1. Introduction
We start by introducing the basic notation. Suppose we are given a simple graph with (n is called the order of graph G) and (m is called the size of graph G). Given () and vertex , let be the set of all vertices from set D, adjacent to vertex v (also called the neighbors of vertex v from set D); we will use for the set of vertices in set D which are not neighbors of vertex v (). We let , and we call the degree of vertex v in set D. We denote by the cardinality of set (). We will use more compact notation , , , and instead of , , , and , respectively, when this will cause no confusion. The minimum (the maximum, respectively) degree in graph G is traditionally denoted by (, respectively). and , respectively, will stand for the subgraph of graph G induced by and the complement of graph G, respectively.
Let X and Y be subsets of set V. We denote by the set of all the edges in graph G joining a vertex with a vertex . Let u and v be vertices from set V. Then the distance between these two vertices is the length (the number of edges) of a minimum -path. The length of the longest path, for any u and v, is called the diameter of graph G, denoted by . The girth of graph G is the length of the shortest cycle in that graph and is denoted by .
Let
be a nonempty subset of set
V. Then
D is called a total
k-dominating set for graph
G if there are at least
k vertices in set
D adjacent to every vertex
(we will also say that vertex
v is totally
k-dominated by set
D). The cardinality of a total
k-dominating set in graph
G with the minimum cardinality is called the total
k-domination number of graph
G and is denoted by
. We will refer to a total
k-dominating set with cardinality
as a
-set. A total 1-dominating set is normally referred to as a total dominating set, and the total 1-domination number is referred to as the total domination number, denoted by
. We refer the reader to [
1,
2,
3,
4,
5,
6,
7,
8,
9] for more detail on these definitions.
Given again a non-empty set
,
D is called a global total
k-dominating set of graph
G (GT
kD set for short) if
D is a total
k-dominating set of both graphs
G and
. The global total
k-domination number of
G, denoted by
, is the cardinality of a global total
k-dominating set with the minimum cardinality. A global total
k-dominating set of cardinality
will be referred to as a
-set. Again, if
, a global total 1-dominating set is a global total dominating set (see [
10,
11]).
As it is well-known and also easily be seen,
for any graph
G with order
n. Here we shall exclusively deal with the connected graphs due to a known fact that if
(
) are the connected components in graph
G, then
(see [
12]).
The main goal of this paper is to complete the current study of global total k-domination number in graphs. First, we give upper and lower bounds on , and then we develop a method that generates a GTkD-set from a GTD-set for the successively increasing values of k. Based on this method, we establish a relationship between and , which, in turn, provides another upper bound on .
The rest of the paper is organized as follows. In the next section, we present known results and give some remarks. In
Section 3 and
Section 4, we derive upper and lower bounds, respectively, for global total
k-domination number. In the
Section 5, we provide our method that obtains a global total
-dominating set from a global total
k-dominating set.
2. Relations between and
Clearly, the definition of a GTkD set gives us an implicit lower bound for the parameter :
Observation 1. Let G be a graph; then .
The above lower bound is not necessarily attainable, as we illustrate in the following figure: we depict graph
G and its complement
, and the corresponding minimum total 2-dominating set in both graphs (black vertices); see
Figure 1.
The following proposition was proved in [
12].
Proposition 1. Let G be a graph,
- (i)
If , then .
- (ii)
If , then .
Corollary 1. Let G be a graph with maximum degree Δ. Then, .
Proposition 2. Let G be a graph with order n and maximum degree Δ. If , then .
Proof. If , then ; consequently, . By Corollary 1 we have . □
Theorem 1. For any graph G, if and only if there exists a minimum total k-dominating set D such that any subset of D with vertices is not included in any star in the graph—that is, and only if there is not a vertex such that .
Proof. Let D be a minimum total k-dominating set which is also a global total k-dominating set, and let be a subset of D with cardinality . If there exists a vertex such that , then and it is adjacent to vertices in , so v has less than k non-adjacent vertices in D, or , and it is adjacent to vertices in , so v has less than k non-adjacent vertices in D. In both cases we have a contradiction with the fact that D is a global total k-dominating set.
