Upper bound on the diameter of a total domination vertex critical graph Michitaka Furuya1,∗ Naoya Kato1 1

Department of Mathematical Information Science, Tokyo University of Science

1-3 Kagurazaka, Shinjuku-ku, Tokyo 162-8601, Japan

Abstract A vertex of a graph is said to be total domination critical if its deletion decreases the total domination number. A graph is said to be total domination vertex critical if all of its vertices except the supporting vertices are total domination vertex critical. We show that if G is a connected total domination vertex critical graph with total domination number k ≥ 4, then the diameter of G is at most ⌋. ⌊ 5k−7 3

Key words and phrases. total domination vertex critical graph, total domination, diameter. AMS 2010 Mathematics Subject Classification. 05C69.

1

Introduction

In this paper, we consider only finite, simple, undirected graphs with no loops and no multiple edges. ∗ e-mail:[email protected]

1

Let G be a graph. We let V (G) and E(G) denote the vertex set and the edge set of G, respectively. For v ∈ V (G), we let dG (v), NG (v), and NG [v] denote the degree of v, the open neighborhood of v, and the closed neighborhood of v, respectively; thus dG (v) = |NG (v)| and NG [v] = NG (v)∪{v}. The minimum degree of G is denoted by δ(G). We let G denote the complement of G. For v, u ∈ V (G), we let dG (v, u) denote the distance between v and u. For v ∈ V (G) and a non-negative integer i, we let (i)

NG (v) = {u ∈ V (G)|dG (v, u) = i}.

For v ∈ V (G), we define the

eccentricity eccG (v) of v in G by eccG (v) = max{dG (v, u)|u ∈ V (G)}; thus (i)

eccG (v) is the maximum integer i for which NG (v) ̸= ∅. The diameter of G is defined to be the maximum of eccG (v) as v ranges over V (G), and is denoted by diam(G). A vertex is called an endvertex if it has degree one, and a vertex is called a supporting vertex if it is adjacent to an endvertex. We let S(G) denote the set of supporting vertices of G. For terms and symbols not defined here, we refer the reader to [1]. Let G be a graph with no isolated vertex. For two subsets X, Y of V (G), we say that X totally dominates Y (or X γt -dominates Y for short) ∪ if Y ⊆ v∈X NG (v). A subset of V (G) which totally dominates V (G) is called a total dominating set of G. The minimum cardinality of a total dominating set of G is called the total domination number of G, and is denoted by γt (G). We have γt (G) ≥ 2 unless G is empty (note that we discuss the total domination number of a graph only when the graph under consideration has no isolated vertex). A total dominating set of G having cardinality γt (G) is called a γt -set of G. A vertex v ∈ V (G) − S(G) is said to be total domination critical (γt -ciritical) if γt (G − v) < γt (G). We say that G is total domination vertex critical (γt -critical) if every vertex of G − S(G) is γt -critical in G. If G is γt -critical and γt (G) = k, G is said to be k-γt -critical. Various properties of γt -critical graphs were explored in [3, 4, 5]. Goddard et al. [2] proved the following result. Theorem A Let k ≥ 3 be an integer, and let G be a connected k-γt -critical graph.

2

(i) If k = 3, then the diameter of G is at most 3. (ii) If 4 ≤ k ≤ 8, then the diameter of G is at most ⌊ 5k−7 3 ⌋. (iii) If k ≥ 9, then the diameter of G is at most 2k − 3. They showed that the bound in Theorem A is best possible for each 3 ≤ k ≤ 8 by constructing a k-γt -critical graph attaining the bound. They also constructed a k-γt -critical graph with diameter

5k−7 3

for each k ≡ 2(mod 3),

and conjectured that for each k ≥ 4, every connected k-γt -critical graph has diameter at most ⌊ 5k−7 3 ⌋. In this paper, we prove the following theorem, which shows that the conjecture of Goddard et al. is true. Theorem 1.1 Let k ≥ 3 be an integer, and let G be a connected k-γt critical graph. (i) If k = 3, then the diameter of G is at most 3. (ii) If k ≥ 4, then the diameter of G is at most ⌊ 5k−7 3 ⌋. In Section 3, we show that the bound is best possible for all k ≥ 4 by constructing infinitely many k-γt -critical graphs with diameter ⌊ 5k−7 3 ⌋ for each fixed k. In our proof of Theorem 1.1, we make use of the following lemma, which is proved in [2]. Lemma 1.2 Let k ≥ 3 be an integer, and let G be a connected k-γt -critical graph with δ(G) = 1. Then the diameter of G is at most k. The following observation is useful for our arguments. Observation 1.3 Let G be a γt -critical graph with δ(G) ≥ 2, and let v ∈ V (G). Then the following hold. (i) γt (G − v) = γt (G) − 1. (ii) There exists a γt -set which contains v.

3

2

Proof of Theorem 1.1

Let k, G be as in Theorem 1.1, and let d denote the diameter of G. We may assume that k ≥ 4. By way of contradiction, suppose that d >

5k−7 3 (>

4).

By Lemma 1.2, δ(G) ≥ 2. Let x ∈ V (G) be a vertex with eccG (x) = d. Let (i)

Xi = NG (x) for each i ≥ 0, and let Uj = X0 ∪ · · · ∪ Xj for each j ≥ 0. For j ≥ 2, a γt -set S of G is called j-sufficient if |S ∩ Uj | ≥ (3j + 10)/5. Claim 2.1 For some j ≥ 2, there exists a γt -set which is j-sufficient. Proof. Suppose that for all j ≥ 2, no γt -set is j-sufficient. Take x1 ∈ X1 , and let Sx1 be a γt -set of G − x1 . Since Sx1 γt -dominates U1 − {x1 }, |Sx1 ∩ U2 | ≥ 2. Let S = Sx1 ∪ {x}. Note that S is a γt -set of G. Since S is not 3-sufficient, |S ∩ U3 | < (3 · 3 + 10)/5 = 19/5 < 4. Since |S ∩ U3 | ≥ |Sx1 ∩ U2 | + |{x}| ≥ 3, this forces |S ∩ U3 | = 3 and Sx1 ∩ X3 = ∅. Take x4 ∈ X4 , and let Sx4 be a γt -set G − x4 . Since Sx4 γt -dominates U1 , |Sx4 ∩ U2 | ≥ 2. Take x3 ∈ X3 ∩ NG (x4 ), and let S ′ = Sx4 ∪ {x3 }. Note that S ′ is a γt -set of G. Since S ′ is not 3-sufficient, |S ′ ∩ U3 | < 19/5. Since |S ′ ∩ U3 | ≥ |Sx4 ∩ U2 | + |{x3 }| ≥ 3, this forces |S ′ ∩ U3 | = 3, |Sx4 ∩ U3 | = 2 and Sx4 ∩ X3 = ∅. Since S ′ is not 5-sufficient, |S ′ ∩ U5 | < (3 · 5 + 10)/5 = 5, and hence |Sx4 ∩ (X4 ∪ X5 )| = |S ′ ∩ U5 | − |S ′ ∩ U3 | ≤ 4 − 3 = 1. Now if Sx4 ∩ X4 ̸= ∅, then since Sx4 ∩ X3 = ∅, the unique vertex in Sx4 ∩ X4 cannot be γt -dominated by Sx4 , a contradiction. Thus Sx4 ∩ X4 = ∅. Subclaim 2.1.1 The set (Sx4 ∩ U3 ) ∪ (S − U3 ) is a total dominating set of G. Proof. Since Sx4 γt -dominates V (G) − {x4 } and Sx4 ∩ X4 = ∅, each vertex in U3 is adjacent to a vertex in Sx4 ∩ U3 . Since S γt -dominates V (G) and S ∩ X3 = ∅, each vertex in V (G) − U3 is adjacent to a vertex in S − U3 . 

