The Scientific World Journal
Volume 2016 (2016), Article ID 1029024, 7 pages
http://dx.doi.org/10.1155/2016/1029024

Research Article

On Vertex Covering Transversal Domination Number of Regular Graphs

R. Vasanthi¹ and K. Subramanian²

¹Department of Mathematics, Alagappa Chettiar College of Engineering and Technology, Karaikudi, Tamil Nadu 630 004, India
²Department of Mathematics, Alagappa Government Arts College, Karaikudi, Tamil Nadu 630 003, India

Received 19 December 2015; Revised 22 February 2016; Accepted 6 March 2016

Academic Editor: Kinkar Ch Das

Copyright © 2016 R. Vasanthi and K. Subramanian. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

A simple graph is said to be -regular if each vertex of is of degree . The vertex covering transversal domination number is the minimum cardinality among all vertex covering transversal dominating sets of . In this paper, we analyse this parameter on different kinds of regular graphs especially for and . Also we provide an upper bound for of a connected cubic graph of order . Then we try to provide a more stronger relationship between and .

1. 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 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 for regular graphs. Also we try to provide a more stronger relationship between and .

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 . 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 . 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 .

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 , then .

2. Notations

We use the following notations throughout the paper:

  -set to denote minimum vertex covering set,  -set to denote maximum independent set,  -set to denote a dominating set of minimum cardinality,  -set to denote a vertex covering transversal dominating set of minimum cardinality,   to denote domination number of ,   to denote vertex covering transversal domination number of ,   to denote independent domination number of ,   to denote the order of ,   to denote the degree of a vertex in .

3. 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 for certain family of regular graphs defined in [3].

Example 2. is a ()-regular graph and for .

Example 3. is a 2-regular graph of order and

Example 4. is a complete bipartite -regular graph and .

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 vertices which is -regular, then

Proof. The -dimensional cube or hypercube contains 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 is the -tuple binary number and its complement is also an -tuple binary number obtained by replacing 0 by 1 and 1 by 0 in . The weight of a vertex is the number of 1’s occurring in it. There are exactly vertices of odd weight and vertices of even weight. Each edge of consists of a vertex of even weight and a vertex of odd weight. The vertices of even weight form an independent set and so do the vertices of odd weight. Therefore is bipartite with bipartitions and where is the set of all -tuples of even weight and is the set of all -tuples of odd weight with .
Also and are the only -sets of . Since they are complements of each other, and are the only -sets of .
For , is as shown in Figure 1. Obviously, and are -sets of . Then is a -set of and so = 2.
Now suppose .
For , is the hypercube on 8 vertices which is 3-regular and is represented as in Figure 1. The only two - sets of are and . Then since every two-element set of the form where and is a -set.
If , the hypercube contains vertices and is 4-regular as shown in Figure 2. is bipartite with bipartitions and . Also and are the only -sets of .
Let . Then is a dominating set of . Clearly it intersects both and . Therefore is a vertex covering transversal dominating set and so . Hence it remains to show that is of minimum cardinality.
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 and . Suppose . Since each vertex is of degree 4, all the three vertices in may dominate at most 12 vertices. But there are 16 vertices in 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 , say, 1100, 0011, and the other two mutually complementary vertices from , say, 1000, 0111, form a -set. Thus where and is a -set of .
If , the hypercube contains = 32 vertices and the bipartition contains vertices and contains vertices. Let where and . Then is a -set which intersects both and . Hence = .
Thus in general, is a -set of . In particular, if is odd, and . If is even, and . Hence .

Figure 1

Figure 2

Theorem 6. If is a connected regular graph of degree and , then .

Proof. Choose any vertex . Then ; that is, is adjacent to vertices in . Then there remains exactly one vertex, say, , which is not adjacent to . Therefore is an independent set of . Also is adjacent to 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 . Since dominates every vertex in except , and dominates vertices including , it is obvious that is a dominating set which intersects every minimum vertex covering set of . Also is of minimum cardinality in . Hence .
It is noted that is also a -set.

Remark 7. In the above theorem, should be even. For otherwise, if is odd, then is odd which is impossible as the number of vertices of odd degree in a graph is even.

Lemma 8. Given positive integers and , let be the family of graphs such that and with

  , , ,  ,  ,  ,  .

Then

(i),(ii) is connected and -regular,(iii).

Proof. contains subgraphs which we shall call blocks each containing vertices and isomorphic to each other. By the edge set , we observe that they are connected to each other.
Thus (i) and (ii) are obvious.
For , two connected blocks of each consisting of 11 vertices are as shown in Figure 3.
Now is a maximum independent set of . Then its complement is a minimum vertex covering set of .
Also and , are maximum independent sets in . Let and be the complement of each and . Then and , are minimum vertex covering sets in . Now the subgraph induced by in each block is a complete bipartite graph .
Since if , we have . Also each for and is a vertex covering transversal dominating set of . Then , and , is a dominating set for each block. Therefore , and , is a -set which intersects the only -sets , and of for each .
Hence .

Figure 3

Theorem 9. If , then there exists a connected -regular graph with where is the order of .

Proof. Let be defined as in Lemma 8. Then . Thus .

