Biconnected graph - GeeksforGeeks

Biconnected graph - GeeksforGeeks
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.

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.

// 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;

}

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