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CONNECTIVITY OF
GRAPHS
ASWATHY.A
2
CONNECTIVITY IN GRAPHS
ASWATHY.A
DEPARTMENT OF MATHEMATICS
MOUNT TABOR TRAINING COLLEGE
PATHANAPURAM
3
CONTENTS
PAGE NO
PREFACE 3
INTRODUCTION 4
Chapter 1 CONNECTIVITY OF GRAPHS 6
Chapter2 APPLICATIONS 27
REFERENCE 34
4
PREFACE
In the present times Mathematics occupies an important place in
curriculum. The โConnectivity in graphsโ is one of the important subject at
MSc level. In this book an attempt has been made to cover up most of topics
included in the MSc syllabus by Indian Universities. This book covers the
syllabi of Regular and Correspondence courses of Indian Universities.
The book has been written in a simple and lucid manner and is
up-to-date in its contents. To illustrate theory some examples have been given.
It is hoped that the book will be appreciated by teachers and
students alike. While preparing the book, material has been drawn from works of
different authors, periodicals and journals and the author is indebted to all such
persons and their publishers.
All suggestions for improvement of the book shall be thankfully
accepted.
Author
5
INTRODUCTION
Graph Theory is regarded as one of the areas of Applied Mathematics. Graph Theory
has been independently discovered many times. Leonhard Euler is known as โFather of Graph
Theoryโ. Subsequent rediscoveries of Graph Theory has been made by Gustav Kirchhoff and
Arthur Cayley. Another approach was of Hamiltonโs. The origin of graph theory can be traced
back to Euler's work on the Konigsberg bridges problem (1735), which subsequently led to the
concept of an Eulerian graph. The study of cycles on polyhedra by the Thomas P. Kirkman (1806
- 95) and William R. Hamilton (1805-65) led to the concept of a Hamiltonian graph. The concept
of a tree, a connected graph without cycles, appeared implicitly in the work of Gustav Kirchhoff
(1824-87), who employed graph-theoretical ideas in the calculation of currents in electrical
networks or circuits. Later, Arthur Cayley (1821-95), James J. Sylvester (1806-97), George
Polya (1887-1985), and others use 'tree' to enumerate chemical molecules. The study of planar
graphs originated in two recreational problems involving the complete graph 5K and the
complete bipartite graph 3,3K . These graphs proved to be planarity, as was subsequently
demonstrated by Kuratowski. First problem was presented by A. F. Mobius around the year
1840.
The development of graph theory is very similar the development of probability theory,
where much of the original work was motivated by efforts to understand games of chance. The
large portions of graph theory have been motivated by the study of games and recreational
mathematics. Generally speaking, we use graphs in two situations. Firstly, since a graph is a very
convenient and natural way of representing the relationships between objects we represent
objects by vertices and the relationship between them by lines. In many situations (problems)
such a pictorial representation may be all that is needed. Secondly, we take the graph as
mathematical model, solve the appropriate graph-theoretic problem, and then interpret the
solution in terms of the original problem.
In the past few years, Graph Theory has been established as an important mathematical tool
in wide variety of subjects. Graph Theory has application to some areas of Physics, Chemistry,
Communicative Science, Computer Technology, Electrical and Civil Engineering, Architecture,
Operation Research, Genetics, Psychology, Sociology, Economics, Anthropology and
Linguistics. The theory had also emerged as a worthwhile mathematical discipline in its own
right. Graph Theory is intimately related to many branches of Mathematics including Graph
Theory, Matrix Theory, Numerical Analysis, Probability and Topology. In fact Graph Theory
serves as a mathematical model for any system involving a binary relation.
In this text book, discuss the connectivity of graphs. This textbook is divided into two
chapters. In the first chapter, discussed about definitions and theorems on connectivity. The
second chapter deals with some applications of connectivity.
6
CHAPTER โ1
CONNECTIVITY OF GRAPHS
Definition (2.1)
An edge of a graph ๐บ is called a bridge or a cut edge if the subgraph ๐บ โ ๐ has
more connected components than ๐บ has.
Figure (2.1)
7
Figure (2.2)
A graph and two bridge deletions
Theorem (2.1)
An edge ๐ of a graph ๐บ is a bridge if and only if ๐ is not part of any cycle in ๐บ.
