# The diameter of total domination vertex critical graphs.

**ABSTRACT** A graph G with no isolated vertex is total domination vertex critical if for any vertex v of G that is not adjacent to a vertex of degree one, the total domination number of G v is less than the total domination number of G. These graphs we call t-critical. If such a graph G has total domination number k, we call it k-t-critical. We characterize the connected graphs with minimum degree one that are t-critical and we obtain sharp bounds on their maximum diameter. We calculate the maximum diameter of a k-t-critical graph for k � 8 and provide an example which shows that the maximum diameter is in general at least 5k/3 O(1).

**0**Bookmarks

**·**

**63**Views

- [Show abstract] [Hide abstract]

**ABSTRACT:**Let G be a graph of order n and maximum degree Δ(G) and let γt(G) denote the minimum cardinality of a total dominating set of a graph G. A graph G with no isolated vertex is the total domination vertex critical if for any vertex v of G that is not adjacent to a vertex of degree one, the total domination number of G−v is less than the total domination number of G. We call these graphs γt-critical. For any γt-critical graph G, it can be shown that n≤Δ(G)(γt(G)−1)+1. In this paper, we prove that: Let G be a connected γt-critical graph of order n (n≥3), then n=Δ(G)(γt(G)−1)+1 if and only if G is regular and, for each v∈V(G), there is an A⊆V(G)−{v} such that N(v)∩A=0̸, the subgraph induced by A is 1-regular, and every vertex in V(G)−A−{v} has exactly one neighbor in A.Discrete Mathematics 01/2009; 309:991-996. · 0.58 Impact Factor - SourceAvailable from: ArXiv[Show abstract] [Hide abstract]

**ABSTRACT:**Let $\gamma_t(G)$ be the total domination number of graph $G$, a graph $G$ is $k$-total domination vertex critical (or\ just\ $k$-$\gamma_t$-critical) if $\gamma_t(G)=k$, and for any vertex $v$ of $G$ that is not adjacent to a vertex of degree one, $\gamma_t(G-v)=k-1$. Mojdeh and Rad \cite{MR06} proposed an open problem: Does there exist a 3-$\gamma_t$-critical graph $G$ of order $\Delta(G)+3$ with $\Delta(G)$ odd? In this paper, we prove that there exists a 3-$\gamma_t$-critical graph $G$ of order $\Delta(G)+3$ with odd $\Delta(G)\geq 9$.03/2011; - SourceAvailable from: ngwa.org[Show abstract] [Hide abstract]

**ABSTRACT:**This article describes a computer automated, hydrologic analysis system designed to allow the collection of high quality, long-term pumping test data. The instrument solves two of the major problems encountered in the field during aquifer tests: insufficient data, particularly during the early part of a test when drawdown is rapid; and high labor costs associated with long-term monitoring.To illustrate the system's application, results are presented from the test of a highly transmissive aquifer. The aquifer's drawdown response was rapid; thus the time-drawdown curve was essentially flat after the first two minutes of the test, and correspondingly rapid data acquisition was essential for a unique solution of the aquifer's three-dimensional hydraulic conductivities.Ground Water Monitoring and Remediation 02/1984; 4(1):21 - 25. · 1.05 Impact Factor

Page 1

Discrete Mathematics 286 (2004) 255–261

www.elsevier.com/locate/disc

Note

The diameter of total domination vertex critical graphs

Wayne Goddarda,1, Teresa W. Haynesb, MichaelA. Henningc,2,

Lucas C. van der Merwed

aDepartment of Computer Science, University of KwaZulu-Natal, Durban 4041, South Africa

bDepartment of Mathematics, East Tennessee State University, Johnson City, TN 37614-0002, USA

cSchool of Mathematics, Statistics, & Information Technology, University of KwaZulu-Natal, Pietermaritzburg 3209, South Africa

dDepartment of Mathematics, University of Tennessee in Chattanooga, Chattanooga, TN 37403, USA

Received 23 May 2002; received in revised form 25 May 2003; accepted 21 May 2004

Available online 23August 2004

Abstract

A graph G with no isolated vertex is total domination vertex critical if for any vertex v of G that is not adjacent to a vertex of

degree one, the total domination number of G−v is less than the total domination number of G.These graphs we call?t-critical.

