inverse domination in - dskpdfugcdskpdf.unipune.ac.in/journal/uploads/ma/ma12-130005-a-1.pdf · t...

T Tamizh ChelvamT Asir

G. S. Grace Prema

Inverse Domination inGraphs

Inverse Domination, General Properties, Grid Graphs

The concept of dominating sets introduced by Ore and Berge, is currentlyreceiving much attention in the literature of graph theory. Several types ofdomination parameters have been studied by imposing several conditionson dominating sets. Ore observed that the complement of every minimaldominating set of a graph with minimum degree at least one is also adominating set. This implies that every graph with minimum degree at leastone has two disjoint dominating sets. Recently several authors initiated thestudy of the cardinalities of pairs of disjoint dominating sets in graphs. Theinverse domination number is the minimum cardinality of a dominating setwhose complement contains a minimum dominating set. Motivated by theinverse domination number, there are studies which deals about twodisjoint domination number of a graph.

T Tamizh Chelvam

T. TAMIZH CHELVAM is the Professor and Chairmanof Department of Mathematics at ManonmaniamSundaranar University, India. He has authored 85referred papers on Algebra & Graph Theory. T. ASIRis D.S. Kothari Post Doctoral Fellow and a Doctoratein Algebraic Graph Theory. G.S. Grace Prema isAssociate Professor at St. John's College,Palayamkottai

978-3-659-36290-3

Inverse D

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Tamizh

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a

Preface

The concept of dominating sets introduced by Ore and Berge, iscurrently receiving much attention in the literature of graph the-ory. Several types of domination parameters have been studied byimposing several conditions on dominating sets. Ore observed thatthe complement of every minimal dominating set of a graph withminimum degree at least one is also a dominating set. This impliesthat every graph with minimum degree at least one has two disjointdominating sets. Recently several authors initiated the study of thecardinalities of pairs of disjoint dominating sets in graphs. Kulli andSigarkanti introduced the inverse domination number which is theminimum cardinality of a dominating set whose complement con-tains a minimum dominating set. Motivated by the inverse domina-tion number, Hedetniemi et al. defined the two disjoint dominationnumber of a graph.

It is a natural question why to devote special attention to the caseof two disjoint dominating sets rather than k disjoint dominatingsets for a general k. The reason is that, by Ore’s observation, thetrivial necessary minimum degree condition is also sufficient for theexistence of two disjoint dominating sets. For all fixed k ≥ 3, it isNP-complete to decide the existence of k disjoint dominating setsand no minimum degree condition is sufficient for the existence ofthree disjoint dominating sets. One can find applications for twodisjoint dominating sets in networks. In any network (or graphs),dominating sets are central sets and they play a vital in routingproblems in parallel computing. Also finding efficient dominatingsets is always concern in finding optimal central sets in networks.

i

ii

Suppose S is a dominating set in a graph (or network) G, when thenetwork fails in some nodes in S, the inverse dominating set in V −S

will take care of the role of S. In this aspect, it is worthwhile toconcentrate on dominating and inverse dominating sets in graphs.

This book deals about the inverse domination in graphs and dealtthe following problems:

• characterization of graphs G with γ(G) = γ ′(G) =⌊

|G|2

⌋

.

• characterization of graphs G with γ(G) + γ ′(G) = |G| − 1.

• the inverse domination number of the grid graph Pk × Pn.

T. Tamizh Chelvam, T. Asir and G.S. Grace Prema

Acknowledgments

We thank the authorities of Manonmaniam Sundaranar University, India forgranting permission to publish this book which is an outcome of research car-ried out in the Department of Mathematics with the partial financial assistancethrough UGC-Special Assistance Programme(DRS-II) sanctioned to Departmentof Mathematics by the University Grants Commission, Government of India. Ithank this funding agency at this moment. I also thank the research scholars S.Raja and N. Mohamed Rilwan for their support in type setting the manuscript.I would be failing in my duty if I do not thank my wife Dr. R. Ezily and my sonMaster T. Kuralamudhan for their cooperation.

T. Tamizh Chelvam

I am thankful to the University Grants Commission, Government of India forproviding me with the necessary financial support through Dr. D. S. KothariPostdoctoral Fellowship. I would be thankful to my parents Mr. R. Thangarajand Mrs. D. Victoria, my well wishers and my friends for their encouragementand constant support throughout my life.

T. Asir

I thank the authorities of Manonmaniam Sundaranar University, St. John’sCollege for providing and University Grants Commission, India for providing mea chance to pursue research under FDP programme. I would be thankful tomy parents Mrs. and Mr. Gurupatham Selvaraj and husband Mr. G. SudhakarDavid. I also would like to put on record the patience of my daughter S. Nilastessiand son S. Anban Monickam who stared me of all my motherly responsibilitiesduring the period of research.

G.S. Grace Prema

iii

Contents

1 Introduction 1

1.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Results on Domination . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Inverse Domination Number 11

2.1 Graphs with γ(G) = γ′

(G) . . . . . . . . . . . . . . . . . . . . . 13

2.2 Basic results on γ ′(G) . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Bounds for γ′

(G) . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.4 Graphs with γ(G) = γ′

(G) =⌊

n

2

⌋

. . . . . . . . . . . . . . . . . . 25

3 Sum of γ and γ ′ 35

3.1 Graphs with δ ≥ 2 and γ + γ ′ = n − 1 . . . . . . . . . . . . . . . 37

3.2 Graphs with δ = 1 and γ + γ ′ = n − 1 . . . . . . . . . . . . . . . 45

4 Inverse Domination in Grid Graphs 63

4.1 Inverse Domination in 2 × n Grid Graphs . . . . . . . . . . . . . . 65




4.5 Inverse Domination in 6 × n grid graphs . . . . . . . . . . . . . . 73

4.6 Inverse Domination in 7 × n grid graphs . . . . . . . . . . . . . . 76

4.7 Inverse Domination in P2 × Cn grid graphs . . . . . . . . . . . . . 77

5 Inverse Domination Saturation Number 81

5.1 Classification of Inverse Domination Saturation . . . . . . . . . . 82

5.2 Classification of Inverse Domination Unsaturation . . . . . . . . . 91

Bibiliography 97

v

vi CONTENTS

List of Figures

2.1 Class A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.2 Class B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.3 Class Q2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.4 Class Q3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.5 Class Q4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.6 Class Q5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.1 Graph with γ < γ ′ and γ + γ ′ = n − 1 . . . . . . . . . . . . . . . 43

3.2 Graphs in family C . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.3 Graphs in family D . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.4 Class G1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.5 Class G2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.6 Class R1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.7 Class R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.8 Class R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.9 Class R4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.10 Class R5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.11 Class R6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.12 Class R7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.1 Class S and S ′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.2 2 × 7 grid graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.3 Blocks A2, AB1, AB2, AB3 and AB4 for 2 × n grid . . . . . . . . . 66

4.4 Blocks A,B1, B2, B3 and B4 for 3 × n grid . . . . . . . . . . . . . 67

4.5 Blocks A2, AB1, AB2, AB3 and AB4 for 3 × n grid . . . . . . . . . 68

4.6 Blocks A1, A2, A3, A4 and A5 for 4 × n grid . . . . . . . . . . . . . 69

4.7 Blocks A,B and B1 for 4 × n grid . . . . . . . . . . . . . . . . . . 69

4.8 Blocks AB and BB1 for 4 × n grid . . . . . . . . . . . . . . . . . 69

vii

viii LIST OF FIGURES

4.9 Blocks A, B, B1, B2, B3, B4 and B5 for 5 × n grid . . . . . . . . . 71

4.10 Blocks C, D, D1, D2, D3, D4, D5, F2, F3, F4 and F5 for 5 × n

grid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

4.11 Blocks A, B, B1, B2, B3, B4 and B5 for 6 × n grid . . . . . . . 74

4.12 Blocks B6 and B7 for 6 × n grid . . . . . . . . . . . . . . . . . . . 74

4.13 Blocks A, C1 and C3 for 7 × n grid . . . . . . . . . . . . . . . . . 75

4.14 Blocks C4 and C5 for 7 × n grid . . . . . . . . . . . . . . . . . . . 76

4.15 Blocks C6 and B for 7 × n grid . . . . . . . . . . . . . . . . . . . 77

5.1 Peterson graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.2 Inverse domination saturated graph . . . . . . . . . . . . . . . . . 86

5.3 Twin star graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

5.4 A 3-regular graph . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

Notations

Throughout this book, unless otherwise stated, our notations will be as follows:

⌊x⌋ The largest integer less than or equal to x.

⌈x⌉ The smallest integer greater than or equal to x.

G Graph

n Number of vertices in G

V = V (G) Vertex set of G

E = E(G) Edge set of G

〈S〉 Subgraph induced by S ⊆ V (G)

d(v) Degree of the vertex v

∆(G) Maximum degree of G

δ(G) Minimum degree of G

Kn Complete graph on n vertices

Km,n Complete bipartite graph

Cn Cycle on n vertices

Pn Path on n vertices

G Complement of the graph G

N(S) Open neighborhood of S

N [S] Closed neighborhood of S

G1oG2 Corona of the graphs G1 and G2

β0(G) Independence number of G

γ(G) Domination number of G

γ ′(G) Inverse domination number of G

d(G) Domatic number of G

i(G) Independent domination number of G

G1 ∪ G2 Union of the graphs G1 and G2

G1 + G2 Join of the graphs G1 and G2.

ix

x LIST OF FIGURES

Chapter 1

Introduction

One of the fastest growing areas in graph theory is the study of

domination and related subset problems such as independence, cov-

ering, matching and inverse domination. Several types of domina-

tion parameters have been studied by imposing several conditions

on dominating sets. Though substantial work has been carried out

on domination parameters and related topics in graphs, there are

only a few results concerning inverse domination in graphs. The

purpose of this book is to study about the inverse domination in

graphs.

By a graph G = (V, E), we mean a finite undirected graph with-

out loops or multiple edges. For graph theoretic terminology, we

refer G. Chatrand et al. [8] and F. Harary [21]. A subset S of V is

called a dominating set if for every vertex v ∈ V −S, there exists a

vertex u in S such that u is adjacent to v. The smallest cardinality

of a minimal dominating set in G is called the domination number

of G and is denoted by γ(G) or simply γ when there is no possibil-

ity of confusion. Any dominating set with γ(G) vertices is called a

γ-set of G. An excellent treatment of fundamentals of domination

in graphs is given by Haynes et al. in [23]. Let D be a γ-set of G. A

1

2 CHAPTER 1. INTRODUCTION

dominating set D ′ contained in V −D is called an inverse dominat-

ing set of G with respect to D. The smallest cardinality among all

minimal dominating set in V − D is called the inverse domination

number of G and it is denoted by γ ′(G). Any inverse dominating

set of G which has γ ′(G) vertices is called a γ′

-set of G. By O.

Ore’s Theorem [32], if a graph G has no isolated vertices, then the

complement V − D of every minimal dominating set D contains a

dominating set. Thus every graph without isolated vertices contains

an inverse dominating set with respect to a minimum dominating

set and so every graph has an inverse domination number. Due to

this fact, throughout the book we restrict ourselves to graphs with

no isolated vertices. Hereafter G denotes a simple graph on n ver-

tices with no isolated vertices. This concept of inverse domination

was introduced by V. R. Kulli [28].

By the definition of an inverse domination number γ ′(G), for

any graph G without isolated vertex, we have γ(G) + γ ′(G) ≤ n

and γ ′(G) ≥ γ(G). Also G. S. Domke et al. [13] characterized the

graphs which satisfy γ(G) + γ ′(G) = n. In their work, they give a

lower bound for the inverse domination number for trees. They also

provided a constructive characterization for the trees which achieve

their lower bound.

In this chapter, we present certain basic definitions and graph

theoretical results which are essential for later reference in this book.

In chapter 2, we consider graphs G for which γ(G) = γ ′(G). We

derive some bounds for γ ′(G) and γ(G) through n and characterize

the graphs with γ(G) = γ′

(G) = n2 . Also we characterize the graphs

with γ(G) = γ ′(G) = n−12 and we give the inverse domination

number for some classes of graphs. In chapter 3, we characterize

1.1. BASIC DEFINITIONS 3

graphs with minimum degree at least two for which the sum of

their domination number and inverse domination number is n − 1.

Further, we construct classes of graphs with minimum degree one for

which the sum of their domination number and inverse domination

number is n− 1 and finally we characterize all graphs for which the

sum of domination number and inverse domination number is n−1.

In chapter 4, we study about inverse domination in grid graphs. The

domination number of grid graphs Pk×Pn for 1 ≤ k ≤ 10 and n ≥ 1

have been previously established by M. S. Jacobson et al. [26, 27]

and T. Y. Chang et al. [7]. In this study, we obtain the inverse

domination number and an inverse dominating set for grids Pk ×Pn

where 1 ≤ k ≤ 7 and n ≥ 1. In chapter 5, we indicate a motivation

to define inverse domination saturation and classify graphs into type

I and type II inverse domination saturated graphs. Further we

introduce inverse domination unsaturation and we classify type I

and type II inverse domination unsaturated graphs.

1.1 Basic Definitions

In this section, we collect some basic definitions and theorems which

are needed for the subsequent chapters. For basic graph theoretic

terminology, we refer to F. Harary [21]. Some of the important

definitions in graph theory which are used in this book are listed

here.

A graph G = (V, E) is a finite nonempty set of objects called

vertices together with a set of unordered pairs of distinct vertices

of G called edges. The vertex set and the edge set of G are denoted

by V (G) and E(G) respectively. If e = {u, v} is an edge, then we


write e = uv and we say that u and v are adjacent vertices. If two

vertices are not joined by an edge, then they are nonadjacent. If

two distinct edges are incident with a common vertex, then they

are said to be adjacent to each other. The edge e = uu is called a

loop. The edges ei = ej = uv are called multiple edges or parallel

edges. A graph without loops and parallel edges is called a simple

graph.

The cardinality of the vertex set of a graph G is called the order

of G and is denoted by n where as the cardinality of the edge set is

called the size of G and is denoted by m. A graph with n vertices

and m edges is called a (n, m) graph.

A graph H is called a subgraph of G if V (H) ⊆ V (G) and E(H) ⊆

E(G). A spanning subgraph H of G is a subgraph of G with V (H) =

V (G). For any subset S of vertices of G, the induced subgraph 〈S〉

is the maximal subgraph of G with vertex set S. Thus two vertices

of S are adjacent in 〈S〉 if and only if they are adjacent in G. Note

that 〈S〉 is also denoted by G[S]. For v ∈ V (G), the subgraph

induced by V − {v} is denoted by G − v and it is the subgraph

obtained from G by the removal of v.

The degree of a vertex v in a graph G is the number of edges

of G incident with v and is denoted by d(v). A vertex of degree 0

in G is called an isolated vertex and a vertex of degree 1 is called

a pendant vertex. A vertex that is adjacent to a pendant vertex v

is called a support of v. The minimum and maximum degrees of

vertices of G are denoted by δ(G) and ∆(G) respectively. A graph

is said to be regular if all its vertices are of the same degree k and

it is said to be k-regular graph.


The complement G of a graph G is the graph with vertex set

V (G) such that two vertices are adjacent in G if and only if they

are not adjacent in G. A graph is said to be a complete graph if

every two of its vertices are adjacent. The complete graph on n

vertices is denoted by Kn. A bipartite graph G is a graph whose

vertex set V (G) can be partitioned into two subsets X and Y such

that every edge of G has one end in X and the other end in Y .

The pair (X, Y ) is called a bipartition of G. Further, if G contains

every edge joining any vertex of X to any vertex of Y , then G is

called a complete bipartite graph. The complete bipartite graph with

bipartition (X, Y ) such that |X| = m and |Y | = n is denoted by

Km,n. The graph K1,n−1 is called a star graph.

Let u and v (not necessarily distinct) be vertices of a graph G. A

u−v walk of G is a finite, alternating sequence u = u0, e1, u1, e2, . . . ,

un−1, en, un = v of vertices and edges, beginning with vertex u and

ending with vertex v such that ei = ui−1ui for all i = 1, 2, . . . n.

The number n is called the length of the walk. A u − v walk is

determined by the sequence u = u0, u1, . . . un−1, un = v of its vertices

and hence we specify a walk simply by u = u0, u1, . . . un−1, un. A

walk in which all the vertices are distinct is called a path. A walk

u0, u1, . . . un−1, un is called a closed walk if u0 = un. A closed walk

in which u0, u1, . . . un−1, un are distinct is called a cycle. A path on

n vertices is denoted by Pn and a cycle on n vertices is denoted by

Cn.

A graph G is said to be connected if any two vertices of G are

joined by a path. A maximal connected subgraph of G is called

a component of G. Thus a disconnected graph has at least two

components. The number of components of G is denoted by ω(G).


A graph G is called acyclic if it has no cycles. A connected acyclic

graph is called a tree. A spanning subgraph of a graph which is also

a tree is called a spanning tree of the graph. A wheel is a graph

obtained from a cycle by adding a new vertex and edges joining it

to all the vertices of the cycle. A wheel with n vertices is denoted by

Wn. A path that contains every vertex of G is called a Hamiltonian

path of G. A Hamiltonian cycle is a cycle that contains every vertex

of G. If a graph contains a Hamiltonian cycle, then it is called a

Hamiltonian graph.

Two graphs G1 and G2 have disjoint vertex sets V1, V2 and edge

sets E1, E2 respectively. Their join is denoted by G1 + G2 and it

consists of G1 ∪ G2 and all edges joining every vertex of V1 with

every vertex of V2. The corona of two graphs G1 and G2 is the

graph G = G1oG2 formed from one copy of G1 and |V (G1)| copies

of G2 where the ith vertex of G1 is adjacent to every vertex in the

ith copy of G2.

The open neighborhood N(v) of a vertex v is the set of all vertices

adjacent to v in G. Also N [v] = N(v) ∪ {v} is called the closed

neighborhood of v. The open neighborhood N(S) of a set S of

vertices is the set of all vertices adjacent to some vertex in S. N [S] =

N(S)∪S is called the closed neighborhood of S. Let S be a subset

of vertices of a graph G and u ∈ S. We say that a vertex v is a

private neighbor of u (with respect to S) if N [v] ∩ S = {u}. The

private neighbor set of u with respect to S is defined as pn[u, S] =

{v : N [v] ∩ S = {u}}. Notice that u ∈ pn[u, S] if u is an isolate in

〈S〉, in which case we say that u is its own private neighbor. The

private neighbor set of a set S is defined as pn(S) = {v : v ∈ pn[u, S]

for every u ∈ S}.


A subset S of V in a graph G is said to be independent if no

two vertices in S are adjacent. The maximum cardinality of an

independent set is called the independence number of G and it is

denoted by β0(G). A status is a set S of vertices in a graph which

has the property that for any two vertices u, v ∈ S, N(u)∩V −S =

N(v)∩V −S. In other words, the set of vertices in V −S dominated

by u equals the set of vertices in V − S dominated by v.

A set S ⊆ V of vertices in a graph G = (V, E) is called a domi-

nating set if every vertex v ∈ V is either an element of S or adjacent

to an element of S. The domination number of G is the smallest

cardinality of all minimal dominating sets in G and is denoted by

γ(G) or simply by γ. Moreover such a dominating set of G is called

a γ-set of G. If D = {x} is a dominating set of G, then x is called a

dominating vertex of G. A vertex v ∈ V (G) is said to be a γ-required

vertex of G if v lies in every γ-set of G.

Let D be a γ-set of a graph G. A dominating set D ′ ⊆ V −D is

called an inverse dominating set of G with respect to D. The inverse

domination number γ ′(G) of G is the cardinality of a smallest

inverse dominating set of G. An inverse dominating set D′ is called

a γ ′-set if |D′| = γ ′(G).

A dominating set S is called a perfect dominating set if every

vertex in V −S is adjacent to exactly one vertex in S. A dominating

set S is called an independent dominating set if no two vertices of S

are adjacent. A subset S of V is called a total dominating set if every

vertex in V is adjacent to some other vertex in S. A dominating

set S is called a connected dominating set if the subgraph 〈S〉 is

connected. A dominating set S is called an inverse dominating set

if S ⊆ V − S ′ for some γ-set S ′. A dominating set S is called a


split dominating set if the subgraph 〈V − S〉 is disconnected. The

domatic number d(G) of a graph G is defined to be the maximum

number of elements in a partition of V (G) into dominating sets. A

graph G is called domatically full if d(G) = δ(G) + 1, which is the

maximum possible order of a domatic partition of G.

