J Comb Optim (2008) 16: 68–80DOI 10.1007/s10878-007-9099-8

On the upper total domination number of Cartesianproducts of graphs

Paul Dorbec · Michael A. Henning · Douglas F. Rall

Published online: 15 August 2007© Springer Science+Business Media, LLC 2007

Abstract In this paper we continue the investigation of total domination in Cartesianproducts of graphs first studied in (Henning, M.A., Rall, D.F. in Graphs Comb. 21:63–69, 2005). A set S of vertices in a graph G is a total dominating set of G if every vertexin G is adjacent to some vertex in S. The maximum cardinality of a minimal totaldominating set of G is the upper total domination number of G, denoted by Γt (G).We prove that the product of the upper total domination numbers of any graphs G

and H without isolated vertices is at most twice the upper total domination numberof their Cartesian product; that is, Γt (G)Γt (H) ≤ 2Γt(G�H).

Keywords Graph products · Upper domination number · Upper total dominationnumber

Research of M.A. Henning supported in part by the South African National Research Foundation andthe University of KwaZulu-Natal.

P. Dorbec (�) · M.A. HenningSchool of Mathematical Sciences, University of KwaZulu-Natal, Pietermaritzburg 3209, SouthAfricae-mail: paul.dorbec@ujf-grenoble.fr

M.A. Henninge-mail: henning@ukzn.ac.za

P. DorbecInstitut Fourier Organization UJF-ERTé “Maths à modeler”, 100 rue des Maths, 38402,St Martin d’Hères, France

D.F. RallDepartment of Mathematics, Furman University, Greenville, SC 29613, USAe-mail: drall@herky.furman.edu

J Comb Optim (2008) 16: 68–80 69

1 Introduction

Total domination in graphs was introduced by Cockayne et al. (1980) and is nowwell studied in graph theory (see, for example, Archdeacon et al. 2004; Favaron andHenning 2003; Henning 2000). The literature on this subject has been surveyed anddetailed in the two books by Haynes et al. (1998a, 1998b).

Let G = (V ,E) be a graph with vertex set V and edge set E, and with no isolatedvertex. For sets S,T ⊆ V , S dominates T if every vertex in T \ S is adjacent to avertex of S, while S totally dominates T if every vertex in T is adjacent to a vertex ofS. If S dominates V , then S is called a dominating set, abbreviated DS, of G, while ifS totally dominates V , then S is called a total dominating set, abbreviated TDS, of G.Every graph without isolated vertices has a TDS, since S = V is such a set. The totaldomination number of G, denoted by γt (G), is the minimum cardinality of a TDS,while the upper total domination number of G, denoted by Γt (G), is the maximumcardinality of a minimal TDS of G. A minimal TDS of cardinality Γt(G) we call aΓt(G)-set. The domination number γ (G) and the upper domination number Γ (G)

of G are defined analogously.For notation and graph theory terminology we in general follow (Haynes et al.

1998a). Specifically, let G = (V ,E) be a graph, and let v be a vertex in V . The openneighborhood of v is N(v) = {u ∈ V | uv ∈ E} and the closed neighborhood of v

is N [v] = {v} ∪ N(v). For a set S ⊆ V , its open neighborhood is the set N(S) =∪v∈SN(v) and its closed neighborhood is the set N [S] = N(S) ∪ S. Hence for setsS,T ⊆ V , the set S dominates T if T ⊆ N [S], while S totally dominates T if T ⊆N(S).

Let G = (V ,E) be a graph and let S ⊆ V and v ∈ S. A vertex w ∈ V \ S isan S-external private neighbor of v if N(w) ∩ S = {v}; and the S-external privateneighborhood of v, denoted epn(v, S), is the set of all S-external private neighborsof v. A vertex w ∈ S is an S-internal private neighbor of v if N(w) ∩ S = {v};and the S-internal private neighborhood set of v, denoted ipn(v, S), is the set of allS-internal private neighbors of v. The S-private neighborhood pn(v, S) of v ∈ S isdefined by pn(v, S) = {u ∈ V | N(u)∩S = {v}}. Thus, epn(v, S) = pn(v, S)∩(V \S)

and ipn(v, S) = pn(v, S) ∩ S, while pn(v, S) = epn(v, S) ∪ ipn(v, S). Each vertex inpn(v, S) is called an S-private neighbor of v. The following property of minimalTDSs is established in (Cockayne et al. 1980).

Observation 1 (Cockayne et al. 1980) Let S be a TDS in a graph G with no isolatedvertices. Then, S is a minimal TDS of G if and only for each v ∈ S, epn(v, S) �= ∅ orpn(v, S) = ipn(v, S) �= ∅.

We denote the degree of a vertex v in G by dG(v), or simply by d(v) if the graph G

is clear from context. For a subset S ⊆ V , we let dS(v) denote the number of verticesin S that are adjacent to v. In particular, dV (v) = dG(v). The subgraph induced bythe set S is denoted by G[S].