Proof:
Let ๐ have end vertices ๐ข and ๐ฃ. If ๐ is not a bridge then, it is either a loop or
there is a path ๐ = ๐ข๐ข1๐ข2 . . .๐ขn๐ฃ, from ๐ข to ๐ฃ, different from the edge ๐. If it is a
loop then it forms a cycle. If there is such a path ๐ then ๐ถ = ๐ข๐ข1๐ข2โฆ๐ขn๐ฃ๐ข, the
contradiction of ๐ with , is a cycle in ๐บ.This shows that if ๐ is not a bridge then it
is part of a cycle. This is equivalent to saying that if ๐ is not part of any cycle then
๐ must be a bridge.
8
Conversely, suppose that e is part of some cycle ๐ถ = ๐ข0๐ข1๐ข2...๐ขm in ๐บ. Let ๐ =
๐ขi๐ขi+1. In the case where ๐ = 1, ๐ถ = ๐ข0๐ข1, and so ๐ถ is just the edge ๐ and ๐ is a
loop. On the other hand , ๐๐ ๐ > 1 then ๐ = ๐ขi๐ขi-1โฆ.๐ข0๐ขm-1โฆ๐ขi+1 is a path
from ๐ข to ๐ฃ different from ๐. Thus ๐ is not a bridge .This shows that if ๐ is a bridge
then it is not part of any cycle in ๐บ , completing the proof.
Theorem (2.2)
Let ๐บ be a connected graph .Then ๐บ is a tree if and only if every edge of ๐บ is a
bridge i.e., if and only if for every edge ๐ of ๐บ the subgraph ๐บ โ ๐ has two
components.
Proof:-
Suppose that ๐บ is a tree. Then ๐บ is acyclic i.e., it has no cycle, and so no edge
of ๐บ belongs to a cycle. In other words, if ๐ is any edge of ๐บ then it is a bridge.
Conversely, suppose that ๐บ is connected and that every edge ๐ of ๐บ is a bridge.
Then ๐บ can have no cycles since any edge belonging to a cycle is not a bridge.
Hence ๐บ is acyclic and so is a tree.
Problem (2.1)
Find all bridges in the graph.
9
Figure (2.3)
Solution :-
Bridges in the graph are ๐2,๐3, ๐4, ๐5,๐6,๐7,๐13and ๐14.
Definition (2.2)
A vertex ๐ฃ of a graph ๐บ is called a cut vertex of ๐บ ๐๐ (๐บ โ ๐ฃ) > (๐บ).In
other words , a vertex ๐ฃ is a cut vertex of ๐บ if its deletion disconnects some
connected components of ๐บ , there by producing a subgraph having more
connected components than ๐บ has.
10
Figure (2.4)
๐ is a cut vertex of ๐บ1,since (๐บ1) = 1 while 1G โ๐ฃ) = 3.The graph ๐บ2
has no cut vertices.
Theorem (2.3)
Let ๐ฃ be a vertex of the connected graph ๐บ. Then ๐ฃ is a cut vertex of ๐บ if and
only if there are two vertices ๐ข and ๐ค of ๐บ, both different from ๐ฃ, such that ๐ฃ is on
every ๐ข โ ๐ค path in ๐บ.
Proof:-
First let ๐ฃ be a cut vertex of ๐บ .Then ๐บ โ ๐ฃ is disconnected and so there are
vertices ๐ข and ๐ค of ๐บ which lie in different components of ๐บ โ ๐ฃ .Thus, although
there is a path in ๐บ from ๐ข and ๐ค, there is no such path in ๐บ โ ๐ฃ.This implies that
every path in ๐บ from ๐ข to ๐ค contains the vertex ๐ฃ.
Conversely , suppose that ๐ข and ๐ค are two vertices of ๐บ , different from ๐ฃ,
such that every path in ๐บ from ๐ข to ๐ค contains ๐ฃ.Then there can be no path from ๐ข
11
to ๐ค in ๐บ โ ๐ฃ.Thus ๐บ โ ๐ฃ disconnected (with ๐ข and ๐ค lying in different
components).hence ๐ฃ is a cut vertex.
Note:
No vertex of a complete graph is a cut vertex.
Theorem (2.4)
Let ๐บ be a graph with ๐ vertices, where ๐ โฅ 2.Then ๐บ has at least two vertices
which are not cut vertices.
Proof:-
We may suppose that ๐บ is a connected graph. We proceed by assuming the
result is false for our ๐บ and so the proof will be complete if we derive a
contradiction from this.