1.2 Results on Domination

In this section, we collect certain fundamental results connecting the

domination number and the inverse domination number. First of

all we start with characterization of minimal dominating sets given

by O. Ore [32].

Theorem 1.2.1. [32] A dominating set S is a minimal dominating

set if and only if for each vertex u ∈ S, one of the following two

conditions holds.

(a) u is an isolate of S.

(b) There exists a vertex v ∈ V − S for which N(v) ∩ S = {u}.

Theorem 1.2.2. [32] If a graph G has no isolated vertex and D is

a minimal dominating set of G, then V − D is a dominating set of

G. In particular, γ(G) ≤⌊

n2

⌋

.

Theorem 1.2.3. [4, 23] Let G be a (m, n) graph. Then

(i) n − m ≤ γ(G) ≤ n − ∆.

(ii) ⌈ n1+∆(G)⌉ ≤ γ(G) ≤ n − ∆(G).

1.2. RESULTS ON DOMINATION 9

Theorem 1.2.4. [23] For any graph G, γ(G) + ǫf(G) = n where

ǫf(G) denotes the maximum number of pendant edges in any span-

ning forest of G.

Theorem 1.2.5. For any tree T with n ≥ 2, there exists a vertex

v ∈ V such that γ(T − v) = γ(T ).

Theorem 1.2.6. If G and G have no isolated vertices, then γ(G)+

γ(G) ≤⌊

n2

⌋

+ 2.

The following two theorems give a characterization for graphs G

with γ(G) + γ ′(G) = n.

Theorem 1.2.7. [13] Let G be a connected graph with δ(G) ≥ 2.

Then γ(G) + γ ′(G) = n if and only if G = C4.

Theorem 1.2.8. [13] Let G be a connected graph with n ≥ 3 and

δ(G) = 1. Let L ⊆ V be the set of all degree one vertices and

S = N(L). Then γ(G)+ γ ′(G) = n if and only if the following two

conditions hold:

(i) V − S is independent.

(ii) For every vertex x ∈ V −(S∪L), every stem in N(x) is adjacent

to at least two leaves.

Theorem 1.2.9. [13] For any tree T of order n ≥ 2, γ′

(T ) ≥ n+13 .

Lemma 1.2.10. [20] If G has a unique γ-set D, then every vertex

in D that is not an isolated vertex has at least two private neighbors

other than itself.


Lemma 1.2.11. [20] Let D be a γ-set of a graph G. Suppose for

every x ∈ D, γ(G − x) > γ(G), then D is the unique γ-set of G.

Lemma 1.2.12. [20] Let G be a graph which has a unique γ-set D.

Then for any x ∈ G − D, γ(G − x) = γ(G).

Lemma 1.2.13. [20] Let G be a graph with a unique γ-set D. Then

γ(G − x) ≥ γ(G) for all x ∈ D.

Lemma 1.2.14. [23] If G is a connected graph and γ(G) =⌊

n2

⌋

,

then there is at most one end vertex adjacent to each v ∈ V , except

possibly for one vertex which may be adjacent to exactly two end

vertices when n is odd.

Theorem 1.2.15. [23] A connected graph G satisfies γ(G) =⌊

n2

⌋

if and only if G ∈ G = ∪6i=1Gi.

Theorem 1.2.16. [38] For any two arbitrary graphs G and H and

for k ≥ 1, |V (G)| ≤ γ(kGoH) ≤ 2|V (G)|.

Theorem 1.2.17. [38] Let G and H be a connected graphs. If H

has a dominating vertex, then γ(kGoH) = 2|V (G)|.

Corollary 1.2.18. [38] Let D be the γ(kGoH)-set with |D| = n =

|V (G)| and γ(H) ≥ 2. Then k ≤ d(G), where d(G) is the domatic

number of G.

Theorem 1.2.19. [23] For any tree T, γ(T ) = n − ∆(G) if and

only if T is a wounded spider.

Chapter 2

Inverse Domination Number

The concept of inverse domination was introduced by V.R. Kulli and

A. Sigarkanti [28]. After introducing the concept, they attempted

to prove that for any graph G, γ ′(G) ≤ β0(G) where γ ′(G) is

the inverse domination number and β0(G) is the vertex indepen-

dence number of G. However the proof given by V.R. Kulli and A.

Sigarkanti was incorrect. Up to date, no proof of this result is known

and no counter example is known. This is known as Kulli-Sigarkanti

conjecture in the literature.

G. S. Domke, J. E. Dunbar and L. R. Markus [14] characterized

the graphs for which γ(G) + γ ′(G) = n. Further, they obtained a

lower bound for the inverse domination number of a tree and also

characterized trees which achieve this lower bound. Note that every

inverse dominating set lies in the complement of a dominating set.

From this, a dominating set and an inverse dominating set are dis-

joint dominating sets of G. For many years, researchers have studied

the existence of graphs having disjoint minimum dominating sets.

In this regard, most notable one is the study of various kinds of

domatic numbers, in which one seeks to partition the vertices of G

into a maximum number of various kinds of dominating sets. Also

11

12 CHAPTER 2. INVERSE DOMINATION NUMBER

the existence of two disjoint minimum dominating sets in trees was

first studied by D.W. Bange et al. [3]. Specifically the existence of

two disjoint minimum dominating sets was first studied by D.W.

Bange, A.E. Barkauskas and P.J. Slater [3] in 1978, who actually

studied the existence of two disjoint minimum dominating sets in

trees. In a related paper, T.W. Haynes and M.A. Henning [24] stud-

ied the existence of two disjoint minimum independent dominating

sets in a tree. For a survey of this literature, one can refer to B.

Zelinka [37].

It is well known by Ore’s Theorem [32] that if a graph G has

no isolated vertices, then the complement V − D of every minimal

dominating set D contains a dominating set. Thus every graph

without isolated vertices contains an inverse dominating set with

respect to a minimum dominating set and so the inverse domination

number exists. In this context, throughout this chapter, we assume

that all graphs have no isolated vertices and we consider classes of

graphs G for which γ(G) = γ ′(G). Hereafter G denotes a simple

graph on n vertices with no isolated vertices.

This chapter is organized as follows. In section 3.1, we are in-

terested in graphs G for which γ(G) = γ′

(G). In section 3.2, we

find the inverse domination number for some classes of graphs. In

section 3.3, we given some bounds for γ′

(G). In the final section

of this chapter, we characterize graphs G with γ(G) = γ′

(G) = n2

and also we characterize graphs G for which γ(G) = γ′

(G) = n−12 .

2.1. GRAPHS WITH γ(G) = γ′

(G) 13

2.1 Graphs with γ(G) = γ′(G)

In this section, we give some graphs for which γ(G) = γ ′(G) and

obtain some results on this aspect.

The following result can be proved without any difficulty.

Proposition 2.1.1. The following hold:

(i) For any integer n ≥ 2, γ(Kn) = γ′

(Kn) = 1;

(ii) For integers m, n ≥ 2, γ(Km,n) = γ ′(Km,n) = 2;

(iii) For integers m, n ≥ 2, γ(Km,n) = γ ′(Km,n) = 2.

Lemma 2.1.2. Let G be a graph with γ(G) = γ ′(G). Then G has

no γ- required vertex.

Proof. Let G be a graph with γ(G) = γ ′(G). Let D be a γ-set and

D′ be a γ′

-set of G. Suppose G contains a γ-required vertex u.

Then u lies in every γ-set of G and hence u ∈ D and u ∈ D′, which

is a contradiction to D′ ⊆ V − D.

Proposition 2.1.3. Let x be a dominating vertex of a graph G.

Then γ ′(G) = γ(G − x).

Proof. Since x is a dominating vertex of G, {x} is a γ-set of G.

Hence any γ′

-set of G lies in G−{x} and is a minimum dominating

set of G − {x}. Therefore γ ′(G) = γ(G − x).

Theorem 2.1.4. Let G be a graph such that G and G are connected

with at least two pendant vertices in G. Then γ(G) = γ′

(G) = 2.


Proof. Since G and G are connected graphs, ∆(G) ≤ n − 2 and

∆(G) ≤ n − 2. Therefore γ(G) ≥ 2 and γ′

(G) ≥ 2. Let {a, b}

be two pendant vertices in G. Let a′, b′ be the supports of a and b

respectively.

Case 1. When a′ = b′, D = {a, a′} is a γ-set of G. Since G

is connected, a′ is adjacent to some other vertex c in G. Hence

D′ = {b, c} is a γ′

- set of G. Thus γ(G) = γ′

(G) = 2.

Case 2. When a′ 6= b′, D = {a, a′} is a γ-set of G and D′ =

{b, b′} is a γ′

- set of G. Thus γ(G) = γ′

(G) = 2.

The following theorem gives a necessary and sufficient condition

for γ(Pn) = γ′

(Pn) where Pn is the path on n vertices.

Theorem 2.1.5. For any integer n ≥ 4, γ(Pn) = γ′

(Pn) = ⌈n3⌉ if

and only if n 6≡ 0(mod 3).

Proof. Let V (Pn) = {1, 2, . . . , n}. Assume that n ≥ 4 and n 6≡

0(mod 3). Then n = 3k + 1 or n = 3k + 2 for some positive integer

k > 0. When n = 3k + 1, the set D = {2, 5, 8, . . . , 3(k − 1) + 2, 3k}

is a γ-set with k + 1 = ⌈n3⌉ elements and D′ = {1, 4, 7, . . . , 3k + 1}

is a γ′

-set with k + 1 = ⌈n3⌉ elements. When n = 3k + 2, the set

D = {2, 5, 8, . . . , 3k+2} is a γ-set with k+1 = ⌈n3⌉ elements and the

set D′ = {1, 4, 7, . . . , 3k + 1} is a γ′

-set with k + 1 = ⌈n3⌉ elements.

Conversely assume that γ(Pn) = γ′

(Pn) = ⌈n3⌉. Suppose n ≡ 0(

mod 3). Then n = 3k for some positive integer k. The set D =

{2, 5, 8, . . . , 3k − 1} is a γ-set with k = n3 elements and D is the

only γ-set in Pn. The set D′ = {1, 4, 7, . . . , 3k − 2, 3k} is a γ′

-set

with k + 1 = n3 + 1 elements, which contradicts γ = γ

′

= ⌈n3⌉.

Hence n 6≡ 0(mod 3). Therefore γ(Pn) = γ′

(Pn) = ⌈n3⌉ if and only

if n 6≡ 0(mod 3).


(G) 15

Proposition 2.1.6. For an integer n ≥ 4, γ(Pn) = γ′

(Pn) = 2.

Proof. Since Pn has two pendant vertices, by Theorem 2.1.4, we

have γ(Pn) = γ′

(Pn) = 2.

Corollary 2.1.7. For an integer n ≥ 4, γ′

(Pn) = γ′

(Pn) if and

only if n = 4, 5.

Proof. By Proposition 2.1.6, we have γ′

(Pn) = 2. Again, by Theo-

rem 2.1.5, γ′

(Pn) = ⌈n3⌉ for n 6≡ 0 (mod 3) and γ

′

(Pn) = ⌈n3⌉ + 1

for n ≡ 0 (mod 3). Thus γ′

(Pn) = 2 if and only if n = 4, 5. Hence

γ′

(Pn) = γ′

(Pn) if and only if n = 4, 5.

Theorem 2.1.8. For an integer n ≥ 3, γ(Cn) = γ′

(Cn) = ⌈n3⌉.

Proof. Assume that V (Cn) = {1, 2, . . . , n}. When n = 3k, for some

k > 0, the set D = {1, 4, 7, . . . , 3k − 2} is a γ-set and the set

D′ = {2, 5, 8, . . . , 3k−1} is a γ′

- set both containing k = n3 elements.

When n = 3k + 1, the set D = {1, 4, 7, . . . , 3(k − 1) + 1, 3k} is

a γ-set and the set D′ = {2, 5, 8, . . . , 3k − 1, 3k + 1} is a γ′

-set

both containing k + 1 = ⌈n3⌉ elements. When n = 3k + 2, the set

D = {1, 4, 7, . . . 3k + 1} is a γ-set with k + 1 = ⌈n3⌉ elements and

D′ = {2, 5, 8, . . . 3k + 2} is a γ′

-set with k + 1 = ⌈n3⌉ elements each.

Thus in all the cases γ(Cn) = γ ′(Cn) = ⌈n3⌉.

Proposition 2.1.9. For an integer n ≥ 4, γ(Cn) = γ′

(Cn) = 2.

Proof. Let V (Cn) = {1, 2, . . . , n}. Then each vertex i in Cn is ad-

jacent to i − 1 and i + 1 modulo n. Therefore each vertex i in Cn

is adjacent to the remaining n − 3 vertices. Also i − 1 and i + 1

are adjacent in Cn. Hence D = {i, i + 1} is a γ-set of (Cn) and

D′ = {j, j + 1} is a γ′

-set of (Cn). Thus γ(Cn) = γ′

(Cn) = 2.


Corollary 2.1.10. γ′

(Cn) = γ′

(Cn) if and only if n = 4, 5, 6.

Proof. From Proposition 2.1.9 and Theorem 2.1.8, γ′

(Cn) = γ′

(Cn)

= ⌈n3⌉ = 2. Hence γ

′

(Cn) = γ′

(Cn) if and only if n = 4, 5, 6.

Theorem 2.1.11. Let T be a tree in which all the vertices are either

pendant vertices or their supports. Then T is a wounded spider if

and only if γ′

(T ) = ∆(T ).

Proof. Let T be a tree with n vertices and whose vertices are either

pendant vertices or their supports and let γ′

(T ) = ∆(T ). Let L

be the set of all pendant vertices of T and S be its neighbor set.

Then |S| ≤ |L|. By the assumption of T , V (T ) = S ∪ L. Hence

S is a γ-set of T and L is a γ′

-set of T . Since |S| + |L| = n, we

get γ(T ) + γ′

(T ) = n. This implies that γ(T ) = n − ∆(T ). By

Theorem 1.2.19, T is a wounded spider.

Conversely assume that T is a wounded spider and so γ(T ) = n−

∆(T )(Theorem 1.2.19). As discussed above, we have V (T ) = S ∪L

and so by Theorem 1.2.8, γ(T ) + γ′

(T ) = n. Hence γ′

(T ) =

∆(T ).

Corollary 2.1.12. Let G be a wounded spider on n vertices. Then

γ(G) = γ ′(G) if and only if ∆(G) = n2 .

Proof. Assume that γ(G) = γ ′(G). By Theorem 2.1.11, for a

wounded spider G, we have γ(G) + γ ′(G) = n and γ′

(G) = ∆(G).

Hence ∆(G) = n2 . On the reverse assume that ∆(G) = n

2 . We have

γ′

(G) = ∆(G). Then by Theorem 1.2.19, γ(G) = n−∆(G). Hence

γ(G) = γ ′(G).

2.2. BASIC RESULTS ON γ ′(G) 17

Theorem 2.1.13. Let d be a positive divisor of a positive integer

n. Then there exists a regular graph G on n vertices, for which

γ(G) = γ ′(G) = d.

Proof. Assume that d is a positive integer and divides n ≥ 1. i.e,

n = kd. Let V = ∪ki=1Vi, where Vi = {vi1, vi2, . . . , vid} for 1 ≤ i ≤ d.

Let G be the graph with vertex set as V and each vertex of Vi is

adjacent to exactly one vertex of Vj for j 6= i. Then d(v) = k − 1

for all v ∈ V and so G is k− 1 regular. Also each Vi is a γ-set of G.

Hence γ(G) = |Vi| = d and γ ′(G) = |Vj| = d.

Proposition 2.1.14. If a γ-set of a connected graph G of order

n ≥ 4 is a status, then γ(G) = γ ′(G) = 2.

Proof. Let S be a γ-set as well as a status in G. Trivially |S| ≥ 2

and any vertex of S is adjacent to every other vertex in V −S. Let

S = {u1, u2, ....uk} and let V − S = {v1, v2, . . . , vn−k}. Then D =

{u1, v1} is a γ-set of G and D′ = {u2, v2} is an inverse dominating

set of G with respect to D with |D′| = γ(G). i.e., D′ is a γ′

-set of

G. Hence γ(G) = γ ′(G) = 2.

2.2 Basic results on γ ′(G)

In this section, we find the inverse domination number for some

classes of graphs such as generalized corona, union and join of

graphs. Recall that, a k-corona kG ◦ H contains k copies of G

and |V (G)| copies of H with appropriate edges between each vertex

xji of the copy Gj and all of the vertices of the copy Hi[38].


Lemma 2.2.1. Let H be any graph. Then the following hold.

(i) γ(K1 ◦ H) = 1 and γ ′(K1 ◦ H) = γ(H);

(ii) γ(H ◦ K1) = γ ′(H ◦ K1) = |V (H)|;

(iii) If |V (H)| ≥ 2, then γ(mK1 ◦ H) = γ ′(mK1 ◦ H) = 2 for any

positive integer m ≥ 2;

(iv) If |V (H)| = n, γ(H ◦ Km) = γ ′(H ◦ Km) = n for any positive

integer m ≥ 1.

Proof. (i) Let V (H) = {h1, h2, . . . , hn} and V (K1) = {v}. Then {v}

is a γ- set of K1 ◦ H. Further, any γ′

-set of K1 ◦ H is a minimum

dominating set of (K1 ◦ H) − {v}, which is a γ-set of H. Hence

γ(K1 ◦ H) = 1 and γ′

(K1 ◦ H) = γ(H).

(ii) Let V (H ◦K1) = {h1, h2, . . . , hn, k1, k2, . . . , kn} where hi ∈ H

and ki is the vertex of K1 adjacent to hi for 1 ≤ i ≤ n. Since

each vertex in H ◦ K1 is either an end vertex or its support, D =

{h1, h2, . . . , hn} is a γ-set of H ◦ K1 and {k1, k2, . . . , kn} is a γ ′-set

of H ◦ K1. Thus γ(H ◦ K1) = γ ′(H ◦ K1) = |V (H)|.

(iii) Let V (H) = {h1, h2, . . . , hn} and {v1, v2, . . . , vm} be the

vertices of m copies of K1. Each vertex vi is adjacent to every

vertex of H and no two vertices in {v1, v2, . . . , vm} are adjacent.

Hence D = {h1, v1} is a γ-set of mK1 ◦ H and D′ = {h2, v2} is a

γ ′-set of mK1 ◦ H. Thus γ(mK1 ◦ H) = γ ′(mK1 ◦ H) = 2.

(iv) Let V (H ◦ Km) = {h1, h2, . . . hn} ∪ (⋃n

i=1{ki1, ki2, . . . , kim})

where hi ∈ H and kij’s are the vertices of the ith copy of Km for

1 ≤ i ≤ n and 1 ≤ j ≤ m. Note that the vertex kij dominates

the ith copy of Km and hi ∈ V (H) for i = 1 to n. Hence D =

{k1j, k2j, k3j, . . . , knj} is a dominating set of H ◦Km. Since |D| = n

and |V (H)| = n ≤ γ(H ◦ Km), D is a γ-set of G. Similarly, D′ =


{k1p, k2p, k3p, . . . , knp} is an inverse dominating set of H ◦ Km for

j 6= p and so a γ′

-set of G. Thus γ(H ◦ Km) = γ′

(H ◦ Km) = n

where n = |V (H)|.

Theorem 2.2.2. Let G be a graph and H be any graph with γ(H)

= 1. For a positive integer k, |V (G)| ≤ γ′

(kG ◦ H) ≤ 2|V (G)|.