70 J Comb Optim (2008) 16: 68–80

2 Vizing-like conjectures

For graphs G and H , the Cartesian product G�H is the graph with vertex setV (G) × V (H) where two vertices (u1, v1) and (u2, v2) are adjacent if and only ifeither u1 = u2 and v1v2 ∈ E(H) or v1 = v2 and u1u2 ∈ E(G). In 1968 Vizing (1968)made the following conjecture which he first posed as a question in 1963.

Vizing’s Conjecture For any graphs G and H , γ (G)γ (H) ≤ γ (G�H).

Vizing’s Conjecture hasn’t been settled (although the conjecture has been proventrue for large classes of graphs) and is the most famous open problem involving dom-ination in graphs. The best general upper bound to date on the product of the domi-nation numbers of two graphs in terms of their Cartesian product is due to Clark andSuen (2000) who showed that for any graphs G and H , γ (G)γ (H) ≤ 2γ (G�H).

Several variations of Vizing’s Conjecture have been studied. In 1996, Nowakowskiand Rall (1996) made the following Vizing-like conjecture for the upper dominationof Cartesian products of graphs.

Nowakowski–Rall Conjecture For any graphs G and H , Γ (G)Γ (H) ≤ Γ (G�H).

A beautiful proof of the Nowakowski–Rall Conjecture was recently found byBrešar (2005). In (Henning and Rall 2005), a version of Vizing’s conjecture for totaldomination is studied. For certain classes of graphs G without isolated vertices (in-cluding the class of trees of order at least 2) and for any graph H without isolatedvertices, it is shown that γt (G)γt (H) ≤ 2γt (G�H). However it remains an openquestion (see Henning and Rall 2005) to determine whether the product of the totaldomination numbers of two general graphs without isolated vertices is at most twicethe total domination number of their Cartesian product.

Open Question (Henning and Rall 2005) For any graphs G and H without isolatedvertices, is it true that γt (G)γt (H) ≤ 2γt (G�H)?

Excellent surveys of domination in Cartesian products have been written by Hart-nell and Rall (1998) and Imrich and Klavžar (2000).

3 Main results

Our aim in this paper is to prove a Vizing-like bound for the upper total dominationnumber of Cartesian products of graphs. We shall prove:

Theorem 1 Let G and H be connected graphs of order at least 3 with Γt (G) ≥Γt(H). Then,

Γt(G)(Γt (H) + 1) ≤ 2Γt(G�H),

and this bound is sharp.

J Comb Optim (2008) 16: 68–80 71

Theorem 2 For any graphs G and H with no isolated vertices,

Γt (G)Γt (H) ≤ 2Γt (G�H),

with equality if and only if both G and H are a disjoint union of copies of K2.

3.1 Proofs of Theorems 1 and 2

We begin with the following trivial result.

Lemma 1 If G is any graph with no isolated vertex, then Γt (G)Γt (K2) ≤2Γt(G�K2), with equality if and only if G is a disjoint union of copies of K2.

Proof Let V (K2) = {u,v}. Then, V (G) × {v} is a minimal TDS of G�K2, andso Γt(G�K2) ≥ |V (G)| ≥ Γt(G) = 1

2Γt (G)Γt (K2). Further, if Γt (G�K2) =12Γt (G)Γt (K2), then we must have equality throughout this inequality chain. In par-ticular, Γt (G) = |V (G)|, implying that G is a disjoint union of copies of K2. �

Our key lemma is the following result.

Lemma 2 Every Γt (G)-set contains as a subset a minimal dominating set S suchthat |S| ≥ 1

2Γt (G) and epn(v, S) ≥ 1 for each v ∈ S.

Proof Let D be a Γt (G)-set. Let

A = {v ∈ D : |epn(v,D)| ≥ 1},B = {v ∈ D \ A : dA(v) ≥ 1},C = D \ (A ∪ B).

Then, D = A ∪ B ∪ C. Let v ∈ B ∪ C. Then, epn(v,D) = ∅, and so by Observa-tion 1, |ipn(v,D)| ≥ 1. Let v′ ∈ ipn(v,D), and so v′ ∈ D and N(v′)∩D = {v}. Thus,dD(v′) = 1. Since v′ is adjacent only to v and v /∈ A, we have that v′ /∈ B .

Suppose v ∈ C. Then, v′ /∈ A since dA(v) = 0, and so v′ ∈ C. Hence, epn(v′,D) =∅, and so, by Observation 1, |ipn(v′,D)| ≥ 1. However, the vertex v is the only neigh-bor of v′ in G[D], implying that ipn(v′,D) = {v}; that is, N(v) ∩ D = {v′}, and sodD(v) = 1. Hence if C �= ∅, then G[C] = |C|

2 K2 and for each v ∈ C, dD(v) = 1. Wecall two adjacent vertices in G[C] partners in C. Let (X,Y ) be partite sets in thegraph G[C], and so each vertex in X (resp., in Y ) is adjacent in G[D] only to itspartner in Y (resp., in X). For each x ∈ X, let yx be the partner of x in C.

