- Email: [email protected]
On the global total k -domination number of graphs
Discrete Applied Mathematics (
)
–
Contents lists available at ScienceDirect
Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
On the global total k-domination number of graphs Sergio Bermudo a , Abel Cabrera Martínez b, *, Frank A. Hernández Mira b , José M. Sigarreta b a b
Department of Economics, Quantitative Methods and Economic History, Universidad Pablo de Olavide, ES-41013 Seville, Spain Faculty of Mathematics, Autonomous University of Guerrero. Carlos E. Adame 5, Col. La Garita, 39350 Acapulco, Guerrero, Mexico
article
info
Article history: Received 30 September 2017 Received in revised form 23 April 2018 Accepted 13 May 2018 Available online xxxx Keywords: Global total k-domination Global total domination Total k-domination
a b s t r a c t A subset D of vertices of a graph G is a global total k-dominating set if D is a total k-dominating set of both G and G. The global total k-domination number of G is the g minimum cardinality of a global total k-dominating set of G and it is denoted by γkt (G). In this paper we introduce this concept and we begin the study of its mathematical properties. Specifically, we prove that the complexity of the decision problem associated to g the computation of the value γkt (G) is NP-complete. Moreover, we present tight bounds for this parameter and we obtain relationships between the global total k-domination number of G and the total k-domination numbers of G and G, and other parameters of the graph G. © 2018 Elsevier B.V. All rights reserved.
1. Introduction Let G = (V , E) be a simple graph of order |V | = n and size |E | = m. For a nonempty set D ⊆ V , and a vertex v ∈ V , ND (v ) denotes the set of neighbors of v in D and ND [v] = ND (v ) ∪ {v}. The degree of v in D will be denoted by δD (v ) = |ND (v )|. Analogously, N D (v ) denotes the set of non neighbors of v in D, N D [v] = N D (v ) ∪ {v} and δ D (v ) denotes the cardinality of the set N D (v ) (i.e. δ D (v ) = |N D (v )|). For short, we will often use N(v ), N [v], δ (v ), δ (v ), N(v ) and N [v] instead of NV (G) (v ), NV (G) [v], δV (G) (v ), δ V (G) (v ), N V (G) (v ) and N V (G) [v], respectively. The minimum and maximum degree of G will be denoted by δ (G) and ∆(G), respectively. The subgraph induced by D ⊆ V will be denoted by G[D] and the complement of the graph G will be denoted by G . If X and Y are two subsets of V , then we denote the set of all edges of G that join a vertex of X and a vertex of Y by E(X , Y ). A nonempty set D ⊆ V is a total k-dominating set in G if every vertex in V is adjacent to at least k vertices in D (or every vertex of V is totally k-dominated by D). The total k-domination number of G is the minimum cardinality among all total k-dominating sets in G and it is denoted by γkt (G). A γkt (G)-set is a total k-dominating set of cardinality γkt (G) (see [2–7]). A total dominating set is a total 1-dominating set, and the total domination number is the total 1-domination number, which is also denoted by γt (G). A nonempty set D ⊆ V is a global total k-dominating set (or GTkD set for short) if D is a total k-dominating set of both G g and G. The global total k-domination number of a graph G, denoted by γkt (G), is the minimum cardinality of a GTkD set of G. g g A γkt (G)-set is a global total k-dominating set of cardinality γkt (G). When k = 1, a global total 1-dominating set is a global total dominating set (see [1,8]). In order to have a GTkD set D, any vertex of V must be totally k-dominated by D in G and G, so 1 ≤ k ≤ min{δ (G), n − (∆(G) + 1)}. Hence, this is a necessary and sufficient condition to guarantee the existence of a GTkD set.
*
Corresponding author. E-mail addresses: [email protected] (S. Bermudo), [email protected] (A. Cabrera Martínez), [email protected] (F.A. Hernández Mira), [email protected] (J.M. Sigarreta). https://doi.org/10.1016/j.dam.2018.05.025 0166-218X/© 2018 Elsevier B.V. All rights reserved.
Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
2
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
Fig. 1. Complementary graphs C5 and C5 .
Notice that, if G1 , G2 , . . . , Gr , with r ≥ 2, are the connected components of a graph G, then any minimum GTkD set in G is formed by a minimum total k-dominating set in each subgraph Gi . That is stated in the following result. Proposition 1.1. Let G1 , G2 , . . . , Gr , with r ≥ 2, be the connected components of a graph G. Then
γktg (G) =
r ∑
γkt (Gi ).
i=1
⋃r ∑r g Proof. Let Di be a γkt (Gi )-set for i ∈ {1, . . . , r }. It is not difficult to see that i=1 Di is a GTkD set in G, so γkt (G) ≤ i=1 |Di | = ∑ r γ (G ). kt i i=1 g Now, let A be a γkt (G)-set. We observe that, for every i ∈ {1, . . . , r }, the vertex set V (Gi ) must be totally k-dominated by A, so it follows that |A ∩ V (Gi )| ≥ γkt (Gi ). Hence, γktg (G) = |A| ≥
r r ∑ ∑ |A ∩ V (Gi )| ≥ γkt (Gi ), i=1
i=1
which completes the proof. □ The following observation is an immediate consequence of the previous proposition. g
Observation 1.2. Let Kn1 ,n2 ,...,nm be a complete m-partite graph and let k + 1 ≤ n1 ≤ n2 . . . ≤ nm . Then, γkt (Kn1 ,n2 ,...,nm ) = m(k + 1). g
Clearly, the definition of a GTkD set gives us a direct lower bound for the parameter γkt (G). g
Observation 1.3. Let G be a graph, then γkt (G) ≥ max{γkt (G), γkt (G)}. As an example, we can observe the complementary graphs C5 and C5 in Fig. 1. We can check that the sets {v1 , v2 , v3 } and {v1 , v3 , v5 } are γt (C5 )-set and γt (C5 )-set, respectively. Hence γt (C5 ) = γt (C5 ) = 3. The set {v1 , v2 , v3 , v5 } is a GT1D set of C5 and we can check that there is not a set of vertices D, with cardinality three, such that D is a total dominating set of both C5 g g and C5 , thus {v1 , v2 , v3 , v5 } is a γ1t (C5 )-set and, consequently, γ1t (C5 ) = 4. If we take k = 2, V (C5 ) is an unique γ2t (C5 )-set g
and γ2t (C5 )-set. So γ2t (C5 ) = γ2t (C5 ) = γ2t (C5 ) = 5. Since two complementary graphs are tightly related by their edges, we can characterize one of them using the information about the other. Given a nonempty set D ⊆ V , D is a total k-dominating set of G if and only if every vertex in V is not adjacent to at least k vertices in D. This simple quotation shows a hint for the next observation.
Observation 1.4. Let G = (V , E) be a graph and let D ⊆ V . The set D is a GTkD if and only if δD (v ) ≥ k and δ D (v ) ≥ k, for every vertex v ∈ V . In this paper, we initiate the study of the global total k-domination number of a graph G. 2. Bounds for the global total k-domination number In concordance with Proposition 1.1, from now on, we only consider connected graphs such that their complements are also connected graphs. Given a graph G = (V , E), for every k ≤ min{δ (G), n − (∆(G) + 1)}, V is a GTkD set. Moreover, by Observation 1.4 we g deduce the following lower bound for γkt (G). g
Proposition 2.1. Let G be a graph with order n, then 2k + 1 ≤ γkt (G) ≤ n. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
3
The next theorems characterize graphs G having global total k-domination number equal to this lower and upper bound, respectively. For this purpose, given two graphs G and H, we define the k-join operation of G with H, as the disjoint union of G and H by adding edges joining each vertex of G with k or k + 1 vertices of H. We denote the k-join of G with H by G ◦k H and denote T [G ◦k H ] as the family of graphs obtained by this operation. g
Theorem 2.2. Let G be a graph, then γkt (G) = 2k + 1 if and only if G ∈ T [F ◦k H ], where H is a k-regular graph of order 2k + 1 and F is any graph. g
g
Proof. Firstly, we assume that γkt (G) = 2k + 1. Let D be a γkt (G)-set and we consider H = G[D] and F = G[V \ D]. By Observation 1.4, it is not difficult to check that H is a k-regular induced subgraph of G, and every υ ∈ V \ D satisfies that k ≤ δD (υ ) ≤ k + 1, so G ∈ T [F ◦k H ], where H is a k-regular graph of order 2k + 1. Now, we assume that G ∈ T [F ◦k H ], where H is a k-regular graph of order 2k + 1 and F is any graph. By Observation 1.4, g it is easy to see that V (H) is a GTkD set in G, so γkt (G) ≤ |V (H)| = 2k + 1, and using Proposition 2.1 it follows that V (H) is a g g γkt (G)-set, hence γkt (G) = 2k + 1. □ In order to present other results about the global total k-domination number we need to introduce some new terminology and notation. Given a graph G, we consider the set Ai (G) = {v ∈ V (G) : δ (v ) = i}, and we define the set Rk (G) as follows:
⎛ Rk (G) = ⎝
⎞ ⋃ v∈Ak (G)
⎛
⎞ ⋃
N(v )⎠ ∪ ⎝
(V \ N [w])⎠ .