On the other hand, we take a minimum total k-dominating set D such that for any subset of D with vertices and every vertex , we have . Then, for any vertex we have , so v has, at least, k non-neighbors in D. If we have , so v has, at least, k non-neighbors in D. Therefore, D is a global total k-dominating set. □
3. Upper Bounds for the Global Total k-Domination Number
In this section, we obtain some upper bounds for the global total
k-domination number in a graph. Bermudo et al. in [
12] showed a characterization when the global total
k-domination number is equal to the order of the graph, but we give here that characterization in a more specific way. To do that, in the following proposition we give a condition to guarantee that the global total
k-domination number is less than
n.
Proposition 3. Let G be a graph with order n, minimum degree δ and maximum degree Δ. If , then .
Proof. Let us see that, for any , the set is a GTkD set of G. We have that and . For every we have and . Therefore, D is a GTkD set of G. □
Proposition 3 is not an equivalence, as we can see if we consider a triangle and we add a leaf to every vertex of the triangle. In such a case and .
Now, in order to present the characterization of all graphs having a global total k-domination number equal to the number of vertices, we need to define the following set. Given a graph G and an integer i, let (i.e., the set of vertices in graph G with the degree i).
Theorem 2. Given graph G with order n and the minimum and the maximum degrees δ and Δ, if and only if one of the conditions (a)–(c) below hold
- (a)
and .
- (b)
and .
- (c)
and .
Proof. (a) If and , we consider for any . We note that there exists such that ; this implies that . Thus, D is not a GTkD set of G. Hence, .
(b) If and , for any there exists such that and . If we consider , then ; thus, D is not a GTkD set of G. Therefore, .
(c) If and , using (a) or (b), we obtain that is not a GTkD set of G, for any . Consequently, .
Finally, if we assume that , by Proposition 3 we have that . For every vertex , we note that is not a GTkD set of G, so there exists such that or . If , since , then we have that ; this implies that and . If , since , then we have that ; that is, , so and . If , since or , we have that and , or and . □
The following corollary was directly obtained from Theorem 2.
Corollary 2. Let G be a graph with minimum degree δ, maximum degree Δ and order . Then if and only if one of the following condition holds.
- (a)
and .
- (b)
and .
Corollary 3. Let G be a graph of order n, minimum degree δ and maximum degree Δ. If one of the following conditions holds:
- (a)
and .
- (b)
and
- (c)
and or ,
then .
Proof. Since , , and , it is enough to check that implies . However, for any vertex , if , then there exists a vertex which is not a neighbor of v, so . □
It was proved in [
12] that
. It would be convenient to characterize the graphs
G such that
, and the graphs
G such that
. On the other hand, the invariants of a graph are important when characterizing them; below we use some of them such as diameter and girth. The following proofs use the ideas showed in [
11].
Theorem 3. If G is a graph such that , every total k-dominating set is a GTkD set of G.
Proof. Let D be a total k-dominating set and such that . Since and , there exist and . For any vertex we know that . If for some , then ; that means, . Therefore, D is a GTkD set of G. □
Corollary 4. If G is a graph such that , then .
According to the idea given in [
11], we obtain the following result.
Proposition 4. If G is a graph such that and there exist such that for every , then .
Proof. Let D be a minimum total k-dominating set of a graph; then there exists the vertex set such that . For any vertex and , w cannot be adjacent to both and , so is a global total k-dominating set. □
In
Figure 2 we can see an example where the equality in Proposition 4 for
is attained. Taking into account that any neighbor of a vertex of degree 2 must belong to any total 2-dominating set (grey vertices), we show in that figure the minimum total 2-dominating set (b) and the minimum global total 2-dominating set (c).
For a graph G, we let .
Proposition 5. Let G be a graph of order n and minimum degree δ; then .
Proof. Let us see that every set such that is a global total k-dominating set. Since , every vertex v satisfies , . Since , every vertex v satisfies , so . □
4. Lower Bounds for the Global Total k-Domination Number
We know that any graph
G satisfies
, and a characterization for graphs satisfying the equality was given in [
12]. Additionally, in that work the authors showed the following inequality.
Remark 1. Let G be a graph with order n, minimum degree δ and maximum degree Δ
. Then, For example, the lower bound given above can be reached in the graph shown in
Figure 3.
