`International Journal of CombinatoricsVolumeΒ 2012Β (2012), Article IDΒ 831489, 7 pageshttp://dx.doi.org/10.1155/2012/831489`
Research Article

Graphs with Constant Sum of Domination and Inverse Domination Numbers

Department of Mathematics, Manonmaniam Sundaranar University, Tamil Nadu, Tirunelveli 627 012, India

Received 1 March 2012; Accepted 10 July 2012

Copyright Β© 2012 T. Tamizh Chelvam and T. Asir. 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 subset D of the vertex set of a graph G, is a dominating set if every vertex in is adjacent to at least one vertex in D. The domination number is the minimum cardinality of a dominating set of G. A subset of , which is also a dominating set of G is called an inverse dominating set of G with respect to D. The inverse domination number is the minimum cardinality of the inverse dominating sets. Domke et al. (2004) characterized connected graphs G with , where n is the number of vertices in G. It is the purpose of this paper to give a complete characterization of graphs G with minimum degree at least two and .

1. Introduction

Let be a simple graph. For , if every vertex in is adjacent to at least one vertex in , then is said to be a dominating set of [1].A dominating set is said to be a minimal dominating set if no proper subset of is a dominating set of . The minimum cardinality among all dominating sets of is called domination number of , and it is denoted by . Any dominating set of with cardinality is noted as a set of [1]. Let be a -set of . If contains a dominating set of , then is called an inverse dominating set with respect to . The minimum cardinality of all inverse dominating sets is called the inverse domination number [2] and is denoted by . An inverse dominating set is called a - set if . By virtue of the definition of the inverse domination number, . The concept of the inverse domination number was introduced by Kulli and Sigarkanti [2]. It is well known by Oreβs Theorem [3] that if a graph has no isolated vertices, then the complement of every -set contains a dominating set. Thus any graph with no isolated vertices contains an inverse dominating set. However, for graphs with isolated vertices, one cannot find an inverse dominating set. For this reason, hereafter, we restrict ourselves to graphs with no isolated vertices.

A Gallai-type theorem has the form , where and are parameters defined on the graph , and is the number of vertices in . Cockayne et al. [4] proved certain Gallai-type theorems for graphs. In the year 1996, Cockayne et al. characterized graphs with and . Since then Baogen et al. [5] and Randerath and Volkmann [6] independently characterized all graphs satisfying . Next in the year 2004, Domke et al. [7] characterized graphs for which . Later on, in the year 2010, Tamizh Chelvam and Grace Prema [8] characterized graphs with where is an odd positive integer. Now, in this paper we characterize all graphs with for which .

Motivated by the inverse domination number, Hedetniemi et al. [9] defined and studied the disjoint domination number of a graph . A pair of disjoint sets of vertices is said to dominate a vertex , if and dominate . Further is a dominating pair, if dominates all vertices in .The total cardinality of a pair is , and the minimum cardinality of a dominating pair is the disjoint domination number of . As mentioned earlier, by Oreβs observation, for every graph without isolated vertices and Hedetniemi et al. characterized all extremal graphs for this bound. In this connection, the existence of two disjoint minimum dominating sets in trees was first studied by Bange et al. [10]. In a related paper, Haynes and Henning [11] studied the existence of two disjoint minimum independent dominating sets in a tree.

Another application of finding two disjoint sets is the one in respect of networks. In any network (or graphs), dominating sets are central sets, and they play a vital role in routing problems in parallel computing [12]. Also finding efficient dominating sets is always concern in finding optimal central sets in networks [13]. Suppose that is a -set in a graph (or network) , when the network fails in some nodes in , the inverse dominating set in will take care of the role of . In this aspect, it is worthwhile to concentrate on dominating and inverse dominating sets. Note that . From the point of networks, one may demand , where as many graphs do not enjoy such a property. For example consider the star graph . Clearly where as . If we consider the graph with , then and . In both the cases if is large, then is sufficiently large compare to .