w∈An−(k+1) (G)
Lemma 2.3. Let G be a graph and D be a GTkD set of G. Then Rk (G) ⊆ D. Proof. On one hand, if v ∈ Ak (G), since v must be k-dominated by D, then N(v ) ⊆ D. On the other hand, if w ∈ An−(k+1) (G), then |V \ N [w]| = k, since there must be k vertices in D which are not adjacent to w , we have that V \ N [w] ⊆ D. Consequently, Rk (G) ⊆ D. □ g
Corollary 2.4. Let G be a graph. Then, γkt (G) = |Rk (G)| if and only if Rk (G) is a GTkD set of G. In the following result we show a characterization when the global total k-domination number is equal to the order of the graph. g
Theorem 2.5. Let G be a graph of order n, then γkt (G) = n if and only if Rk (G) = V . g
Proof. Firstly, we assume that Rk (G) = V . Notice that, as V is a GTkD set in G, by Corollary 2.4 it follows that γkt (G) = n. g Now, kt (G) = n and we suppose that there exists a vertex x such that x ̸ ∈ Rk (G). This implies that ⋃ we assume that γ⋃ x ̸ ∈ v∈A (G) N(v ) and x ̸ ∈ w∈A (G) V \ N [w], so for every y ∈ N(x) it follows that δ (y) > k and for every z ∈ V \ N [x] k
n−(k+1)
it follows that δ (z) < n − (k + 1), consequently, δ (z) > k. Thus, by Observation 1.4, V \ {x} is a GTkD set in G, which is a contradiction. Hence, Rk (G) = V . □ The following corollary is directly obtained by Theorem 2.5. g
Corollary 2.6. If G is a k-regular or n − (k + 1)-regular graph of order n, then γkt (G) = n. g
Another interesting problem could consist in characterizing graphs G which satisfy that γkt (G) = γkt (G) = γkt (G). By g Corollary 2.6 we directly obtain that every k-regular graph G of order 2k + 1 satisfies γkt (G) = γkt (G) = γkt (G). Nevertheless, there exist many non-regular graphs satisfying the same property. To show this, we need the following result from [2]. Theorem 2.7. Let D be a total k-dominating set in a graph G. If δD (v ) = k for every v ∈ D and then |D| = γkt (G).
v∈V \D δD (v )
∑
≤ k|V \ D| + k − 1,
g
Proposition 2.8. For every even number k, there exists a non-regular graph G satisfying that γkt (G) = γkt (G) = γkt (G). Proof. Let H be a k-regular graph of order 2k + 1 and let G be a graph obtained from the graph H and another vertex, namely v , by joining the vertex v to exactly k vertices of V (H). On one hand, notice that V (H) is a GTkD set in G, so γktg (G) ≤ 2k + 1. g Also, using the lower bound in Proposition 2.1 it follows that γkt (G) = 2k + 1. On the other hand, using Theorem 2.7, we g obtain that V (H) is also a minimum total k-dominating set in G and G, so γkt (G) = γkt (G) = γkt (G). □ Given a graph G, it is natural to think that the global total k-domination number of G, the total k-domination number of G and the total k-domination number of G are related. The next result shows a relationship between these parameters. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
4
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
g
Fig. 2. Graph G2 which satisfies γ2t (G2 ) = γ2t (G2 ) + γ2t (G2 ).
Proposition 2.9. Let G be a graph, then
γkt (G) + γkt (G) 2
≤ γktg (G) ≤ γkt (G) + γkt (G).
Proof. The lower bound follows directly using Observation 1.3. On the other hand, let D be a γkt (G)-set and S be a γkt (G)-set, g note that D ∪ S is a GTkD set of G. Therefore, γkt (G) ≤ |D ∪ S | ≤ |D| + |S | = γkt (G) + γkt (G). □ Let us see that, for any k ≥ 2 there exists a graph Gk attaining the upper bound given in the above proposition. We consider the graph Gk with order n = (k + 1)2 formed by a complete graph Kk+1 of k + 1 vertices and k + 1 copies of the null graph Nk of k vertices. We denote V (Kk+1 ) = {v1 , v2 , . . . , vk+1 } and we denote the ith copy of Nk by Nki for every i ∈ {1, . . . , k + 1}. The edges of Gk are formed by joining the vertices of Nki with all vertices of V (Kk+1 ) except the vertex vi , for every i ∈ {1, . . . , k + 1} (see Fig. 2). the By Theorem 2.7, we have that V (Kk+1 ) is a γkt (Gk )-set, so γkt (Gk ) = k + 1. In Gk , each vertex vi of Kk+1 is only adjacent ⋃k+to 1 k vertices of the copy Nki , hence any total k-dominating set of Gk contains all copies Nki for i ∈ {1, . . . , k + 1}. Since i=1 Nki is a total k-dominating set of Gk , it is the unique γkt (Gk )-set, so γkt (Gk ) = k(k + 1). Finally, every vertex vi ∈ V (Kk+1 ) is adjacent j to a vertex u ∈ Nk (j ̸ = i) such that δ (u) = k, and every vertex u ∈ Nki satisfies u ∈ V (Gk ) \ N [vi ], where δ (vi ) = n − (k + 1), g then V (Gk ) = Rk (Gk ), consequently, by Theorem 2.5, it follows that γkt (Gk ) = (k + 1) + k(k + 1) = γkt (Gk ) + γkt (Gk ). The lower bound in Proposition 2.9 is attained by the graph given by Proposition 2.8. g We continue giving a relationship between γkt (G) and γkt (G), in which we use the maximum degree of the graph. Proposition 2.10. Let G be a graph, g
(i) If γkt (G) > ∆(G) + k, then γkt (G) = γkt (G). g (ii) If γkt (G) ≤ ∆(G) + k, then γkt (G) ≤ ∆(G) + k + 1. Proof. (i) Let D be a γkt (G)-set of G and we assume that |D| > ∆(G) + k. For every v ∈ V , since k ≤ δD (v ) ≤ ∆(G), there are g at least k vertices in D which are not adjacent to v , hence D is a GTkD set of G and, consequently, γkt (G) ≤ γkt (G). Therefore, g γkt (G) = γkt (G). (ii) Let D be a γkt (G)-set of G and we assume that |D| ≤ ∆(G) + k. Then we add some arbitrary vertices to D until we have a set D′ (D ⊂ D′ ) of cardinality ∆(G) + k + 1. Notice that, as D is a γkt (G)-set then D′ is a total k-dominating set of G, that is, δD′ (v ) ≥ k for every v ∈ V (G). Thus δD′ (v ) ≥ k for every v ∈ V (G). On the other hand, since δD′ (w) ≤ δ (w ) ≤ ∆(G) for every w ∈ V (G) and |D′ | = ∆(G) + k + 1, then δ D′ (w) ≥ k. Therefore, by Observation 1.4 it follows that D′ is a GTkD set, which g implies γkt (G) ≤ |D′ | = ∆(G) + k + 1. □ The upper bound given in the second part of this proposition is attained, for instance, in the graph Gk given after Proposition 2.9. g
Theorem 2.11. Let G be a graph, then γkt (G) ≤ γkt (G) ≤ γkt (G) + ∆(G). Proof. The lower bound follows directly from Observation 1.3. g g On the other hand, we suppose that γkt (G) > γkt (G) + ∆(G). Since γkt (G) ≥ k + 1 we have γkt (G) > ∆(G) + k + 1 and, g by Proposition 2.10, we obtain γkt (G) > ∆(G) + k. Using again Proposition 2.10, we conclude that γkt (G) = γkt (G), which is a contradiction. □ The upper bound given in this proposition is also attained, for instance, in the graph Gk given after Proposition 2.9, where
∆(Gk ) = k(k + 1) = γkt (Gk ).
Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
S. Bermudo et al. / Discrete Applied Mathematics (
g
Fig. 3. A graph G such that γ2t (G) =
)
–
5
n+1 . 2
g
Proposition 2.12. Let G be a graph, if γkt (G) = γkt (G) + ∆(G) then G ∈ T [H ◦k Kk+1 ], where H is some graph such that |V (H)| ≥ ∆(G). g
Proof. If γkt (G) = γkt (G) + ∆(G), by Proposition 2.10(i), it follows that γkt (G) ≤ ∆(G) + k. Moreover, by Proposition 2.10(ii) g we have γkt (G) = γkt (G) + ∆(G) ≤ ∆(G) + k + 1, that is, γkt (G) ≤ k + 1. Therefore, γkt (G) = k + 1 and every γkt (G)-set g D must satisfy G[D] = Kk+1 . Finally, since |V (G)| ≥ γkt (G) = k + 1 + ∆(G) and every vertex v ∈ V (G) \ D satisfies that k ≤ δD (v ) ≤ k + 1, then G ∈ T [H ◦k Kk+1 ] for some graph H such that |V (H)| ≥ ∆(G). □ g
Let us observe that the equality γkt (G) = γkt (G) + ∆(G) is satisfied, for instance, by the graph Gk given after Proposition 2.9, g but there exist graphs G ∈ T [H ◦k Kk+1 ], such that H is a graph with order |V (H)| ≥ ∆(G), satisfying γkt (G) < γkt (G) + ∆(G). i i For instance, the graph Gk given after Proposition 2.9 changing Nk by Nk+1 for every i ∈ {1, 2, . . . , k + 1}. g Now, we continue with some other bounds for γkt (G). Theorem 2.13. Let G be a graph of order n and size m. Then
γktg (G) ≥ max
{
2m + 2nk − n∆(G) −2m + 2nk + nδ (G) 2∆(G)
,
2(n − (δ (G) + 1))
}
.