Suppose v ∈ B . Since G[C] = |C|2 K2 and each vertex of C has degree 1 in G[D],

the vertex v is not adjacent to any vertex of C. In particular, v′ ∈ A ∪ B . As shownearlier, v′ /∈ B . Hence, v′ ∈ A. Thus, for each v ∈ B , pn(v,D) ⊆ A. This in turnimplies that |A| ≥ ∑

v∈B |ipn(v,D)| ≥ |B|.Let U = V (G) \ (D ∪ N(A) ∪ N(X)) be the set of vertices in V (G) \ D not

dominated by A or X in G. Since D is a TDS of G, the set U is dominated by B ∪Y .

72 J Comb Optim (2008) 16: 68–80

Let BY be a minimum subset of B ∪ Y that dominates U . Thus for each v ∈ BY , wehave that |epn(v,BY ) ∩ U | ≥ 1.

We now consider the set S = A ∪ BY ∪ X. Since B ⊆ N(A) and Y ⊆ N(X), theset S dominates D. By construction, the set S also dominates V (G) \ D. Thus, S

dominates V (G), but S is not necessarily a minimal DS of G. We now construct aminimal DS of G from the DS S as follows. We consider the vertices in X in turn,and for each vertex x ∈ X we systematically delete x from S if epn(x, S) = ∅ ateach stage in the resulting set S. (Observe that if the partner yx ∈ Y of x in C is notin S, then yx ∈ epn(x, S), and so x is not deleted from S.) Let X∗ be the resultingsubset of vertices of X that belong to the set S upon the completion of this process.Thus, S = A ∪ BY ∪ X∗ and, by construction, |epn(v, S)| ≥ 1 for each v ∈ S. Ifx ∈ X \ X∗, then the partner of x in C is in the set S, implying that S dominates C.Since B ⊆ N(A), the set B is dominated by S. Therefore the set S dominates D. Byconstruction, the set S also dominates V (G) \ D. Hence, S is a minimal DS of G.

It remains to show that |S| ≥ 12Γt (G). For every vertex x ∈ X, the set S contains

at least one of x and its partner in C, whence |S ∩C| ≥ |X| = 12 |C|. As shown earlier,

|A| ≥ |B|. Hence,

|S| ≥ |A| + |S ∩ C| ≥ |A| + 1

2|C| ≥ 1

2(|A| + |B| + |C|) = 1

2|D| = 1

2Γt (G), (1)

as claimed. �

Recall Theorem 1.

Theorem 1 Let G and H be connected graphs of order at least 3 with Γt (G) ≥Γt(H). Then, Γt (G)(Γt (H) + 1) ≤ 2Γt(G�H), and this bound is sharp.

Proof By Lemma 2, there exists a minimal dominating set S of G such that |S| ≥12Γt (G) and epn(v, S) ≥ 1 for each v ∈ S. For each w ∈ V (G), let Hw be the copyof H in G�H corresponding to w; that is, Hw = (G�H)[{w} × V (H)]. For eachz ∈ V (H), let Gz be the copy of G in G�H corresponding to z; that is, Gz =(G�H)[V (G) × {z}].

We now consider the set D = S × V (H). Since S dominates V (G), the set D

dominates G�H . Further, for every vertex w ∈ S, the vertices in {w} × V (H) aretotally dominated by their neighbors in Hw , and so D is a TDS of G�H . We showthat D is in fact a minimal TDS of G�H . Let v ∈ D. Then, v = (w, z) for somevertices w ∈ S and z ∈ V (H). Let v′ = (w′, z) where w′ ∈ epn(w,S) in G. Then,v′ ∈ epn(v,D) in G�H . Hence in G�H , |epn(v,D)| ≥ 1 for every v ∈ D. Thus, byObservation 1, D is a minimal TDS of G�H , and so Γt(G�H) ≥ |D|. Since H is aconnected graph of order at least 3, we have that |V (H)| ≥ Γt (H)+1. Consequently,

Γt (G�H) ≥ |D| = |S| × |V (H)| ≥ 1

2Γt (G)(Γt (H) + 1),

which establishes the desired upper bound in the statement of the theorem. That thebound in Theorem 1 is sharp is shown by Lemma 3. �

J Comb Optim (2008) 16: 68–80 73

Fig. 1 A daisy with k petals

Theorem 2 is an immediate consequence of Lemma 1 and Theorem 1.For k ≥ 2, we call a connected graph that can be constructed from k ≥ 2 disjoint

copies of K3 by identifying a set of k vertices, one from each K3, into one vertex adaisy with k petals (see Fig. 1).

Lemma 3 For all k ≥ 2, if G and H are both daisies with k petals, then

Γt (G)(Γt (H) + 1) = 2Γt (G�H).

