TOTAL DOMINATION IN GRAPHS AND GRAPH MODIFICATIONS

TOTAL DOMINATION IN GRAPHS AND GRAPH

MODIFICATIONS

by

Wyatt Jules Desormeaux

THESIS

presented in fulfillment of the requirements

for the degree of

PHILOSOPHIAE DOCTOR

in

MATHEMATICS

in the

FACULTY OF SCIENCE

of the

UNIVERSITY OF JOHANNESBURG

Promoter: Prof. Michael A. Henning

Co-Promoter: Prof. Teresa W. Haynes

DECEMBER 2011

Dedication

I would like to dedicate this thesis to my father Robert Jules Desormeaux for always

stressing that education comes before anything else, and for the wisdom that he taught

me to always question those things that I want to believe harder than those things

which I don’t. To my mother Dr. Lorraine V. Desormeaux for always supporting me

in my educational endeavors, and for being the best mom in the World. Finally, to

the memory of my late brother Robert John Desormeaux.

2

Acknowledgments

First, I would like to thank Dr. Michael A. Henning of the University of Johannesburg

for his guidance, wisdom and help throughout all of the research in this thesis. I would

also like to thank Dr. Teresa W. Haynes of East Tennessee State University for all of

her patience and the time and guidance she has given me throughout this research.

Finally, I wish to thank the University of Johannesburg for their generous financial

support.

3

Contents

1 Introduction 13

1.1 Basic Terminology of Graph Theory . . . . . . . . . . . . . . . . . . . 14

1.2 Domination Parameters . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.3 Graph Modifications . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Critical and Stable Upon edge Removal 23

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.2 Total Domination Edge Critical Graphs . . . . . . . . . . . . . . . . . 25

2.3 Total Domination Edge Stable Graphs . . . . . . . . . . . . . . . . . 27

2.3.1 Disjoint Minimum Total Dominating Sets . . . . . . . . . . . 35

2.3.2 Bounds on the Diameter . . . . . . . . . . . . . . . . . . . . . 38

2.4 Open Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 An Extremal Problem 41

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4

3.2 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.2.1 Proof of Theorem 18 . . . . . . . . . . . . . . . . . . . . . . . 43

3.2.2 Proof of Theorem 19 . . . . . . . . . . . . . . . . . . . . . . . 48

4 Stable Graphs Upon Edge Addition 53

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.2 Total Domination Edge Addition Stable Graphs . . . . . . . . . . . . 54

4.2.1 A Characterization . . . . . . . . . . . . . . . . . . . . . . . . 54

4.2.2 Upper Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.2.3 Claw-Free Graphs . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.2.4 Realizability Results . . . . . . . . . . . . . . . . . . . . . . . 69

5 Changing and Stable Graphs 74

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

5.2 Preliminary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.2.1 Effects of Vertex Removal . . . . . . . . . . . . . . . . . . . . 75

5.2.2 Properties of Vertices in V −(G) ∪ V +(G) . . . . . . . . . . . . 77

5.3 γt-Changing Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

5.4 γt-Stable Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

6 Edge Lift Critical Trees 90

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

6.1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

5

6.2 γ−L -critical Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

6.3 γ+L -critical Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

6.3.1 The Family F . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

6.3.2 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

6.3.3 Key Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

6.3.4 Proof of Theorem 66 . . . . . . . . . . . . . . . . . . . . . . . 102

7 Edge Lifting and Total Domination 109

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

7.2 Effects of Lifting on the Total Domination Number . . . . . . . . . . 110

7.3 Classes of Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

7.3.1 The Family F1 . . . . . . . . . . . . . . . . . . . . . . . . . . 114

7.3.2 The Family F2 . . . . . . . . . . . . . . . . . . . . . . . . . . 115

7.3.3 The Families F3 and F4 . . . . . . . . . . . . . . . . . . . . . 116

7.3.4 The Family F5 . . . . . . . . . . . . . . . . . . . . . . . . . . 117

7.4 Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

8 Total Domination and Diameter Two 127

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

8.1.1 Known Results and Observations . . . . . . . . . . . . . . . . 127

8.2 Upper Bounds on the Total Domination Number . . . . . . . . . . . . 129

8.2.1 Upper Bounds in Terms of Order . . . . . . . . . . . . . . . . 129

6

8.2.2 Upper Bounds in Terms of Minimum Degree . . . . . . . . . . 133

8.2.3 Upper Bounds for Planar Graphs . . . . . . . . . . . . . . . . 136

8.3 Exact Values of γt(G) . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

8.4 Closing Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

9 The Annihilation Number 141

9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

9.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

9.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

9.3.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

9.4 Known Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

9.5 Proof of Theorem 112 . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

9.6 Proof of Theorem 113 . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

9.7 Characterization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

7

List of Figures

1.1 A graph. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.2 A graph G and G. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.3 Graphs G and G − ab. . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.4 Graphs G and G + ac. . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.5 Graphs G and G − a. . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.6 Graphs G and Guvx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1 A tree T19 in the family T . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2 A triangle-free example. . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.1 The graph G3,6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.2 The circulant graph G8 = C8〈1, 4〉. . . . . . . . . . . . . . . . . . . . 46

3.3 The graph H5,3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.1 The graph G. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.2 The graph G when n = 12 and k = 5. . . . . . . . . . . . . . . . . . . 72

8

5.1 A graph in the family H. . . . . . . . . . . . . . . . . . . . . . . . . . 85

5.2 The graph G when k = 2 and H = P3. . . . . . . . . . . . . . . . . . 89

7.1 Graph G. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

7.2 The graph Hi. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

7.3 The graph G when k = 6. . . . . . . . . . . . . . . . . . . . . . . . . 120

7.4 The graph F when k = 5. . . . . . . . . . . . . . . . . . . . . . . . . 120

8.1 The graphs F1, F2 and F3. . . . . . . . . . . . . . . . . . . . . . . . . 130

8.2 The graph G9 with γt(G9) = 4. . . . . . . . . . . . . . . . . . . . . . 131

9.1 Operations to build Γ . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

9

Summary

In this thesis, our primary objective is to investigate the effects that various graph

modifications have on the total domination number of a graph. In Chapter 1, we

introduce basic graph theory concepts and preliminary definitions.

In Chapters 2 and 3, we investigate the graph modification of edge removal. In

Chapter 2, we characterize graphs for which the removal of any arbitrary edge in-

creases the total domination number. We also begin the investigation of graphs for

which the removal of any arbitrary edge has no effect on the total domination number.

In Chapter 3, we continue this investigation and determine the minimum number of

edges required for these graphs. The contents of Chapters 2 and 3 have been published

in Discrete Applied Mathematics [15] and [16].

In Chapter 4, we investigate the graph modification of edge addition. In particular,

we focus our attention on graphs for which adding an edge between any pair of

nonadjacent vertices has no effect on the total domination number. We characterize

these graphs, determine a sharp upper bound on their total domination number and

determine which combinations of order and total domination number are attainable.

10

11

We also study claw-free graphs which have this property. The contents of this chapter

were published in Discrete Mathematics [20].

In Chapter 5, we investigate the graph modification of vertex removal. We char-

acterize graphs for which the removal of any vertex changes the total domination

number and find sharp upper and lower bounds on the total domination number of

these graphs. We also characterize graphs for which the removal of an arbitrary vertex

has no effect on the total domination number and we further show that they have no

forbidden subgraphs. The contents of this chapter were published in Discrete Applied

Mathematics [14].

In Chapters 6 and 7, we investigate the graph modification of edge lifting. In

Chapter 6, we show that there are no trees for which every possible edge lift decreases

the domination number, and we characterize trees for which every possible edge lift

increases the domination number. The contents of Chapter 6 were published in the

journal Quaestiones Mathematicae [17]. In Chapter 7, we show that there are no

trees for which every possible edge lift decreases the total domination number and

that there are no trees for which every possible edge lift leaves the total domination

number unchanged. We characterize trees for which every possible edge lift increases

the total domination number. At the time of the writing of this thesis, the contents of

Chapter 7 have been published online in the Journal of Combinatorial Optimization

[18] and will appear in print in a future issue.

In Chapter 8, we determine upper bounds on the total domination number of

12

diameter two graphs. We also give exact values of the total domination number for

families of these graphs with certain forbidden subgraphs. Many results related to the

work of chapter 8 have recently been discovered after the conclusion of the research

for this thesis. Some of the work in chapter 8 will appear in a future joint publication.

In Chapter 9, we determine upper and lower bounds on the total domination

number of trees in terms of their annihilation number. The contents of Chapter 9

were submitted to Discrete Applied Mathematics [19] for publication.

Chapter 1

Introduction

The primary purpose of this thesis is to study the effect that various graph modifi-

cations have on the total domination number of a graph. In addition, we study the

effect that the operation of edge lifting has on the domination number of a graph.

We finish this work with a study of the total domination number of diameter two

graphs and the relationship between the total domination number of a graph and its

annihilation number. In Section 1.1, we introduce the basic terminology of graph

theory used in this work. In Section 1.2, we introduce the definitions of domination

and total domination. In Section 1.3, we describe each of the graph modifications

considered in this work.

13

CHAPTER 1. INTRODUCTION 14

G

Figure 1.1: A graph.

1.1 Basic Terminology of Graph Theory

As defined in [5], a graph G = (V (G), E(G)) is a nonempty, finite set of elements

called vertices together with a (possibly empty) set of unordered pairs of distinct

vertices of G called edges. The vertex set of G is denoted by V (G) and the edge set

of G is denoted by E(G). When there is no risk of ambiguity, these are denoted V

and E respectively. In Figure 1.1, we have an example of a graph.

In this thesis, we will be studying simple graphs, which are graphs for which there

exists at most one edge between any two vertices, and for which the endpoints of any

edge are distinct. The graphs we will examine are finite and undirected. Given any

graph G, the order of G, denoted n(G) = |V (G)|, is the number of vertices in G. The

size of G, denoted m(G) = |E(G)|, is the number of edges in G. If there is no risk of

ambiguity, these are written as n and m respectively. For example, for the graph G

in Figure 1.1, the order n(G) = 10 and the size m(G) = 15. The complement of G,

denoted G, is a graph with V (G) = V (G) and E(G) = {ab|ab 6∈ E(G)}.

G GFigure 1.2: A graph G and G.


For example, consider the graphs G and G seen in Figure 1.2. For any vertices

v, u ∈ V (G), u and v are adjacent if uv ∈ E(G). A u-v path is a finite alternating

sequence {u = v0, e1, v1, e2, . . . , ek, vk = v} of vertices and edges such that ei = vi−1vi

for i = 1...k and ei = ej if and only if i = j. Among all u-v paths, the number of edges

in a shortest length u-v path is known as the distance from u to v, denoted by dG(u, v)

(or d(u, v) if there is no risk of ambiguity). The eccentricity eG(v) (or e(v) if there is

no ambiguity) of a vertex v ∈ V is the distance between v and a vertex farthest from

v in G. The maximum eccentricity among the vertices of G is its diameter, which is

denoted by diam(G). A vertex v with e(v) = diam(G) is called a diametrical vertex

or equivalently it is called a peripheral vertex. A path of length diam(G) between

diametrical vertices is called a diametrical path. We say that a graph G is a diameter

k graph if diam(G) = k. The girth of G, denoted g(G), is the length of a shortest

cycle of G.

For any vertex v ∈ V (G), the open neighborhood of v is NG(v) = {u ∈ V (G) | uv ∈

E(G)}, and the closed neighborhood NG[v] = NG(v) ∪ {v}. For a set S ⊆ V (G),

its open neighborhood NG(S) = ∪v∈SNG(v), and its closed neighborhood NG[S] =

NG(S) ∪ S. These are sometimes denoted N(v), N [v], N(S) or N [S] respectively if

there is no risk of ambiguity. The degree of a vertex v is degG(v) = |N(v)|. This is

sometimes denoted as dG(v) or d(v) if there is no risk of ambiguity. The minimum

degree of G is δ(G) = min{d(v)|v ∈ V (G)}. The maximum degree of G is ∆(G)

=max{d(v)|v ∈ V (G)}. A vertex of degree zero is an isolated vertex. A vertex

of degree one is called a leaf or an endvertex, its neighbor a support vertex and its

incident edge a pendant edge. A support vertex adjacent to two or more leaves is


called a strong support vertex.

Given S ⊆ V (G), and v ∈ S, a vertex w ∈ V (G) is an open S-private neighbor of

v if NG(w) ∩ S = {v} and a closed S-private neighbor of v if NG[w] ∩ S = {v}.

The open S-private neighborhood of v, denoted by pn(v, S), is the set of all vertices

in the open neighborhood of v but not in the open neighborhood of S \ {v}; that is,

pn(v, S) = N(v) \ N(S \ {v}). Thus if u ∈ pn(v, S), then N(u) ∩ S = {v}. The set

epn(v, S) = pn(v, S)∩ (V \ S) is called the open external S-private neighbor set of v,

while a vertex u ∈ epn(v, S) is called an open external S-private neighbor of v. We

define the open internal S-private neighbor set of v to be the set ipn(v, S) = pn(v, S)∩

S and call a vertex u ∈ ipn(v, S) an open internal S-private neighbor of v. For a subset

A ⊆ S, we define pn(A, S) =⋃

v∈A pn(v, S) and epn(A, S) = pn(A, S) ∩ V \ S.

For ease of notation, when discussing the concepts of open S private neighbor,

open S private neighborhood, open external S private neighbor, open internal S-

private neighbor, open external S private neighborhood and open internal S-private

neighborhood, we will drop the use of the word open throughout this work. Further-

more, if no ambiguity will result, we may at times drop reference to the set S as

well.

If S ⊆ V (G), then we denote the subgraph of G induced by S as G[S]. If uv ∈

E(G) for every u, v ∈ S, then S forms a clique of order |S|, and G[S] is called a

complete graph of order |S|. If uv 6∈ E(G) for every u, v ∈ S, then S is an independent

set of order |S| and G[S] is called an empty graph of order |S|. A set P ⊆ V (G) is a

packing if N [u] ∩ N [v] = ∅ for every u, v ∈ P . If X and Y are two subsets of V (G),

then we denote the set of all edges of G that join a vertex of X and a vertex of Y by


[X, Y ].

For graphs G and H, the Cartesian product G2H of G and H is the graph with

vertex set V (G) × V (H) where two vertices (ui, vj) and (uh, vk) are adjacent if and

only if either ui = uh and vjvk ∈ E(H) or vj = vk and uiuh ∈ E(G).

The corona of graphs G and H, denoted G ◦ H, is formed from G by adding for

each v ∈ V (G), a copy of H and the edges from v to every vertex in H. In particular,

the corona G ◦ K1 is the graph formed from G by adding a new vertex v′ and the

pendant edge vv′ for each v ∈ V (G). The 2-corona of a graph G, denoted G ◦ K2,

is constructed from G by adding for each vertex v ∈ V (G), two new vertices v′ and

v′′ and the two pendant edges vv′ and vv′′. Hence, G ◦ K1 has order 2|V (G)|, while

G ◦ K2 has order 3|V (G)|.

We refer to a nontrivial tree as a tree with at least two vertices. A caterpil-

lar is a tree with the property that removing all its leaves forms a path called its

spine. The code of the caterpillar having spine Pk = v1v2 . . . vk is the ordered k-tuple

(`1, `2, . . . , `k), where `i is the number of leaves adjacent to vi. A cycle on n vertices

is denoted by Cn and a path on n vertices by Pn. A star is a tree with at most one

non-leaf vertex. For r, s ≥ 1, a double star S(r, s) is a tree with exactly two vertices

that are not leaves, with one adjacent to r leaves and the other to s leaves. A galaxy

is a union of stars. For k ≥ 2, a subdivided star, denoted K∗1,k, is the tree obtained

from a star K1,k by subdividing each edge exactly once.

If a graph G does not contain an induced subgraph that is isomorphic to some

graph F , then we say that G is F -free. In particular, if F = C3, we say that G is

triangle-free, while if F = C4, we say that G is quadrilateral-free. More generally,


if F1, F2, . . . , Fk are k graphs and G does not contain an induced subgraph that is

isomorphic to Fi for all i, 1 ≤ i ≤ k, then we say that G is (F1, F2, . . . , Fk)-free. For

a set P ⊆ V (G), let P be the corresponding set of vertices in V (G). For a vertex

v ∈ V (G), let v represent the corresponding vertex in V (G).

Given a graph G with vertex set V (G), a proper coloring of G is a partitioning of

V (G) into independent sets. These sets are called color classes. A proper coloring

of G that has a minimum number of color classes is called a χ(G)-coloring and the

number of color classes in such a coloring is χ(G). By a weak partition of a set,

we mean a partition of the set in which some of the subsets may be empty. Other

definitions will be given as needed throughout this work. For other definitions and

terminology related to graph theory, the interested reader is referred to [5, 30, 31].

1.2 Domination Parameters

A set S ⊆ V (G) is a dominating set (abbreviated DS) if N [S] = V (G). In other

words, every vertex in V (G) \ S is adjacent to a vertex in S. Every graph has a DS

since V (G) is such a set. Among all DS of G, a DS with minimum cardinality is said

to be a γ(G)-set. Its cardinality is known as the domination number of G and it is

denoted by γ(G).

While we will consider the domination number in detail in Chapter 6 of this work,

our primary attention in this work will be on the total domination number. In [10],

Cockayne et al. defined a total dominating set (abbreviated TDS) of a graph G as a

set S ⊆ V (G) such that every vertex of G is adjacent to a vertex of S. Stated more


formally, S is a TDS if V (G) = N(S). Note that total domination is undefined for

graphs with isolated vertices; however, every graph G for which δ(G) ≥ 1 has a TDS

since V (G) is such a set. Among all TDS of G, a TDS with minimum cardinality is

said to be a γt(G)-set. Its cardinality is known as the total domination number of G

and it is denoted by γt(G). We note that a TDS of G contains all the support vertices

of G.

For subsets S, T ⊆ V , we say the set S dominates the set T if T ⊆ N [S] and S

totally dominates T if T ⊆ N(S).

We will also make passing reference to two other domination parameters in this

work. An independent dominating set, abbreviated IDS, of a graph G is a set S ⊆

V (G) that is a DS of G and is an independent set. The independent domination

number of G, denoted by i(G), is the minimum cardinality of an IDS of G. An IDS

of G of cardinality i(G) is called an i(G)-set.

In [26], Harary and Haynes defined a set S ⊆ V (G) to be a double dominating

set, abbreviated DDS, of G if for every vertex v ∈ V (G), |N [v] ∩ S| ≥ 2. The double

domination number γ×2(G) is the minimum cardinality of a DDS of G and a DDS of

cardinality γ×2(G) is called a γ×2(G)-set. We note that γt(G) ≤ γ×2(G).

The literature on domination parameters has been surveyed and detailed in the

two books by Haynes, Hedetniemi, and Slater [30, 31]. A recent survey of total

domination in graphs can be found in [38].


1.3 Graph Modifications

We will primarily investigate four different graph modifications in this work.

The first graph modification we will examine is edge removal. Given any graph G

and edge ab ∈ E(G), the graph formed from G by removing the edge ab is denoted

G − ab. As an illustration, consider the graphs G and G − ab seen in Figure 1.3

a

b

d

c

Ga

b

d

c

G − ab

Figure 1.3: Graphs G and G − ab.

The second graph modification we will consider is edge addition. Given any edge

ac ∈ E(G), the graph formed by adding the edge ac to G is denoted by G + ac. As

an illustration, consider the graphs G and G + ac seen in Figure 1.4

a

b

d

c

Ga

b

d

c

G + ac

Figure 1.4: Graphs G and G + ac.

The third graph modification we will consider is vertex removal. Given any vertex

a ∈ V (G), the graph formed from G by removing the vertex a and all of its incident

edges is denoted by G − a. As an illustration, consider the graphs G and G − a seen

in Figure 1.5

The fourth graph modification we will consider is edge lifting. The process of

edge lifting (also called edge splitting) was introduced by Lovasz in [44, 45] to study


a

b

d

c

G

b

d

c

G − a

Figure 1.5: Graphs G and G − a.

u x

v

G

u x

v

Guvx

Figure 1.6: Graphs G and Guvx .

edge connectivity in graphs. Let vertices u and v be at distance two apart in a

graph G, and let x be a common neighbor of u and v. We denote the induced path

between u and v via x as uxv. We say that ux and vx are lifted off x and call the

operation of removing ux and vx and adding uv to E(G) an edge lift. The graph

formed from G by an edge lift on uxv is denoted as Guvx , where V (Guv

x ) = V (G) and

E(Guvx ) = E(G) \ {ux, vx} ∪ {uv}. The action of edge lifting is a hybrid between

edge removal and edge addition, in that two edges are removed from a graph and a

new edge is added to it. As an illustration, consider the graphs G and Guvx seen in

Figure 1.6.

When a graphical parameter is of interest in an application, it is often important

to know how the parameter behaves when the graph is modified. For instance, the

effects of removing or adding an edge, or removing a vertex have been considered


on parameters such as connectivity, chromatic number and domination number. Van

der Merwe [53] initiated the study of those graphs where the total domination num-

ber decreases upon the addition of any edge. This concept was further investigated

in [25, 34, 35, 36, 50, 51, 52] and elsewhere. Goddard, Haynes, Henning, and van der

Merwe [23] began the study of the graphs whose total domination number decreases

upon the removal of any vertex. Further properties of these graphs were explored in

[7, 43, 54, 55, 57, 58]. A search of the literature did not find any prior work done

on relating the operation of edge lifting to either the domination number or total

domination number of a graph.

Chapter 2

Total domination critical and

stable graphs upon edge removal

2.1 Introduction

In [3] and [56], the authors characterized graphs for which the domination number

changes upon the removal of any edge. In [56], the authors briefly investigated graphs

where the domination number remains unchanged on the removal of any edge. These

graphs were extensively studied by Haynes et.al. in [28] and [29]. An extensive review

of the literature showed that similar problems were not addressed relative to the total

domination number of a graph. Accordingly, in this chapter, we study those graphs

where the total domination number increases after the removal of any edge. We also

study those graphs where the total domination number remains unchanged after the

removal of any edge.

23

CHAPTER 2. CRITICAL AND STABLE UPON EDGE REMOVAL 24

Total domination is undefined for graphs with isolated vertices. Consequently, to

aid in our investigation of graphs that are stable or critical after the deletion of an

edge, we say that γt(G) = ∞ if the graph G has an isolated vertex. For instance if

we delete a pendant edge e ∈ E(G), then γt(G − e) = ∞.

We say that a graph G is total domination edge stable, or γt-stable for short, if

the removal of any edge of G does not change the total domination number, that is,

γt(G − e) = γt(G) for every edge e ∈ E(G). If γt(G) = k and G is γt-stable, we say

that G is kt-stable. We say that G is total domination edge critical, or γt-critical for

short, if the removal of any edge in the graph changes the total domination number,

that is, γt(G − e) 6= γt(G) for every edge e ∈ E(G). We note that removing an edge

from a graph cannot decrease the total domination number. Hence if G is γt-critical,

then γt(G − e) > γt(G) for every edge e ∈ E(G).

An edge e ∈ E(G) is a stable edge of G if γt(G − e) = γt(G), while e is a critical

edge of G if γt(G− e) > γt(G). Thus every edge in a γt-stable graph is a stable edge,

while every edge in a γt-critical graph is a critical edge.

A vertex v in G is called a γt-good vertex if v is in some γt(G)-set, and we define

Tt(G) to be the set of all γt-good vertices of G.

We will need the following property of minimal TDSs established by Cockayne et

al. [10].

Proposition 1 ([10]) If S is a minimal TDS of a connected graph G, then for each

vertex v ∈ S, |epn(v, S)| ≥ 1 or G[S \ {v}] contains an isolated vertex.


2.2 Total Domination Edge Critical Graphs

Our aim in this section is to present a characterization of γt-critical graphs. For this

purpose, we define a family of trees T as follows.

Definition 2.2.1 A tree T ∈ T if T is a nontrivial star, or a double star, or if T

can be obtained from a subdivided star K∗1,k, where k ≥ 2, by adding zero or more

pendant edges to the non-leaf vertices of K∗1,k.

The tree T19, for example, shown in Figure 2.1 belongs to the family T and can

be obtained from a subdivided star K∗1,5.

u u u u u u u u u u

u u u u u

u

u u u

��

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�

@@

@

HHHHHH

@@

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��

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AAA

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Figure 2.1: A tree T19 in the family T .

We shall need the following lemma.

Lemma 2 If G is a γt-critical graph, then for every γt(G)-set S, G[S] is a galaxy of

nontrivial stars.

Proof. Let S be any γt(G)-set in the γt-critical graph G, and let GS = G[S]. Let e

be an arbitrary edge in GS . If both ends of e have degree at least 2 in GS , then S

is a TDS in G − e, and so γt(G − e) ≤ |S| = γt(G), contradicting the fact that G is


γt-critical. Hence at least one end of the edge e is a leaf in GS , implying that GS is

a galaxy of nontrivial stars. 2

We are now ready to present our characterization of γt-critical graphs.

Theorem 3 A connected graph G is γt-critical if and only if G ∈ T .

Proof. Assume that G = (V, E) is γt-critical. Let S be any γt(G)-set. By Lemma 2,

G[S] is a galaxy of nontrivial stars. If v is a leaf in G[S] and v is adjacent to a vertex

of degree at least two in G[S], then by Proposition 1, |epn(v, S)| ≥ 1. Thus, v has an

external private neighbor and is therefore adjacent to at least one vertex in V \S. For

every edge e of G, if S is a TDS in G− e, then γt(G− e) ≤ |S| = γt(G), contradicting

the fact that G is γt-critical. Hence for every edge e of G, the set S is not a TDS in

G − e. This implies that V \ S is an independent set and that each vertex in V \ S

is adjacent to exactly one vertex of S and is therefore a leaf in G. Thus since G is

connected, the subgraph G[S] is connected. Hence, G[S] is a star. If G[S] = K2, then

G is either a star or a double star, and so G ∈ T . Hence we may assume that G[S] is

a star K1,k where k ≥ 2. As observed earlier, each leaf in the star G[S] is adjacent to

at least one vertex in V \ S. Let L denote a set of k vertices in V \ S that dominate

the set of k leaves in G[S]. Then, G[S ∪ L] = K∗1,k and G can be obtained from this

subdivided star by adding zero or more pendant edges to each vertex of S. Thus,

G ∈ T .

Now, assume that G ∈ T . Let G = (V, E) and let e ∈ E. If e is incident with a

leaf in G, then γt(G−e) = ∞, and so e is a critical edge. Hence, we may assume that

e is not incident with a leaf in G. In particular, G is not a star. If G is a double star,


then e joins the two central vertices of G. Thus, γt(G − e) = 4 while γt(G) = 2, and

so e is a critical edge. Hence we may assume that G is not a double star. Thus, G is

obtained from a star T = K1,k, for some k ≥ 2, by adding at least one pendant edge

to each leaf of T and adding zero or more pendant edges to the center v of the star

T . Every edge in the set E \ E(T ) is incident with a leaf in G. Hence, by our earlier

assumptions, e ∈ E(T ). But then γt(G− e) = k + 2 while γt(G) = k + 1 (irrespective

of whether v is a support vertex of G). Hence, once again, the edge e is a critical

edge. Therefore, G is γt-critical. 2

2.3 Total Domination Edge Stable Graphs

In this section, we consider γt-stable graphs. Recall that, by convention, if G is a

graph with an isolated vertex, then γt(G) = ∞. Thus if e is an edge of a graph

G incident with a leaf, then γt(G − e) = ∞. We begin with the following simple

observation.

Observation 4 Let G be a graph. Then the following properties hold.

(a) If G is γt-stable, then δ(G) ≥ 2.

(b) If e is a stable edge in G, then every γt(G − e)-set is a γt(G)-set.

We shall need the following proposition.

Proposition 5 A graph G is γt-stable if and only if δ(G) ≥ 2, and for each e =

uv ∈ E(G), there exists a γt(G)-set S such that one of the following conditions are

satisfied.


(a) u, v /∈ S.

(b) u, v ∈ S, |N(u) ∩ S| ≥ 2 and |N(v) ∩ S| ≥ 2.

(c) Without loss of generality, if u ∈ S and v /∈ S, then |N(v) ∩ S| ≥ 2.

Proof. Assume that G is γt-stable. By Observation 4, δ(G) ≥ 2. Let e = uv be an

arbitrary edge of G. Let G′ = G− uv and let S be any γt(G′)-set. By Observation 4,

the set S is a γt(G)-set. If u, v /∈ S, then condition (a) holds. Hence we may assume,

renaming vertices if necessary, that u ∈ S. If v ∈ S, then since S is a TDS for G′,

|NG′(u) ∩ S| ≥ 1 and |NG′(v) ∩ S| ≥ 1, and so |NG(u) ∩ S| ≥ 2 and |NG(v) ∩ S| ≥ 2.

Thus condition (b) holds. If v /∈ S, then since S is a TDS for G′, |NG′(v) ∩ S| ≥ 1,

and so |NG(v)∩S| ≥ 2. Thus condition (c) holds. Hence the set S is a γt(G)-set such

that one of the three conditions (a), (b) and (c) are satisfied.

Assume, next, that δ(G) ≥ 2 and that for every edge e = uv ∈ E(G), there exists

a γt(G)-set S satisfying the hypothesis. In all three conditions (a), (b) and (c), the

set S is also a TDS for G−e. Hence, γt(G) ≤ γt(G−e) ≤ |S| = γt(G). Consequently,

γt(G) = γt(G − e). Therefore, the graph G is γt-stable. 2

As a consequence of Proposition 5, we have the following corollary. Recall that

Tt(G) is the set of γt-good vertices of G; that is, Tt(G) = {v ∈ V (G) | v belongs to

some γt(G)-set }.

Corollary 6 Let G be a γt-stable graph. Then, G has the following properties.

(a) G has at least two distinct γt(G)-sets.

(b) γ×2(G) ≤ |Tt(G)|.


Proof. Let G = (V, E) be a γt-stable graph. Assume that S ⊆ V is the only γt(G)-

set. Let u be an arbitrary vertex in S and let v be a vertex in S that is adjacent to

u. Then the unique γt(G)-set S satisfies condition (b) of Proposition 5. In particular,

|N(u) ∩ S| ≥ 2. Hence since u is an arbitrary vertex in S, we have that δ(G[S]) ≥ 2.