Thus we are assuming that there is at most one vertex in ๐บ which is not cut
vertex. Now let ๐ข , ๐ฃ be vertices in ๐บ such that the distance ๐ (๐ข , ๐ฃ)between them
is the greatest of distances between pairs of vertices in ๐บ. i.e., ๐(๐ข , ๐ฃ) =
๐๐๐๐(๐บ).since ๐บ is connected and has at least two vertices, ๐ข โ ๐ฃ. Thus by our
assumption, one of these vertices must be a cut vertex, say ๐ฃ. Then ๐บ โ ๐ฃ is
disconnected and so there is a vertex ๐ค in ๐บ which does not belong to the same
component as ๐ข does in ๐บ โ ๐ฃ. This implies that every ๐ข๐ค path in ๐บ contains the
vertex ๐ฃ. It follows from this that the shortest path in ๐บ from ๐ข to ๐ค contains the
shortest path from ๐ข to ๐ฃ and this contradiction completes our proof.
12
Definition (2.3)
A vertex cut of ๐บ is a subset V of ๐ such that ๐บ โ V is disconnected.
A k-vertex cut is a vertex cut of k elements.
Definition (2.4)
Let ๐บ be a simple graph .The connectivity (vertex connectivity) of ๐บ, denoted
by ๐(๐บ),is the minimum number of vertices in ๐บ whose deletion from ๐บ leaves
either a disconnected graph or trivial graph.
Note:-
For ๐ โฅ 2, the deletion of any vertex from ๐พn,results in ๐พn-1 and in general the
deletion of ๐ก vertices (๐ก < ๐)results in ๐พ๐ โ ๐ก.
This shows that ๐(๐พ๐) = ๐ โ 1.
A connected graph ๐บ has ๐(๐บ) = 1 if and only if either ๐บ =2K or ๐บ has a
cut vertex.๐(๐บ) = 0 if and only if either ๐บ=๐พ1 or ๐บ is disconnected.
Figure (2.5)
13
๐(๐บ1) = 3 and ๐(๐บ2) = 4.
Theorem (2.5)
The vertex connectivity of any graph G can never exceed the edge
connectivity of G.
Proof:-
Let ฮฑ denote the edge connectivity of G. Therefore, there exists a cut set S
in G with ฮฑ edges. Let S partition the vertices of G into subsets 1V and
2V . By
removing at most ฮฑ vertices from 1V ( or
2V ) on which the edges in S are incident,
we can effect the removal of S (together with all other edges incident on these
vertices) from G. Hence the theorem.
Theorem (2.6)
A graph of order n is k-connected (where 11 nk ) if the degree of each vertex
is at least 2/)2( kn .
Proof:-
If the graph is complete, it is k-connected for 1 nk . Assume that the graph is
not complete and not k-connected. So there is a disconnecting set S of s vertices
such that G-S is a disconnected graph. Let H be a component of G-S with as few
vertices as possible. If the order of H is r, rsnr , which gives the upper
bound 2/sn for the order r. If v is any vertex of H, the degree of v in the graph
14
G cannot exceed sr 1 . Thus 2/22/)2(12/)(deg knsnssnv ,
violating the given inequality.
Theorem (2.7)
A graph is n-edge connected if and only if the number of edges in any cut is
at least k.
Theorem (2.8)
A vertex of a tree is a cut vertex if and only if it is not a pendant vertex.
Proof:-
Let v be a pendant vertex of a tree T. The removal of v results in the removal
of v and the only edge incident on it so that T-v is still connected. Therefore v is
not a cut vertex of T.
Figure (2.6)
Let u be a non pendant vertex of a tree T so that the degree of u is greater than 1.
Therefore there exists at least two edges uv and uw incident on u. Since T is a tree
vuw is the unique path connecting v and w. The removal of u deletes the edges uv
and uw. Therefore there is no path connecting the vertices v and w in T-u and so T-
u is disconnected. Hence u is a cut vertex of the tree T.
15
Definition (2.5)
A simple graph ๐บ is called ๐-connected if ๐(๐บ) โฅ ๐.
Note:-
๐บ is 1-connected if and only if ๐บ is connected and has atleast two vertices.
๐บ is 2- connected if and only if ๐บ is connected with atleast 3 vertices but
no cut vertices.
In figure (2.4), ๐บ1 is 3-connected and ๐บ2 is 4-connected.
Definition (2.6)
An edge cut of ๐บ is a subset of ๐ of the form [๐ , ๐] where ๐ is a non-empty
proper subset of ๐.