Proof. By Theorem 1.2.17, we have γ(kG◦H) = |V (G)| = n. Since

γ(H) = 1, there exists a γ-set Di = {hi1} of the ith copy of H for

1 ≤ i ≤ n. Then D = ∪ni=1{hi1} is a dominating set of kG◦H. Since

γ(kG ◦ H) = |V (G)|, D is a γ-set of kG ◦ H. Since γ(kG ◦ H) ≤

γ′

(kG ◦H), we have |V (G)| ≤ γ′

(kG ◦H). Now, for each i, choose

hi2 ∈ Hi, and xji ∈ Gj. Then D′ = (∪n

i=1{hi2}) ∪ (∪ni=1{x

ji}) is

an inverse dominating set with respect to D with 2|V (G)| vertices

and so γ′

(kG ◦ H) ≤ 2|V (G)|. Thus |V (G)| ≤ γ′

(kG ◦ H) ≤

2|V (G)|.

Remark 2.2.3. Theorem 2.2.2 gives an upper and lower bound for

γ′

(kG ◦H) where G is any graph, H is a graph with γ(H) = 1 and

k is a positive integer. We shall observe that both the bounds are

sharp.

Actually note that the lower bound is obtained when k = 1.

That is γ(G ◦ H) = γ′

(G ◦ H) = |V (G)|. For, when k = 1,

D = (∪ni=1{hi1}) is a γ- set of G ◦ H where {h1} is the γ- set of H.

and D′ = V (G) is an inverse dominating set of G ◦ H. Therefore

γ′

(G ◦H) ≤ |V (G)|. Also we have |V (G)| ≤ γ′

(G ◦H), and hence

γ′

(G ◦ H) = |V (G)|. Therefore we get D′ = V (G) is the γ′

- set of

G ◦ H. Hence γ(G ◦ H) = γ′

(G ◦ H) = |V (G)|.


Theorem 2.2.4. Let G and H be two connected graphs with γ′

(H)

= 1. Then γ(kG ◦ H) = γ′

(kG ◦ H) = |V (G)|.

Proof. Since γ(H) ≤ γ′

(H), we have 1 ≤ γ(H) ≤ γ′

(H) = 1,

and so γ(H) = 1. Let V (H) = {h1, h2, . . . , hm} and V (Hi) =

{hi1, hi2, . . . , him} for i = 1 to n. When D = {h1} is a γ-set of H

and D′ = {h2} is a γ′

-set of H, D = ∪ni=1{hi1} is a dominating set

of (kG ◦ H). This gives that γ(kG ◦ H) ≤ |V (G)| and by Theorem

1.2.16, we have |V (G)| ≤ γ(kG ◦H). Hence D is a γ set of kG ◦H

and D ′ = ∪ni=1{hi2} is an inverse dominating set of kG ◦ H. From

this we have, γ′

(kG ◦ H) ≤ |V (G)|. By Theorem 2.2.2, we have

|V (G)| ≤ γ′

(kG ◦ H) and hence D′ is a γ′

-set of kG ◦ H, with

|V (G)| vertices. Hence γ(kG ◦ H) = γ′

(kG ◦ H) = |V (G)|.

Theorem 2.2.5. Let n ≥ 2 and m ≥ 1. For any integer k ≥

2, γ′

(kKn ◦ K1,m) = 2|V (G)| = 2n.

Proof. Let V (Kn) = {x1, x2, . . . xn}, n ≥ 2. Let Kjn be the jth copy

of the graph Kn and V (Kjn) = {xj

1, xj2, . . . x

jn}, where xj

i corresponds

to the vertex xi ∈ V (Kn). Let V (K1,m) = {v, h1, h2, . . . hm} where v

is the vertex with d(v) = m. Note that γ(K1,m) = 1 and {v} is the

γ-set of K1,m. Suppose Di = {vi} is a γ-set of ith copy of K1,m for

1 ≤ i ≤ n. Then D = ∪ni=1{vi} is a dominating set of kKn ◦ K1,m.

By Theorem 1.2.17, γ(kG ◦ H) = |V (G)| = n. Hence D is a γ-set

of kKn ◦K1,m and so D′ = {x11, x

12, . . . x

1n}∪ (∪n

i=1{hi1}) is an inverse

dominating set of kKn ◦K1,m. By the definition of kKn ◦K1,m and

γ′

(H) 6= 1, we need at least n vertices to dominate the vertices

of the ith copy of K1,m. Since E(Gj) = ∅, we need another set

of at least n vertices to dominate the vertices of the Kjn’s. Hence

γ′

(kKn ◦ K1,m) ≥ 2n. From this D′ is a γ′

-set of kKn ◦ K1,m and

hence γ′

(kG ◦ H) = 2|V (G)| = 2n.


Suppose D1 and D2 are γ-sets of G1 and G2 respectively. Then

D = D1 ∪D2 is a γ-set of G1 ∪G2. In view of this fact, we have the

following theorem.

Theorem 2.2.6. For any two graphs G1 and G2,

γ′

(G1 ∪ G2) = γ′

(G1) + γ′

(G2).

Proposition 2.2.7. Let G1 and G2 be two graphs and G = G1+G2.

Then γ(G) = γ ′(G) = 1, if at least one of the following holds.

(i) γ(G1) = γ(G2) = 1;

(ii) γ(G1) = γ′

(G1) = 1;

(iii) γ(G2) = γ′

(G2) = 1.

Proof. Let {u1} and {u2} be the γ-sets of G1 and G2 respectively.

Then {u1} is a γ-set of G and {u2} is the γ′

-set of G. Hence

γ(G) = γ(G) = 1. Let {u1} and {v1} be the γ and γ′

- sets of of

G1. Then {u1} is a γ-set of G and {v1} is a γ′

-set of G. Hence

γ(G) = γ ′(G) = 1. Similar is the proof for Case (iii).

Proposition 2.2.8. Let G1 and G2 be two graphs with γ(Gi) ≥ 2

for i = 1, 2. Then γ(G1 + G2) = γ ′(G1 + G2) = 2.

Proof. In G, each vertex of G1 is adjacent to every vertex of G2 and

vice versa. Hence a set containing one vertex of G1 and one vertex

of G2 is a γ-set of G and also a γ′

-set of G. Thus γ(G) = γ ′(G) =

2.

Now we have the following corollary.

Corollary 2.2.9. Let Gi be a graph with ni ≥ 2 vertices for 1 ≤

i ≤ r. Let G = G1 + G2 + · · ·Gr. Then γ(G) = γ ′(G) = 2.


Theorem 2.2.10. Let G be a complete k-partite graph with partite

sets V1, V2, . . . , Vk and |Vi| ≥ 2 for i = 1 to k. Then γ(G) =

γ ′(G) = 2.

Proof. Since |Vi| ≥ 2 for i = 1 to k, let x1, y1 ∈ V1 and x2, y2 ∈ V2.

Then D = {x1, x2} is a γ-set of G and D′ = {y1, y2} is a γ′

-set of

G. Therefore γ(G) = γ ′(G) = 2.

2.3 Bounds for γ′(G)

In this section, we given some bounds for γ′

(G). Moreover, we

prove an important result which says that γ′

(G + e) ≤ γ′

(G) for

e ∈ E(G) (refer Theorem 2.3.7). First of all, we start with a lower

bound for γ ′(G) whenever γ(G) = γ ′(G).

Theorem 2.3.1. Let G be a connected graph with n vertices, m

edges and γ(G) = γ ′(G). Then γ ′(G) ≥ 13(2n − m).

Proof. Let D be a γ-set and D′ be a γ′

-set of G. Note that every

vertex in V − (D ∪ D′) has at least one neighbor in D and one

neighbor in D′, every vertex in D has a neighbor in D′, and every

vertex in D′ has a neighbor in D. Thus the number of edges in G

is at least 2(V − (D ∪ D′)) + |D′|. That is m ≥ 2(n − 2γ′

) + γ′

and so m ≥ 2n − 3γ ′(G). Therefore γ ′(G) ≥ 13(2n − m).

Corollary 2.3.2. Let T be a tree with n vertices. Then γ′

(T ) ≥n+1

3 .

Proof. Note that T has m = n− 1 edges. Applying Theorem 2.3.1,

we get that γ′

(T ) ≥ n+13 .

2.3. BOUNDS FOR γ′

(G) 23

Proposition 2.3.3. For any (m, n) graph G, n−m ≤ γ ′(G) ≤ m.

Proof. For any (m, n) graph G, we have n − m ≤ γ(G) ≤ γ ′(G).

Also γ ′(G) ≤ n − γ(G). Thus γ ′(G) ≤ m.

Proposition 2.3.4. For any graph G,

⌈n

1 + ∆(G)⌉ ≤ γ ′(G) ≤ ⌊

n∆(G)

1 + ∆(G)⌋.

Proof. By Theorem 1.2.3, ⌈ n1+∆(G)⌉ ≤ γ(G) ≤ γ ′(G). On the other

hand γ ′(G) ≤ n − γ(G). These imply that γ ′(G) ≤ ⌊ n∆(G)1+∆(G)⌋.

Proposition 2.3.5. Let G be a graph with (d1, d2, . . . dn) and di ≥

di+1 for 1 ≤ i ≤ n. Then γ ′(G) ≤ max {n− k : k + (d1 + d2 + . . . +

dk) ≥ n}.

Proof. Let G be a graph with degree sequence (d1, d2, . . . dn) with

di ≥ di+1 for 1 ≤ i ≤ n. Then by Theorem 2.12 [23], we have γ(G) ≥

min {k : k + (d1 + d2 + . . . + dn) ≥ n}. Now γ ′(G) ≤ n − γ(G)

implies that γ ′(G) ≤ max {n−k : k +(d1 +d2 + . . .+dk) ≥ n}.

Theorem 2.3.6. Let G be a graph and uv = e ∈ E(G). If γ(G +

e) < γ(G), then there exists a γ-set D of G containing both u and

v such that at least one of them has no private neighbor other than

itself.

Proof. Assume that there exists no γ-set G containing both u and

v. Let D be a γ-set such that u /∈ D. Then there exists a vertex

u1 ∈ N(u) which has a private neighbor other than u that lies in

D to dominate u. Hence u1 lies in a γ-set of G + e also. Therefore


γ(G + e) is in no way smaller than γ(G), which is a contradiction.

Hence there exists a γ-set containing both u and v.

Suppose both of them have a private neighbor other than itself

with respect to any γ-set D, then neither D−{u} nor D−{v} is a γ-

set of G+e. Also no set with fewer than γ(G) elements can dominate

G + e. Therefore γ(G + e) ≥ γ(G), which is a contradiction.

Theorem 2.3.7. Let G be a graph and uv = e ∈ E(G). Then

γ′

(G + e) ≤ γ ′(G).

Proof. Let D be a γ-set and D′ be a γ′

-set of G with respect to D.

Then γ(G + e) ≤ γ(G) for any e ∈ E(G).

Case 1. Let uv = e ∈ E(G) be such that γ(G + e) = γ(G).

Then D is a γ-set of G + e. Hence D′ is an inverse dominating set

of G + e with respect to D. Therefore γ′

(G + e) ≤ |D′| = γ ′(G).

Case 2. Let xy = e ∈ E(G) be such that γ(G + e) < γ(G).

Then by Theorem 2.3.6, there exists a γ-set D of G containing both

x and y such that at least one of them, say x, has no private neighbor

other than itself. Note that D1 = D − {x} is a γ-set of G + e and

D1′

= D′ is an inverse dominating set of G + e with respect to D1.

Therefore γ′

(G + e) ≤ |D′| = γ ′(G).

Remark 2.3.8. (i). Let uv = e ∈ E(G). Then γ ′(G) ≤ γ′

(G− e).

(ii). Let H be a spanning sub-graph of a graph G. Then γ ′(G) ≤

γ′

(H).

Proposition 2.3.9. Let G be a Hamiltonian graph on n vertices.

Then γ ′(G) ≤ ⌈n3⌉.


(G) =⌊

N

2

⌋

25

Proof. Let G be Hamiltonian. Then G contains a spanning cycle

say Cn. By Theorem 2.1.8, we have γ(Cn) = γ′

(Cn) = ⌈n3⌉. By

Remark 2.3.8, we have γ ′(G) ≤ γ′

(Cn) = ⌈n3⌉.

Corollary 2.3.10. Let G be a graph on n ≥ 3 vertices. If δ(G) ≥ n2 ,

then γ ′(G) ≤ ⌈n3⌉.

Proof. Let G be a connected graph on n ≥ 3 vertices. If δ(G) ≥ n2 ,

then by Theorem 4.3 [5], we have G is Hamiltonian and thus by

Remark 2.3.8, γ ′(G) ≤ ⌈n3⌉.

Corollary 2.3.11. Let G be a connected graph on n ≥ 3 vertices.

If C(G) is complete, then γ ′(G) ≤ ⌈n3⌉.

Proof. Let G be a connected graph on n ≥ 3 vertices. If C(G)

is complete, then by Corollary 4.4[5], G is Hamiltonian. By Re-

mark 2.3.8, γ ′(G) ≤ ⌈n3⌉.

Proposition 2.3.12. For any graph G with no isolated vertices,

γ ′(G) ≤ ǫf(G), where ǫf(G) denotes the maximum number of pen-

dant edges in a spanning forest of G.

Proof. By Theorem 1.2.4, γ(G) + ǫf(G) = n for any graph G. Thus

ǫf(G) = n − γ(G) ≥ γ ′(G).

2.4 Graphs with γ(G) = γ′(G) =

⌊

n2

⌋

In this section, we present the characterization of graphs for which

γ(G) = γ′

(G) =⌊

n2

⌋

. First of all, we present characterization of

graphs with γ(G) = γ′

(G) = n2 . Recall that T.W Haynes et al. [23]

have characterized the class of graphs with γ(G) = n2 .


Theorem 2.4.1. Let G be a connected graph with n vertices. As-

sume that n is even and δ(G) = 1. Then γ(G) = γ ′(G) = n2 if and

only if G = H ◦ K1 for some connected graph H on n2 vertices.

Proof. Let G = H ◦K1, where H is a connected graph on n2 vertices.

Then trivially V (H) is a γ-set for G and hence γ(G) = n2 . Note that

the set of all pendant vertices in G forms an inverse dominating set

of G with minimum cardinality. Hence γ′

(G) = n2 .

On the converse, assume that γ(G) = γ′

(G) = n2 . Let L = {v :

d(v) = 1}. Since δ(G) = 1, we get that L 6= ∅ and let S = N(L).

Since γ(G) + γ′

(G) = n, by Theorem 1.2.8, we have (i) V − S

independent and (ii) for every x ∈ V − (S ∪L), each vertex in N(x)

is adjacent to at least two pendant vertices.

Let S1 be the set of all vertices in S, which are adjacent to exactly

one pendant vertex in L and let S2 = S − S1.

Case 1. Assume that V − (S ∪ L) = ∅. Then every vertex of

G is either a pendant vertex or adjacent to a pendant vertex. Thus

S is a dominating set of G and no set with fewer than |S| elements

can dominate S. Also |S| ≤ |L|. Hence S is a γ-set of G. Since

γ(G) = γ ′(G) = n2 , we have |S| = n

2 and |L| = n2 . Hence each

vertex of S is adjacent to exactly one pendant vertex of L. Take

H = 〈S〉. Then G = H ◦ K1.

Case 2. Assume that V − (S∪L) 6= ∅. Let N1 = N(S)− (S∪L)

and N2 = V − (N1 ∪ (S ∪ L)).

Claim 1. N2 = ∅. By Theorem 1.2.8, we have V − S is inde-

pendent. Hence every vertex of G is either in S or adjacent to some

vertex in S. Hence there is no vertex in N2 and so S is a γ-set of

G.


(G) =⌊

N

2

⌋

27

Claim 2. N1 = ∅. For let L1 = N(S1) ∩ L and L2 = N(S2) ∩ L.

Then |S1| = |L1| and |S2| < |L2|. If N1 6= ∅, by Theorem 1.2.8,

for each x ∈ N1, each stem in N(x) is adjacent to at least two

pendant vertices. Suppose S2 6= ∅. Since S is a γ-set and V − S is

a γ′

-set, we have L is contained a γ′

-set with |L| > |S|. This is a

contradiction to γ(G) = γ ′(G). Hence S2 = ∅ and N1 = ∅. Thus

V − (S ∪L) = ∅. Hence by Case (i) we get that G = H ◦K1 where

H = 〈S〉.

Theorem 2.4.2. For a connected graph G with even number of

vertices n and δ(G) ≥ 2, γ(G) = γ′

(G) = n2 if and only if G = C4.

Proof. When G = C4, trivially we have γ(G) = γ′

(G) = n2 .

Conversely assume that γ(G) = γ′

(G) = n2 . Then γ(G) +

γ ′(G) = n as n is even. Since G is a graph with δ(G) ≥ 2, by

Theorem 1.2.7, G is nothing but C4.

Remark 2.4.3. Let G1, G2, . . . Gk be the k connected components

of a graph G. Let D1, D2, . . . Dk be γ-sets and D′

1, D′

2, . . . D′

k be

γ′

-sets of G1, G2, . . . , Gk. Then D1∪D2∪. . .∪Dk is a γ-set of G and

D1′

∪D2′

∪ . . .∪Dk′

is a γ′

-set of G. Therefore γ(G) =∑k

i=1 γ(Gi)

and γ ′(G) =∑k

i=1 γ′

(Gi).

Proposition 2.4.4. Let G1, G2, . . . , Gk be the k connected compo-

nents of a graph G. Then γ(G) = γ′

(G) if and only if γ(Gi) =

γ′

(Gi) for i = 1 to k.

Proof. Let G1, G2, . . . , Gk be the k connected components of G.

Then we have γ(G) =∑k

i=1 γ(Gi) and γ ′(G) =∑k

i=1 γ′

(Gi). Thus

trivially, γ(G) = γ′

(G) if γ(Gi) = γ′



Conversely assume that γ(G) = γ′

(G). We have γ(Gi) ≤ γ′

(Gi)

for i = 1 to k. Suppose γ(Gi) < γ′

(Gi) for some i, then we must

have γ(Gj) > γ′

(Gj) for some j 6= i, which is impossible. Hence

γ(Gi) = γ′


Corollary 2.4.5. Let G be a graph with n vertices with connected

components G1, G2, . . . , Gk and |V (Gi)| = ni for 1 ≤ i ≤ k. Then

γ(G) = γ′

(G) =⌊

n2

⌋

if and only if γ(Gi) = γ′

(Gi) =⌊

ni

2

⌋

for

i = 1, 2, . . . , k. In particular, n is even if and only if each ni is

even.

Proof. First assertion follows from Proposition 2.4.4.

Let n be even. Since γ(G) = γ′

(G) =⌊

n2

⌋

, we have γ(G) +

γ′

(G) = n and so we must have γ(Gi) + γ′

(Gi) = ni and γ(Gi) =

γ′

(Gi) =⌊

ni

2

⌋

, for each i. This implies that each ni must be even.

Conversely if each ni is even, then n is even.

Corollary 2.4.6. Let G be a graph with even number of vertices

and no isolated vertices. Then γ(G) = γ ′(G) = n2 if and only if the

connected components Gi of G are either the cycle C4 or the corona

Gi = Hi ◦ K1 for some connected graph Hi.

Proof. Assume that γ(G) = γ ′(G) = n2 . Let G1, G2, . . . Gk be the

connected components of G and |V (Gi)| = ni for 1 ≤ i ≤ k. Since

n is even and γ(G) = γ ′(G) = n2 , in view of Corollary 2.4.5, we

get γ(Gi) = γ′

(Gi) = ni

2 for each i. Now by Theorems 2.4.1 and

2.4.2, we see that γ(G) = γ ′(G) if and only if the components Gi

of G are either the cycle C4 or the corona Gi = Hi ◦ K1 for some

connected graph Hi.


(G) =⌊

N

2

⌋

29

Figure 2.1: Class A

Figure 2.2: Class B

Remark 2.4.7. Let G be a graph with even number of vertices, no

isolated vertices and γ(G) = γ ′(G) = n2 . Then the components of

Gi of G, with δ(Gi) ≥ 2 are the cycle C4 and the components Gi of

G, with δ(Gi) = 1 are the corona Gi = Hi ◦ K1 for some connected

graph Hi.

We now turn our attention to graphs G with γ(G) = γ ′(G) =n−1

2 . We give the complete characterization of graphs which satisfy

the above condition. Let A be the set of all graphs given in Fig-

ure 2.1. Let B be the set of all graphs given in Figure 2.2. Let

Q1 = A ∪ B. For any graph H, let S(H) denotes the set of all

connected graphs, each of which can be formed from H ◦ K1 by

adding a new vertex x and edges joining x to two or more vertices

of H. Then define Q2 =⋃

H S(H) where the union is taken over all

graphs H. The graphs in Q2 are of the form given in Figure 2.3.