Proof For notation convenience, we may assume G = H . Let V (G) = {a1, a2,

. . . , ak , b1, b2, . . . , bk, c} where c is the vertex of degree 2k in G and E(G − c) ={a1b1, a2b2, . . . , akbk}. Then, G has order 2k + 1 and Γt (G) = 2k. Thus,Γt(G)(Γt (H) + 1) = 2k(2k + 1). We show that Γt (G�H) ≤ k(2k + 1). For1 ≤ i, j ≤ k, we denote the sets Ri , Cj , Bi,j , Xi and Yj by

Ri = {(ai, v), (bi, v) : v ∈ V (H)},Cj = {(v, aj ), (v, bj ) : v ∈ V (G)},Bi,j = {(ai, aj ), (ai, bj ), (bi, aj ), (bi, bj )},Xi = {(ai, c), (bi, c)},Yj = {(c, aj ), (c, bj )}.

Thus, Xi ⊂ Ri , Yj ⊂ Cj , and Bi,j = Ri ∩ Cj . Further, let

X =k⋃

i=1

Xi and Y =k⋃

i=1

Yi.

Let V = V (G�H) and let S be a minimal TDS of G�H . We show that |S| ≤k(2k + 1). We proceed further by proving four claims about the set S.

Claim 1 For some i and j with 1 ≤ i, j ≤ k, if Xi ⊂ S and |S ∩Bi,j | ≥ 2 or if Yj ⊂ S

and |S ∩ Bi,j | ≥ 2, then |S| < k(2k + 1).

Proof By symmetry, we may assume that Xi ⊂ S and |S ∩ Bi,j | ≥ 2 for some i andj with 1 ≤ i, j ≤ k. By Observation 1, each vertex of S has an S-private neighbor.Since Xi dominates Ri , the only possible S-private neighbors of vertices in S ∩ Bi,j

are in Yj . Thus, |S ∩ Bi,j | = 2 with one vertex u of S ∩ Bi,j having (c, aj ) as its S-private neighbor and with the other vertex v of S ∩Bi,j having (c, bj ) as its S-private

74 J Comb Optim (2008) 16: 68–80

neighbor. Hence the only vertices of S that are adjacent to a vertex of Yj are u and v.Since every vertex in S ∩ (Cj ∪ {(c, c)}) is adjacent to a vertex of Yj , it follows thatS ∩ (Cj ∪ {(c, c)}) = {u,v}. Hence in order to dominate the vertices in Cj \ Bi,j ,we must have that X� ⊂ S for every � = 1,2, . . . , k, and so, X ⊂ S. Since X totallydominates V \ Y , the S-private neighborhood of every vertex in S \ X is containedin the set Y . Since each vertex is the S-private neighbor of at most one vertex in S,it follows that |S \ X| ≤ |Y |, implying that |S| = |X| + |S \ X| ≤ |X| + |Y | = 4k <

k(2k + 1). �

By Claim 1, we may assume that for all i and j with 1 ≤ i, j ≤ k, if Xi ⊂ S, then|S ∩ Bi,j | ≤ 1 and if |S ∩ Bi,j | ≥ 2, then |S ∩ Xi | ≤ 1, for otherwise, |S| ≤ k(2k + 1)

as desired. By symmetry, we may assume an identical statement holds when Xi isreplaced by Yj .

Claim 2 For all pairs (i, j) where 1 ≤ i, j ≤ k, except for possibly one pair, we have|S ∩ Bi,j | ≤ 2.

Proof Suppose that |S ∩ Bi,j | ≥ 3 and |S ∩ Br,s | ≥ 3 for two distinct pairs (i, j)

and (r, s) where 1 ≤ i, j, r, s ≤ k. If Bi,j ⊂ S, then no vertex of Bi,j wouldhave an S-private neighbor, contradicting Observation 1. Hence, |S ∩ Bi,j | = 3and |S ∩ Br,s | = 3. Without loss of generality, we may assume that S ∩ Bi,j ={(ai, aj ), (ai, bj ), (bi, aj )}. Thus the S-private neighbors of (ai, bj ) and (bi, aj ) canonly be (c, bj ) and (bi, c), respectively. Consequently, for every v ∈ V (G) \ {ai, c},we have that (v, bj ) /∈ S, while for every w ∈ V (H) \ {aj , c}, we have that (bi,w) /∈S. It follows that r �= i and s �= j .

We may also assume, without loss of generality, that S ∩Br,s = {(ar , as), (ar , bs),

(br , as)}. Thus the S-private neighbors of (ar , bs) and (br , as) are necessarily (c, bs)

and (br , c), respectively. We now consider the vertex (ai, as), which has degree 4 inG�H . The four neighbors of (ai, as) are (c, as), (ai, bs), (ai, c) and (bi, as). Sinceboth (c, as) and (ai, bs) are adjacent to (c, bs), which is the S-private neighbor of(ar , bs) ∈ S, we have that (c, as) /∈ S and (ai, bs) /∈ S. And since (ai, c) and (bi, as)

are adjacent to (bi, c), which is the S-private neighbor of (bi, aj ) ∈ S, we have that(ai, c) /∈ S and (bi, as) /∈ S. But then the vertex (ai, as) is not totally dominated byS, contradicting the fact that S is a TDS of G�H . �

Claim 3 For some i and j with 1 ≤ i, j ≤ k, if |S ∩ Bi,j | ≥ 3, then |S| ≤ k(2k + 1).