An ๐ โedge cut is an edge cut of k elements.
If ๐บ is non-trivial and ๐ธ is an edge cut of ๐บ, then ๐บ โ ๐ธ is disconnected.
Definition (2.7)
Let ๐บ be a simple graph. The edge connectivity of ๐บ, denoted by k (G), is the
smallest number of edges in ๐บ whose deletion from ๐บ either leaves a disconnected
graph or an empty graph.
Any non empty simple graph ๐บ with a bridge has k (G) =1.
k (G)=0 if and only if either ๐บ is disconnected or an empty graph.
16
All non-trivial connected graphs are 1-edge connected.
Figure (2.7)
Here k (๐บ) =2.
Theorem (2.9)
The edge connectivity of a graph ๐บ cannot exceed the degree of the vertex with
the smallest degree in ๐บ.
Proof:-
Let vertex iv be the vertex with the smallest degree in ๐บ. Let ivd be the
degree of iv . Vertex iv can be separated from ๐บ by removing the ivd edges
incident on vertex iv . Hence the theorem.
Definition (2.8)
A simple graph ๐บ is called n-edge connected if nGk .
17
Theorem (2.10)
A simple graph ๐บ is n-edge connected if and only if given any pair of distinct
vertices ๐ข and ๐ฃ of ๐บ, there are atleast ๐ edge disjoint paths from ๐ข to ๐ฃ.
Proof :-
Suppose that ๐บ is ๐-edge connected and ๐ข and ๐ฃ are two distinct vertices of ๐บ.
Then any ๐ข โ ๐ฃ separating set of edges must be atleast ๐ members. Thus, there
must be atleast ๐-edge disjoint ๐ข โ ๐ฃ paths, as required.
Conversely, suppose that given any pair of distinct vertices ๐ข and ๐ฃ of ๐บ ,
there are atleast ๐ โedge disjoint paths from ๐ข to ๐ฃ. Then, for each pair of vertices
๐ข and ๐ฃ every ๐ข โ ๐ฃ separating set of edges must have atleast ๐ members. Thus it
requires the deletion of atleast ๐ edges from ๐บ inorder to produce a disconnected
graph or an empty graph. In other words, ๐บ is ๐-edge connected, as required .
Theorem (2.11) (Whitneyโs Theorem)
For any graph , kk .
Proof :-
Let ๐บ be a connected graph.
If ๐บ has no edges, then k =0. Otherwise a disconnected graph results when
all the edges incident with a vertex of minimum degree are removed. In either case,
k .
18
To show kk .
If ๐บ is disconnected or trivial, then 0 kk . If ๐บ is connected and has a
bridge ๐ฅ, then k =1. In this case, ๐ = 1 since either ๐บ has a cut vertex incident
with ๐ฅ or ๐บ is ๐พ2.
Finally, suppose ๐บ has 2k edges whose removal disconnects it. Clearly,
the removal of 1k of these lines produces a graph with a bridge uvx . For
each of these 1k edges, select an incident vertex different from ๐ข or ๐ฃ. The
removal of these vertices also removes the 1k edges and quite possibly more. If
the resulting graph is disconnected, then kk , if not ๐ฅ is a bridge, and hence the
removal of ๐ข or ๐ฃ will result in either a disconnected or a trivial graph, so kk
in every case.
Thus for any graph ๐บ, kk .
Theorem (2.12)
For any three integers ๐, ๐ , ๐ก such that tsr 0 there is a graph ๐บ with
skrk , and t .
Proof :-
Take two disjoint copies of ๐พ๐ก+1 . Let ๐ด be a set of ๐ vertices in one of them
and ๐ต be a set of ๐ vertices in the other. Join the vertices of ๐ด and ๐ต by ๐ edges
utilizing all the vertices of ๐ต and all the vertices of ๐ด. Since ๐ด is a vertex cut and
the set of these ๐ edges is an edge cut of the resulting graph ๐บ, it is clear that
19
rGk and sGk . Also there is atleast one vertex which is not in ๐ด โช ๐ต and
it has degree ๐ก, so that ๐ฟ(๐บ) = ๐ก .
Theorem (2.13)
โฅ๐
2 ensures k in a graph.