Figure 2.3: Class Q2


Consider a new vertex a and a copy of C4. For a graph G ∈ Q2,

let θ(G) be the graph obtained by joining G to C4 with the single

edge xa where x is the new vertex added in forming G. Then define

Q3 = {θ(G)/G ∈ Q2}. The graphs in Q3 are of the form given in

Figure 2.4.

Let u, v, w be a vertex sequence of the path P3. For any graph

H, let P (H) be the set of all connected graphs which are formed

from H ◦ K1 by joining each of u and w to one or more vertices of

H. Then define Q4 =⋃

H P (H). Any graph in Q4 is of the form

given in Figure 2.5.



(G) =⌊

N

2

⌋

31


Let H be a graph and X ∈ B. Let R(H, X) be the set of con-

nected graphs which may be formed from H ◦ K1 by joining each

vertex of U ⊆ V (X) to one or more vertices of H such a way that

no set with fewer than γ(X) vertices of X dominates V (X) − U .

Then define Q5 = ∪H,XR(H, X). Any graph G in Q5 is of the form


Lemma 2.4.8. For all graphs G in Q1, Q2, Q3, Q4, Q5, we have

γ(G) = γ ′(G) = n−12 .

Proof. D = {1, 3, 6} and D′ = {2, 5, 7} form the γ-set and γ′

-set

respectively for all the graphs in A with n−12 vertices each. For C3

in B, the set D = {1} and D′ = {2} are the γ and γ′

-sets withn−1

2 vertices each. D = {1, 4} and D′ = {3, 5} are the γ and γ′

-sets for all the other graphs in B with n−12 vertices each. Thus

γ(G) = γ ′(G) = n−12 for all the graphs in Q1.

Consider the class Q2 and let G ∈ Q2. Let x be adjacent to y

and z in H. Now, y and z are adjacent to the pendant vertices

y1 and z1 respectively. Since there are n−12 pendant vertices, which

are adjacent to distinct n−12 vertices of H, any γ-set of G contains

at least n−12 vertices. Further, the set D = V (H) ∪ {y1} − {y} is

a dominating set containing n−12 elements and hence it is a γ-set.

D′ = V (G)−D ∪{x} is a dominating set in V −D, containing n−12


vertices. Hence D′ is a γ′

-set and thus γ(G) = γ ′(G) = n−12 for all

G ∈ Q2.

Similarly D = V (H)∪{b, d} is a γ-set and D′ = V (G)−(D∪{x})

is a γ′

-set for G ∈ Q3. Thus, γ(G) = γ ′(G) = n−12 for all G ∈ Q3.

Again D = V (H) ∪ {u} is a γ-set with n−12 vertices and D′ =

V (G) − (D ∪ {w}) is a γ′

-set containing n−12 vertices. Thus γ =

γ′

= n−12 for all G ∈ Q4.

Let |V (H)| = m and |V (X)| = p. Then |V (R(H, X))| = 2m +

p = n. Let DX and D′X be the γ-set and γ

′

- set of the graph

X respectively. Then D = V (H) ∪ DX is a dominating set and

no set with fewer vertices dominates G and hence D is a γ-set.

D′ = D′X together with all the pendant vertices forms a γ′-set of

R(H, X). Hence |D| = m + p−12 = 2m+p−1

2 = n−12 and similarly

|D′| = n−12 . Hence γ(G) = γ ′(G) = n−1

2 for all G ∈ Q5. Thus

γ(G) = γ ′(G) = n−12 for all graphs in G = ∪5

i=1Qi.

Remark 2.4.9. Note that the classes of graphs Q1, Q3, Q4 and Q5

are the same as the classes of graphs G2, G4, G5 and G6 respectively

and Q2 is a subclass of the class G3 given in Theorem 1.2.15.

Theorem 2.4.10. A connected graph G with n vertices satisfies

γ(G) = γ′

(G) = n−12 if and only if G ∈ Q = ∪5

i=1Qi.

Proof. Let G ∈ Q = ∪5i=1Qi. By Lemma 2.4.8, γ(G) = γ ′(G) =

n−12 . Conversely, suppose that γ(G) = γ ′(G) = n−1

2 . Note that n

is odd. Now by Theorem 1.2.15, G ∈ G = ∪6i=1Gi. Since n is odd,

G cannot be in G1. In view of Remark 2.4.9, G ∈ Q1 or Q3 or Q4

or Q5 and G ∈ G3. It is enough to prove that whenever G ∈ G3,

we get G ∈ Q2. If G ∈ Q3, let S be the set of end vertices in G

and T be the set of neighbors of vertices in S. If |T | = t, then by


(G) =⌊

N

2

⌋

33

Lemma 1.2.14, |S| = t or |S| = t + 1. Hence there is a γ-set of G

containing T . Let G′

= G − (S ∪ T ).

Case 1. If |S| = t + 1, then one can show that G′ = ∅ as in

the course of proof of Theorem 1.2.15. Hence we get γ = t and

γ′

= t + 1, a contradiction. Therefore |S| 6= t + 1.

Case 2. If |S| = t, then either G′

contains isolated vertices or

G′ contains no isolated vertices.

Sub Case 2.1. If G′

contains isolated vertices, then let y be an

isolated vertex of G′. As in the proof of Theorem 1.2.15, one can

prove that G′

− y is empty. Since y is not a pendant vertex of G,

y is adjacent to two or more vertices of T . Hence G ∈ Q2.

Sub Case 2.2. If G′

contains no isolated vertices, then as in the

above mentioned proof G /∈ Q3. If G ∈ G4 then G ∈ Q3. If G ∈ G5

then G ∈ Q4. If G ∈ G6, then G ∈ Q5. Thus G ∈ ∪5i=1Qi.

Corollary 2.4.11. Let G be a graph with odd order n and δ(G) = 1.

Then γ(G) = γ ′(G)= n−12 if and only if the components Gi of G

are either C4 or Gi = Hi ◦K1 for some connected graph Hi together

with exactly one component Gj ∈ ∪5i=1Qi.

Proof. Suppose that the connected components Gi of G are either

the cycle C4 or the corona Gi = Hi ◦ K1 for some connected graph

Hi together with exactly one component of Gj ∈ ∪5i=1Qi. Then

by Corollary 2.4.6, Remark 2.4.7 and Lemma 2.4.8, we get that

γ(G) = γ ′(G) = n−12 .

Conversely assume that γ(G) = γ ′(G) = n−12 . Let G1, . . . , Gk

be the connected components of G and |V (Gi)| = ni. Since γ(G) =

γ ′(G) = n−12 , we see that γ(G) + γ ′(G) = n − 1 and that there is


exactly one vertex which is neither in the γ -set nor in the γ′

-set of

G. Since n is odd, there exists at least one component say Gk such

that nk is odd. In view of Proposition 2.4.4, and nk being odd, we

get that γ(Gk) = γ′

(Gk) = nk−12 and in the component Gk, there is

one vertex which is neither in the γ-set nor in the γ′

- set of G.

Suppose that there exists an odd component say Gt. Then we get

one more vertex which is neither in the γ-set nor in the γ′

-set of G,

which leaves at least two vertices that are neither in the γ-set nor in

the γ′

-set of G, a contradiction. Hence we see that each nj is even

for j 6= k. Then γ(Gk) = γ′

(Gk) = nk−12 and γ(Gj) = γ

′

(Gj) =nj

2 ,

for j 6= k. Thus by Corollary 2.4.6, the even components Gj, j 6= i

are the cycle C4 or the corona H ◦K1. Moreover by Theorem 2.4.10,

the only odd component is one of the graphs in ∪5i=1Qi. That is, we

get Gj ∈ ∪5i=1Qi. Hence the components of G are the cycle C4 or

the corona G = H ◦ K1 for any connected graph H together with

exactly one component Gj ∈ ∪5i=1Qi.

Remark 2.4.12. Let G be a graph with odd number of vertices

and no isolated vertices. Let γ(G) = γ ′(G) = n−12 with δ(G) ≥ 2.

Then the components of G are the cycle C4 together with exactly

one component Gj where Gj ∈ Q1.

Chapter 3

Sum of γ and γ ′

In this chapter, we deal about results concerning sum of domination

and inverse domination numbers. A Gallai-type theorem has the

from α(G)+β(G) = n, where α(G) and β(G) are parameters defined

on the graph G and n is the number of vertices in G. E. J. Cockayne

et al. [9] proved certain Gallai-type theorems for graphs. In the year

1996, E. J. Cockayne et al. characterized graphs with δ(G) ≥ 2 and

γ(G) = ⌊n2⌋. Since then E. J. Cockayne et al. [10] and B. Randerath

et al. [34] independently characterized all graphs satisfying γ(G) =⌊

n2

⌋

. Next in the year 2003, G. S. Domke et al. [14] characterized

graphs for which γ(G) + γ ′(G) = n. Also note that in chapter 3,

we characterized graphs with γ(G) = γ ′(G) = n−12 where n is an

odd positive integer. In this chapter, we characterize all graphs G

with δ(G) ≥ 2 for which γ(G) + γ ′(G) = n − 1.

In analogous to the inverse domination number, S. M. Hedet-

niemi et al. [25] defined and studied the disjoint domination num-

ber γγ(G) of a graph G. A pair (D1, D2) of disjoint sets of ver-

tices D1, D2 ⊆ V is said to dominate a vertex u ∈ V , if D1 and

D2 dominate u. Further (D1, D2) is a dominating pair, if (D1, D2)

dominates all vertices in V . The total cardinality of a pair (D1, D2)

35

36 CHAPTER 3. SUM OF γ AND γ ′

is |D1| + |D2| and the minimum cardinality of a dominating pair is

the disjoint domination number γγ(G) of G. As mentioned earlier,

by Ore’s observation, γγ(G) ≤ |V (G)| for every graph G without

isolated vertices and S. M. Hedetniemi et al. characterized all the

extremal graphs for this bound. In this connection, the existence

of two disjoint minimum dominating sets in trees was first studied

by D. W. Bange et al. [3]. In a related paper, T. W. Haynes and

M. A. Henning [24] studied the existence of two disjoint minimum

independent dominating sets in a tree.

Another application of finding two disjoint γ-sets is the one in

respect of networks. In any network (or graphs), dominating sets are

central sets and they play a vital role in routing problems in parallel

computing [29]. Also finding efficient dominating sets is always

concerned in finding optimal central sets in networks [12]. Suppose

S is a γ-set in a graph (or network) G, when the network fails in

some nodes in S, the inverse dominating set in V −S will take care

of the role of S. In this aspect, it is worthwhile to concentrate on

dominating and inverse dominating sets. Note that γ ′(G) ≥ γ(G).

From the point of networks, one may demand γ ′(G) = γ(G), where

as many graphs do not enjoy such a property. For example consider

the star graph K1,n. Clearly γ(K1,n) = 1 where as γ ′(K1,n) = n.

Suppose if we consider the graph G = K1,n2K2 with n ≥ 3, then

γ(G) = 2 and γ ′(G) = n. In both the cases if n is large, then γ ′(G)

is sufficiently large when compared to γ(G).

In this chapter we characterize all graphs G with δ(G) ≥ 2 for

which γ(G)+γ ′(G) = n−1. In this regard, it may be possible that

γ ′(G) is larger than γ(G) and γ(G)+γ ′(G) = n−1. But we prove

that graphs G with γ(G) + γ ′(G) = n − 1 are having exactly two

3.1. GRAPHS WITH δ ≥ 2 AND γ + γ ′ = N − 1 37

disjoint minimum dominating sets. Let us first recall, the following

characterizations of graphs for which γ(G) + γ ′(G) = n.

Theorem 3.0.13. [14, Theorem 2] Let G be a connected graph on

n vertices with δ(G) ≥ 2. Then γ(G) + γ ′(G) = n if and only if

G = C4.

Theorem 3.0.14. [14, Theorem 3] Let G be a connected graph on

n vertices with n ≥ 1 and δ(G) = 1. Let L ⊆ V be the set of all

degree one vertices and S = N(L). Then γ(G) + γ ′(G) = n if and

only if the following two conditions hold:

(i) V − S is an independent set and

(ii) for every vertex x ∈ V − (S ∪ L), every stem in N(x) is

adjacent to at least two leaves.

3.1 Graphs with δ ≥ 2 and γ + γ ′ = n − 1

Tamizh Chelvam and Grace Prema [35] characterized graphs for

which γ(G) = γ ′(G) = n−12 . In this context, we attempt to charac-

terize graphs G with δ(G) ≥ 2 for which γ(G) + γ ′(G) = n− 1. To

attain this aim, we first present the theorem which is useful in the

further discussion. To prove the following theorem we prefer case

by case analysis, since no better proof technique is available.

Theorem 3.1.1. Let G be a connected graph on n vertices with

δ(G) ≥ 2 and γ(G) + γ ′(G) = n − 1. Then γ(G) = γ ′(G).

Proof. Let D be a γ-set of G and D′ be a γ ′-set of G with respect

to D. Note that γ(G) ≤ γ ′(G). Assume that γ(G)+γ ′(G) = n−1

and let V (G) − (D ∪ D′) = {w}. Let S ⊆ D be those vertices that


are adjacent to more than one vertex in D′. Suppose γ(G) < γ ′(G).

Then |D| < |D′| and so S 6= ∅. Let S ′ = N(S) ∩ D′.

Claim 1. There is at most one vertex in S ′ which is adjacent to

a vertex in D − S.

Suppose not, there exist at least two vertices t′, r′ in S ′ and

t, r ∈ D − S such that t′ is adjacent to t and r′ is adjacent to r.

Then either both t, r ∈ D − S are adjacent to w or at least one of

t, r is not adjacent to w.

Suppose both t, r ∈ D − S are adjacent to w. Since t′, r′ are the

only vertices in V (G)−(D∪{w}) which are adjacent to t, r and t′, r′

are dominated by some vertices in S, D1 = D ∪ {w} − {t, r} ⊂ D

is a dominating set of G which is a contradiction to the fact that D

is a γ-set of G.

Suppose at least one of them, say t, is not adjacent to w. Since

t ∈ D−S and δ(G) ≥ 2, t is adjacent to a vertex u in D. Therefore

D1 = D − {t} is a dominating set of G which is a contradiction to

the fact that D is a γ-set of G. Hence, at most one vertex t′ ∈ S ′

which is adjacent to a vertex in D − S. By similar argument as

given above, one can prove that t′ is adjacent to exactly one vertex

in D − S. Let us take

S ′1 =

S ′ − {t′} if t′ exists

S ′ otherwise

Note that each vertex in S has at least two neighbors in S ′ and

so S ′1 6= ∅.

Claim 2. S ′1 is independent.

Suppose there exists a vertex x′ ∈ S ′1 which is adjacent to y′ ∈ S ′

1.

Suppose w is not adjacent to both x′ and y′. By the fact that each


vertex in S has at least two neighbors in D′ and Claim 1, D′−{x′}

is a γ ′-set of G, a contradiction. If w is adjacent to one of x′ or

y′, say x′, then D′ − {y′} is a γ ′-set of G which is a contradiction.

Hence S ′1 is independent.

Now we have the following three possibilities.

1. w is not adjacent to any of the vertices in S ′1.

2. w is adjacent to exactly one vertex in S ′1.

3. w is adjacent to more than one vertex in S ′1.

Case 1. Suppose w is not adjacent to any of the vertices of S ′1.

If there exists x′ ∈ S ′1 which is adjacent to a vertex in D′−S ′

1, then

D′ −{x′} is a γ ′-set with respect to D, a contradiction. Therefore,

Claim 2 along with δ(G) ≥ 2 together imply that each vertex in S ′1

has at least two neighbors in S.

Suppose there exists a vertex x ∈ S which is adjacent to y ∈ S,

then as in the proof of Claim 2, we get that either D − {x} or

D − {y} is a γ-set of G, a contradiction. Thus S is independent.

Case 1.1. Suppose there exists a pair of vertices u, v ∈ S such

that NS′

1(u) ∩ NS′

1(v) = {u′} for some u′ ∈ S ′

1.

Case 1.1.1. If w is adjacent to a vertex in D − {u, v}, then by

the assumption in Case 1.1, the vertices in S ′1 −{u′}, dominated by

either u or v, are also adjacent to some vertex in S − {u, v} and so

D − {u, v} ∪ {u′} is a γ-set of G which is a contradiction.

Case 1.1.2. If w is adjacent to one of u or v, say u, and let

u′ 6= v′ ∈ NS′

1(v). Note that by assumption in Case 1, w is not

adjacent to u′ as well as v′.

If x ∈ S with NS′

1(x) = {u′, v′}. Suppose there exists no vertex in

S ′1−{u′, v′} which is adjacent to only v and x, then D−{v, x}∪{v′} is

a γ-set of G, which is a contradiction. If there exists x′ ∈ S ′1−{u′, v′}

such that NS(x′) = {v, x}, then D1 = D − {v} ∪ {v′} is a γ-set of


G and D′1 = D′ − {v′, x′} ∪ {v} is a γ ′-set of G with respect to

D1, a contradiction. If there exists y ∈ S − {u, v, x} such that y is

adjacent to v′ and x′ only, then by similar argument one can get a

contradiction in all the cases.

If there is no vertex x ∈ S such that NS′

1(x) = {u′, v′}, then

by Claim 2, D1 = D − {v} ∪ {v′} is a γ-set of G and so D′1 =

D − {u′, v′} ∪ {v} is a γ ′-set of G which is a contradiction.

Case 1.2. Suppose, for each pair of vertices x, y ∈ S, there exist

at least two vertices x′, y′ ∈ S ′1 such that {x′, y′} ⊆ NS′

1(x)∩NS′

1(y).

Case 1.2.1. Suppose, for some u′, v′ ∈ S ′1, there exists at most

one vertex u ∈ S such that NS(u′)∩NS(v′) = {u}. If w is adjacent

to a vertex in D − {u}, then D1 = D − {u} ∪ {u′} is a γ-set and

so D′1 = D′ − {u′, v′} ∪ {u} is a γ ′-set of G, a contradiction. If

ND(w) = {u}, then D1 = D − {u} ∪ {w} is a γ-set and so D′1 =

D′ − {u′, v′} ∪ {u} is a γ ′-set of G, a contradiction.

Case 1.2.2. Suppose, for each pair of vertices x′, y′ ∈ S ′1, there

exist at least two vertices x, y ∈ S such that {x, y} ⊆ NS(x′) ∩

NS(y′). Note that |S| ≥ 2 and |S ′1| ≥ 2. Assume that |S| = k and

S = {u1, u2, . . . , uk}. If w is adjacent to some vertex in S, say u1.

By the assumption in Case 1.2, there exist u2 ∈ S and u′1, u

′2 ∈ S ′

1

such that < {u1, u2, u′1, u

′2} >= K2,2 as S and S ′

1 are independent.

Assume that |S| = k ≥ 3. Suppose NS′

1(u3) = {u′

1, u′2}. Since u′

2

dominates both u2, u3 and u′1, u

′2 are the vertices dominated by u2

and u3, D1 = D − {u3, u2} ∪ {u′2} is a γ-set of G, a contradiction.

Thus u3 is adjacent to a vertex in S − {u′1, u

′2}, say u′

3. Suppose

NS(u′3) ⊆ {u1, u2, u3}. Since each pair of vertices in S has at least

two neighbors, we have D1 = D − {u2} ∪ {u′2} as a γ-set and D′

1 =

D′ − {u′2, u

′3} ∪ {u2} as a γ ′-set of G, a contradiction. Thus u′

3 is

adjacent to some vertex in S − {u1, u2, u3}, say u4. Proceeding like


this up to uk and let u′k ∈ NS′

1(uk). If NS(u′

k) ⊆ S, then D1 =

D−{uk−1}∪{u′k−1} is a γ-set and so D′

1 = D′−{u′k−1, u

′k}∪{uk−1}

is a γ ′-set of G, a contradiction. Hence u′k is adjacent to at least

one vertex in S − {u1, . . . , uk} = ∅ which is not possible.