Hence the desired conclusion holds.

We have |(Sx4 ∩ U3 ) ∪ (S − U3 )| = |Sx4 ∩ U3 | + |S − U3 | = 2 + (k − 3) = k − 1. This together with Subclaim 2.1.1 contradicts the assumption that

4

γt (G) = k, which completes the proof of Claim 2.1.



Having Claim 2.1 in mind, let m ≥ 2 denote the maximum integer such that there exists an m-sufficient γt -set. Let S1 be an m-sufficient γt -set. Then |S1 ∩ Um | ≥ (3m + 10)/5, and we also have |S1 ∩ Um+1 | < (3(m + 1) + 10)/5 by the maximality of m. Since d > (5k − 7)/3 and k ≥ |S1 ∩ Um | ≥ (3m + 10)/5, it follows that d ≥ m + 2. Claim 2.2 S1 ∩ Xm+1 = ∅. Proof. Since |S1 | ≥ (3m + 10)/5 and |S1 ∩ Um+1 | < (3m + 13)/5, it follows that |S1 ∩ Xm+1 | = |S1 ∩ Um+1 | − |S1 ∩ Um | < (3m + 13)/5 − (3m + 10)/5 = 3/5, which implies S1 ∩ Xm+1 = ∅.



Recall that d ≥ m + 2. If |S1 ∩ (Xm+2 ∪ Xm+3 )| ≥ 2, then |S1 ∩ Um+3 | ≥ |S1 ∩ Um | + 2 ≥ (3m + 10)/5 + 2 > (3(m + 3) + 10)/5, which contradicts the maximality of m. Thus |S1 ∩ (Xm+2 ∪ Xm+3 )| ≤ 1. Since S1 γt -dominates Xm+2 and S1 ∩Xm+1 = ∅, S1 ∩(Xm+2 ∪Xm+3 ) ̸= ∅. If S1 ∩Xm+2 ̸= ∅, then S1 does not γt -dominate the vertex in S1 ∩ Xm+2 , a contradiction. Thus S1 ∩ Xm+2 = ∅, and hence |S1 ∩ Xm+3 | = 1. Write S1 ∩ Xm+3 = {wm+3 }. Since S1 ∩ Xm+1 = S1 ∩ Xm+2 = ∅, wm+3 is adjacent to every vertex in Xm+2 . Since S1 γt -dominates wm+3 , |S1 ∩ Xm+3 | = 1 and S1 ∩ Xm+2 = ∅, S1 ∩ Xm+4 ̸= ∅. In particular, d ≥ m + 4. Let S2 be a γt -set of G − wm+3 . Note that S2 is not a γt -set of G. Since wm+3 is adjacent to every vertex in Xm+2 , this implies S2 ∩ Xm+2 = ∅. Claim 2.3 The set (S2 ∩ Um+2 ) ∪ (S1 − Um+2 ) is a total dominating set of G. Proof. Since S2 γt -dominates V (G) − {wm+3 }, each vertex in Um+1 is adjacent to a vertex in S2 ∩ Um+2 . Since S1 γt -dominates V (G) and S1 ∩ Xm+1 = S1 ∩ Xm+2 = ∅, each vertex in V (G) − Um+1 is adjacent to a vertex in S1 − Um+2 . Hence the desired conclusion holds.



If |S2 ∩ Um+2 | ≤ |S1 ∩ Um+2 | − 1, then |(S2 ∩ Um+2 ) ∪ (S1 − Um+2 )| ≤ |S1 ∩Um+2 |+|S1 −Um+2 |−1 = k−1 which, in view of Claim 2.3, contradicts 5

the assumption that γt (G) = k. Thus |S2 ∩ Um+2 | ≥ |S1 ∩ Um+2 |.

(2.1)

Suppose that |Xm+3 | ≥ 2. Since S2 γt -dominates (Xm+3 −{wm+3 })∪Xm+4 and S2 ∩Xm+2 = ∅, |S2 ∩(Xm+3 ∪Xm+4 ∪Xm+5 )| ≥ 2. Let wm+2 ∈ Xm+2 . Then by (2.1), |(S2 ∪{wm+2 })∩Um+5 | ≥ |S1 ∩Um+2 |+3 ≥ (3m+10)/5+3 = (3(m+5)+10)/5. Since S2 ∪{wm+2 } is a γt -set of G, S2 ∪{wm+2 } is (m+5)sufficient, which contradicts the maximality of m. Thus |Xm+3 | = 1. In particular, wm+3 is adjacent to every vertex in Xm+4 . Take wm+4 ∈ Xm+4 , and let S3 be a γt -set of G − wm+4 . Note that S3 ∩Xm+3 = ∅ and S3 ∪{wm+3 } is a γt -set of G. Suppose that S3 ∩Xm+2 ̸= ∅. Claim 2.4 The set (S3 ∩ Um+2 ) ∪ (S2 − Um+2 ) is a total dominating set of G. Proof. Since S3 γt -dominates V (G) − {wm+4 } and S3 ∩ Xm+3 = ∅, each vertex in Um+2 is adjacent to a vertex in S3 ∩ Um+2 . Since S3 ∩ Xm+2 ̸= ∅, Xm+3 = {wm+3 } and wm+3 is adjacent to every vertex in Xm+2 , this implies that S3 ∩ Um+2 γt -dominates Um+3 . Since S2 γt -dominates V (G) − {wm+3 }, each vertex in V (G) − Um+3 is adjacent to a vertex in S2 − Um+2 . 

Hence the desired conclusion holds.