Proof :-
Since it is enough to rule out k , suppose ๐บ is a graph with
2
n and k . Let ๐น be a set of k edges disconnecting ๐บ, and ๐ถ1 and ๐ถ2 be the
components of ๐บ โ ๐น and ๐ด1 and ๐ด2 the end vertices of the edges of ๐น in ๐ถ1 and
๐ถ2 respectively. Let rA 1 and sA 2 and suppose ๐ (๐ถ1) = ๐ด
1. Then each
vertex of ๐ถ1 is adjacent with atleast one edge of ๐น. So the number ๐
1 of edges in
๐ถ1 satisfies the following inequality.
krm 2
11
r2
1 since k .
12
1 r
rr 12
1 since Fr k .
20
But a graph on r vertices cannot have more than rr 12
1 edges. Thus,
11 ACV and similarly, 22 ACV .
Thus each of 1C and
2C contains atleast ๐ฟ + 1 vertices.
GVn
12
1
22
n
= ๐ + 2
This contradiction establishes the theorem.
Problem (2.2)
Show that if ๐ and ๐ are the only two odd degree vertices of a graph ๐บ, then
๐ and ๐ are connected in ๐บ.
Solution:
If ๐บ is connected, nothing to prove.
Let ๐บ be disconnected.
If possible assume that ๐ and ๐ are not connected.
Then ๐ and ๐ lie in the different components of ๐บ. Hence the component of ๐บ
containing ๐ (similarly containing ๐) contains only one odd degree vertex ๐, which
21
is not possible as each component of ๐บ is itself a connected graph and in a graph
number of odd degree vertices should be even.
Therefore ๐ and ๐ lie in the same component of ๐บ.
Hence they are connected.
Problem (2.3)
Prove that a connected graph ๐บ remains connected after removing an edge ๐ from
๐บ if and only if ๐ lie in some circuit in ๐บ.
Solution:
If an edge ๐ lies in a circuit ๐ถ of the graph ๐บ then between the end vertices of ๐,
there exist at least two paths in ๐บ.
e
Figure (2.8)
Hence removal of such an edge e from the connected graph G will not effect
the connectivity of G. Conversely, if e does not lies in any circuit of G then
removal of e disconnects the end vertices of e.
Hence G is disconnected.
22
Definition (2.9)
A connected graph that has no cut vertices is called a block.
Every block with atleast 3 vertices is 2-connected.
Definition (2.10)
A block of a graph ๐บ is a maximal subgraph ๐ป of ๐บ such that ๐ป is a block.
Every graph is the union of its blocks.
Figure (2.9)
23
The blocks of ๐บ are
Definition (2.11)
Let ๐ข and ๐ฃ be two vertices of a graph ๐บ. A collection {๐(1), ๐(2), โฆ , ๐(๐)}
of ๐ข โ ๐ฃ paths is said to be internally disjoint if , given any distinct pair ๐(๐) and
๐(๐) in the collection, ๐ข and ๐ฃ are the only vertices ๐(๐) and ๐(๐)in common.
Theorem (2.14)
Let ๐บ be a simple graph with atleast 3 vertices. Then ๐บ is 2-connected if and
only if for each pair of distinct vertices ๐ข and ๐ฃ of ๐บ there are two internally
disjoint ๐ข โ ๐ฃ paths in ๐บ.
Proof :-
Suppose that any pair of distinct vertices is connected by a pair of internally
disjoint paths. Then clearly ๐บ is connected so it remains to prove that ๐บ has no cut
24
vertices. Assume to the contrary, that ๐ฃ is a cut vertex of ๐บ. Then there are two
vertices ๐ข and ๐ค of ๐บ, both different from ๐ฃ, such that ๐ฃ is on every ๐ข โ ๐ค path in
๐บ. However, by the hypothesis there are two internally disjoint ๐ข โ ๐ค paths in
๐บ and atmost one of these can then pass through ๐ฃ. Thus ๐ฃ is not on every ๐ข โ
๐ค path, a contradiction. Hence ๐บ has no cut vertices.
Conversly, suppose that ๐บ is 2-connected. Let ๐ข and ๐ฃ be a pair of distinct
vertices of ๐บ. We use induction on ๐(๐ข, ๐ฃ), the distance between ๐ข and ๐ฃ, to show
that there is a pair of internally disjoint ๐ข โ ๐ฃ paths. First, if ๐(๐ข, ๐ฃ) = 1 then
๐ข and ๐ฃ are joined by an edge, say ๐. It follows that ๐ is not a bridge, since ๐บ has
no cut vertices. Hence , there is a ๐ข โ ๐ฃ path ๐ different and so internally disjoint
from the ๐ข โ ๐ฃ path ๐ given by the single edge ๐.