Proof As in the proof of Claim 2, |S ∩ Bi,j | = 3 and we may assume that S ∩ Bi,j ={(ai, aj ), (ai, bj ), (bi, aj )}. Thus the S-private neighborhood of (ai, bj ) consists onlyof the vertex (c, bj ), while the S-private neighborhood of (bi, aj ) consists only ofthe vertex (bi, c). Consequently, S ∩ {(c, c), (c, aj ), (ai, c)} = ∅. Further for everyv ∈ V (G) \ {ai, c}, we have that (v, bj ) /∈ S, while for every v ∈ V (H) \ {aj , c}, wehave that (bi, v) /∈ S. In particular, S ∩ (Xi ∪ Yj ) ⊆ {(bi, c), (c, bj )}.

If {(bi, c), (c, bj )} ⊂ S, then {(bi, c), (c, bj )} dominates the four neighbors of(ai, aj ) ∈ S, implying that (ai, aj ) has no S-private neighbor, contradicting Observa-tion 1. Hence, |S ∩ (Xi ∪ Yj )| ≤ 1. Without loss of generality, we may assume thatS ∩ Yj = ∅, and so S ∩ Xi ⊆ {(bi, c)}.

J Comb Optim (2008) 16: 68–80 75

Suppose (bi, c) ∈ S and (c, c) is an S-private neighbor of (bi, c). Then, |S ∩ (X ∪Y)| = |{(bi, c)}| = 1. By Claim 2, |S ∩ Br,s | ≤ 2 for every pair (r, s) distinct from(i, j) where 1 ≤ r, s ≤ k. Since there are k2 − 1 such pairs, and since k ≥ 2, it followsthat |S| ≤ 2k2 + 2 ≤ k(2k + 1), which is the desired bound. Hence we may assumethat if (bi, c) ∈ S, then (c, c) is not an S-private neighbor of (bi, c).

We show first that |S ∩ Ri | ≤ 2k + 1. By Claim 2, |S ∩ Bi,s | ≤ 2 for all s =1,2, . . . , k with s �= j , and so |S ∩Ri | ≤ |S ∩Xi |+ 2(k − 1)+ 3 = |S ∩Xi |+ 2k + 1.Hence if S ∩ Xi = ∅, then |S ∩ Ri | ≤ 2k + 1 as claimed. Thus we may assumethat (bi, c) ∈ S. By an earlier assumption, (c, c) is not an S-private neighbor of(bi, c). Thus the vertex (bi, as) or (bi, bs) is an S-private neighbor of (bi, c) forsome s �= j where 1 ≤ s ≤ k. Without loss of generality, we may assume that(bi, as) is an S-private neighbor of (bi, c). Thus, (ai, as) /∈ S. As observed earlier,(bi, as) /∈ S and (bi, bs) /∈ S. Consequently, |S ∩ Bi,s | ≤ 1, implying that |S ∩ Ri | ≤1 + 2(k − 2) + 3 + 1 = 2k + 1, as claimed.

Suppose that |S ∩ Rr | ≤ 2k for all r �= i where 1 ≤ r ≤ k. By Claim 2,|S ∩ Bi′,j ′ | ≤ 2 for all pairs (i′, j ′) where 1 ≤ i′, j ′ ≤ k and (i′, j ′) �= (i, j). If onthe one hand |S ∩ Ys | ≤ 1 for all s �= j where 1 ≤ s ≤ k, then |S ∩ Y | ≤ k − 1, and so

|S| = |S ∩ Y | +k∑

�=1

|S ∩ R� |

≤ (k − 1) + 2k(k − 1) + (2k + 1)

= k(2k + 1),

which is the desired bound. If on the other hand, Ys ⊂ S for some s �= j where1 ≤ s ≤ k, then the neighborhood of each vertex of Bi,s is dominated by Ys ∪{(ai, aj ), (bi, aj )} ⊂ S, implying that S ∩ Bi,s = ∅. Hence letting k′ = |{s ∈{1,2, . . . , k} | Ys ⊂ S}|, we have that k′ ≥ 1 and |S ∩ Ri | ≤ |S ∩ Xi | + |S ∩ Bi,j | +2(k −k′ −1) ≤ 2(k −k′ +1). Thus, since |S ∩Rr | ≤ 2k for all r �= i where 1 ≤ r ≤ k,

|S| = |S ∩ Y | +k∑

�=1

|S ∩ R� |

≤ (k + k′ − 1) + 2k(k − 1) + 2(k − k′ + 1)

= k(2k + 1) − k′ + 1

≤ k(2k + 1),

which is the desired bound. Hence we may assume that |S ∩ Rr | ≥ 2k + 1 for somer �= i where 1 ≤ r ≤ k, for otherwise the desired bound follows. By Claim 2, |S ∩Br,s | ≤ 2 for all s = 1,2, . . . , k. Hence, |S ∩ Xr | ≥ 1. If Xr ⊂ S, then, by our earlierassumptions, |S ∩ Br,s | ≤ 1 for all s = 1,2, . . . , k. But then since k ≥ 2, |S ∩ Rr | ≤k +2 < 2k +1, a contradiction. Hence, |S ∩Xr | = 1. It follows that |S ∩Rr | = 2k +1and that |S ∩ Br,s | = 2 for all s = 1,2, . . . , k. Without loss of generality, we mayassume that S ∩ Xr = {(ar , c)}. Since (v, bj ) /∈ S for v ∈ {ar, br}, we must have thatS ∩ Br,j = {(ar , aj ), (br , aj )}.