Proof. Let D be a γkt (G)-set. Since every vertex in V \ D has at least k neighbors in D, we have (n − |D|)k ≤ E(D, V \ D) ≤ |D|∆(G)−E(D,V \D) −|D|) |D|(∆(G) − k) and E(V \ D, V \ D) ≤ (∆(G)−k)(n . Now, using that E(D, D) ≤ ≤ |D|∆(G)−2(n−|D|)k we obtain 2 2 m = E(D, D) + E(D, V \ D) + E(V \ D, V \ D) (∆(G) − k)(n − |D|) |D|∆(G) − (n − |D|)k + |D|(∆(G) − k) + , ≤ 2 2 which is equivalent to 2m ≤ 2|D|∆(G) − 2nk + n∆(G). Therefore, we obtain that |D| ≥ a γkt (G)-set S, we have |S | ≥
−2m+2nk+nδ (G) . 2(n−(δ (G)+1))
2m+2nk−n∆(G) . 2∆(G)
Analogously, if consider
Hence, by Observation 1.3 it follows the result. □
The lower bound in the above theorem is tight if we consider that k is an even number and G is a k-regular graph with g (2k+1)k cardinality 2k + 1. In this case, m = and γkt (G) = 2k + 1. 2 Theorem 2.14. Let G be a graph with order n. If k ≥ max g
{ ∆(G) 2
} g +1) , n−(δ(G) , then γkt (G) ≥ 2
n+1 . 2
∆(G)
Proof. Let D be a γkt (G)-set of G and let v ∈ D. Firstly, if k ≥ 2 , then δD (v ) ≥ δ(V \D) (v ), consequently, 2δD (v ) ≥ +1) δ(V \D) (v ) + δD (v ) = δ (v ). Analogously, if k ≥ n−(δ(G) , we obtain that 2δ D (v ) ≥ δ (V \D) (v ) + δ D (v ) = δ (v ). Using both 2 1 inequalities we obtain 2(|D| − 1) = 2(δD (v ) + δ D (v )) ≥ δ (v ) + δ (v ) = n − 1, thus |D| ≥ n+ . □ 2 The lower bound in the theorem above is tight if we consider that k is an even number and G is a 2k-regular graph with order n = 4k + 1 satisfying that G ∈ T [F ◦k H ], where H is a k-regular graph of order 2k + 1 and F is a graph of order 2k. By g (4k+1)+1 Theorem 2.2 we have γkt (G) = 2k + 1 = = n+2 1 . In Fig. 3 we show a 4-regular graph G ∈ T [F ◦2 H ] of order 9, where 2 F is the subgraph induced by the four black vertices and H is the subgraph induced by the white vertices. The following two theorems are well known (see [5]). Theorem 2.15. Let k ∈ N and suppose G = (V , E) be a graph of order n with minimum degree δ (G) ≥ k. Then γkt (G) ≥ this bound is sharp.
kn
∆(G)
and
Theorem 2.16. Let ( ) k ∈ N and suppose G = (V , E) be a graph of order n and size m with minimum degree δ (G) ≥ k. Then γkt (G) ≥ 2 n − mk and this bound is sharp. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
6
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
Using these theorems and Observation 1.3 we obtain the following corollaries. Corollary 2.17. Let G be a graph with order n. Then,
γktg (G) ≥ max
{
kn
}
kn
,
∆(G) n − (δ (G) + 1)
.
Corollary 2.18. Let G be a graph with order n and size m. Then,
{ ( ( )} m) n(n − 1) − 2m γktg (G) ≥ max 2 n − ,2 n − . k
2k
Using now Theorem 2.15 and Proposition 2.10 we obtain another corollary. ∆(G)(∆(G)+k+1)
Corollary 2.19. Let G be a graph with order n. If n ≥
k
g
, then γkt (G) = γkt (G).
3. The global total k-domination number for some particular graphs g
In this section we obtain the exact value of γkt (G) for some specific families of graphs, such as paths, cycles and cubic graphs. In order to do this, we need the following well-known values of the total k-domination number of paths and cycles. It is known (see [2])
γt (Pn ) = γt (Cn ) =
⎧ n ⎪ ⎨ +1 2
n
⎪ ⎩ ⌈ ⌉ 2
n ≡ 2 (mod 4) and
γ2t (Cn ) = n.
otherwise
The following proposition shows the global total k-domination number for paths and cycles. Proposition 3.1. (a) For any path Pn and cycle Cn with order n ≥ 4 it holds
⎧ 4 ⎪ ⎪ ⎨n + 1 γ1tg (Pn ) = γ1tg (Cn ) = 2 ⎪ ⎪ ⎩ ⌈n⌉ 2
if
n = 4, 5
if
n ≡ 2 (mod 4)
otherwise. g
(b) For any cycle Cn with order n ≥ 5 it holds γ2t (Cn ) = n. Proof. (a) The cases n = 4 and n = 5 are trivial. If n ≥ 6, the result is obtained by Proposition 2.10 and the total domination number of paths and cycles. (b) It is obtained by Corollary 2.6. □ The following lemma, which was proved in [1], will be used to get the global total k-domination numbers of cubic graphs. g
Lemma 3.2. For any graph G, γ1t (G) ≥ 4. Proposition 3.3. Let G be a cubic graph of order n and let S3 be the star graph of four vertices. Then, the following global total k-domination numbers hold:
g 1t (G)
γ
{ γt (G) 5 4
=
γ2tg (G) =
{
5
γ2t (G)
if γt (G) > 4 if γt (G) = 4 and for any γt (G)-set D of G is G[D] = S3 otherwise, if γ2t (G) ≤ 5 if γ2t (G) > 5
and
γ3tg (G) = n. g
Proof. First, let us prove the result for k = 1. By Proposition 2.10(i), if γt (G) > 4, then γ1t (G) = γt (G). If γt (G) ≤ 4, by g Proposition 2.10(ii) and Lemma 3.2 it follows that 4 ≤ γ1t (G) ≤ 5. We will consider three cases. Case 1. We assume that γt (G) = 2. By Theorem 2.15 we have n ≤ 6. Since G is a cubic graph, n is even and n ≥ ∆(G) + k + 1 = 5, thus n = 6. Let D = {v, w} be a γt (G)-set and let D′ = D ∪ {x, y}, where x and y are vertices such g that x ∈ N(v ) \ N(w ) and y ∈ N(w ) \ N(v ). It is clear that D′ is a GT1D set in G, so γ1t (G) ≤ 4. By Lemma 3.2 we conclude that g γ1t (G) = 4. Case 2. We assume that γt (G) = 3. By Theorem 2.15 we have n ≤ 9. Since G is a cubic graph, n is even and n ≥ ∆(G) + k + 1 = 5, thus n = 6 or n = 8. If D is a γt (G)-set, then the subgraph induced by D is a path P3 or a cycle Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
7
Fig. 4. Cubic graph C3 □P2 .