If such a graph G has total domination number k, we call it k-?t-critical. We characterize the connected graphs with minimum

degree one that are ?t-critical and we obtain sharp bounds on their maximum diameter.We calculate the maximum diameter of a

k-?t-critical graph for k?8 and provide an example which shows that the maximum diameter is in general at least 5k/3−O(1).

© 2004 Elsevier B.V.All rights reserved.

MSC: 05C69

Keywords: Total domination; Vertex critical; Bounds; Diameter

1. Introduction

Formanygraphparameters,criticalityisafundamentalquestion.Muchhasbeenwrittenaboutthosegraphswhereaparameter

(such as connectedness or chromatic number) goes up or down whenever an edge or vertex is removed or added. For domination

number (the smallest cardinality of a set whose closed neighborhood is the whole graph), Brigham et al. [1] began the study of

those graphs where the domination number decreases on the removal of any vertex. These we call ?-critical. Further properties

of these graphs were explored in [3,4,8–11], but they have not been characterized.

In this paper, we introduce the same concept for total domination. We use the notation of [6]. In particular, if G = (V,E)

denotesagraph,thenthe(open)neighborhoodofvertexv ∈ V isdenotedbyN(v)={u ∈ V | uv ∈ E}whileN[v]=N(v)∪{v}.

For a set S ⊆ V, N(S) =?

set if N(S) = V. For sets S,T ⊆ V, S totally dominates T if T ⊆ N(S). The minimum cardinality of a total dominating set

is the total domination number, denoted ?t(G). A total dominating set of cardinality ?t(G) we call a ?t(G)-set. For a detailed

treatment of this parameter, the reader is referred to [6].

v∈SN(v) and N[S] = N(S) ∪ S. The set S is a dominating set if N[S] = V, and a total dominating

E-mail address: henning@nu.ac.za (M.A. Henning).

1Current address: Department of Computer Science, Clemson University, Clemson SC 29634-1906, USA.

2Research supported in part by the SouthAfrican National Research Foundation and the University of KwaZulu-Natal.

0012-365X/$-see front matter © 2004 Elsevier B.V.All rights reserved.

doi:10.1016/j.disc.2004.05.010

Page 2

256

W. Goddard et al./Discrete Mathematics 286 (2004) 255–261

For a set S ⊆ V, we denote the subgraph of G induced by S by G[S]. The minimum and maximum degrees of the graph G

are denoted by ?(G) and ?(G), respectively.An end-vertex is a vertex of degree one and a support vertex is one that is adjacent

to an end-vertex. Let S(G) be the set of support vertices of G. We say that a vertex v ∈ V is critical if ?t(G−v)<?t(G). Since

total domination is undefined for a graph with isolated vertices, we say that a graph G is total domination vertex critical, or

just ?t-critical, if every vertex of V − S(G) is critical. If G is ?t-critical, and ?t(G) = k, then we say that G is k-?t-critical. For

example, the 5-cycle is 3-?t-critical and the 6-cycle is 4-?t-critical.

Note that a graph is ?t-critical if and only if each component is ?t-critical.Also, K2is trivially 2-?t-critical. So henceforth we

consider only connected graphs of order at least 3. The removal of a vertex can decrease the total domination number by at most

one. Hence:

Observation 1. If G is a ?t-critical graph, then ?t(G−v)=?t(G)−1 for every v ∈ V −S(G). Furthermore, a ?t(G−v)-set

contains no neighbor of v.

Next we observe a sufficient condition for a graph not to be ?t-critical.

Observation 2. If a graph G has nonadjacent vertices u and v with v / ∈S(G) and with N(u) ⊆ N(v), then G is not ?t-critical.

