On Vertex Covering Transversal Domination Number of Regular Graphs

A simple graph G = (V, E) is said to be r-regular if each vertex of G is of degree r. The vertex covering transversal domination number γ vct(G) is the minimum cardinality among all vertex covering transversal dominating sets of G. In this paper, we analyse this parameter on different kinds of regular graphs especially for Q n and H 3,n. Also we provide an upper bound for γ vct of a connected cubic graph of order n ≥ 8. Then we try to provide a more stronger relationship between γ and γ vct.


Introduction
Hamid [1] introduced independent transversal domination in graphs. It was defined using maximum independent set in a graph. Vasanthi and Subramanian [2] introduced vertex covering transversal domination in graphs using minimum vertex covering set in a graph. The vertex covering transversal domination number of some standard graphs such as , , , , , and and trees is dealt with in paper [2]. Bounds of vct are also established through various parameters in [2]. Lam et al. [3] worked on independent domination number of regular graphs. In this paper, we investigate our parameter vct for regular graphs. Also we try to provide a more stronger relationship between and vct .
A simple graph = ( , ) is said to be -regular if each vertex of is of degree . A set ⊆ of vertices in is called an independent set if no two vertices in are adjacent. Also is said to be a maximum independent set if there is no other independent set such that | | > | |. The cardinality of a maximum independent set is called the independence number and is denoted by 0 ( ). A set ⊆ of vertices in is called a vertex covering set (or simply covering set) if every edge of is incident to at least one vertex in . Also is said to be a minimum vertex covering set if there is no other vertex covering set such that | | < | |. The cardinality of a minimum vertex covering set is called the vertex covering number and is denoted by 0 ( ). A set ⊆ of vertices in the graph is called a dominating set if every vertex in − is adjacent to a vertex in . A dominating set which intersects every minimum vertex covering set in is called a vertex covering transversal dominating set. The minimum cardinality of a vertex covering transversal dominating set is called vertex covering transversal domination number of and is denoted by vct ( ).
The parameter independent domination number ( ) was introduced by Cockanye and Hedetniemi in [4]. The independent domination number ( ) is the minimum cardinality among all independent dominating sets of . An independent set is dominating if and only if it is maximal. So ( ) is the minimum cardinality of a maximal independent set in . In paper [3], the following theorem which gives the upper bound for independent domination number of a connected cubic graph has been proved. Theorem 1. If is a connected cubic graph of order where ≥ 8, then ( ) ≤ 2 /5.

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The Scientific World Journal

Notations
We use the following notations throughout the paper: 0 -set to denote minimum vertex covering set, 0 -set to denote maximum independent set, -set to denote a dominating set of minimum cardinality, vct -set to denote a vertex covering transversal dominating set of minimum cardinality, ( ) to denote domination number of , vct ( ) to denote vertex covering transversal domination number of , ( ) to denote independent domination number of , ( ) to denote the order of , deg ( ) to denote the degree of a vertex in .

vct for Regular Graphs
Here, we provide the vertex covering transversal domination number of some standard regular graphs such as complete graphs, complete bipartite regular graphs, cycles, and hypercube . We also establish vct for certain family of regular graphs defined in [3].
The following theorem provides the vertex covering transversal domination number of -dimensional cube or hypercube defined in [5].