C3 . If the subgraph induced by D is a path P3 , v is the vertex in D such that δD (v ) = 2, and we take D′ = D ∪ {x}, where g g x ̸ ∈ D ∩ N(v ), it is easy to see that D′ is a GT1D set in G , so γ1t (G) ≤ 4. Again by Lemma 3.2, we conclude that γ1t (G) = 4. If the subgraph induced by D is a cycle C3 , then n = 6 and G is the Cartesian product C3 □P2 (see Fig. 4). It can be checked that g any set S of four vertices in G, such that the subgraph induced by S is a triangle-free graph, is a GT1D set in G, so γ1t (G) ≤ 4. g Therefore, γ1t (G) = 4. Case 3. We assume that γt (G) = 4. If D is a γt (G)-set of G, then the subgraph induced by D can be a path P4 , a cycle C4 , the union K2 ∪ K2 , a star S3 , or a star S3 with one extra edge (note that this induced subgraph G[D] cannot have two extra edges because, if this happened and v were the center of the star, the set D \ {v} would be a total dominating set). g g In the first three cases, D is also a GT1D set, so γ1t (G) ≤ 4. Again by Lemma 3.2, we have γ1t (G) = 4. If the subgraph induced by D is a star S3 with an extra edge, that is, D = {v1 , v2 , v3 , v4 } with δD (v1 ) = 3, δD (v2 ) = 2 = δD (v3 ) and δD (v4 ) = 1, then there exists a vertex u ∈ N(v4 ) \ {v1 } such that N(u) ∩ D = {v4 } (otherwise the set D \ {v4 } would be a total dominating g set in G, contradiction). In such a case, the set D′ = {v2 , v3 , v4 , u} is a GT1D set, so γ1t (G) = 4. If every γt (G)-set D of G satisfies g that the subgraph induced by D is a star S3 , then D cannot be a γ1t (G)-set because the center of the star S3 has no neighbor in g g g D. Hence, γ1t (G) > 4 and, since γ1t (G) ≤ 5, we conclude that γ1t (G) = 5. Note that this family of cubic graphs G (where every γt (G)-set D of G satisfies that the subgraph induced by D is a star S3 ) is not empty, for example, the Petersen graph belongs to this family. g Now, let us prove the result when k = 2. By Proposition 2.10(i), if γ2t (G) > 5, then γ2t (G) = γ2t (G). If γ2t (G) ≤ 5, by g Proposition 2.10(ii) and Proposition 2.1, it follows that 5 ≤ γ2t (G) ≤ 6. If γ2t (G) = 3, by Theorem 2.15 we obtain that n ≤ 4, contradiction. If γ2t (G) = 4 or γ2t (G) = 5, as G is a cubic graph and n is an even number, using again Theorem 2.15, we obtain g that n = 6. We can see that γ2t (G) < 6, because the graph G does not satisfy the characterization given in Theorem 2.5, then g γ2t (G) = 5. g Finally, the equality γ3t (G) = n is directly obtained by Corollary 2.6. □ 4. Complexity In this section we investigate the complexity of the following decision problem:
GLOBAL TOTAL k-DOMINATION PROBLEM (GTkD-PROBLEM for short) INSTANCE: A non trivial graph G and a positive integer r g PROBLEM: Deciding whether γkt (G) is less than r We will do the complexity analysis of the GTkD-PROBLEM making a reduction from another decision problem.
TOTAL k-DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γkt (G) is less than r The following result was proved in [5]. Theorem 4.1. For all k ≥ 1, the TOTAL k-DOMINATION PROBLEM is NP-complete. In order to present our complexity results we need to introduce a family of graphs. Fixed k, Hk is obtained from k copies of Kk+1 and another vertex, namely u, by joining the vertex u to one vertex of each copy Kk+1 . Given a graph G of order n and (1) (n) minimum degree δ (G) ≥ k, and n graphs Hk , . . . , Hk isomorphic to the graph Hk , the graph Gn,k is constructed by adding (i) (i) edges between the ith-vertex of G and the vertex u of the ith-graph Hk . We show an example in Fig. 5, where k = 2, H2 is the graph (I), G is the cycle C4 , and G4,2 is the graph showed in (II). Proposition 4.2. Let G be a graph of order ) n and k be a positive integer such that k ≤ δ (G). Then there exists a γkt (Gn,k )-set ⋃n ( (i) containing all vertices in i=1 Hk \ {u(i) } and no vertices in {u(1) , u(2) , . . . , u(n) }. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
8
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
Fig. 5. The graph G4,2 where G is a C4 graph.
(i)
Proof. Let A be a γkt (Gn,k )-set. For every i ∈ {1,(. . . , n} the vertices of V (Hk ) \ u(i) must be totally k-dominated by the set A, ) ⋃n (i) (i) so it can be checked that every vertex of i=1 V (Hk ) \ {u } belongs to A. Now, we suppose that u(j) ∈ A and we denote by
{
}
v⏐ j the( neighbor of u)⏐(j) in V (G), that is, N(u(j) ) ∩ V (G) = {vj }. Since the vertex vj must be totally k-dominated by the set A, then ⏐A ∩ N(vj ) \ {u(j) } ⏐ = k − 1. Otherwise, A \ {u(j) } would be a total k-dominating set in Gn,k smaller than A, a contradiction. ( ) Now, using that δV (G) (vj ) ≥ k we have that there exists w ∈ N(vj ) ∩ (V (G) \ A), what implies that A′ = A \ {u(j) } ∪ {w} is a γkt (Gn,k )-set too. Applying this procedure to every vertex u(j) belonging to A, we obtain a γkt (Gn,k )-set containing no vertices in {u(1) , u(2) , . . . , u(n) }. □ Now, we are able to present our complexity result for the GTkD problem of graphs. Theorem 4.3. GTkD-PROBLEM is NP-complete. Proof. Let G be any graph of order n and k ≤ δ (G) be a fixed positive integer. We consider now the graph Gn,k as previously g described. We shall prove that γkt (Gn,k ) = nk(k + ( 1) + γkt (G). ( ) ) Let D be a γkt (G)-set and we consider S = D ∪
⋃n
i=1 V
(i)
Hk
\ {u(i) } ⊆ V (Gn,k ). Clearly, S is a total k-dominating set of
Gn,k , thus, γkt (Gn,k ) ≤ nk(k + 1) + γkt (G). On the other hand, if A is the γkt (Gn,k )-set given by Proposition 4.2, then
⏐ ( n )⏐ ⏐ ⋃ ( (i) ) ⏐⏐ ⏐ V Hk ⏐ = nk(k + 1) ⏐A ∩ ⏐ ⏐ i=1
and A ∩ V (G) is a total k-dominating set in G, thus |A ∩ V (G)| ≥ γkt (G). Consequently, we obtain that
⏐ ( n )⏐ ⏐ ⋃ ( (i) ) ⏐⏐ ⏐ V Hk γkt (Gn,k ) = ⏐A ∩ ⏐ + |A ∩ V (G)| ≥ nk(k + 1) + γkt (G). ⏐ ⏐ i=1
Now, it is not difficult to see that for every vertex v of the graph Gn,k there exist k vertices v1 , . . . , vk in A such that each vi is not adjacent to v , for i ∈ {1, . . . , k}. Therefore, using Observation 1.4, it follows that A is a GTkD set, so |A| = γkt (Gn,k ) ≥ γktg (Gn,k ). Using also Observation 1.3, we conclude that γktg (Gn,k ) = γkt (Gn,k ) = nk(k + 1) + γkt (G). g Finally, if we take s = nk(k + 1) + r, we see that γkt (G) ≤ r if and only if γkt (Gn,k ) ≤ s. Therefore, since TOTAL kDOMINATION PROBLEM is NP-complete, we have that GTkD-PROBLEM is NP-complete. □ 5. Conclusions and open problems In this article we have introduced a new kind of domination parameter in graphs. The main results deal with the complexity of the associated decision problem, which is proved to be NP-complete. Moreover, we have presented tight g lower and upper bounds on the parameter γkt (G) and some characterizations in concordance with these bounds. However, there are several open problems which could be dealt with in order to continue with that investigation. We conclude with some of them.
• Characterize the graphs G for which γkt (G) = γkt (G) = γktg (G). • Characterize the graphs G for which γktg (G) = γkt (G) + ∆(G). • Can we provide the value of γktg (G) for some other specific families of graphs G? Acknowledgments This paper was partly supported by two grants from AEI and FEDER (MTM2016-78227-C2-1-P & MTM2015-69323-REDT) and a grant from CONACYT (FOMIX-CONACyT-UAGro 249818), Mexico. First author’s work was also partially supported by Plan Nacional I+D+I Grant MTM2015-70531 and Junta de Andalucía FQM-260. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
9
References [1] [2] [3] [4] [5] [6]
M.H. Akhbari, C. Eslahchiz, N. Jafari Rad, R. Hasni, Some Remarks on Global Total Domination in Graphs, Appl. Math. E-Notes 15 (2015) 22–28. S. Bermudo, J.C. Hernández-Gómez, J.M. Sigarreta, On the total k-domination in graphs, Discuss. Math. Graph Theory 38 (2018) 301–317. S. Bermudo, D.L. Jalemskaya, J.M. Sigarreta, Total 2-domination in Grid graphs, Util. Math. (2017) (in press). S. Bermudo, J.L. Sánchez, J.M. Sigarreta, Total k-domination in Cartesian product graphs, Period. Math. Hungar. 75 (2017) 255–267. H. Fernau, J.A. Rodríguez-Velázquez, J.M. Sigarreta, Global powerful r-alliances and total k-domination in graphs, Util. Math. 98 (2015) 127–147. J. He, H. Liang, Complexity of Total k-Domination and Related Problems, Joint Conference of FAW 2011(the 5th International Frontiers of Algorithmics Workshop) and AAIM 2011 (the 7th International Conference on Algorithmic Aspects of Information and Management) 2011, pp. 147–155. [7] V.R. Kulli, On n-total domination number in graphs, in: Graph Theory, Combinatorics, Algorithms and Applications, SIAM, Philadelphia, 1991 pp. 319–324. [8] V.R. Kulli, B. Janakiram, The total global domination number of a graph, Indian J. Pure. Appl. Math. 27 (1996) 537–542.
Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
)
–
Contents lists available at ScienceDirect
Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
On the global total k-domination number of graphs Sergio Bermudo a , Abel Cabrera Martínez b, *, Frank A. Hernández Mira b , José M. Sigarreta b a b
Department of Economics, Quantitative Methods and Economic History, Universidad Pablo de Olavide, ES-41013 Seville, Spain Faculty of Mathematics, Autonomous University of Guerrero. Carlos E. Adame 5, Col. La Garita, 39350 Acapulco, Guerrero, Mexico
article
info
Article history: Received 30 September 2017 Received in revised form 23 April 2018 Accepted 13 May 2018 Available online xxxx Keywords: Global total k-domination Global total domination Total k-domination
a b s t r a c t A subset D of vertices of a graph G is a global total k-dominating set if D is a total k-dominating set of both G and G. The global total k-domination number of G is the g minimum cardinality of a global total k-dominating set of G and it is denoted by γkt (G). In this paper we introduce this concept and we begin the study of its mathematical properties. Specifically, we prove that the complexity of the decision problem associated to g the computation of the value γkt (G) is NP-complete. Moreover, we present tight bounds for this parameter and we obtain relationships between the global total k-domination number of G and the total k-domination numbers of G and G, and other parameters of the graph G. © 2018 Elsevier B.V. All rights reserved.
1. Introduction Let G = (V , E) be a simple graph of order |V | = n and size |E | = m. For a nonempty set D ⊆ V , and a vertex v ∈ V , ND (v ) denotes the set of neighbors of v in D and ND [v] = ND (v ) ∪ {v}. The degree of v in D will be denoted by δD (v ) = |ND (v )|. Analogously, N D (v ) denotes the set of non neighbors of v in D, N D [v] = N D (v ) ∪ {v} and δ D (v ) denotes the cardinality of the set N D (v ) (i.e. δ D (v ) = |N D (v )|). For short, we will often use N(v ), N [v], δ (v ), δ (v ), N(v ) and N [v] instead of NV (G) (v ), NV (G) [v], δV (G) (v ), δ V (G) (v ), N V (G) (v ) and N V (G) [v], respectively. The minimum and maximum degree of G will be denoted by δ (G) and ∆(G), respectively. The subgraph induced by D ⊆ V will be denoted by G[D] and the complement of the graph G will be denoted by G . If X and Y are two subsets of V , then we denote the set of all edges of G that join a vertex of X and a vertex of Y by E(X , Y ). A nonempty set D ⊆ V is a total k-dominating set in G if every vertex in V is adjacent to at least k vertices in D (or every vertex of V is totally k-dominated by D). The total k-domination number of G is the minimum cardinality among all total k-dominating sets in G and it is denoted by γkt (G). A γkt (G)-set is a total k-dominating set of cardinality γkt (G) (see [2–7]). A total dominating set is a total 1-dominating set, and the total domination number is the total 1-domination number, which is also denoted by γt (G). A nonempty set D ⊆ V is a global total k-dominating set (or GTkD set for short) if D is a total k-dominating set of both G g and G. The global total k-domination number of a graph G, denoted by γkt (G), is the minimum cardinality of a GTkD set of G. g g A γkt (G)-set is a global total k-dominating set of cardinality γkt (G). When k = 1, a global total 1-dominating set is a global total dominating set (see [1,8]). In order to have a GTkD set D, any vertex of V must be totally k-dominated by D in G and G, so 1 ≤ k ≤ min{δ (G), n − (∆(G) + 1)}. Hence, this is a necessary and sufficient condition to guarantee the existence of a GTkD set.
*
Corresponding author. E-mail addresses: [email protected] (S. Bermudo), [email protected] (A. Cabrera Martínez), [email protected] (F.A. Hernández Mira), [email protected] (J.M. Sigarreta). https://doi.org/10.1016/j.dam.2018.05.025 0166-218X/© 2018 Elsevier B.V. All rights reserved.
Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
2
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
Fig. 1. Complementary graphs C5 and C5 .
Notice that, if G1 , G2 , . . . , Gr , with r ≥ 2, are the connected components of a graph G, then any minimum GTkD set in G is formed by a minimum total k-dominating set in each subgraph Gi . That is stated in the following result. Proposition 1.1. Let G1 , G2 , . . . , Gr , with r ≥ 2, be the connected components of a graph G. Then
γktg (G) =
r ∑
γkt (Gi ).
i=1
⋃r ∑r g Proof. Let Di be a γkt (Gi )-set for i ∈ {1, . . . , r }. It is not difficult to see that i=1 Di is a GTkD set in G, so γkt (G) ≤ i=1 |Di | = ∑ r γ (G ). kt i i=1 g Now, let A be a γkt (G)-set. We observe that, for every i ∈ {1, . . . , r }, the vertex set V (Gi ) must be totally k-dominated by A, so it follows that |A ∩ V (Gi )| ≥ γkt (Gi ). Hence, γktg (G) = |A| ≥
r r ∑ ∑ |A ∩ V (Gi )| ≥ γkt (Gi ), i=1
i=1
which completes the proof. □ The following observation is an immediate consequence of the previous proposition. g
Observation 1.2. Let Kn1 ,n2 ,...,nm be a complete m-partite graph and let k + 1 ≤ n1 ≤ n2 . . . ≤ nm . Then, γkt (Kn1 ,n2 ,...,nm ) = m(k + 1). g
Clearly, the definition of a GTkD set gives us a direct lower bound for the parameter γkt (G). g
Observation 1.3. Let G be a graph, then γkt (G) ≥ max{γkt (G), γkt (G)}. As an example, we can observe the complementary graphs C5 and C5 in Fig. 1. We can check that the sets {v1 , v2 , v3 } and {v1 , v3 , v5 } are γt (C5 )-set and γt (C5 )-set, respectively. Hence γt (C5 ) = γt (C5 ) = 3. The set {v1 , v2 , v3 , v5 } is a GT1D set of C5 and we can check that there is not a set of vertices D, with cardinality three, such that D is a total dominating set of both C5 g g and C5 , thus {v1 , v2 , v3 , v5 } is a γ1t (C5 )-set and, consequently, γ1t (C5 ) = 4. If we take k = 2, V (C5 ) is an unique γ2t (C5 )-set g
and γ2t (C5 )-set. So γ2t (C5 ) = γ2t (C5 ) = γ2t (C5 ) = 5. Since two complementary graphs are tightly related by their edges, we can characterize one of them using the information about the other. Given a nonempty set D ⊆ V , D is a total k-dominating set of G if and only if every vertex in V is not adjacent to at least k vertices in D. This simple quotation shows a hint for the next observation.
Observation 1.4. Let G = (V , E) be a graph and let D ⊆ V . The set D is a GTkD if and only if δD (v ) ≥ k and δ D (v ) ≥ k, for every vertex v ∈ V . In this paper, we initiate the study of the global total k-domination number of a graph G. 2. Bounds for the global total k-domination number In concordance with Proposition 1.1, from now on, we only consider connected graphs such that their complements are also connected graphs. Given a graph G = (V , E), for every k ≤ min{δ (G), n − (∆(G) + 1)}, V is a GTkD set. Moreover, by Observation 1.4 we g deduce the following lower bound for γkt (G). g
Proposition 2.1. Let G be a graph with order n, then 2k + 1 ≤ γkt (G) ≤ n. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
3
The next theorems characterize graphs G having global total k-domination number equal to this lower and upper bound, respectively. For this purpose, given two graphs G and H, we define the k-join operation of G with H, as the disjoint union of G and H by adding edges joining each vertex of G with k or k + 1 vertices of H. We denote the k-join of G with H by G ◦k H and denote T [G ◦k H ] as the family of graphs obtained by this operation. g
Theorem 2.2. Let G be a graph, then γkt (G) = 2k + 1 if and only if G ∈ T [F ◦k H ], where H is a k-regular graph of order 2k + 1 and F is any graph. g
g
Proof. Firstly, we assume that γkt (G) = 2k + 1. Let D be a γkt (G)-set and we consider H = G[D] and F = G[V \ D]. By Observation 1.4, it is not difficult to check that H is a k-regular induced subgraph of G, and every υ ∈ V \ D satisfies that k ≤ δD (υ ) ≤ k + 1, so G ∈ T [F ◦k H ], where H is a k-regular graph of order 2k + 1. Now, we assume that G ∈ T [F ◦k H ], where H is a k-regular graph of order 2k + 1 and F is any graph. By Observation 1.4, g it is easy to see that V (H) is a GTkD set in G, so γkt (G) ≤ |V (H)| = 2k + 1, and using Proposition 2.1 it follows that V (H) is a g g γkt (G)-set, hence γkt (G) = 2k + 1. □ In order to present other results about the global total k-domination number we need to introduce some new terminology and notation. Given a graph G, we consider the set Ai (G) = {v ∈ V (G) : δ (v ) = i}, and we define the set Rk (G) as follows:
⎛ Rk (G) = ⎝
⎞ ⋃ v∈Ak (G)
⎛
⎞ ⋃
N(v )⎠ ∪ ⎝
(V \ N [w])⎠ .