Theorem 4. Let G be a graph of order n, maximum degree Δ
and size m. Then Proof. Let
D be a
-set. Since every vertex in
cannot have more that
neighbors in
D, we have
, so
which implies that
then
□
Theorem 5. Let G be a graph with order n, maximum degree Δ
and size m. Then, Proof. We suppose that D is a -set and for some , and . Since D is minimal, for any vertex there exists a vertex such that one of the following conditions holds.
- (1)
, and ,
- (2)
, and ,
- (3)
, and ,
- (4)
, and .
Now, in cases (1) and (3), we take
, and in cases (2) and (4), we take
, and we know that there exists a vertex
such that one of the above conditions holds. Since
we can obtain
vertices satisfying one of the conditions above. We suppose that there exist
i,
,
s and
vertices satisfying (1), (2), (3) and (4), respectively. Then,
and
therefore,
then
□
Let us see another lower bound using the algebraic connectivity. Given a graph
G, its adjacency matrix
A and the diagonal matrix
D whose entries are the degrees of all vertices in the graph, the Laplacian matrix is defined as
. The algebraic connectivity of
G, denoted by
is the second smallest eigenvalue of the Laplacian matrix. The algebraic connectivity of
with order
n satisfies the following equality given by Fielder [
13].
where
and
.
Theorem 6. Let G be a graph with order n and algebraic connectivity μ. Then, Proof. Let
D be a
-set. It can be found that if we take
in the set given above, since
is the minimum, we have
therefore,
. □
Theorem 7. Let G be a graph of order n and maximum degree Δ
. If , then Proof. Let
D be a
-set. For every
, if we suppose that
, we have
, then
which implies that
, or equivalently, that
; that is,
.
If , since and , we can obtain the same result. □
The lower bound given in Theorem 7 is attained, for instance, in the graph given in
Figure 4.
In graph theory, it is common to analyze graphs obtained by some transformation from an originally given graph. An example of such a transformation is the elimination of one or more edges of the graph. Given a graph G, it is natural to think about what happens if you add or delete edges on the graph. We note that removing an edge in G is equivalent to adding an edge to graph . Therefore, it suffices to study just one of these cases.
Proposition 6. Let G be a graph with order n, minimum degree δ and maximum degree Δ
, and let . Then the following inequalities are satisfied (for an edge e): Proof. Let G be a graph and D be a -set, and we consider . Notice that , or ; we will divide the proof into three cases and we denote .
Case 1: If . Note that every vertex in has at least k neighbors and k non-neighbors in D. Therefore, .
Case 2: If . Let , where and . We note that for every , and . On the other hand, note that in , and if in , then . Now, if in , then there exists such that . Therefore, is a GTkD set of , so .
Case 3: If . Let where . We note that for every , and . In the worst case and ; the others cases are solved as the above; there exists such that and . Now, if then is a GTkD set of and ; otherwise, and then is a GTkD set of ; hence .
Thus, the first inequality is satisfied: . Now, as we say above for this problem, removing an edge in G is analogous to adding an edge in . Since and are complementary graphs and it is known that , it is verified that . Hence, by the first inequality . So, . □
Let S be a subset of set V such that the maximum degree of the subgraph induced by the vertices from set S is no more than . Then set S will be referred to as a k-independent set of vertices. The cardinality of a k-independent set of the maximum cardinality will be referred to as the k-independence number in graph G and will be denoted by . The lower k-independence number is the minimum cardinality of a maximal k-independent set in graph G.
Proposition 7. Let D be a global total k-dominating set in G and let be a maximum -independent. Then, Proof. Since is a maximal -independent set, is a dominating set; thus, . Moreover, . □
5. Deriving Upper Bounds for from
It is intuitively clear that the greater k is, the more difficult is to find a global total k-dominating set of graph with the minimum cardinality. In particular, the following relationship is easy to see: , for every . Ideally, one would wish to have a method that obtains a GTD set of minimum cardinality from a GTkD set with the minimum cardinality. It is clear that this is not an easy task. In this next section we develop a method that generates a GTD set from a GTkD, based on which we establish a relationship between minimum cardinality GTkD and GTD sets—more precisely, between and , which, in turn, provides upper bounds for .