Proof. Let S be a ?t(G − v)-set. In order for u to be totally dominated, there is a vertex x ∈ N(u) ∩ S. Since N(u) ⊆ N(v), S

also totally dominates V(G), and so ?t(G)?|S| = ?t(G − v). Thus, G is not ?t-critical.

?

We proceed as follows. In Section 2, we characterize the connected ?t-critical graphs that have an end-vertex, and we obtain

sharp bounds on their maximum diameter. In Section 3, we consider the maximum diameter of a k-?t-critical graph. Finally, in

Section 4 we briefly discuss links between k-?t-critical graphs and related families and list some open problems.

2. Graphs with end-vertices

In this section, we characterize the ?t-critical graphs with end-vertices. For this purpose, we recall that the corona cor(H) of

a graph H (denote H ◦ K1in [6]) is that graph obtained from H by adding a pendant edge to each vertex of H.

Theorem 3. Let G be a connected graph of order at least 3 with at least one end-vertex. Then, G is k-?t-critical if and only if

G = cor(H) for some connected graph H of order k with ?(H)?2.

Proof. SupposeG=cor(H)forsomeconnectedgraphHoforderkwith?(H)?2.Then?t(G)=|V(H)|=k.Letu ∈ V(G)−S(G).

Then degu = 1 and u is adjacent to a unique vertex v of H. Since ?(H − v)?1, it follows that V(H) − {v} totally dominates

G − u, and so ?t(G − u)?|V(H)| − 1 = ?t(G) − 1. Thus, G is k-?t-critical.

Now, suppose that G is a k-?t-critical graph with ?(G) = 1. Let v?be an end-vertex and let v be its neighbor. Suppose there

exists w ∈ N(v) − {v?} with w / ∈S(G). Then by Observation 1, a ?t(G − w)-set does not contain v, but v is required to totally

dominate v?, a contradiction.Thus each vertex in N(v)−{v?} is a support vertex. It follows that G=cor(H) for some connected

graph H of order k?2. In particular, ?t(G) = |V(H)| = k.

Suppose that H has an end-vertex v. Let w be the neighbor of v in H, and let w?be the end-vertex of G adjacent to w. Let S?be

a ?t(G−w?)-set. Then, V(H)−{w} ⊂ S?. In order to totally dominate v, S?must contain v?or w, and so |S?|?|V(H)|=?t(G),

a contradiction. Hence, ?(H)?2.

?

As a consequence of Theorem 3, we have the following corollaries.

Corollary 4. No tree is ?t-critical.

Corollary 5. If G is a connected k-?t-critical graph with at least one end-vertex, then diam(G)?k if k ∈ {3,4} and

diam(G)?k − 1 if k?5, and these bounds are sharp.

Proof. By Theorem 3, G = cor(H). Hence, diam(G) = 2 + diam(H). If k = 3, then H = K3and so diam(G) = 3. If k = 4,

then C4is a subgraph of H and so diam(G)?4. For k?5, it is a simply exercise to show that the maximum diameter of a graph

H on k vertices with minimum degree 2 is k − 3. For k ∈ {3,4,5}, an extremal H is a cycle, while for k?6, an extremal H is

obtained from two disjoint triangles by joining them with an edge and then subdividing this edge k − 6 times.

?

Page 3

W. Goddard et al./Discrete Mathematics 286 (2004) 255–261

257

The above theorem is also useful in proving a characterization of 3-?t-critical graphs.A graph H is vertex diameter k-critical

if diam(H) = k and diam(H − v)>k for all v ∈ V(H). Hanson and Wang [5] observed the following result.

Theorem 6 (Hanson andWang [5]). For a graph G, ?t(G)=2 if and only if the complement¯G has diameter greater than two.

Theorem 7. A connected graph G is 3-?t-critical if and only if¯G is vertex diameter 2-critical or G is the net, cor(K3).