Thus, by Proposition 1, |epn(u, S)| ≥ 1 for every vertex u ∈ S. Let u ∈ S and let

v ∈ epn(u, S). Then, |N(v) ∩ S| = 1, contradicting the fact that the set S satisfies

condition (c) of Proposition 5. This establishes part (a).

Let v be an arbitrary vertex of G. If v ∈ Tt(G), then v is contained in some

γt(G)-set S. Thus, |N(v) ∩ S| ≥ 1. By definition of the set Tt(G), we note that

S ⊆ Tt(G), and therefore |N(v) ∩ Tt(G)| ≥ 1. If v /∈ Tt(G), then by Proposition 5,

there exists a γt(G)-set D such that |N(v) ∩ D| ≥ 2. Thus, since D ⊆ Tt(G), we

have that |N(v) ∩ Tt(G)| ≥ 2. Hence for every vertex v ∈ V , we have that either

v ∈ Tt(G) and |N(v) ∩ Tt(G)| ≥ 1 or v /∈ Tt(G) and |N(v) ∩ Tt(G)| ≥ 2. Therefore,

|N [v] ∩ Tt(G)| ≥ 2 for all v ∈ V . Thus, Tt(G) is a double dominating set for G, and

so γ×2(G) ≤ |Tt(G)|. 2

As a further consequence of Proposition 5, we have the following property of

kt-stable graphs where k = 3.

Proposition 7 If G is 3t-stable graph, then every edge of G is contained in an in-

duced cycle of length 3, 4 or 5.

Proof. Let G be a 3t-stable graph. Then, γt(G) = 3 and every edge of G is stable.

Let S be a γt(G)-set. We note that G[S] is either a path P3 or a cycle C3. Let e = uv

be an edge of G. By Proposition 5, the set S satisfies one of the three conditions (a),


(b) or (c) in the statement of Proposition 5. If the set S satisfies condition (b), then,

u, v ∈ S, |N(u)∩S| ≥ 2 and |N(v)∩S| ≥ 2, implying that G[S] = K3, and so u and v

belong to a triangle in G. Hence we may assume that the set S satisfies condition (a)

or condition (c).

Suppose the set S satisfies condition (a). Then, u, v /∈ S. If u and v have a

common neighbor in S, then these three vertices induce a triangle in G, as desired.

Hence we may assume that u and v have no common neighbor in G. Let x and y be

neighbors of u and v, respectively, in S. By assumption, x 6= y. If xy ∈ E(G), then

{u, v, x, y} induces a 4-cycle, as desired. Hence we may assume that no neighbor of

u in S is adjacent with a neighbor of v in S. In particular, xy /∈ E(G). But then x

and y are the ends of a path P3 in G[S], and so S ∪ {u, v} induces a 5-cycle in G, as

desired.

Suppose the set S satisfies condition (c). Without loss of generality, we may

assume u ∈ S and v /∈ S, and so |N(v)∩ S| ≥ 2. Thus there exists a neighbor w of v

in S different from u. If uw ∈ E(G), then {u, v, w} induce a triangle in G, as desired.

Hence we may assume that no neighbor of v in S different from u is adjacent with u.

In particular, uw /∈ E(G). But then u and w are the ends of a path P3 in G[S], and

so S ∪ {v} induces a 4-cycle in G, as desired. 2

The total domination number of a path Pn and a cycle Cn on n ≥ 3 vertices is

easy to compute. We shall need the following observation.

Observation 8 ([37]) The following properties hold.

(a) For n ≥ 3, γt(Cn) = γt(Pn) = bn/2c + dn/4e − bn/4c.

(b) For any integer k ≥ 2, there exists an integer n such that γt(Cn) = k.


(c) Every cycle is a γt-stable graph.

(d) For every integer k ≥ 2, there exists a kt-stable graph.

We describe next a construction that builds γt-stable graphs.

Theorem 9 For any positive integer k ≥ 2, and any graph G, there exists a connected

kt-stable graph H such that G is a vertex induced subgraph of H.

Proof. By Observation 8, there exists a kt-stable cycle. Let F be any connected

kt-stable graph (not necessarily a cycle) and let w ∈ V (F ). Let W = NF (w) and

let w′ be an arbitrary vertex in W . By Observation 4, we note that |W | ≥ 2. Let

H be the graph obtained from the disjoint union of the graphs F and G by adding

all edges joining vertices in W and vertices in V (G). Thus, V (H) = V (F ) ∪ V (G)

and E(H) = E(F ) ∪ E(G) ∪ [W, V (G)]. By construction, H is connected and G is a

vertex induced subgraph of H. It remains to show that H is kt-stable.

We show first that γt(H) = k. Let D be a γt(F )-set, and so |D| = k. In order to

totally dominate the vertex w in F , we note that at least one vertex in W belongs

to the set D. Hence, D is a TDS of H, and so γt(H) ≤ k. We show next that

γt(H) = k. For the sake of contradiction, suppose that γt(H) < k. Let S be a γt(H)-

set. If S ⊆ V (F ), then S is a TDS of F , and so γt(F ) ≤ |S| < k, a contradiction.

Hence, |S ∩ V (G)| ≥ 1. If |S ∩ V (G)| ≥ 2, then (S ∩ V (F )) ∪ {w, w′} is a TDS for

F , once again implying that γt(F ) < k, a contradiction. Hence, |S ∩ V (G)| = 1. Let

v be the vertex of G in S. In order to totally dominate v, we note that |S ∩ W | ≥ 1.

But then (S \ {v}) ∪ {w} is a TDS of F , once again implying that γt(F ) < k, a

contradiction.


We show next that every edge of H is a stable edge. Let D be a γt(F )-set, and

so |D| = k. We note that D is also a γt(H)-set. Let e ∈ E(H). Suppose e ∈ E(G).

Then, D is a TDS of H − e, and so γt(H) ≤ γt(H − e) ≤ |D| = γt(H). Consequently,

γt(H) = γt(H − e) and e is a stable edge in H. Suppose e ∈ E(F ). Let S be a

γt(F − e)-set. Since F is a γt-stable graph, the set S is a γt(F )-set. In particular,

|S| = k. In order to totally dominate w in F − e, we note that at least one vertex in

W belongs to the set S. Hence, S is a TDS of H − e, implying as before that e is a

stable edge in H. Suppose, finally, that e ∈ [W, V (G)]. Let v be the vertex in W that

is incident with the edge e. We now consider the graph H − f where f = vw. Let R

be a γt(F − f)-set. Then, R is a γt(F )-set and |R| = k. In order to totally dominate

w in F − f , we note that at least one vertex in W \ {v} belongs to the set S. Hence,

S is a TDS in both H − f and in H − e, implying as before that e is a stable edge in

H. Hence every edge of H is stable. Thus, H is a kt-stable graph. 2

As an immediate consequence of Theorem 9, we have the following result.

Corollary 10 There is no characterization of connected kt-stable graphs in terms of

forbidden subgraphs.

The following result characterizes bipartite γt-stable graphs.

Theorem 11 Let G be a bipartite graph. Then, G is γt-stable if and only if for every

vertex v in G, |N(v) ∩ Tt(G)| ≥ 2.

Proof. Let G be a bipartite γt-stable graph and let v ∈ V (G). Let S be a γt(G)-set.

Then there is a vertex u ∈ S that is adjacent to v. By definition of the set Tt(G), we


note that S ⊆ Tt(G), and so u ∈ N(v) ∩ Tt(G). We now consider the graph G − uv.

Let D be a γt(G−uv)-set and let w be a vertex in D that is adjacent to v in G−uv.

By Observation 4, D is a γt(G)-set, and so D ⊆ Tt(G). Thus, w ∈ N(v) ∩ Tt(G).

Since u 6= w, we have that |N(v)∩Tt(G)| ≥ |{u, w}| = 2, as claimed. This establishes

the necessity.

To prove the sufficiency, let G be a bipartite graph with partite sets A and B, and

suppose that |N(v) ∩ Tt(G)| ≥ 2 for every vertex v in G. For any γt(G)-set D, let

DA = A∩D and DB = B∩D. We note that the set DA dominates B and the set DB

dominates A. Further, if there is a subset A′ of vertices in A that dominates B such

that |A′| < |DA|, then A′ ∪ DB is a TDS of G of cardinality less than |D| = γt(G),

which is impossible. Hence, DA is a minimum set of vertices in A that dominates B.

Similarly, DB is a minimum set of vertices in B that dominates A. We show now

that every edge of G is a stable edge. Let e ∈ E(G). Then, e = ab where a ∈ A and

b ∈ B. Since |N(a) ∩ Tt(G)| ≥ 2, there exists a vertex b′ ∈ (N(a) ∩ Tt(G)) \ {b}. Let

S be a γt(G)-set that contains b′. Then, SB is a minimum set of vertices in B that

dominates A. Since |N(b)∩Tt(G)| ≥ 2, there exists a vertex a′ ∈ (N(b)∩Tt(G))\{a}.

Let D be a γt(G)-set that contains a′. Then, DA is a minimum set of vertices in A

that dominates B. Thus, DA ∪ SB is a γt(G)-set. Further, since DA ∪ SB is a TDS

in G − ab, we have that γt(G) ≤ γt(G − ab) ≤ |DA ∪ SB | = γt(G). Consequently,

γt(G) = γt(G − ab). Hence, G is γt-stable. 2

We note that the converse of Theorem 11 is not true for graphs in general. For

example, if G is the graph obtained from a 6-cycle v1v2 . . . v6v1 by adding the chords

v1v3 and v4v6, then G has exactly two γt(G)-sets, namely the set {v1, v6} and {v3, v4}.


Thus, Tt(G) = {v1, v3, v4, v6} and |N(v)∩Tt(G)| ≥ 2 for every vertex v in G. However,

the edges v1v3 and v4v6 are both critical edges in G, and so G is not a γt-stable graph.

Furthermore, while Theorem 11 is true for bipartite graphs, it is not true for

triangle free graphs in general.

In order to see this, we need to construct a triangle free graph containing a vertex-

induced odd cycle of order at least 5. We need for every w ∈ V (G), |N(w)∩Tt(G)| ≥ 2,

and we need for G to have at least one critical edge.

Consider the following graph seen in Figure 2.2.

vu

x1

x2

x3

x4

x5

x6

Figure 2.2: A triangle-free example.

The vertices (x1, x2, x3, u, x4) induce a 5-cycle; hence G is not bipartite. Also, G

is triangle-free. Assume for the purposes of contradiction that γt(G) = 2. Let S be a

γt(G)-set. In order to totally dominate u, S must contain at least one of x3 and x4;

and to totally dominate v, S must contain at least one of x1 and x6. We may assume,

without loss of generality, that x1 ∈ S. Since the second vertex in S must be adjacent

to x1, x4 ∈ S. This leaves x3 un-dominated by S, a contradiction. If x3 ∈ S, then x5

is not totally dominated by S, a contradiction. Thus, γt(G) ≥ 3. The set {x1, x2, x3}

is a TDS of G. Hence γt(G) ≤ 3, and so γt(G) = 3.


By symmetry {x4, x5, x6} is a TDS for G as well. Hence |Tt(G) ∩ N(w)| ≥ 2 for

every w ∈ V (G).

We claim that x2x5 is a critical edge. To see this, let G′ = G − x2x5, and S ′ be a

γt(G′)-set. Assume to the contrary that x2x5 is a stable edge, that is, γt(G

′) = 3. As

before, we may assume, without loss of generality, that x1 ∈ S ′. Moreover, at least

one of x3 and x4 is in S ′ to totally dominate u. First, let x3 ∈ S ′. Since S ′ is a TDS,

it follows that x2 ∈ S ′. But then x5 is not totally dominated by S ′, a contradiction.

Hence x4 ∈ S ′. But now the remaining vertex of S ′ must dominate both x3 and x6,

a contradiction. Therefore x2x5 is a critical edge in G, so G is a triangle-free graph

satisfying the necessary conditions of Theorem 11 and G is not γt-stable.

2.3.1 Disjoint Minimum Total Dominating Sets

In this section, we study the properties of graphs G with disjoint γt(G)-sets. As an

immediate consequence of Theorem 11, we have the following result.

Corollary 12 If a bipartite graph G has two disjoint γt(G)-sets, then G is γt-stable.

The remark after the proof of Theorem 11 shows that Corollary 12 is not true for

general graphs. However if G is a graph with three pairwise disjoint γt(G)-sets and

if e ∈ E(G), then at least one of these three γt(G)-sets does not contain an end of

e and is therefore also a TDS in G − e. Consequently, γt(G − e) = γt(G) for every

e ∈ E(G), implying that G is a γt-stable graph. Hence we have the following simple

observation.


Observation 13 If a graph G has at least three pairwise disjoint γt(G)-sets, then G

is γt-stable.

We show next that a graph G with exactly two disjoint γt(G)-sets cannot have

too may critical edges.

Proposition 14 If a graph G has exactly two disjoint γt(G)-sets, then G has at most

γt(G) critical edges.

Proof. Let G be a graph with two disjoint γt(G)-sets A and B. If every edge of

G is a stable edge, then the result is immediate. Hence we may assume that G has

at least one critical edge. Let e be a critical edge in G. If e is not incident with a

vertex in A, then the set A is a TDS in G − e, implying that γt(G − e) = γt(G), a

contradiction. Hence, at least one end of e is in A. Similarly, at least one end of e

is in B. Thus, e = ab where a ∈ A and b ∈ B. If |N(b) ∩ A| ≥ 2, then A is a TDS

in G − e, a contradiction. Hence, N(b) ∩ A = {a}. Similarly, N(a) ∩ B = {b}. This

implies that G has at most γt(G) critical edges. 2

The sharpness of the upper bound of Proposition 14 is established in the following

proposition.

Proposition 15 For any integer r ≥ 2, there exists a graph G with exactly two

disjoint γt(G)-sets such that γt(G) = r and G has exactly r critical edges.

Proof. Let F and H be two vertex-disjoint copies of a star K1,r−1, where V (F ) =

{u1, u2, . . . , ur} and V (H) = {v1, v2, . . . , vr} and where u1 and v1 are the central

vertices of the stars F and H, respectively. Let G be obtained from the union of F


and H as follows: For i = 1, 2, . . . , r, add the edge uivi, and add at least one new

vertex of degree-2 that is joined to both ui and vi. Let W denote the resulting set of

vertices of degree-2 in G. We note that the set W is an independent set in G.

We show first that γt(G) = r and that G has exactly two γt(G)-sets. Further, we

show these two γt(G)-sets are vertex disjoint. Since V (F ) is a TDS in G, we note

that γt(G) ≤ r. Let S be a γt(G)-set. Then, |S| ≤ r. In order to totally dominate

the added common neighbors of ui and vi (that belong to the set W ), we note that

|S ∩ {ui, vi}| ≥ 1 for each i = 1, 2, . . . , r. Consequently, |S| = r and |S ∩ {ui, vi}| = 1

for each i = 1, 2, . . . , r. Suppose u1 ∈ S. Then, v1 /∈ S. If vi ∈ S for some i, 2 ≤ i ≤ r,

then ui /∈ S. But then vi is isolated in G[S], a contradiction. Hence, vi /∈ S for all

i = 1, 2, . . . , r. Thus, S = V (F ). Similarly, if v1 ∈ S, then S = V (H). Hence, G has

exactly two γt(G)-sets, namely the disjoint sets V (F ) and V (H).

We show next that G has exactly r critical edges. As shown in the proof of

Proposition 14, the only possibly critical edges in G are those edges joining a vertex

in the one γt(G)-set to a vertex in the other γt(G)-set. Hence, every critical edge in

G has one end in V (F ) and the other end in V (H). By construction, G has exactly r

such edges in [V (F ), V (H)], namely the edges uivi for i = 1, 2, . . . , r. We show that

each such edge is a critical edge. For the sake of contradiction, assume to the contrary

that e = uivi is a stable edge for some i ∈ {1, 2, . . . , r}. Let D be a γt(G− e)-set. By

Observation 4(b), the set D is a γt(G)-set. However, G has exactly two γt(G)-sets,

namely the disjoint sets V (F ) and V (H). Thus either D = V (F ) or D = V (H). If

D = V (F ), then vi is not totally dominated by D in G − e. If D = V (H), then ui is

not totally dominated by D in G− e. In both cases, we contradict the fact that D is


a TDS in G − e. Hence, every edge uivi, 1 ≤ i ≤ r, is a critical edge. Thus, G has

exactly r critical edges, as claimed. 2

2.3.2 Bounds on the Diameter

In this section, we determine the bounds on the diameter of a γt-stable graph G

in terms of γt(G). For this purpose, if G is a graph and S is a γt(G)-set, then we

denote the number of components in G[S] by c(S) and we define c(G) = min{c(S) |

S is a γt(G)-set}.

Lemma 16 For any connected graph G, diam(G) ≤ γt(G) + 2c(G) − 1.

Proof. Let x and y be two diametrical vertices of G, and so d(x, y) = diam(G). Let

P be a diametrical path between x and y. Let S be a γt(G)-set such that c(S) = c(G).

Let G1, . . . , Gc(G) denote the components of G[S]. For i = 1, . . . , c(G), let Si = V (Gi).

Further let xi be the first vertex on P in N [Si] and let yi be the last vertex of P in

N [Si]. We note that the xi-yi subpath Pi of P has length at most 2 + diam(Gi) ≤

|Si| + 1. Hence since S dominates V (P ), and since there are at most c(G) − 1 edges

of P that do not belong to one of these subpaths Pi, 1 ≤ i ≤ c(G), we have that the

number of edges on P is at most c(G)− 1 +∑c(G)

i=1 (|Si|+1) = |S|+2c(G)− 1. Hence,

diam(G) = |E(P )| ≤ |S| + 2c(G) − 1. 2

The sharpness of the upper bound of Lemma 16 for γt-stable graphs is established

in the following proposition.

Proposition 17 There exist infinite families of connected γt-stable graphs G that

satisfy diam(G) = γt(G) + 2c(G) − 1.


Proof. Let k ≥ 1 be an integer and let n = 4k. Let a1a2 . . . an, b1b2 . . . bn, and

c1c2 . . . cn be three vertex disjoint paths on n vertices. Let A = {ai | i ≡ 2, 3, (mod 4)},

B = {bi | i ≡ 2, 3 , (mod 4)} and C = {ci | i ≡ 2, 3 , (mod 4)}. For i = 1, . . . , n, let

Vi = {ai, bi, ci}.

Let G be the graph obtained from the disjoint union of the above these three

paths as follows: For each i = 1, . . . , k, add at least one new vertex of degree 3 that is

joined to the three vertices of V4(i−1)+2 and call the resulting set of added vertices the

set Xi. For each i = 1, . . . , k, add at least one new vertex of degree 3 that is joined

to the three vertices of V4(i−1)+3 and call the resulting set of added vertices the set

Yi. For each i = 1, . . . , k, add nine edges so that each vertex in V4(i−1)+2 dominates

the set V4(i−1)+1 ∪ V4(i−1)+2. Finally for each i = 1, . . . , k, add nine edges so that each

vertex in V4(i−1)+3 dominates the set V4(i−1)+3 ∪ V4i.

We show first that γt(G) = 2k and that c(G) = k. Let D be a set of 2k vertices,

one from each of the sets X1, . . . , Xk and one from each of the sets Y1, . . . , Yk. We

note that these 2k vertices have disjoint neighborhoods in G. Hence any TDS in G

has cardinality at least 2k in order to totally dominate the set D. Hence, γt(G) ≥

2k. However, the set A is a TDS of G, and so γt(G) ≤ |A| = 2k. Consequently,

γt(G) = 2k. Furthermore, if S is a γt(G)-set, then S contains exactly one vertex from

each of the sets V4(i−1)+2, 1 ≤ i ≤ k, in order to totally dominate the set Xi and S

contains exactly one vertex from each of the sets V4(i−1)+3, 1 ≤ i ≤ k, in order to

totally dominate the set Yi. Thus, G[S] consists of k components, each of which is a

path P2. In particular, c(G) = k.


We show next that G is a γt-stable graph. Since each of the sets A, B and C is a

γt(G)-set, we note that the graph G has three vertex-disjoint γt(G)-sets. Hence, by

Observation 13, the graph G is γt-stable. By construction, the graph G is connected

and diam(G) ≤ n − 1. Further, the path a1a2 . . . an is a shortest a1-an path in G,

implying that diam(G) = n − 1 = 4k − 1 = γt(G) + 2c(G) − 1. 2

2.4 Open Problem

We close this chapter with the following open question that we have yet to settle: Is

the complexity question to recognize γt-stable graphs NP-hard or polynomial? We

believe it is an NP-hard problem but have yet to find a proof.

Chapter 3

An Extremal Problem for Total

Domination Stable Graphs Upon

Edge Removal

3.1 Introduction

In [22], the authors studied graphs whose domination number remains unchanged

after the removal of an arbitrary edge. In [22], the authors solved the extremal

problem of determining the minimum number of edges necessary in these graphs.

A search of the literature showed that the equivalent problem had not been solved

for total domination stable graphs upon edge removal. Accordingly, in this chapter

we continue the study of total domination stable graphs upon edge removal first

41

CHAPTER 3. AN EXTREMAL PROBLEM 42

introduced in Chapter 2. We establish lower bounds on the number of edges necessary

in a γt-stable graph. As we did in Chapter 2, to aid in our investigation, we say that

γt(G) = ∞ if the graph G has an isolated vertex. If γt(G) = k and G is γt-stable, we

say that G is kt-stable. We denote the set of vertices of G contained in every γt(G)-set

by At(G). For a subset X ⊆ V and a vertex v ∈ V , we define dX(v) to be the number

of neighbors of v that belong to X.

3.2 Main Result

Our aim in this chapter is to determine the minimum number of edges necessary in a

connected γt-stable graph with n vertices and total domination number γt. We shall

prove:

Theorem 18 If G is a connected γt-stable graph of order n, size m, with total dom-

ination number γt and with minimum degree δ ≥ 2, then

m ≥

12((δ + 1)n − δγt) if γt is even

12((δ + 1)n − δγt + 1) if γt is odd.

Further for every given integer δ ≥ 2 and every given integer γt ≥ 2, there exists a

connected γt-stable graph that is δ-regular of order n, size m and with total domination

number γt that achieves the above lower bound.

Theorem 19 Let G be a connected γt-stable graph of order n, size m, and with total

domination number γt. Then the following hold.

(a) If γt is even, then m ≥ 3n/2 − γt with equality if and only if G = Cn and

n ≡ 0 (mod 4).


(b) If γt is odd, then m ≥ (3n + 1)/2 − γt with equality if and only if G = Cn and n

is odd.

3.2.1 Proof of Theorem 18

In this section, we present a proof of Theorem 18. We begin with the following result

which gives a lower bound on the size of a graph in terms of its order, total domination

number and minimum degree.

Proposition 20 If G is a graph of order n, size m, with total domination number γt

and with minimum degree δ ≥ 2, then

m ≥

12((δ + 1)n − δγt) if γt is even

12((δ + 1)n − δγt + 1) if γt is odd.

Proof. Let G = (V, E). Let A be a γt(G)-set and B = V \ A. For i = 1, . . . , δ − 1,

let Bi consist of all vertices in B that have exactly i neighbors in A, and let Bδ

consist of all vertices in B that have at least δ neighbors in A. Hence if v ∈ Bi where

1 ≤ i ≤ δ − 1, then dA(v) = i while if v ∈ Bδ, then dA(v) ≥ δ. For i = 1, 2, . . . , δ,

let |Bi| = bi. Since A is a TDS in G, B = (B1, B2, . . . , Bδ) is a weak partition of

B. Suppose that γt is even. Since every vertex in A is adjacent to at least one other

vertex of A, we have that G[A] contains at least |A|/2 edges. Counting the edges in

G, we have that


m = |[A, A]|+ |[A, B]|+ |[B, B]|

≥ 1

2|A|+

δ∑

i=1

ibi +1

2

δ∑

i=1

(δ − i)bi

=1

2|A|+ 1

2

δ∑

i=1

(δ + i)bi

≥ 1

2|A|+

(

δ + 1

2

)

δ∑

i=1

bi

=1

2|A|+

(

δ + 1

2

)

(n − |A|)

=1

2((δ + 1)n − δ|A|) .

Since |A| = γt, we have that m ≥ 12((δ + 1)n − δγt), as desired. If γt is odd, then

G[A] contains at least (|A| + 1)/2 edges. An identical counting argument as above

now shows that m ≥ 12((δ + 1)n − δγt + 1). 2

The lower bound in Theorem 18 is an immediate consequence of Proposition 20.

We shall need the following observation before proceeding.

Observation 21 ([37]) If G ∈ {Cn, Pn} for n ≥ 3, then γt(G) = bn/2c + dn/4e −

bn/4c.

We are now in a position to discuss the realizability of the lower bounds established

in Theorem 18. Let δ ≥ 2 and γt ≥ 2 be fixed, given integers.

Suppose δ = 2 and γt is even. Then, γt = 2k for some integer k ≥ 1, and

we let G = C4k. Then, G is a δ-regular graph of order n = 4k and size m =


G1 G2 G3

b1,2

a1,1

b1,1

a2,2 a2,3

a2,1

b2,1

a3,2

b2,3 b3,2

a1,2 a1,3

b1,3 b2,2 b3,3

a3,3

a3,1

b3,1

Figure 3.1: The graph G3,6.

4k. By Observation 21, γt(G) = γt, and so m = 12((δ + 1)n − δγt). Further, by

Observation 8, G is a γt-stable graph.

Suppose δ = 2 and γt is odd. Then, γt = 2k + 1 for some integer k ≥ 1, and we

let G = C4k+1. Then, G is a δ-regular graph of order n = 4k +1 and size m = 4k +1.

By Observation 21, γt(G) = γt, and so m = 12((δ + 1)n − δγt + 1). Further, by

Observation 8, G is a γt-stable graph.

Suppose δ ≥ 3 and γt is even. Then γt = 2k for some integer k ≥ 1. For 1 ≤ i ≤ k,

let Gi = P2 2Kδ be the cartesian product of P2 and Kδ . Let V (Gi) = Ai ∪ Bi where

Ai = {ai,1, ai,2, . . . , ai,δ}, Bi = {bi,1, bi,2, . . . , bi,δ}, Gi[Ai] = Gi[Bi] = Kδ and where

ai,jbi,j ∈ E(Gi) for j = 1, 2, . . . , δ. Let G = Gδ,γtbe the graph obtained from

the disjoint union of the graphs Gi − ai,1ai,δ, 1 ≤ i ≤ k, by adding the k edges

{a1,1a2,δ, a2,1a3,δ, . . . , ak−1,1ak,δ, ak,1a1,δ}. Thus,

G = Gδ,γt=

(

k⋃

i=1

(Gi − ai,1ai,δ)

)

∪(

k⋃

i=1

{ai,1ai+1(modk),δ})

.

The graph G3,6 is illustrated Figure 3.1.

Then, G is a δ-regular graph of order n = 2kδ and size m = kδ2. Since G

is δ-regular, γt(G) ≥ n/δ = 2k. The set ∪ki=1{ai,1, bi,1} is a TDS for G, and so


γt(G) ≤ 2k. Consequently, γt(G) = 2k = γt, and so m = 12((δ + 1)n − δγt). Let

D1 = ∪ki=1{ai,1, bi,1}, D2 = ∪k

i=1{ai,2, bi,2} and Dδ = ∪ki=1{ai,δ, bi,δ}. Since the sets D1,

D2 and Dδ are vertex disjoint γt(G)-sets, the graph G is γt-stable by Observation 13.

If δ = 3 and γt = 3, we take G to be the circulant graph G8 = C8〈1, 4〉 (shown in

Figure 3.2), i.e., the graph with vertex set {v0, v1, . . . , v7} and edge set {vivi+j (mod8) |

i ∈ {0, 1, . . . , 7} and j ∈ {1, 4}}. Then, G is a 3-regular graph of order n = 8,

size m = 12 with γt(G) = 3, and so m = 12((δ + 1)n − δγt + 1). Since any three

consecutive vertices on the cycle v0v1 . . . v7v0 totally dominate V (G), the graph G is

γt-stable.

t t

t t

t t

t t

��

@@@

BB

BB

BBB

��

��

@@@

��PPPPPPP

v4

v7

v3

v0

v5

v6

v2

v1

Figure 3.2: The circulant graph G8 = C8〈1, 4〉.

Suppose that δ = 3 and γt ≥ 5 is odd. Thus γt = 2k + 3 for some integer

k ≥ 1. Let G = F3,γtbe the graph obtained from the disjoint union of the graph

G3,γt−3 = G3,2k and the circulant graph G8 (in Figure 3.2) by deleting the edges

ak,1a1,3 and v0v4 and adding the edges v0a1,3 and v4ak,1. Then, G is a 3-regular graph

of order n = n(G3,2k) + n(G8) = 6k + 8 and size m = 3(3k + 4). Since G is 3-regular,

γt(G) ≥ dn/3e = 2k+3. Let D1, D2 and D3 be the three vertex disjoint γt(G3,2k)-sets

defined earlier. Since the set {v1, v2, v3}∪D2 is a TDS for G, we have γt(G) ≤ 2k +3.

Consequently, γt(G) = 2k + 3 = γt, and so m = 12((δ + 1)n − δγt + 1). It remains to

show that G is γt-stable. Assume, to the contrary, that G contains a critical edge e.


We now consider the three γt(G)-sets A = D1 ∪{v0, v1, v2}, B = D3 ∪{v4, v5, v6} and

C = D2 ∪ {v5, v6, v7}. Since the sets A and B are vertex disjoint, it follows from the

proof of Proposition 14 that e ∈ [A, B]. If either endpoint of e is in D1 ∪ D3, then

C is a TDS for G − e. If both endpoints of e are in {v0, v1, v2, v4, v5, v6}, then either

e = v1v5 or e = v2v6. If e = v1v5, then D3∪{v2, v3, v4} is a TDS for G−e. If e = v2v6,

then D3 ∪ {v3, v4, v5} is a TDS for G − e. Hence, γt(G − e) ≤ γt(G3,2k) + γt(G8) =

2k + 3 = γt(G), contradicting our assumption that γt(G − e) > γt(G).