76 J Comb Optim (2008) 16: 68–80

Suppose that (c, c) is not an S-private neighbor of (ar , c). Since (br , aj ) ∈ S

and (br , aj ) is adjacent to (br , c), for some s �= j where 1 ≤ s ≤ k the ver-tex (ar , as) or (ar , bs) is an S-private neighbor of (ar , c). Without loss of gen-erality, we may assume that (ar , as) is an S-private neighbor of (ar , c). Then,(ar , bs) /∈ S and (br , as) /∈ S, and so S ∩ Br,s = {(ar , as), (br , bs)}. The S-privateneighbors of (ar , as) and (br , bs) can only be (c, as) and (c, bs), respectively, imply-ing that S ∩ Cs = {(ar , as), (br , bs)}. But then the vertex (ai, as) is not dominatedby S, a contradiction. Hence, (c, c) is the only S-private neighbor of (ar , c). Thus,|S ∩ (X∪Y)| = |{(ar , c)}| = 1. By Claim 2, |S ∩Br,s | ≤ 2 for every pair (r, s) distinctfrom (i, j) where 1 ≤ r, s ≤ k. Since there are k2 − 1 such pairs, and since k ≥ 2, itfollows that |S| ≤ 2k2 + 2 ≤ k(2k + 1), which is the desired bound. �

By Claim 3, we may assume that for all pairs (i, j) where 1 ≤ i, j ≤ k, we have|S ∩Bi,j | ≤ 2, for otherwise |S| ≤ k(2k +1) as desired. Hence by our earlier assump-tions, |S ∩ Ri | ≤ 2k + 1 and |S ∩ Cj | ≤ 2k + 1.

Claim 4 For some i with 1 ≤ i ≤ k, if |S ∩ Ri | ≥ 2k + 1, then |S| ≤ k(2k + 1).

Proof Suppose |S ∩ Ri | ≥ 2k + 1 for some i with 1 ≤ i ≤ k. On the one hand, ifXi ⊂ S, then, by our earlier assumptions, |S ∩ Bi,j | ≤ 1 for all j with 1 ≤ j ≤ k,whence |S ∩Ri | ≤ k+2. On the other hand, if S ∩Xi = ∅, then |S ∩Ri | ≤ 2k. In bothcases (recall that k ≥ 2), |S ∩ Ri | < 2k + 1, a contradiction. Hence, |S ∩ Xi | = 1. If|S ∩ Bi,j | ≤ 1 for some j with 1 ≤ j ≤ k, then |S ∩ Ri | ≤ 2k, a contradiction. Hence,|S ∩ Bi,j | = 2 for all j where 1 ≤ j ≤ k. Therefore, by our earlier assumptions,|S ∩ Yj | ≤ 1, and so |S ∩ Y | ≤ k.

Without loss of generality, we may assume that (ai, c) ∈ S. Assume (c, c) is anS-private neighbor of (ai, c). Then, |S ∩ (X ∪ Y)| ≤ 2. By assumption, |S ∩ Br,s | ≤ 2for every pair (r, s) where 1 ≤ r, s ≤ k. Since there are k2 such pairs, and since k ≥ 2,it follows that |S| ≤ 2k2 + 2 ≤ k(2k + 1), which is the desired bound. Hence we mayassume that (c, c) is not an S-private neighbor of (ai, c).

Assume (bi, c) is an S-private neighbor of (ai, c). Then, (c, c) /∈ S and S ∩{(bi, v) | v ∈ V (H)} = ∅, implying that for all j with 1 ≤ j ≤ k, S ∩ Bi,j ={(ai, aj ), (ai, bj )}. Let j ∈ {1,2, . . . , k}. We show that |S ∩ Cj | ≤ 2k. If S ∩ Yj = ∅,then this is immediate. Hence, assume |S ∩ Yj | = 1. Without loss of generality, wemay assume that (c, aj ) ∈ S. Then the S-private neighbor of (ai, aj ) is the vertex(c, aj ), implying that S ∩{(v, aj ) | v ∈ V (G)} = {(ai, aj ), (c, aj )}. For some r wherer �= i and 1 ≤ r ≤ k, the vertex (ar , aj ) or (br , aj ), say (ar , aj ), is an S-private neigh-bor of (c, aj ). But then (ar , bj ) /∈ S, implying that |S ∩ Br,j | ≤ 1 and therefore that|S ∩ Cj | ≤ 2k, as claimed.