The purpose of this paper is to characterize all graphs with for which . In this regards, it may be possible that is larger than and . But we prove that graphs with having exactly two disjoint minimum dominating sets. Hereafter denotes a simple graph on vertices with no isolated vertices. The minimum degree of a graph is denoted by . The set of neighbors of a vertex in a graph is denoted by , and the set of neighbors of in an induced subgraph of induced by is denoted by . Also and denote the path and cycle on vertices, respectively. The Cartesian product of graphs and is the graph whose vertex set is and whose edge set is the set of all pairs such that either and or and .

Let us first recall the following characterizations of graphs for which .

Theorem 1.1 (see [7, Theorem 2],). Let be a connected graph on vertices with . Then if and only if .

Theorem 1.2 (see [7, Theorem 3],). Let be a connected graph on vertices with and . Let be the set of all degree one vertices and . Then if and only if the following two conditions hold:(1) is an independent set and (2)for every vertex , every stem in is adjacent to at least two leaves.

2. Graphs with πΎ(πΊ)+πΎξ(πΊ)=πβ1

Tamizh Chelvam and Grace Prema [8] characterized graphs for which . In this context, we attempt to characterize graphs with for which . To attain this aim, we first present the theorem which is useful in the further discussion. To prove the following theorem, since no better proof technique is available, authors prefer case by case analysis.

Theorem 2.1. Let be a connected graph on vertices with . Then implies that .

Proof. Let be a -set of and be a -set of with respect to . Note that . Assume that , and let . Let be those vertices that are adjacent to more than one vertex in . Suppose that . Then and so . Let .
Claimββ1. There is at most one vertex in which is adjacent to a vertex in .
Suppose not, there are at least two vertices , in and such that is adjacent to and is adjacent to . Then either both are adjacent to or at least one of is not adjacent to .
Suppose that both are adjacent to . Since , are the only vertices in which are adjacent to , and are dominated by some vertices in , is a dominating set of , which is a contradiction to the fact that is a -set of .
When at least one of them, say , is not adjacent to . Since and , is adjacent to a vertex in . Therefore is a dominating set of , which is a contradiction to the fact that is a -set of . Hence, at most one vertex which is adjacent to a vertex in . By similar argument as given above, one can prove that is adjacent to exactly one vertex in . Let us take
Note that each vertex in has at least two neighbors in and so .
Claimββ2. ( is independent.)
Suppose that there exists a vertex which is adjacent to . Suppose that is not adjacent to both and . By the fact that each vertex in has at least two neighbors in and Claim 1, is a -set of , a contradiction. If is adjacent to one of or , say , then is a -set of , which is a contradiction. Hence is independent.
Now we have the following three possibilities. (1) is not adjacent to any of the vertices in .(2) is adjacent to exactly one vertex in .(3) is adjacent to more than one vertex in .
Caseββ1. Suppose that is not adjacent to any of the vertices of . If there exists which is adjacent to a vertex in , then is a -set with respect to , a contradiction. Therefore, Claim 2 along with together implies that each vertex in has at least two neighbors in . Suppose that there exists a vertex which is adjacent to , then, as in the proof of Claim 2, we get that either or is a -set of , a contradiction. Thus is independent.
Caseββ1.1. Suppose that there exists a pair of vertices such that for some .
Caseββ1.1.1.β βIf is adjacent to a vertex in , then, by the assumption in Case 1.1, the vertices in , dominated by either or , are also adjacent to some vertex in , and so is a -set of , which is a contradiction.
Caseβ1.1.2. If is adjacent to one of or , say , and let . Note that by assumption in Case 1, is not adjacent to as well as .
If there exists such that . Suppose that there exists no vertex in which is adjacent to only and , and then is a -set of , which is a contradiction. If there exists such that , then is a -set of and is a -set of with respect to , a contradiction. If there exists such that is adjacent to and only, then by similar argument one can get a contradiction in all the cases.
If there is no vertex such that , then, by Claim 2, is a -set of , and so is a -set of , which is a contradiction.
Caseββ1.2. Suppose that, for each pair of vertices , there exist at least two vertices such that .
Caseββ1.2.1. Suppose that, for some , there exists at most one vertex such that . If is adjacent to a vertex in , then is a -set, and so is a -set of , a contradiction. If , then is a -set and so is a -set of , a contradiction.
Caseββ1.2.2. Suppose that, for each pair of vertices , there exist at least two vertices such that . Note that and . Assume that and . If is adjacent to some vertex in , say , then by the assumption in Case 1.2, there exist and such that as , and are independent.
Assume that . Suppose that . Since dominates both and are the vertices dominated by and , is a -set of , a contradiction. Thus is adjacent to vertex in , say . Suppose that . Since each pair of vertices in has at least two neighbors, we have as a -set and as a -set of , a contradiction. Thus is adjacent to some vertex in , take . Proceed like this up to , and let . If , then is a -set, and so is a -set of , a contradiction. Hence is adjacent to at least one vertex in , which is not possible.
Let . If , then is adjacent to and only. Therefore is a -set, and is a -set of , a contradiction. If , then , which is a contradiction.
Caseββ2. Suppose that is adjacent to exactly one vertex . If is adjacent to a vertex in , then is a -set of , a contradiction. Thus every vertex in has at least two neighbors in .
If , then, as in Case 1, replacing by , we get contradiction in all the possibilities.
Let and . Since has at least two neighbors in , and so , which is contradiction to .
If , then since , , a contradiction to .
Caseββ3. Suppose is adjacent to more than one vertex in , say . If no vertex in is adjacent to only , then is a -set of , a contradiction. Thus there exists a vertex such that . If is adjacent to , then is a -set and is a -set of , a contradiction. Now let and let .
Suppose that there exists such that . Suppose there exists such that . Then is a -set of , a contradiction. Otherwise, is a -set and is a -set of , a contradiction.
Suppose that . If is adjacent to , then, is a -set of . If is not adjacent to , then as , is adjacent to at least a vertex say . Since , and , we get is a -set of , which is a contradiction.
Hence, .