Proof. Assume G is 3-?t-critical. Since ?t(G)>2, Theorem 6 implies that diam(¯G) = 2. If G has a pendant edge, then by

Theorem 3, G is the net. If G has no pendant edge, then ?t(G − v) = 2 for all v ∈ V, and so, by Theorem 6, diam(¯G − v)>2.

Thus,¯G is vertex diameter 2-critical. Conversely, if¯G is vertex diameter 2-critical, then Theorem 6 implies that ?t(G)?3 and

?t(G − v) = 2. Since G is connected, ?t(G)?1 + ?t(G − v), and so G is 3-?t-critical.

?

For example, the Petersen graph is vertex diameter 2-critical and so the complement is 3-?t-critical.

3. Bounds on the diameter

In this section, we establish bounds on the diameter of a connected k-?t-critical graph. We first determine which cycles are

?t-critical. To this end we recall the total domination numbers of path Pnand cycle Cnon n vertices.

Observation 8 (Henning [7]). For n?3, ?t(Pn) = ?t(Cn) = ?n/2? + ?n/4? − ?n/4?.

Proposition 9. A cycle Cnis ?t-critical if and only if n ≡ 1,2(mod4).

Proposition 9 shows that the diameter of a k-?t-critical graph can be linear in k. We provide next a trivial upper bound on the

diameter of a k-?t-critical graph G. Throughout this section, for x ∈ V, we let Sxdenote a ?t(G − x)-set.

Proposition 10. The diameter of a k-?t-critical graph G is at most 2k − 3.

Proof. Let v be a diametrical vertex of G. Let d = diam(G). For i = 0,1,...,d, let Videnote the set of all vertices of G

at distance i from v. In particular, V0= {v} and V1= N(v). Then, |Sv| = k − 1. By Observation 1, Sv∩ V1= ∅. Hence

to totally dominate V1, |Sv∩ V2|?1. In fact by Observation 2, |Sv∩ V2|?2. Thus, S = Sv∪ {v1} is a ?t(G)-set for any

v1∈ V1and |S ∩ (V1∪ V2)|?3. For any i?3, |S ∩ (Vi∪ ··· ∪ Vi+3)|?2. It follows that if d = 2 + 4j + r where 0?r?3,

then k = |S|?3 + 2j if r ∈ {0,1} while k?3 + 2j + ?r/2? if r ∈ {2,3}. In all cases, d?2k − 3 with inequality if d / ≡ 3

(mod4).

?

Next we establish a sharp upper bound on the diameter of a connected k-?t-critical graph for small k.

Theorem 11. For k?8, the diameter of a k-?t-critical graph is at most the value given by the following table:

k

3

3

4

4

5

6

6

7

7

9

8

11 diam

Proof. If ?(G) = 1, then the upper bounds follow from Corollary 5. If k = 3, then the upper bound follows from Proposition

10. Hence we may assume ?(G)?2 and k?4. Let v be a diametrical vertex of G. Let d = diam(G). For i = 0,1,...,d, let Vi

denote the set of all vertices of G at distance i from v. In particular, V0= {v} and V1= N(v). For i ∈ {0,1,...,d}, let

i?

j=0

LetV =V(G).ForsubsetsSandTofV,wewriteS?tT ifStotallydominatesTinG.Furthermore,wewriteS?→tT ifS ∩ T?tT.

As before, for u ∈ V, let Sube a ?t(G − u)-set.

Case 1: k = 4.

Let u ∈ V1. Then |Su| = 3. To totally dominate V0, |Su∩ V1|?1. Since G[Su] must be connected, it follows that Su⊂ V?3,

and so d?4.

Case 2: k = 5.

V?i=

Vj

and

V?i=

d?

j=i

Vj.

Page 4

258

W. Goddard et al./Discrete Mathematics 286 (2004) 255–261

Suppose d?7. Let u ∈ V1; then |Su| = 4. To totally dominate V0∪ V4∪ V7, it follows that d = 7 and |Su∩ Vj| = 1

for j ∈ {1,2,5,6}. Then Su?→tV?4. By symmetry, for w ∈ V6it follows that |Sw∩ V?3| = 2 and Sw?→tV?3. Therefore,

(Su∩ V?4) ∪ (Sw∩ V?3)?tV, which contradicts ?t(G) being 5.