Theorem 5.
If is a hypercube containing 2 vertices which is -regular, then Proof. The -dimensional cube or hypercube contains 2 vertices and is -regular. Each vertex in is represented by a -tuple with 0's and 1's. Two vertices in are adjacent if and only if the -tuples differ in exactly one position. Also any V ∈ is the -tuple binary number and its complement V is also an -tuple binary number obtained by replacing 0 by 1 and 1 by 0 in V. The weight of a 0, 1 vertex is the number of 1's occurring in it. There are exactly 2 −1 vertices of odd weight  Suppose there exists a vertex covering transversal dominating set of cardinality less than 4. It must have at least 2 vertices as it intersects both 1 and 2 . Suppose | | = 3.
Since each vertex is of degree 4, all the three vertices in may dominate at most 12 vertices. But there are 16 vertices in 4 and so do not dominate at least 1 vertex. This is a contradiction to the assumption that is a vertex covering transversal dominating set.
Also any set containing two mutually complementary vertices from 1 , say, 1100, 0011, and the other two mutually complementary vertices from 2 , say, 1000, 0111, form a vctset.
Thus in general, Proof. Choose any vertex ∈ ( ). Then deg ( ) = − 2; that is, is adjacent to − 2 vertices in . Then there remains exactly one vertex, say, V, which is not adjacent to . Therefore = { , V} is an independent set of . Also V is adjacent to −2 vertices in except . Hence no other vertex may be included in . Therefore is a maximum independent set of . Now let ∈ − . Then is adjacent to both and V. Since dominates every vertex in except V, and dominates − 2 vertices including V, it is obvious that = { , } is a dominating set which intersects every minimum vertex covering set of . Also is of minimum cardinality in . Hence vct ( ) = 2.
It is noted that {V, } is also a vct -set.
Remark 7. In the above theorem, should be even. For otherwise, if is odd, then − 2 is odd which is impossible as the number of vertices of odd degree in a graph is even.
Lemma 8. Given positive integers ≥ 2 and ≥ 3, let ( , ) be the family of graphs such that = ⋃ =1 ( ∪ ∪ ) and Proof. ( , ) contains subgraphs which we shall call blocks each containing 3 −1 vertices and isomorphic to each other. By the edge set 4 , we observe that they are connected to each other.
Theorem 11. For every ≥ 4, there exists a connected -regular graph of order such that V ( ) = / .
Remark 12. Theorems 9 and 11 hold good if vct is replaced by .

vct for Regular Cubic Graphs
In this section, we provide the vertex covering transversal domination number of some regular cubic graphs especially Harary graph 3, defined in [6]. We also obtain an upper bound for the vertex covering transversal domination number of a connected cubic graph.
Example 13. Consider the triangular prism graph 3 shown in Figure 5. It is a regular cubic graph.
3 has 6 vertices and 9 edges. Assume that the graph 3 is labelled as shown in the diagram. It is Example 14. Consider Peterson graph which is cubic regular shown in Figure 6.

Remark 16.
In most of the graphs considered by us, it is observed that vct = .
Proof. Let be an independent dominating set of cardinality ( ). Then is a maximal independent set of minimum cardinality. Since is independent, no two vertices of are adjacent in . Let = − . Then the vertices in are adjacent only to the vertices in .

Case 2. Suppose
is not a 0 -set. But is a maximal independent dominating set of minimum cardinality. We claim that intersects every 0 -set of .
Suppose that does not intersect an 0 -set of . Then ⊂ − where − is a 0 -set of . This is a contradiction to the maximality of .

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Relation between ( ) and vct ( )
In this section, we prove a more stronger relationship between and vct than that proved in [2]. In view of the results and theorems dealt with in the previous sections, we try to characterize graphs for which = vct and < vct . Proof. Let be a minimum dominating set which is not an independent set of . Then at least two vertices, say, , V in , are adjacent to each other. Therefore V is an edge in and hence either or V lies in every minimum vertex covering set of . So intersects every 0 -set of . Therefore itself is a vct -set. Hence vct ( ) = ( ).
Remark 20. The converse is not true. If vct ( ) = ( ), then there may exist a -set which is independent also. For example, consider 6 , the cycle on 6 vertices as shown in Figure 8.
Obviously that this happens if there exists a -set which is also a -set. It obviously produces the result that "If vct ( ) = ( ) = ( ), then there exists at least one -set in which is not a vct -set." The next general question is that "What happens if all thesets of are 0 -sets?". The following theorem provides the answer to it.
Theorem 22. Let be a simple connected graph. If every -set of is a 0 -set, then V ( ) = ( ) + 1.
Proof. Since every -set of is a 0 -set, choose a vertex V in its complement. This is possible since ̸ = ( ) as is a 0 -set of a connected graph . Obviously is not a vct -set as it does not intersect the 0 -set ( ) − . Let = ∪ {V}. We claim that intersects every 0 -set of . Suppose that ∩ = Φ for some 0 -set in . Then ⊆ where = ( ) − is a 0 -set. This implies that 0 ( ) + 1 ≤ 0 ( ) which is a contradiction. Hence intersects every 0 -set of . Also is a vct -set of as it contains exactly one vertex more than that of the -set . Thus vct ( ) = ( ) + 1.
Remark 23. It is easy to conclude that even though vct = , there are graphs in which -sets do not become vct -sets. This implies that the collection of vct -sets in such graphs is contained in the collection of -sets. So this may lead to consider vct -sets in the graphs for which vct = when we are in a situation to select a minimum number of -sets