Assume that (ai, as) or (ai, bs), say (ai, as), is an S-private neighbor of (ai, c)

for some s with 1 ≤ s ≤ k. Then, S ∩ Bi,s = {(ai, as), (bi, bs)} and (c, as) /∈ S. TheS-private neighbor of (bi, bs) is (c, bs), while (ai, c) or (c, as) are the only possibleS-private neighbors of (ai, as). Assume (c, as) is an S-private neighbor of (ai, as).Then, S ∩ Cs = {(ai, as), (bi, bs)}, and so

|S| = |S ∩ X| +k∑

j=1

|S ∩ Cj |

J Comb Optim (2008) 16: 68–80 77

≤ (2k − 1) + |S ∩ Cs | + (k − 1)(2k + 1)

≤ (2k − 1) + 2 + (k − 1)(2k + 1)

= k(2k + 1),

which is the desired bound. Hence we may assume that (ai, c) is the unique S-private neighbor of (ai, as), for otherwise the desired bound follows. This impliesthat for all j �= s with 1 ≤ j ≤ k, we have that S ∩ Bi,j = {(bi, aj ), (bi, bj )}, whichin turn implies that S ∩ Yj = ∅ and |S ∩ Cj | ≤ 2k. We show that |S ∩ Cs | ≤ 2k.If (c, bs) /∈ S, then S ∩ Ys = ∅ and the result is immediate. Assume, then, that(c, bs) ∈ S. Since (c, bs) is the S-private neighbor of (bi, bs), S ∩ {(v, bs) | v ∈V (G)} = {(bi, bs), (c, bs)}. For some r where r �= i and 1 ≤ r ≤ k, the vertex (ar , bs)

or (br , bs), say (ar , bs), is the S-private neighbor of (c, bs). But then (ar , as) /∈ S,implying that |S ∩ Br,s | ≤ 1 and therefore that |S ∩ Cs | ≤ 2k, as claimed. Hence, forall j with 1 ≤ j ≤ k, we have |S ∩ Cj | ≤ 2k.

Hence with our assumption that (c, c) is not an S-private neighbor of (ai, c), wehave shown that we may assume that for all j with 1 ≤ j ≤ k, |S ∩ Cj | ≤ 2k andthat |S ∩ {(ai, aj ), (bi, aj )}| = 1 and |S ∩ {(ai, bj ), (bi, bj )}| = 1. Hence if Xm ⊂ S

for some m �= i with 1 ≤ m ≤ k and x ∈ (S ∩ Rm) \ Xm, then x would have noS-private neighbor, a contradiction. Consequently, if Xm ⊂ S for some m �= i, then|S ∩ Rm| = |Xm| = 2, and so

|S| = |S ∩ Y | +k∑

�=1

|S ∩ R� |

≤ k + |S ∩ Rm| + (k − 1)(2k + 1)

= k + 2 + (k − 1)(2k + 1)

< k(2k + 1),

which is the desired bound. Therefore we may assume that |S ∩ Xm| ≤ 1 for all m

with 1 ≤ m ≤ k. Thus, |S ∩ X| ≤ k, and so

|S| = |S ∩ X| +k∑

j=1

|S ∩ Cj | ≤ k + k(2k) = k(2k + 1),

which is the desired bound. �

By Claim 4, for all i with 1 ≤ i ≤ k, we may assume that |S ∩ Ri | ≤ 2k for other-wise |S| ≤ k(2k + 1) as desired. By symmetry, for all j with 1 ≤ j ≤ k, we may alsoassume that |S ∩ Cj | ≤ 2k.

Claim 5 For some r and s with 1 ≤ r, s ≤ k, if Xr ∪ Ys ⊂ S, then |S| ≤ k(2k + 1).

Proof Suppose Xr ∪ Ys ⊂ S for some r and s with 1 ≤ r, s ≤ k. Let x = |{i : 1 ≤i ≤ k,Xi ⊂ S}|, and let y = |{ j : 1 ≤ j ≤ k,Yj ⊂ S}|. Then, x ≥ 1 and y ≥ 1. By

78 J Comb Optim (2008) 16: 68–80

assumption, for all i and j with 1 ≤ i, j ≤ k, if Xi ⊂ S or if Yj ⊂ S, then |S ∩Bi,j | ≤ 1. Further, if Xi ⊂ S and Yj ⊂ S and x ∈ S ∩ Bi,j , then x would have noS-private neighbor, a contradiction. Consequently, if Xi ∪ Yj ⊂ S, then S ∩ Bi,j = ∅.By assumption, |S ∩Bi,j | ≤ 2 for every pair (i, j) where 1 ≤ i, j ≤ k. Since there arek2 such pairs and k ≥ 2, it follows that if (c, c) /∈ S, then

|S| = |S ∩ X| + |S ∩ Y | +∑

1≤i,j≤k

|S ∩ Bi,j |

≤ (k + x) + (k + y) + (k − x)y + (k − y)x + 2(k − x)(k − y)

≤ k(2k + 1) + (x + y − kx − ky + k)