Case 3: k = 6.

Suppose d?8. Let u ∈ V1; then |Su|=5. To totally dominate V0∪V4∪V5∪V8, it follows that d =8, and |Su∩Vj|=1 for

either j ∈ {1,2,3,6,7} or j ∈ {1,2,5,6,7}.

Letw ∈ V7.Asbefore,|Sw∩Vj|=1foreitherj ∈ {1,2,5,6,7}orj ∈ {1,2,3,6,7}.ThenSw?→tV?3whileSu∩V?3?tV?4.

So, if Sw∩V3=∅, then ?t(G)?5, a contradiction. Hence we may assume that Sw∩V3is nonempty; similarly we may assume

that Su∩ V5is nonempty.

Let x ∈ V4. To totally dominate V0and V8, it follows that |Sx∩ V?2|,|Sx∩ V?6|?2. Since no vertex of G[Sx] is isolated,

Sx∩ V4= ∅. Without loss of generality we may assume that |Sx∩ V?3| = 2. Then, Sx?→tV?3while Su?→tV?4; so ?t(G)?5,

a contradiction.

Case 4: k = 7.

Suppose d?10. Let u ∈ V1; then |Su| = 6.As before, |Su∩ V?2|,|Su∩ V?8|?2. Hence, to totally dominate V4∪ V5∪ V6,

|Su∩V5|?1 and |Su∩(V4∪V5∪V6)|?2. Hence, |Su∩V?2|=2, |Su∩V?8|=2, |Su∩V5|?1, and |Su∩(V4∪V5∪V6)|=2.

In particular, Su?→tV?4.

Let w ∈ V9. By symmetry, |Sw∩ V?2| = 2, |Sw∩ V?8| = 2, |Sw∩ V5|?1, and |Sw∩ (V4∪ V5∪ V6)| = 2. In particular,

Sw?→tV?6. If Su∩V6=∅, then Su?→tV?7, and (Sw∩V?6)∪(Su∩V?8)?tV which contradicts ?t(G) being 7. Thus we may

assume that |Su∩ V6| = 1, and so |Su∩ V5| = 1; similarly, |Sw∩ V4| = |Sw∩ V5| = 1 .

Let x ∈ V5. Then, as before, |Sx∩ V?2| = 2 and |Sx∩ V?8| = 2. Suppose there is another vertex in V5. Then |Sx∩ V5|?1,

and |Sx∩ (V4∪V5∪V6)|=2.Without loss of generality, Sx∩V4=∅. So, Sx?→tV?3.Therefore (Sx∩V?2)∪(Su∩V?4)?tV

which contradicts ?t(G) being 7. Hence there is no other vertex in V5, and so |Sx∩Vj|=1 for j ∈ {1,2,3,7,8,9} and d =10.

Let y ∈ V6. By Observation 1, since x dominates V6, x / ∈Sy.As before, |Sy∩ V?2|,|Sy∩ V?8|?2. If |Sy∩ V4| = 0, then to

totallydominateV?4itfollowsthat(sincex / ∈Sy)|Sy∩V?3|?3,whiletototallydominate{x}∪V10itfollowsthat|Sy∩V?6|?4,

and so |Sy|?7, a contradiction. Hence, |Sy∩V4|?1 and thus |Sy∩V?4|?4. In particular, this implies that V6={y} and that the

two vertices of Sy∩V?8totally dominate V?7. But then (Sw∩V?6)∪(Sy∩V?8) is a total dominating set of G of cardinality

6, a contradiction.

Case 5: k = 8.