w∈An−(k+1) (G)
Lemma 2.3. Let G be a graph and D be a GTkD set of G. Then Rk (G) ⊆ D. Proof. On one hand, if v ∈ Ak (G), since v must be k-dominated by D, then N(v ) ⊆ D. On the other hand, if w ∈ An−(k+1) (G), then |V \ N [w]| = k, since there must be k vertices in D which are not adjacent to w , we have that V \ N [w] ⊆ D. Consequently, Rk (G) ⊆ D. □ g
Corollary 2.4. Let G be a graph. Then, γkt (G) = |Rk (G)| if and only if Rk (G) is a GTkD set of G. In the following result we show a characterization when the global total k-domination number is equal to the order of the graph. g
Theorem 2.5. Let G be a graph of order n, then γkt (G) = n if and only if Rk (G) = V . g
Proof. Firstly, we assume that Rk (G) = V . Notice that, as V is a GTkD set in G, by Corollary 2.4 it follows that γkt (G) = n. g Now, kt (G) = n and we suppose that there exists a vertex x such that x ̸ ∈ Rk (G). This implies that ⋃ we assume that γ⋃ x ̸ ∈ v∈A (G) N(v ) and x ̸ ∈ w∈A (G) V \ N [w], so for every y ∈ N(x) it follows that δ (y) > k and for every z ∈ V \ N [x] k
n−(k+1)
it follows that δ (z) < n − (k + 1), consequently, δ (z) > k. Thus, by Observation 1.4, V \ {x} is a GTkD set in G, which is a contradiction. Hence, Rk (G) = V . □ The following corollary is directly obtained by Theorem 2.5. g
Corollary 2.6. If G is a k-regular or n − (k + 1)-regular graph of order n, then γkt (G) = n. g
Another interesting problem could consist in characterizing graphs G which satisfy that γkt (G) = γkt (G) = γkt (G). By g Corollary 2.6 we directly obtain that every k-regular graph G of order 2k + 1 satisfies γkt (G) = γkt (G) = γkt (G). Nevertheless, there exist many non-regular graphs satisfying the same property. To show this, we need the following result from [2]. Theorem 2.7. Let D be a total k-dominating set in a graph G. If δD (v ) = k for every v ∈ D and then |D| = γkt (G).
v∈V \D δD (v )
∑
≤ k|V \ D| + k − 1,
g
Proposition 2.8. For every even number k, there exists a non-regular graph G satisfying that γkt (G) = γkt (G) = γkt (G). Proof. Let H be a k-regular graph of order 2k + 1 and let G be a graph obtained from the graph H and another vertex, namely v , by joining the vertex v to exactly k vertices of V (H). On one hand, notice that V (H) is a GTkD set in G, so γktg (G) ≤ 2k + 1. g Also, using the lower bound in Proposition 2.1 it follows that γkt (G) = 2k + 1. On the other hand, using Theorem 2.7, we g obtain that V (H) is also a minimum total k-dominating set in G and G, so γkt (G) = γkt (G) = γkt (G). □ Given a graph G, it is natural to think that the global total k-domination number of G, the total k-domination number of G and the total k-domination number of G are related. The next result shows a relationship between these parameters. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
4
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
g
Fig. 2. Graph G2 which satisfies γ2t (G2 ) = γ2t (G2 ) + γ2t (G2 ).
Proposition 2.9. Let G be a graph, then
γkt (G) + γkt (G) 2
≤ γktg (G) ≤ γkt (G) + γkt (G).
Proof. The lower bound follows directly using Observation 1.3. On the other hand, let D be a γkt (G)-set and S be a γkt (G)-set, g note that D ∪ S is a GTkD set of G. Therefore, γkt (G) ≤ |D ∪ S | ≤ |D| + |S | = γkt (G) + γkt (G). □ Let us see that, for any k ≥ 2 there exists a graph Gk attaining the upper bound given in the above proposition. We consider the graph Gk with order n = (k + 1)2 formed by a complete graph Kk+1 of k + 1 vertices and k + 1 copies of the null graph Nk of k vertices. We denote V (Kk+1 ) = {v1 , v2 , . . . , vk+1 } and we denote the ith copy of Nk by Nki for every i ∈ {1, . . . , k + 1}. The edges of Gk are formed by joining the vertices of Nki with all vertices of V (Kk+1 ) except the vertex vi , for every i ∈ {1, . . . , k + 1} (see Fig. 2). the By Theorem 2.7, we have that V (Kk+1 ) is a γkt (Gk )-set, so γkt (Gk ) = k + 1. In Gk , each vertex vi of Kk+1 is only adjacent ⋃k+to 1 k vertices of the copy Nki , hence any total k-dominating set of Gk contains all copies Nki for i ∈ {1, . . . , k + 1}. Since i=1 Nki is a total k-dominating set of Gk , it is the unique γkt (Gk )-set, so γkt (Gk ) = k(k + 1). Finally, every vertex vi ∈ V (Kk+1 ) is adjacent j to a vertex u ∈ Nk (j ̸ = i) such that δ (u) = k, and every vertex u ∈ Nki satisfies u ∈ V (Gk ) \ N [vi ], where δ (vi ) = n − (k + 1), g then V (Gk ) = Rk (Gk ), consequently, by Theorem 2.5, it follows that γkt (Gk ) = (k + 1) + k(k + 1) = γkt (Gk ) + γkt (Gk ). The lower bound in Proposition 2.9 is attained by the graph given by Proposition 2.8. g We continue giving a relationship between γkt (G) and γkt (G), in which we use the maximum degree of the graph. Proposition 2.10. Let G be a graph, g
(i) If γkt (G) > ∆(G) + k, then γkt (G) = γkt (G). g (ii) If γkt (G) ≤ ∆(G) + k, then γkt (G) ≤ ∆(G) + k + 1. Proof. (i) Let D be a γkt (G)-set of G and we assume that |D| > ∆(G) + k. For every v ∈ V , since k ≤ δD (v ) ≤ ∆(G), there are g at least k vertices in D which are not adjacent to v , hence D is a GTkD set of G and, consequently, γkt (G) ≤ γkt (G). Therefore, g γkt (G) = γkt (G). (ii) Let D be a γkt (G)-set of G and we assume that |D| ≤ ∆(G) + k. Then we add some arbitrary vertices to D until we have a set D′ (D ⊂ D′ ) of cardinality ∆(G) + k + 1. Notice that, as D is a γkt (G)-set then D′ is a total k-dominating set of G, that is, δD′ (v ) ≥ k for every v ∈ V (G). Thus δD′ (v ) ≥ k for every v ∈ V (G). On the other hand, since δD′ (w) ≤ δ (w ) ≤ ∆(G) for every w ∈ V (G) and |D′ | = ∆(G) + k + 1, then δ D′ (w) ≥ k. Therefore, by Observation 1.4 it follows that D′ is a GTkD set, which g implies γkt (G) ≤ |D′ | = ∆(G) + k + 1. □ The upper bound given in the second part of this proposition is attained, for instance, in the graph Gk given after Proposition 2.9. g
Theorem 2.11. Let G be a graph, then γkt (G) ≤ γkt (G) ≤ γkt (G) + ∆(G). Proof. The lower bound follows directly from Observation 1.3. g g On the other hand, we suppose that γkt (G) > γkt (G) + ∆(G). Since γkt (G) ≥ k + 1 we have γkt (G) > ∆(G) + k + 1 and, g by Proposition 2.10, we obtain γkt (G) > ∆(G) + k. Using again Proposition 2.10, we conclude that γkt (G) = γkt (G), which is a contradiction. □ The upper bound given in this proposition is also attained, for instance, in the graph Gk given after Proposition 2.9, where
∆(Gk ) = k(k + 1) = γkt (Gk ).
Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
S. Bermudo et al. / Discrete Applied Mathematics (
g
Fig. 3. A graph G such that γ2t (G) =
)
–
5
n+1 . 2
g
Proposition 2.12. Let G be a graph, if γkt (G) = γkt (G) + ∆(G) then G ∈ T [H ◦k Kk+1 ], where H is some graph such that |V (H)| ≥ ∆(G). g
Proof. If γkt (G) = γkt (G) + ∆(G), by Proposition 2.10(i), it follows that γkt (G) ≤ ∆(G) + k. Moreover, by Proposition 2.10(ii) g we have γkt (G) = γkt (G) + ∆(G) ≤ ∆(G) + k + 1, that is, γkt (G) ≤ k + 1. Therefore, γkt (G) = k + 1 and every γkt (G)-set g D must satisfy G[D] = Kk+1 . Finally, since |V (G)| ≥ γkt (G) = k + 1 + ∆(G) and every vertex v ∈ V (G) \ D satisfies that k ≤ δD (v ) ≤ k + 1, then G ∈ T [H ◦k Kk+1 ] for some graph H such that |V (H)| ≥ ∆(G). □ g
Let us observe that the equality γkt (G) = γkt (G) + ∆(G) is satisfied, for instance, by the graph Gk given after Proposition 2.9, g but there exist graphs G ∈ T [H ◦k Kk+1 ], such that H is a graph with order |V (H)| ≥ ∆(G), satisfying γkt (G) < γkt (G) + ∆(G). i i For instance, the graph Gk given after Proposition 2.9 changing Nk by Nk+1 for every i ∈ {1, 2, . . . , k + 1}. g Now, we continue with some other bounds for γkt (G). Theorem 2.13. Let G be a graph of order n and size m. Then
γktg (G) ≥ max
{
2m + 2nk − n∆(G) −2m + 2nk + nδ (G) 2∆(G)
,
2(n − (δ (G) + 1))
}
.