If |S3 ∩ Um+2 | ≥ |S1 ∩ Um+2 | + 1, then |(S3 ∪ {wm+3 }) ∩ Um+3 | ≥ |S1 ∩ Um+2 | + 2 ≥ (3m + 10)/5 + 2 > (3(m + 3) + 10)/5, which contradicts the maximality of m. Thus |S3 ∩ Um+2 | ≤ |S1 ∩ Um+2 |. By (2.1), |S2 − Um+2 | = (k − 1) − |S2 ∩ Um+2 | ≤ (k − 1) − |S1 ∩ Um+2 |. Consequently |(S3 ∩ Um+2 ) ∪ (S2 − Um+2 )| ≤ |S1 ∩ Um+2 | + ((k − 1) − |S1 ∩ Um+2 |) = k − 1 which, in view of Claim 2.4, contradicts the assumption that γt (G) = k. Thus S3 ∩ Xm+2 = ∅. Since S3 γt -dominates Xm+3 and S3 ∩ Xm+2 = S3 ∩ Xm+3 = ∅, S3 ∩ Xm+4 ̸= ∅ and |S3 ∩ (Xm+4 ∪ Xm+5 )| ≥ 2. Claim 2.5 The set (S3 ∩ Um+2 ) ∪ (S1 − Um+2 ) is a total dominating set of G. 6

Proof. Since S3 γt -dominates V (G) − {wm+4 }, each vertex in Um+1 is adjacent to a vertex in S3 ∩ Um+2 . Since S1 γt -dominates V (G) and S1 ∩ Xm+1 = S1 ∩ Xm+2 = ∅, each vertex in V (G) − Um+1 is adjacent to a vertex in S1 − Um+2 . Hence the desired conclusion holds.



If |S3 ∩ Um+2 | ≥ |S1 ∩ Um+2 |, then |(S3 ∪ {wm+3 }) ∩ Um+5 | ≥ |S1 ∩ Um+2 | + 3 ≥ (3m + 10)/5 + 3 = (3(m + 5) + 10)/5, which contradicts the maximality of m. Thus |S3 ∩ Um+2 | ≤ |S1 ∩ Um+2 | − 1. Therefore |(S3 ∩ Um+2 ) ∪ (S1 − Um+2 )| ≤ |S1 ∩ Um+2 | + |S1 − Um+2 | − 1 = k − 1 which, in view of Claim 2.5, contradicts the assumption that γt (G) = k. This completes the proof of Theorem 1.1.

3



Examples

In this section, we prove a theorem concerning the construction of γt -critical graphs (Theorem 3.8), and then use the theorem to construct examples which show the sharpness of the bound in Theorem 1.1. In our construction, we make use of the coalescence of graphs. Let A1 and A2 be graphs. For i = 1, 2, let xi be a vertex of Ai . Under this notation, we let (A1 • A2 )(x1 , x2 : x) denote the graph obtained from A1 and A2 by identifying x1 and x2 into a vertex labelled x. We call (A1 • A2 )(x1 , x2 : x) the coalescence of A1 and A2 via x1 and x2 . We first give some properties of the total domination number of the coalescence of graphs. Lemma 3.1 For each i = 1, 2, let Ai be a graph with δ(Ai ) ≥ 2, and xi be a vertex of Ai . Let G = (A1 • A2 )(x1 , x2 : x). (i) If xi is γt -critical in Ai for some i ∈ {1, 2}, then γt (G) ≤ γt (A1 ) + γt (A2 ) − 1. (ii) We have γt (G) ≥ γt (A1 ) + γt (A2 ) − 2. Further, if γt (G) = γt (A1 ) + γt (A2 ) − 2, then γt (Ai − NAi [xi ]) = γt (Ai ) − 2 for some i ∈ {1, 2}. (iii) If γt (G) = γt (A1 ) + γt (A2 ) − 1 and both A1 and A2 are γt -critical, then G is γt -critical. 7

Proof. (i) Let Si be a γt -set of Ai − xi , and let S3−i be a γt -set of A3−i . If x3−i ∈ S3−i , let S = ((S1 ∪ S2 ) − {x3−i }) ∪ {x}; if x3−i ̸∈ S3−i , let S = S1 ∪ S2 . Then S is a total dominating set of G. We also have |S| = |Si | + |S3−i | = γt (Ai − xi ) + γt (A3−i ) ≤ γt (A1 ) + γt (A2 ) − 1. Hence γt (G) ≤ γt (A1 ) + γt (A2 ) − 1. (ii) It suffices to show that γt (G) = γt (A1 ) + γt (A2 ) − 2 and γt (Ai − NAi [xi ]) = γt (Ai ) − 2 for some i ∈ {1, 2} under the assumption that γt (G) ≤ γt (A1 )+γt (A2 )−2. Thus assume γt (G) ≤ γt (A1 )+γt (A2 )−2. Let S be a γt -set of G. Since S γt -dominates x, S ∩ NG (x) ̸= ∅. We may assume that S ∩ NA1 (x1 ) ̸= ∅. Suppose that x ̸∈ S. Then S ∩ V (A1 ) is a total dominating set of A1 and S ∩ V (A2 ) is a total dominating set of A2 − x2 . Hence |S ∩ V (A1 )| ≥ γt (A1 ) and |S ∩ V (A2 )| ≥ γt (A2 − x2 ). Since removing a vertex can decrease the total domination number at most by one, we get γt (G) = |S| = |S ∩ V (A1 )|+|S∩V (A2 )| ≥ γt (A1 )+γt (A2 −x2 ) ≥ γt (A1 )+γt (A2 )−1. This contradicts the assumption that γt (G) ≤ γt (A1 ) + γt (A2 ) − 2. Thus x ∈ S. Note that ((S − {x}) ∩ V (A1 )) ∪ {x1 } is a total dominating set A1 , and hence |((S −{x})∩V (A1 ))∪{x1 }| ≥ γt (A1 ). If S ∩NA2 (x2 ) ̸= ∅, then ((S − {x}) ∩ V (A2 )) ∪ {x2 } is a total dominating set A2 , and hence |((S − {x}) ∩ V (A2 )) ∪ {x2 }| ≥ γt (A2 ), which implies that γt (G) = |S| = |((S − {x}) ∩ V (A1 )) ∪ {x1 }| + |((S − {x}) ∩ V (A2 )) ∪ {x2 }|−1 ≥ γt (A1 )+γt (A2 )−1, a contradiction. Thus S∩NA2 (x2 ) = ∅. Consequently (S − {x}) ∩ V (A2 ) is a subset of V (A2 ) − NA2 [x2 ] and γt -dominates V (A2 ) − NA2 [x2 ]. Since γt (G) ≤ γt (A1 ) + γt (A2 ) − 2, we now obtain γt (A2 −NA2 [x2 ]) ≤ |(S−{x})∩V (A2 )| = |S|−|((S−{x})∩ V (A1 )) ∪ {x1 }| ≤ γt (G) − γt (A1 ) ≤ (γt (A1 ) + γt (A2 ) − 2) − γt (A1 ) = γt (A2 ) − 2. Since we clearly have γt (A2 ) ≤ γt (A2 − NA2 [x2 ]) + 2, this forces γt (A2 − NA2 [x2 ]) = γt (A2 ) − 2 and γt (G) = γt (A1 ) + γt (A2 ) − 2, as desired. (iii) Let v ∈ V (G). We prove that γt (G − v) ≤ γt (G) − 1.