Suppose d?12. Let u ∈ V1; then |Su| = 7. As before, |Su∩ V?2|,|Su∩ V?10|?2. Hence |Su∩ Vj| = 1 for all j in either

{1,2,5,6,7,10,11}, {1,2,5,6,9,10,11}, or {1,2,3,6,7,10,11}. So Su∩ V?3?tV?4.

Let w ∈ V11. Then, by symmetry, |Sw∩ Vj| = 1 for all j in one of the three possibilities given above. Then (Sw∩ V?3) ∪

(Su∩ V?3)?tV, and so contradicts ?t(G) being 8 unless |Sw∩ V?3| = 3. Thus we may assume that |Sw∩ Vj| = 1 for all j in

{1,2,3,6,7,10,11}. By symmetry, |Su∩ Vj| = 1 for all j in {1,2,5,6,9,10,11}. In particular, Sw?→tV?8and Su?→tV?4.

Nowlety ∈ V4.TototallydominateV6itfollowsthat|Sy∩(V5∪V6∪V7)|?2.Henceeither|Sy∩V?4|=2or|Sy∩V?8|=2.

In the former case, let S =(Sy∩V?2)∪(Su∩V?5), while in the latter case, let S =(Sw∩V?7)∪(Sy∩V?10). In both cases,

S is a total dominating set of G of cardinality 7, a contradiction.

?

3.1. Constructions

First, we give a way of constructing a critical graph from two smaller critical graphs.

Lemma 12. Let F and H be j-?t-critical and k-?t-critical graphs, respectively, with minimum degrees at least two and let G be

a graph formed by identifying a vertex of F with a vertex of H. If ?t(G) = j + k − 1, then G is ?t-critical.

Proof. Note that since ?(F)?2 and ?(H)?2, S(G)=∅. Label the identified vertex v. Let u ∈ V(G).Without loss of generality,

u ∈ V(F). Since F is j-?t-critical, ?t(F − u) = j − 1. If u ?= v, then any ?t(F − u)-set dominates v implying that only

?t(H − v) = k − 1 vertices are needed to totally dominate H. Hence, ?t(G − u)?j − 1 + k − 1<?t(G). If u = v, then

?t(G − v) = ?t(F − v) + ?t(H − v) = j − 1 + k − 1<?t(G). Thus, ?t(G − u)<?t(G) and G is ?t-critical.

We define a graph as pointed if there are two designated diametrical vertices called LEFT and RIGHT. Then for two pointed

graphs G and H, we define G◦H as the pointed graph obtained by identifying and undesignating the RIGHT-vertex from G and

the LEFT-vertex from H. Note that the operator ◦ is associative.

Now we define our building blocks. Let H1be a copy of P4and let H2be a copy of ¯ H1. Let F be the pointed graph obtained

from H1∪H2by adding all edges between H1and H2except for a perfect matching between corresponding vertices of H1and

H2, and then adding two new vertices LEFT and RIGHT such that LEFT is joined to every vertex in H1and RIGHT is joined to

every vertex in H2. The graph is shown in Fig. 1 where for clarity we omit the edges between H1and H2. It is straightforward

to check that F is 3-?t-critical with diameter 3.

?

Page 5

W. Goddard et al./Discrete Mathematics 286 (2004) 255–261

259

Fig. 1. The 3-?t-critical graph F of diameter 3.

P4

P4

compcomp

Fig. 2. The pointed graph Q.

LetRbethepointedgraphon17verticesdefinedasfollows.LetS={s1,s2,s3,s4},T ={t1,t2,t3,t4}andU={u1,u2,u3,u4}.

Add edges such that s1,s2,s3,s4,s1induces a cycle, and t3,t1,t4,t2and u3,u1,u4,u2induce P4s. Add all edges between S

and T except for a perfect matching between corresponding vertices; similarly with S and U. Add all edges between T and U.

Add two new vertices a and a?such that a is adjacent to {s2,s3,s4,t2,t3,u2,u3} and a?is adjacent to {s1,s2,s3,t2,t3,u2,u3}.