Proof. Let D be a γkt (G)-set. Since every vertex in V \ D has at least k neighbors in D, we have (n − |D|)k ≤ E(D, V \ D) ≤ |D|∆(G)−E(D,V \D) −|D|) |D|(∆(G) − k) and E(V \ D, V \ D) ≤ (∆(G)−k)(n . Now, using that E(D, D) ≤ ≤ |D|∆(G)−2(n−|D|)k we obtain 2 2 m = E(D, D) + E(D, V \ D) + E(V \ D, V \ D) (∆(G) − k)(n − |D|) |D|∆(G) − (n − |D|)k + |D|(∆(G) − k) + , ≤ 2 2 which is equivalent to 2m ≤ 2|D|∆(G) − 2nk + n∆(G). Therefore, we obtain that |D| ≥ a γkt (G)-set S, we have |S | ≥
−2m+2nk+nδ (G) . 2(n−(δ (G)+1))
2m+2nk−n∆(G) . 2∆(G)
Analogously, if consider
Hence, by Observation 1.3 it follows the result. □
The lower bound in the above theorem is tight if we consider that k is an even number and G is a k-regular graph with g (2k+1)k cardinality 2k + 1. In this case, m = and γkt (G) = 2k + 1. 2 Theorem 2.14. Let G be a graph with order n. If k ≥ max g
{ ∆(G) 2
} g +1) , n−(δ(G) , then γkt (G) ≥ 2
n+1 . 2
∆(G)
Proof. Let D be a γkt (G)-set of G and let v ∈ D. Firstly, if k ≥ 2 , then δD (v ) ≥ δ(V \D) (v ), consequently, 2δD (v ) ≥ +1) δ(V \D) (v ) + δD (v ) = δ (v ). Analogously, if k ≥ n−(δ(G) , we obtain that 2δ D (v ) ≥ δ (V \D) (v ) + δ D (v ) = δ (v ). Using both 2 1 inequalities we obtain 2(|D| − 1) = 2(δD (v ) + δ D (v )) ≥ δ (v ) + δ (v ) = n − 1, thus |D| ≥ n+ . □ 2 The lower bound in the theorem above is tight if we consider that k is an even number and G is a 2k-regular graph with order n = 4k + 1 satisfying that G ∈ T [F ◦k H ], where H is a k-regular graph of order 2k + 1 and F is a graph of order 2k. By g (4k+1)+1 Theorem 2.2 we have γkt (G) = 2k + 1 = = n+2 1 . In Fig. 3 we show a 4-regular graph G ∈ T [F ◦2 H ] of order 9, where 2 F is the subgraph induced by the four black vertices and H is the subgraph induced by the white vertices. The following two theorems are well known (see [5]). Theorem 2.15. Let k ∈ N and suppose G = (V , E) be a graph of order n with minimum degree δ (G) ≥ k. Then γkt (G) ≥ this bound is sharp.
kn
∆(G)
and
Theorem 2.16. Let ( ) k ∈ N and suppose G = (V , E) be a graph of order n and size m with minimum degree δ (G) ≥ k. Then γkt (G) ≥ 2 n − mk and this bound is sharp. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
6
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
Using these theorems and Observation 1.3 we obtain the following corollaries. Corollary 2.17. Let G be a graph with order n. Then,
γktg (G) ≥ max
{
kn
}
kn
,
∆(G) n − (δ (G) + 1)
.
Corollary 2.18. Let G be a graph with order n and size m. Then,
{ ( ( )} m) n(n − 1) − 2m γktg (G) ≥ max 2 n − ,2 n − . k
2k
Using now Theorem 2.15 and Proposition 2.10 we obtain another corollary. ∆(G)(∆(G)+k+1)
Corollary 2.19. Let G be a graph with order n. If n ≥
k
g
, then γkt (G) = γkt (G).
3. The global total k-domination number for some particular graphs g
In this section we obtain the exact value of γkt (G) for some specific families of graphs, such as paths, cycles and cubic graphs. In order to do this, we need the following well-known values of the total k-domination number of paths and cycles. It is known (see [2])
γt (Pn ) = γt (Cn ) =
⎧ n ⎪ ⎨ +1 2
n
⎪ ⎩ ⌈ ⌉ 2
n ≡ 2 (mod 4) and
γ2t (Cn ) = n.
otherwise
The following proposition shows the global total k-domination number for paths and cycles. Proposition 3.1. (a) For any path Pn and cycle Cn with order n ≥ 4 it holds
⎧ 4 ⎪ ⎪ ⎨n + 1 γ1tg (Pn ) = γ1tg (Cn ) = 2 ⎪ ⎪ ⎩ ⌈n⌉ 2
if
n = 4, 5
if
n ≡ 2 (mod 4)
otherwise. g
(b) For any cycle Cn with order n ≥ 5 it holds γ2t (Cn ) = n. Proof. (a) The cases n = 4 and n = 5 are trivial. If n ≥ 6, the result is obtained by Proposition 2.10 and the total domination number of paths and cycles. (b) It is obtained by Corollary 2.6. □ The following lemma, which was proved in [1], will be used to get the global total k-domination numbers of cubic graphs. g
Lemma 3.2. For any graph G, γ1t (G) ≥ 4. Proposition 3.3. Let G be a cubic graph of order n and let S3 be the star graph of four vertices. Then, the following global total k-domination numbers hold:
g 1t (G)
γ
{ γt (G) 5 4
=
γ2tg (G) =
{
5
γ2t (G)
if γt (G) > 4 if γt (G) = 4 and for any γt (G)-set D of G is G[D] = S3 otherwise, if γ2t (G) ≤ 5 if γ2t (G) > 5
and
γ3tg (G) = n. g
Proof. First, let us prove the result for k = 1. By Proposition 2.10(i), if γt (G) > 4, then γ1t (G) = γt (G). If γt (G) ≤ 4, by g Proposition 2.10(ii) and Lemma 3.2 it follows that 4 ≤ γ1t (G) ≤ 5. We will consider three cases. Case 1. We assume that γt (G) = 2. By Theorem 2.15 we have n ≤ 6. Since G is a cubic graph, n is even and n ≥ ∆(G) + k + 1 = 5, thus n = 6. Let D = {v, w} be a γt (G)-set and let D′ = D ∪ {x, y}, where x and y are vertices such g that x ∈ N(v ) \ N(w ) and y ∈ N(w ) \ N(v ). It is clear that D′ is a GT1D set in G, so γ1t (G) ≤ 4. By Lemma 3.2 we conclude that g γ1t (G) = 4. Case 2. We assume that γt (G) = 3. By Theorem 2.15 we have n ≤ 9. Since G is a cubic graph, n is even and n ≥ ∆(G) + k + 1 = 5, thus n = 6 or n = 8. If D is a γt (G)-set, then the subgraph induced by D is a path P3 or a cycle Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
7
Fig. 4. Cubic graph C3 □P2 .