Bange et al. [10] characterized trees with two disjoint minimum dominating sets. In the following corollary, we give the necessary condition for graphs with minimum degree at least two having two disjoint minimum dominating sets.

Corollary 2.2. Let be a connected graph on vertices with . If , then has two disjoint -sets.

The following example shows that in general Theorem 2.1 is not true whenever .

Example 2.3. Consider the graph , the path on vertices. Then and . Therefore, but .
Consider the graph in Figure 1. Clearly, and . Therefore whereas .

Figure 1: Graph .

Lemma 2.4. Let be a connected graph with . Then if and only if , and is odd.

Proof. If , then, by Theorem 2.1, . Therefore , and hence is odd. Conversely, assume that and is odd. Since is an odd integer, we get .

Let and be the families of graphs given in Figures 2 and 3, respectively.

Figure 2: Graphs in family .
Figure 3: Graphs in family .

Note that the class is a subclass of the class , and the class is same as the class where and are classes given in Theorem 2.6 [1, Page -45].

The next theorem characterizes all connected graphs with for which . By Lemma 2.4 and Lemma 2.4 [1], we get the main theorem of this paper.

Theorem 2.5. Let be a connected graph with . Then if and only if .

It may be worth noting that, for any graph , the disjoint domination number . Due to this, we get an upper bound for and which is better than the Oreβs observation for disjoint domination number and sum of domination number and inverse domination number [7].

Corollary 2.6. Let be a connected graph with and . Then .

We suggest the following problems for further study in this direction.

Open Problems
(1) Find a necessary and sufficient condition for a graph with and .
(2) Characterize all connected graphs with for which .

Acknowledgment

The work reported here is supported by the UGC Major Research Project F. no. 37-267/2009(SR) awarded to the first author by the University Grants Commission, Government of India. Also the work is supported by the INSPIRE program (IF 110072) of Department of Science and Technology, Government of India for the second author.

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