Finallyaddthreenewvertices, LEFT,r?,and RIGHT,suchthat LEFTisadjacenttoallofS, RIGHTisadjacenttoallofT ∪U ∪{a},

and r?is adjacent to all of T ∪ U ∪ {a?}. It is straightforward, though tedious, to check that R is 3-?t-critical with diameter 3.

The graph R has more properties than F, and so in later discussion we can replace F by R. However, wherever possible we

use F because it is simpler.

Let J1and J3be disjoint copies of 2K2and let J2be a copy of¯ J1. Let J be the pointed graph obtained from J1∪J2∪J3by

adding all edges between J1and J2(respectively, J2and J3) except for a perfect matching between corresponding vertices of

J1and J2(respectively, J2and J3), and then adding two new vertices LEFT and RIGHT such that LEFT is adjacent to all of J1

and RIGHT is adjacent to all of J3. It is straightforward to check that J is 4-?t-critical with diameter 4.

Finally, let Q be the pointed graph obtained from F ◦ F by deleting the cut-vertex v and adding all edges joining the four

neighbors of v in the one copy of F to the four neighbors of v in the other copy. See Fig. 2. It is straightforward to check that

?t(Q) = 4 and Q has diameter 5. (The graph Q is not 4-?t-critical.)

Observe that each of the pointed graphs F, R, J, and Q has a ?t-set containing the vertex LEFT and a ?t-set containing the

vertex RIGHT. With these constructions, one can show that the bounds in Theorem 11 are best possible. Examples are given in

the following table, where the last line of the table (when k = 8) will be proved in Theorem 13.

k

Graph with maximum diameter

3

4

5

6

7

8

F

J

F ◦ F

F ◦ J

F ◦ R ◦ F

F ◦ Q ◦ F

In general we have the following.

Theorem 13. For all k ≡ 2(mod3), there exists a k-?t-critical graph of diameter (5k − 7)/3.

Proof. Forq?0,definethepointedgraphYq=F ◦Q◦···◦QforqcopiesofQanddefineZq=Yq◦F.Thendiam(Zq)=5q+6.

For a pointed graph G, we define Li(G) as the vertices at distance i from LEFT and Ri(G) as the vertices at distance i from

RIGHT.

Page 6

260

W. Goddard et al./Discrete Mathematics 286 (2004) 255–261

Claim 1. (a) ?t(Yq)?3q + 3; if T totally dominates Yq− {RIGHT}, then |T|?3q + 2.

(b) ?t(Zq)?3q + 5.

(c) ?t(Zq− v)?3q + 4 for all v ∈ V(Zq).

Proof. (a) By induction on q. Note that Y0=F; so the base case is true.Assume then q?1. Label the ith copy of Q by Qi. Note

that Yq= Yq−1◦ Qq, and let x be the vertex so identified.

Let S be a total dominating set of Yq. Define S?as the intersection of S with V(Qq)−{x}.Then it can be checked that |S?|?3.

Furthermore, if |S?| = 3, then S?does not dominate x; and so |S − S?|??t(Yq−1)?3q. On the other hand, in general S − S?

totally dominates Yq−1− {x}, and so has at least 3q − 1 elements. In either case, |S|?3q + 3.

LetTbeatotaldominatingsetofYq−{RIGHT}anddefineT?astheintersectionofTwithV(Qq)−{x}.Bysimilararguments,

|T?|?2, and if equality holds, then T − T?totally dominates Yq−1. Thus, |T|?3q + 2.

(b) Note that Zq=Yq−1◦F. Then using the same approach as in (a), except with S?defined as the intersection of S with the

non-cut-vertices of the final copy of F, the desired result follows readily.

(c) Assume v was from a copy of F (without loss of generality, the left one). Construct a set Svas follows. Since F is 3-?t-

critical, there exist two vertices which totally dominate F − v.Also, there exist two vertices in L2(Q1) which totally dominate

L1(Q1) ∪ L2(Q1) ∪ L3(Q1). Thereafter take one vertex from each of the first three levels of each block; that is, a vertex from

each of L0(B), L1(B) and L2(B) for B = Q2,...,Qq,F. These can be chosen such that Svdominates Zq− v. The set has

cardinality 3q + 4.