We first need to introduce some necessary definitions. Given , a subset of the set of vertices V, let be the set of vertices from having at least one neighbor in D; that is, |∃ such that . Similarly, we denote by the set of vertices from having at least one non-neighbor in D.
Now let A and B be subsets of set V. We will say that a subset is a relative dominating set of B from set A if for every there exists at least one vertex such that or . Correspondingly, we call the minimum cardinality of such a relative dominating set the relative domination number of set B from set A and denote it by . We abbreviate by -set a relative dominating set of B from set A of cardinality .
Finally,
is the relative domination number of
B from set
A in graph
and
-set is a relative dominating set of
B from set
A in graph
with cardinality
; see an example in
Figure 5.
Lemma 1. Let G be a graph with diam(G) = 2 and g(G) = 4, and let S be an induced subgraph isomorphic to . Let and . Then .
Proof. Let be a -set, and . Note that since , . Thus, we can see that , or . If , then it has at least one neighbor in S and hence also in D. On the other hand, if , then v must have at least one neighbor in and hence also in D. If , then v has at least one neighbor in S, and hence also in D. Therefore, D is a total 1-dominating set of G.
If , then there exists one non-neighbor vertex of v in S, and hence also in D. If , then the four vertices in S are non-neighbors of v, and hence vertex v has at least one non-neighbor in set D. If , since , v it has at most two neighbors in S; thus, it has at least two non-neighbors in S and hence also in D. Therefore, D is a global 1-dominating set of G. Finally, D is a global total 1-dominating set of G, so . □
Corollary 5. Let G be a graph with diam = 2 and g = 4; let S be an induced subgraph isomorphic to , and . Then the following conditions hold.
If , then .
Since , .
If , , then .
Let k be a positive integer with , and D be a -set for graph G. Below we define special sets of vertices that will be used in future derivations.
.
| and are all vertices in H which are global total (k + 1)-dominated.
are all vertices in D with only k neighbors.
are all vertices in D with only k non-neighbors.
are all the vertices in H which have at least one neighbor in set X.
-set, a relative dominating set of X from set .
are all the vertices in set H which have at least one non-neighbor in set Y.
-set, a relative dominating set of X from set in .
are all the vertices in set H which are not yet global total ()-dominated.
-set ∪-set;
;
Now we show that the set S obtained as above is a global total -dominating set given a -set D.
Theorem 8. Let G be a graph and D be an arbitrary -set. Then the set S obtained as above is a global total -dominating set of graph G.
Proof. Let D be an arbitrary -set, , : and , and . Further, let , E be a -set, F be a -set and (all these sets being constructed as above specified). If and , then every vertex from has at least adjacent and non-adjacent vertices in set D. Besides, note that every vertex has at least adjacent and non-adjacent vertices in set . Additionally, since , is a global total -dominating set of graph G.
Assume now that and , and let and N be a -set (notice that by the construction of the set , there always exists the set N). Observe that every vertex from set has at least adjacent and non-adjacent vertices in set . Besides, every vertex has at least adjacent and non-adjacent vertices in set . Since , is a global total -dominating set of G.
The case and is analogous to the above case. We obtain that is a global total -dominating set of G, where and R is a -set.
Finally, assume that and . Let , , N be a -set and R be a -set. Using a similar arguments as above, we again obtain that S is a global total -dominating set of graph G. □
In the next proposition we derive an upper bound on the cardinality of the global total -domination number. In the same lemma, we give a necessary condition when the global total -domination number is equal to the total -domination number.
Proposition 8. Let G be a graph with and D be a -set. Then the following conditions hold:
- (a)
.
- (b)
If , then .
Proof. (a) By Theorem 8, S is a global total -dominating set of G; hence, the bound trivially holds.
(b) Recall that
. Additionally, it is easy to see that
is a total
-dominating set of
G. In [
12] it is proved that if
, then
(see Proposition 2.10). Hence, if
, then
. Hence, if
then
. □
Using the definition of the above introduced sets and Theorem 8 and Proposition 8, we can obtain a global total
k-domination set for any
. As a side-result, we also obtain the corresponding upper bounds to a global total
k-domination number. Finally, we note that this procedure provides a global total
k-dominating set of minimum cardinality,
, for some graphs; see
Figure 6.