Lemma 10. Given positive integers and , let be the graph with and with

  , , ,  ,  ,  ,  .

Then

(i),(ii) is connected and ()-regular,(iii).

Proof. (i) and (ii) are obvious. If and , the graph is as shown in Figure 4.
It is clear that each , to , is a maximum independent set in . Therefore its complement , to , is an set in . Further and are the only -sets of . Now each , to , to is a dominating set intersecting and and also of minimum cardinality .
Hence .

Figure 4

Theorem 11. For every , there exists a connected -regular graph of order such that .

Proof. Let be defined as in Lemma 10. Then is a connected -regular graph with . Also .
Thus .

Remark 12. Theorems 9 and 11 hold good if is replaced by .

4. for Regular Cubic Graphs

In this section, we provide the vertex covering transversal domination number of some regular cubic graphs especially Harary graph 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 shown in Figure 5. It is a regular cubic graph.

Figure 5

has 6 vertices and 9 edges. Assume that the graph is labelled as shown in the diagram. It is clear that and for are -sets of . Then their complements and for are -sets of . Now each , is a -set for . Clearly it intersects each and . Therefore .

Example 14. Consider Peterson graph which is cubic regular shown in Figure 6.

Figure 6

Assuming that the graph is labelled as shown in Figure 6, it is obvious that , are -sets of . Then their complements , , are -sets of . Now are -sets intersecting each . Hence .

Note that are also -sets in .

Theorem 15. If is a Harary graph with , then .

Proof. is a 3-regular graph and so is even. By the definition of , every vertex is adjacent to the vertices , , and where .
Let . The graphs and are shown in Figure 7.
Case  1. Suppose where is odd.
Then and are the only -sets of .
Subcase  1. Let ().
Then is a -set which intersects and and .
Subcase  2. Suppose ().
Then is a -set which intersects and and .
Subcase  3. Suppose ().
Then is a -set which intersects and with .
Thus in all the subcases of Case  1, .
Case  2. Suppose where is even.
Then , , is a -set for each .
Therefore , , , is an -set for each .
Subcase  1. Let ().
Then , is a -set which intersects each for .
Subcase  2. Let ().
Then , is a -set which intersects each for .
Subcase  3. Let ().
Then , is a -set which intersects each for .
The -sets mentioned in all the subcases of Case  2 are also of cardinality .
Thus .

Figure 7

Remark 16. In most of the graphs considered by us, it is observed that .

Theorem 17. If is a connected cubic graph of order with , then .

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  1. Suppose itself is a -set. Then is an -set. Let where . Then is a vertex covering transversal dominating set of . Therefore . Hence (by Theorem 1 proved in [3]).
Case  2. Suppose is not a -set. But is a maximal independent dominating set of minimum cardinality. We claim that intersects every -set of .
Suppose that does not intersect an -set of . Then where is a -set of . This is a contradiction to the maximality of .
Hence itself is a vertex covering transversal dominating set of . Therefore .
Thus Cases  1 and 2 imply that .

5. Relation between and

In this section, we prove a more stronger relationship between and 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 and .

Theorem 18. If is a simple connected graph, then .

Proof. Let be a minimum dominating set. If , then obviously . If not, then and . Let . Then is dominated by some vertex in . Let . Since is an edge in , either or is included in every minimum vertex covering set of . This implies that intersects every minimum vertex covering set in . Hence .

Theorem 19. Let be a simple connected graph. If there exists a -set which is not independent, then .

Proof. Let be a minimum dominating set which is not an independent set of . Then at least two vertices, say, , in , are adjacent to each other. Therefore is an edge in and hence either or lies in every minimum vertex covering set of . So intersects every -set of . Therefore itself is a -set. Hence .

Remark 20. The converse is not true. If , then there may exist a -set which is independent also. For example, consider , the cycle on 6 vertices as shown in Figure 8.
Obviously and are the only -sets of . Also is a -set which is independent. Further, it is a -set as it intersects both the -sets of . Thus there exists a -set which is independent in even though .

Figure 8

Remark 21. Now, the obvious question is “If , is every -set of a -set?” The answer is “not always.” The -sets and -sets in the graphs and discussed in the previous sections are the best examples for it. So it is noted that this happens if there exists a -set which is also a -set. It obviously produces the result that “If , then there exists at least one -set in which is not a -set.” The next general question is that “What happens if all the -sets of are -sets?”. The following theorem provides the answer to it.

Theorem 22. Let be a simple connected graph. If every -set of is a -set, then .

Proof. Since every -set of is a -set, choose a vertex in its complement. This is possible since as is a -set of a connected graph . Obviously is not a -set as it does not intersect the -set . Let . We claim that intersects every -set of . Suppose that for some -set in . Then where is a -set. This implies that which is a contradiction. Hence intersects every -set of . Also is a -set of as it contains exactly one vertex more than that of the -set . Thus .

Remark 23. It is easy to conclude that even though , there are graphs in which -sets do not become -sets. This implies that the collection of -sets in such graphs is contained in the collection of -sets. So this may lead to consider -sets in the graphs for which when we are in a situation to select a minimum number of -sets in such graphs. This approach may affect a new variation in domination theory.

Competing Interests

The authors declare that there are no competing interests regarding the publication of this paper.

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