Bridges in Graph
An edge in an undirected connected graph is a scaffold if eliminating it disengages it. For a disengaged undirected graph, a definition is comparable, an extension is an edge eliminating which builds a number of detached segments.
Like Articulation Points, spans speak to weaknesses in an associated network and are helpful for planning dependable organisations. For instance, in a wired PC organisation, an explanation point shows the basic PCs and an extension demonstrates the basic wires or associations.
How to Discover All Bridges in a Given Graph?
A straightforward methodology is to individually eliminate all edges and check whether evacuation of an edge causes disengaged diagram. Following are steps of basic methodology for associated with it.
- For each edge (u, v), do following
- Remove (u, v) from a diagram
- See if the diagram stays associated (We can either utilise BFS or DFS)
- Add (u, v) back to the chart.
Time complexity of above method is O(E*(V+E)) if we are using the adjancy matrix.
We are going to do better this complexity, how?
The thought is like O(V+E) calculation for Articulation Points. We do DFS crossing of the given graph. In DFS tree an edge (u, v) (u is a parent of v in DFS tree) is connected if there doesn’t exist some other choice to arrive at u or a progenitor of u from subtree established with v. As examined in the past post, the worth low[v] demonstrates soonest visited vertex reachable from subtree established with v. The condition for an edge (u, v) to be a bridge is, “low[v] > disc[u]”.
Articulation Points
A vertex in an undirected associated chart is an articulation point if disposing of it (and edges through it) isolates the graph. Explanation focuses speak to weaknesses in an associated network – single focus whose disappointment would part the organisation into at least 2 segments. They are helpful for planning dependable organisations.
For a confined undirected chart, an explanation point is a vertex taking out which grows number of associated portions.
How to Discover All Articulation Points in a Given Diagram?
A straightforward methodology is to individually eliminate all vertices and check whether the expulsion of a vertex causes disengaged diagram. Following are steps of basic methodology for the associated chart.
- For each vertex v, do the following
- Remove v from the graph
-
See if the diagram stays associated (We can either utilise BFS or DFS Algorithm)
- Add v back to the graph
The time complexity of the above technique is O(V*(V+E)) for a chart spoke to utilising contiguousness list. Would we be able to improve?
An O(V+E) calculation to discover all Articulation Points (APs). The thought is to utilise DFS (Depth First Search). In DFS, we follow vertices in a tree structure called the DFS tree. In DFS tree, a vertex u is a parent of another vertex v if v is found by u (clearly v is nearby of u in the diagram). In DFS tree, a vertex u is the exclamation point in the event that one of the accompanying two conditions is valid.
- u is the foundation of DFS tree and it has at any rate two kids.
- u isn’t the foundation of DFS tree and it has a youngster v to such an extent that no vertex in subtree established with v has a back edge to one of the precursors (in DFS tree) of u.
Way to Write a C++ Program to Find if a Given Undirected Graph is Bi-Connected
#include <iostream>
#include <list>
#define NIL -1
using namespace std;
// A class for representing an undirected graph
class Graph
{
// Number of vertices
int V;
// A dynamic array for adjacency list
list *adj;
public:
// Constructor
Graph(int V);
// to add an edge into the graph
void addEdge(int v, int w);
// returns true if the graph is fully Biconnected
bool isBC();
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
// the graph is undirected
adj[w].push_back(v);
}
/* A recursive function is that returns true if there is an articulation point
in given graph, otherwise it returns false. This function almost same as
isAPUtil() */
// u –> The vertex to be visited next in the graph
// visited[] –> keeps tract of visited vertices in graph
// disc[] –> Stores discovery times of visited vertices in graph
// parent[] –> Stores the parent vertices in DFS tree
/* A static variable that is used for simplicity, we can avoid use of static
variable by passing the pointer. */
static int time = 0;
// Counts the children in DFS Tree
int children = 0;
// Marks the current node as visited in code
visited[u] = true;
// Initialize the discovery time and low value
disc[u] = low[u] = ++time;
// Going through the all vertices adjacent to this
list<int>::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
{
// v is a current adjacent of u
int v = *i;
//If v is not visited yet, then we make it a child of u in DFS tree and recursion for it
if (!visited[v])
{
children++;
parent[v] = u;
// check if subgraph rooted with v has an articulation point in it
if(isBCUtil(v, visited, disc, low, parent))
return true;
// Check the subtree rooted with v has a connection to one of ancestors of u
low[u] = min(low[u], low[v]);
// u is articulation point in following case:
// (1) u is root of DFS tree and can have two or more children.
if (parent[u] == NILL && children > 1)
return true;
/* (2) If u is not the root and low value of one of its child is more than the
discovery value of u. */
if (parent[u] != NIL && low[v] >= disc[u])
return true;
}
// Update the low value of u for parent function call.
else if (v != parent[u])
low[u] = min(low[u], disc[v]);
}
return false;
}
/* The function that returns true if graph is Biconnected, otherwise it is false.
It uses the recursion function isBCUtil() */
bool Graph::isBC()
{
// Mark all the vertices as not visited in graph
bool *visited = new bool[V];
int *disc = new int[V];
int *low = new int[V];
int *parent = new int[V];
// Initialising the parent and visited, and ap(articulation point) array
For (int i=0; i<V; i++)
{
parent[j] = NIL;
visited[j] = false;
}
/* Call the function recursive helper function to find if there is an
articulation point in given graph. We do the DFS traversal of graph
starring from vertex 0 */
if ( isBCUtil (0, visited, disc, low,parent)==true)
return false;
/* Now check whether the given graph is connected or not in manner.
An undirected graph is connected if all vertices are reachable and also
from any starting point (we have taken this 0 as starting point) */
for (int i=0; i<V; i++)
if (visited[j] == false)
return false;
return true;
}
// Main function program to test above function
int main()
{
// Creating the graphs given in above diagrams
Graph g1(2);
g1.addEdge(0, 1);
g1.isBC()? cout << “done\n” : cout << “Not\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 << "done\n" : cout << "Not\n";
Graph g3(3);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.isBC()? cout << "done\n" : cout << "Not\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 << "done\n" : cout << "Not\n";
Graph g5(3);
g5.addEdge(0, 1);
g5.addEdge(1, 2);
g5.addEdge(2, 0);
g5.isBC()? cout << "done\n" : cout << "Not\n";
return 0;
}
Frequently Asked Questions
What do you mean by bi-connectivity in an undirected graph?
In an undirected graph, a vertex is biconnected if its removal does not disconnect the graph. A graph is biconnected if it has no articulation points. An articulation point is a vertex whose removal increases the number of connected components in a graph. In other words, an undirected graph is biconnected if there are two vertex-disjoint paths between any two vertices in the graph.
What do you mean by a graph?
A graph is a data structure that is used to model relationships between objects or entities. It is a collection of vertices, also known as nodes, and edges that connect these vertices. The edges can be directed or undirected, depending on whether they have a specific direction or not.
What do you understand by a cyclic graph?
A cyclic graph is a graph in which at least one cycle, or closed path, exists. This means that there is a sequence of edges that connects a set of vertices in such a way that the first and last vertices in the sequence are the same.
Conclusion
In this article, we have discussed bi-connectivity in undirected graph. We have also discussed its implementation in C++. You can refer to our Guided Path on Coding Ninjas Studio to upskill yourself in Data Structures and Algorithms, Competitive Programming, JavaScript, and many more! If you wish to test your competency in coding, you may check out the mock test series and participate in the contests hosted on Coding Ninjas Studio!
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