We now assume that 2, kvud and that if ๐ฅ and ๐ฆ are any pair of
vertices with ๐(๐ฅ, ๐ฆ) < ๐ then there are two internally disjoint ๐ฅ โ ๐ฆ paths. Let
๐ be a path of length ๐ from ๐ข to ๐ฃ and let ๐ค be the second last vertex of ๐. Then
๐(๐ข, ๐ค) = ๐ โ 1 and there are two internally disjoint ๐ข โ ๐ค paths say ๐1 and ๐2 .
Since ๐บ is 2-connected, ๐ค is not a cut vertex, i.e., ๐บ โ ๐ค is connected and so
there is a ๐ข โ ๐ฃ path P which does not pass through ๐ค. Let ๐ฅ be the last vertex of
p which is also a vertex of either ๐1 or ๐2 .
25
Figure (2.10)
Suppose that 1Qx . Let ๐1 be the ๐ข โ ๐ฃ path given by the ๐ข โ ๐ฅ section of ๐1
followed by the ๐ฅ โ ๐ฃ section of P . Let ๐2 be the ๐ข โ ๐ฃ path given by the path
๐2 followed by the edge ๐ค๐ฃ. Then ๐1 and ๐2 are internally disjoint by the
definition of ๐1, ๐2, ๐ฅ and P . The proof now follows by induction and is
complete.
Corollary (2.15)
If ๐บ is 2-connected, then any two vertices of ๐บ lie on a common cycle.
Definition (2.12)
An edge ๐ is said to be subdivided when it is deleted and replaced by a path of
length two connecting its ends, the internal vertex of this path being a new vertex.
26
Figure (2.11)
Subdivision of an edge
Note :-
The class of blocks with atleast 3 vertices is closed under the operation of
subdivision.
Corollary (2.16)
If ๐บ is a block with ๐ โฅ 3, then any two edges of ๐บ lie on a common cycle.
Proof :-
Let ๐บ be a block with ๐ โฅ 3, and let ๐1 and ๐2 be two edges of ๐บ. Form a new
graph G by subdividing ๐1 and ๐2 and denote the new vertices by ๐ฃ1 and ๐ฃ2.
Clearly G is a block with atleast five vertices, and hence is 2-connected. It follows
that ๐ฃ1 and ๐ฃ2 lie on a common cycle of G . Thus ๐1 and ๐2 lie on a common cycle
of ๐บ.
27
CHAPTER-2
APPLICATIONS
Application (3.1)
If we think of graph as representing a communication network, the
connectivity (or edge connectivity) becomes the smallest number of
communication stations (or communication links) whose break down would
jeopardize communication in the system. The higher the connectivity and edge
connectivity, the more reliable the network. From this point of view, a tree
network, such as the one obtained by Kruskalโs Algorithm, is not very reliable, and
one is led to consider the following generalization of the connector problem.
Let k be a given positive integer and let ๐บ be a weighted graph. Determine a
minimum-weight k-connected spanning subgraph of ๐บ.
For k=1, this problem reduces to the connector problem, which can be solved
by Kruskalโs Algorithm. For values of k greater than one, the problem is unsolved
and is known to be difficult. However if ๐บ is a complete graph in which each edge
is assigned unit weight, then the problem has a simple solution which we now
present.
Observe that, for a weighted complete graph on ๐-vertices in which edge is
assigned unit weight, a minimum-weighted ๐-connected spanning subgraph is
simply an m-connected graph on ๐ vertices with as few edges as possible. We shall
denote by ๐(๐, ๐) the least number of edges that an ๐-connected graph on ๐
vertices can have (It is, of course, assumed that ๐ < ๐).
๐(๐, ๐) โฅ [๐๐
2] (1)
28
We shall show that equality holds in (1) by constructing an ๐-connected graph
Hm ,n on ๐ vertices that has exactly [๐๐
2] edges. The structure of Hm ,n depends on
the pattern of ๐ and ๐, there are 3 cases.
Case 1 : ๐ even. Let ๐ = 2๐. Then H2r,m is constructed as follows. It has vertices
0,1, โฆ . , ๐ โ 1 and two vertices ๐ and ๐ are joined if ๐ โ ๐ โค ๐ โค ๐ + ๐ . H8,n is
shown in figure 3.1 (a).