8

Case 1: v = x. For each i, let Si be a γt -set of Ai − xi . Then S1 ∪ S2 is a total dominating set of G − x. We also have |S1 ∪ S2 | = |S1 | + |S2 | = γt (A1 − x1 ) + γt (A2 − x2 ) ≤ γt (A1 ) + γt (A2 ) − 2 = γt (G) − 1. Hence γt (G − v) ≤ γt (G) − 1. Case 2: v ̸= x. Without loss of generality, we may assume that v ∈ V (A1 )−{x1 }. Let S1 be a γt -set of A1 − v, and let S2 be a γt -set of A2 − x2 . If x1 ∈ S1 , let S = (S1 − {x1 }) ∪ S2 ∪ {x}; if x1 ̸∈ S1 , let S = S1 ∪ S2 . Then S is a total dominating set of G−v. We also have |S| ≤ γt (A1 )+γt (A2 )−2 = γ(G) − 1. Hence γt (G − v) ≤ γt (G) − 1, as desired.



As a preparation for our construction, we describe the definition of the graph Q constructed by Goddard et al. in [2]. (i,j) (i,j) (i,j) (i,j)

For each i = 1, 2 and j = 1, 2, let P (i,j) = x1 x2 x3 x4 be a ∪ (i,1) | 1 ≤ l ≤ 4}, path of order 4. Let U = {u1 , u2 }. Let E1 = i=1,2 {ui xl ∪ (i,1) (i,2) ′ | (l, l ) ̸∈ {(1, 2), (2, 4), (3, 1), (4, 3)}} and E3 = E2 = i=1,2 {xl xl′ {xx′ | x ∈ V (P (1,2) ), x′ ∈ V (P (2,2) )}. Let Q be the graph defined by ∪ ∪ V (Q) = ( ( V (P (i,j) ))) ∪ U i=1,2 j=1,2

and E(Q) = (

∪

(

∪

E(P (i,j) ))) ∪ (

i=1,2 j=1,2

∪

Ei )

1≤i≤3

(see Figure 1). Note that Q − u1 ≃ Q − u2 , Q − NQ [u1 ] ≃ Q − NQ [u2 ] and Q − ({u1 } ∪ V (P1,1 ) ∪ V (P1,2 )) ≃ Q − ({u2 } ∪ V (P2,1 ) ∪ V (P2,2 )). By (2,2)

inspection, we see that γt (Q[V (P (2,1) ) ∪ V (P (2,2) )]) = 2, and {x2 and

(2,1) (2,1) {x2 , x3 }

are the only γt -sets of Q[V (P

(2,1)

) ∪ V (P

(2,2)

(2,2)

, x3

)].

Lemma 3.2 ([2]) We have γt (Q) = 4 and diam(Q) = dQ (u1 , u2 ) = 5. Lemma 3.3

(i) γt (Q − NQ [u1 ]) = 3.

(ii) There exists a γt -set of Q − NQ [u1 ] which contains u2 . 9

}

(1,2)

x1

(2,2)

x1

(2,1)

(1,1) x1

x1

(2,2)

(1,2)

(1,1) x2

x2

x2

(2,1)

x2

u2

u1 (2,1)

(1,1)

x3

(2,2)

x3

(1,2) x3

x3

(2,1)

(1,1)

x4

x4

(1,2)

x4

Figure 1: Graph Q

10

(2,2)

x4

(iii) Every vertex of V (Q)−(NQ [u1 ]∪V (P (1,2) )) is γt -critical in Q−NQ [u1 ]. Proof. We first prove (i) and (ii). Suppose that γt (Q −NQ [u1 ]) = 2. Then there exists a total dominating set S of Q − NQ [u1 ] with |S| = 2. Since eccQ−NQ [u1 ] (u2 ) = 3, we see that |S ∩ V (P (2,1) )| = |S ∩ V (P (2,2) )| = 1. This implies that S is a total dominating set of Q[V (P (2,1) ) ∪ V (P (2,2) )], (2,2)

which contradicts the fact that {x2 only γt -sets of Q[V (P

(2,1)

) ∪ V (P

together with the fact that

(2,2)

(2,2)

, x3

(2,1)

} and {x2

(2,1)

, x3

} are the

)]. Thus γt (Q − NQ [u1 ]) ≥ 3. This

(2,1) (2,2) {u2 , x1 , x1 }

is a total dominating set of

Q − NQ [u1 ] yields (i) and (ii). We next prove (iii). Let x ∈ V (Q)−(NQ [u1 ]∪V (P (1,2) )). We show that x is γt -critical in Q − NQ [u1 ]. Without loss of generality, we may assume (2,i)

that x ∈ {u2 }∪{xj

(2,2)

| 1 ≤ i ≤ 2, 1 ≤ j ≤ 2}. If x = u2 , then {x2

is a total dominating set of (Q − NQ [u1 ]) − x. Similarly (2,1)

{x4

(2,2)

, x4

(2,1)

}, {x3

(2,2)

, x3

(2,1)

} or {x1

(Q − NQ [u1 ]) − x according as x = is proved.

(2,2)

, x4

(2,1) x1 ,

(2,2)

, x3

}

(2,1) (2,2) {x3 , x2 },

} is a total dominating set of

(2,1) x2 ,

(2,2)

x1

(2,2)

or x2

. Thus (iii)



For each i = 1, 2, let Bi be a γt -critical graph with δ(Bi ) ≥ 2, and let bi ∈ V (Bi ) be a vertex with γt (Bi − NBi [bi ]) ≥ γt (Bi ) − 1 (the construction of such graphs will be given later; see Propositions 3.9 through 3.16). Let a ≥ 0 be an integer. Let G(0) be a graph isomorphic to B1 , (0)

and let z2

be the vertex of G(0) corresponding to b1 in B1 . Let G(a+1) (a+1)

be a graph isomorphic to B2 , and let z1

be the vertex of G(a+1) cor-

responding to b2 in B2 . For each 1 ≤ i ≤ a, let G(i) be a graph iso(i)

morphic to Q, and let z1

(i)

and z2

be the vertices of G(i) correspond(a)

ing to u1 and u2 , respectively. Let Z1 (B1 ; b1 ) be the graph obtained by concatenating G(0) , G(1) , · · · , G(a) by letting G(i) and G(i+1) coalesce (i)

via z2

(i+1)

and z1

(a)

(we let z (i) denote the vertex of Z1 (B1 ; b1 ) arising

(i) (i+1) through their identification) for each 0 ≤ i ≤ a − 1. and z1 (a) (a) (a) (a+1) Let Z2 (B1 , B2 ; b1 , b2 ) = (Z1 (B1 ; b1 ) • G(a+1) )(z2 , z1 : z (a) ). Note (0) (0) that Z1 (B1 ; b1 ) = G(0) , Z2 (B1 , B2 ; b1 , b2 ) ≃ (B1 • B2 )(b1 , b2 : b) and Z (a) (B1 , B2 ; b1 , b2 ) ≃ Z (a) (B2 , B1 ; b2 , b1 ).