Suppose δ ≥ 4 and γt = 3. Construct the graph G = Hδ,3 as follows. Begin with

the disjoint union of a copy of P3 with central vertex y and end-vertices x and z, two

copies of Kδ−1 labeled X and Z and a copy of Kδ−2 labeled Y . Label the vertices

of X as {x1, . . . , xδ−1}, the vertices of Y as {y1, . . . , yδ−2}, and the vertices of Z as

{z1, . . . , zδ−1}. Add edges from x to every vertex in X, from y to every vertex in Y

and from z to every vertex in Z. For 1 ≤ i ≤ δ − 2, add the edge xiyi and the edge

yizi. Finally, add the edge xδ−1zδ−1. The graph H5,3 is illustrated in Figure 3.3.

x y z

x4

x3

x1

x2

y1y2 y3

z4z1

z2z3

Figure 3.3: The graph H5,3.

Then, G is a δ-regular graph of order n = 3δ − 1 and size m = δ(3δ − 1)/2.

Since G is δ-regular, γt(G) ≥ dn/δ e = 3. However {x, y, z} is a TDS for G, implying

that γt(G) = 3 = γt, and so m = 12((δ + 1)n − δγt + 1). Since the sets {x, y, z},


{x1, y1, z1} and {x2, y2, z2} are vertex disjoint γt(G)-sets, the graph G is γt-stable by

Observation 13.

Suppose, finally, that δ ≥ 4 and γt ≥ 5 is odd. Then, γt = 2k + 3 for some

integer k ≥ 1. Let G = Lδ,γtbe the graph obtained from the disjoint union of the

graph Gδ,γt−3 = Gδ,2k and the graph Hδ,3 by deleting the edges ak,1a1,δ and x1x2 and

adding the two edges ak,1x2 and a1,δx1. Then, G is a δ-regular graph of order n =

n(Gδ,2k)+n(Hδ,3) = 2δk+3δ−1 and size m = kδ 2 +(3δ 2−δ)/2. Since G is δ-regular,

γt(G) ≥ dn/δ e = 2k+3. Let D1, D2 and Dδ be the three vertex disjoint γt(Gδ,2k)-sets

defined earlier. Since the set {x, y, z} ∪ D2 is a TDS for G, we have γt(G) ≤ 2k + 3.

Consequently, γt(G) = 2k + 3 = γt, and so m = 12((δ + 1)n − δγt + 1). Since the sets

D1 ∪{x1, y1, z1}, D2 ∪{x, y, z} and Dδ ∪{x2, y2, z2} are vertex disjoint γt(G)-sets, the

graph G is γt-stable by Observation 13. 2


In this section, we present a proof of Theorem 19. We first consider the case when γt

is even. Recall that At(G) denotes the set of vertices in a graph G contained in every

γt(G)-set.

Proposition 22 If G is a connected γt-stable graph of order n, size m, and with even

total domination number γt, then m ≥ 3n/2 − γt with equality if and only if G = Cn

where n ≡ 0 (mod 4).

Proof. Let G = (V, E) be a connected γt-stable graph of order n, size m, and

with even total domination number γt. By Observation 4, δ(G) ≥ 2. Thus applying


Theorem 18, we have that m ≥ 3n/2− γt, which establishes the desired lower bound.

It remains for us to characterize the γt-stable graphs achieving equality in the bound.

If G = Cn where n ≡ 0 (mod 4), then, by Observation 21, γt = n/2 and so m = n =

3n/2 − γt. Further, by Observation 8, G is a γt-stable graph. This establishes the

sufficiency.

Conversely, suppose that m = 3n/2 − γt. Since m is an integer, we note that n is

even. We show that G is a cycle. Assume, to the contrary, that ∆(G) ≥ 3. We shall

adopt the notation introduced in the proof of Proposition 20 where in our case δ = 2.

In particular, A is a γt(G)-set and B = V \ A, while B1 is the set of vertices in B

that have exactly one neighbor in A and B2 = B \B1 is the set of vertices in B that

have at least two neighbors in A. Further, |B1| = b1 and |B2| = b2. Let |A| = a and

|B| = b, and so b = n − a and b = b1 + b2. We note that in our case when δ = 2, the

inequality chain presented in the proof of Proposition 20 simplifies to the following.

m ≥ 1

2a + (b1 + 2b2) +

1

2b1 ≥

1

2(a + 3b) =

1

2(3n − 2a) =

1

2(3n − 2γt). (3.1)

Since m = 3n/2 − γt, we must have equality throughout the above inequality

chain (3.1). This implies that G[A] consists of |A|/2 disjoint copies of P2. Further,

B = B1 and each vertex in B has degree 2 in G and is adjacent to one vertex in A

and to one vertex in B. Thus, G[B] consists of |B|/2 disjoint copies of P2 and each

vertex in B is adjacent to exactly one vertex in A. These properties hold for every

γt(G)-set A and for the set B = V \ A. Hence every vertex of degree greater than 2

in G must belong to the set At(G).

Let v be a vertex of maximum degree in G. By assumption, dG(v) ≥ 3, and so

v ∈ A. Let u be the neighbor of v in A. Let x ∈ N(u) \ {v}. Then, x ∈ B. Let y be


the neighbor of x in B, and let w be the neighbor of y in A. Assume v 6= w. Since

G is a γt-stable graph, every edge of G is a stable edge. In particular, wy is a stable

edge. Let S be a γt(G − wy)-set. We note that in the graph G − wy, we have that

uxy is a path of length 2 with d(y) = 1 and d(x) = 2. Thus, x ∈ S in order to totally

dominate y. If y ∈ S, then we can simply replace y in S by the vertex u. Hence we

may assume that {u, x} ⊆ S. By Observation 4, the set S is a γt(G)-set. Note that

dG(v) ≥ 3. As seen earlier, this implies that v is in every possible γt(G)-set. Hence

v ∈ At(G) ⊆ S, and so G[S] contains a path P3 induced by {u, v, x}. Note that if

v = w, then the same argument implies that {w, u, x} ⊆ S and G[S] contains the P3

path induced by {w, u, x}. Either scenario contradicts our earlier observation that

every γt(G)-set induces disjoint copies of P2. We, deduce, therefore, that G = Cn. As

observed earlier, n is even. If n ≡ 2 (mod 4), then by Observation 21 we note that

m = 3n/2 − γt + 1, a contradiction. Hence, n ≡ 0 (mod 4), as desired. 2

Next we consider the case when γt is odd.

Proposition 23 If G is a connected γt-stable graph of order n, size m, and with odd

total domination number γt, then m ≥ (3n + 1)/2 − γt with equality if and only if

G = Cn where n is odd.

Proof. Let G = (V, E) be a connected γt-stable graph of order n, size m, and with odd

total domination number γt. By Observation 4, δ(G) ≥ 2. Thus applying Theorem 18,

we have that m ≥ (3n + 1)/2 − γt, which establishes the desired lower bound. It

remains for us to characterize the γt-stable graphs achieving equality in the bound.

If G = Cn where n is odd, then by Observation 21 we have that m = (3n +1)/2− γt.

Further by Observation 8, G is a γt-stable graph. This establishes the sufficiency.


Conversely, suppose that m = (3n+1)/2−γt . Since m is an integer, we note that

n is odd. We show that G is a cycle. Assume, to the contrary, that ∆(G) ≥ 3. Once

again, we follow the notation introduced in the proof of Proposition 20 where in our

case δ = 2. Since γt is odd, we note that G[A] contains at least (a + 1)/2 edges. As

shown in the proof of Proposition 20, we have the following inequality chain.

m ≥ 1

2(a+1)+(b1+2b2)+

1

2b1 ≥

1

2(a+3b+1) =

1

2(3n−2a+1) =

1

2(3n+1−2γt). (3.2)

Since m = (3n + 1)/2 − γt, we must have equality throughout the inequality

chain (3.2). In particular, G[A] contains (a+1)/2 edges, and so G[A] = (a−32

)K2∪P3.

Further, B = B1 and each vertex in B has degree 2 in G and is adjacent to one vertex

in A and to one vertex in B. Thus, G[B] = b2K2 and each vertex in B is adjacent to

exactly one vertex in A. These properties hold for every γt(G)-set A and for the set

B = V \ A. Hence every vertex of degree greater than 2 in G must belong to the set

At(G).

Let v be a vertex of maximum degree in G. By assumption, dG(v) ≥ 3, and so

v ∈ A. Recall that every vertex in B has degree 2 in G and has one neighbor in A and

the other in B. Let x be a neighbor of v in B, and let y be the neighbor of x in B. Let

z be the neighbor of y in A. Assume v 6= z. Since G is a γt-stable graph, every edge

of G is a stable edge. In particular, vx is a stable edge. Let S be a γt(G−vx)-set. We

note that in the graph G − vx, we have that xyz is a path of length 2 with d(x) = 1

and d(y) = 2. Thus, y ∈ S in order to totally dominate x. If x ∈ S, then we can

simply replace x in S by the vertex z. Hence we may assume that {y, z} ⊆ S. By

Observation 4, the set S is a γt(G)-set. Note that dG(v) ≥ 3. As seen earlier, this

implies that v is in every possible γt(G)-set. Hence v ∈ At(G) ⊆ S, and so both


neighbors of x belong to S and x /∈ S. This contradicts our earlier observation that

for every γt(G)-set S, each vertex not in S has degree 2 in G and has one neighbor in

S and the other in V \ S. A similar argument follows if v = z. We deduce therefore,

that G = Cn. As observed earlier, n is odd. 2

Theorem 19 is an immediate consequence of Proposition 22 and Proposition 23.

Chapter 4

Total Domination Stable Graphs

Upon Edge Addition

4.1 Introduction

In this chapter, we study those graphs where the total domination number remains

unchanged upon the addition of any edge. We say that a graph G is total domination

edge addition stable, or γ+t -stable for short, if the addition of any edge to E(G) does

not change the total domination number. In other words, γt(G+ e) = γt(G) for every

edge e ∈ E(G). We note that adding an edge to a graph cannot increase the total

domination number. Hence γt(G+ e) ≤ γt(G) for every edge e ∈ E(G). It was shown

in [36] that adding an edge to a graph decreases the total domination number by at

most two.

Proposition 24 ([36]) For every edge uv ∈ E(G), γt(G) − 2 ≤ γt(G + uv) ≤ γt(G).

53

CHAPTER 4. STABLE GRAPHS UPON EDGE ADDITION 54

An edge e ∈ E(G) is a stable edge if γt(G + e) = γt(G), while e is a critical edge

if γt(G + e) < γt(G). Thus in a γ+t -stable graph G, every edge in E(G) is a stable

edge. For example, Hattingh, Joubert, and Van der Merwe showed in [27] that every

edge in E(Pn) is a stable edge of the path Pn for n ≡ 0 (mod 4).

Proposition 25 ([27]) For every integer k ≥ 1, the path P4k is γ+t -stable.

For any γt(G)-set S, and any A ⊆ S, we define the external S-private neighborhood

of A as epn(A, S) =⋃

u∈A epn(u, S). We define the external S-private neighbor count

of A as |epn(A, S)|. We will use the following result of Cockayne et al. [10] throughout

this chapter.


vertex v ∈ S, |epn(v, S)| ≥ 1 or |ipn(v, S)| ≥ 1.

4.2 Total Domination Edge Addition Stable Graphs

4.2.1 A Characterization

Since γt(G + e) ≤ γt(G) for every e ∈ E(G), and for every isolate free graph G,

γt(G) ≥ 2, we note that if γt(G) = 2, then G is γ+t -stable. Accordingly, we restrict

our attention in this chapter to graphs G for which γt(G) ≥ 3. For our first result, we

construct γ+t -stable graphs having a specified total domination number and induced

subgraph.

Proposition 27 For every positive integer k ≥ 2 and any graph H, there exists a

γ+t -stable graph G such that γt(G) = k and H is a vertex induced subgraph of G.


Proof. Begin with a caterpillar Gk with spine Pk = (v1, v2, ..., vk) and code (2, 2, ..., 2).

Form the graph G from Gk and a copy of H by adding an edge from v1 to every vertex

in H. Let S be any γt(G)-set. Since the k vertices of the spine of Gk are support

vertices in G, {v1, v2, ..., vk} ⊆ S. Hence, γt(G) ≥ k. Clearly the set of vertices on

the spine of Gk is a total dominating set of G, so γt(G) ≤ k. Hence, γt(G) = k.

To see that G is γ+t -stable, let uv ∈ E(G) be any arbitrary edge of G. If u or

v is a non-leaf vertex of G, then G′ = G + uv has k support vertices, implying that

γt(G′) = k, that is, uv is a stable edge. Assume that both u and v are leaves in G.

In this case, either G′ still has k support vertices and again uv is a stable edge or u

and v are both incident to the same support vertex, say vi. But then any TDS of G′

includes the remaining k−1 support vertices and at least one of u, v, and vi, implying

that uv is a stable edge. Thus we conclude that G is γ+t -stable with γt(G) = k and

H as a vertex induced subgraph. 2

As a consequence of Proposition 27 we obtain the following result.

Corollary 28 There exists no forbidden subgraph characterization of γ+t -stable graphs.

Next, we characterize γ+t -stable graphs. For this purpose, we will need the follow-

ing key lemma which will also prove useful in establishing upper bounds on the total

domination number of γ+t -stable graphs in terms of their order.

Lemma 29 If G is a γ+t -stable graph and γt(G) ≥ 3, then for every γt(G)-set S and

for every v ∈ S, one of the following properties hold.

(a) |pn(v, S)| ≥ 2 and |epn(v, S)| ≥ 1.

(b) If epn(v, S) = ∅, then |ipn(v, S)| ≥ 3.


Proof. Let G be a γ+t -stable graph with γt(G) ≥ 3. Suppose that S is a γt(G)-set

and that v ∈ S. By Proposition 26, |pn(v, S)| ≥ 1. If pn(v, S) = {u}, then the set

S \ {v} is a TDS for G + uw, where w ∈ S \ {u, v}, and so G is not γ+t -stable, a

contradiction. Therefore, |pn(v, S)| ≥ 2. If |epn(v, S)| ≥ 1, we are finished. Hence we

may assume that epn(v, S) = ∅. If ipn(v, S) = {x, y}, then the set S \ {v} is a TDS

for G + xy of cardinality γt(G)− 1 implying that G is not γ+t -stable, a contradiction.

Hence, |ipn(v, S)| ≥ 3, as desired. 2

As a consequence of Lemma 29, we show that removing a vertex from a γ+t -stable

graph cannot decrease the total domination number.

Corollary 30 If a graph G is γ+t -stable, then γt(G − v) ≥ γt(G) for every vertex

v ∈ V (G) such that G − v has no isolated vertex.

Proof. Let G be a γ+t -stable graph. Assume, for purposes of contradiction, that

there exists a vertex v ∈ V (G) such that γt(G − v) < γt(G). Adding any neighbor of

v to a γt(G− v)-set produces a TDS of G, and so γt(G − v) = γt(G)− 1. Let T be a

γt(G − v)-set, and let x ∈ N(v). If T contains a neighbor of v, then T is a TDS for

G of cardinality γt(G) − 1, a contradiction. Hence, T ∩ N(v) = ∅. We now consider

the γt(G)-set S = T ∪ {x} and note that pn(x, S) = {v}, contradicting Lemma 29.

Hence, γt(G − v) ≥ γt(G). 2

We remark that the necessary conditions of Lemma 29 for a graph G to be γ+t -

stable are by no means sufficient. To see this, consider the graph G shown in Fig-

ure 4.1.


a b c

d e f

g

h

i

Figure 4.1: The graph G.

Let S be any γt(G)-set. Since a, b, c, and f are support vertices in G, it follows that

{a, b, c, f} ⊆ S. The set {a, b, c, d, e, f} is a TDS for G, and so γt(G) ≤ 6. It is easy to

see, that γt(G) = 6 and e ∈ S. Hence the only possible γt(G)-sets are {a, b, c, d, e, f}

and {a, b, c, e, f, i}. Both of these sets satisfy the conditions of Lemma 29. Hence,

G satisfies the conditions of Lemma 29. However, γt(G + fh) = 5. Thus, G is not

γ+t -stable.

First we give a characterization of γ+t -stable graphs G with γt(G) = 3.

Theorem 31 Let G be a graph with γt(G) = 3. The graph G is γ+t -stable if and

only if γ(G) = 3 and for every γt(G)-set S and for every v ∈ S, one of the following

properties hold.

(a) |pn(v, S)| ≥ 2 and |epn(v, S)| ≥ 1.


Proof. Let G be a graph with γt(G) = 3. Suppose that G is a γ+t -stable graph.

By Lemma 29, one of properties (a) and (b) holds for every vertex in a γt(G)-set.

Since γt(G) = 3, we note that γ(G) ≥ 2. If γ(G) = 2, then let S = {x, y} be any

γ(G)-set. The set S is a TDS for G + xy, implying that xy is a critical edge and

therefore G is not γ+t -stable, a contradiction. Hence, γ(G) = 3. This proves the


necessity. To prove the sufficiency, suppose that γ(G) = 3 and that properties (a)

and (b) hold for every γt(G)-set S and v ∈ S. Assume, for purposes of contradiction,

that γt(G + xy) < γt(G) for some edge xy ∈ E(G). Let T be a γt(G + xy)-set, and

so |T | = 2. If T = {x, y}, then γ(G) = 2, a contradiction. If neither x nor y is in T ,

then T is a TDS of G and γt(G) = 2, a contradiction. Hence, T contains exactly one

of x and y. Assume, without loss of generality, that x ∈ T . Let T = {x, w}. In G,

the set T totally dominates V (G) \ {y}. We now consider the set S = T ∪{v}, where

v ∈ N(y). Then the set S is a γt(G)-set satisfying ipn(v, S) = ∅ and epn(v, S) = {y},

contradicting property (a). Hence, G is γ+t -stable. 2

Next we give a characterization of γ+t -stable graphs with total domination number

at least 4. For this purpose, for a graph G and for any two vertices u, v ∈ V (G) such

that {u, v} does not dominate G, we define Guv to be the graph

Guv = G[V (G) \ (N [u] ∪ N [v])].

Theorem 32 Let G be a graph with γt(G) ≥ 4. The graph G is γ+t -stable if and only

if the following two conditions hold.

1. For every γt(G)-set S and for every v ∈ S, one of the following is true.

(a) |pn(v, S)| ≥ 2 and |epn(v, S)| ≥ 1.


2. For every two vertices u, v ∈ V (G), such that dG(u, v) ≥ 3 and Guv is isolate-

free, we have γt(Guv) > γt(G) − 3.

Proof. Let G be a graph with γt(G) ≥ 4. Suppose that G is a γ+t -stable graph. By

Lemma 29, Condition 1 holds. Assume, for purposes of contradiction, that Condi-


tion 2 is not true. Then there exist vertices u, v ∈ V (G) such that dG(u, v) ≥ 3 and

Guv is isolate-free, but γt(Guv) ≤ γt(G)−3. Let Suv be a γt(Guv)-set. Then Suv∪{u, v}

is a TDS of G+uv, and so γt(G+uv) ≤ |Suv|+2 = γt(Guv)+2 ≤ γt(G)−1, implying

that uv is a critical edge and therefore G is not γ+t -stable, a contradiction. Thus,

Condition 2 holds. This proves the necessity.

For the sufficiency, we will prove the contrapositive. Assume that G is not γ+t -

stable. Then there exists an edge uv ∈ E(G) such that γt(G + uv) < γt(G). Let S∗

be a γt(G+uv)-set. If S∗∩{u, v} = ∅, then S∗ is a TDS of G, whence γt(G) ≤ |S∗| =

γt(G+uv), a contradiction. Hence at least one of u and v is in S∗. By Proposition 24,

either γt(G + uv) = γt(G) − 2 or γt(G + uv) = γt(G) − 1.

Suppose γt(G + uv) = γt(G) − 2. If exactly one of u and v is in S∗, say without

loss of generality, u ∈ S∗ and v /∈ S∗, then S∗∪{w} where w ∈ N(v) is a TDS of G of

cardinality γt(G)− 1, a contradiction. Thus, {u, v} ⊆ S∗. If either u or v, say u, has

a neighbor in S∗ \ {u, v}, then S = S∗ ∪ {w} where w ∈ NG(v) is a TDS of G with

cardinality less than γt(G), a contradiction. Hence NG(u) ∩ S∗ = NG(v) ∩ S∗ = ∅.

Let x ∈ N(u) and y ∈ N(v). Then the set S = S∗ ∪ {x, y} is a γt(G)-set. However,

epn(x, S) = ∅ and |ipn(x, S)| ≤ 1. Hence, Condition 1 does not hold. Therefore we

may assume that γt(G + uv) = γt(G) − 1, for otherwise Condition 1 does not hold

and we are finished.

Suppose that exactly one of u and v is in S∗. Renaming vertices if necessary,

we may assume that u ∈ S∗ and v /∈ S∗. Let w be any neighbor of v. The set

S = S∗ ∪ {w} is a γt(G)-set. However, the vertex v is the only possible S-private

neighbor of w, and so |pn(w, S)| ≤ 1. Therefore Condition 1 does not hold. Hence


we may assume that {u, v} ⊆ S∗.

If NG(u) ∩ S∗ 6= ∅ and NG(v) ∩ S∗ 6= ∅, then S∗ is a γt(G)-set of cardinality

γt(G) − 1, a contradiction. Hence renaming vertices if necessary, we may assume

that NG(u) ∩ S∗ = ∅. If NG(v) ∩ S∗ 6= ∅, then S = S∗ ∪ {w} where w ∈ NG(u)

is a γt(G)-set. However, epn(w, S) = ∅ and |ipn(w, S)| ≤ 1, and so Condition 1

does not hold. Hence we may assume that NG(v) ∩ S∗ = ∅. If u and v share a

common neighbor w, then the set S = S∗ ∪ {w} is a γt(G)-set where epn(w, S) = ∅

and |ipn(w, S)| ≤ 2, and so Condition 1 does not hold. Hence we may assume that

dG(u, v) ≥ 3. Since γt(G + uv) = γt(G) − 1 ≥ 3, we note that S∗ 6= {u, v}. Thus,

the set {u, v} does not dominate G, and so the graph Guv is well-defined. Since

NG(u)∩S∗ = NG(v)∩ S∗ = ∅, the set Suv = S∗ \ {u, v} is a TDS for Guv. Thus, Guv

is isolate-free and γt(Guv) ≤ |Suv| = |S∗| − 2 = γt(G + uv) − 2 = γt(G) − 3. Hence,

Condition 2 does not hold. This proves the sufficiency. 2

We note that if G is a graph with diam(G) = 2, then Condition 2 in Theorem 32 is

vacuously satisfied. Additionally, Condition 2 is also vacuously satisfied if γt(G) = 4.

Hence as an immediate consequence of Theorem 32, we have the following result.

Corollary 33 Let G be a graph with γt(G) = 4 or with diam(G) = 2 and γt(G) ≥ 5.

The graph G is γ+t -stable if and only if for every γt(G)-set S and for every v ∈ S,

one of the following properties hold.

(a) |pn(v, S)| ≥ 2 and |epn(v, S)| ≥ 1.



4.2.2 Upper Bounds

The decision problem Total Dominating Set is NP-complete, so it is of interest to

determine upper bounds on the total domination number of a graph in terms of its

minimum degree. The known upper bounds on the total domination number of a

graph G in terms of its order n are summarized in Theorem 34.

Theorem 34 Let G be a graph of order n. Then the following holds.

(a) δ(G) ≥ 1 ⇒ γt(G) ≤ 2n/3 if n ≥ 3 and G is connected ([10])(b) δ(G) ≥ 2 ⇒ γt(G) ≤ 4n/7 if n ≥ 11 and G is connected ([37])

(c) δ(G) ≥ 3 ⇒ γt(G) ≤ n/2 ([2, 9, 49])(d) δ(G) ≥ 4 ⇒ γt(G) ≤ 3n/7 ([48])

In this section, we determine upper bounds on the total domination number of

γ+t -stable graphs. We show that if we add the condition that a graph G is γ+

t -stable,

then the bounds in Theorem 34 can be improved.

Theorem 35 Let G be a connected γ+t -stable graph of order n with γt(G) ≥ 3. Then

γt(G) ≤ n/2. Further if δ(G) ≥ 2, then equality holds if and only if G = Cn where

n ≡ 0 (mod 4).

Proof. Let G = (V, E) be a connected γ+t -stable graph of order n with γt(G) ≥ 3.

Let S be a γt(G)-set. We define a weak partition (S0, S1, S2) of the set S as follows.

Let S0 consist of all vertices of S that have no S-external private neighbor. Let S1

consist of all vertices of S that have exactly one S-external private neighbor. Let S2

consist of all vertices of S that have at least two S-external private neighbors. That

is,S0 = { v ∈ S : |epn(v, S)| = 0 }S1 = { v ∈ S : |epn(v, S)| = 1 }S2 = { v ∈ S : |epn(v, S)| ≥ 2 }.


By Lemma 29, if v ∈ S0, then |ipn(v, S)| ≥ 3. For each S-internal private neighbor

u ∈ ipn(v, S), we note that N(u) ∩ S = {v} and ipn(u, S) = ∅. Therefore by

Lemma 29, |epn(u, S)| ≥ 2 for every vertex u ∈ ipn(v, S). For each v ∈ S0, we let Av

consist of exactly three vertices in ipn(v, S), and we let

A =⋃

v∈S0

(Av ∪ {v}).

Then |A| = 4|S0| and

|epn(A, S)| =∑

v∈S0

∑

u∈Av

|epn(u, S)| ≥∑

v∈S0

2|Av| = 6|S0| = 3|A|/2.

By Lemma 29, if v ∈ S1, then |ipn(v, S)| ≥ 1. Let B ′ = { v ∈ S1 : |ipn(v, S)∩S2| ≥

1 }. For each v ∈ B ′, we select a vertex v′ ∈ ipn(v, S) ∩ S2, and we let

B =⋃

v∈B′

{v, v′}.

Then |B| = 2|B ′|, and since |epn(v, S)| = 1 and |epn(v′, S)| ≥ 2 for each v ∈ B ′, we

have that

|epn(B, S)| =∑

v∈B′

(|epn(v′, S)| + |epn(v, S)|) ≥ 3|B ′| = 3|B|/2.

Let C = { v ∈ S1 : ipn(v, S) ∩ S2 = ∅ }. Then (B ′, C) is a weak partition of the

set S1. Let v ∈ C and let u ∈ ipn(v, S). Then u ∈ S1 and N(u) ∩ S = {v}. If

v /∈ ipn(u, S), then by Lemma 29, |epn(u, S)| ≥ 2, and so u ∈ S2, a contradiction.

Hence, ipn(u, S) = {v}. This in turn implies that N(v)∩S = {u} and ipn(v, S) = {u}.

Thus the subgraph G[C ] of G induced by the set C consists of disjoint copies of K2.

Further,

|epn(C, S)| =∑

v∈C

|epn(v, S)| = |C|.


Let D = S2 \ (A ∪B). Since |epn(v, S)| ≥ 2 for every vertex v ∈ D, we have that

|epn(D, S)| =∑

v∈D

|epn(v, S)| ≥ 2|D|.

By construction, we note that (A, B, C, D) is a weak partition of the set S and

|S| = |A|+ |B|+ |C|+ |D|.

By construction, the sets epn(A, S), epn(B, S), epn(C, S) and epn(D, S) are pair-

wise disjoint sets in V (G) \ S. Hence,

n − |S| ≥ |epn(A, S)| + |epn(B, S)|+ |epn(C, S)| + |epn(D, S)|

≥ 32|A| + 3

2|B| + |C|+ 2|D|

≥ |A|+ |B|+ |C| + |D|

= |S|,

and so, γt(G) = |S| ≤ n/2.

We show next that if δ(G) ≥ 2 and γt(G) = n/2, then G = Cn where n ≡

0 (mod 4). Suppose that δ(G) ≥ 2 and γt(G) = n/2. Then we have equality through-

out the above inequality chain. In particular, A ∪ B ∪ D = ∅, and so S = C , and

V \ S = epn(C, S). Recall that for each vertex v ∈ C , |epn(v, S)| = 1. Thus,

n−|S| = |epn(C, S)| = |C|, and so n = 2|C|. Let C = kK2 where k ≥ 2 is an integer.

Then n = 4k. For every vertex v ∈ C , we note that dG(v) = 2. Let epn(v, S) = {v′}

for each v ∈ C . Let H = G[V \ S] = G[epn(C, S)] denote the subgraph of G induced

by the 2k vertices in V \ S. Since δ(G) ≥ 2 and each vertex in epn(C, S) is adjacent

to exactly one vertex in C , we have that δ(H) ≥ 1. Suppose that ∆(H) ≥ 2. Then H

contains a vertex v′ such that δ(H − v′) ≥ 1. Let v denote the vertex in C such that

epn(v, S) = {v′}, and let w be an arbitrary neighbor of v′ in H. Then V (H) \ {v′}


is a TDS in G + vw, and so γt(G + vw) ≤ 2k − 1 < γt(G), implying that vw is a

critical edge and therefore G is not γ+t -stable, a contradiction. Hence, H is a 1-regular

graph. Therefore, G is a 2-regular graph on 4k vertices. Thus since G is connected,

G = C4k. 2

As an immediate consequence of Theorem 35, we have the following result.

Corollary 36 Let G be a connected γ+t -stable graph of order n with γt(G) ≥ 3. If

δ(G) ≥ 2 and G 6= C4k for some integer k, then γt(G) < n/2.

The following result is a consequence of the proof of Theorem 35.

Corollary 37 Let G be a connected γ+t -stable graph of order n with γt(G) ≥ 3 and

γt(G) odd. Then γt(G) ≤ (n − 2)/2.

Proof. Let G be a connected γ+t -stable graph of order n with γt(G) ≥ 3 and γt(G)

odd. We shall adopt the same notation as in the proof of Theorem 35 and follow

the proof of Theorem 35 identically. Since G[C ] consists of disjoint copies of K2,

the fact that γt(G) is odd implies that at least one component of G[S] has order at

least 3. Every vertex in such a component belongs to the set A ∪ B ∪ D. Hence,

|A|+ |B| + |D| ≥ 3, and so

n − |S| ≥ 32|A|+ 3

2|B|+ |C|+ 2|D|

≥ |A|+ |B| + |C|+ |D| + 12(|A|+ |B| + |D|)

≥ |S| + 32.