Assume v was from a copy of Q; by symmetry, without loss of generality, v ∈ L0(Qi) ∪ L1(Qi) ∪ L2(Qi). Construct a set

Svas follows. Since F is 3-?t-critical, there exist two vertices in L1(Qi)∪L2(Qi) which totally dominate L0(Qi)∪L1(Qi)∪

L2(Qi) ∪ L3(Qi) except for v. To the right, take triples at the start of blocks as before (that is, a vertex from each of L0(B),

L1(B) and L2(B) for B =Qi+1,...,Qq,F). To the left, take two vertices from R2(Qi−1) (or R2(F) if q =1), and thereafter

triples as before (that is, a vertex from each of R0(B), R1(B) and R2(B) for B =Qi−2,...,Q1,F). These can be chosen such

that Svdominates Zq− v. The set has cardinality 3q + 4. This completes the proof of the claim.

?

By Claim 1, Zqis (3q + 5)-?t-critical. For k = 2 the desired graph is K2. For k?5, the graphs Z(k−5)/3have the desired

properties.

?

4. Open questions

We close with a list of open problems and questions.

1. Characterize the 3-?t-critical graphs with diameter 3. Does there exist a 4-?t-critical graph with diameter 2?

2. Consider the connection between ?-critical and ?t-critical graphs. For example, K3× K3is ?-critical but not ?t-critical. The

cycle C5is ?t-critical, but not ?-critical. So, which graphs are vertex domination critical and total domination vertex critical

(or one but not the other)?

3. Determine the maximum diameter of a k-?t-critical graph.

4. If G is a ?t-critical graph of order n, then it can be shown that n??(G)(?t(G)−1)+1. Characterize those graphs achieving

equality.

5. Cockayne et al. [2] showed that if G is a connected graph of order n?2, then ?t(G)?max(n − ?(G),2). Characterize

?t-critical graphs G with ?t(G) = n − ?(G).

Acknowledgements

The authors are very grateful to the referees and wish to thank them for their helpful comments and insight.

References

[1] R.C. Brigham, P.Z. Chinn, R.D. Dutton, Vertex domination-critical graphs, Networks 18 (1988) 173–179.

[2] E. Cockayne, R. Dawes, S. Hedetniemi, Total domination in graphs, Networks 10 (1980) 211–219.

[3] O. Favaron, D. Sumner, E. Wojcicka, The diameter of domination-critical graphs, J. Graph Theory 18 (1994) 723–734.

[4] J. Fulman, D. Hanson, G. MacGillivray, Vertex domination-critical graphs, Networks 25 (1995) 41–43.

[5] D. Hanson, P. Wang,A note on extremal total domination edge critical graphs, Utilitas Math. 63 (2003) 89–96.

Page 7

W. Goddard et al./Discrete Mathematics 286 (2004) 255–261

261

[6] T.W. Haynes, S.T. Hedetniemi, P.J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, Inc., NewYork, 1998.

[7] M.A. Henning, Graphs with large total domination number, J. Graph Theory 35 (2000) 21–45.

[8] D.P. Sumner, Critical concepts in domination, Discrete Math. 86 (1990) 33–46.

[9] D.P. Sumner, P. Blitch, Domination critical graphs, J. Combin. Theory Ser. B 34 (1983) 65–76.

[10] D.P. Sumner, E. Wojcicka, Graphs critical with respect to the domination number, in: T.W. Haynes, S.T. Hedetniemi, P.J. Slater (Eds.),

Domination in Graphs:Advanced Topics, Marcel Dekker, Inc., NewYork, 1998(Chapter 16).

[11] E. Wojcicka, Hamiltonian properties of domination-critical graphs, J. Graph Theory 14 (1990) 205–215.