n" : cout << "No\n"; Graph g5(3); g5.addEdge(0, 1); g5.addEdge(1, 2); g5.addEdge(2, 0); g5.isBC()? cout << "Yes\n" : cout << "No\n"; return 0; }

## Java

// A Java program to find if a given undirected graph is // biconnected import java.io.*; import java.util.*; import java.util.LinkedList; // This class represents a directed graph using adjacency // list representation class Graph { private int V; // No. of vertices // Array of lists for Adjacency List Representation private LinkedList<Integer> adj[]; int time = 0; static final int NIL = -1; // Constructor 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].add(w); //Note that the graph is undirected. adj[w].add(v); } // A recursive function that returns true if there is an articulation // point in given graph, otherwise returns false. // This function is almost same as isAPUtil() @ http://goo.gl/Me9Fw // u --> The vertex to be visited next // visited[] --> keeps tract of visited vertices // disc[] --> Stores discovery times of visited vertices // parent[] --> Stores parent vertices in DFS tree boolean isBCUtil(int u, boolean visited[], int disc[],int low[], int parent[]) { // Count of children in DFS Tree int children = 0; // Mark the current node as visited visited[u] = true; // Initialize discovery time and low value disc[u] = low[u] = ++time; // Go through all vertices aadjacent to this Iterator<Integer> i = adj[u].iterator(); while (i.hasNext()) { int v = i.next(); // v is current adjacent of u // If v is not visited yet, then make it a child of u // in DFS tree and recur for it if (!visited[v]) { children++; parent[v] = u; // check if subgraph rooted with v has an articulation point if (isBCUtil(v, visited, disc, low, parent)) return true; // Check if the subtree rooted with v has a connection to // one of the ancestors of u low[u] = Math.min(low[u], low[v]); // u is an articulation point in following cases // (1) u is root of DFS tree and has two or more chilren. if (parent[u] == NIL && children > 1) return true; // (2) If u is not root and low value of one of its // child is more than discovery value of u. if (parent[u] != NIL && low[v] >= disc[u]) return true; } // Update low value of u for parent function calls. else if (v != parent[u]) low[u] = Math.min(low[u], disc[v]); } return false; } // The main function that returns true if graph is Biconnected, // otherwise false. It uses recursive function isBCUtil() boolean isBC() { // Mark all the vertices as not visited boolean visited[] = new boolean[V]; int disc[] = new int[V]; int low[] = new int[V]; int parent[] = new int[V]; // Initialize parent and visited, and ap(articulation point) // arrays for (int i = 0; i < V; i++) { parent[i] = NIL; visited[i] = false; } // Call the recursive helper function to find if there is an // articulation/ point in given graph. We do DFS traversal // starring from vertex 0 if (isBCUtil(0, visited, disc, low, parent) == true) return false; // Now check whether the given graph is connected or not. // An undirected graph is connected if all vertices are // reachable from any starting point (we have taken 0 as // starting point) for (int i = 0; i < V; i++) if (visited[i] == false) return false; return true; } // Driver method public static void main(String args[]) { // Create graphs given in above diagrams Graph g1 =new Graph(2); g1.addEdge(0, 1); if (g1.isBC()) System.out.println("Yes"); else System.out.println("No"); Graph g2 =new Graph(5); g2.addEdge(1, 0); g2.addEdge(0, 2); g2.addEdge(2, 1); g2.addEdge(0, 3); g2.addEdge(3, 4); g2.addEdge(2, 4); if (g2.isBC()) System.out.println("Yes"); else System.out.println("No"); Graph g3 = new Graph(3); g3.addEdge(0, 1); g3.addEdge(1, 2); if (g3.isBC()) System.out.println("Yes"); else System.out.println("No"); Graph g4 = new Graph(5); g4.addEdge(1, 0); g4.addEdge(0, 2); g4.addEdge(2, 1); g4.addEdge(0, 3); g4.addEdge(3, 4); if (g4.isBC()) System.out.println("Yes"); else System.out.println("No"); Graph g5= new Graph(3); g5.addEdge(0, 1); g5.addEdge(1, 2); g5.addEdge(2, 0); if (g5.isBC()) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by Aakash Hasija