Let |S| = k = 2. If |S ′1| ≥ 3, then u′

3 is adjacent to u1 and

u2 only. Therefore D1 = D − {u2} ∪ {u′2} is a γ-set and D′

1 =

D′ − {u′2, u

′3} ∪ {u2} is a γ ′-set of G, a contradiction. If |S ′

1| = 2,

then D = D′, which is a contradiction.

Case 2. Suppose w is adjacent to exactly one vertex x′ ∈ S ′1.

If u′ ∈ S ′1 − {x′} is adjacent to a vertex in D′, then D′ − {u′} is a

γ ′-set of G, a contradiction. Thus every vertex in S ′1 − {x′} has at

least two neighbors in S.

If |S ′1| ≥ 3, then as in case 1, replace S ′

1 by S ′1 − {x′}, we get

contradiction in all the possibilities.

Let |S ′1| = 2 and S ′

1 = {x′, y′}. Since y′ has at least two neighbors

in S, |S| ≥ 2 and so |S| ≥ |S ′1| which is contradiction to |D| < |D′|.

If |S ′1| = 1, then since |D| < |D′|, S = ∅, a contradiction to

S 6= ∅.

Case 3. Suppose w is adjacent to more than one vertex in S ′1,

say u′, v′ ∈ S ′1. If no vertex in S is adjacent to only u′, v′, then

D′ − {u′, v′} ∪ {w} is a γ-set of G, a contradiction. Thus there

exists a vertex u ∈ S such that NS′

1(u) = {u′, v′}. If w is adjacent

to u, then D1 = D−{u}∪{w} is a γ-set and D′1 = D′−{u′, v′}∪{u}

is a γ ′-set of G, a contradiction. Now let u 6= x ∈ ND(w) and let

x′ ∈ ND′(x).

Suppose there exists y ∈ D − {x} such that y ∈ N(x′). Suppose

there exists z ∈ D − {u, x} such that ND′(z) ⊆ {u′, v′, x′}. Then

D − {u, z} ∪ {u′} is a γ-set of G, a contradiction. Otherwise, D1 =


D − {u, x} ∪ {u′, w} is a γ-set and D′1 = D′ − {u′, v′, x′} ∪ {x, u} is

a γ ′-set of G, a contradiction.

Suppose ND(x′) = {x}. If x′ is adjacent to w, then D′−{u′, x′}∪

{w} is a γ ′-set of G. If x′ is not adjacent to w, then as δ(G) ≥ 2,

x′ is adjacent to at least one vertex, say y′ ∈ D′. Since x′ ∈ N(y′),

x, u′ ∈ N(w) and ND(x′) = {x}, we get D′ − {u′, x′} ∪ {w} is a

γ ′-set of G which is a contradiction.

Hence γ(G) = γ ′(G).

D.W. Bange et al. [3] characterized trees with two disjoint mini-

mum dominating sets. In the following corollary, we give the neces-

sary condition for graphs with minimum degree at least two having

two disjoint minimum dominating sets.

Corollary 3.1.2. Let G be a connected graph on n vertices with

δ(G) ≥ 2. If γ(G) + γ ′(G) = n− 1, then G has two disjoint γ-sets.

The following example shows that, in general Theorem 3.1.1 is

not true whenever δ(G) = 1.

Example 3.1.3. (i) Consider the graph P6, the path on 6 vertices.

Then γ(P6) = 2 and γ ′(P6) = 3. Therefore γ(P6) + γ ′(P6) = 5 but

γ(P6) 6= γ ′(P6).

(ii) Consider the graph G in Figure 3.1. Clearly, γ(G) = 3

and γ ′(G) = 4. Therefore γ(G) + γ ′(G) = 7 = n − 1 whereas

γ(G) 6= γ ′(G).

Lemma 3.1.4. Let G be a connected graph with δ(G) ≥ 2. Then

γ(G) + γ ′(G) = n − 1 if and only if γ(G) = γ ′(G) = ⌊n2⌋ and n is

odd.


84

2

6

7

1

5

3

Figure 3.1: Graph with γ < γ ′ and γ + γ ′ = n − 1

Proof. If γ(G) + γ′

(G) = n − 1, then by Theorem 3.1.1, γ(G) =

γ ′(G). Therefore γ(G) = γ ′(G) = n−12 and hence n is odd. Con-

versely, assume that γ(G) = γ ′(G) = ⌊n2⌋ and n is odd. Since n is

an odd integer, we get γ(G) + γ ′(G) = n − 1.

Let C and D be the families of graphs given in Figures 3.2 and

3.3 respectively.

H1 H2 H3

H4 H5 H6

Figure 3.2: Graphs in family C

Note that, the class C is a subclass of the class A and the class D

is same as the class B where A and B are classes given in Theorem

2.6 [23, p-45].

The next theorem characterizes all connected graphs G with

δ(G) ≥ 2 for which γ(G) + γ ′(G) = n − 1. By Lemma 3.1.4


H7 H8 H9 H10 H11

Figure 3.3: Graphs in family D

and Lemma 2.4 [23], we get the main theorem of this chapter.

Theorem 3.1.5. Let G be a connected graph with δ(G) ≥ 2. Then

γ(G) + γ ′(G) = n − 1 if and only if G ∈ C ∪ D.

It may be worth noting that for any graph G, the disjoint domi-

nation number γγ(G) ≤ γ(G)+γ ′(G). Due to this, we get an upper

bound for γγ(G) and γ(G) + γ ′(G) which is better than Ore’s ob-

servation for disjoint domination number and sum of domination

number and inverse domination number [14].

Corollary 3.1.6. Let G be a connected graph with δ(G) ≥ 2 and

G /∈ {H1, . . . , H11}. Then γγ(G) ≤ γ(G) + γ ′(G) ≤ n − 2.

We suggest the following problems for further study in this di-

rection.

Open problems:

1. Find a necessary and sufficient condition for a graphs G with

δ(G) = 1 and γ(G) + γ ′(G) = n − 1.

2. Characterize all connected graphs G with δ(G) ≥ 2 for which

γ(G) + γ ′(G) = n − 2.

3.2. GRAPHS WITH δ = 1 AND γ + γ ′ = N − 1 45

3.2 Graphs with δ = 1 and γ + γ ′ = n − 1

Up to this point, we have identified the graphs with δ(G) ≥ 2 which

satisfy γ(G) + γ ′(G) = n − 1. In order to identify the graphs with

δ(G) = 1 which satisfy γ(G) + γ ′(G) = n − 1, we construct a set

of classes of graphs as detailed below. Further, we prove that these

are the only graphs which exactly fulfill our requirement.

Construction C1: Let H be any connected graph and S =

V (H). Let P(H) be the set of all connected graphs obtained from

H by adding a set of new vertices and edges such that each vertex

in V (H) is adjacent to one or more new pendant vertices. Define

G1 = ∪ P(H) for any connected graph H.

Let G ∈ G1 and S1 be the set of all vertices v ∈ V (G) such that

v is adjacent to exactly one pendant vertex in G. Clearly S1 ⊆ S

and let S2 = S − S1. Let L be the set of all pendant vertices in G.

Let L1 = N(S1) ∩ L and let L2 = N(S2) ∩ L. Hereafter whenever

we use S, S1, S2, L, L1 and L2, they mean only these six sets. The

class of graphs G1 are given in Figure 3.4.

r

r

r

r

r

r

rrr

r

r

rr

r

r

rr

rr

S1S2

r

r

��

��

r

Figure 3.4: Class G1


BBBBBB

s

s s

s

s

s

s

ss

s

s

s

s

s

s

s

s

ss

s

S1S2

s

s

Figure 3.5: Class G2

Construction C2: Let Q(G) be the set of connected graphs

obtained from G ∈ G1 by adding a set of vertices say I and a set

of edges joining each vertex of I with two or more vertices in S2.

Define G2 = ∪ Q(G) where G ∈ G1. The class of graphs G2 are


Construction C3: Let S(K) be the set of all connected graphs

obtained from K ∈ G2 by adding a new vertex x and a set of edges

joining x to two or more vertices in S with at least one vertex in

S1. Then define R1 = ∪ S(K) where K ∈ G2. The class of graphs

R1 are given in Figure 3.6.

Construction C4: Let T(K) be the set of all connected graphs

obtained from K ∈ G2 by adding two vertices x and y with an edge

xy and joining each of x, y to vertices in S such that when both x

and y are adjacent to vertices in S1. Then either |N(x) ∩ S| = 1

or |N(y) ∩ S| = 1. Then define R2 = ∪T(K), where K ∈ G2. The

class of graphs R2 are given in Figure 3.7.


BBBBBB

s

s s

s

s

s

s

ss

s

s

s

s

s

s

s

s

ss

s

S1S2

sx�

��

��

��

Figure 3.6: Class R1

BBBBBB

s

s s

s

s

s

s

ss

s

s

s

s

s

s

s

s

ss

s

S1S2

x s

y



BBBBBB

s

s s

s

s

s

s

s

s

s

ss

s

s

s

s

s

s

s

s

ss

s

S1S2

DDDDDDDDDD

wx

y


Construction C5: Let Z(K) be the set of all connected graphs

obtained from K ∈ G2 by adding a new path P (w, x, y) and edges

joining each of w and y to one or more vertices in S. Then define

R3 = ∪ Z(K), where K ∈ G2. The class of graphs R3 are given in

Figure 3.8.

Construction C6: Let V(K) be the set of all connected graphs

obtained from K ∈ G2 by adding a new path P4 with end vertices x

and w and internal vertices y and u. Also add the edges joining each

of x and w to one or more vertices in S2 together with or without

edges joining one of y, u to vertices in S. Then define R4 = ∪V(K),

where K ∈ G2. The class of graphs R4 are given in Figure 3.9.

Construction C7: Let W(K) be the set of all connected graphs

which may be obtained from any graph K ∈ G2 by adding a cycle

C4 consisting of vertices p, q, r, s and another new vertex v, an edge

joining v and p and edges joining v to one or more vertices in S.


BBBBBB

s

s s

s

s

s

s

s

s

s

s

ss

s

s

s

s

s

s

s

s

ss

s

S1S2

y

uw

x


Then define R5 = ∪ W(K), where K ∈ G2. The class of graphs R5

are given in Figure 3.10.

Construction C8: Let X(K) be the set of all connected graphs

which may be formed from a graph K ∈ G2 by adding a cycle

C4 with vertices u, v, w, x and edges joining one vertex of C4 to

vertices in S with at least one edge in S1 or edges joining at most

three vertices of C4 to vertices in S2. Then define R6 = ∪ X(K),

where K ∈ G2. The class of graphs R6 are given in Figure 3.11.

Construction C9: Let Y(K, M) be the set of all connected

graphs obtained from K ∈ G2 and M ∈ B, the collection of graphs

given in Figure 2.2 by joining each vertex of F ⊆ V (M) to one or

more vertices in S such that no set with fewer than γ(M) vertices

of M dominates V (M) − F . Then define R7 = ∪Y(K, M) where

K ∈ G2 and M ∈ B. The class of graphs R7 are given in Figure 3.12.


BBBBB

r

r r

r

r

r

r

r

r

r

r

rrr

r

r

rr

r

r

rr

rr

S1S2

��rvp s

rq


BBBBBB

��

s

s s

s

s

s

s

s

s

s

s

s

ss

s

s

s

s

s

s

s

s

ss

s

S1S2

w

yu

x



s

s

s

s

s

s

ss

s

s

s

s

s

s

s

s

ss

s

S1S2

��

��

��

s

s s

ss

s

y

uv w

x


Theorem 3.2.1. If a graph G ∈ Ri for some i = 1, 2, . . . , 7, then

γ(G) + γ ′(G) = n − 1.

Proof. Let us prove the result by the way of giving the necessary γ

and γ′

-sets in each of the classes separately.

Case 1 Let G ∈ R1. By construction, let x ∈ V (G) be the

new vertex that is adjacent to a vertex u in S1 and let w be the

pendant vertex adjacent to u. Clearly D = S is a dominating set

of G. Therefore γ(G) ≤ |S|. Also since γ ≥ number of supports

= |S|, we get γ(G) = |S|. Then D = S ∪ {w} − {u} is a γ-set of

G and D′ = V (G) − (D ∪ {x}) is an inverse dominating set of G

with respect to D. Therefore γ(G) + γ ′(G) ≤ (|D|+ |D′|) = n− 1.

Note that each vertex in S2 lies in every γ-set of G and hence the

vertices that are adjacent to only the vertices of S2 lie in every γ′

-

set of G. Also observe that each vertex in (S1 ∪ L1) lies either in

a γ-set of G or in a γ′

-set of G. Hence γ(G) + γ ′(G) ≥ n − 1.


Therefore γ(G) + γ ′(G) = n − 1.

Case 2 Let G ∈ R2. Then as in case 1, we see that D = S is a

γ-set and D′ = V (G)− (D ∪ {y}) is an inverse dominating set of G

with respect to D. Therefore γ(G) + γ ′(G) ≤ (|D|+ |D′|) = n− 1.

Note that each vertex in S2 lies in every γ-set of G and hence the

vertices that are adjacent to only the vertices of S2 together with

the vertex {x} or {y} lie in every γ′

-set of G. Also observe that

each vertex in (S1∪L1) lies either in a γ-set of G or in a γ′

-set of G.

Hence γ(G) + γ ′(G) ≥ n − 1. Therefore γ(G) + γ ′(G) = n − 1.

Case 3 Let G ∈ R3, D be a γ-set and D′ be a γ′

-set of G. Then

as discussed earlier, we see that ((S ∪ L) ∪ I) ⊆ (D ∪ D′). Since at

least one of the vertices of P3, say x is not adjacent to any of the

vertices of S, x or one of the neighbors say w ∈ D and the other

neighbor say y or x itself must lie in D′. Thus the only vertex that

is neither in the γ-set nor in the γ′

set of G is either x or y. Thus

D = S ∪ {w} is a γ-set and D′ = V (G) − (D ∪ {x}) is a γ′

-set of

G. Hence γ(G) + γ ′(G) = n − 1.

Case 4 Let G ∈ R4. Let D be a γ-set and D′ be a γ′

-set of G.

Then as discussed earlier, we see that ((S∪L)∪I) ⊆ (D∪D′). Since

at least one of the vertices of P4, say y is not adjacent to any of the

vertices of S, y ∈ D and since none of the vertices of the path P4

are adjacent to any of the vertices of S1, at least two other vertices

say x, w must lie in D′. Thus the only vertex that is neither in the

γ-set nor in the γ′

set of G is u. Thus D = S ∪ {y}, where y is not

adjacent to vertices in S is a γ-set of G and D′ = V (G)− (D∪{u})

is a γ′

-set of G. Hence γ(G) + γ ′(G) = n − 1.


-set of

G. Then as discussed earlier, we see that ((S ∪L)∪ I) ⊆ (D ∪D′).


Since none of the vertices of C4 is adjacent to any of the vertices

of S, two of the vertices of C4 say r, s must lie in a γ-set and two

other vertices say p, q must lie in the γ′

set of G. Since S2 ⊆ D, v

is dominated by a vertex v1 ∈ D and is dominated by p ∈ D′. Thus

D = S ∪ {q, s} is a γ-set and D′ = V (G)− (D ∪ {v}) is a γ′

-set of

G. Thus the only vertex that is neither in the γ-set nor in the γ′

set of G is v. Hence γ(G) + γ ′(G) = n − 1.


-set of

G. Then as discussed earlier, we see that ((S ∪L)∪ I) ⊆ (D ∪D′).

Since at least one of the vertices of C4, say w, is not adjacent to

any of the vertices of S, w ∈ D and since no vertex of the cycle C4

is adjacent to any of the vertices of S1, at least two other vertices,

say x, w, must lie in D′. Thus D = S ∪ {w}, is a γ-set of G and

D′ = V (G)−(D∪{u}) is a γ′

-set of G. Hence γ(G)+γ ′(G) = n−1.

Thus the only vertex that is neither in the γ-set nor in the γ′

set

of G is v. Hence γ(G) + γ ′(G) = n − 1.

Case 7 Let G ∈ R7. Let D1 be the γ-set and D1′

be the γ′

-set

of M ∈ B. Then |D1′

∪ D1′

| = |V (M)| − 1. Let t be the vertex

which is neither in the γ-set nor in the γ′

-set of M . Then by our

construction, D = S∪ D1 is a γ-set and D′ = V (G)− (D∪{t}) is a

γ′

-set of G. Hence γ(G)+γ′

(G) = n−1. Thus γ(G)+γ ′(G) = n−1,

for all G in ∪ Ri, i = 1 to 7.

Theorem 3.2.2. Let G be a connected graph with δ(G) ≥ 1. Let

L ⊆ V (G) be the set of all pendant vertices in G and S = N(L).

Let V − S be independent. Then γ(G) + γ ′(G) = n− 1 if and only

if G ∈ R1.

Proof. If G ∈ R1, then by Theorem 3.2.1, γ(G) + γ ′(G) = n − 1.


Conversely, assume that γ(G)+γ ′(G) = n−1. Let M1 = V −(S∪L).

If M1 = ∅, then we have V = S ∪ L. Hence γ(G) + γ ′(G) = n, a

contradiction. Therefore M1 is non-empty. By Theorem 1.2.8, we

have at least one vertex x ∈ M1 such that at least one vertex in

N(x) is adjacent to only one pendant vertex.

We claim that there exists exactly one vertex u in M1 such that

at least one vertex in N(u) is adjacent to only one pendant vertex.

Suppose there are two vertices x and y in M1 such that N(x) and

N(y) each contain at least one vertex which is adjacent to only

one pendant vertex. Let x1 ∈ N(x) and y1 ∈ N(y) be the vertices

in S that are adjacent to only one pendant vertex, say x2 and y2

respectively. Since V −S is independent, S is a γ-set of G. Therefore

D = S ∪ {x2, y2} − {x1, y1} is a γ-set of G and D′ = V (G) − (D ∪

{x, y}) is an inverse dominating set of G with respect to D. Now

neither x nor y are in D ∪ D′. Therefore γ(G)+γ ′(G) ≤ |D+D′| ≤

n− 2, a contradiction. Therefore there exists an unique vertex x in

M1, such that at least one vertex in N(x) is adjacent to only one

pendant vertex. Hence by construction C3, G ∈ R1.

So far, we have constructed a class of graphs with minimum

degree one which satisfy γ(G) + γ ′(G) = n − 1. Now we are going

to prove that they are the only graphs with such property.

Notation: Let G be a connected graph with δ(G) = 1. Let

L ⊆ V be the set of all vertices of degree one in G. Let S = N(L).

Let S1 be the set of all vertices in S, which are adjacent to only one

pendant vertex and let S2 = S−S1. Let I be the set of independent

vertices in G but not in S ∪ L that are adjacent to vertices in S2.


Theorem 3.2.3. Let G be a connected graph with δ(G) = 1. Then

γ(G) + γ ′(G) = n − 1 if and only if G ∈⋃7

i=1 Ri.

Proof. One part of the theorem follows from Theorem 3.2.1. Con-

versely assume γ(G) + γ ′(G) = n − 1. Let S, L and I denote the

sets as mentioned above.

Case 1 When V (G) − S is independent, by Theorem 3.2.2, we

get G ∈ R1.

Case 2 When V (G) − S is not independent, let K = 〈I ∪

( S ∪ L) 〉. By construction C2, K ∈ G2. Let N = 〈V (G)−V (K)〉.

Note that δ(N) ≥ 1.

Sub case 2.1 When δ(N) = 1, let X be the set of pendant

vertices in N . No vertex of N is a pendant vertex, as a vertex of G.

Hence each vertex of X, as a vertex of G is adjacent to at least

one vertex in S. Now there are two possibilities. They are either

N − X = ∅ or N − X 6= ∅.