Since n − |S| is an integer, we have that n − |S| ≥ |S| + 2, whence γt(G) ≤ |S| ≤

(n − 2)/2. 2


We show next that if the minimum degree is at least 3, then the bound of Theo-

rem 35 can be improved.

Theorem 38 Let G be a connected γ+t -stable graph of order n with γt(G) ≥ 3 and

with maximum degree ∆. Then the following hold.

(a) If δ ≥ 3, then γt(G) ≤(

∆2∆+1

)

n.

(b) If δ ≥ 4, then γt(G) ≤(

∆2∆+2

)

n.

Proof. Let G be a connected γ+t -stable graph of order n with γt(G) ≥ 3 and with

minimum degree δ and maximum degree ∆. We shall adopt the same notation as

in the proof of Theorem 35. Let S be a γt(G)-set. We define the weak partitions

(S0, S1, S2) and (A, B, C, D) of the set S exactly as in the proof of Theorem 35. As

before, |S| = |A|+ |B| + |C|+ |D|.

We now let R be the set of vertices in V \ S that are not S-external private

neighbors of any vertex in S but are adjacent to at least one vertex in C . We note

that every vertex v ∈ C has one S-external private neighbor and is adjacent to one

other vertex of C , with all other dG(v) − 2 ≥ 1 neighbors in the set R. Recall that

δ(G) = δ and ∆(G) = ∆. Counting the edges between the sets C and R, we have

that

(δ − 2)|C| ≤ ∆|R|,

and so |R| ≥ (δ−2)|C|/∆. By construction, the sets epn(A, S), epn(B, S), epn(C, S),

epn(D, S) and R are pairwise disjoint sets in V (G) \ S. Hence,

n − |S| ≥ |epn(A, S)|+ |epn(B, S)|+ |epn(C, S)|+ |epn(D, S)| + |R|

≥ 32|A|+ 3

2|B|+ |C|+ 2|D| + δ−2

∆|C|

= 32|A|+ 3

2|B|+ 2|D| +

(

∆+δ−2∆

)

|C|.


Suppose δ ≥ 3. Then ∆+δ−2∆

≥ ∆+1∆

. Since ∆ ≥ 3, we note that 32≥ ∆+1

∆. Hence,

n − |S| ≥ 32|A|+ 3

2|B|+ 2|D| +

(

∆+1∆

)

|C|

≥(

∆+1∆

)

(|A|+ |B|+ |C| + |D|)

=(

∆+1∆

)

|S|,

and so, γt(G) = |S| ≤(

∆2∆+1

)

n. This establishes part (a).

To prove part (b), suppose δ ≥ 4. Then ∆+δ−2∆

≥ ∆+2∆

. Since ∆ ≥ 4, we note that

32≥ ∆+2

∆. Hence,

n − |S| ≥ 32|A|+ 3

2|B|+ 2|D| +

(

∆+2∆

)

|C|

≥(

∆+2∆

)

(|A|+ |B|+ |C| + |D|)

=(

∆+2∆

)

|S|,

and so, γt(G) = |S| ≤(

∆2∆+2

)

n. This establishes part (b). 2

We remark that the upper bound of Theorem 34(c) is attained only for cubic

graphs, and the (infinite family of) cubic graphs achieving this upper bound are char-

acterized in [40]. The incidence bipartite graph of the complement of the Fano plane

(or, equivalently, the relative complement of the Heawood graph) achieves equality

in the Thomasse-Yeo bound in Theorem 34(d). Hence there is a 4-regular graph

that achieves the bound of Theorem 34(d). However both the upper bounds in The-

orem 34(c) and 34(d) for cubic graphs and 4-regular graphs, respectively, can be

improved if we restrict our attention to γ+t -stable graphs. As an immediate conse-

quence of Theorem 38, we have the following result.

Corollary 39 Let G be a connected γ+t -stable graph of order n with γt(G) ≥ 3. Then

the following holds .


(a) If G is a cubic graph, then γt(G) ≤ 3n/7.

(b) If G is a 4-regular graph, then γt(G) ≤ 2n/5.

4.2.3 Claw-Free Graphs

A graph is claw-free if it does not contain K1,3 as an induced subgraph. Our aim

in this section is twofold: First to establish properties of γ+t -stable claw-free graphs

and second to determine upper and lower bounds for the total domination number of

γ+t -stable claw-free cubic graphs. We begin with the following property of γ+

t -stable

claw-free graphs.

Lemma 40 Let G be a γ+t -stable claw-free graph with γt(G) ≥ 3, and let S be a

γt(G)-set. Then the following properties hold.

(a) For every v ∈ S, |pn(v, S)| ≥ 2 and |epn(v, S)| ≥ 1.

(b) Every component of G[S] is a complete graph.

(c) For every v ∈ S, G[epn(v, S)] is a complete graph.

Proof. Part (a) is an immediate consequence of Lemma 29. To prove part (b), assume

to the contrary that G[S] contains a component C that is not complete. Then there

exist a pair u and v of vertices of C at distance two apart in C . Let w ∈ V (C) be a

common neighbor of u and v. By part(a), |epn(w, S)| ≥ 1. Let x ∈ epn(w, S). But

then G[u, v, w, x] forms a claw in G centered at w, a contradiction. This establishes

part (b). To prove part (c), assume to the contrary that G[epn(v, S)] is not complete.

Then epn(v, S) contains two vertices a and b that are not adjacent. Since S is a TDS

of G, there is a vertex z ∈ S adjacent to v. But then G[a, b, v, z] forms a claw in G


centered at v, a contradiction. This establishes part (c). 2

We are now in a position to determine upper and lower bounds for the total

domination number of γ+t -stable claw-free cubic graphs. For this purpose, we define

the boundary of a set D of vertices in a graph G to be the set ∂(D) = NG(D) \ D.

Theorem 41 If G is a γ+t -stable claw-free cubic graph of order n with γt(G) ≥ 3,

then n/3 ≤ γt(G) ≤ 2n/5.

Proof. Let G = (V, E) be a γ+t -stable claw-free cubic graph of order n with γt(G) ≥ 3,

and let S be a γt(G)-set. We show first that G[S] consists of disjoint copies of

K2. Assume, to the contrary, that G[S] contains a component C that is not a K2-

component. By Lemma 40(b), C is a complete graph on at least three vertices. But

then for each vertex v ∈ V (C) we note that ipn(v, S) = ∅, and so by Lemma 29 we

have that |epn(v, S)| ≥ 2, implying that dG(v) ≥ 4, contradicting the fact that G is

a cubic graph. Hence, G[S] consists of disjoint copies of K2.

We show next that every vertex in V \ S is dominated by vertices in exactly one

component of G[S]. Assume, to the contrary, that there exists a vertex v ∈ V \S that

is dominated by vertices from more than one component of G[S]. Since G is claw-free,

the vertex v is dominated by vertices from at most two components of G[S]. Hence,

v is dominated by vertices from exactly two components of G[S]. Let u and w be

vertices from different components of G[S] that are both adjacent to v. Let x and y

be the neighbors of u and w, respectively, in G[S]. By Lemma 29, |epn(u, S)| ≥ 1

and |epn(w, S)| ≥ 1. Let u′ ∈ epn(u, S), and let w′ ∈ epn(w, S). Since G is a cubic

graph, we may assume, renaming vertices if necessary, that u′v /∈ E. If vx /∈ E,


then G[{u, u′, v, x}] is a claw centered at u, a contradiction. Hence, vx ∈ E, and so

N(v) = {u, w, x}. But then G[{v, w, w′, y}] is a claw centered at w, a contradiction.

Hence every vertex in V \S is dominated by vertices in exactly one component of G[S].

In particular, this implies that ∂(C) ∩ ∂(D) = ∅ for any two distinct components C

and D of G[S].

We now consider an arbitrary component C of G[S]. Since G[S] consists of dis-

joint copies of K2, the component C consists of two adjacent vertices u and v. By

Lemma 29, |epn(u, S)| ≥ 1 and |epn(v, S)| ≥ 1. Let N(u) = {u1, u2, v} where

u1 ∈ epn(u, S), and let N(v) = {v1, v2, v} where v1 ∈ epn(v, S). Since every ver-

tex in V \ S is dominated by vertices in exactly one component of G[S], we note

that if u2 /∈ epn(u, S), then N(u2) ∩ S = {u, v}, and so u2 = v2. In this case,

|∂(C)| = |{u1, u2, v1}| = 3. Further if epn(u, S) = {u1, u2}, then epn(v, S) = {v1, v2}.

In this case, |∂(C)| = |{u1, u2, v1, v2}| = 4. Hence either |∂(C)| = 3 or |∂(C)| = 4.

Since n − |S| =∑ |∂(C)| where the sum is taken over all components C of G[S], we

have that n − |S| ≥ 3|S|/2, and so γt(G) = |S| ≤ 2n/5. Further, n − |S| ≤ 4|S|/2,

and so γt(G) = |S| ≥ n/3. 2

4.2.4 Realizability Results

In this section we determine, for γ+t -stable graphs, which combinations of order and

total domination number are attainable. Since we are restricting our attention in

this chapter to graphs whose total domination number is at least three, we note that

by Corollary 37, all such graphs have order at least 7. We first show that given any

integer n ≥ 8 and even integer k ≥ 4 with k ≤ n/2, there exists a connected γ+t -stable


graph G with γt(G) = k. The total domination number of a path Pn on n vertices is

easy to compute (or see [37]).

Proposition 42 For every path Pn, n ≥ 3, the following holds.

(a) ([37]) γt(Pn) = bn/2c + dn/4e − bn/4c. Further if n ≡ 0 (mod 4), then the

γt(Pn)-set is unique.

(b) For any set of integers {n1, ..., nk} such that n1 + · · ·+ nk = n and ni ≥ 2 for

all i, γt(Pn) ≤ γt(Pn1) + · · · + γt(Pnk).

For integers r ≥ 2 and s ≥ 1, we define the comet graph Cr,s to be the graph

of order r + s obtained from the disjoint union of a star K1,r−1 and a path Ps on s

vertices by adding an edge joining the central vertex of the star with an end-vertex

of the path.

Proposition 43 For every integer n ≥ 8 and every even integer k such that 4 ≤ k ≤

n/2, there exists a γ+t -stable graph G of order n such that γt(G) = k.

Proof. If k = n/2, then we simply take G to be the path P2k. Since k is an even

integer, we note that 2k ≡ 0 (mod 4), and so, by Propositions 25 and 42, G is a γ+t -

stable graph of order n with γt(G) = k. Hence we may assume that 4 ≤ k ≤ (n−1)/2.

We now consider the comet graph G = Cn−2k+2,2k−2. Then a longest path in G has

2k vertices. Let P : v1, v2, . . . , v2k be a longest path in G where v2 is the central vertex

of the star used to build the comet. Thus, dG(v2) = n−2k +2 while all other vertices

in G have degree at most 2. Since k is an even integer, we note that 2k ≡ 0 (mod 4).

Let S be a γt(G)-set. If S contains a leaf of v2, then we can simply replace this

vertex of S by the vertex v3. Hence we may assume that S ⊆ V (P ), and so S is a TDS


of the path P . By Proposition 42, k = γt(P ) ≤ |S| = γt(G). However the unique

γt(P )-set, namely the set D =⋃(k−2)/2

i=0 {v4i+2, v4i+3}, is also a TDS of G. Hence,

γt(G) ≤ γt(P ) = k. Consequently, γt(G) = k (and the set D is the unique γt(G)-set).

Suppose that there is an edge e = uv ∈ E(G) such that γt(G + e) < γt(G).

Let S∗ be any γt(G + e)-set. At most one of u and v is a leaf-neighbor of v2, for

otherwise, γt(G + e) = γt(G). Renaming the leaf-neighbors of v2, if necessary, we

may assume that e ∈ E(P ). But then every γt(G + e)-set is a TDS in P + e, and so

γt(P + e) ≤ γt(G + e) < k = γt(P ), contradicting the fact that the path Pn where

n ≡ 0 (mod 4) is a γ+t -stable graph. Hence, every edge in E(G) is a stable edge,

whence G is a γ+t -stable graph of order n with γt(G) = k. 2

Corollary 37 shows that if G is a γ+t -stable graph with order n ≥ 7 with γt(G) = k

where k is odd, then k ≤ (n − 2)/2. We show next that given any odd integer k ≥ 3

and any integer n ≥ 2k + 2, there exists a connected γ+t -stable graph G of order n

with γt(G) = k.

Proposition 44 For every odd integer k ≥ 3 and any integer n ≥ 2k+2, there exists

a connected γ+t -stable graph G of order n with γt(G) = k.

Proof. Let k ≥ 3 be an odd integer, and let n ≥ 2k + 2, and so 3 ≤ k ≤ (n −

2)/2. To construct G, we begin with a caterpillar G1 having spine (y, x, z) and code

(2, 1, n − 2k). If k = 3, then we let G = G1. Then γt(G) = 3 and S = {x, y, z} is

the unique γt(G)-set. Further, γ(G) = 3 and every vertex v ∈ S has property (a) in

the statement of Theorem 31. Hence, by Theorem 31, G is γ+t -stable with γt(G) = 3.

Hence we may assume in what follows that k ≥ 5, for otherwise we are finished.


Let y1 and y2 be the two leaf neighbors of y in G1, and let x1 be the leaf adjacent

to x. Add a path G2 = P2k−6, and add an edge from a leaf w of the path P2k−6 to y1.

Let G denote the resulting graph. In the special case when n = 12 and k = 5, the

graph G is shown in Figure 4.2.

xy z

G2

G1

y1

w

y2 x1

Figure 4.2: The graph G when n = 12 and k = 5.

Since k is odd, we note that 2k−6 ≡ 0 (mod 4). By Proposition 42, γt(G2) = k−3.

Further, we note that there is a unique γt(G2)-set which we shall denote by S ′. The

set S ′ ∪ {x, y, z} is a TDS of G, and so γt(G) ≤ |S ′| + 3 = (k − 3) + 3 = k. Let S be

a γt(G)-set. Since x, y and z are support vertices in G, we have that {x, y, z} ⊂ S.

If y1 ∈ S, then we can simply replace y1 in S by the neighbor of w on the path G2.

Hence we may assume that S∗ = S \ {x, y, z} ⊆ V (G2). Since S∗ is a TDS of G2, we

have that k − 3 = γt(G2) ≤ |S∗| = |S| − 3, and so γt(G) = |S| ≥ k. Consequently,

γt(G) = k. In particular, this implies that γt(G2) = |S∗|, and so S∗ is the unique

γt(G2)-set and S = S∗ ∪ {x, y, z} is the unique γt(G)-set. Hence if w1, w2, . . . , w2k−6

denotes the path G2 where w = w1, then S∗ =⋃(k−5)/2

i=0 {w4i+2, w4i+3}.

It remains for us to show that G is γ+t -stable. For the sake of contradiction,

assume that G is not γ+t -stable. Since every vertex v ∈ S satisfies Condition 1(a)

in the statement of Theorem 32, there exist two vertices u, v ∈ V (G) such that

dG(u, v) ≥ 3, Guv is isolate-free and γt(Guv) ≤ γt(G)− 3 = k − 3. We note that since


Guv is isolate-free, neither u nor v belongs to the set V (G1) \ {x1, y2}. In particular,

we note that z and all its leaf neighbors in G belong to Guv . Let D be a γt(Guv)-set,

and so |D| ≤ k − 3. Let D1 = D ∩ V (G1) and D2 = D ∩ V (G2). In order to totally

dominate z and its leaf neighbors in Guv, we note that |D1| ≥ 2. If {u, v} = {x1, y2},

then we note that by Proposition 42(a), |D2| ≥ γt(G2) = k − 3. Consequently,

|D| = |D1|+|D2| ≥ (k−3)+2 = k−1, a contradiction. If u ∈ V (G2) and v ∈ {x1, y2},

then by Proposition 42, we have |D2| ≥ γt(G[V (G2) \ N [u]]) ≥ γt(P2k−9) = k − 4.

Thus, |D| = |D1| + |D2| ≥ (k − 4) + 2 = k − 2, a contradiction. Finally, assume

that u, v ∈ V (G2). We note that {x, y, z} ⊆ D1. Further by Proposition 42, |D2| ≥

γt(G[V (G2) \ (N [u] ∪ N [v])]) ≥ γt(P2k−12) = k − 5. Therefore, |D| = |D1| + |D2| ≥

(k − 5) + 3 = k − 2, a contradiction. Hence G is a γ+t -stable graph. 2

Chapter 5

Total Domination Changing and

Stable Graphs Upon Vertex

Removal

5.1 Introduction

Goddard et al. [23] began the study of the graphs whose total domination number

decreases upon the removal of any vertex. Further properties of these graphs were

explored in [7, 43, 54, 55, 57, 58]. In this chapter, we continue the study of the effects

of vertex removal on the total domination number. For a graph G, we let S(G) denote

the set of support vertices of G and L(G) the set of leaves of G. Furthermore, we let

At(G) denote the set of vertices of G which are contained in every γt(G)-set.

74

CHAPTER 5. CHANGING AND STABLE GRAPHS 75

5.2 Preliminary Results

In this section, we consider the effects that removing a vertex from a graph has on

its total domination number. We begin with the remark that the total domination

number of a path Pn or a cycle Cn on n vertices is easy to compute (see [37]).

Observation 45 ([37]) For n ≥ 3, γt(Pn) = γt(Cn) = bn/2c + dn/4e − bn/4c.

We will employ the following result of Cockayne et al. [10] throughout this chapter.


vertex v ∈ S, |epn(v, S)| ≥ 1 or |ipn(v, S)| ≥ 1.

5.2.1 Effects of Vertex Removal

Removing a vertex from a graph can cause its total domination number to increase,

to decrease, or to remain the same. For an example showing increase, let r ≥ 2

and consider a subdivided star G = K∗1,r. If v is the central vertex of G, then

γt(G) = r + 1 while γt(G − v) = 2r. For an example showing decrease, let G = Cn

where n ≡ 1, 2 (mod 4). Then by Observation 45, we have γt(G−v) < γt(G) for every

vertex v of G. For an example where the total domination number remains unchanged

upon removal of a vertex, let G = Cn for n ≡ 0, 3 (mod 4). Then by Observation 45,

γt(G − v) = γt(G) for every vertex v of G.

Accordingly, for any graph G, we define the following weak partition of its vertex

set V (G).

Definition 5.2.1 For a graph G, we define a weak partition V (G) = V 0(G)∪V +(G)∪

V −(G) of its vertex set, where


• V 0(G) = {v ∈ V (G) | γt(G − v) = γt(G)}

• V +(G) = {v ∈ V (G) | γt(G − v) > γt(G)}

• V −(G) = {v ∈ V (G) | γt(G − v) < γt(G)}

In [23], a graph G is defined to be total domination vertex removal critical, or

γt-critical for short, if γt(G − v) < γt(G) for every vertex v ∈ V (G) \ S(G). In other

words, G is γt-critical if V (G) \ S(G) = V −(G). Since total domination is undefined

for a graph with isolated vertices, the definition of γt-critical graphs does not allow

removing a support vertex. We too consider those graphs whose total domination

number changes upon the removal of a vertex. However, we consider both types of

changes, that is, the total domination number can increase or decrease. Furthermore,

to extend the definition to include all vertices of G, we define γt(G) = ∞ if the graph

G has an isolated vertex. We observe that if v is a support vertex of a graph G

without isolates, then γt(G − v) = ∞, and so S(G) ⊆ V +(G). We also note that if

G is a connected graph on at least three vertices, then L(G) ⊆ V 0(G) ∪ V −(G). We

are now in a position to define our two main concepts in this chapter.

Definition 5.2.2 A graph G is γt-changing if γt(G − v) 6= γt(G) for every vertex

v ∈ V (G), while a graph G is γt-stable if γt(G−v) = γt(G) for every vertex v ∈ V (G).

Thus a graph G is γt-changing if the removal of any vertex from G either increases

or decreases the total domination number, that is, V (G) = V −(G)∪V +(G). A graph

G is γt-stable if V (G) = V 0(G). Since S(G) ⊆ V +(G), it follows that the γt-critical

graphs are a subset of the γt-changing graphs. For a simple example of a γt-changing


graph, consider the path P5, where the leaves of P5 are in V −(P5) and the remaining

vertices are in V +(P5).

5.2.2 Properties of Vertices in V−(G) ∪ V

+(G)

We begin with the following lemma from [23].

Lemma 47 ([23]) Let G be a graph, and let v ∈ V −(G). For every γt(G − v)-set S,

|S| = γt(G) − 1 and S ∩ N(v) = ∅.

As a consequence of Lemma 47, we have the following characterization of the set

V −(G) of vertices in a graph G.

Proposition 48 Let G be a graph without isolated vertices. A vertex v is in V −(G)

if and only if there exists some γt(G)-set S and a vertex u ∈ S such that v /∈ S and

pn(u, S) = {v}.

Proof. Let G be a graph without isolated vertices, and let v ∈ V −(G). Let S∗ be

an arbitrary γt(G − v)-set, and let u ∈ N(v). By Lemma 47, |S∗| = γt(G) − 1 and

S∗∩N(v) = ∅. Then S = S∗∪{u} is a γt(G)-set such that v /∈ S and pn(u, S) = {v}.

Conversely, assume that there exists a γt(G)-set S such that v /∈ S and pn(u, S) = {v}

for some u ∈ S. The set S \ {u} is a TDS for G − v of cardinality γt(G) − 1. Hence,

v ∈ V −(G). 2

Next we give a characterization of the set V +(G) of vertices in a graph G. Recall

that At(G) denotes the set of vertices of G which are contained in every γt(G)-set.


Proposition 49 Let G be a graph and v ∈ V (G). Then v ∈ V +(G) if and only if

v ∈ At(G) and either v ∈ S(G) or no subset of V (G) \N [v] with cardinality γt(G) is

a TDS for G − v.

Proof. Let G be a graph, and let v ∈ V +(G). If there exists a γt(G)-set D such that

v /∈ D, then the set D is a TDS for G − v. Hence, γt(G − v) ≤ |D| = γt(G), and

so v /∈ V +(G), a contradiction. Therefore, v ∈ At(G). If v ∈ S(G), we are finished.

Hence we may assume that v /∈ S(G). If there exists some set S ⊆ V (G) \ N [v] of

cardinality γt(G) which total dominates G − v, then γt(G − v) ≤ |S| = γt(G), and

so v /∈ V +(G), a contradiction. Therefore no subset S ⊆ V (G) \ N [v] of cardinality

γt(G) is a TDS for G − v. This proves the necessity.

For the sufficiency, suppose that v ∈ At(G) and either v ∈ S(G) or no subset of

V (G)\N [v] with cardinality γt(G) is a TDS for G−v. If v ∈ S(G), then γt(G−v) = ∞,

and so v ∈ V +(G) as desired. Hence we may assume that v /∈ S(G). We now consider

an arbitrary γt(G− v)-set S. On one hand, if S ⊆ V (G) \N [v], then by assumption,

|S| > γt(G). On the other hand, if S 6⊆ V (G) \ N [v], then S contains a vertex in

N(v). But then S is a TDS of G. However since v /∈ S and v ∈ At(G), the set S is

not a minimum TDS of G, implying that |S| > γt(G). In both cases, we have that

γt(G − v) = |S| > γt(G), and so v ∈ V +(G) as desired. 2

As a consequence of Proposition 49, we have the following result.

Corollary 50 Let G be a graph of order n. Then the following holds.

(a) |V +(G)| ≤ γt(G).

(b) If G is a γt-changing graph, then |V −(G)| ≥ n − γt(G).

(c) If G is a γt-changing graph and V (G) = V +(G), then G = n2K2.


Proof. Part (a) follows from Proposition 49 and the observation that γt(G) ≥

|At(G)|. Part (b) follows from Part (a) and the observation that V (G) = V −(G) ∪

V +(G) if G is a γt-changing graph. If G is a γt-changing graph with V (G) = V +(G),

then by Proposition 49, At(G) = V (G), implying that γt(G) = |V (G)| and G = kK2

for some k ≥ 1, thereby establishing Part (c). 2

We conclude this section with the following property of edges that join a vertex

in V +(G) and a vertex in V −(G).

Lemma 51 If uv is an edge in a graph G, where u ∈ V +(G) and v ∈ V −(G), then

v is a leaf of G.

Proof. Assume, for purposes of contradiction, that uv is an edge in G, where u ∈

V +(G) and v ∈ V −(G), but that v is not a leaf of G. Thus, dG(v) ≥ 2. Let S be

any γt(G − v)-set. By Lemma 47, |S| = γt(G) − 1 and S ∩ N(v) = ∅. In particular,

u /∈ S. Let w be a neighbor of v different from u. Then the set S ∪ {w} is a γt(G)-

set not containing u, and so u /∈ At(G). Hence by Proposition 49, u /∈ V +(G), a

contradiction. Therefore, v is a leaf of G. 2

As an immediate consequence of Lemma 51 and Corollary 50(c), we have the

following result.

Corollary 52 If G is a connected γt-changing graph with δ(G) ≥ 2, then V (G) =

V −(G).


5.3 γt-Changing Graphs

In this section, we study the γt-changing graphs; that is, we study graphs G with

V (G) = V −(G) ∪ V +(G). By Corollary 50(c), the only connected γt-changing graph

G with V (G) = V +(G) is G = K2. Further the γt-changing graphs G with V (G) =

V −(G) are precisely the γt-critical graphs (defined in Section 5.2.1) with minimum

degree at least two. A descriptive characterization of these graphs can be deduced

from Proposition 48.

Corollary 53 A graph G is a γt-changing graph with V +(G) = ∅ if and only if

δ(G) ≥ 2, and for every v ∈ V (G), there exists some γt(G)-set S and some vertex

u ∈ S such that v /∈ S and pn(u, S) = {v}.

In what follows, we therefore consider the γt-changing graphs where V −(G) 6= ∅

and V +(G) 6= ∅. As observed earlier, if G is a connected graph on at least three

vertices, then L(G) ⊆ V 0(G) ∪ V −(G) (where we recall that L(G) denotes the set of

leaves in G).

We begin by defining a family of graphs F . Recall that a strong support vertex

in a graph is a vertex that is adjacent to two or more leaves in the graph.

Definition 5.3.1 Let F be the family of all graphs G that can be obtained from a

connected graph H, where every support vertex of H is a strong support vertex, by

adding a new vertex v′ and an edge vv′ to every vertex v in H that is not a support

vertex.

We are now in a position to characterize the connected γt-changing graphs G with

V +(G) 6= ∅.


Theorem 54 Let G be a connected γt-changing graph. Then V +(G) 6= ∅ if and only

if G ∈ F .

Proof. Let G be a connected γt-changing graph. Suppose that V +(G) 6= ∅. If

V −(G) = ∅, then by Corollary 50(c), we have that G = K2. Since G = K2 is formed

from H = K1 by adding a new vertex joined by an edge to the vertex of H, the graph

G ∈ F . Hence we may assume that V −(G) 6= ∅, for otherwise the desired result

follows. Since G is connected, Lemma 51 implies that L(G) 6= ∅. Let H = G − L(G)

be the graph obtained from G by deleting all leaves of G. Since G is connected, so is

H. We proceed further with the following claim which establishes properties of the

graph G.

Claim: The following properties hold in the graph G.

(a) V −(G) = L(G) and V +(G) = V (H).

(b) V (H) is the unique γt(G)-set.

(c) Every support vertex of G is adjacent to exactly one leaf.

(d) No support vertex of G is a support vertex of H.

(e) Every vertex of H that is not a support vertex of G is a strong support vertex

of H.

Proof. (a) Since L(G) ⊆ V 0(G) ∪ V −(G) and V 0(G) = ∅, we note that L(G) ⊆

V −(G). We show next that V −(G) ⊆ L(G). Let v ∈ V −(G) and consider a shortest

path P : v = v0, v1, . . . , vk from v to a vertex in V +(G). By our choice of P , the vertex

vk−1 ∈ V −(G). By Lemma 51, the vertex vk−1 is a leaf of G, implying that v = vk−1

and therefore that v is a leaf. Hence, V −(G) ⊆ L(G). Consequently, V −(G) = L(G).

Thus, V (H) = V (G) \ V −(G) = V +(G).


(b) If H = K1, then G is a star K1,r for some r ≥ 2. But then removing an

arbitrary leaf from G leaves the total domination unchanged, and so L(G) ⊆ V 0(G),

a contradiction. Hence, |V (H)| ≥ 2. Therefore since H is connected, the set V (H)

is a TDS of G, and so γt(G) ≤ |V (H)|. By Part (a), V (H) = V +(G). Therefore by

Corollary 50, γt(G) ≥ |V (H)|. Consequently, γt(G) = |V (H)| and V (H) is a γt(G)-

set. Further by Proposition 49, V (H) is in every γt(G)-set, implying that V (H) is

the unique γt(G)-set.

(c) Suppose that G contains a strong support vertex. Let v be such a vertex of G.

Since G is not a star, we note that v ∈ V (H). Removing a leaf adjacent to v does not

change the total domination number of G, contradicting the fact that V −(G) = L(G)

from Part (a). Hence every support vertex of G is adjacent to exactly one leaf.

(d) Suppose that v is a support vertex in G and also in H. Let x be the leaf-

neighbor of v in G and let w be a leaf-neighbor of v in H. Since w is a leaf in H but

not in G, the vertex w is adjacent to a leaf, say w′, in G. Since every support vertex

of G is adjacent to exactly one leaf, we note that dG(w) = 2 and NG(w) = {v, w′}.

We now consider the graph G − x. Let Sx be a γt(G − x)-set. Since w is a support

vertex of G, w ∈ Sx. If w′ ∈ Sx, then we simply replace w′ in Sx with the vertex

v. Hence we may assume that v ∈ Sx. But then Sx is a TDS in G, implying that

γt(G) ≤ γt(G− x) and that the leaf x ∈ V 0(G), contradicting our earlier observation

that V −(G) = L(G). Hence no support vertex of G is a support vertex of H.