≤ k(2k + 1) + (x + y)(1 − k) + k

≤ k(2k + 1) + 2(1 − k) + k

= k(2k + 1) + 2 − k

≤ k(2k + 1),

which is the desired bound. Assume, then, that (c, c) ∈ S. Then for all i and j with1 ≤ i, j ≤ k, if Xi ⊂ S or if Yj ⊂ S and x ∈ S ∩ Bi,j , then x would have no S-privateneighbor, a contradiction. Consequently, if Xi ⊂ S or if Yj ⊂ S, then S ∩ Bi,j = ∅.We may assume that x ≤ y. Since (c, c) ∈ S must have an S-private neighbor, weknow that x < k. Since x + y ≥ 2, it follows that

|S| ≤ 1 + (k + x) + (k + y) + 2(k − x)(k − y)

≤ 2k2 + 2k + (x + y)(1 − k) + x(y − k) + y(x − k) + 1

≤ k(2k + 1) + k + 2(1 − k) − y + 1

≤ k(2k + 1) + 2 − k

≤ k(2k + 1),

which is the desired bound. �

By Claim 5, we may assume that there is no pair (r, s) with 1 ≤ r, s ≤ k such thatXr ∪ Ys ⊂ S, for otherwise |S| ≤ k(2k + 1) as desired. In particular, |S ∩ X| ≤ k or|S ∩ Y | ≤ k. We may assume that |S ∩ Y | ≤ |S ∩ X|. Thus, |S ∩ Y | ≤ k. If |S ∩ Y | ≤k − 1, or if |S ∩ Y | = k and (c, c) /∈ S, then

|S| = |S ∩ (Y ∪ {(c, c)}| +k∑

i=1

|S ∩ Ri | ≤ k + 2k2 = k(2k + 1),

which is the desired bound. Thus we may assume that (c, c) ∈ S, and that k = |S ∩Y | ≤ |S ∩ X| and |S ∩ Yj | = 1 for all j with 1 ≤ j ≤ k. Let x = |{i : 1 ≤ i ≤ k,Xi ⊂S}|, and let x′ = |{i : 1 ≤ i ≤ k,S ∩ Xi = ∅}|. Since |S ∩ X| ≥ k, we have that x ≥ x′.For all i and j with 1 ≤ i, j ≤ k, if Xi ⊂ S, then S ∩Bi,j = ∅. Further, if |S ∩Xi | = 1and |S ∩ Bi,j | = 2, then there would be a vertex in S ∩ Bi,j that has no S-private

J Comb Optim (2008) 16: 68–80 79

neighbor, a contradiction. Consequently, if |S∩Xi | = 1, then |S∩Bi,j | ≤ 1. It followsthat

|S| = |{(c, c)}| + |S ∩ X| + |S ∩ Y | +∑

1≤i,j≤k

|S ∩ Bi,j |

≤ 1 + (k + x − x′) + k + 2x′k + k(k − x − x′)

≤ k(2k + 1) + k + 1 − k2 + (x − x′)(1 − k)

≤ k(2k + 1) + k + 1 − k2

≤ k(2k + 1),

as desired. Hence we have shown that for every minimal TDS S of G�H , we have|S| ≤ k(2k + 1), whence Γt (G�H) ≤ k(2k + 1) = 1

2Γt (G)(Γt (H) + 1). Howeveras established in the proof of the upper bound of Theorem 1, Γt (G)(Γt (H) + 1) ≤2Γt(G�H). Consequently, Γt (G)(Γt (H) + 1) = 2Γt (G�H). �

4 Closing remark

For all graphs G with no isolated vertices, γ (G) ≤ γt (G). However, for each pos-itive integer k, there exist graphs G and H such that Γ (G) − Γt(G) = k andΓt(H) − Γ (H) = k. As an immediate consequence of Lemma 2, we have the fol-lowing relationship between the upper total domination number of a graph with noisolated vertex and its upper domination number.

Corollary 1 For any graph G of order n with no isolated vertex,

(2

n − 1

)

Γ (G) ≤ Γt (G) ≤ 2Γ (G).

Proof Let D be a Γt(G)-set. By Lemma 2, there is a minimal DS S of G such thatS ⊆ D and |S| ≥ 1

2Γt (G). Hence, Γ (G) ≥ |S| ≥ 12Γt(G). This establishes the upper

bound. The lower bound follows from the observations that Γ (G) ≤ n − 1 (withequality if and only if G is a star K1,n−1) and Γt(G) ≥ 2. �

We close with the remark that if G is a graph achieving equality in the bound ofCorollary 1, then it can be readily deduced from the proof of Lemma 2 that the graphG has the following three properties: (i) Every Γt (G)-set induces a subgraph thatconsists of disjoint copies of K2; (ii) Γ (G) = β(G) (where β(G), the independencenumber of G, is the maximum size of an independent set of vertices of G); (iii) Forevery Γt (G)-set S, every vertex of V (G) \ S is contained in a common triangle withtwo vertices of S. However we have yet to find a nice characterization of graphs G

satisfying Γt (G) = 2Γ (G).

80 J Comb Optim (2008) 16: 68–80

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