Sub case 2.1.1 If N − X = ∅, then N = X, and each vertex of

N is a pendant vertex. Hence each component of N is a K2. Let

the number of components in N be m. Note that any vertex of G is

either in S or adjacent to vertices in S. Therefore D = S is a γ-set

of G. Also D′ = (L∪l)∪ {one vertex in each of the m components of

N} is an inverse dominating set of G with respect to D. Therefore

γ(G)+ γ ′(G) ≤ n−m = n− 1, by the assumption. This gives that

m = 1. Therefore N = K2. Assume that V (K2) = {x, y}. Suppose

x and y are adjacent to vertices only in S2. Since N = {x, y}, by

construction of C4, G = T (K), K ∈ G2 and so G ∈ R2. Similarly

if only one of the vertices {x, y} is adjacent to vertices in S1, then

also by construction C4, G ∈ R2. If each of x and y are adjacent


to vertices in S1, then we claim that either |N(x) ∩ S| = 1 or

|N(y) ∩ S| = 1. If not, let x1, x2 ∈ N(x) ∩ S where x1 ∈ S1 and let

y1, y2 ∈ (N(y) ∩ S) where y1 ∈ S1. Let N(x1) = x1′

, N(y1) = y1′

in L. Since S is a γ-set, D = S ∪ {x1′

, y1′

} − {x1, y1} is a γ-set of

G and D′ = V (G) − (D ∪ {x, y}) is an inverse dominating set with

respect to D, so that γ(G) + γ ′(G) ≤ n− 2, a contradiction. Thus

when both the ends x and y of K2 are adjacent to vertices in S1,

we have either |N(x) ∩ S| = 1 or |N(y) ∩ S| = 1. By construction

C4, G = T (K), K ∈ G2 and so G ∈ R2.

Sub case 2.1.2 When N − X 6= ∅, there are three possibilities

as discussed below. They are N −X has only isolates or no isolates

or both isolates and non-isolates.

Sub case 2.1.2.1 If N − X has only isolates, then each isolate

u ∈ N −X is adjacent to at least two vertices in X. Let |N −X| =

k. Let V (N − X) = {u1, u2, . . . , uk}. Let ui be adjacent with

vi1, vi2, . . . , viki. Then D = S ∪ {v11

, v21, . . . , vk1

} is a γ-set of G and

D′ = (L ∪ I) ∪ {u1, u2, . . . , uk} is an inverse dominating set of G

with respect to D. Hence γ(G) + γ′

G ≤ n − k. Since we have

γ(G) + γ′

(G) = n − 1, we get that k = 1. Hence there is only one

isolate in N −X. Let u be the isolate in N −X. Next we claim that

u is adjacent to exactly two vertices in X. Suppose u is adjacent

to three or more vertices in X, say {u1, u2, . . . , um}, m ≥ 3. Then

D = S∪{u1} is a γ-set of G and D′ = V (G)−(D∪{u2, u3, . . . , um})

is an inverse dominating set of G with respect to D. Therefore

γ(G) + γ ′(G) ≤ n − 2, a contradiction. Hence m ≤ 2. Since

dG(u) ≥ 2, m = 2. Therefore u is adjacent to exactly two vertices,

say v and w and so N is the path P3 with vertices v, u and w. Since

v, w are the vertices in X, they are adjacent to vertices in S.


We claim that u is not adjacent with vertices in S. Suppose

u is adjacent with vertices in S, then D = S is a γ-set of G and

D′ = V (G) − (D ∪ {v, w}) is an inverse dominating set of G with

respect to D. Therefore γ(G) + γ ′(G) ≤ n − 2, a contradiction.

Hence u is not adjacent with vertices in S. Hence by construction

C5, G = Z(K) and so G ∈ R3.

Sub case 2.1.2.2 If N −X has no isolates, then let N1 = 〈N −

X〉. Note that each vertex in X is adjacent to at least one vertex

in S and to exactly one vertex in N1. So D = S ∪D1 is a γ-set of G

where D1 is a γ-set of N1 and D′ = V (G) − (D ∪ {u}) with u ∈ X

is an inverse dominating set of G.

Let D1′

be a γ′

-set of N1. We claim that |N1 ∩ D′| ≤ |D′

1|.

If not, |N1 ∩ D′| > |D′

1|. That is there are some vertices in N1

which are not in D′

1 but in D′

to dominate the vertices of X. But

by the choice of D′

, all the vertices of X except u (∈ X) are in D′.

Thus D′ is not a γ′

- set of G. Hence γ(G) + γ ′(G) < n − 1, a

contradiction. Hence the claim. Therefore |N1 ∩ D| ∪ |N1 ∩ D′| =

|N1∩(D∪D′)| ≤ |D1∪D1′

| ≤ |V (N1)|. Since γ(G)+γ ′(G) = n − 1

and I ∪ (S ∪ L) ⊆ D ∪ D′, and since u ∈ X is the only vertex

which is neither in the γ- set of G nor in the γ′

- set of G, we have

|N1 ∩ (D ∪ D′)| = |V (N1)|. Therefore |D1 ∪ D1′

| = |V (N1)|. That

is γ(N1) + γ′

(N1) = |V (N1)|.

Sub case 2.1.2.2.1 When δ(N1) = 1, we claim that |V (N1)| =

n1 = 2. Since δ(N1) = 1, trivially n1 ≥ 2. If n1 ≥ 3, then by

Theorem 1.2.8, there are at least two pendant vertices in N1 and

each pendant vertex in N1 is adjacent with one vertex in X. Let

{x1, x2} be two vertices in X that are adjacent with the pendant

vertices of N1. Let H be the subgraph obtained from N1 by deleting


the pendant vertices of N1. Let H1 be the γ-set of N1 with no

pendant vertices of N1. Then D = S ∪ H1 is a γ-set of G and

D′ = V (G) − (D ∪ {x1, x2}) is an inverse dominating set of G with


Therefore n1 = 2. Since δ(N1) = 1, N1 = K2 and so N = P4.

Let (x1, x2, x3, x4) be the path P4. We claim that all the vertices

of P4 are not adjacent to vertices in S. Suppose all the 4 vertices

of P4 are adjacent to vertices in S, then D = S is a γ-set of G and

D′ = V (G) − (D ∪ {x1, x4}) is an inverse dominating set of G with


Therefore all the vertices of P4 are not adjacent to vertices in S.

Next we claim that both x1 and x4 are adjacent to vertices in S2

only. Since x1 and x4 are pendant vertices in N , they are adjacent

to vertices in S. Suppose at least one of x1 and x4, say x1 is adjacent

to vertices in S1. Let x1 be adjacent with y1 in S1 where z1 is the

pendant vertex adjacent to y1. Then D = (S ∪ {z1, x2})− {y1} is a

γ-set of G and D′ = V (G)− (D∪{x1, x4}) is an inverse dominating

set of G with respect to D. Therefore γ(G) + γ ′(G) ≤ n − 2, a

contradiction. Therefore both x1 and x4 are adjacent to vertices in

S2 only. Hence by the above claims and if one of x2 or x3 is not

adjacent with vertices in S, we get G = V (K) by construction C6.

Thus G ∈ R4.

Sub case 2.1.2.2.2 When δ(N1) ≥ 2. Since γ(N1) + γ′

(N1) =

|V (N1)|, by Theorem 1.2.7, we observe that N1 = C4. We claim

that |X| = 1. If not, |X| ≥ 2. Let {x1, x2} be two vertices in X

which are adjacent to vertices of C4. Let {a, b, c, d} be the vertices

of C4. Without loss of generality, let ax1 and bx2 be the edges in N .

Then D = S∪{c, d} is a γ-set of G and D′ = V (G)− (D∪{x1, x2})


is an inverse dominating set of G with respect to D. Hence γ(G) +

γ ′(G) ≤ n − 2, a contradiction. Therefore |X| = 1. Hence assume

ax is an edge in N . Since x is a pendant vertex in N, x is adjacent

to vertices in S.

We claim that no vertex of C4 is adjacent with vertices in S.

If not, at least one vertex say a (or b) is adjacent to vertices in

S. Then D = S ∪ {c} (or D = S ∪ {d} accordingly) is a γ-set of

G and D′ = V (G) − (D ∪ {x, d}), (or D′ = V (G) − (D ∪ {x, b})

accordingly) is an inverse dominating set of G with respect to D.

Hence γ(G) + γ ′(G) ≤ n − 2, a contradiction. Therefore no vertex

of C4 is adjacent with vertices in S. Thus only x is adjacent with

vertices in S. Hence by construction C7, G = W (K) and G ∈ R5.

Sub case 2.1.2.2.3 From the above two cases we see that the

components of N − X shall be either with only isolates or with no

isolates, since in each case we have one vertex which is neither in

the γ-set of G nor in the γ′

-set of G.

Sub case 2.2 When δ(N) ≥ 2. Let D and D1 be the γ-sets of

G and N respectively. Also let D′ and D1′

be the γ′

-sets of G and

N respectively. Since either the vertices of S or the vertices of L

lie in any dominating set of G, we need no vertex of N to dominate

vertices in S. Therefore |N ∩ (D ∪ D′)| ≤ |D1 ∪ D1′

| ≤ |V (N)|.

Thus either |N ∩(D∪D′)| ≤ |D1∪D′

1| or |N ∩(D∪D′)| = |D1∪D′

1|

are the two possibilities.

Sub case 2.2.1 Suppose |N ∩ (D ∪ D′)| < |D1 ∪ D1′

|. Then

|N ∩ (D∪D′)| < |D1 ∪D1′

| ≤ |V (N)|. Since γ(G)+γ ′(G) = n− 1

and S∪L ⊆ D∪D′, we have |N∩(D∪D′)| = |V (N)|−1. Therefore

|D1 ∪ D1′

| = |V (N)|. That is |D1| + |D1′

| = |V (N)|. Thus

γ(N) + γ′

(N) = |V (N)|. Note that δ(N) ≥ 2 and by Theorem

1.2.7, we get N = C4.


Sub case 2.2.1.1 Only one vertex of C4 is adjacent to vertices

in S1. If not, there are at least two vertices of C4 that are adjacent

to vertices in S1 or one vertex of C4 adjacent to vertices in S1 and

at least one or more of the other vertices is adjacent to vertices in

S2. Let e, f, g and h be the vertices of C4. Let e, f be the vertices

adjacent to x1 and y1 in S1 respectively. Let x2 and y2 be the

neighbors of x1 and y1 in L respectively. Then D = (S ∪ {y2, g}) −

{y1} is a γ-set of G and D′ = V (G) − (D ∪ {e, f}) is an inverse

dominating set of G with respect to D. Hence γ(G)+γ ′(G) ≤ n−2,

a contradiction. Similarly we get a contradiction in the other case

too. Hence, if the vertices of C4 are adjacent to vertices in S1, then

only one vertex of C4 is adjacent to vertices in S1. Therefore by

construction C8, G ∈ R6.

Sub case 2.2.1.2 At most three vertices of C4 are adjacent to

vertices in S2. If not, all the vertices of C4 are adjacent to vertices

in S2. Then D = S is a γ-set of G and D′ = V (G) − (D ∪ {e, f})

is an inverse dominating set of G with respect to D. Hence γ(G) +

γ ′(G) ≤ n − 2, a contradiction. Thus if the vertices of C4 are

adjacent to vertices in S2, then at most three vertices of C4 are

adjacent to vertices in S2. Hence by construction C8, G ∈ R6.

Sub case 2.2.2 Suppose |N∩(D∪D′)| = |D1∪D1′

|. Then |N∩

(D∪D′)| = |D1∪D1′

| ≤ |V (N)|. But |N ∩ (D∪D′)| = |V (N)|−1.

Therefore |D1∪D1′

| = |V (N)|−1. That is |D1|+|D1′

| = |V (N)|−1.

Also δ(N) ≥ 2. Therefore N ∈ (A∪B). However N may be further

restricted. It can be easily verified that if N ∈ A, then for any

U 6= ∅, V (N) − U is dominated by less than γ(N) vertices so that

γ(G) + γ ′(G) ≤ n − 2. Therefore for N ∈ B, there exists at least

one suitable non-empty subset U of V (N) such that V (N) − U


is dominated by exactly γ(N) vertices. Thus each vertex in U is

adjacent with one or more vertices in S and so by construction C9,

we get G ∈ R7. Hence G ∈⋃7

i=1 Ri.

Chapter 4

Inverse Domination in Grid

Graphs

This chapter deals about the inverse domination in grid graphs.

Actually we obtain the inverse domination number for grid graphs of

order up to seven. The domination number of k×n grid graphs Pk×

Pn have been previously established by Jacobson and Kinch [26, 27]

and, T. Y. Chang, W. Edwin Clark and E. O. Hare [7]. More

specifically, they find the domination number of grid graphs for

1 ≤ k ≤ 10 and n ≥ 1. In tune with the methodology established

in [7], we obtain the inverse domination number as well as an inverse

dominating set for the graphs Pk × Pn for 1 ≤ k ≤ 7 and n ≥ 1.

Even though γ(Pm) is known for any positive integer m, one could

not able to determine γ(Pk × Pn) using γ(Pk) and γ(Pn). In this

regard, Jacobson and Kinch [26] took the first step towards finding

the domination number for Pk × Pn. Actually, in their work they

have obtained the domination number for the grid graphs Pk × Pn

for a positive integer k and n with 1 ≤ k ≤ 4 and n ≥ 1. Later on,

in the year 1994, T.Y. Chang, W. Edwin Clark and E. O. Hare [7]

identified the dominating sets for the grid graphs Pk × Pn for 5 ≤

k ≤ 10 and n ≥ 1 through the smaller grids Pk × Pm with m < n.

63

64 CHAPTER 4. INVERSE DOMINATION IN GRID GRAPHS

As in chapter 3, γ(Pn) = γ ′(Pn) = ⌈n3⌉ if and only if

n 6≡ 0 (mod 3). Since P1 × Pn is nothing but Pn, we have γ ′(P1 ×

Pn) = γ ′(Pn). It may be worthwhile to note that γ ′(P2 × P3) =

γ ′(P2) × γ ′(P3) = 2. However this is not the case always.

For example γ(P2 × P4) = γ ′(P2 × P4) = 3 whereas γ ′(P2) = 1

and γ ′(P4) = 2. Thus γ ′(P2×P4) 6= γ ′(P2)×γ ′(P4). As mentioned

earlier, even though γ ′(Pn) is obtained in chapter 3, as it is, one

could not able to determine γ ′(Pk × Pn).

Now, we attempt to find the inverse domination number and an

inverse dominating set for the grid graphs Pk×Pn for 1 ≤ k ≤ 7 and

n ≥ 1 using the smaller grids Pk×Pm (m < n). First of all, consider

the graph P1 × Pn for any positive integer n. Let n = 3k + 1 and

V (Pn) = {1, 2, . . . n}. Then by [35], S = {2, 5, 8, . . . 3k − 1, 3k + 1}

and S ′ = {1, 4, 7, . . . 3k − 2, 3k} are the γ-set and γ ′-set of Pn

respectively and |S| = |S ′| = k + 1. If n = 3k + 2, then S =

{2, 5, 8, . . . 3k + 2} and S ′ = {1, 4, 7, . . . 3k + 1} are the γ-set and

γ ′-sets of Pn respectively. Also each S and S ′ has k + 1 vertices.

In contrast to the values of γ(Pn) and γ ′(Pn) for n 6≡ 0( mod 3),

it may be noted that γ(P3k) = n3 and γ ′(P3k) = n

3 + 1. Further

D = {2, 5, 8, . . . , 3k−1} is the γ-set and D′ = {1, 4, 7, . . . , 3k−2, 3k}

is a γ ′-set for P3k. Hence in this case γ(P3k) 6= γ ′(P3k). In general

one can conclude that γ ′1,n =

⌈

n3

⌉

+ 1 for all n.

First we find the inverse domination number for P2 × Pn grid

graphs. In the subsequent sections, we determine inverse domina-

tion number and an inverse dominating set for grid graphs Pk × Pn

for 3 ≤ k ≤ 7. Finally, we determine the inverse domination num-

ber for cylinder graphs P2×Cn. Hereafter we denote the grid graph

Pk × Pn by k × n, γ(Pk × Pn) by γk,n and γ ′(Pk × Pn) by γ ′k,n.

4.1. INVERSE DOMINATION IN 2 × N GRID GRAPHS 65

Figure 4.1: Class S and S ′

Figure 4.2: 2 × 7 grid graph

4.1 Inverse Domination in 2 × n Grid Graphs

In this section, we give the inverse domination number and an in-

verse dominating set for the 2×n grid graphs. It is established in [26]

that γ2,n = ⌈n+12 ⌉ for n ≥ 1 and is restated in [7] as γ2,n =

⌊

n+22

⌋

for

n ≥ 1. As obtained in [7], we make use of the blocks A, B1, B2, B3

and B4 given in Figure 4.1 to construct a dominating set S and

an inverse dominating set S ′ for P2 × Pn with n ≥ 1. Note that

the vertices of the grid graphs Pk × Pn, are precisely the points of

intersection of the lines and hence |V (Pk × Pn)| = kn.

We construct S and S ′ by concatenating suitable blocks given

in Figure 4.1. For example, if we concatenate A and B3 we get the

block AB3 as in Figure 4.2. Note that AB3 is the 2× 7 grid graph.

The vertices with the symbol ‘•’ in each of the blocks in the

figures represent the vertices to be included for a dominating set

S and the vertices with the symbol ‘×’ represent the vertices to

be included for an inverse dominating set S ′. The vertices with

symbol ‘o’(or ‘⊗’) in the blocks indicate those vertices that are not


Figure 4.3: Blocks A2, AB1, AB2, AB3 and AB4 for 2 × n grid

dominated by a dominating set S (or by S ′) constructed up to a

particular stage and to be considered while concatenation. However

a dominating set or an inverse dominating set constructed after con-

catenation shall dominate all the vertices of the newly constructed

block. The concatenated blocks A2, AB1, AB2, AB3 and AB4 are


After concatenation, the set of vertices with the symbol ‘•’ give

a dominating set and the set of vertices with the symbol ‘×’ give

an inverse dominating set. Now B1, B2, B3 and B4 will give S and

S ′ for P2 × Pn for 1 ≤ n ≤ 4. Now for n ≥ 5, let n = 4q + r

with 1 ≤ r ≤ 4 and q ≥ 1. Then AqBr gives a dominating set S

and an inverse dominating set S ′ for P2 ×Pn. By Aq, we mean the

concatenation of A with itself for q times. Let a = |V (A)∩S|, a ′ =

|V (A) ∩ S ′|, br = |V (Br) ∩ S| and b ′r = |V (Br) ∩ S ′|. Note

that a = a ′ = 2, b1 = b1′ = 1, b2 = b ′

2 = 2, b3 = b ′3 = 2

and b4 = b ′4 = 3. Note that the dominating set S as well as an

inverse dominating set S ′ in AqBr contain 2q + br vertices. Hence

|S ′| = |S| = 2q + br =⌊

n+22

⌋

.

In view of γ2,n = ⌊n+22 ⌋ [7], we see that S ′ is a γ

′

-set and hence

γ2,n = γ ′2,n, for all n. Thus we have the following theorem:

Theorem 4.1.1.

γ ′2,n =

⌊

n + 2

2

⌋

for all n ≥ 1.


Figure 4.4: Blocks A,B1, B2, B3 and B4 for 3 × n grid



verse dominating set for 3× n grid graphs. In [7], it is proved that,

γ3,n =

⌊

3n + 4

4

⌋

, n ≥ 1.

To construct a dominating set and an inverse dominating set of

P3 × Pn, we use the grid blocks A, B1, B2, B3 and B4 given in Fig-

ure 4.4. We give the concatenation of A with A, B1, B2, B3, and B4

as A2, AB1, AB2, AB3 and AB4 in Figure 4.5.

Note that for n = 1, 2, 3 and 4 the dominating and inverse domi-

nating sets are as given by the grid blocks B1, B2, B3 and B4 respec-

tively. Note that a = |V (A)∩S| = 3 and a ′ = |V (A)∩S ′| = 3. As

mentioned earlier, let br = |V (Br)∩S| and b ′r = |V (Br)∩S ′|. Note

that b1 = 1, b ′1 = 2, b2 = 2, b ′

2 = 3, b3 = b ′3 = 3 and b4 = b4

′ = 4.

For n ≥ 3, let n = 4q + r with 1 ≤ r ≤ 4 and q ≥ 0. Then AqBr

for q ≥ 0 and 1 ≤ r ≤ 4 gives a dominating set S and an inverse

dominating set S ′ with 3q + br and 3q + b ′r vertices respectively.

Also, note that for n ≥ 3,

3q + br′ =

3q + br if r = 3, 4

3q + br + 1 if r = 1, 2.