(e) Suppose that v is a vertex of H that is not a support vertex of G. Then

N(v) ⊆ V (H) and every neighbor of v has degree at least two in G. Let S = V (H).

By Part (b), the set S is the unique γt(G)-set. Since N(v) ∩ (V \ S) = ∅, we note


that epn(v, S) = ∅. Hence by Proposition 46, |ipn(v, S)| ≥ 1. Let u ∈ ipn(v, S).

Then u ∈ S and v is the only neighbor of u in H. Since dG(u) ≥ 2, the vertex u is a

support vertex of G. Thus by Part (c), dG(u) = 2. Let u′ be the leaf-neighbor of u in

G. If |ipn(v, S)| = 1, then (S \ {v})∪ {u′} is a γt(G)-set, contradicting Part (b) (and

contradicting the fact that v ∈ At(G) by Proposition 49). Hence, |ipn(v, S)| ≥ 2. Let

w ∈ ipn(v, S) \ {u}. An identical argument as shown with the vertex u shows that w

is a support vertex of degree 2 in G. Hence both u and w are leaf-neighbors of v in

the graph H, and so v is a strong support vertex of H. This completes the proof of

Part (e) and of the claim. 2

The properties of the graph G established in the above claim imply that G ∈ F .

This proves the necessity. For the sufficiency, assume that G ∈ F . If G = K2, then

V (G) = V +(G). Hence we may assume that G is constructed from a connected graph

H, where H contains at least two vertices and where every support vertex of H is a

strong support vertex, by adding to every vertex v in H that is not a support vertex

a new vertex v′ and the edge vv′.

We first show that V (H) is a γt(G)-set. By construction, V (H) is a TDS of G,

and so γt(G) ≤ |V (H)|. Let D be a γt(G)-set. By construction, every vertex of H

that is not a support vertex in H is a support vertex in G, and so V (H) \ S(H) ⊆ D

since every TDS in a graph contains all its support vertices. Let v ∈ S(H). By

construction, v is a strong support vertex of H. Let u and w be two leaf-neighbors of

v in H. Then both u and w are support vertices of degree 2 in G. Let u′ and w′ be

the leaf-neighbors of u and w, respectively, in G. If v /∈ D, then {u, u′, w, w′} ⊆ D.

But then (D \{u′, w′})∪{v} is a TDS of G of cardinality less than |D|, contradicting


the minimality of D. Hence, v ∈ D. Since v was an arbitrary vertex in S(H), we

have that S(H) ⊆ D. Therefore, V (H) ⊆ D, implying that γt(G) = |D| ≥ |V (H)|.

Consequently, γt(G) = |V (H)| and V (H) is a γt(G)-set.

We show next that G is γt-changing. Let v be an arbitrary vertex in G. Suppose

first that v is a leaf in G, and so v ∈ L(G). Let u be the support vertex adjacent to v.

By construction, v is the only leaf-neighbor of u in G. Further, u ∈ V (H) and u is not

a support vertex of H. Hence every vertex w ∈ NH(u) has degree at least two in H.

Therefore the set V (H)\{u} is a TDS of G−v, and so γt(G−v) ≤ |V (H)|−1 < γt(G).

Thus, v ∈ V −(G). Suppose next that v is not a leaf of G. Then by construction,

v ∈ V (H). If v is a support vertex of G, then since S(G) ⊆ V +(G), we have that

v ∈ V +(G). Finally, suppose that v ∈ V (H) \ S(G). Then by construction, v is a

strong support vertex in the graph H. Let u and w be two leaf-neighbors of v in

H. Then both u and w are support vertices of degree 2 in G. Let u′ and w′ be the

leaf-neighbors of u and w, respectively, in G. Let Sv be a γt(G − v)-set. Since u and

u′ induce a K2-component in G− v, as do w and w′, we note that {u, u′, w, w′} ⊆ Sv.

But then (Sv \ {u′, w′}) ∪ {v} is a TDS of G, and so γt(G) ≤ |Sv| − 1 < γt(G − v).

Thus, v ∈ V +(G). Therefore, V (G) = V +(G) ∪ V −(G), and so G is a γt-changing

graph. 2

As a consequence of Theorem 54, we obtain bounds on the total domination

number of a γt-changing graph in terms of its order. For this purpose, we define a

family H of graphs.

Definition 5.3.2 Let H be the family of all graphs G that can be obtained from a

connected graph F , by adding to every vertex v in F two disjoint copies of K2 and


adding an edge from v to one vertex in each copy of K2.

A graph in the family H is illustrated in Figure 5.1. Here the graph F is a 4-cycle

C4.

t t t tt t t tt t t tt t t tt t t t

��

AA

��

AA

��

AA

��

AA

F��

�

Figure 5.1: A graph in the family H.

Theorem 55 If G is a connected γt-changing graph of order n with V +(G) 6= ∅ and

V −(G) 6= ∅, then n/2 ≤ γt(G) ≤ 3n/5. Furthermore, the following holds.

(a) γt(G) = n/2 if and only if G = H ◦ K1 for some connected graph H with

δ(H) ≥ 2.

(b) γt(G) = 3n/5 if and only if G ∈ H.

Proof. Let G be a graph as defined in the hypothesis. By Theorem 54, G ∈ F .

Thus, G is constructed from a connected graph H, where every support vertex of H

is a strong support vertex, by adding to every vertex v in H that is not a support

vertex a new vertex v′ and the edge vv′. If H = K1, then G = K2 = K1 ◦ K1 and

V +(G) = V (G). Hence we may assume that |V (H)| ≥ 2. With this assumption,

the graph G satisfies the five properties (a)–(e) listed in the claim of the proof of

Theorem 54.

We now define a weak partition V (H) = A∪LA∪B of the vertex set of the graph

H, where A = V (H) \ S(G), LA is the set of leaf-neighbors of vertices of A in H,


and B = V (H) \ (A ∪ LA). Thus, A is the set of vertices of H that are not support

vertices in G. By the construction of the familyF , the vertices of A are strong support

vertices of H, while S(G) = LA ∪ B. Thus, |LA| ≥ 2|A| and |L(G)| = |LA| + |B|.

Hence,

n = |V (H)| + |L(G)|= |V (H)| + |LA| + |B|= 2|V (H)| − |A|= 2γt(G) − |A|,

and so,

γt(G) =1

2(n + |A|). (5.1)

Further,

γt(G) = |V (H)| = |A|+ |LA| + |B| ≥ 3|A|, (5.2)

and so

0 ≤ |A| ≤ 1

3γt(G). (5.3)

By Equation (5.1) and (5.3), we deduce that

n

2≤ γt(G) ≤ 3n

5. (5.4)

Suppose that we have equality in the lower bound of Equation (5.4), and so

γt(G) = n/2. This requires equality in the lower bound of Equation (5.3), and so

|A| = 0. Therefore H contains no support vertices. Further, every vertex of H is

a support vertex in G. Hence, δ(H) ≥ 2 and by the construction, G = H ◦ K1.

Conversely, suppose that G = H ◦ K1 for some connected graph H with δ(H) ≥ 2.

Then G ∈ F and as shown in the proof of the sufficiency of Theorem 54, G is

a γt-changing graph with V +(G) = V (H) and V −(G) = L(G). This establishes


Part (a) which characterizes the graphs achieving equality in the lower bound of

Equation (5.4).

Suppose that we have equality in the upper bound of Equation (5.4), and so

γt(G) = 3n/5. This requires equality in the upper bound of Equation (5.3), and so

|A| = γt(G)/3 = |V (H)|/3. Further this requires equality throughout the Inequality

Chain (5.2), and so |B| = 0 and |LA| = 2|A|. Thus, H = F ◦ K2 is the 2-corona

of some connected graph F . Moreover, G is obtained from H by adding for each

vertex v ∈ V (H) \ V (F ), a new vertex v′ and the pendant edge vv′. Equivalently, G

is obtained from F by adding to every vertex v in F two disjoint copies of K2 and

adding an edge from v to one vertex in each copy of K2. Thus, G ∈ F . Conversely,

suppose that G ∈ H. Then G can be obtained from a connected graph F by adding to

every vertex v in F two disjoint copies of K2 and adding an edge from v to one vertex

in each copy of K2. In particular, we note that G ∈ F . Therefore as shown in the proof

of the sufficiency of Theorem 54, G is a γt-changing graph with V +(G) = V (H) and

V −(G) = L(G). This establishes Part (b) which characterizes the graphs achieving

equality in the upper bound of Equation (5.4). 2

5.4 γt-Stable Graphs

Recall that a graph G is γt-stable if and only if V +(G) ∪ V −(G) = ∅. Hence as an

immediate consequence of Proposition 48 and Proposition 49, we have the following

characterization of γt-stable graphs.

Corollary 56 A graph G is γt-stable if and only if δ(G) ≥ 2 and for every v ∈ V (G),


both of the following conditions hold.

(a) There is no γt(G)-set S such that v /∈ S and pn(u, S) = {v} for some vertex

u ∈ S.

(b) Either v /∈ At(G) or v ∈ At(G) and there exist a TDS S in G − v such that

|S| = γt(G) and S ⊆ V (G) \ N [v].

We close with the following result which constructs γt-stable graphs having a

specified total domination number and induced subgraph.

Proposition 57 For every positive integer k ≥ 2 and every nontrivial graph H, there

exists a γt-stable graph G such that γt(G) = k and H is a vertex induced subgraph of

G.

Proof. Let G1, . . . , Gk−1 be k − 1 vertex disjoint copies of the complete graph K4.

For i = 1, . . . , k − 1, let V (Gi) = {ai, bi, ci, di}. Let A = {ai | 1 ≤ i ≤ k − 1},

B = {bi | 1 ≤ i ≤ k − 1}, C = {ci | 1 ≤ i ≤ k− 1}, and D = {di | 1 ≤ i ≤ k − 1}. Let

H be an arbitrary graph on at least two vertices, and let G be the graph constructed

from the disjoint union (⋃k−1

i=1 Gi) ∪ H of the graphs G1, . . . , Gk−1 and H, by adding

all edges joining vertices in V (H) to vertices in A ∪ B. The graph G for the case

when k = 2 and H = P3 is illustrated in Figure 5.2.

Let S be an arbitrary γt(G)-set. In order for S to totally dominate C ∪ D, it

is necessary that S ∩ V (Gi) 6= ∅ for all i, 1 ≤ i ≤ k − 1. Thus, |S| ≥ k − 1.

If |S| = k − 1, then |S ∩ V (Gi)| = 1 for all i, 1 ≤ i ≤ k − 1. But then G[S]

consists only of isolated vertices, contradicting the fact that S is a TDS of G. Hence,

γt(G) = |S| ≥ k. However a set D that contains exactly one vertex from the set


a1

b1

c1d1 HG1

Figure 5.2: The graph G when k = 2 and H = P3.

{ai, bi} for all i, 1 ≤ i ≤ k − 1, and contains an arbitrary vertex from H is a TDS of

G of cardinality k, and so γt(G) ≤ k. Consequently, γt(G) = k.

If k = 2, then the graph G is easily seen to be γt-stable. Hence we may assume

that k ≥ 3. Let S be an arbitrary γt(G)-set. Then S contains exactly one vertex

vi ∈ {ai, bi} for all i, 1 ≤ i ≤ k−1, and contains an arbitrary vertex h ∈ H. However,

|pn(h, S)| = |A ∪ B| = 2(k − 1) ≥ 4, while |pn(vi, S)| = |{ci, di}| = 2 for every

vertex u ∈ S \ {h}. Therefore there is no vertex v /∈ S such that pn(u, S) = {v}

for some vertex u ∈ S. Hence Condition (a) in Corollary 56. If h1 and h2 are two

distinct vertices in H, then S1 = A∪ {h1} and S2 = B ∪ {h2} are two vertex disjoint

γt(G)-sets. Hence no vertex belongs to every γt(G)-set, and so At(G) = ∅. Hence

Condition (b) in Corollary 56. Thus by Corollary 56, the graph G is a γt-stable graph.

By construction, H is a vertex induced subgraph of G. 2

Chapter 6

Domination Edge Lift Critical

Trees

6.1 Introduction

In this chapter, we will shift our attention away from the study of total domination

and we will consider the domination number of a graph. Particularly, we are going

to study the effects of a graph operation known as edge lifting on the domination

numbers of trees. The operation of edge lifting is used in a host of algorithmic

applications. For instance, Booth et.al. investigated the utility of edge lifting in

immersion order testing [4] and in the edge disjoint paths problem. A graph H is

immersed in G if and only if H can be obtained from G by a series of edge lifts. Also

a graph H is immersed in G if and only if there is an injection from V (H) to V (G)

so that the edges of H are represented by edge disjoint paths in G. Edge lifting was

90

CHAPTER 6. EDGE LIFT CRITICAL TREES 91

studied in [13] relative to its effects on the domination number of graphs. They [13]

showed that an edge lift can leave the domination number the same, or increase or

decrease it by one. Our research showed that while the operation of edge lifting was

used in various algorithmic applications, until [13] no one had considered its effects

on the domination or the total domination number of a graph.

We say that a graph G is domination edge lift critical if any arbitrary edge lift in

G changes the domination number. In this chapter, we characterize the domination

edge lift critical trees.

6.1.1 Background

The set A(G) of a graph G is defined in [13] to be the set of all vertex induced copies

of P3 in G. Throughout this chapter, we let uxv denote any arbitrary P3 path in

A(G) with vertex set {u, x, v} and central vertex x. It is shown in [13] that an edge

lift can increase (respectively, decrease) the domination number by at most one.

Theorem 58 ([13]) For a graph G and path uxv ∈ A(G), γ(G) − 1 ≤ γ(Guvx ) ≤

γ(G) + 1.

In [13], a weak partition A(G) = (A+(G),A−(G),A0(G)) is defined on A(G),

according to the effect that the edge lift on each path in A(G) has on the domination

number, as follows:

A+(G) = {uxv ∈ A(G) | γ(Guvx ) = γ(G) + 1}

A−(G) = {uxv ∈ A(G) | γ(Guvx ) = γ(G) − 1}

A0(G) = {uxv ∈ A(G) | γ(Guvx ) = γ(G)}.

The graphs G are then categorized according to how every edge lift in G affects

the domination number.


Definition 6.1.1 ([13])

(a) A graph G is γ−L -critical if A(G) = A−(G).

(b) A graph G is γ+L -critical if A(G) = A+(G).

(c) A graph G is γL-changing if A+(G) 6= ∅, A−(G) 6= ∅, and A0(G) = ∅.

(d) A graph G is γL-stable if A(G) = A0(G).

(e) A graph G is γL-mixed if A+(G) 6= ∅, A−(G) 6= ∅, and A0(G) 6= ∅.

It is shown in [13] that there exists a connected graph G with A(G) 6= ∅ in classes

(a), (b), (d), and (e) for any given domination number larger than two. In addition,

they establish the following result.

Theorem 59 ([13]) There are no connected γL-changing graphs.

Corollary 60 ([13]) If T is a domination edge lift critical tree, then T is either γ−L -

critical or γ+L -critical.

In Section 6.2, we show that, in fact, if T is a domination edge lift critical tree of

order n ≥ 3, then T is γ+L -critical. Equivalently, we show that no tree of order n ≥ 3

is γ−L -critical. Finally in Section 6.3, we characterize the γ+

L -critical, and hence the

domination edge lift critical trees.

We shall use the following two lemmas from [13]:

Lemma 61 ([13]) Let G be a graph and let uxv ∈ A(G). Then, uxv ∈ A−(G) if and

only if there exists some γ(G)-set S, such that, without loss of generality, {u, x} ⊆ S,

|N(x) ∩ S| ≥ 2, and epn(x, S) = {v}.


Lemma 62 ([13]) Let G be a graph, and let uxv ∈ A(G). If the path uxv ∈ A+(G),

then for every γ(G)-set S, one of the following conditions holds:

(a) x ∈ S, S ∩ {u, v} = ∅, and epn(x, S) ∩ {u, v} 6= ∅.

(b) x 6∈ S, S ∩ {u, v} 6= ∅, and N(x) ∩ (S \ {u, v}) = ∅.

Corollary 63 ([13]) The set A+(G) contains at most γ(G) vertex-disjoint copies of

P3.

6.2 γ−L -critical Trees

Graphs of order at most two are vacuously γ−L -critical. In this section we show that

there are no γ−L -critical trees with order at least three. First we observe the following

immediate consequence of Lemma 61.

Corollary 64 ([13]) If G is a γ−L -critical graph, then the central vertex of every

induced P3 in G has degree at least 3 in G.

Theorem 65 No tree on at least three vertices is γ−L -critical.

Proof. Let T be a tree on n ≥ 3 vertices, and assume, for purposes of contradiction,

that T is γ−L -critical. Let P : v1v2 . . . vk be a longest path in T . Since v1v2v3 is an

induced P3, the vertex v2 has degree at least 3 in T by Corollary 64. Let u1 be a

neighbor of v2 not on P . By our choice of P , the vertex u1 is a leaf in T . We now

consider the graph T v1v3v2

. Let S be a γ(T v1v3v2

)-set. If u1 ∈ S, then we simply replace

u1 in S with the vertex v2. Hence we may choose S so that v2 ∈ S in order to

dominate u1. But then S is a DS in T , and so γ(T ) ≤ |S| = γ(T v1v3v2

). However since

T is γ−L -critical, we have that γ(T v1v3

v2) < γ(T ), a contradiction. 2


6.3 γ+L -critical Trees

Our aim in this section is to give a constructive characterization of γ+L -critical trees.

For this purpose, we first define a family F of trees.

6.3.1 The Family F

We begin with a star K1,t, where t ≥ 1. If t ≥ 2, then we call the center of the star

an A-center and call its leaves A-leaves. In the special case when t = 1, we have

K1,t = P2 and we designate both vertices as B-leaves (and not A-leaves or A-centers).

Let Tw denote a star of order at least three with a leaf named w. We define F as the

set of trees containing K1,t, which is closed under the following two operations.

Operation O1: Extend T ′ ∈ F by adding Tw and the edge wy where y is an

A-leaf of T ′. We call the center of Tw an A-center and the leaves of Tw −w, we

call A-leaves. Remove y from the set of A-leaves.

Operation O2: Extend T ′ ∈ F by adding Tw and the edge wy where y is either

a B-leaf or an A-center of T ′. We call the center of Tw a B-center. If Tw = P3,

we call the leaf of Tw − w different from the center of Tw a B-leaf; otherwise,

if Tw 6= P3, we call all leaves of Tw − w A-leaves. If y is an A-center of T ′, we

rename it as a B-center; otherwise, y is a B-leaf of T ′, and we remove y from

the set of B-leaves.

For k ≥ 1, we denote the subfamily Fk of F as the set of all trees in F formed

from a star K1,t, where t ≥ 1, by applying k − 1 operations of type O1 or O2. In


particular, F1 = {K1,t | t ≥ 1} and

F =⋃

k≥1

Fk.

For T ∈ F , we call a star that is either the initial nontrivial star or a star added

in by an operation of type O1 or O2 to form T an underlying star of T . Thus, T ∈ F

has k underlying stars if and only if T ∈ Fk. We denote an underlying star with

center x by S∗x. By construction each underlying star, except possibly for the initial

star, has order at least three. Let U(T ) be the set of centers of the underlying stars

of T , where if the initial underlying star is a P2, then we select exactly one of its

vertices (centers) to be in the set U(T ). The set U(T ) therefore contains exactly one

center from each of the underlying stars of T .

6.3.2 Main Result

We shall prove the following result.

Theorem 66 A nontrivial tree T is γ+L -critical if and only if T ∈ F .

6.3.3 Key Lemmas

We will use the concept of strong equality that was introduced in [33]. If every γ(G)-

set is also an independent set, then γ(G) is said to strongly equal i(G), denoted γ(G) ≡

i(G). A characterization of trees with strongly equal domination and independent

domination numbers is given in [33]. In order to state this characterization, the

following operation and family F∗ are defined in [33] as follows.


Operation O∗: Let w be a vertex of a tree Tw such that every leaf of Tw,

except possibly for w itself, is at a distance two from w, and let y be a vertex

in a nontrivial tree T ′. Let T be formed from Tw ∪ T ′ by adding the edge wy.

The family F∗ is the set of trees T such that T = K1 or T is obtained from a

nontrivial star by a finite sequence of operations of type O∗.

Theorem 67 ([33]) For every nontrivial tree T , γ(T ) ≡ i(T ) if and only if T ∈ F∗.

By construction, the family F ⊂ F∗. Hence we have the following corollary of

Theorem 67.

Corollary 68 If T ∈ F , then γ(T ) ≡ i(T ).

In order to prove our main result, Theorem 66, we proceed with a series of lemmas.

Lemma 69 If T is a γ+L -critical tree, then γ(T ) ≡ i(T ).

Proof. Let T be a γ+L -critical tree, and let S be a γ(T )-set. Assume, for purposes of

contradiction, that S is not an independent set. Accordingly, let u and x be adjacent

vertices in S. By the minimality of S, epn(x, S) 6= ∅ and epn(u, S) 6= ∅. Let v ∈

epn(x, S), and consider the induced path uxv. Since S is a dominating set of T and

{u, x} ⊆ S, we note that S is a dominating set of T uvx , and so γ(T uv

x ) ≤ |S| = γ(T ),

contradicting the fact that uxv ∈ A+(G). 2

Next we establish the domination number of a tree in the family F .

Lemma 70 For k ≥ 1, if T ∈ Fk, then γ(T ) = k.


Proof. We proceed by induction on k. If k = 1, then T is a star and γ(T ) = 1 = k.

This establishes the base case. Suppose that k ≥ 2, and that if Tj ∈ F where

1 ≤ j < k, then γ(Tj) = j. Since T ∈ Fk, the tree T is formed from a tree T ′ ∈ Fk−1

by one application of either Operation O1 or O2 by adding a tree Tw. Applying the

inductive hypothesis to the tree T ′ ∈ Fk−1, we have that γ(T ′) = k − 1. Since every

γ(T ′)-set can be extended to a DS of T by adding to it the center of the star Tw, we

have that γ(T ) ≤ γ(T ′) + 1 ≤ k. To show that γ(T ) ≥ k, let S be a γ(T )-set. If S

contains a leaf of Tw different from w, then we may replace this leaf in S with the

central vertex of the star Tw. Further if w ∈ S, then we may replace w in S with the

vertex y ∈ V (T ′) that is adjacent to w in T . Hence we may assume that S ∩ V (Tw)

consists only of the central vertex of the star Tw. But then the set S ∩ V (T ′) is a

DS in T ′ of cardinality |S| − 1, and so k − 1 = γ(T ′) ≤ |S| − 1 = γ(T ) − 1, whence

γ(T ) ≥ k. Consequently, γ(T ) = k. 2

Recall that the set U(T ) is the set of centers of the underlying stars of a tree

T ∈ F . In particular, we note that if T ∈ Fk, then U(T ) is a DS of T of cardinality k.

As a consequence of Lemma 70, we therefore have the following result.

Corollary 71 If T ∈ F , then U(T ) is a γ(T )-set.

We proceed further by establishing properties of trees that belong to the family

F .

Lemma 72 Every B-leaf of T ∈ F belongs to some γ(T )-set.

Proof. We proceed by induction on k, where T ∈ Fk. If k = 1, then T = P2, and the

result follows readily. This establishes the base case. Let k ≥ 2 and assume that every


B-leaf of a tree in Fk−1 belongs to some minimum DS in the tree. We now consider

a tree T ∈ Fk. Let T be constructed from a tree T ′ ∈ Fk−1 by adding a star Tw on

at least three vertices with leaf w and central vertex x, and applying Operation O1

or Operation O2 to extend the tree T ′ to the tree T by adding the edge wy where

y ∈ V (T ′). Let z be an arbitrary B-leaf of T .

On the one hand, suppose z ∈ V (T ′). By construction, the vertex z is a B-leaf

in T ′. Applying the inductive hypothesis to T ′, there exists a γ(T ′)-set S ′ containing

z. By Lemma 70, |S ′| = k − 1 and S ′ ∪ {x} is γ(T )-set of cardinality k containing z.

On the other hand, suppose that z ∈ V (Tw). Then, z is a leaf of Tw different from w.

By construction, Tw = P3 and T is obtained from T ′ by applying Operation O2. Now

either y is a B-leaf in T ′ or an A-center of T ′. If y is B-leaf in T ′, then, applying the

inductive hypothesis to T ′, there exists a γ(T ′)-set Sy containing y. By Lemma 70,

|Sy| = k− 1 and Sy ∪{z} is γ(T )-set of cardinality k containing z. If y is an A-center

in T ′, then y ∈ U(T ) and, by Corollary 71, U(T ) is a γ(T )-set containing the center

x of Tw. But then (U(T ) \ {x}) ∪ {z} is a γ(T )-set containing z. In all cases, there

exists a γ(T )-set which contains z, as desired. Hence every B-leaf of T belongs to

some γ(T )-set. 2

Lemma 73 No A-leaf of T ∈ F belongs to some γ(T )-set.

Proof. We proceed by induction on k ≥ 1, where T ∈ Fk. If k = 1, then T is

a star of order at least three and T has a unique γ(T )-set consisting of the central

vertex of the star. This establishes the base case. Let k ≥ 2, and assume that no

A-leaf of a tree in Fk−1 belongs to some minimum DS in the tree. We now consider

a tree T ∈ Fk. Let T be constructed from a tree T ′ ∈ Fk−1 by adding a star Tw on


at least three vertices with leaf w and central vertex x, and applying Operation O1

or Operation O2 to extend the tree T ′ to the tree T by adding the edge wy where

y ∈ V (T ′). Let z be an arbitrary A-leaf of T . We show that z is in no γ(T )-set.

Suppose z ∈ V (Tw). If Tw has order at least 4, then the central vertex of Tw has

at least two leaf-neighbors in T and is therefore in every γ(T )-set, implying that z

is in no γ(T )-set. Hence we may assume that Tw has order exactly 3, and so Tw is

the path wxz. By construction, T is obtained from T ′ by Operation O1, and so y

is an A-leaf in T ′. Assume that there does exist a γ(T )-set Dz that contains z. In

order to dominate the vertex w, the set Dz contains w, x, or y. If x or w belong to

Dz, we can simply replace it with the vertex y. Hence we may choose the set Dz so

that y ∈ Dz and Dz ∩ V (Tw) = {z}. But then D′z = Dz ∩ V (T ′) is a DS of T ′ of

cardinality |Dz| − 1 = k − 1. Thus by Lemma 70, D′z is a γ(T ′)-set containing y,

which is an A-leaf in T ′. However applying the inductive hypothesis to T ′, the vertex

y is no γ(T ′)-set, a contradiction. Hence there exists no γ(T )-set that contains the

leaf z.

We may therefore assume that z ∈ V (T ′), for otherwise the desired result follows.

By construction, z is also an A-leaf in T ′. Assume that there does exist a γ(T )-set

Dz that contains z. If a leaf-neighbor of x in T belongs to Dz, we can simply replace

it with the vertex x. Hence we may choose Dz so that x ∈ Dz and no leaf-neighbor

of x in T belongs to Dz. If w ∈ Dz , we can replace it with the vertex y. Hence we

may further choose Dz so that w /∈ Dz. But then D′z = Dz ∩ V (T ′) is a DS of T ′

of cardinality |Dz | − 1 = k − 1. Thus by Lemma 70, D′z is a γ(T ′)-set containing

z, which is an A-leaf in T ′. However applying the inductive hypothesis to T ′, there


is no γ(T ′)-set that contains an A-leaf of T ′, a contradiction. Hence there exists no

γ(T )-set that contains the leaf z. 2

By construction, every leaf of a tree in the family F is either an A-leaf or a B-leaf

in the tree. Hence as an immediate consequence of Lemma 72 and Lemma 73, we

have the following result.

Lemma 74 Let T ∈ F and let z be a leaf of T . Then, z is a B-leaf of T if and only

if there exists a γ(T )-set that contains z.

Lemma 75 No neighbor of an A-center in a tree T ∈ F belongs to some γ(T )-set.

Proof. We proceed by induction on k ≥ 1, where T ∈ Fk. If k = 1, then T is

a star of order at least two. If T = P2, then T has no A-center, and the result is

vacuously true. Hence we may assume that T has order at least three. Thus, T has a

unique γ(T )-set consisting of the central vertex of the star, which is the A-center of

T . Therefore no neighbor of the A-center belongs to some γ(T )-set. This establishes

the base case. Let k ≥ 2, and assume that no neighbor of an A-center in a tree in

Fk−1 belongs to some minimum DS in the tree. We now consider a tree T ∈ Fk. Let

T be constructed from a tree T ′ ∈ Fk−1 by adding a star Tw on at least three vertices

with leaf w and central vertex x, and applying Operation O1 or Operation O2 to

extend the tree T ′ to the tree T by adding the edge wy where y ∈ V (T ′).

Assume, to the contrary, that there exists a vertex u that is adjacent to an A-

center in T and belongs to some γ(T )-set S. Let S ′ = S ∩ V (T ′). By construction,

no neighbor of an A-center is itself an A-center. Further every leaf neighbor of an

A-center is an A-leaf.


Suppose u = y. Then the vertex y is an A-leaf in T ′ and T is obtained from T ′

by applying Operation O1. If a leaf-neighbor of x in T belongs to S, we can simply

replace it with the vertex x. Hence we may choose S so that x ∈ S and no leaf-

neighbor of x in T belongs to S. Since y ∈ S, this in turn implies that w /∈ S, and so

S∩V (Tw) = {x}. Thus, S ′ is a DS of T ′ of cardinality |S|−1 = k−1. By Lemma 70,

S ′ is therefore a γ(T ′)-set containing the A-leaf u in T ′, contradicting Lemma 73.

Hence, u 6= y.

Suppose u ∈ V (T ′)\{y}. Since γ(T ) = k and at least one vertex of Tw is in S, we

have |S ′| ≤ γ(T ) − 1 = k − 1 = γ(T ′). Applying our inductive hypothesis to T ′, no

γ(T ′)-set contains a vertex adjacent to an A-center in T ′. The set S ′ dominates all

vertices in T ′, except possibly for the vertex y. If S ′ is a DS of T ′, then S ′ is a γ(T ′)-

set that contains a vertex, namely u, adjacent to an A-center in T ′, a contradiction.