from z2

11

(a)

Lemma 3.4 There exists a total dominating set of Z1 (B1 ; b1 ) of cardi(a)

nality γt (B1 ) + 3a which contains z2 . (0)

Proof. By Observation 1.3(ii), there exists a γt -set S (0) of G(0) with z2 ∈ (i)

S (0) . For each 1 ≤ i ≤ a, there exists a γt -set S (i) of G(i) − NG(i) [z1 ] with ∪ (i) (i) z2 ∈ S (i) by Lemma 3.3(ii). Then S = (( 0≤i≤a S (i) ) − {z2 | 0 ≤ i ≤ (a)

a − 1}) ∪ {z (i) | 0 ≤ i ≤ a − 1} is a total dominating set of Z1 (B1 ; b1 ). ∑ (i) We also have |S| = γt (G0 ) + 1≤i≤a γt (G(i) − NG(i) [z1 ]) = γt (B1 ) + 3a 

by Lemma 3.3(i). Hence S is a desired set.

(a)

(i) We have γt (Z1 (B1 ; b1 )) = γt (B1 ) + 3a.

Lemma 3.5

(a)

(a)

(ii) There exists a γt -set of Z1 (B1 ; b1 ) which contains z2 . (a)

(a)

(iii) We have γt (Z1 (B1 ; b1 ) − NZ (a) (B1 ;b1 ) [z2 ]) ≥ γt (B1 ) + 3a − 1. 1

(i)

Proof. For each i, let Z (i) = Z1 (B1 ; b1 ). We proceed by induction on a. If a = 0, then we get the desired results by the choice of B1 and Observation 1.3. Thus we may assume that a ≥ 1. Note that Z (a) ≃ (a−1)

(Z (a−1) • Q)(z2

, u1 : x). By the induction assumption, (a−1)

γt (Z (a−1) − NZ (a−1) [z2

]) ≥ γt (Z (a−1) ) − 1.

(3.1)

By Lemma 3.3, γt (Q − NQ [u1 ]) ≥ 3 = γt (Q) − 1. Hence by (3.1) and Lemma 3.1(ii), γt (Z (a) ) ≥ γt (Z (a−1) ) + γt (Q) − 1. Consequently γt (Z (a) ) ≥ (γt (B1 ) + 3(a − 1)) + 4 − 1 = γt (B1 ) + 3a by the induction assumption. This together with Lemma 3.4 implies that (i) and (ii) hold. Note that (a−1)

(a)

Z (a) −NZ (a) [z2 ] ≃ (Z (a−1) •(Q−NQ [u2 ]))(z2

, u1 : x). We clearly have

γt ((Q − NQ [u2 ]) − NQ [u1 ]) ≥ 2 = γt (Q − NQ [u2 ]) − 1 by Lemma 3.3(i). Hence by (3.1), Lemma 3.1(ii) and the induction assumption, γt (Z (a) − (a)

NZ (a) [z2 ]) ≥ γt (Z (a−1) ) + γt (Q − NQ [u2 ]) − 1 = γt (B1 ) + 3(a − 1) + 3 − 1 = γt (B1 ) + 3a − 1. This proves (iii).



(a)

Lemma 3.6 We have γt (Z2 (B1 , B2 ; b1 , b2 )) ≥ γt (B1 ) + γt (B2 ) + 3a − 1.

12

(a)

(a)

Proof. Let Z (a) = Z1 (B1 ; b1 ). Recall that Z2 (B1 , B2 ; b1 , b2 ) = (Z (a) • (a)

(a+1)

G(a+1) )(z2 , z1

(a)

: z (a) ) ≃ (Z (a) • B2 )(z2 , b2 : x). By Lemma 3.5,

(a)

γt (Z (a) −NZ (a) [z2 ]) ≥ γt (Z (a) )−1. By the choice of B2 , γ(B2 −NB2 [b2 ]) ≥ (a)

γt (B2 ) − 1. Hence by Lemmas 3.1(ii) and 3.5, γt (Z2 (B1 , B2 ; b1 , b2 )) ≥ γt (Z (a) ) + γt (B2 ) − 1 = γt (B1 ) + 3a + γt (B2 ) − 1, as desired.



(a)

Lemma 3.7 Let Z (a) = Z2 (B1 , B2 ; b1 , b2 ). Then γt (Z (a) −v) ≤ γt (B1 )+ γt (B2 ) + 3a − 2 for every v ∈ V (Z (a) ). (0)

Proof. Recall that Z2 (B1 , B2 ; b1 , b2 ) ≃ (B1 • B2 )(b1 , b2 : b). By Lem(0)

mas 3.1(i) and 3.6, γt (Z2 (B1 , B2 ; b1 , b2 )) = γt (B1 ) + γt (B2 ) − 1. Hence (0)

Z2 (B1 , B2 ; b1 , b2 ) is γt -critical by Lemma 3.1(iii). Consequently the lemma holds for a = 0. Thus we may assume that a ≥ 1. (a+1)

(0)

Case 1: v ∈ (V (G(0) ) − {z2 }) ∪ {z (0) } or v ∈ (V (G(a+1) ) − {z1 {z

(a)

}) ∪

}. (0)

Without loss of generality, we may assume that v ∈ (V (G(0) )−{z2 })∪ {z (0) }. If v ̸= z (0) , let S0′ be a γt -set of G(0) − v; if v = z (0) , let S0′ (0)

(0)

be a γt -set of G(0) − z2 . If z2 (0) if z2 ̸∈ (1) {z1 }).

∈ S0′ , let S0 = (S0′ − {z2 }) ∪ {z (0) }; (0)

S0′ , let S0 = S0′ . Let S1 be a γt -set of G(1) − (NG(1) [z2 ] ∪ (1)

By Lemma 3.3(i),(iii), |S1 | = 2. For each 2 ≤ i ≤ a, there (i)

(i)

exists a γt -set Si of G(i) − NG(i) [z2 ] which contains z1 by Lemma 3.3(ii). By Observation 1.3(ii), there exists a γt -set Sa+1 of G(a+1) which contains ∪ (a+1) (i) z1 . Then S = (( 0≤i≤a+1 Si )−{z1 | 2 ≤ i ≤ a+1})∪{z (i) | 1 ≤ i ≤ a} is a total dominating set of Z (a) − v. We also have |S| = (γt (B1 ) − 1) + 2 + ∑ (i) (i) − NG(i) [z2 ]) + γt (B2 ) = γt (B1 ) + γt (B2 ) + 3a − 2. Hence 2≤i≤a γt (G γt (Z (a) − v) ≤ γt (B1 ) + γt (B2 ) + 3a − 2. (0)

(a+1)

Case 2: v ̸∈ (V (G(0) ) − {z2 }) ∪ (V (G(a+1) ) − {z1

}) ∪ {z (0) , z (a) }.