Figure 4.5: Blocks A2, AB1, AB2, AB3 and AB4 for 3 × n grid

Hence for n = 4q + r, we get |S ′| = |S| whenever r = 3, 4 and

|S ′| ≤ |S| + 1 whenever r = 1, 2. Thus we obtain the following

result:

Theorem 4.2.1.

γ ′3,n

≤⌊

3n+44

⌋

+ 1 if n ≡ 1, 2(mod 4)

=⌊

3n+44

⌋

if n ≡ 0, 3(mod 4).


This section gives the inverse domination number and an inverse

dominating set for 4 × n grid graphs. It is proved that [7],

γ4,n =

n + 1 if n = 1, 2, 3, 5, 6, 9

n otherwise.

The blocks A1, A2, A3, A4 and A5 in Figure 4.6 gives a dominating

set and an inverse dominating set for P4×Pk, where k = 1, 2, 3, 4, 7.

One can see that A4A1, A4A2, A24 and A5A2 give a dominating set

and an inverse dominating set for P4 × Pk whenever k = 5, 6, 8, 9.

We use the blocks A, B and B1 given in Figure 4.7 to construct a

dominating set and an inverse dominating set for P4 × Pn; n ≥ 10.

The blocks AB and BB1 in Figure 4.8 gives a γ-set and a γ′

-set for

P4 × P6 and P4 × P4 grids respectively. Let n = 3q + r; 1 ≤ r ≤ 3.

Since n ≥ 10, we have q ≥ 3. Consider q − r. There are two


Figure 4.6: Blocks A1, A2, A3, A4 and A5 for 4 × n grid

Figure 4.7: Blocks A,B and B1 for 4 × n grid

possibilities. They are either q − r is even or odd.

Figure 4.8: Blocks AB and BB1 for 4 × n grid


Case 1. When q − r is even say 2t, then (BA)t(BB1)r gives

dominating and inverse dominating sets for P4 × Pn grid graph.

Case 2. When q−r is odd, say 2t+1, then A(BA)t(BB1)r gives

dominating and inverse dominating sets for P4 × Pn grid graph.

Let a = |V (A) ∩ S|, a ′ = |V (A) ∩ S ′|, b = |V (B) ∩ S| and

b ′ = |V (B) ∩ S ′|, b1 = |V (B1) ∩ S| and b ′1 = |V (B1) ∩ S ′|. Note

that a = a ′ = 3, b = b ′ = 3 and b1 = b ′1 = 1. Note that S and

S ′ contain only one vertex from each column. Hence we see that

γ4,n = γ ′4,n = n for all n ≥ 10. As seen earlier, γ 4,n = γ ′

4,n for

all n ≤ 10. Therefore γ 4,n = γ ′4,n for all n. Thus we proved the

following theorem.

Theorem 4.3.1.

γ ′4,n =

n + 1 if n = 1, 2, 3, 5, 6, 9

n otherwise.


This section concerns with the inverse domination number and in-

verse dominating set for 5 × n grid graphs. By [7], we have

γ5,n =

⌊

6n+65

⌋

n = 2, 3, 7⌊

6n+85

⌋

otherwise.

Tony Yu Chang, Edwin Clark and E.O. Hare [7] constructed the

blocks A, B, B1, B2, B3, B4 and B5 for constructing a dominating

set for a 5 × n grid graph. The blocks are given in Figure 4.9.

Further, we use the blocks C, D, D1, D2, D3, D4, D5, F2, F3, F4

and F5 for constructing an inverse dominating set for a 5 × n grid


Figure 4.9: Blocks A, B, B1, B2, B3, B4 and B5 for 5 × n grid

Figure 4.10: Blocks C, D, D1, D2, D3, D4, D5, F2, F3, F4 and F5 for 5 × ngrid

graph and the blocks are given in Figure 4.10.

In contrast to the procedure mentioned in the previous sections,

we use different sets of blocks to determine a dominating set and

an inverse dominating set. For n ≤ 4 dominating and inverse dom-

inating sets are obtained from 1 × 5, 2 × 5, 3 × 5 and 4 × 5 grids.

Even though we make use of different sets of blocks, the final con-

catenation gives same graph.

Let a = |V (A) ∩ S|, b = |V (B) ∩ S ′|, c = |V (C) ∩ S|, d =

|V (D) ∩ S ′|, br = |V (Br) ∩ S| and b ′r = |V (Br) ∩ S ′|. Also

let dr = |V (Dr) ∩ S|, d ′r = |V (Dr) ∩ S ′|, fr = |V (Fr) ∩ S| and

f ′r = |V (Fr) ∩ S ′|. Note that a = b = c = d = 6, b1 = d1 = 2,

b2 = d2 = f2 = 4, b3 = d3 = f3 = 5, b4 = d4 = f4 = 6 and

b5 = d5 = f5 = 7. Let n = 5q + r for 1 ≤ r ≤ 5.

Case 1. When n is even, we deal with 3 cases.

(1) n ≡ 1(mod 5).

(2) n 6≡ 1(mod 5) and q is even.

(3) n 6≡ 1(mod 5) and q is odd.


Sub case 1.1. When n is even and n ≡ 1(mod 5), then in this

case r = 1 and n = 5q + 1. Since n is even, q must be odd and

let q = 2k + 1 for some k with k ≥ 0. Then A(BA)kB1 gives a

dominating set S as in [26], hence a γ- set and D5(CD)k D1 gives

an inverse dominating set S ′ with respect to S. Also, one can note

that |S| = a + k(b + a) + b1 = 12k + 8 and similarly |S ′| = 12k + 9.

Thus |S ′| = |S| + 1. Hence in this case, γ ≤ γ ′ ≤ γ + 1.

Sub case 1.2. When n is even with n 6≡ (1 mod 5) and q is

even, then n = 5q + r where 2 ≤ r ≤ 5. Let q = 2k. Note that

k ≥ 1. Now (BA)kBr gives a dominating set S and Dr(CD)k gives

an inverse dominating set S ′ with respect to S. Also note that

|S ′| = |S| = 12k + br and hence in this case γ ′ = γ.

Sub case 1.3. When n is even with n 6≡ (1 mod 5) and q is

odd, then in this case, n = 5q + r and 2 ≤ r ≤ 5. Let q = 2k + 1;

k ≥ 0. Then A(BA)kBr gives a dominating set S and Dr(CD)kC

gives an inverse dominating set S ′ with respect to S. Note that

|S ′| = |S| = 12k + 6 + br and hence in this case also γ ′ = γ.

Case 2. When n is odd, we deal with 3 cases.

(1) n ≡ 1(mod 5).

(2) n 6= 1(mod 5) and q is even.

(3) n 6= 1(mod 5) and q is odd.

Sub case 2.1. When n is odd and n ≡ 1 (mod 5), then n = 5q+1

with q is even. Let q = 2k for k ≥ 1. Then (BA)kB1 gives a

dominating set S. And F5(DC)k−lDD1 gives an inverse dominating

set S ′ with respect to S. We shall verify that |S| = 12k + 2 and

that |S ′| = 12k + 3. Hence in this case γ ≤ γ ′ ≤ γ + 1.


Sub case 2.2. When n is odd with n 6≡ 1 (mod 5) and q is even,

then n = 5q + r and 2 ≤ r ≤ 5. Let q = 2k. Note that k ≥ 1. Then

(BA)kBr gives a dominating set S and Fr(DC)k gives an inverse

dominating set S ′ with respect to S. Also, one can verify that

|S| = |S ′| = 12k + br. From this, again we have γ ′ = γ.

Sub case 2.3. When n is odd with n 6≡ 1 (mod 5) and q is odd,

then n = 5q + r and 2 ≤ r ≤ 5. Let q = 2k + 1. Note that k ≥ 0.

Then A(BA)kBr gives a dominating set S and Fr(DC)kD gives an

inverse dominating set S ′ with respect to S and one can note that

|S| = |S ′| = 12k + 6 + br. Hence γ ′ = γ in this case too. Thus

γ ′5,n =

γ5,n or γ5,n + 1 if n ≡ 1(mod 5)

γ5,n otherwise.

Hence we have proved the following result:

Theorem 4.4.1.

γ ′5,n =

⌊

6n+65

⌋

if n = 2, 3, 7⌊

6n+85

⌋

or⌊

6n+85

⌋

+ 1 if n ≡ 1 (mod 5)⌊

6n+85

⌋

otherwise.

4.5 Inverse Domination in 6 × n grid graphs


verse dominating set for 6×n grid graphs. It was proved that in [7],


Figure 4.11: Blocks A, B, B1, B2, B3, B4 and B5 for 6 × n grid

Figure 4.12: Blocks B6 and B7 for 6 × n grid

for every n ≥ 4

γ6,n =

⌊

10n + 10

7

⌋

n ≡ 1(mod 7)

⌊

10n + 12

7

⌋

otherwise.

T. Y. Chang, Edwin Clark and E. O. Hare [7] gave a set of

blocks to construct a dominating set for a 6 × n grid graph. By

our assumption n ≥ 6. The case 6 × 6 has been dealt by the block

B6 of Figure 4.12. We give a set of blocks for the construction of

dominating and inverse dominating sets for 6 × n (for n ≥ 6) grid

graphs in Figure 4.11 and in Figure 4.12.

Here we give a dominating and an inverse dominating set for

6× n, n ≥ 6 and those for n ≤ 5 are obtained from the k × 6 grids

where 1 ≤ k ≤ 5. The blocks A, B, Br, 1 ≤ r ≤ 7 are used for

finding the dominating and inverse dominating sets for the 6 × n

grid graphs. Let n = 7q + r for 1 ≤ r ≤ 7. Let br = |V (Br) ∩ S|,


Figure 4.13: Blocks A, C1 and C3 for 7 × n grid

b ′r = |V (Br)∩S ′|, a = a ′ = |V (A)∩S| and b = b ′ = |V (B)∩S ′|.

Then note that a = b = a′

= b′

= 10, b1 = b′

1 = 2, b2 = b′

2 = 4,

b3 = b′

3 = 6, b4 = b′

4 = 7 and b5 = 8 and b′

5 = 9, b6 = b′

6 = 10 and

b7 = b′

7 = 11. We consider two cases according as q is even or q is

odd.

Case 1. If q is even then q = 2k. Note that k ≥ 1. Then

(BA)kBr gives a dominating set S and an inverse dominating set

S ′ with |S| = |S′

| = 12k + br.

Case 2. If q is odd, then q = 2k + 1 and note that k ≥ 0. Then

A(BA)kBr gives a dominating set S and an inverse dominating set

S ′ with |S| = |S′

| = 12k + 6 + br.

Note that the number of vertices contributed by each block to a

dominating set and an inverse dominating set are the same except

for B5, and that it differs by one for B5. Hence we see that γ ≤

γ ′ ≤ γ + 1 when n = 7q + 5; and that γ = γ ′ otherwise. Thus we

have proved the following theorem:


Figure 4.14: Blocks C4 and C5 for 7 × n grid

Theorem 4.5.1. For n ≥ 6,

γ ′6,n =

⌊

10n + 10

7

⌋

if n ≡ 1(mod 7)

⌊

10n + 12

7

⌋

or

⌊

10n + 12

7

⌋

+ 1 if n ≡ 5(mod 7)

⌊

10n + 12

7

⌋

otherwise.

4.6 Inverse Domination in 7 × n grid graphs


verse dominating set for the 7 × n grid graphs. From [7], we have

γ7,n =

⌊

5n + 3

3

⌋

for every 6 ≤ n ≤ 500.

We use the blocks A, C1, C3 given in Figure 4.13 and C4, C5, C6

and B given in Figure 5.14 and in Figure 5.15 to construct a dom-

inating set S and an inverse dominating set S ′ for a 7 × n graph.

Let a = |V (A) ∩ S|, a ′ = |V (A) ∩ S ′|, cr = |V (Cr) ∩ S| and

b = |V (B) ∩ S ′|. Note that a = a ′ = b = 10, c1 = c′

1 = 2,

c3 = c′

3 = 5, c4 = c′

4 = 7 and c5 = c′

5 = 9, and c6 = c′

6 = 11.

Let n = 6q + r; 1 ≤ r ≤ 6. As earlier, we give a dominating

4.7. INVERSE DOMINATION IN P2 × CN GRID GRAPHS 77

Figure 4.15: Blocks C6 and B for 7 × n grid

set and an inverse dominating set for 7 × n; n ≥ 7 and those for

n ≤ 6 are obtained from the k × 7 grids where 1 ≤ k ≤ 6. Since

n ≥ 7, we have q ≥ 1. We handle the cases r = 2, r = 3 and

r = 6 separately. When r = 2, Aq−1C4C4 gives a dominating set

S and an inverse dominating set S ′ and each has 10q + 4 vertices.

When r = 3, Aq−1C1C3C5 gives a dominating set S and an inverse

dominating set S ′ and each has 10q+6 vertices. When r = 6, AqC6

gives a dominating set S and C6Bq gives an inverse dominating set

S ′ and each has 11q + 1 vertices. AqCr gives a dominating set S

and an inverse dominating set S ′ for r= 1,4,5 with 10q + 2, 10q + 7

and 10q + 9 vertices respectively. Thus we see that γ ′7,n = γ 7,n for

all n. Therefore we have the following theorem:

Theorem 4.6.1.

γ ′7,n = γ 7,n =

⌊

5n + 3

3

⌋

for every 6 ≤ n ≤ 500.

4.7 Inverse Domination in P2 × Cn grid graphs

In this section, we give the domination number, dominating set, the

inverse domination number and an inverse dominating set for the

cylinder grid graphs P2 × Cn with n ≥ 3. We shall observe that

P2 × Cn can be obtained from P2 × Pn, by concatenating the ends

of it. We make use of the basic blocks A, B1, B2 and B3 given for


the P2 × Pn grid graphs in Figure 4.1. We now split into four cases

according to n ≡ i(mod 4) for i = 0, 1, 2, 3.

Case 1 When n ≡ 0(mod 4), n = 4q for some integer q ≥ 1.

We observe that as in P2 × Pn, Aq gives a dominating set with

2q vertices. Hence γ(P2 × Cn) ≤ 2q. By Theorem 1.2.3, we have

⌈ n1+∆(G)⌉ ≤ γ(G). Here ∆(G) = 3 and n = 2× 4q. Therefore we get

⌈ 8q1+3⌉ ≤ γ(G) ≤ 2q. Thus γ(P2 × C4q) = 2q. Note that the same

block Aq gives an inverse dominating set with 2q vertices. Since

γ(P2 × Cn) ≤ γ′

(P2 × Cn), we get γ′

(P2 × C4q) = 2q.

Case 2 When n ≡ 1(mod 4), n = 4q + 1 for some integer

q ≥ 1. Here, the block AqB1 gives a dominating set with 2q + 1

vertices. Hence γ(P2 × C4q+1) ≤ 2q + 1. As in Case 1, we get

⌈8q+21+3 ⌉ ≤ γ(G) ≤ 2q + 1. Thus γ(P2 × C4q+1) = 2q + 1. Note that

the same block AqB1 gives an inverse dominating set with 2q + 1

vertices. As in Case 1, we get γ′

(P2 × C4q+1) = 2q + 1.

Case 3 When n ≡ 2(mod 4), n = 4q + 2 for some integer q ≥ 1.

Here the block AqB2 gives a dominating set with 2q + 2 vertices.

Hence γ(P2×C4q+2) ≤ 2q+2. As in Case 1, we get ⌈8q+41+3 ⌉ ≤ γ(G) ≤

2q +2. Thus γ(P2 ×C4q+2) = 2q +1 or 2q +2. We now claim that

γ(P2×C4q+2) = 2q+2. We prove by induction on q. When q = 1, we

observe that γ(P2×C4q+2) = 4 = 2q+2. Assume the result for q−1.

Thus we have γ(P2×C4(q−1)+2) = 2(q−1)+2. That is γ(P2×C4q−2) =

2q. Consider the vertices of the γ-set D on the cylinder. Let V (P2×

C4q−2) = {u1, u2, u3, ......u4q−2, v1, v2, v3, ......v4q−2}. Without loss of

generality assume that ui ∈ D. Now sub-divide the edge uiui+1 four

times so as to get the new vertices u′

1, u′

2, u′

3, u′

4. Similarly sub-divide

the edge vivi+1 four times so as to get the new vertices v′

1, v′

2, v′

3, v′

4.

Now join the vertices u′

iv′

i for i = 1, 2, 3, 4. we shall note that the

4.7. INVERSE DOMINATION IN P2 × CN GRID GRAPHS 79

new graph obtained is nothing but P2×C4q+2. Let D1 = D∪{u′

4, v′

2}.

The vertices u′

4, v′

2 dominates all the new vertices except u′

1, which

is dominated by ui and the vertex u′

4 ∈ D1 dominates ui+1 which is

dominated by ui. Hence D1 is a dominating set of P2 × C4q+2 and

no set with fewer vertices than D1 can dominate P2 × C4q+2. Thus

D1 is the required γ-set of P2×C4q+2. Hence γ(P2×C4q+2) = 2q+2.

Therefore AqB2 gives a γ-set. Note that the same block gives an

inverse dominating set with 2q+2 vertices. Hence γ′

(P2×C4q+2) =

2q + 2.

Case 4 When n ≡ 3(mod 4), n = 4q + 3 for some integer

q ≥ 0. Here, the block AqB3 gives a dominating set with 2q + 2

vertices. Hence γ(P2 × C4q+3) ≤ 2q + 2. As mentioned earlier, we

get ⌈8q+61+3 ⌉ ≤ γ(G) ≤ 2q+2. Thus γ(P2×C4q+3) = 2q+2. Here also

the same block AqB3 gives an inverse dominating set with 2q + 2

vertices. Hence we get γ′

(P2 × C4q+3) = 2q + 2. Thus in all the

cases, we see that γ(P2 × Cn) = γ′

(P2 × Cn).

Hence we have proved the following theorem.

Theorem 4.7.1. γ′

(P2 × Cn) =

⌈n2⌉ if n ≡ 0 (mod 4)

⌈n+12 ⌉ otherwise.

Chapter 5

Inverse Domination Saturation

Number

In this chapter, we discuss about the inverse domination saturation

number and the corresponding inverse domination saturation set of

a graph. The concepts of domination saturation and domination

saturation number ds(G) were introduced by B.D. Acharya [1]. It

was noted that γ(G) ≤ ds(G) ≤ γ(G) + 1. A graph G is said to be

of class I with respect to domination saturation if ds(G) = γ(G),

where as G is said to be of class II if ds(G) = γ(G) + 1 [2].

In this chapter, we introduce the concept of inverse domination

saturation number. As in domination saturation, we classify graphs

into type I and type II inverse domination saturated graphs. Also

we introduce inverse domination unsaturation and we classify type

I and type II inverse domination unsaturated graphs.

In section 6.2, we give a motivation to define inverse domination

saturation number of a graph. We define and give some examples

for inverse domination saturation, inverse domination unsaturation

and inverse domination saturation number with respect to a vertex

and that to a graph. Also we give some results related to inverse

81

82 CHAPTER 5. INVERSE DOMINATION SATURATION NUMBER

domination saturation and classify into type I and type II inverse

domination saturated graphs. In section 6.3, we give some results

related to inverse domination unsaturation and classify into type I

and type II inverse domination unsaturated graphs.

5.1 Classification of Inverse Domination Satu-

ration

In this section, we classify the inverse domination saturation into

two classes.

Example 5.1.1. Let G = C5 be the cycle on five vertices and let

V (C5) = {1, 2, . . . , 5}. Here D = {2, 5} is a γ- set of G. D1′

=

{1, 4} and D2′

= {1, 3} are γ′

-sets of G with respect to D. One

can note that D is an inverse dominating set with respect to a γ-set

D1′

of G. Hence D is also a γ′

-set of G. Thus each vertex of G is

contained in a γ′

-set of G.

Example 5.1.2. Consider the path P5 with V (P5) = {1, 2, . . . , 5}.

Here D = {2, 5} is a γ-set of G and D′ = {1, 4} is a γ′

-set of G and

vice versa. Note that there is no γ′

-set of P5 containing the vertex

3. However {1, 3, 4} is an inverse dominating set with respect to D

which contains 3.