## Python

# Python program to find if a given undirected graph is # biconnected from collections import defaultdict #This class represents an undirected graph using adjacency list representation class Graph: def __init__(self,vertices): self.V= vertices #No. of vertices self.graph = defaultdict(list) # default dictionary to store graph self.Time = 0 # function to add an edge to graph def addEdge(self,u,v): self.graph[u].append(v) self.graph[v].append(u) '''A recursive function that returns true if there is an articulation point in given graph, otherwise returns false. This function is almost same as isAPUtil() u --> The vertex to be visited next visited[] --> keeps tract of visited vertices disc[] --> Stores discovery times of visited vertices parent[] --> Stores parent vertices in DFS tree''' def isBCUtil(self,u, visited, parent, low, disc): #Count of children in current node children =0 # Mark the current node as visited and print it visited[u]= True # Initialize discovery time and low value disc[u] = self.Time low[u] = self.Time self.Time += 1 #Recur for all the vertices adjacent to this vertex for v in self.graph[u]: # If v is not visited yet, then make it a child of u # in DFS tree and recur for it if visited[v] == False : parent[v] = u children += 1 if self.isBCUtil(v, visited, parent, low, disc): return True # Check if the subtree rooted with v has a connection to # one of the ancestors of u low[u] = min(low[u], low[v]) # u is an articulation point in following cases # (1) u is root of DFS tree and has two or more chilren. if parent[u] == -1 and children > 1: return True #(2) If u is not root and low value of one of its child is more # than discovery value of u. if parent[u] != -1 and low[v] >= disc[u]: return True elif v != parent[u]: # Update low value of u for parent function calls. low[u] = min(low[u], disc[v]) return False # The main function that returns true if graph is Biconnected, # otherwise false. It uses recursive function isBCUtil() def isBC(self): # Mark all the vertices as not visited and Initialize parent and visited, # and ap(articulation point) arrays visited = [False] * (self.V) disc = [float("Inf")] * (self.V) low = [float("Inf")] * (self.V) parent = [-1] * (self.V) # Call the recursive helper function to find if there is an # articulation points in given graph. We do DFS traversal starting # from vertex 0 if self.isBCUtil(0, visited, parent, low, disc): return False '''Now check whether the given graph is connected or not. An undirected graph is connected if all vertices are reachable from any starting point (we have taken 0 as starting point)''' if any(i == False for i in visited): return False return True # Create a graph given in the above diagram g1 = Graph(2) g1.addEdge(0, 1) print "Yes" if g1.isBC() else "No" g2 = Graph(5) g2.addEdge(1, 0) g2.addEdge(0, 2) g2.addEdge(2, 1) g2.addEdge(0, 3) g2.addEdge(3, 4) g2.addEdge(2, 4) print "Yes" if g2.isBC() else "No" g3 = Graph(3) g3.addEdge(0, 1) g3.addEdge(1, 2) print "Yes" if g3.isBC() else "No" g4 = Graph (5) g4.addEdge(1, 0) g4.addEdge(0, 2) g4.addEdge(2, 1) g4.addEdge(0, 3) g4.addEdge(3, 4) print "Yes" if g4.isBC() else "No" g5 = Graph(3) g5.addEdge(0, 1) g5.addEdge(1, 2) g5.addEdge(2, 0) print "Yes" if g5.isBC() else "No" #This code is contributed by Neelam Yadav

Output:

Yes Yes No No Yes

**Time Complexity:** The above function is a simple DFS with additional arrays. So time complexity is same as DFS which is O(V+E) for adjacency list representation of graph.

**References:**

http://www.cs.purdue.edu/homes/ayg/CS251/slides/chap9d.pdf

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