C++ programming Biconnected graph - There is a simple cycle through any two vertices.A connected graph is Biconnected if it is connected and doesn’t have

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.

**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.

```
// 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;
}
```

Output:

Yes
Yes
No
No
Yes

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