Example 5.1.3. Consider the graph G = K1,n with V (K1,n) =

{ 0, 1, 2, . . . , n}. Let 0 be the vertex of degree n. Then D = {0}

is the γ-set of G and D′ = {1, 2, . . . n} is the γ′

-set of G. One can

realize that 0 is not contained in any inverse dominating set of G

and so 0 is not contained in any γ′

-set of G.

5.1. CLASSIFICATION OF INVERSE DOMINATION SATURATION 83

After realizing these examples, we consider the following cases:

(i) There are graphs for which each vertex is contained in some

γ′

-set.

(ii) There are graphs for which some vertices are contained in some

inverse dominating set but not contained in any γ′

-set of G.

(iii) There are graphs in which some vertices are neither contained

in any γ′

-set of G nor in any inverse dominating set of G.

Hence we define inverse domination saturation of a vertex and

inverse domination saturation of a graph.

Definition 5.1.4. Let G be a connected graph and u ∈ V (G). Then

u is said to be inverse domination saturated if there exists an inverse

dominating set containing u, and is said to be inverse domination

unsaturated if there exists no inverse dominating set containing u.

Example 5.1.5. Every vertex of C5 is inverse domination satu-

rated.

Example 5.1.6. Let u be a vertex of a tree T , which is adjacent to

at least two pendent vertices. Then u lies in every γ-set of T . Hence

there exists no inverse dominating set containing u. Therefore u is

inverse domination unsaturated.

Definition 5.1.7. Let G be a connected graph and let u ∈ V (G).

Define d′S(u) by

d′S(u) =

Min{|S′

| : where S ′ is an inverse dominating set of G

containing u}

0 if no such set exists.


Example 5.1.8. Consider G = K1,n. Let u be the vertex of degree

n. Then D′S(u) = 0 and D′S(v) = n for every vertex v ∈ V (G) −

{u}.

Example 5.1.9. Consider the graph P8. Let V (P8) = {1, 2, . . . , 8}

and let {1, 8} be the pendant vertices. Then D = {2, 5, 8} is a γ-set

and D′ = {1, 4, 7} is a γ′

-set of P8 and vice versa. Hence D′S(i) = 3

for i = 1, 2, 4, 5, 7, 8. Note that there is no γ′

-set containing 3 or

6. But {1, 3, 6, 7} is an inverse dominating set containing 3 and 6.

Hence D′S(i) = 4 for i = 3, 6.

Definition 5.1.10. Let G be a connected graph. Then G is said

to be an inverse domination saturated, if each vertex u ∈ V (G) is

inverse domination saturated and is said to be an inverse domination

unsaturated if at least one vertex u ∈ V (G) is inverse domination

unsaturated.

Example 5.1.11. (i) All cycles Cn are inverse domination satu-

rated.

(ii) The corona H ◦ K1 is inverse domination saturated for any

graph H.

(iii) K1,n is inverse domination unsaturated.

Definition 5.1.12. Let G be an inverse domination saturated graph.

Then the inverse domination saturation number D′S(G) is defined

by

D′S(G) = Maxu∈V (G)

{D′S(u)}.


Example 5.1.13. Consider the Peterson graph given in Figure 5.1.

s1

s7

s6 s5

s2

s8

s4 s3

s

10

s

9

@@

@@

@@�

��

��

�AA

AA

AA�

��

��

�

��

CCCCCC

QQ

QQQ

��

��

��

AA

A

Figure 5.1: Peterson graph

Here D1 = {1, 2, 10}, D2 = {4, 5, 8} and D3 = {1, 5, 9} are γ-

sets. Also D′

1 = {3, 4, 7}, D′

2 = {2, 3, 6} and D′

3 = {3, 4, 7} are

the γ′

-sets. Therefore for each i (1 ≤ i ≤ 10) there exists a γ′

-set

containing i. Thus for 1 ≤ i ≤ 10, D′S(i) = 3 and so D′S(G) = 3.

Also note that γ′

(G)=3.

Remark 5.1.14. A connected graph G is inverse domination sat-

urated if and only if there exists a γ-set not containing v for some

v ∈ V (G). Also for a connected graph G, if γ(G) = γ ′(G), then G

is inverse domination saturated.

The following example shows that the converse of Remark 5.1.14

is not true.

Example 5.1.15. Consider the graph given in Figure 5.2.


t 8

t7

t

4

t

5

t

3

t

1

t

2

t

9t

10

t

11

t

12

t 6 @@

@@

��

��

��

��

��

��

@@

@@

@@

@@

Figure 5.2: Inverse domination saturated graph

Then D1 = {1, 4, 7, 11}, D2 = {2, 5, 6, 11}, D3 = {2, 7, 9, 12} are

γ-sets of G and D′

1,1 = {2, 5, 6, 9, 10}, D′

1,2 = {2, 5, 8, 10, 12}, D′

2,1 =

{1, 4, 7, 10, 12}, D′

2,2 = {1, 3, 7, 9, 10, }, D′

3,1 = {3, 4, 6, 8, 11} and

D′

3,2 = {1, 4, 5, 6, 11} are the corresponding γ′

- sets of G. Thus

each vertex in G is inverse domination saturated. Hence the graph

G is inverse domination saturated. But γ(G) 6= γ ′(G).


Lemma 5.1.16. Let G be a connected graph with γ(G) = γ ′(G).

Then γ ′(G) ≤ D′S(G) ≤ γ ′(G) + 1.

Proof. Let G be a connected graph with γ(G) = γ ′(G). Then by

Remark 5.1.14, G is an inverse domination saturated graph.

Case 1 Suppose that Dv′

is a γ ′-set containing v, for each

v ∈ V (G), then D′S(v) = γ ′(G) for each vertex v ∈ V (G). Then

D′S(G) = γ ′(G).

Case 2 Suppose that there exists a vertex, say u ∈ V (G), such

that there exists no γ′

-set containing u. Let T and T′

be a γ-

set and a γ′

-set of G respectively. Then u /∈ T ∪ T′

. Therefore

T′

∪{u} is the smallest inverse dominating set containing u. Hence

D′S(G) = Max{D′S(u)} for u ∈ V (G), implies that D′S(G) =

γ ′(G) + 1. Thus D′S(G) = γ ′(G) or γ ′(G) + 1. Hence γ ′(G) ≤

D′S(G) ≤ γ ′(G) + 1.

Proposition 5.1.17. Let G(6= Kn) be a connected graph with at

least two dominating vertices. Then D′S(G) = γ ′(G) + 1=2.

Proof. Since G has at least two dominating vertices, say u, v, we

have γ(G) = γ ′(G)=1. Then D = {u} is a γ-set of G and D′ = {v}

is a γ′

-set of G. Since G 6= Kn, there exists at least one vertex say

w ∈ V (G), such that d(w) < n− 1. Hence D′ ∪ {w} is the smallest

inverse dominating set containing w. Therefore D′S(w) = γ ′(G)+1.

Thus D′S(G) = γ ′(G) + 1=2.

Definition 5.1.18. An inverse domination saturated graph G is

said to be type I inverse domination saturated if D′S(G) = γ ′(G)

and it is said to be type II inverse domination saturated otherwise.


Example 5.1.19. Kn (n ≥ 2) and P4 are type I inverse domination

saturated.

Example 5.1.20. Note that γ′

(P5) = 2 and D′S(P5) = 3. Hence

P5 is type II inverse domination saturated.

Theorem 5.1.21. For all n ≥ 3, Cn is type I inverse domination

saturated.

Proof. In view of Theorem 1.2.3, there exists a γ′

-set containing

each vertex of Cn. Hence Cn is type I inverse domination saturated

for all n.

Lemma 5.1.22. For n ≥ 3, Pn is inverse domination saturated if

and only if n 6≡ 0(mod 3). Also Pn is type I inverse domination sat-

urated if n ≡ 1(mod 3) and is type II inverse domination saturated

if n ≡ 2(mod 3).

Proof. Assume that Pn is inverse domination saturated and let

V (Pn) = {1, 2, . . . 3k}. Suppose n ≡ 0(mod 3). Then n = 3k

for some k. Also D = {2, 5, 8, . . . 3k − 1} is a γ-set of G, and

D1′

= {1, 4, 7, . . . 3k − 2, 3k} and D2′

= {1, 3, 6, 9, . . . 3k} are γ′

-

sets of P3k. Note that γ(G) = k and γ ′(G) = k + 1 and that there

exists an inverse dominating set containing each element of V −D.

But there exists no inverse dominating set containing any element of

D, since D is the only γ-set of G. Hence P3k is inverse domination

unsaturated. Therefore n 6≡ 0(mod 3). Conversely, assume that

n 6≡ 0(mod 3). Then either n ≡ 1(mod 3) or n ≡ 2(mod 3).

Case (i) When n ≡ 1(mod 3), n = 3k + 1 for some k ≥ 1. Then

D = {1, 4, 7, . . . 3k − 2, 3k + 1} is a γ-set of G, with k + 1 elements


and D1′

= {2, 5, 8, . . . 3k − 1, 3k}, D2′

= {2, 3, 6, 9, . . . 3k} are

γ′

-sets of G. Note that D is an inverse dominating set of G with

respect to D1′

. Since |D| = |D1′

| = k + 1, D is also a γ′

-set of G.

Hence for each i ∈ V (G), there exists a γ′

-set containing i. Thus

P3k+1 is type I inverse domination saturated.

Case (ii) When n ≡ 2(mod 3), n = 3k + 2 for some k ≥ 1.

Then D = {1, 4, 7, . . . , 3k + 1} is a γ-set of G, with k + 1 elements

and D1′

= {2, 5, 8, . . . , 3k + 2}, is a γ′

-set of G. Note that D is

an inverse dominating set of G with respect to D1′

and also that

|D| = |D1′

| = k + 1. Therefore D is also a γ′

-set of G. Hence for

each i ∈ (D ∪ D′), there exists a γ′

-set containing i. But for the

vertices of the form 3i, there exists no γ′

-set containing it. However

D2′

= {2, 3, 6, 9, . . . , 3k, 3k + 2} is an inverse dominating set with

respect to D containing the vertices of the form 3i. Also note that

|D2′

| = k + 2 = γ ′(G) + 1. Therefore P3k+2 is type II inverse

domination saturated.

Theorem 5.1.23. Let G be a connected graph with γ(G).d(G) = n,

where d(G) is the domatic number of G. Then G is type I inverse

domination saturated.

Proof. Let D1, D2, D3, . . . Dd be a domatic partition of V (G) into

dominating sets. Therefore |Di| ≥ γ(G), for each i. Hence n =∑d

i=1 |Di| ≥ |Di|d(G) ≥ γ(G)d(G) = n. This implies that |Di| =

γ(G), for all i. Hence each Di is a γ-set of G. Then each Di is an

inverse dominating set with respect to Dj, for i 6= j, and hence a

γ′

-set of G. Thus each vertex in G lies in a γ′

- set of G. Hence G

is type I inverse domination saturated.


Remark 5.1.24. The converse of the Theorem 5.1.23 is not true

in general. For, consider the cycle C5. Let V (C5) = {1, 2, 3, 4, 5}.

Then D = {1, 4} is a γ-set and D1′

= {2, 5} and D2′

= {3, 5} are

γ′

sets containing each vertex of V (G)−D. Also note that D is an

inverse dominating set of G with respect to D′

1 and hence a γ′

-set of

G. Thus there exists a γ′

-set containing each vertex of C5. Hence

C5 is type I inverse domination saturated. But γ(G).d(G) 6= n since

γ(G) = 2 and d(G) = 2.

Theorem 5.1.25. Let G be a regular graph and domatically full.

Then G is type I inverse domination saturated.

Proof. Let G be a k-regular graph which is domatically full. Then

d(G) = δ(G) + 1 = k + 1. Let D1, D2, . . . , Dδ(G)+1 be a domatic

partition of G. Let u ∈ V (G) be any vertex. Then either u ∈ Di

or exactly one of its neighbors is in Di, for each i, since each Di is

a dominating set. Also for i 6= j, each vertex in Di is adjacent to

exactly one vertex in Dj. For if, a vertex in Di is adjacent to more

than one vertex in Dj, then since d(u) = k and there are k other

sets Di, u cannot be adjacent to any vertex in some set Dk. Also

note that |Di| = |Dj| for all i, j. For if, |Di| < |Dj| for some i 6= j,

then there are at least two vertices in Dj that are adjacent to one

vertex in Di, which gives a contradiction.

Next we claim that |Di| = γ, for all i. Clearly |Di| ≥ γ. Sup-

pose |Di| ≥ γ + 1 for some i, then n ≥ kγ + (γ + 1). That is

n ≥ (k + 1).γ(G) + 1, which implies γ(G) < n(k+1) = n

(∆(G)+1) , a

contradiction to γ ≥ n(∆(G)+1) . Hence |Di| = γ, for each i. Thus

∑

|Di| = n implies that γ(G)d(G) = n, where d(G) is the domatic

number of G. Then by Theorem 5.1.23, G is type I inverse domi-

nation saturated.

5.2. CLASSIFICATION OF INVERSE DOMINATION UNSATURATION 91

Remark 5.1.26. The converse need not be true in general. Con-

sider the graph G = C5. Clearly G is a regular graph which is type

I inverse domination saturated. But here, d(G) = 2 6= δ(G) + 1.

Therefore G is not domatically full.

5.2 Classification of Inverse Domination Unsat-

uration

In this section, we classify the inverse domination unsaturation into

type I inverse domination unsaturation and type II inverse domina-

tion unsaturation.

Example 5.2.1. Consider a twin star graph G in Figure 5.3 with

V (G) = {1, 2, . . . , 8}.

s

1

s

2

s

3

s4

s5

s6

s7

s

8

@@

@

��

�

@@

@

��

�

Figure 5.3: Twin star graph

Here D = {4, 5} is the γ-set of G and D′ = {1, 2, 3, 6, 7, 8} is

the γ′

−set of G. One can note that the vertices of D is neither

contained in a γ′

−set of G nor in an inverse dominating set of G.

Hence G is inverse domination unsaturated.

Example 5.2.2. Consider the 3-regular graph given in Figure 5.4.

Then D = {1, 4, 9, 14} is a γ-set of G. Also D1′

= {2, 5, 7, 8, 13, 15}

and D2′

= {3, 6, 7, 10, 12, 16} are γ′

-sets of G. Also note that


s3

s

2

s6

s4 s5

@@

@

@@

@

@@

@@

@@

s

16s

15

s

13s

14

s

12s

1s

11

��

��

��

��

�

��

�s

7s

8

s

9s

10

��

�

��

�

@@

@��

��

��

@@

@@

@@

��

��

��

@@

@@

@@

Figure 5.4: A 3-regular graph

T = {1, 5, 8, 15} is a γ-set of G, T1′

= {2, 4, 7, 9, 13, 14} and

T2′

= {3, 6, 7, 11, 12, 16} are γ′

-sets of G. Thus D′S(i) = 6, for

i = 2 to 16. But D′S(1) = 0. Hence G is inverse domination

unsaturated.


From these examples we infer that the class of inverse domina-

tion unsaturated graphs are of two types. This motivates to define

type I inverse domination unsaturated graphs and type II inverse

domination unsaturated graphs.

Definition 5.2.3. An inverse domination unsaturated graph is said

to be type I inverse domination unsaturated, if there exists no in-

verse dominating set containing any vertex of a γ-set of G, and is

said to be of type II inverse domination unsaturated if there exists

an inverse dominating set containing at least one vertex of every

γ-set of G.

Example 5.2.4. K1,n is type I inverse domination unsaturated.

Example 5.2.5. P6 is type I inverse domination unsaturated. For,

D = {2, 5, } is a γ-set of P6 and D1′

= {1, 4, 6} and D2′

= {1, 3, 6}

are the γ′

-sets of P6. Hence there is no inverse dominating set

containing the vertices of the γ-set.

Example 5.2.6. The 3-regular graph cited above in Figure 5.4

is type II inverse domination unsaturated. For, as seen in exam-

ple 5.2.2, the vertex 1 is the only vertex in the γ-set of G, that is

not contained in any inverse dominating set of G.

Remark 5.2.7. In view of Lemma 5.1.22, Pn is type I inverse dom-

ination unsaturated if and only if n ≡ 0(mod 3).

Theorem 5.2.8. A wounded spider G is type II inverse domination

unsaturated if and only if γ(G) 6= ∆(G).


Proof. Let G be a wounded spider which is inverse domination

unsaturated. Since G is a wounded spider, by Theorem 2.1.11

we have γ ′(G) = ∆(G). Suppose γ(G) = ∆(G), then we get

γ(G) = γ ′(G). By Remark 5.1.14, G is inverse domination sat-

urated, which is a contradiction. Hence γ(G) 6= ∆(G). Con-

versely, let G be a wounded spider with γ(G) 6= ∆(G). Since

γ ′(G) = ∆(G) for a wounded spider and γ(G) 6= ∆(G), we see

that at most ∆ − 2 edges of K1,∆(G) are sub-divided. Let V (G) =

{v, 1, 2, . . .∆, v1, v2, . . . vk}, where 1, 2, . . . , ∆ are the vertices inci-

dent with v and each vi is adjacent with i for i = 1, 2, 3, . . . , k ≤

∆ − 2. Then we see that D = {v, v1, v2, . . . , vk} is a γ-set and

D′ = {1, 2, . . . ∆} is a γ′

-set of G. Also T = {v, 1, 2, . . . , k} is a

γ-set and T′

= {v1, v2, . . . , vk, k + 1, . . . , ∆} is a γ′

-set of G. How-

ever there is no inverse dominating set containing v. Hence G is

type II inverse domination unsaturated.

Theorem 5.2.9. Let G be a connected graph. Then G is type I

inverse domination unsaturated if and only if G has a unique γ-set.

Proof. Let G be a connected graph such that G is type I inverse

domination unsaturated. Then there exists no inverse dominating

set containing any vertex of at least one γ-set say D of G. Suppose

G has more than one γ-set, then we have the following cases.

Case(i) Suppose that there exists at least two disjoint γ-sets

say D1 and D2. Since Dj ⊆ V (G) − Di, for i, j = 1, 2 and i 6= j,

each vertex of Di, for i = 1, 2 and lies in the inverse dominating set

V (G) − Dj. Also each vertex of V (G) − Di, for i = 1, 2 lies in the

inverse dominating set Dj ∪ {x}. Hence G is inverse domination

saturated, a contradiction.


Case (ii) Let D1 and D2 be any two γ-sets with D1 ∩ D2 6= ∅.

Then each vertex of D1 − D2 lies in the inverse dominating set

V (G)−D2 and each vertex of D2−D1 lies in the inverse dominating

set V (G) − D1. Thus at least one vertex in each Di, for i = 1, 2

is inverse domination saturated. Hence G is not type I inverse

domination unsaturated, a contradiction. Therefore G has a unique

γ - set.

Conversely, assume that G has an unique γ - set, say D. Then

trivially there exists no inverse dominating set containing any vertex

of D. Hence G is type I inverse domination unsaturated.

Theorem 5.2.10. Let G be a connected graph which is type I in-

verse domination unsaturated. Then there exists an inverse domi-

nating set D′ in which at least one vertex has no private neighbor

other than itself with respect to D′.

Proof. Let G be a connected graph which is type I inverse domi-

nation unsaturated. Then by Theorem 5.2.9, G has a unique γ-set

say D. Then by Lemma 1.2.10, every vertex in D has at least two

private neighbors other than itself. Let D′ = V (G) − D. Then any

vertex x ∈ D is not a private neighbor of any vertex in D′. Hence

at least one vertex in D′ has no private neighbors other than itself

with respect to D′.

Corollary 5.2.11. Let G be a connected graph which is type I in-

verse domination unsaturated. Then there exists a γ-set D in which

no vertex is a private neighbor of any vertex with respect to an in-

verse dominating set D′.


Proposition 5.2.12. Let G be a connected graph which is type I

inverse domination unsaturated. Then there exists a γ-set D, such

that γ(G − x) ≥ γ(G), for all x ∈ D.

Proof. Let G be a connected graph which is type I inverse domi-

nation unsaturated. Then by Theorem 5.2.9, G has a unique γ-set

say, D. Now by Lemma 1.2.13, γ(G−x) ≥ γ(G), for all x ∈ D.


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