Biconnected Graph

An undirected graph is said to be a biconnected graph, if there are two vertex-disjoint paths between any two vertices are present. In other words, we can say that there is a cycle between any two vertices.

We can say that a graph G is a bi-connected graph if it is connected, and there are no articulation points or cut vertex are present in the graph.

To solve this problem, we will use the DFS traversal. Using DFS, we will try to find if there is any articulation point is present or not. We also check whether all vertices are visited by the DFS or not, if not we can say that the graph is not connected.

Input and Output

Input:
The adjacency matrix of the graph.
0 1 1 1 0
1 0 1 0 0
1 1 0 0 1
1 0 0 0 1
0 0 1 1 0

Output:
The Graph is a biconnected graph.

Algorithm

isArticulation(start, visited, disc, low, parent)

Input: The start vertex, the visited array to mark when a node is visited, the disc will hold the discovery time of the vertex, and low will hold information about subtrees. The parent will hold the parent of the current vertex.

Output − True if any articulation point is found.

Begin
   time := 0      //the value of time will not be initialized for next function calls
   dfsChild := 0
   mark start as visited
   set disc[start] := time+1 and low[start] := time + 1
   time := time + 1

   for all vertex v in the graph G, do
      if there is an edge between (start, v), then
         if v is visited, then
            increase dfsChild
            parent[v] := start

            if isArticulation(v, visited, disc, low, parent) is true, then
               return ture
            low[start] := minimum of low[start] and low[v]

            if parent[start] is φ AND dfsChild > 1, then
               return true

            if parent[start] is φ AND low[v] >= disc[start], then
               return true
         else if v is not the parent of start, then
            low[start] := minimum of low[start] and disc[v]
   done
   return false
End

isBiconnected(graph)

Input: The given graph.

Output − True if the graph is bi-connected.

Begin
   initially set all vertices are unvisited and parent of each vertices are φ
   if isArticulation(0, visited, disc, low, parent) = true, then
      return false

   for each node i of the graph, do
      if i is not visited, then
         return false
   done
   return true
End

Example

#include
#define NODE 5
using namespace std;

int graph[NODE][NODE] = {
   {0, 1, 1, 1, 0},
   {1, 0, 1, 0, 0},
   {1, 1, 0, 0, 0},
   {1, 0, 0, 0, 1},
   {0, 0, 0, 1, 0}
};
                               
int min(int a, int b) {
   return (a 1) {    //when u have 2 or more children
               return true;
            }

            if(parent[start] != -1 && low[v]>= disc[start])
               return true;
         } else if(v != parent[start])    //update low of start for previous call
            low[start] = min(low[start], disc[v]);
      }
   }
   return false;
}

bool isBiConnected() {
   bool *vis = new bool[NODE];
   int *disc = new int[NODE];
   int *low = new int[NODE];
   int *parent = new int[NODE];
   
   for(int i = 0; i
Output
The Graph is a biconnected graph.