This document discusses bi-connected components in graphs. It defines an articulation point as a vertex in a connected graph whose removal would disconnect the graph. A bi-connected component is a maximal subgraph that contains no articulation points. The document presents algorithms for identifying articulation points and bi-connected components in a graph using depth-first search (DFS). It introduces the concepts of tree edges, back edges, forward edges and cross edges in a DFS tree and explains how to use these edge types to determine if a vertex is an articulation point based on the minimum discovery time of its descendant vertices.

1. “Bi-connected Components”
Prepared by:-
Soni Axay R. (130240116056)
Guided by:-
Prof. Nimisha Sharma

2. Sub-Topics
 Introduction
 Articulation Point
 Identification of Articulation Point
 Identification of bi-connected components
 Finding Articulation point by DFS
 Example

3. Introduction
 Basically it is a Graph theory.
 A graph is biconnected if it contains no
‘articulation’ points.
 A components of graph G is maximal
“biconnected subgraph” .That means it is
not contained any larger biconnected
subgraph of G.

4. Articulation Point
 Let G = (V,E) be a connected undirected graph.
Articulation Point: is any vertex of G whose removal results in a disconnected
graph.

5. Articulation Point
Articulation Point: is any vertex of G whose removal results
in a disconnected graph.

6. Biconnected components
 A graph is biconnected if it contains no articulation points.

7. Identification of Articulation
point
 Keep track of removal vertex from each subtree?
 Too expensive
 Time complexity O(V(V+E))
 Another method is to use the DFS to find the articulation point in the graph.
 By using the DFS we will get the DFS Tree.
 Classification of graph categories:-
 Tree edge
 Back edge
 Forward edge
 Cross edge

8. Conditions:-
 Consider an internal vertex u
 Not a leaf,
 Assume it is not the root
 Let v1, v2,…, vk denote the children of u
 Each is the root of a subtree of DFS
 If for some child, there is no back edge from any node in
this subtree going to a proper ancestor of u, then u is an
articulation point
u
Here u is an
articulation point

9. internal vertex u
 Here u is not an articulation point
 A back edge from every subtree of u to proper
ancestors of u exists
u

10. What if u is a leaf
 A leaf is never an articulation point
 A leaf has no subtrees..

11. What about the root?
 the root is an articulation point if an only if it has two or more
children.
 Root has no proper ancestor
 There are no cross edges between its subtrees
This root is an
articulation point

12. Calculation By DFS
 Define Low[u]
 Low[u]: minimum of d[u]
and
{d[w] | where (v,w) is a back edge and v is a
(nonproper) descendent of u.}

13. Computing Low[u]
 Initialization:
Low[u] = d[u]
 When a new back edge (u, v) is detected:
Low[u] = min( Low[u], d[v])
 Tree edge (u, v):
Low[u] = min(Low[u], Low[v])

14.  Once Low[u] is computed for all vertices u, we can test
whether a nonroot vertex u is an articulation point
 u is an articulation point iff it has a child v for which
Low[v] >= d[u]

15. Example
a
j
ie
c
b
g
d h
f
a
b e
c
d
g
f
h
i
j
Low = 1
1
1
13
3
3
8
8
8
d=1
2
3
8
4
5
6
7
9
10

16. Biconnected components
 A graph is biconnected if and only if it consists of a single
biconnected component
 No articulation points

17. Thank
You!!!