Case 2 : ๐ odd, ๐ even. Let ๐ = 2๐ + 1. Then H2r+1,n is constructed by first
drawing H2r,n and then adding edges joining vertex ๐ to vertex ๐ +๐
2 for 1 โค ๐ โค
๐
2 .
H9,8 is shown in figure 3.1(b).
Case 3 : ๐ odd, ๐ odd. Let = 2๐ + 1 . Then H2r+1, n is constructed by first drawing
H2r,n and then adding edges joining vertex 0 to vertices ๐โ1
2 and
๐+1
2 and vertex ๐ to
vertex ๐ +๐+1
2 for 1 โค ๐ <
๐โ1
2 . H9,9 is shown in figure 3.1 (c).
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Figure (3.1)
Theorem (3.1) (Harary Theorem)
The graph nmH , is ๐-connected.
Proof :-
Consider the case ๐ โ 2๐. We shall show that nmH , has no vertex cut of
fewer than 2๐ vertices. If possible, let V be a vertex cut with |V | < 2๐ . Let ๐ and
๐ be vertices belonging to different components of VH nr,2 . Consider the two
sets of vertices ๐ = {๐, ๐ + 1, โฆ , ๐ โ 1, ๐} and ๐ = {๐, ๐ + 1, โฆ , ๐ โ 1, ๐} where
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addition is taken modulo ๐. Since |V | < 2๐, we may assume, without loss of
generality, that |V โฉ ๐ | < ๐. Then there is clearly a sequence of distinct vertices in
๐| V which starts with ๐, ends with ๐, and is such that the difference between any
two consecutive terms is atmost ๐. But such a sequence is an (๐, ๐) path in
VH nr,2 , a contradiction. Hence nrH ,2 is 2๐-connected. The case ๐ = 2๐ + 1 is
also 2๐ + 1 -connected.
It is easy to see that
2,
mnnmf (2)
It now follows from (1) and (2) that
2,
mnnmf and that nmH , is an ๐-
connected graph on ๐-vertices with a few edges as possible.
Application (3.2)
Suppose we are given n stations that are to be connected by means of e lines
(telephone lines, bridges, rail roads, tunnels, or highways) where 1 ne . What is
the best way of connecting? By โbestโ we mean that the network should be as
invulnerable to destruction of individual stations and individual lines as possible.
In other words, construct a graph with n vertices and e edges that has the maximum
possible edge connectivity and vertex connectivity.
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Figure (3.2)
For example, the graph in figure (3.2) has n=8, e=16, and has vertex
connectivity of one and edge connectivity of three. Another graph with the same
number of vertices and edges (8 and 16, respectively) can be drawn as shown in
figure (3.3).
Figure (3.3)
Graph with 8 vertices and 16 edges
It can easily be seen that the edge connectivity as well as the vertex connectivity
of this graph is four. Consequently, even after any three stations are bombed, or
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any three lines destroyed, the remaining stations can still continue to
โcommunicateโ with each other. Thus the network of figure (3.3) is better
connected than that of figure (3.2) (although both consist of the same number of
lines-16).
Theorem (3.2)
The maximum vertex connectivity one can achieve with a graph G of n
vertices and e edges 1 ne is the integral part of the number n
e2
Proof :-
Every edge in G contributes two degrees. The total (2e degrees) is divided
among n vertices. Therefore, there must be at least one vertex in G whose degree is
equal to or less than the number n
e2 . The vertex connectivity of G cannot exceed
this number.
To show that this value can actually be achieved, one can first construct an n-
vertex regular graph of degree equal to the integral part of the number n
e2 and
then add the remaining 2
ne .(integral part of the number n
e2 ) edges
arbitrarily.
Thus we can summarize as follows:
vertex connectivity โค edge connectivity โค n
e2,
and
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maximum vertex connectivity possible = integral part of the number n
e2 .
Thus for a graph with 8 vertices and 16 edges, we can achieve a vertex
connectivity(and therefore edge connectivity) s high as four (8
162 ).
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REFERENCES
1. Diestel, R., Graph Theory, Electronic Edition,
2. Chartrand, Gary (1985), Introductory Graph Theory, Dover.
3. Bondy, J.A.; Murty, U.S.R. (2008), Graph Theory, Springer.
4. Biggs, N.; Lloyd, E.; Wilson, R. (1986), Graph Theory, 1736โ1936, Oxford
University Press.
5. Harary, Frank (1969), Graph Theory, Reading, MA: Addison-Wesley.