Let 0 ≤ i0 ≤ a−1 be the integer such that dZ (a) (z (0) , z (i0 ) ) ≤ dZ (a) (z (0) , v) < dZ (a) (z (0) , z (i0 +1) ). Replacing the roles of G(0) and G(a+1) by each other if necessary, we may assume dZ (a) (v, z (i0 ) ) < dZ (a) (v, z (i0 +1) ). If v ̸= z (i0 ) , let (i +1)

Si0 +1 be a γt -set of (G(i0 +1) − NG(i0 +1) [z2 0 be a γt -set of

]) − v; if v = z (i0 ) , let Si0 +1

(i +1) (i +1) (G(i0 +1) − NG(i0 +1) [z2 0 ]) − z1 0 .

13

By Lemma 3.3(i),(iii),

(i +1)

|Si0 +1 | = 2. If v ̸= z (i0 ) , then the eccentricity of z1 0 (i +1) NG(i0 +1) [z2 0 ]) − v (i +1) clearly have z1 0 ̸∈

i0 + 2 ≤ i ≤ a, there

in (G(i0 +1) −

(i +1) is 3, and hence z1 0 ̸∈ Si0 +1 ; if v = z (i0 ) , we (i0 +1) Si0 +1 . Thus z ̸∈ Si0 +1 in either case. For each (i) exists a γt -set Si of G(i) − NG(i) [z2 ] which contains

(i)

z1 by Lemma 3.3(ii). By Observation 1.3(ii), there exists a γt -set Sa+1 of (a+1)

G(a+1) which contains z1

.

Subcase 2.1: i0 = 0. (0)

Let S0 be a γt -set of G(0) − z2 . Then S = ((

∪

(i)

0≤i≤a+1

Si ) − {z1 | 2 ≤

i ≤ a + 1}) ∪ {z (i) | 1 ≤ i ≤ a} is a total dominating set of Z (a) − v. We ∑ (i) also have |S| = (γt (B1 ) − 1) + 2 + 2≤i≤a γt (G(i) − NG(i) [z2 ]) + γt (B2 ) = γt (B1 ) + γt (B2 ) + 3a − 2. Hence γt (Z (a) − v) ≤ γt (B1 ) + γt (B2 ) + 3a − 2. Subcase 2.2: 1 ≤ i0 ≤ a − 1. By Observation 1.3(ii), there exists a γt -set S0 of G(0) which con(0)

tains z2 . For each 1 ≤ i ≤ i0 − 1, there exists a γt -set Si of G(i) − (i)

(i)

NG(i) [z1 ] which contains z2

by Lemma 3.3(ii). Let Si0 be a γt -set of

(i ) (G − NG(i0 ) [z1 0 ]) − ∪ (i) ( 0≤i≤a+1 Si − ({z2 | 0

By Lemma 3.3(iii), |Si0 | = 2. Then S =

(i0 )

(i ) z2 0 .

(i)

≤ i ≤ i0 − 1} ∪ {z1 | i0 + 2 ≤ i ≤ a + 1})) ∪ {z (i) |

0 ≤ i ≤ i0 − 1 or i0 + 1 ≤ i ≤ a} is a total dominating set of Z (a) − v. ∑ (i) We also have |S| = γt (B1 ) + 1≤i≤i0 −1 γt (G(i) − NG(i) [z1 ]) + 2 + 2 + ∑ (i) (i) − NG(i) [z2 ]) + γt (B2 ) = γt (B1 ) + γt (B2 ) + 3a − 2. i0 +2≤i≤a γt ((G Hence γt (Z (a) − v) ≤ γt (B1 ) + γt (B2 ) + 3a − 2, as desired.



By Observation 1.3(i) and Lemmas 3.6 and 3.7, we get the following theorem. Theorem 3.8 Let a ≥ 0 be an integer. For each i = 1, 2, let Bi be a connected γt -critical graph with δ(Bi ) ≥ 2, and let bi ∈ V (Bi ) be a (a)

vertex with γt (Bi − NBi [bi ]) ≥ γt (Bi ) − 1. Then Z2 (B1 , B2 ; b1 , b2 ) is a (γt (B1 ) + γt (B2 ) + 3a − 1)-γt -critical graph with diameter max{eccB1 (b1 ) + eccB2 (b2 ) + 5a, diam(B1 ), diam(B2 )}. In the remainder of this section, we construct candidates for B1 and B2 , and apply Theorem 3.8 to them. We first construct 3-γt -critical graphs

14

with diameter 3. Let m ≥ 2 be an integer and, for i = 1, 2, let C (i) = (i) (i)

(i)

(i)

(1)

x1 x2 · · · x5m x1 be a cycle of order 5m. For each 1 ≤ j ≤ 5, let Xj (1) {xl | l ≡ j(mod (2) (2) X1 = {xl | l ≡

l ≡ 5(mod 5)},

5)} and 1(mod

(2) X4

(2) Xj

=

(2) {xl

(2) 5)}, X2 (2) {xl | l

=

=

| l ≡ 2j − 1(mod 5)}. Note that

(2) {xl

(2)

| l ≡ 3(mod 5)}, X3 (2) X5 ′

(2)

= {xl

|

(2) {xl | l (1) ∈ Xi , x′

≡ 2(mod 5)} and = ≡ ∪ ∪ 4(mod 5)}. Let Y = {y1 , y2 }. Let E1 = 1≤i≤5 ( j̸=i {xx | x ∈ ∪ (2) (i) Xj }) and E2 = i=1,2 {yi x | x ∈ V (C )}. Let Gm be the graph defined =

by V (Gm ) = V (C (1) ) ∪ V (C (2) ) ∪ Y and E(Gm ) = E(C (1) ) ∪ E(C (2) ) ∪ E1 ∪ E2 . The graph Gm is depicted in Figure 2. In the figure, two solid lines indicate that for each i = 1, 2, all edges between yi and C (i) are present; dotted lines indicate that no edge between the two sets joined by a dotted line is present, and all other edges between C (1) and C (2) are present; dashed lines indicate that for each i = 1, 2, all edges inside C (i) are present except for a perfect matching between the two sets joined by a dashed line. Proposition 3.9 Let m ≥ 2 be an integer. Then Gm is 3-γt -critical, and eccGm (y1 ) = diam(Gm ) = 3. Proof. By the construction of Gm , eccGm (y1 ) = diam(Gm ) = dGm (y1 , y2 ) = 3. (1)

(1)

(2)

First we prove that γt (Gm ) = 3. Since {x1 , x4 , x1 } is a total dominating set of Gm , γt (Gm ) ≤ 3. Suppose that γt (Gm ) = 2, and let S be a γt -set. Since dGm (y1 , y2 ) = 3 and S γt -dominates Y , |S ∩ V (C (1) )| = (1)