C3 . If the subgraph induced by D is a path P3 , v is the vertex in D such that δD (v ) = 2, and we take D′ = D ∪ {x}, where g g x ̸ ∈ D ∩ N(v ), it is easy to see that D′ is a GT1D set in G , so γ1t (G) ≤ 4. Again by Lemma 3.2, we conclude that γ1t (G) = 4. If the subgraph induced by D is a cycle C3 , then n = 6 and G is the Cartesian product C3 □P2 (see Fig. 4). It can be checked that g any set S of four vertices in G, such that the subgraph induced by S is a triangle-free graph, is a GT1D set in G, so γ1t (G) ≤ 4. g Therefore, γ1t (G) = 4. Case 3. We assume that γt (G) = 4. If D is a γt (G)-set of G, then the subgraph induced by D can be a path P4 , a cycle C4 , the union K2 ∪ K2 , a star S3 , or a star S3 with one extra edge (note that this induced subgraph G[D] cannot have two extra edges because, if this happened and v were the center of the star, the set D \ {v} would be a total dominating set). g g In the first three cases, D is also a GT1D set, so γ1t (G) ≤ 4. Again by Lemma 3.2, we have γ1t (G) = 4. If the subgraph induced by D is a star S3 with an extra edge, that is, D = {v1 , v2 , v3 , v4 } with δD (v1 ) = 3, δD (v2 ) = 2 = δD (v3 ) and δD (v4 ) = 1, then there exists a vertex u ∈ N(v4 ) \ {v1 } such that N(u) ∩ D = {v4 } (otherwise the set D \ {v4 } would be a total dominating g set in G, contradiction). In such a case, the set D′ = {v2 , v3 , v4 , u} is a GT1D set, so γ1t (G) = 4. If every γt (G)-set D of G satisfies g that the subgraph induced by D is a star S3 , then D cannot be a γ1t (G)-set because the center of the star S3 has no neighbor in g g g D. Hence, γ1t (G) > 4 and, since γ1t (G) ≤ 5, we conclude that γ1t (G) = 5. Note that this family of cubic graphs G (where every γt (G)-set D of G satisfies that the subgraph induced by D is a star S3 ) is not empty, for example, the Petersen graph belongs to this family. g Now, let us prove the result when k = 2. By Proposition 2.10(i), if γ2t (G) > 5, then γ2t (G) = γ2t (G). If γ2t (G) ≤ 5, by g Proposition 2.10(ii) and Proposition 2.1, it follows that 5 ≤ γ2t (G) ≤ 6. If γ2t (G) = 3, by Theorem 2.15 we obtain that n ≤ 4, contradiction. If γ2t (G) = 4 or γ2t (G) = 5, as G is a cubic graph and n is an even number, using again Theorem 2.15, we obtain g that n = 6. We can see that γ2t (G) < 6, because the graph G does not satisfy the characterization given in Theorem 2.5, then g γ2t (G) = 5. g Finally, the equality γ3t (G) = n is directly obtained by Corollary 2.6. □ 4. Complexity In this section we investigate the complexity of the following decision problem:
GLOBAL TOTAL k-DOMINATION PROBLEM (GTkD-PROBLEM for short) INSTANCE: A non trivial graph G and a positive integer r g PROBLEM: Deciding whether γkt (G) is less than r We will do the complexity analysis of the GTkD-PROBLEM making a reduction from another decision problem.
TOTAL k-DOMINATION PROBLEM INSTANCE: A non trivial graph G and a positive integer r PROBLEM: Deciding whether γkt (G) is less than r The following result was proved in [5]. Theorem 4.1. For all k ≥ 1, the TOTAL k-DOMINATION PROBLEM is NP-complete. In order to present our complexity results we need to introduce a family of graphs. Fixed k, Hk is obtained from k copies of Kk+1 and another vertex, namely u, by joining the vertex u to one vertex of each copy Kk+1 . Given a graph G of order n and (1) (n) minimum degree δ (G) ≥ k, and n graphs Hk , . . . , Hk isomorphic to the graph Hk , the graph Gn,k is constructed by adding (i) (i) edges between the ith-vertex of G and the vertex u of the ith-graph Hk . We show an example in Fig. 5, where k = 2, H2 is the graph (I), G is the cycle C4 , and G4,2 is the graph showed in (II). Proposition 4.2. Let G be a graph of order ) n and k be a positive integer such that k ≤ δ (G). Then there exists a γkt (Gn,k )-set ⋃n ( (i) containing all vertices in i=1 Hk \ {u(i) } and no vertices in {u(1) , u(2) , . . . , u(n) }. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
8
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
Fig. 5. The graph G4,2 where G is a C4 graph.
(i)
Proof. Let A be a γkt (Gn,k )-set. For every i ∈ {1,(. . . , n} the vertices of V (Hk ) \ u(i) must be totally k-dominated by the set A, ) ⋃n (i) (i) so it can be checked that every vertex of i=1 V (Hk ) \ {u } belongs to A. Now, we suppose that u(j) ∈ A and we denote by
{
}
v⏐ j the( neighbor of u)⏐(j) in V (G), that is, N(u(j) ) ∩ V (G) = {vj }. Since the vertex vj must be totally k-dominated by the set A, then ⏐A ∩ N(vj ) \ {u(j) } ⏐ = k − 1. Otherwise, A \ {u(j) } would be a total k-dominating set in Gn,k smaller than A, a contradiction. ( ) Now, using that δV (G) (vj ) ≥ k we have that there exists w ∈ N(vj ) ∩ (V (G) \ A), what implies that A′ = A \ {u(j) } ∪ {w} is a γkt (Gn,k )-set too. Applying this procedure to every vertex u(j) belonging to A, we obtain a γkt (Gn,k )-set containing no vertices in {u(1) , u(2) , . . . , u(n) }. □ Now, we are able to present our complexity result for the GTkD problem of graphs. Theorem 4.3. GTkD-PROBLEM is NP-complete. Proof. Let G be any graph of order n and k ≤ δ (G) be a fixed positive integer. We consider now the graph Gn,k as previously g described. We shall prove that γkt (Gn,k ) = nk(k + ( 1) + γkt (G). ( ) ) Let D be a γkt (G)-set and we consider S = D ∪
⋃n
i=1 V
(i)
Hk
\ {u(i) } ⊆ V (Gn,k ). Clearly, S is a total k-dominating set of
Gn,k , thus, γkt (Gn,k ) ≤ nk(k + 1) + γkt (G). On the other hand, if A is the γkt (Gn,k )-set given by Proposition 4.2, then
⏐ ( n )⏐ ⏐ ⋃ ( (i) ) ⏐⏐ ⏐ V Hk ⏐ = nk(k + 1) ⏐A ∩ ⏐ ⏐ i=1
and A ∩ V (G) is a total k-dominating set in G, thus |A ∩ V (G)| ≥ γkt (G). Consequently, we obtain that
⏐ ( n )⏐ ⏐ ⋃ ( (i) ) ⏐⏐ ⏐ V Hk γkt (Gn,k ) = ⏐A ∩ ⏐ + |A ∩ V (G)| ≥ nk(k + 1) + γkt (G). ⏐ ⏐ i=1
Now, it is not difficult to see that for every vertex v of the graph Gn,k there exist k vertices v1 , . . . , vk in A such that each vi is not adjacent to v , for i ∈ {1, . . . , k}. Therefore, using Observation 1.4, it follows that A is a GTkD set, so |A| = γkt (Gn,k ) ≥ γktg (Gn,k ). Using also Observation 1.3, we conclude that γktg (Gn,k ) = γkt (Gn,k ) = nk(k + 1) + γkt (G). g Finally, if we take s = nk(k + 1) + r, we see that γkt (G) ≤ r if and only if γkt (Gn,k ) ≤ s. Therefore, since TOTAL kDOMINATION PROBLEM is NP-complete, we have that GTkD-PROBLEM is NP-complete. □ 5. Conclusions and open problems In this article we have introduced a new kind of domination parameter in graphs. The main results deal with the complexity of the associated decision problem, which is proved to be NP-complete. Moreover, we have presented tight g lower and upper bounds on the parameter γkt (G) and some characterizations in concordance with these bounds. However, there are several open problems which could be dealt with in order to continue with that investigation. We conclude with some of them.
• Characterize the graphs G for which γkt (G) = γkt (G) = γktg (G). • Characterize the graphs G for which γktg (G) = γkt (G) + ∆(G). • Can we provide the value of γktg (G) for some other specific families of graphs G? Acknowledgments This paper was partly supported by two grants from AEI and FEDER (MTM2016-78227-C2-1-P & MTM2015-69323-REDT) and a grant from CONACYT (FOMIX-CONACyT-UAGro 249818), Mexico. First author’s work was also partially supported by Plan Nacional I+D+I Grant MTM2015-70531 and Junta de Andalucía FQM-260. Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.
S. Bermudo et al. / Discrete Applied Mathematics (
)
–
9
References [1] [2] [3] [4] [5] [6]
M.H. Akhbari, C. Eslahchiz, N. Jafari Rad, R. Hasni, Some Remarks on Global Total Domination in Graphs, Appl. Math. E-Notes 15 (2015) 22–28. S. Bermudo, J.C. Hernández-Gómez, J.M. Sigarreta, On the total k-domination in graphs, Discuss. Math. Graph Theory 38 (2018) 301–317. S. Bermudo, D.L. Jalemskaya, J.M. Sigarreta, Total 2-domination in Grid graphs, Util. Math. (2017) (in press). S. Bermudo, J.L. Sánchez, J.M. Sigarreta, Total k-domination in Cartesian product graphs, Period. Math. Hungar. 75 (2017) 255–267. H. Fernau, J.A. Rodríguez-Velázquez, J.M. Sigarreta, Global powerful r-alliances and total k-domination in graphs, Util. Math. 98 (2015) 127–147. J. He, H. Liang, Complexity of Total k-Domination and Related Problems, Joint Conference of FAW 2011(the 5th International Frontiers of Algorithmics Workshop) and AAIM 2011 (the 7th International Conference on Algorithmic Aspects of Information and Management) 2011, pp. 147–155. [7] V.R. Kulli, On n-total domination number in graphs, in: Graph Theory, Combinatorics, Algorithms and Applications, SIAM, Philadelphia, 1991 pp. 319–324. [8] V.R. Kulli, B. Janakiram, The total global domination number of a graph, Indian J. Pure. Appl. Math. 27 (1996) 537–542.
Please cite this article in press as: S. Bermudo, et al., On the global total k-domination number of graphs, Discrete Applied Mathematics (2018), https://doi.org/10.1016/j.dam.2018.05.025.