Hence, S ′ dominates all vertices in T ′ except for the vertex y. In order to dominate

y, we therefore have that w ∈ S. If a leaf-neighbor of x in T belongs to S, we can

simply replace it with the vertex x. Hence we may choose S so that x ∈ S. But then

{x, w} ⊆ S, and so the γ(T )-set S is not independent, contradicting Corollary 68.

Hence, u ∈ V (Tw).

Since u ∈ V (Tw), the central vertex x of Tw is an A-center. By construction, T

is obtained from T ′ by applying Operation O1. Thus every leaf-neighbor of x in T is

an A-leaf and therefore, by Lemma 73 does not belong to S. Hence u = w, and we

may assume that x ∈ S in order to dominate the leaf neighbors of x in T . But then

{x, w} ⊆ S, and so the γ(T )-set S is not independent, contradicting Corollary 68. 2



We are now in a position to prove our main result. Recall the statement of Theo-

rem 66.

Theorem 66. A nontrivial tree T is γ+L -critical if and only if T ∈ F .

Proof. Suppose that T is a nontrivial γ+L -critical tree. We proceed by induction on

the order n of T to show that T ∈ F . If T is a nontrivial star, then T ∈ F1 ⊂ F .

Hence we may assume that diam(T ) ≥ 3 and n ≥ 4. This establishes the base case.

For the inductive hypothesis, assume that if T ′ is a nontrivial γ+L -critical tree of order

less than n, then T ′ ∈ F . Let T be a nontrivial γ+L -critical tree of order n. We

now root T at a vertex r on a longest path P , and let w be the vertex at distance

diam(T ) − 2 from r on P . Let v be the child of w on P and u the child of v on P .

Necessarily, u is a leaf. Let y be the parent of w. Let D be a γ(T )-set containing the

set of support vertices of T .

Suppose there exists a γ(T )-set Sw that contains the vertex w. If a leaf-neighbor

of v in T belongs to Sw, we can simply replace it with the vertex v. Hence we may

choose Sw so that v ∈ Sw. But then {v, w} ⊆ Sw, and so the γ(T )-set Sw is not

independent, contradicting Lemma 69. Hence there is no γ(T )-set that contains w.

The vertex w therefore has no leaf-neighbor.

Suppose that dT (w) ≥ 3. Let v′ be a child of w different from v. Then, dT (v′) ≥ 2

and every neighbor of v′ different from w is a leaf. We now consider the tree T v′yw .

Since the support vertices v and v′ both belong to the γ(T )-set D, the set D is also a

DS of T v′yw , and so γ(T v′y

w ) ≤ |D| = γ(T ), contradicting the fact that T is γ+L -critical.

Hence, dT (w) = 2 and N(w) = {v, y}. Let Tw be the subtree of T rooted at w; that


is, Tw is the subtree of T induced by the set N [v]. Then Tw is a star of order at least

three with center v.

Let T ′ be the component of T −wy containing y. Every γ(T ′)-set can be extended

to a DS of T by adding to it the vertex v, and so γ(T ) ≤ γ(T ′) + 1. Since the

γ(T )-set D does not contain the vertex w, the set D \ {v} is a DS of T ′, and so

γ(T ′) ≤ |D| − 1 = γ(T )− 1. Consequently, γ(T ) = γ(T ′) + 1.

If y is the only vertex of T ′, then γ(T vyw ) = |{v, w}| = 2 = γ(T ), contradicting the

fact that T is γ+L -critical. Hence, T ′ has order at least two. Let x denote the parent

of y in T . If T ′ = P2, then T can be obtained from the star Tw − w by adding the

star wyx and applying Operation O2 by adding the edge wv, showing that T ∈ F , as

desired. Hence we may assume that T ′ has order at least three.

Assume that T ′ is not γ+L -critical. Then there exists an edge lift in T ′ that does

not increase the domination number. After such an edge lift of a P3 in T ′, there is

therefore a DS D′ of cardinality at most γ(T ′) that dominates the resulting graph.

But then D′ ∪ {v} is a DS of the graph resulting from the same edge lift in T with

cardinality at most |D′| + 1 ≤ γ(T ′) + 1 = γ(T ), contradicting the fact that T is

γ+L -critical. Hence, T ′ is a γ+

L -critical tree. Applying our inductive hypothesis to the

tree T ′, we have that T ′ ∈ F .

Suppose y is a leaf of T ′. By construction, every leaf of a tree in the family F

is either an A-leaf or a B-leaf in the tree. If y is an A-leaf in T ′, then T can be

constructed from T ′ by adding the star Tw and applying Operation O1 to extend T ′

to T by adding the edge wy. If y is a B-leaf in T ′, then T can be constructed from T ′

by adding the star Tw and applying Operation O2. In both cases, T ∈ F , as desired.


Hence we may assume that y is not a leaf in T ′.

Suppose y is a leaf in an underlying star of T ′ ∈ F . Since y is not a leaf in T ′,

there exists a neighbor z of y in T ′ not in the underlying star of T ′ that contains y.

We now consider the set U(T ′) of centers of the underlying stars of a tree T ′ ∈ F .

By Corollary 71, U(T ′) is a γ(T ′)-set. The set U(T ′) ∪ {v} is a DS of T wzy , and so

γ(T wzy ) ≤ |U(T ′)|+1 = γ(T ′)+ 1 = γ(T ), contradicting the fact that T is γ+

L -critical.

Hence, y is a center of an underlying star of T ′.

Every center of an underlying star of a tree in F is either an A-center or a B-

center. We show that y is an A-center. Assume, to the contrary, that y is a B-center.

By construction, the centers of the underlying stars form an independent set. Suppose

that y is at a distance 2 from another B-center, say p, in T ′. Let q be the common

neighbor of y and p. Then, {p, y} ⊆ U(T ′). The set U(T ′) ∪ {v} is a DS of T qwy , and

so γ(T qwy ) ≤ |U(T ′)|+ 1 = γ(T ′) + 1 = γ(T ), a contradiction. Hence, y is at distance

at least 3 from every other center in T ′.

Let T ′ ∈ Fk, and let T ′1, T

′2, . . . , T

′k = T ′ be the sequence of trees used to construct

T ′, where T ′1 is a nontrivial star and for i = 2, . . . , k, the tree T ′

i is obtained from the

tree T ′i−1 ∈ F by applying Operation O1 or O2. Let j be the smallest integer such

that y ∈ V (T ′j). Since y is a B-center and is at distance at least 3 from every other

center in T ′, it follows from the way in which trees in the family F are constructed

that j ≥ 2 and that y is at a distance 2 from a B-leaf q of the tree T ′j−1. Let z

be the common neighbor of y and q. By Lemma 72, there exists a γ(T ′j−1)-set Dq

that contains the B-leaf q. For i = j, j + 1, . . . , k, let ci denote the center of the star

added to T ′i−1 when constructing the tree T ′

i . In particular, we note that cj = y. Let


D∗ = {cj, cj+1, . . . , ck}. Then the set D′ = Dq ∪D∗ is a DS of T ′ of cardinality |D′| =

|Dq|+|D∗| = (j−1)+(k−j+1) = k = γ(T ′). In particular, we note that {q, y} ⊆ D′.

Thus the set D′ ∪ {v} is a DS of T wzy , and so γ(T wz

y ) ≤ |D′| + 1 = γ(T ′) + 1 = γ(T ),

a contradiction. Hence, y is an A-center. Therefore the tree T can be constructed

from the tree T ′ by adding the star Tw and applying Operation O2 and adding the

edge yw. Thus, T ∈ F , as desired. This establishes the necessity.

Next we consider the sufficiency. Let T ∈ F , and so T ∈ Fk for some k ≥ 1.

We prove by induction on k that T is a γ+L -critical tree. If k = 1, then T is a

nontrivial star, and hence T is γ+L -critical. This establishes the base case. For the

inductive hypothesis, let k ≥ 2 and assume that every tree in Fj, where j < k, is

a γ+L -critical tree. Let T ∈ Fk, and let T be constructed from a tree T ′ ∈ Fk−1 by

adding a star Tw on at least three vertices with leaf w and center x, and applying

Operation O1 or Operation O2 to extend the tree T ′ to the tree T by adding the edge

wy where y ∈ V (T ′). By our inductive hypothesis, T ′ ∈ Fk−1 is a γ+L -critical tree.

By Lemma 70, γ(T ) = k and γ(T ′) = k − 1.

Let uqv be a path in T . We show that uqv ∈ A+(T ). Let S be a γ(T uvq )-set, and

let S ′ = S ∩ V (T ′). Assume, for purposes of contradiction, that uqv /∈ A+(T ). Thus,

|S| = γ(T uvq ) ≤ γ(T ). In order to dominate x, the set S contains at least one vertex

in Tw, and so |S ′| ≤ |S| − 1 ≤ γ(T ) − 1 = γ(T ′).

Suppose that uqv is a path in T ′. If S ′ is a DS in the graph T′uvq , then γ(T

′uvq ) ≤

|S ′| ≤ γ(T ′), contradicting the fact that T ′ is γ+L -critical. Hence, S ′ is not a DS in

T′uvq . The only possible vertex not dominated by S ′ in T

′uvq is the vertex y. Hence,

w ∈ S in order to dominate y in T uvq . In order to dominate the leaf-neighbors of x


in T uvq , we may choose S to contain x. Hence, |S ′| ≤ |S| − 2 ≤ γ(T ′) − 1, and so the

set S ′ ∪ {y} is a DS in the graph T′uvq , implying that γ(T

′uvq ) ≤ |S ′| + 1 ≤ γ(T ′), a

contradiction. Hence, uqv is not a path in T ′.

Suppose that uqv is a path in Tw. Then, q = x. Renaming u and v, if necessary,

we may assume that u 6= w. In order to dominate u and x in T uvq , the set S contains

at least two vertices in V (Tw), implying that |S ′| ≤ |S| − 2 ≤ γ(T ′) − 1. Hence,

S ′ cannot be a DS of T ′. The only possible vertex not dominated by S ′ in T ′ is the

vertex y. Hence, w ∈ S in order to dominate y in T uvq . Since S ′∪{y} is a DS of T ′, we

have γ(T ′) ≤ |S ′| + 1. Consequently, γ(T ′) = |S ′| + 1, |S ′| = |S| − 2 and |S| = γ(T ).

Let w′ be the vertex of S different from w that belongs to Tw. But then the set

(S \ {w′}) ∪ {x} is a γ(T )-set that is not independent, contradicting Corollary 68.

Hence, uqv is not a path in Tw. Thus, {w, y} ⊂ {u, q, v} and either q = w or q = y.

Suppose that q = w. Then, uqv is the path ywx since by construction of T we

note that NT (w) = {y, x}. Since w is isolated in T uvq , we have w ∈ S. Further since x

is a support vertex in T uvq , we may choose S so that x ∈ S. But then S is a γ(T )-set

that is not independent, a contradiction. Hence, q = y and uqv is the path zyw for

some neighbor z of y in T ′. In order to dominate the leaves of x in T zwy , we may

choose S so that x ∈ S.

Suppose T is constructed from T ′ by applying Operation O1. Then, y is an A-leaf

of T ′. Therefore, dT (y) = 2, and so y is isolated in T wzy . Hence, y ∈ S and S ′ is

a DS of T ′. Since |S ′| ≤ γ(T ′), the set S ′ is necessarily a γ(T ′)-set that contains

the A-leaf y, contradicting Lemma 73. Hence, T is constructed from T ′ by applying

Operation O2. Thus, y is either a B-leaf or an A-center in T ′.


Suppose y is a B-leaf in T ′. Then, dT (y) = 2 and y ∈ S in order to dominate itself.

Thus, S ′ is a γ(T ′)-set which in turn implies that S is a γ(T )-set. By Corollary 68,

neither w nor z belongs to S. In order to dominate the vertex z in T wzy , the set S

contains a neighbor z′ of z in T wzy . Since y is isolated in T wz

y , we note that y 6= z′. But

then (S \{y})∪{z} is a γ(T )-set that is not independent, contradicting Corollary 68.

Hence, y must be an A-center in T ′. By construction of the family F , we note

that dT ′(y) ≥ 2, and so |V (T ′)| ≥ 3. Further all the neighbors of y in T ′ have degree

at most 2 in T ′ and are leaves of the underlying star of T ′ containing y.

Suppose that z is a leaf of T ′. Then, z is a leaf in T zwy with support vertex w.

If z ∈ S, we can replace it with the vertex w. Hence we may choose S so that

w ∈ S. If y ∈ S, then S \ {w} is a DS of T with cardinality less than γ(T ), a

contradiction. Hence, y /∈ S. In order to dominate y in T zwy , the vertex y has a

neighbor in S \ {w}. But then (S \ {w}) ∪ {y} is a γ(T )-set that is not independent,

contradicting Corollary 68. Hence, z is not a leaf in T ′, and so dT ′(z) = 2.

Let z′ be a neighbor of z in T ′ different from y. Suppose w ∈ S. Recall that

x ∈ S. If y ∈ S, then S \ {w} is a DS of T with cardinality less than γ(T ), a

contradiction. Hence, y /∈ S. In order to dominate y in T zwy , the vertex y has a

neighbor in S \ {w}. But then (S \ {w}) ∪ {y} is a γ(T )-set that is not independent,

contradicting Corollary 68. Hence w /∈ S. Thus, S ′ is a γ(T ′)-set. Moreover, S ′ \ {y}

dominates z, so either z or z′ is in S. If z ∈ S, we can replace it with the vertex z′.

Hence we may choose S so that z′ ∈ S.

Since y is an A-center, by our construction of the family F , no center in T ′ is at

distance 2 from y. Hence, z′ is a leaf in an underlying star, say S∗u, in T ′ with center


u. Since y is an A-center, by construction we have that S∗u 6= P2. Assume that u has

two neighbors, a and b, in T ′ other than z′. Since S ′ is a γ(T ′)-set, by Corollary 68,

S ′ is independent. Hence, u /∈ S ′. Thus considering the path aub, Lemma 62 implies

that S ′ ∩ {a, b} 6= ∅ and that N(u) ∩ (S ′ \ {a, b}) = ∅ . But since z′ ∈ S ′ ∩ N(u), we

have a contradiction. Hence, S∗u = P3 and dT (u) = 2.

Let {z′, u, a} be the vertex set of S∗u. Since S ′ is a γ(T ′)-set containing z′,

Lemma 75 implies that u is a B-center in T . By construction, the edge zz′ was

added during some operation between leaves of the underlying stars S∗u and S∗

y when

building the tree T ′. Suppose the vertex u was in the tree when S∗y was added. Then

since y is an A-center, Operation O1 was used when adding the star S∗y . This implies

that z′ was an A-leaf before the star S∗y was added. However since u is a B-center, S∗

u

must have been added to the tree in a previous application using Operation O2. But

then since S∗u = P3, the vertex z′ was added as a B-leaf, a contradiction. Hence the

vertex y was in the tree when S∗u was added. Since y is an A-center in T ′ and no B-

center ever changes to an A-center, y was an A-center when zz′ was added. Hence, z

was an A-leaf before the star S∗u was added. Thus, S∗

u was added using Operation O1,

implying that u was an A-center when S∗u was first added. Since dT (u) = 2, no oper-

ation can change u to a B-center in T , and so we have a contradiction. We deduce,

therefore, that for every path uqv in T , we must have uqv ∈ A+(T ). Consequently,

T is a γ+L -critical tree. This completes the proof of Theorem 66. 2

Chapter 7

Edge Lifting and Total Domination

in Graphs

7.1 Introduction

In this chapter, we study the effect of edge lifting on the total domination number

of a graph. We restrict our attention to connected graphs that contain at least one

induced P3. Further, since total domination is undefined for a graph with isolated

vertices, we restrict our attention to graphs G having no isolated vertices after any

arbitrary edge lift. In other words, no center of an induced P3 in G has degree 2, and

so every vertex of degree 2 in G, if any, is on a triangle. Let L be the family of such

graphs.

109

CHAPTER 7. EDGE LIFTING AND TOTAL DOMINATION 110

7.2 Effects of Lifting on the Total Domination Num-

ber

An edge lift in a graph can cause the total domination number to increase, decrease,

or remain the same. For example, in the graph G shown in Figure 7.1 we observe

that γt(G) = 5 and that γt(Guvx ) = 4, γt(G

vzx ) = 5 and γt(G

abv ) = 6.

w z vu

a b

x

Figure 7.1: Graph G.

For a graph G, we let A(G) denote the set of all vertex induced copies of P3 in G

where the center vertex has degree at least 3. Throughout this chapter, we let uxv

denote any arbitrary P3 path in A(G).

To facilitate our study of the effects that edge lifts have on the total domination

number of a graph, we define the following weak partition A(G) = (A+(G),A−(G),A0(G)),

according to the effect that the edge lift on each path in A(G) has on the total dom-

ination number, as follows:

A+(G) = {uxv ∈ A(G) | γt(Guvx ) > γt(G)}

A−(G) = {uxv ∈ A(G) | γt(Guvx ) < γt(G)}

A0(G) = {uxv ∈ A(G) | γt(Guvx ) = γt(G)}.

We first show that an edge lift off the center of a path in A(G) can decrease the

total domination number by at most one and increase it by at most two.

Theorem 76 For every isolate-free graph G ∈ L and every path uxv ∈ A(G), γt(G)−

1 ≤ γt(Guvx ) ≤ γt(G) + 2.


Proof. Every γt(Guvx )-set is either a TDS of G or can be extended to a TDS of

G by adding to it the vertex x. Hence, γt(G) ≤ γt(Guvx ) + 1, which establishes the

lower bound. To prove the upper bound, let S be an arbitrary γt(G)-set and let

w ∈ N(x) \ {u, v}. If x /∈ S or if x ∈ S and {u, v} ⊂ S, let S ′ = S ∪ {w}. If

x ∈ S and S ∩ {u, v} = {u}, let S ′ = S ∪ {v, w}. If x ∈ S and S ∩ {u, v} = {v}, let

S ′ = S ∪ {u, w}. If x ∈ S and S ∩ {u, v} = ∅, let S ′ = S ∪ {u, v}. In all cases, the set

S ′ is a TDS of Guvx and |S ′| ≤ |S|+2. Hence, γt(G

uvx ) ≤ |S ′| ≤ |S|+2 = γt(G)+ 2. 2

That the bounds of Theorem 76 are sharp may be seen as follows. For the lower

bound, consider the corona G = H ◦K1 where H = Kk, k ≥ 3. Every vertex of V (H)

is a support vertex in G, and so γt(G) ≥ |V (H)| = k. Since V (H) is a TDS for G,

γt(G) ≤ k. Consequently, γt(G) = k. Let uxv ∈ A(G). We note that the vertex x

and exactly one of u and v belong to H. The set V (H) \ {x} forms a TDS of Guvx ,

and so γt(Guvx ) ≤ k − 1. Since Guv

x contains k − 1 support vertices, γt(Guvx ) ≥ k − 1.

Thus γt(Guvx ) = k − 1 = γt(G) − 1. That the upper bound is achieved may be seen

by simply taking a star G = K1,n−1 on n ≥ 4 vertices. Then, γt(G) = 2 and for all

paths uxv ∈ A(G), we have γt(Guvx ) = 4 = γt(G) + 2.

The following result characterizes the paths in A−(G).

Theorem 77 Let G be any isolate-free graph and let uxv ∈ A(G). Then uxv ∈

A−(G) if and only if there exists a γt(G)-set S such that the following hold.

(i) x ∈ S.

(ii) |S ∩ {u, v}| ≥ 1.

(iii) |N(x) ∩ (S \ {u, v})| ≥ 1.

(iv) pn(x, S) ⊆ {u, v}.


(v) If u ∈ S and v /∈ S, then pn(x, S) = {v}.

(vi) If u /∈ S and v ∈ S, then pn(x, S) = {u}.

Proof. Let uxv ∈ A−(G). For notational convenience, we let H = Guvx . Then

γt(H) + 1 = γt(G). Let S∗ be an arbitrary γt(H)-set. If x ∈ S∗ or S∗ ∩ {u, v} = ∅,

then S∗ is a TDS for G of cardinality γt(G) − 1. If v has a neighbor in S∗ different

from u and u has a neighbor in S∗ different from v, then once again S∗ is a TDS for

G of cardinality γt(G) − 1. In both cases, we have a contradiction. Hence, x /∈ S∗,

|S∗ ∩ {u, v}| ≥ 1, and renaming vertices, if necessary, we may assume that u is the

only vertex in S∗ adjacent to v; that is, NH(v) ∩ S∗ = {u}. Since S∗ is a TDS of H

and x is adjacent to neither u nor v in H, we have that N(x) ∩ (S∗ \ {u, v}) 6= ∅.

We now consider the set S = S∗ ∪ {x}. The set S is a TDS of G, and so

γt(G) ≤ |S| = |S∗| + 1 = γt(H) + 1 = γt(G). Consequently, γt(G) = |S|. Therefore

the set S is a γt(G)-set and, by properties of the set S∗ established earlier, the set

S satisfies properties (i), (ii) and (iii) in the statement of the theorem. Since S∗ is a

TDS for H and x /∈ S∗, the only possible vertices totally dominated by x in G but

by no other vertex in S are the vertices u and v. Hence in the graph G, we have that

pn(x, S) ⊆ {u, v}, and so property (iv) holds. By assumption, NH(v) ∩ S∗ = {u},

implying that u ∈ S and v ∈ pnG(x, S). Thus, property (vi) holds vacuously. If

v ∈ S, then property (v) holds vacuously. If v /∈ S, then the vertex u is totally

dominated by a vertex in H, and therefore in G, different from both v and x. Hence,

u /∈ pnG(x, S). Thus since pnG(x, S) ⊆ {u, v}, we have that pnG(x, S) = {v}, and so

once again property (v) holds. This establishes the necessity.

To prove the sufficiency, suppose that there exists a γt(G)-set S such that proper-


ties (i)–(vi) in the statement of the theorem hold in the graph G. By properties (i) and

(ii), x ∈ S and at least one of u and v belong to S. Renaming vertices, if necessary,

we may assume that u ∈ S. We now consider the set Sx = S \{x}. By properties (iii)

and (iv), the only possible vertices not totally dominated by Sx in the graph H are

the vertices u and v. If v ∈ S, then u is totally dominated by v in H. If v /∈ S, then,

by property (v), the vertex u is totally dominated by a vertex in G, and therefore

in H, different from both v and x. In both cases, u and v are totally dominated by

Sx in H. Hence, Sx is a TDS of H, and so γt(H) ≤ |Sx| = |S| − 1 = γt(G) − 1.

Consequently by Theorem 76, γt(H) = γt(G) − 1. Therefore, uxv ∈ A−(G). 2

7.3 Classes of Graphs

Our purpose in this chapter is to explore the effect that edge lifting has on the total

domination number of a graph. Recall that L is the family of connected graphs G

with A(G) 6= ∅ such that every vertex of degree 2 in G, if any, is contained in a

triangle (and is therefore not the center of an induced P3). We classify families of

graphs in L as follows.

Definition 7.3.1 For every graph G ∈ L with γt(G) = k, we define the following.

(a) A graph G is k−L -critical if A(G) = A−(G).

(b) A graph G is kL-stable if A(G) = A0(G).

(c) A graph G is kL-changing if A+(G) 6= ∅, A−(G) 6= ∅, and A0(G) = ∅.

(d) A graph G is kL-mixed if A+(G) 6= ∅, A−(G) 6= ∅, and A0(G) 6= ∅.

(e) A graph G is k+L -critical if A(G) = A+(G).


We show that there exists a graph in each of the five classes for any given total

domination number at least four, and hence the five classes of graphs in L are not

empty. We also establish properties of graphs in certain classes in L. Since every

vertex of degree 2 in a graph that belongs to the family L is contained in a triangle,

we have the following observation.

Observation 78 No bipartite graph in L has a vertex of degree 2.

For notational convenience, let F1 be the k−L -critical graphs, F2 the kL-stable

graphs, F3 the kL-changing graphs, F4 the kL-mixed graphs, and F5 the k+L -critical

graphs in L.

7.3.1 The Family F1

In this section, we consider the family F1 of k−L -critical graphs.

Proposition 79 For every integer k ≥ 3, there exists a graph G ∈ F1 with γt(G) = k.

Proof. As observed earlier, for every integer k ≥ 3 if G is the corona Kk ◦ K1, then

γt(G) = k and G is a k−L -critical graph, and so G ∈ F1. 2

Proposition 80 If G is a bipartite graph in F1, then δ(G) ≥ 3.

Proof. Suppose that G ∈ F1 is a bipartite graph. By Observation 78, the bipartite

graph G has no vertex of degree 2. Hence it suffices for us to show that G has no


leaf. Assume, to the contrary, that G has a leaf w. Let x be the neighbor of w. Since

A(G) 6= ∅, we note that dG(x) ≥ 2. Let u be a neighbor of x distinct from w. By

Theorem 77(iii), the vertex x has degree at least 3 in G. Let v ∈ N(x)\{u, w}. Since

G is a bipartite graph, the set N(x) is independent in G. We now consider the path

uxv ∈ A(G). Let S be a γt(Guvx )-set. Since w is a leaf neighbor of x in Guv

x , the vertex

x is a support vertex in Guvx , implying that x ∈ S. But then S is also a TDS in G,

and so γt(G) ≤ |S| = γt(Guvx ). However since G ∈ F1, we have that γt(G

uvx ) < γt(G),

a contradiction. Therefore, δ(G) ≥ 3. 2

As a special case of Proposition 80, we have that the family F1 contains no tree.

7.3.2 The Family F2

In this section, we consider the family F2 of kL-stable graphs.


Proof. For k ≥ 4, consider the corona G = Kk ◦ K2. Then, γt(G) = k and V (Kk)

is the unique γt(G)-set. Let uxv ∈ A(G). Renaming u and v if necessary, we may

assume that ux ∈ E(Kk) and v is a neighbor of x of degree 2 in G. Thus, γt(Guvx ) =

γt(G) = k and there are two γt(Guvx )-sets, namely V (Kk) and (V (Kk) \ {x}) ∪ {v}.

Therefore, G ∈ F2 and γt(G) = k. 2

We show next that there does not exist any kL-stable trees in the family L. We

begin with the following lemma.

Lemma 82 If G ∈ F2, then G has no strong support vertex.


Proof. Let G ∈ F2 and assume, to the contrary, that G has a strong support vertex x.

Let u and v be two leaf neighbors of x, and consider the lift on uxv. Let S be any

γt(Guvx )-set. Since the edge uv forms a K2-component in Guv

x , we have that {u, v} ⊂ S.

The set (S\{x, y})∪{v} is a TDS of G, and so γt(G) ≤ |S|−1 = γt(Guvx )−1. However,

G ∈ F2, implying that γt(G) = γt(Guvx ), a contradiction. Therefore, G has no strong

support vertex. 2

Proposition 83 There is no tree in the family F2.

Proof. Assume, for purposes of contradiction, that there is a tree T in the family F2.

Let v1v2 . . . vr be a longest path in T . By Lemma 82, the tree T has no strong support

vertex. In particular, v2 is not a strong support vertex, implying that dT (v2) = 2,

contradicting Observation 78. 2

7.3.3 The Families F3 and F4

In this section, we show the family F3 of kL-changing graphs and the family F4 of

kL-mixed graphs are not empty.


Proof. For k ≥ 4, consider the corona G = Ck ◦ K1. Since V (Ck) is the set of

support vertices in G, every TDS in G contains the set V (Ck), and so γt(G) ≥ k.

However V (Ck) is a TDS for G, implying that γt(G) ≤ k. Consequently, γt(G) = k.

Let uxv ∈ A(G). There are two possibilities. Either uxv is a path in the cycle Ck


or, renaming u and v if necessary, ux ∈ E(Ck) and v is the leaf neighbor of x in G.

In the first case, γt(Guvx ) = γt(G) + 1, and in the second case, γt(G

uvx ) = γt(G) − 1.

Thus, A+(G) 6= ∅, A−(G) 6= ∅, and A0(G) = ∅. Therefore, G ∈ F3 and γt(G) = k. 2


Proof. For k ≥ 4, let G be obtained from the corona Kk ◦ K3 by deleting two leaf

neighbors from one strong support vertex of G. Let V (Kk) = {v1, v2, . . . , vk}, where

v1 is the vertex in G with only one leaf neighbor, u1 say. Let u2 and w2 be two leaf

neighbors of v2 in G. Then, γt(G) = k and V (Kk) is the unique γt(G)-set. Since

γt(Gu1v2v1

) = γt(G) − 1, γt(Gu2v1v2

) = γt(G), and γt(Gu2w2v2

) = γt(G) + 2, we note that

A−(G) 6= ∅, A0(G) 6= ∅ and A+(G) 6= ∅. Therefore, G ∈ F4 and γt(G) = k. 2

7.3.4 The Family F5

In this section, we consider the family F5 of k+L -critical graphs in L.


Proof. We first consider the case when k is even. Let k = 2`, where ` ≥ 2, and let

G be the graph constructed as follows. For 1 ≤ i ≤ `, let Hi be the graph shown

in Figure 7.2 with the vertices of degree at least 2 in Hi labeled as indicated as in

Figure 7.2. Let G be obtained from the disjoint union ∪`i=1Hi of H1, H2, . . . , H` by

adding the edges {eiai+1(mod `) | 1 ≤ i ≤ `}. The graph G for the case when k = 6 is


illustrated in Figure 7.3. Let

A =⋃

i=1

{ai} and B =⋃

i=1

{bi}.

Let S be any γt(G)-set. Since A∪B is the set of support vertices in G, and since

every TDS in a graph contains all its support vertices, we have that A ∪ B ⊆ S.

However A ∪ B is a TDS for G, implying that S = A ∪ B and that S is the unique

γt(G)-set. Thus, γt(G) = |A ∪ B| = 2` = k. Since G has no isolated vertices and no

vertices of degree 2, we have that G ∈ L. Let uxv ∈ A(G). Let S∗ be an arbitrary

γt(Guvx )-set. Regardless of the choice of uxv, the vertices of S are support vertices in

Guvx . Therefore, S ⊆ S∗ and γt(G

uvx ) ≥ γt(G). Assume, for purposes of contradiction,

that γt(Guvx ) = γt(G). Then, S = S∗ and the set S is the unique γt(G

uvx )-set. However,

it is a simple exercise to check that for every possible choice of uxv, the set S is not

a TDS for Guvx , a contradiction. Thus, γt(G

uvx ) > γt(G). This is true for every path

uxv ∈ A(G). Therefore, G ∈ F5 and γt(G) = k.