(2)

|S ∩ V (C (2) )| = 1. Write S ∩ V (C (1) ) = {xj1 } and S ∩ V (C (2) ) = {xj2 }. Let j1′ and j2′ be integers with xj1 ∈ Xj ′ and xj2 ∈ Xj ′ . By the defi(1)

(1) 1

(2)

(2) 2

nition of S, xj1 xj2 ∈ E(Gm ), and hence j1′ ̸= j2′ . If j2′ ≡ j1′ − 1(mod 5) (1) (2)

or j2′ ≡ j1′ + 1(mod 5), then one of the vertices in NC (1) (xj1 ) is not γt (1)

dominated by S; if j2′ ≡ j1′ − 2(mod 5) or j2′ ≡ j1′ + 2(mod 5), then one of (2)

the vertices in NC (2) (xj2 ) is not γt -dominated by S. Consequently S is not a total dominating set of Gm , which is a contradiction. Thus γt (Gm ) = 3. 15

(2)

(1)

X1

(1)

X2

(1)

X3

(1)

X4

X5

(1)

X5

C (1)

C (2)

X1

(2)

X2 y1

(2)

X3

(2)

X4

(2)

Figure 2: Graph Gm

16

y2

Next we prove that γt (Gm −v) = 2 for any v ∈ V (Gm ). Let v ∈ V (Gm ). Case 1: v ∈ Y . (3−i)

Write v = yi . Then {x1

(3−i)

, x4

} is a total dominating set of Gm −yi .

Hence γt (Gm − v) = 2. Case 2: v ∈ (V (C (1) ) ∪ V (C (2) )). (i)

Let i ∈ {1, 2} and j ∈ {1, · · · , 5} be integers with v ∈ Xj . Let v ′ ∈ NC (i) (v) and v ′′ ∈

(3−i) . Xj

Then {v ′ , v ′′ } is a total dominating set of

Gm − v. Hence γt (Gm − v) = 2. 

Therefore Gm is a 3-γt -critical graph.

Proposition 3.10 Let m ≥ 2 be an integer. Then γt (Gm − NGm [y1 ]) ≥ γt (Gm ) − 1. Proof. We have γt (Gm − NGm [y1 ]) ≥ 2 = γt (Gm ) − 1.



Next we construct 4-γt -critical graphs with diameter 4. Let m ≥ 2 be (i)

an integer. For each 1 ≤ i ≤ 3, let Xi = {xj,l | 1 ≤ j ≤ 2, 1 ≤ l ≤ m}. ∪ ∪ (i) (3) Let Y = {y1 , y2 }. Let E1 = i=1,2 {xyi | x ∈ Xi }, E2 = i=1,2 {xj,l xj ′ ,l′ | ∪ (i) (i) (3) (3) (j, l) ̸= (j ′ , l′ )}, E3 = i=1,2 {x1,l x2,l | 1 ≤ l ≤ m} and E4 = {xj,l xj ′ ,l′ | l ̸= l′ }. Let Hm be the graph defined by V (Hm ) = (

∪

Xi ) ∪ Y

1≤i≤3

and

∪

E(Hm ) =

Ei

1≤i≤4

(see Figure 3). Proposition 3.11 Let m ≥ 2 be an integer. Then Hm is 4-γt -critical, and eccHm (y1 ) = diam(Hm ) = 4. Proof. By the construction of Hm , eccHm (y1 ) = diam(Hm ) = dHm (y1 , y2 ) = 4.

17

x1,1

(1)

x1,1

(3)

x1,1

x2,1

(1)

x2,1

(3)

x2,1

x1,2

(1)

x1,2

(3)

x1,2

x2,2

(1)

x2,2

(3)

x2,2

x1,m

(1)

x1,m

(3)

x1,m

(1)

x2,m

(3)

x2,m

y1

x2,m

X1

X3 Figure 3: Graph Hm

18

(2)

(2)

(2)

y2

(2)

(2)

(2)

X2

(1)

(2)

(3)

(3)

First we prove that γt (Hm ) = 4. Since {x1,1 , x1,1 , x1,2 , x2,2 } is a total dominating set of Hm , γt (Hm ) ≤ 4. Suppose that γt (Hm ) ≤ 3, and let S be a γt -set. Recall that dHm (y1 , y2 ) = 4. Since S γt -dominates Y , we have S ∩ X1 ̸= ∅ and S ∩ X2 ̸= ∅. Since |S| ≤ 3, there exists a vertex in S which γt -dominates both a vertex in S ∩ X1 and a vertex in S ∩ X2 , (1)

and hence |S ∩ X1 | = |S ∩ X2 | = |S ∩ X3 | = 1. Write S ∩ X1 = {xj,l } (3)

(3)

(1)

(1) (3)

and S ∩ X3 = {xj ′ ,l′ }. Since xj ′ ,l′ γt -dominates xj,l , xj,l xj ′ ,l′ ∈ E(Hm ), and hence (j, l) ̸= (j ′ , l′ ). If l ̸= l′ , then ′

(1) xj ′ ,l′

is not γt -dominated by S, a

′

contradiction. Thus l = l , and hence j = 3 − j. By the same argument, (2)

(3)

we get S ∩ X2 = {xj,l }. This implies that xj,l is not γt -dominated by S, a contradiction. Thus γt (Hm ) = 4. Next we prove that γt (Hm −v) ≤ 3 for any v ∈ V (Hm ). Let v ∈ V (Hm ). Case 1: v ∈ Y . (3−i)

(3)

(3)

Write v = yi . Then {x1,1 , x1,2 , x2,2 } is a total dominating set of Hm − yi . Hence γt (Hm − v) ≤ 3. Case 2: v ∈ X1 ∪ X2 . Write v = xj,l . Let l′ ∈ {1, · · · , m} − {l}. Then {x1,l′ , x3−j,l , xj,l } is a (i)

(i)

(3−i)

(3)

(i)

total dominating set of Hm − xj,l . Hence γt (Hm − v) ≤ 3. Case 3: v ∈ X3 . (3)

(1)

(2)

(3)

Write v = xj,l . Then {xj,l , xj,l , x3−j,l } is a total dominating set of Hm −

(3) xj,l .

Hence γt (Hm − v) ≤ 3, as desired.



Proposition 3.12 Let m ≥ 2 be an integer. Then γt (Hm − NHm [y1 ]) ≥ γt (Hm ) − 1. Proof. Suppose that γt (Hm − NHm [y1 ]) = 2. Let S be a γt -set of Hm − NHm [y1 ]. Since S γt -dominates y2 , S ∩ X2 ̸= ∅. Since there is no vertex in X2 which is adjacent to every vertex in X3 , we see that S ⊆ X2 ∪ X3 . Since each vertex in X2 γt -dominates only one vertex in X2 , it follows that (2)

(3)

|S ∩ X2 | = |S ∩ X3 | = 1. Write S = {xj,l , xj ′ ,l′ }. By the definition of S, xj,l xj ′ ,l′ ∈ E(Hm − NHm [y1 ]), and hence (j, l) ̸= (j ′ , l′ ). If l ̸= l′ , then (2) (3)

xj ′ ,l′ is not γt -dominated by S, a contradiction. Thus l = l′ and hence (2)

19

j ′ = 3 − j. However, xj,l is not γt -dominated by S, a contradiction. Thus (3)

γt (Hm − NHm [y1 ]) ≥ 3 = γt (Hm ) − 1.