We next consider the case when k is odd. Let k = 2` + 1, where ` ≥ 2, and let F

be the graph constructed as follows. For 1 ≤ i ≤ ` − 1, let Hi be the graph shown

in Figure 7.2, as before. Let H∗ be a caterpillar with a spine P3 = (x, y, z) and code

(3, 3, 3). Let F be obtained from the disjoint union H∗∪(∪`−1i=1Hi) of H∗, H1, . . . , H`−1

by adding the edges {eiai+1(mod `) | 1 ≤ i ≤ ` − 2} ∪ {e`−1x} if ` ≥ 3 or adding the

edge e1x if ` = 2. The graph F for the case when k = 5 is illustrated in Figure 7.4.

Let X = {x, y, z} and let

Y =`−1⋃

i=1

{ai} and Z =`−1⋃

i=1

{bi}.


ai bi

di ci

ei

Figure 7.2: The graph Hi.

Let D be any γt(F )-set. Since X∪Y ∪Z is the set of support vertices in F , we have

that X∪Y ∪Z ⊆ D. However X∪Y ∪Z is a TDS for F , implying that D = X∪Y ∪Z

and that D is the unique γt(F )-set. Thus, γt(F ) = |X ∪ Y ∪ Z| = 2` + 1 = k. Since

F has no isolated vertices and no vertices of degree 2, we have that F ∈ L. Let

uxv ∈ A(F ). Let D∗ be an arbitrary γt(Duvx )-set. Regardless of the choice of uxv,

the vertices of D are support vertices in F uvx . Therefore, D ⊆ D∗ and γt(F

uvx ) ≥ γt(F ).

Assume, for purposes of contradiction, that γt(Fuvx ) = γt(F ). Then, D = D∗ and the

set D is the unique γt(Fuvx )-set. However, it is a simple exercise to check that for

every possible choice of uxv, the set D is not a TDS for F uvx , a contradiction. Thus,

γt(Fuvx ) > γt(F ). This is true for every path uxv ∈ A(F ). Therefore, F ∈ F5 and

γt(F ) = k. 2

The following result establishes properties of graphs in the family F5. Recall that

a claw in a graph is an induced copy of K1,3.

Lemma 87 Let G ∈ F5 and let V = V (G). Let S be an arbitrary γt(G)-set. Then

the following properties hold.

(a) If u and v are adjacent vertices in G[S] of degree at least 2, then neither u


a1 b1 a2 b2 a3 b3

d1 c1d2 c2 d3 c3

e1 e2e3

Figure 7.3: The graph G when k = 6.

a1 b1

x y z

d1 c1

e1

Figure 7.4: The graph F when k = 5.


nor v has an S-external private neighbor.

(b) Every component of G[S] is in {P2, P3, P4}.

(c) The components of G[V \ S] are complete graphs.

(d) Every vertex in V \ S has at most two neighbors in S.

(e) If a vertex v ∈ V \ S has two neighbors in S, then every neighbor of v not in

S is also adjacent to these two vertices of S.

Proof. (a) Suppose that u and v are adjacent vertices in G[S] of degree at least 2

but u or v has an S-external private neighbor. We may assume that epn(u, S) 6= ∅

and that w ∈ epn(u, S). Then, wuv ∈ A(G). Since S is a TDS of Gvwu , we have

that γt(Gvwu ) ≤ |S| = γt(G), contradicting the fact that G is k+

L -critical. Hence,

epn(u, S) = epn(v, S) = ∅.

(b) Let C be an arbitrary component of G[S]. Assume that C contains a cycle

C∗. By Part (a), no vertex of V (C∗) has an S-external private neighbor. Hence by

the minimality of S, every vertex in C∗ has an S-internal private neighbor. Let u and

v be adjacent vertices on the cycle C∗ and let w ∈ ipn(u, S). Then, wuv ∈ A(G).

Since S is a TDS of Gvwu , we have that γt(G

vwu ) ≤ γt(G), a contradiction. Hence the

component C is a tree. Assume that C contains a vertex x with dC(x) ≥ 3. Let u

and v be two neighbors of x in C . Then uxv ∈ A(G). Since S is a TDS of Guvx , we

have that γt(Guvx ) ≤ γt(G), a contradiction. Therefore, ∆(C) = 2, implying that C is

a path. If C is a path on five or more vertices, then by the minimality of S, the third

vertex on the path must have an S-external private neighbor, contradicting Part (a).

Hence, C ∈ {P2, P3, P4}.


(c) Let B be a component of G[V \ S]. Assume that uxv ∈ A(G) and {u, x, v} ⊆

V (B). Since S is a TDS of Guvx , we have that γt(G

uvx ) ≤ γt(G), a contradiction. Hence

the component B contains no induced path P3, implying that B is complete.

(d) Let v ∈ V \ S and assume that |N(v) ∩ S| ≥ 3. By Part (b), at least two

neighbors of v, say x and y, in S are not adjacent. Since S is a TDS of Gxyv , we have

that γt(Gxyv ) ≤ γt(G), a contradiction. Hence |N(v) ∩ S| ≤ 2.

(e) Let v ∈ V \ S and assume |N(v) ∩ S| = 2. Let u ∈ V \ S and w ∈ S be

neighbors of v. Assume that u and w are not adjacent. Since S is a TDS of Guwv , we

have that γt(Guwv ) ≤ γt(G), a contradiction. Hence uw ∈ E(G). 2

As an immediate consequence of Lemma 87, we have the following result.

Corollary 88 If G ∈ F5 and S is a γt(G)-set, then no claw in G has a center in

V (G) \ S.

Corollary 89 Let G ∈ F5 be a triangle-free graph and let S be an arbitrary γt(G)-

set. Then the following properties hold in the graph G.

(a) Every component of G[S] is a P2-component or a P3-component.

(b) V (G) \ S is an independent set.

Proof. (a) Let C be a component of G[S]. By Lemma 87(b), the component C

belongs to {P2, P3, P4}. Assume that C = P4. Let xyzw be the path in C . Recall

that by definition, for any graph G ∈ F5, every degree two vertex is contained in a

triangle. Since G is triangle-free, it follows that G has no vertices of degree 2. In

particular, dG(y) ≥ 3. Let y′ ∈ V \ S be a neighbor of y. By the triangle-freeness


of G, the vertex y′ is not adjacent to z. Thus, y′yz ∈ A(G). Since S is a TDS of

Gy′zy , we have that γt(G

y′zy ) ≤ |S| = γt(G), a contradiction. Hence either C = P2 or

C = P3.

(b) Assume to the contrary, that there is an edge xy with {x, y} ⊆ V \ S. Since

G is triangle-free, Lemma 87(c) implies the vertices x and y form a K2-component in

G[V \ S]. Since x and y are dominated by S, both x and y have degree at least 2 in

G. Hence by the triangle-freeness of G, dG(x) ≥ 3 and dG(y) ≥ 3. By Lemma 87(d)

and 87(e), x and y are both adjacent to the same two vertices in S, implying that G

contains a triangle, a contradiction. Therefore, V \ S is an independent set. 2

The following result characterizes triangle-free graphs in the family F5.

Theorem 90 Let G be a triangle-free graph in L. Then G ∈ F5 if and only if one

of the following hold.

(a) G is a star of order at least four.

(b) G is a caterpillar with a spine P2 and code (i, j), where i, j ≥ 2.

(c) G is a caterpillar with a spine P3 and code (i, j, h), where i, h ≥ 2 and j ≥ 1.

Proof. The sufficiency is straightforward to check. To prove the necessity, suppose

that G ∈ F5 and that G is not a star. By the triangle-freeness of G, the graph G has no

vertices of degree 2. Let S be a γt(G)-set. Assume that G[S] is disconnected. Let C1

and C2 be two components of G[S]. Since G is connected, at least one vertex, say u, of

C1 is adjacent to a vertex, say v, in V \S with d(v) ≥ 3. By Corollary 89, N(v) ⊆ S,

and so v is adjacent to at least three vertices of S, contradicting Lemma 87(d). Hence,

G[S] is connected. By Corollary 89, either G[S] = P2 or G[S] = P3. Further, V \ S


is an independent set. Thus by the triangle-freeness of G and since G has no vertex

of degree 2, every vertex in V \ S is a leaf in G. If G[S] = P2, then since G is not a

star, the graph G is a caterpillar with a spine P2 and code (i, j), where i, j ≥ 2. If

G[S] = P3, then let G[S] be the path xyz. By the minimality of S, |epn(x, S)| ≥ 1

and |epn(z, S)| ≥ 1. Since G has no vertices of degree 2, the vertices in S all have

degree at least 3. Hence, G is a caterpillar with a spine P3 and code (i, j, h), where

i, h ≥ 2 and j ≥ 1. 2

As a special case of Theorem 90, we have that if G is a bipartite graph in L, then

G ∈ F5 if and only if G satisfies one of the conditions (a), (b) or (c) in the statement

of Theorem 90.

We conclude this chapter by showing that there are no claw-free, regular graphs

in the family F5.

Theorem 91 There exists no claw-free, regular graphs in F5.

Proof. Assume, for purposes of contradiction, that G = (V, E) is a claw-free, r-

regular graph and G ∈ F5. Note that if r ≤ 2, then A(G) = ∅. Hence, r ≥ 3. Let S

be an arbitrary γt(G)-set.

Claim 1 We may assume the graph G satisfies the following properties.

(a) Each component of G[V \ S] is a Kr−1 or a Kr.

(b) Each component of G[S] is a K2.

(c) If a vertex v ∈ S has a neighbor in a component B of G[V \S], such that B = Kr−1,

then N(v) \ S = V (B).


Proof. (a) Let B be a component of G[V \ S]. By Lemma 87(c), B is a complete

graph. By Lemma 87(d), every vertex in V \S has at most two neighbors in S. Since

S is a TDS in G, every vertex in V \ S has at least one neighbor in S. Hence by the

regularity of G, it follows that B = Kr−1 or B = Kr .

(b) Let A be a component of G[S]. By Lemma 87(b), A ∈ {P2, P3, P4}. Suppose

that A ∈ {P3, P4}, and let b be a support vertex of A. Since r ≥ 3, b has a neighbor x

in V \S. Let Bx be the component of G[V \S] containing x. By Part(a), Bx = Kr−1

or Bx = Kr. Since G is claw-free, x is adjacent to a neighbor of b in A. Hence x has

two neighbors in S, implying that Bx = Kr−1. By Lemma 87(e), every vertex of Bx

is adjacent to b, and so dG(b) ≥ r + 1, a contradiction.

(c) Let v ∈ S have a neighbor in a component B of G[V \S] such that B = Kr−1.

Since G is r-regular, every vertex of B is adjacent to exactly two vertices of S and v

is adjacent to r − 1 vertices of V \ S. By Lemma 87(e), every vertex of B is adjacent

to the same two vertices of S. Hence N(v) \ S = V (B). (2)

We now return to the proof of Theorem 91. We first show that no component

of G[V \ S] is a Kr. For the purpose of a contradiction, assume that B = Kr is

a component of G[V \ S]. Since G is r-regular, each vertex in V (B) is adjacent to

exactly one vertex of S and Claim 1(b) implies that each vertex of S is adjacent to

exactly r − 1 vertices of V \ S. Since there are r edges between S and V (B), at least

two vertices in S have neighbors in B and some vertex of S, say u, has fewer than

r − 1 neighbors in B. Let v be a neighbor of u in B. Since u has r − 1 neighbors in

V \ S, there is a neighbor of u, say x, in another component Bx 6= B of G[V \ S].

We deduce from Claim 1(c) that Bx = Kr. Since G is r-regular, each of v and x has


exactly one neighbor in S. But then {u, v, x, y}, where y is the neighbor of u in S,

induces a claw in G, a contradiction. Therefore every component of G[V \ S] is a

Kr−1.

Let B be an arbitrary component of G[V \S], and so B = Kr−1. By Lemma 87(e),

the vertices of B are adjacent to the same two vertices of S. If these two vertices are

in the same component of G[S], then G is either a complete graph and A(G) = ∅

or G is disconnected, contradicting the assumption that G ∈ F5. Hence Claim 1(c)

implies that there exists nonadjacent vertices u and v in S, such that N(u) \ S =

N(v) \ S = V (B). Let {x, y} ⊆ V (B) and N(u) ∩ S = {w}. By Claim 1(c),

N(w) \ {u} = V (Bw) for some component Bw of G[V \ S] different from B. Note

by Lemma 87(e), V (Bw) is dominated by S \ {w, u, v}. We now consider the edge

lift on uxv. Then, (S \ {u, w}) ∪ {y, z}, where z ∈ V (Bw), is a TDS for Guvx . Hence

γt(Guvx ) ≤ γt(G), contradicting the assumption that G ∈ F5. 2

7.4 Open Problems

We finish this chapter with some possible open problems to consider.

1. Is it true that Kk ◦ K1 are the only k−L -critical graphs with degree 1 vertices?

2. Does there exist k−L -critical graphs with all possible values of minimum degree?

Chapter 8

Total Domination Numbers of

Graphs with Diameter Two

8.1 Introduction

In this chapter, we determine upper bounds on the total domination number of

diameter-2 graphs and the exact value of the total domination number of some specific

families of diameter-2 graphs.

8.1.1 Known Results and Observations

Known upper bounds on the total domination number of a graph G in terms of its

order n are summarized in Table 1.

127

CHAPTER 8. TOTAL DOMINATION AND DIAMETER TWO 128

δ(G) ≥ 1 ⇒ γt(G) ≤ 2n/3 if n ≥ 3 and G is connected ([10])

δ(G) ≥ 2 ⇒ γt(G) ≤ 4n/7 if n ≥ 11 and G is connected ([38])

δ(G) ≥ 3 ⇒ γt(G) ≤ n/2 ([2, 9, 49])

δ(G) ≥ 4 ⇒ γt(G) ≤ 3n/7 ([48])

Table 1. Upper bounds on the total domination number of a graph G.

Henning and Yeo [40] characterized the connected graphs G of order n with δ(G) ≥

3 satisfying γt(G) = n/2. However none of the extremal graphs have diameter two,

so we have the following result.

Theorem 92 ([40]) If G is a diameter-2 graph of order n with δ(G) ≥ 3, then

γt(G) < n/2.

Yeo [59] showed that the incidence bipartite graph of the complement of the

Fano plane is the unique connected graph G of order n with δ(G) ≥ 4 satisfying

γt(G) = 3n/7. Since this graph has diameter 3, we have the following result.

Theorem 93 ([59]) If G is a diameter-2 graph of order n with δ(G) ≥ 4, then

γt(G) < 3n/7.

If G is a diameter-2 graph and v is an arbitrary vertex in G, then N(v) is a DS

of G and N [v] is a TDS of G. In particular, choosing v to be a vertex of minimum

degree in G, we have the following useful observation.

Observation 94 If G is a graph with diam(G) = 2, then γt(G) ≤ δ(G) + 1.


Let G = (V, E) be a graph with no isolated vertex, and let D be a γt(G)-set.

Then V = ∪v∈DN(v), implying that |V | ≤ ∑

v∈D dG(v) ≤ ∆(G) · |D|; or equivalently,

γt(G) = |D| ≥ |V |/∆(G). Hence we have the following observation.

Observation 95 If G is an isolate-free graph of order n, then γt(G) ≥ n/∆(G).

8.2 Upper Bounds on the Total Domination Num-

ber

Our aim in this section is to determine upper bounds on the total domination number

of diameter-2 graphs. In Section 8.2.1 we establish upper bounds in terms of the

order of the graph, while in Section 8.2.2 we give upper bounds in terms of the

minimum degree of the graph. We conclude this section by giving bounds on the

total domination number for diameter-2 graphs whose complements are planar.

8.2.1 Upper Bounds in Terms of Order

Here we establish upper bounds on the total domination number of a diameter-2

graph in terms of its order. For this purpose, we define a family of graphs F .

Definition 8.2.1 Let F = {C4, K4−e, K1,3, F1, F2, F3}, where the graph F1 is formed

by attaching a leaf to a vertex of a K3, F2 is formed from a 5-cycle by adding a new

vertex and joining it to two nonadjacent vertices of the C5, and F3 is formed from a

5-cycle by adding a new vertex and joining it to three consecutive vertices on the C5.

The graphs F1, F2, and F3 are illustrated in Figure 8.1.


F1 F2 F3

Figure 8.1: The graphs F1, F2 and F3.

Theorem 96 If G /∈ {P3, C5} is a diameter-2 graph of order n, then γt(G) ≤ n/2

with equality if and only if G ∈ F .

Proof. Since G has diameter 2, we note that n ≥ 3. Further since G 6= P3, we

note that n ≥ 4. If n = 4, then γt(G) = 2 ≤ n/2. A simple exhaustive check

of all diameter-2 graphs on four vertices yields the four graphs of order 4 in the

family F . Hence we may assume that n ≥ 5. If δ(G) = 1, then by Observation 94,

γt(G) ≤ 2 < n/2. If δ(G) ≥ 3, then by Theorem 92, γt(G) < n/2. Hence we may

assume that δ(G) = 2, for otherwise γt(G) < n/2. Therefore by Observation 94,

γt(G) ≤ 3. If n ≥ 7, then γt(G) < n/2. Hence we may assume that n ∈ {5, 6}. If G is

a cycle, then the diameter-2 condition implies that G = C5, a contradiction. Hence,

∆(G) ≥ 3. Let v be a vertex of maximum degree in G. If n = 5 or if ∆(G) ≥ 4, then

the vertex v and one of its neighbors is a TDS of G, and so γt(G) = 2 < n/2. Hence

we may assume that n = 6 and ∆(G) = 3 (and still δ(G) = 2).

Let N(v) = {v1, v2, v3} and V \ N [v] = {u1, u2}. Since diam(G) = 2, both u1

and u2 have a neighbor in N(v). If u1 and u2 have a common neighbor vi in N(v),

then {v, vi} is a TDS of G, and so γt(G) < n/2. Hence we may assume that u1 and

u2 have no common neighbor in N(v). Since δ(G) = 2, this implies that u1u2 ∈ E.

Renaming vertices, if necessary, we may assume that for i ∈ {1, 2}, we have uivi ∈ E


and uiv3−i /∈ E. Further since at most one of u1 and u2 is adjacent to v3, we may

assume that u1v3 /∈ E. On the one hand if u2v3 /∈ E, then the diameter-2 constraint

implies that v1v3 ∈ E and v2v3 ∈ E. In this case since ∆(G) = 3, no other edges are

present and hence, G = F3. On the other hand suppose that u2v3 ∈ E. If v1vi ∈ E

for some i ∈ {2, 3}, then {vi, u2} is a TDS for G, and so γt(G) < n/2. Hence we may

assume that dG(v1) = 2. The only possible remaining edge in G is the edge v2v3. If

v2v3 /∈ E, then G = F2, while if v2v3 ∈ E, then G = F3. We have therefore shown

that γt(G) ≤ n/2 with equality only if G ∈ F . If G ∈ F has order n, then it is a

simple exercise to verify that diam(G) = 2 and γt(G) = n/2. 2

The result of Theorem 96 can be improved when the order of the diameter 2 graph

is large. For this purpose, let G9 be the diameter-2 graph of order n = 9 shown in

Figure 8.2.

Figure 8.2: The graph G9 with γt(G9) = 4.

Theorem 97 If G 6= G9 is a diameter-2 graph of order n ≥ 7, then γt(G) ≤ 3n/7

with strict inequality if n ≥ 8.

Proof. Let G be a diameter-2 graph of order n ≥ 7. As an immediate consequence of

Theorem 96, we have that γt(G) < n/2. In particular if n = 7, then γt(G) ≤ 3 = 3n/7


while if n = 8, then γt(G) ≤ 3 < 3n/7. Further if n = 9, then γt(G) ≤ 4. We have

proven, using an exhaustive computer search,1 that the graph G = G9, shown in

Figure 8.2, is the only diameter-2 graph of order 9 having γt(G) = 4. Since G 6= G9

by assumption, we deduce that γt(G) ≤ 3 < 3n/7. Hence we may assume that n ≥ 10,

for otherwise the desired result follows. For purposes of contradiction, assume that

γt(G) ≥ 3n/7. By Observation 94, δ(G) ≥ γt(G) − 1. Hence since n ≥ 10 and

γt(G) ≥ 3n/7, we have that δ(G) ≥ 3n/7 − 1 ≥ 30/7 − 1 > 3. Since δ(G) is an

integer, δ(G) ≥ 4. But then by Theorem 93, γt(G) < 3n/7, a contradiction. Thus,

γt(G) < 3n/7, as desired. 2

As a consequence of Theorems 96 and 97, we have the following result.

Corollary 98 Let G be a diameter-2 graph of order n ≤ 11. Then γt(G) ≤ 1 +√

n

with equality if and only if G = G9.

Proof. If G = P3, then γt(G) = 2 < 1 +√

n, while if G = C5, then γt(G) = 3 <

1 +√

n. If n ≤ 7 and G /∈ {P3, C5}, then by Theorem 96, γt(G) ≤ bn/2c < 1 +√

n.

Hence we may assume that n ≥ 8, for otherwise the desired result follows. If G = G9,

then n = 9 and γt(G) = 4 = 1 +√

n. If G 6= G9 and n ≤ 11, then by Theorem 97,

γt(G) ≤ b3n/7c < 1 +√

n. 2

1We remark that we also have a mathematical proof of this fact, but it is relatively lengthy and

technical and we omit it here.


8.2.2 Upper Bounds in Terms of Minimum Degree

Throughout this subsection for notational simplicity, for a graph G and its comple-

ment G, we denote δ(G), δ(G), γt(G) and γt(G) by δ, δ, γt and γt, respectively.

Further, we let δ∗ = δ∗(G) = min{δ, δ}. We establish upper bounds on the total

domination number of a diameter-2 graph in terms of its minimum degree. If G is a

graph with diam(G) ≥ 3, then the two peripheral vertices on a diametrical path in G

form a TDS in the complement G, and so γt = 2. Conversely, if γt = 2 and if {u, v}

is a γt(G)-set, then dG(u, v) ≥ 3. Hence we have the following result first observed

in [21].

Observation 99 ([21]) Let G and G be isolate-free graphs. Then γt ≥ 3 and γt ≥ 3

if and only if diam(G) = diam(G) = 2.

Theorem 100 If G and G are diameter-2 graphs, then

γt ≤ 1 +

⌈

δ

γt − 2

⌉

and γt ≤ 1 +

⌈

δ

γt − 2

⌉

.

Proof. Since both G and G have diameter two, by Observation 99 we have γt ≥ 3 and

γt ≥ 3. Let v be a vertex of G having degree δ. Let A = NG(v) and B = V (G)\N [v].

We note that |A| = δ and that if any set A′ ⊂ A with cardinality at most γt − 2

dominates B in G, then A′ ∪ {v} is a TDS of G with cardinality less than γt, a

contradiction. Hence for every subset A′ ⊂ A with |A′| ≤ γt − 2, there exists a vertex

b′ ∈ B that is not adjacent to any vertex of A′ in G, implying that b′ dominates A′ in G.

Let k = dδ/(γt−2)e and for 1 ≤ j ≤ k, select the sets Aj so that Aj ⊂ A, |Aj| = γt−2,

and⋃k

j=1 Aj = A. For each set Aj, 1 ≤ j ≤ k, we select one vertex bj ∈ B that is not


dominated by Aj in G, and let B ′ =⋃k

j=1{bj}. Then |B ′| ≤ k and B ′ dominates A in

G. Therefore B ′ ∪{v} is a TDS of G, and so γt ≤ |B ′|+1 ≤ k +1 = 1+ dδ/(γt − 2)e.

By a symmetric argument, γt ≤ 1 + d δ/(γt − 2)e. 2

As a consequence of Theorem 100, we have the following results.

Corollary 101 If G and G are isolate-free graphs, then (γt − 2)(γt − 2) ≤ δ∗ − 1.

Proof. Since G and G are isolate-free, we have that δ∗ ≥ 1. Further, diam(G) ≥ 2

and diam(G) ≥ 2. If diam(G) ≥ 3 or diam(G) ≥ 3, then γt = 2 or γt = 2 and the

desired result follows immediately. Hence we may assume that diam(G) = diam(G) =

2. By Theorem 100, γt ≤ 1 + dδ/(γt − 2)e and γt ≤ 1 + dδ/(γt − 2)e. Thus,

γt < 2 + δ/(γt − 2) and γt < 2 + δ/(γt − 2); or, equivalently, (γt − 2)(γt − 2) < δ and

(γt − 2)(γt − 2) < δ. Hence, (γt − 2)(γt − 2) < min{δ, δ} = δ∗. 2

Paths of order at least four and the self-complementary cycle C5 are examples of

graphs attaining the bound of Corollary 101.

Corollary 102 If G is a diameter-2 graph with γt > 1 +√

n, then γt < 1 +√

n.

Proof. Let G be a diameter-2 graph with γt > 1 +√

n. By Corollary 98, n ≥ 12. If

diam(G) = 1, then G consists of isolated vertices, while if diam(G) ≥ 3, then γt(G) =

2 < 1 +√

n. Both cases produce a contradiction. Hence diam(G) = 2. In particular,

both G and G are isolate-free, and so, by Corollary 101, (γt−2)(γt−2) < δ∗. Therefore

since√

n−1 < γt−2 and δ∗ = min{δ, δ} < n/2, we have that (√

n−1)(γt−2) < n/2,

or, equivalently, γt < 2 + n2(√

n−1). For all n ≥ 12, we note that 2 + n

2(√

n−1)< 1 +

√n,

implying that γt < 1 +√

n. 2


Corollary 103 For any graph G of order n ≥ 4, γt < 1 +√

n or γt < 1 +√

n.

Proof. Let G be any graph of order n ≥ 4. It is evident that if either G or G is

disconnected, then γt = 2 < 1 +√

n or γt = 2 < 1 +√

n, and we are finished. Also,

if diam(G) ≥ 3 or diam(G) ≥ 3, then γt = 2 < 1 +√

n or γt = 2 < 1 +√

n, and we

are finished. Accordingly, assume that diam(G) = diam(G) = 2. By Corollary 102,

γt < 1 +√

n or γt < 1 +√

n. 2

Corollary 104 If G and G are isolate-free graphs with γt ≤ γt, then γt ≤ 2+ b√

δ∗c.

Proof. By Corollary 101, (γt − 2)(γt − 2) ≤ δ∗. Since γt ≤ γt, we have that

(γt − 2)2 ≤ δ∗; or, equivalently, γt ≤ 2 +√

δ∗. Since γt is an integer, it follows that

γt ≤ 2 + b√

δ∗c. 2

We will need the following result from [39] before proceeding further.

Theorem 105 ([39]) Let G and G be connected graphs. Then the following holds.

(a) If diam(G) = diam(G) = 2, then γt ≤ δ∗ + 1 and γt ≤ δ∗ + 1.

(b) If γt, γt ≥ 3, then γt + γt ≤ δ∗ + 4.

As a consequence of Corollaries 101 and 104 and Theorem 105, we have the fol-

lowing result.

Corollary 106 Let G and G be diameter-2 graphs. Then the following holds.

(a) γtγt ≤ 3δ∗ + 3.

(b) If γt ≤ γt, then γt ≤ 2 + b√

δ∗c and γt ≤ 1 + δ∗.


Proof. (a) By Observation 99, we have γt ≥ 3 and γt ≥ 3. Hence applying

Theorem 105(b), we have γt + γt ≤ δ∗ + 4. Therefore by Corollary 101, we have

δ∗ − 1 ≥ (γt − 2)(γt − 2) ≥ γtγt − 2(δ∗ + 4) + 4, or, equivalently, γtγt ≤ 3δ∗ + 3. This

proves Part (a). Part (b) is a restatement of Corollary 104 and Theorem 105(a). 2

For an example of a graph G obtaining the bounds of Corollary 106(b), consider

the complement of the Petersen graph. In this case δ = 6 and δ = 3, and so δ∗ = 3.

Also, γt = 3 = 2 + b√

δ∗c and γt = 4 = 1 + δ∗. We note that bound of Part (a) is

sharp for the 5-cycle.

8.2.3 Upper Bounds for Planar Graphs

Goddard and Henning [24] showed that the total domination number of a diameter-2

planar graph is at most 3.

Theorem 107 ([24]) If G is a diameter-2 planar graph, then γt(G) ≤ 3.

In this section, we show that the total domination number of a diameter-2 graph

whose complement is planar is at most 4.

Theorem 108 If G is a diameter-2 graph and G is planar, then γt(G) ≤ 4.

Proof. Let G = (V, E) be a diameter-2 graph, and let u ∈ V and U = V \ N [u]. If

dG(u) ≤ 3, then the result follows immediately by Observation 94, while if U = ∅,

then γt(G) = 2. Hence we may assume that δ(G) ≥ 4 and that U 6= ∅.

Suppose that G is triangle-free. Since G is planar, it follows by the Four Color

Theorem (see [1]) that V (G) can be partitioned into at most four color classes. Each


color class is an independent set in G and hence, by the triangle-freeness of G, has at

most two vertices. Thus n ≤ 8, and by Corollary 98 we have γt(G) ≤ 1 + b√nc ≤ 3.

Hence we may assume that G contains a triangle, for otherwise γt(G) ≤ 3.

Without loss of generality, we may choose the vertex u to belong to a triangle in

G. Let uvwu be a triangle in G. Let Q be the set of vertices in U not dominated by

{v, w} in G. If |Q| ≥ 3, then the subgraph induced by Q ∪ {u, v, w} in G contains

a subgraph isomorphic to K3,3, implying by Kuratowski’s Theorem (see [42]) that G

is not planar, a contradiction. Hence, |Q| ≤ 2. If there is a neighbor q of u that

dominates Q, then γt(G) ≤ |{q, u, v, w}| ≤ 4. If Q = ∅, then {u, v, w} is a TDS of

G and γt(G) ≤ |{u, v, w}| = 3. Hence we may assume that |Q| = 2 and that the

two vertices of Q have no common neighbor in N(u), for otherwise γt(G) ≤ 4. Let

Q = {q1, q2} and for i = 1, 2, let ui be a common neighbor of qi and u.

If v dominates U \Q, then {u, u1, u2, v} is a TDS of G and γt(G) ≤ 4, as desired.

Hence we may assume that there exists a vertex y ∈ U \ Q such that vy /∈ E(G)

and wy ∈ E(G). Let A = {u, v, w} and B = {q1, q2, y}. If there exists a vertex

x ∈ V (G) \ (A ∪ B) that is adjacent to neither w nor y in G, then the subgraph

induced by A ∪ B ∪ {x} in G contains a subdivided K3,3 (with partite sets A and B

and where x is the vertex of degree two in the subdivided K3,3 that is adjacent to

w ∈ A and y ∈ B in G), implying by Kuratowski’s Theorem that G is not planar, a

contradiction. Hence, {w, y} dominates V (G) \ Q in G. Consequently, {u1, u2, w, y}

is a TDS of G, and so γt(G) ≤ 4. 2


8.3 Exact Values of γt(G)

Our aim in this section is to determine the total domination number of graphs with

forbidden subgraphs. In particular, our main result shows that if G 6= K1,n−1 is

a triangle-free and quadrilateral-free diameter-2 graph of order n, then γt(G) = 1 +

√n − 1. Note that if G is a diameter-2 graph, then G is either a star or 3 ≤ g(G) ≤ 5.

Thus our main result can be reworded as: A diameter-2 graph G of girth 5 and order

n has γt(G) = 1 +√

n − 1. We begin by showing that γt(G) is small for several

families of diameter-2 graphs.

Theorem 109 Let G be a diameter-2 graph. If G is

(a) (C3, C5)-free or

(b) (C4, C5, K4 − e)-free,

then γt(G) = 2.

Proof. Let G = (V, E) be a diameter-2 graph. Let u ∈ V , and let U = V \ N [u].

Assume that U 6= ∅, dG(u) ≥ 2, and that no vertex in N(u) dominates U . We show

that this assumption produces a contradiction by considering Part (a) and Part (b)

in turn.

(a) Suppose that G is (C3, C5)-free, and let x ∈ N(u). Then there exists a vertex

w ∈ U not adjacent to x. By the diameter-2 constraint, dG(u, w) = 2 and dG(x, w) =

2. Let y be a common neighbor of u and w, and let z be a common neighbor of x and

w. Since G is triangle-free, the set N(u) is an independent set, implying that z ∈ U .

If yz ∈ E, then wyzw is a 3-cycle in G. If yz /∈ E, then uxzwyu is an induced 5-cycle

in G. Both cases produce a contradiction, proving Part (a).


(b) Suppose that G is (C4, C5, K4 − e)-free. By the diameter-2 constraint, every

vertex in U has at least one neighbor in N(u). By the constraint that G is (C4, K4−e)-

free, every vertex in U has at most one neighbor in N(u). Consequently, every vertex

in U has exactly one neighbor in N(u). By assumption, U 6= ∅ and |N(u)| ≥ 2. Let

w ∈ U and x be the common neighbor of w and u. Let y ∈ N(u) \ {x}. By the

diameter-2 constraint, w and y have a common neighbor, say y′. If xy 6∈ E, then

y′ 6= x and y′ ∈ U . But then either xy′ ∈ E, in which case xy′yux is an induced

4-cycle in G, or xy′ /∈ E, in which case uxwy′yu is an induced 5-cycle in G. Both

cases produce a contradiction. Hence, xy ∈ E, implying that x is adjacent to every

vertex in N(u) \ {x}. Since u is an arbitrary vertex in V , we may chose u to be

a vertex of maximum degree ∆(G) and keep the established properties. But then

dG(x) ≥ 2 + |N(u)| − 1 > ∆(G), a contradiction.

Since both Part (a) and Part (b) produce a contradiction, we therefore have that

U = ∅ or dG(u) = 1 or some vertex in N(u) dominates U . In all three cases,

γt(G) = 2. 2

We remark that the diameter-2 graphs of girth 5 are precisely the diameter-2

Moore graphs. It is shown (see [41, 47]) that Moore graphs are r-regular and that

diameter-2 Moore graphs have order n = r2 + 1 and exist for r = 2, 3, 7 and possibly

57, but for no other degrees. The Moore graphs for the first three values of r are

unique, namely

• The 5-cycle (2-regular graph on n = 5 vertices)

• The Petersen graph (3-regular graph on n = 10 vertices)

• The Hoffman-Singleton graph (7-regular on n = 50 vertices).


We now determine the total domination number of diameter-2 Moore graphs.

Theorem 110 If G is a diameter-2 graph of order n and girth 5, then γt(G) =

1 +√

n − 1.

Proof. Since G is a diameter-2 graph with g(G) = 5, G is a diameter-2 Moore graph

and n = r2 + 1. Hence r =√

n − 1. By Observations 94 and 95, we therefore have

that n/√

n − 1 ≤ γt(G) ≤ 1+√

n − 1, or, equivalently,√

n − 1+1/√

n − 1 ≤ γt(G) ≤

1 +√

n − 1. Since both γt(G) and√

n − 1 are integers, γt(G) = 1 +√

n − 1. 2

8.4 Closing Conjecture

The results obtained in this chapter, seemed to hint at the following conjecture.

Conjecture 8.4.1 If G is a diameter-2 graph of order n, then γt(G) ≤ 1 +√

n.

Subsequent research shows that this conjecture is untrue. In a future joint pub-

lication, it will be shown that diameter-2 graphs of order n exist which have a total

domination number greater that 1 +√

n. In general, the following will be shown

regarding diameter-2 graphs.

Theorem 111 If G is a diameter-2 graph of order n ≥ 2, then γt(G) ≤ 1+√

n ln(n).

Chapter 9

Relating the Annihilation Number

and the Total Domination Number

of a Tree

9.1 Introduction

In this final chapter, we study the relationship between the annihilation number of a

tree and its total domination number.

9.2 Background

Let d1, d2, . . . , dn be the degree sequence of a graph G arranged in non-decreasing

order, and so d1 ≤ d2 ≤ · · · ≤ dn. Pepper [46] defined the annihilation number of G,

141

CHAPTER 9. THE ANNIHILATION NUMBER 142

denoted a(G), to be the largest integer k such that the sum of the first k terms of the

degree sequence is at most half the sum of the degrees in the sequence. Equivalently,

the annihilation number is the largest integer k such that the

k∑

i=1

di ≤n∑

i=k+1

di.

We observe that if G has m edges and annihilation number k, then∑k

i=1 di ≤ m.

Motivated by a Graffiti.pc [11] conjecture involving an upper bound on the total

domination number of a graph in terms of its annihilation number, we investigate

relationships between the total domination and annihilation numbers of trees.

9.3 Main Results

We begin by posing a conjecture that can be found in a slightly different form in

Graffiti.pc [11].

Conjecture 9.3.1 If G is a connected nontrivial graph, then γt(G) ≤ a(G) + 1.

Let G be a graph of order n ≥ 2. As an immediate consequence of the definition

of the annihilation number, we observe that a(G) ≥ bn/2c. If G has minimum degree

at least 3, then it is known ([2, 9, 48, 49]) that γt(G) ≤ n/2. Hence if G is a graph

with minimum degree at least 3, then γt(G) ≤ a(G). Thus Conjecture 9.3.1 holds

for graphs with minimum degree at least 3. This suggests that the restriction to

trees may be the most interesting case. Our next aim in this chapter is to prove that

Conjecture 9.3.1 is true for trees.


Theorem 112 If T is a nontrivial tree, then γt(T ) ≤ a(T ) + 1, and this bound is

sharp.

A proof of Theorem 112 is given in Section 9.5. Our final aim is to characterize the

trees achieving equality in the bound of Theorem 112. We shall prove the following

result in Section 9.6.

Theorem 113 Let T be a nontrivial tree of order n with n1 vertices of degree 1.

Then, γt(T ) = a(T ) + 1 if and only if γt(T ) = (n + n1)/2.

In Section 9.7, we give a constructive characterization of these trees based on a

result of Chen et al. [8].

9.3.1 Notation

For 1 ≤ i ≤ n − 1, we let ni(G) and n≥i(G) denote the number of vertices in G of

degree i and of degree at least i, respectively. For a subset S ⊆ V , we let

Σ(S, G) =∑

v∈S

dG(v).

For a set S ⊆ V , the graph G− S denotes the graph obtained from G by deleting

the vertices in S and all edges incident with vertices in S. If S = {v}, we simply

denote G − S by G − v.

9.4 Known Results

Let d1, d2, . . . , dn be the degree sequence of a graph G arranged in non-decreasing

order. If S is an independent set in G, then Pepper [46] observed that the vertex


degrees in S can be canceled out by the degrees of vertices outside S; that is,

|S|∑

i=1

di ≤ Σ(S, G) ≤ Σ(V \ S, G) ≤n∑

i=|S|+1

di,

and so, a(G) ≤ |S|. Hence we have the following result, where α(G) denotes the

independence number in G.

Observation 114 ([46]) For every graph G, we have a(G) ≤ α(G).

We shall need the following upper bound on the total domination number of a

graph with minimum degree at least two.

Theorem 115 ([10]) If G is a connected graph of order n ≥ 3, then γt(G) ≤ 2n/3.

We denote the number of support vertices in a graph G by s(G). Further, we

let s∗(G) denote the number of isolated vertices in the subgraph of G induced by its

support vertices. We will need the following result from [12].

Theorem 116 ([12]) If T is a tree of order n ≥ 3, then γt(T ) ≤ (n + s∗(T ))/2.

As a consequence of Theorem 116, we have the following result from [6].

Corollary 117 ([6]) If T is a tree of order n ≥ 3, then γt(T ) ≤ (n + s(T ))/2.

9.5 Proof of Theorem 112

Recall the statement of Theorem 112.

Theorem 112. If T is a nontrivial tree, then γt(T ) ≤ a(T ) + 1, and this bound is

sharp.


Proof. We proceed by induction on the order n ≥ 2. The base case when n = 2 is

trivial. For the inductive hypothesis, let n ≥ 3 and assume that for every nontrivial

tree T ′ on order less than n we have γt(T′) ≤ a(T ′)+ 1. Let T be a tree of order n. If

diam(T ) = 2, then T is a star K1,n−1. In this case, a(T ) = n − 1 and γt(T ) = 2, and

so γt(T ) ≤ a(T ) + 1. If diam(T ) = 3, then T is a double star S(r, s). In this case,

a(T ) = r + s and γt(T ) = 2, and so γt(T ) < a(T ) + 1. Hence we may assume that

diam(T ) ≥ 4, for otherwise the desired result follows.

Throughout the remainder of this proof, we will consider trees T ′ formed from T

by removing a set of vertices. For such a tree T ′ of order n′, let d′1, d

′2, . . . , d

′n′ be the

non-decreasing degree sequence of T ′, and let S ′ be a set of vertices corresponding to

the first a(T ′) terms in the degree sequence of T ′. We denote the size of T ′ by m′,

and so m′ = m(T ′). We proceed further with a series of claims that we may assume

are satisfied by the tree T .

Claim A There is no strong support vertex in T .

Proof. Suppose that T has a strong support vertex u. Let v be a leaf neighbor

of u and let T ′ = T − v. If u 6∈ S ′, then Σ(S ′, T ) = Σ(S ′, T ′). If u ∈ S ′, then

Σ(S ′, T ) = Σ(S ′, T ′) + 1. Thus, Σ(S ′, T ) − 1 ≤ Σ(S ′, T ′) ≤ m′ = m − 1, and so

Σ(S ′, T ) ≤ m. Therefore, a(T ) ≥ |S ′| = a(T ′). Applying our inductive hypothesis to

T ′, we have that γt(T′) ≤ a(T ′) + 1. Since u is a strong support vertex of T , it is a

support vertex of T ′ and is therefore in every γt(T′)-set, implying that γt(T

′) = γt(T ).

Hence, γt(T ) ≤ a(T ′) + 1 ≤ a(T ) + 1, as desired. Hence we may assume that there is

no strong support vertex in T , for otherwise the desired result follows. (2)


Let r be an end of a longest path in T and let u be a vertex at distance diam(T )

from r. Root T at r. Necessarily, u is a leaf. Let v be the parent of u, w the parent

of v, x the parent of w and y the parent of x. By Claim A, we note that d(v) = 2.

Claim B d(w) = 2.

Proof. Suppose that d(w) ≥ 3. Let T ′ = T − {u, v} and D be a γt(T′)-set. If

w is a support vertex in T ′, then w ∈ D. If w is not a support vertex in T ′, then

every child of w is a support vertex (of degree 2, by Claim A). If a leaf-neighbor of

a child of w in T ′ belongs to D, then we can simply replace such a leaf in D with

the vertex w. Hence we may assume that w ∈ D. Thus the set D ∪ {v} is a TDS

of T , and so γt(T ) ≤ |D| + 1 = γt(T′) + 1. If w ∈ S ′, then Σ(S ′, T ) = Σ(S ′, T ′) + 1.

If w 6∈ S ′, then Σ(S ′, T ) = Σ(S ′, T ′). In both cases, Σ(S ′, T ) ≤ Σ(S ′, T ′) + 1. Let

S = S ′ ∪ {u}. Then, Σ(S, T ) = Σ(S ′, T ) + 1 ≤ Σ(S ′, T ′) + 2 ≤ m′ + 2 = m, implying

that a(T ) ≥ |S| = |S ′| + 1 = a(T ′) + 1. Applying our inductive hypothesis to T ′, we

have that γt(T′) ≤ a(T ′) + 1. Therefore, γt(T ) ≤ γt(T

′) + 1 ≤ a(T ′) + 2 ≤ a(T ) + 1,

as desired. (2)

Claim C d(x) = 2.

Proof. Suppose that d(x) ≥ 3. We now consider the tree T ′ = T − {u, v, w}, and

so m′ = m − 3. Every γt(T′)-set can be extended to a TDS of T by adding to it

both v and w, and so γt(T ) ≤ γt(T′) + 2. Suppose that x /∈ S ′. In this case, let

S = S ′ ∪ {u, v}. Then, Σ(S, T ) = Σ(S ′, T ′) + dT (u) + dT (v) ≤ m′ + 3 = m, implying




′) + 2 ≤ a(T ′) + 3 ≤ a(T ) + 1,

as desired.

Hence we may assume that x ∈ S ′, for otherwise the desired result follows. In this

case, let S = (S ′\{x})∪{u, v, w}. Since dT (w) = 2 ≤ dT ′(x), we have that Σ(S, T ) =

Σ(S ′, T ′) − dT ′(x) + dT (u) + dT (v) + dT (w) ≤ Σ(S ′, T ′) + 3 ≤ m′ + 3 = m, implying



′) + 2 ≤ a(T ′) + 3 ≤ a(T ) + 1,

as desired. (2)

We now return to the proof of Theorem 112. Let T ′ = T − {u, v, w, x}, and so

m′ = m− 4. Every γt(T′)-set can be extended to a TDS of T by adding to it both v

and w, and so γt(T ) ≤ γt(T′) + 2.

If y ∈ S ′, then Σ(S ′, T ) = Σ(S ′, T ′) + 1. If y 6∈ S ′, then Σ(S ′, T ) = Σ(S ′, T ′).

In both cases, Σ(S ′, T ) ≤ Σ(S ′, T ′) + 1. Let S = S ′ ∪ {u, v}. Then, Σ(S, T ) =

Σ(S ′, T ) + dT (u) + dT (v) ≤ Σ(S ′, T ′) + 4 ≤ m′ + 4 = m, implying that a(T ) ≥

|S| = |S ′| + 2 = a(T ′) + 2. Applying our inductive hypothesis to T ′, we have that

γt(T′) ≤ a(T ′) + 1. Therefore, γt(T ) ≤ γt(T

′) + 2 ≤ a(T ′) + 3 ≤ a(T ) + 1, as desired.

This establishes the desired upper bound.

That the bound is sharp may be seen as follows. Let F denote the family of trees

T which are formed from any nontrivial tree T ′ by attaching a path of length 2 to

each vertex of V (T ′) so that the resulting paths are vertex-disjoint. Note that the

resulting tree T will have order 3|V (T ′)|. Let T ∈ F and T ′ have order k. Hence T has

order n = 3k, size m = 3k − 1 and total domination number γt(T ) = 2k. Let T have

n1 vertices of degree 1 and n2 vertices of degree 2. Then, n1 = k and n2 ≥ k+2. Let s


denote the degree sequence of T with the degree sequence arranged in non-decreasing

order. If S is a set of vertices in T corresponding to the first 2k terms in the sequence

s, then∑

v∈S d(v) = k + 2k = 3k > m, implying that a(T ) < |S| = 2k. Conversely,

if S is a set of vertices in T corresponding to the first 2k − 1 terms in the sequence

s, then∑

v∈S d(v) = k + 2(k − 1) = 3k − 2 < m, implying that a(T ) ≥ |S| = 2k − 1.

Consequently, a(T ) = 2k − 1 = γt(T ) − 1. Hence for every tree T ∈ F , we have

γt(T ) = a(T ) + 1. This completes the proof of Theorem 112. 2

9.6 Proof of Theorem 113

In this section, we present a proof of Theorem 113. For this purpose, we shall need

some preliminary observations and lemmas. We begin with the following observation.

Observation 118 If T is a nontrivial tree that is not a star, then there exists a

γt(T )-set that contains no leaves of T .

Lemma 119 Let T be a tree of order n ≥ 2. Then the following holds.

(a) If n ≥ 3, then γt(T ) < n1(T ) + n2(T ).

(b) n1(T ) + n2(T ) > bn/2c.

(c) If T has no strong support vertices, then a(T ) < (n + n1(T ))/2.

(d) If n ≥ 3, then γt(T ) ≤ (n + 2s(T ) − n1(T ) + 1)/2.

Proof. Let T = (V, E) have size m.

(a) If T is a star, then γt(T ) = 2. Further since n ≥ 3, we have that n1(T ) +

n2(T ) ≥ 3, and the result holds. Hence we may assume that T is not a star. By


Observation 118, there exists a γt(T )-set S that contains no leaves of T . Let s2 and

s≥3 be the number of vertices of S of degree 2 and of degree at least 3, respectively, in

T . We note that n = n1(T )+n2(T )+n≥3(T ) and that s2 ≤ n2(T ) and s≥3 ≤ n≥3(T ). If

n≥3(T ) ≥ n1(T ), then∑

v∈V d(v) ≥ n1(T )+2n2(T )+3n≥3(T ) = 2n−n1(T )+n≥3(T ) ≥

2n > 2m, a contradiction. Hence, n1(T ) > n≥3(T ). Therefore, γt(T ) = |S| =

s2 + s≥3 ≤ n2(T ) + n≥3(T ) < n2(T ) + n1(T ).

(b) Assume, for purposes of contradiction, that n1(T ) + n2(T ) ≤ bn/2c. This

implies that n≥3(T ) ≥ n − bn/2c = dn/2e. Thus,∑

v∈V d(v) ≥ n1(T ) + 2n2(T ) +

3n≥3(T ) ≥ n1(T ) + n2(T ) + 3n≥3(T ) = (n − n≥3(T )) + 3n≥3(T ) = n + 2n≥3(T ) ≥

n + 2dn/2e ≥ 2n > 2m, a contradiction.

(c) Let T ′ be the tree obtained from T by removing all leaves. Let T ′ have order n′,

and so n′ = n− n1(T ). By Part (b), the tree T ′ has at least d(n− n1(T ))/2e vertices

of degree 1 or 2. By assumption, T has no strong support vertices. Hence every leaf

in T ′ has degree 2 in T , while every vertex of degree 2 in T ′ has either degree 2 or

degree 3 in T . Thus, n2(T ) + n3(T ) ≥ d(n − n1(T ))/2e = d(n + n1(T ))/2e − n1(T ).

Let S be a set of vertices in T corresponding to the first d(n + n1(T ))/2e terms in

the degree sequence of T with the degree sequence arranged in non-decreasing order.

Then, n1(T ) vertices in S have degree 1 in T , while the remaining vertices in S all have

degree at least 2 in T . Hence, Σ(S, T ) ≥ n1(T )+2(d(n+n1(T ))/2e−n1(T )) ≥ n > m,

implying that a(T ) < |S| = d(n + n1(T ))/2e. Therefore, a(T ) < (n + n1(T ))/2.

(d) Let T ′ be the tree obtained from T by deleting leaves until each support vertex

of T is adjacent to exactly one leaf. Let T ′ have order n′. Then, n′ = n−n1(T )+s(T ).

Further, T ′ has s(T ) leaves, and so n1(T′) = s(T ). By Theorem 112 and by Part (c),


we have that γt(T′) ≤ a(T ′)+1 ≤ (n′+n1(T

′)−1)/2+1 = (n+2s(T )−n1(T )+1)/2.

The desired result now follows from the observation that every γt(T′)-set is a TDS of

T , and so γt(T ) ≤ γt(T′). 2

We remark that a star on n ≥ 3 vertices is an example of a tree that achieves

equality in the upper bound in Lemma 119(d).

By Lemma 119(a), if T is a tree of order n ≥ 3, then γt(T ) < n1(T ) + n2(T ).

Further if γt(T ) = a(T ) + 1, then this implies that the first a(T ) + 2 terms in

the non-decreasing degree sequence of T are 1s and 2s. Hence as a consequence

of Lemma 119(a), we have the following observation.

Observation 120 Let T be a tree of order n ≥ 3 satisfying γt(T ) = a(T ) + 1. Let u

be an arbitrary vertex in T that is not a leaf. Then we can choose a set S of vertices

in T corresponding to the first a(T ) + 1 terms in the degree sequence of T with the

degree sequence arranged in non-decreasing order so that u /∈ S.

The following lemma gives necessary conditions for a tree T to have γt(T ) =

a(T ) + 1.

Lemma 121 Let T be a tree of order n ≥ 3 satisfying γt(T ) = a(T ) + 1. Then the

following holds.

(a) diam(T ) ≥ 4.

(b) T has no strong support vertex.

(c) γt(T ) = (n + n1(T ))/2 and n ≡ n1(T ) (mod 2).

(d) n1(T ) ≤ n/3 and n2(T ) > n/3.

(e) The set of support vertices of T is an independent set.


Proof. Let T = (V, E) have size m. By assumption, γt(T ) = a(T ) + 1. Let

n1 = n1(T ), n2 = n2(T ) and n≥3 = n≥3(T ).

(a) If diam(T ) = 2, then T is a star K1,n−1. In this case, a(T ) = n − 1 ≥ 2

and γt(T ) = 2, and so γt(T ) < a(T ) + 1, a contradiction. If diam(T ) = 3, then

T is a double star S(r, s). In this case, a(T ) = r + s ≥ 2 and γt(T ) = 2, and so

γt(T ) < a(T ) + 1, a contradiction. Hence, diam(T ) ≥ 4.

(b) Suppose that T has a strong support vertex u. Let v be a leaf neighbor of

u, and let T ′ = T − v. Let S ′ be a set of vertices in T ′ corresponding to the first

a(T ′) terms in the degree sequence of T ′ with the degree sequence arranged in non-

decreasing order. As shown in the proof of Claim A in Theorem 112, we have that

a(T ) ≥ |S ′| = a(T ′) and γt(T ) ≤ γt(T′). By Theorem 112, γt(T

′) ≤ a(T ′)+ 1. Hence,

a(T ) + 1 = γt(T ) ≤ γt(T′) ≤ a(T ′) + 1 ≤ a(T ) + 1. Therefore we must have equality

throughout this inequality chain. In particular, a(T ) = a(T ′) and γt(T′) = a(T ′) + 1.

By construction, diam(T ′) = diam(T ), and so by Part (a) we note that diam(T ′) ≥ 4,

and so certainly n′ ≥ 3. Since u is not a leaf in T ′, by Observation 120 we can

choose the set S ′ so that u /∈ S ′. Thus, Σ(S ′, T ) = Σ(S ′, T ′) ≤ m′ = m − 1. Hence,

Σ(S ′ ∪ {v}, T ) = Σ(S ′, T ) + 1 ≤ m, implying that a(T ) ≥ |S ′| + 1 = a(T ′) + 1, a

contradiction. Hence, T has no strong support vertex.

(c) As observed earlier, γt(T ) = a(T )+1 implies that the first a(T )+2 terms in the

non-decreasing degree sequence of T are 1s and 2s. If the first a(T ) + 1 terms of this

sequence are all 1s, then a(T ) = m = n− 1, and so γt(T ) = n, implying that T = P2,

a contradiction. Hence, n1 ≤ a(T ). Let S be a set of vertices in T corresponding to

the first a(T ) terms in the degree sequence of T with the degree sequence arranged


in non-decreasing order and let z be a vertex corresponding to the (a(T ) + 1)st term

in the degree sequence. Then, Σ(S, T ) = n1 + 2(a(T ) − n1) ≤ m = n − 1, while

Σ(S ∪ {z}, T ) = n1 + 2(a(T )− n1) + 2 ≥ m + 1 = n, implying that (n + n1 − 2)/2 ≤

a(T ) ≤ (n + n1 − 1)/2. If a(T ) = (n + n1 − 1)/2, then γt(T ) = (n + n1 + 1)/2.

However since T has no strong support vertex, Corollary 117 implies that γt(T ) ≤

(n + s(T ))/2 = (n + n1)/2, a contradiction. Hence, a(T ) = (n + n1 − 2)/2, and so

γt(T ) = (n + n1)/2. Since a(T ) and γt(T ) are integers, this in turn implies that n

and n1 have the same parity. Thus, n ≡ n1 (mod 2).

(d) If n1 > n/3, then by Part (c) we have γt(T ) = (n + n1)/2 > (n + n/3)/2 =

2n/3, contradicting Theorem 115. Hence, n1 ≤ n/3. As shown in the proof of

Lemma 119(a), we have n1 > n≥3. Thus, n = n1 + n2 + n≥3 < 2n1 + n2 ≤ 2n/3 + n2,

and so n2 > n/3.

(e) By Part (c), γt(T ) = (n + n1)/2. Since T has no strong support vertex,

s(T ) = n1. If s∗(T ) < s(T ), then by Theorem 116, γt(G) ≤ (n + s∗(T ))/2 <

(n + s(T ))/2 = (n + n1)/2, a contradiction. Hence s∗(T ) = s(T ), that is, the set of

support vertices of T is an independent set. 2

We are now in a position to prove Theorem 113. Recall the statement of the

theorem.

Theorem 113. Let T be a nontrivial tree of order n with n1 vertices of degree 1.

Then, γt(T ) = a(T ) + 1 if and only if γt(T ) = (n + n1)/2.

Proof. Suppose that γt(T ) = a(T )+1. If T = P2, then γt(T ) = 2 = (n+n1)/2. Hence

we may assume that n ≥ 3. The necessity now follows from Lemma 121(c). Next we

consider the sufficiency. Suppose that γt(T ) = (n + n1)/2. Then by Corollary 117,


n1 ≤ s(T ). However, n1(G) ≥ s(G) for every graph G. Consequently, n1 = s(T ),

implying that T has no strong support vertex. Hence by Lemma 119(c), we have that

a(T ) < (n + n1)/2. By assumption, γt(T ) = (n + n1)/2. Thus a(T ) ≤ γt(T ) − 1. By

Theorem 112, a(T ) ≥ γt(T ) − 1. Consequently, γt(T ) = a(T ) + 1. 2

9.7 Characterization

We finish this work with a constructive characterization of the trees achieving the

upper bound of Theorem 112. By Theorem 113, it suffices to characterize the non-

trivial trees T satisfying γt(T ) = (n + n1(T ))/2. As we will see, this characterization

is a corollary to the results of Chen et al. in [8]. For continuity, we need the following

definitions. A locating total dominating set, abbreviated LTDS, of a graph G is a TDS

of G with the additional property that for every pair of vertices u and v in V (G) \ S,

NG(u) ∩ S 6= NG(v) ∩ S. The locating total domination number of G, denoted by

γLt (G), is the minimum cardinality of a LTDS of G. A LTDS of G of cardinality

γLt (G) is called a γL

t (G)-set. Locating total domination was introduced in [32].

We will also need the following four results from [8].

Definition 9.7.1 ([8]) We describe a procedure to build a family Γ of labeled trees.

The label of a vertex v is called its status and is denoted by sta(v). Let Γ be the family

of labeled trees T = Tk that can be obtained as follows. Let T0 be a P6 in which the

two leaves have status C, the two support vertices have status A and the remaining

two vertices have status B. If k ≥ 1, then Tk can be obtained recursively from Tk−1

by one of the following operations.


• Operation τ1. For any y ∈ V (Tk−1), if sta(y) = C and y is a leaf of Tk−1, then

add a path xwvz and edge xy. Let sta(x) = sta(w) = B, sta(v) = A, and

sta(z) = C.

• Operation τ2. For any y ∈ V (Tk−1), if sta(y) = B, then add a path xwv and

edge xy. Let sta(x) = B, sta(w) = A, and sta(v) = C.

C

T0

A B B A C

Cy

B

xOperation τ1

B

w

A

v

C

z

By

B

xOperation τ2

A

w

C

v

Figure 9.1: Operations to build Γ

Theorem 122 ([8]) Let T be a tree of order n ≥ 3. Then the following holds.

(a) γLt (T ) ≤ (n + n1(T ))/2.

(b) If T ∈ Γ, then γt(T ) = (n + n1(T ))/2.

(c) γLt (T ) = (n + n1(T ))/2 if and only if T ∈ Γ.

We conclude by giving a constructive characterization of trees achieving the upper

bound of Theorem 112 .

Corollary 123 For any tree T of order n ≥ 3, γt(T ) = (n + n1(T ))/2 if and only if

T ∈ Γ.


Proof. Assume that γt(T ) = (n + n1(T ))/2. Since any LTDS is also a TDS, it

follows that γt(T ) ≤ γLt (T ). By Theorem 122(a), γL

t (T ) ≤ (n + n1(T ))/2, and so

γt(T ) = γLt (T ) = (n + n1(T ))/2. By Theorem 122(c), it follows that T ∈ Γ. This

proves the necessity, and the sufficiency follows from Theorem 122(b). 2

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TOTAL DOMINATION IN GRAPHS AND GRAPH MODIFICATIONS

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