We construct one more family of candidates Lm for B1 and B2 . The construction of the graph Lm is based on the following graph R defined in [2]. Let C = s1 s2 s3 s4 s1 be a cycle of order 4, and let U = {a1 , a2 , u1 , u2 , v}. (i) (i) (i) (i)

For each i = 1, 2, let P (i) = t3 t1 t2 t4 be a path of order 4. Let E1 = ∪ (i) ′ ′ (1) ), t′ ∈ V (P (2) )}, E3 = i=1,2 {sj tj ′ | j ̸= j }, E2 = {tt | t ∈ V (P ∪ ∪ (1) (1) (2) (2) i=1,2 {ai x | x ∈ {si , s3 , s4 , t3 , t4 , t3 , t4 }}, E4 = i=1,2 {ui x | x ∈ V (P (1) ) ∪ V (P (2) ) ∪ {ai }} and E5 = {vs | s ∈ V (C)}. Let R be the graph defined by V (R) = V (C) ∪ V (P (1) ) ∪ V (P (2) ) ∪ U and E(R) = E(C) ∪ E(P (1) ) ∪ E(P (2) ) ∪ (

∪

Ei )

1≤i≤5

(see Figure 4). Proposition 3.13 ([2]) The graph R is 3-γt -critical, and eccR (u1 ) = diam(R) = dR (u1 , v) = 3. Proposition 3.14 We have γt (R − NR [u1 ]) ≥ γt (R) − 1. Proof. We have γt (R − NR [u1 ]) ≥ 2 = γt (R) − 1.



Let Lm = (R • Gm )(u1 , y1 : u′ ). Proposition 3.15 Let m ≥ 2 be an integer. Then Lm is 5-γt -critical, and eccLm (v) = diam(Lm ) = 6. Proof. It is easy to see that eccLm (v) = diam(Lm ) = dLm (v, y2 ) = 6. By Lemma 3.1(i),(ii) and Propositions 3.9, 3.10, 3.13 and 3.14, γt (Lm ) = γt (R) + γt (Gm ) − 1 = 5. Hence Lm is 5-γt -critical by Lemma 3.1(iii) and Propositions 3.9 and 3.13.



20

s1

a1 (1)

t3

u1

(1)

t1

(1)

s4

t2

(1)

v

t4

(2)

t3

(2)

s3

t1

(2)

u2

t2

(2)

t4 s2

a2 Figure 4: Graph R

21

Proposition 3.16 Let m ≥ 2 be an integer. Then γt (Lm − NLm [v]) ≥ γt (Lm ) − 1. Proof. Since the distance between y2 and u2 in Lm − NLm [v] is 5, there is no total dominating set of Lm − NLm [v] having cardinality at most 3. Hence γt (Lm − NLm [v]) ≥ 4 = γt (Lm ) − 1.



We are now ready to construct examples which show the sharpness of the bound in Theorem 1.1. (a)

(a)

(a)

For m ≥ 2 and a ≥ 0, let Z3,1 (m) = Z2 (Gm , Gm ; y1 , y1 ), Z3,2 (m) = (a)

(a)

(a)

Z2 (Gm , Hm ; y1 , y1 ) and Z3,3 (m) = Z2 (Gm , Lm ; y1 , v). Proposition 3.17 Let m ≥ 2 and a ≥ 0 be integers. (a)

(i) The graph Z3,1 (m) is a (3a+5)-γt -critical graph with diameter ⌊ 5(3a+5)−7 ⌋. 3 (a)

(ii) The graph Z3,2 (m) is a (3a+6)-γt -critical graph with diameter ⌊ 5(3a+6)−7 ⌋. 3 (a)

(iii) The graph Z3,3 (m) is a (3a+7)-γt -critical graph with diameter ⌊ 5(3a+7)−7 ⌋. 3 Proof. (a)

(i) By Theorem 3.8 and Propositions 3.9 and 3.10, Z3,1 (m) is a (3a + 5)γt -critical graph with diameter 5a + 6 = ⌊ 5(3a+5)−7 ⌋. 3 (a)

(ii) By Theorem 3.8 and Propositions 3.9, 3.10, 3.11 and 3.12, Z3,2 (m) is a (3a + 6)-γt -critical graph with diameter 5a + 7 = ⌊ 5(3a+6)−7 ⌋. 3 (a)

(iii) By Theorem 3.8 and Propositions 3.9, 3.10, 3.15 and 3.16, Z3,3 (m) is a (3a + 7)-γt -critical graph with diameter 5a + 9 = ⌊ 5(3a+7)−7 ⌋. 3  Proposition 3.18 Let k ≥ 3 be an integer. Then there exist infinitely many connected graphs G such that G is k-γt -critical and such that G has diameter 3 if k = 3, and has diameter ⌊ 5k−7 3 ⌋ if k ≥ 4.

22

Proof. Note that Propositions 3.9, 3.11 and 3.17 hold for all integers m ≥ 2. Therefore if k = 3, then graphs Gm are desired graphs by Proposition 3.9. If k = 4, then graphs Hm are desired graphs Proposition 3.11. Thus we may assume that k ≥ 5. If k ≡ 2(mod 3), then letting k = 3a + 5 (a)

(a ≥ 0), we see that graphs Z3,1 (m) are desired graphs Proposition 3.17(i); if k ≡ 0(mod 3), then letting k = 3a + 6 (a ≥ 0), we see that graphs (a)

Z3,2 (m) are desired graphs by Proposition 3.17(ii); if k ≡ 1(mod 3), then (a)

letting k = 3a + 7 (a ≥ 0), we see that graphs Z3,3 (m) are desired graphs by Proposition 3.17(iii).



References [1] R. Diestel, “Graph Theory” (4th edition), Graduate Texts in Mathematics 173, Springer (2010). [2] W. Goddard, T.W. Haynes, H.A. Henning and L.C. van der Merwe, The diameter of total domination vertex critical graphs, Discrete Math. 286 (2004) 255–261. [3] M.A. Henning and N.J. Rad, On total domination vertex critical graphs of high connectivity, Discrete Appl. Math. 157 (2009) 1969– 1973. [4] M.A. Henning and A. Yeo, Perfect matchings in total domination critical graphs, Graphs Combin. 27 (2011) 685–701. [5] C. Wang, Z. Hu and X. Li, A constructive characterization of total domination vertex critical graphs, Discrete Math. 309 (2009) 991–996.

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