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International Journal of Combinatorics
VolumeΒ 2012Β (2012), Article IDΒ 831489, 7 pages
http://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

Academic Editor: MartinΒ Kochol

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 π‘‰βˆ’π· is adjacent to at least one vertex in D. The domination number 𝛾(𝐺) is the minimum cardinality of a dominating set of G. A subset of π‘‰βˆ’π·, which is also a dominating set of G is called an inverse dominating set of G with respect to D. The inverse domination number π›Ύξ…ž(𝐺) is the minimum cardinality of the inverse dominating sets. Domke et al. (2004) characterized connected graphs G with 𝛾(𝐺)+π›Ύξ…ž(𝐺)=𝑛, 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 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1.

1. Introduction

Let 𝐺=(𝑉,𝐸) be a simple graph. For π·βŠ†π‘‰, if every vertex in π‘‰βˆ’π· is adjacent to at least one vertex in 𝐷, then 𝐷 is said to be a dominating set of 𝐺 [1].A dominating set 𝐷 is said to be a minimal dominating set if no proper subset of 𝐷 is a dominating set of 𝐺. The minimum cardinality among all dominating sets of 𝐺 is called domination number of 𝐺, and it is denoted by 𝛾(𝐺). Any dominating set of 𝐺 with cardinality 𝛾(𝐺) is noted as a 𝛾 set of 𝐺 [1]. Let 𝐷 be a 𝛾-set of 𝐺. If π‘‰βˆ’π· contains a dominating set 𝐷′ of 𝐺, then 𝐷′ is called an inverse dominating set with respect to 𝐷. The minimum cardinality of all inverse dominating sets is called the inverse domination number [2] and is denoted by 𝛾′(𝐺). An inverse dominating set 𝐷′ is called a π›Ύξ…ž- set if |π·ξ…ž|=π›Ύξ…ž. By virtue of the definition of the inverse domination number, 𝛾(𝐺)β‰€π›Ύξ…ž(𝐺). The concept of the inverse domination number was introduced by Kulli and Sigarkanti [2]. It is well known by Ore’s Theorem [3] that if a graph 𝐺 has no isolated vertices, then the complement π‘‰βˆ’π· of every 𝛾-set 𝐷 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 𝛼(𝐺)+𝛽(𝐺)=𝑛, where 𝛼(𝐺) and 𝛽(𝐺) are parameters defined on the graph 𝐺, and 𝑛 is the number of vertices in 𝐺. Cockayne et al. [4] proved certain Gallai-type theorems for graphs. In the year 1996, Cockayne et al. characterized graphs with 𝛿(𝐺)β‰₯2 and 𝛾(𝐺)=βŒŠπ‘›/2βŒ‹. Since then Baogen et al. [5] and Randerath and Volkmann [6] independently characterized all graphs 𝐺 satisfying 𝛾(𝐺)=βŒŠπ‘›/2βŒ‹. Next in the year 2004, Domke et al. [7] characterized graphs for which 𝛾(𝐺)+π›Ύξ…ž(𝐺)=𝑛. Later on, in the year 2010, Tamizh Chelvam and Grace Prema [8] characterized graphs with 𝛾(𝐺)=π›Ύξ…ž(𝐺)=(π‘›βˆ’1)/2 where 𝑛 is an odd positive integer. Now, in this paper we characterize all graphs 𝐺 with 𝛿(𝐺)β‰₯2 for which 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1.

Motivated by the inverse domination number, Hedetniemi et al. [9] defined and studied the disjoint domination number 𝛾𝛾(𝐺) of a graph 𝐺. A pair (𝐷1,𝐷2) of disjoint sets of vertices 𝐷1,𝐷2βŠ†π‘‰ is said to dominate a vertex π‘’βˆˆπ‘‰, if 𝐷1 and 𝐷2 dominate 𝑒. Further (𝐷1,𝐷2) is a dominating pair, if (𝐷1,𝐷2) dominates all vertices in 𝑉.The total cardinality of a pair (𝐷1,𝐷2) is |𝐷1|+|𝐷2|, and the minimum cardinality of a dominating pair is the disjoint domination number 𝛾𝛾(𝐺) of 𝐺. As mentioned earlier, by Ore’s observation, 𝛾𝛾(𝐺)≀|𝑉(𝐺)| for every graph 𝐺 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 𝛾 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 𝑆 is a 𝛾-set in a graph (or network) 𝐺, when the network fails in some nodes in 𝑆, the inverse dominating set in Vβˆ’π‘† will take care of the role of 𝑆. In this aspect, it is worthwhile to concentrate on dominating and inverse dominating sets. Note that 𝛾′(𝐺)β‰₯𝛾(𝐺). From the point of networks, one may demand π›Ύξ…ž(𝐺)=𝛾(𝐺), where as many graphs do not enjoy such a property. For example consider the star graph 𝐾1,𝑛. Clearly 𝛾(𝐾1,𝑛)=1 where as 𝛾′(𝐾1,𝑛)=𝑛. If we consider the graph 𝐺=𝐾1,𝑛░𝐾2 with 𝑛β‰₯3, then 𝛾(𝐺)=2 and π›Ύξ…ž(𝐺)=𝑛. In both the cases if 𝑛 is large, then π›Ύξ…ž(𝐺) is sufficiently large compare to 𝛾(𝐺).

The purpose of this paper is to characterize all graphs 𝐺 with 𝛿(𝐺)β‰₯2 for which 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1. In this regards, it may be possible that π›Ύξ…ž(𝐺) is larger than 𝛾(𝐺) and 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1. But we prove that graphs 𝐺 with 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1 having exactly two disjoint minimum dominating sets. Hereafter 𝐺 denotes a simple graph on 𝑛 vertices with no isolated vertices. The minimum degree of a graph 𝐺 is denoted by 𝛿(𝐺). The set of neighbors of a vertex 𝑣 in a graph 𝐺 is denoted by 𝑁𝐺(𝑣), and the set of neighbors of 𝑣 in an induced subgraph of 𝐺 induced by π΄βŠ†π‘‰(𝐺) is denoted by 𝑁𝐴(𝑣). Also 𝑃𝑛 and 𝐢𝑛 denote the path and cycle on 𝑛 vertices, respectively. The Cartesian product of graphs 𝐺1 and 𝐺2 is the graph 𝐺1░𝐺2 whose vertex set is 𝑉(𝐺1)×𝑉(𝐺2) and whose edge set is the set of all pairs (𝑒1,𝑣1)(𝑒2,𝑣2) such that either 𝑒1𝑒2∈𝐸(𝐺1) and 𝑣1=𝑣2 or 𝑣1𝑣2∈𝐸(𝐺2) and 𝑒1=𝑒2.

Let us first recall the following characterizations of graphs for which 𝛾(𝐺)+π›Ύξ…ž(𝐺)=𝑛.

Theorem 1.1 (see [7, Theorem 2],). Let 𝐺 be a connected graph on 𝑛 vertices with 𝛿(𝐺)β‰₯2. Then 𝛾(𝐺)+π›Ύξ…ž(𝐺)=𝑛 if and only if 𝐺=𝐢4.

Theorem 1.2 (see [7, Theorem 3],). Let 𝐺 be a connected graph on 𝑛 vertices with 𝑛β‰₯1 and 𝛿(𝐺)=1. Let πΏβŠ†π‘‰ be the set of all degree one vertices and 𝑆=𝑁(𝐿). Then 𝛾(𝐺)+𝛾′(𝐺)=𝑛 if and only if the following two conditions hold:(1)π‘‰βˆ’π‘† is an independent set and (2)for every vertex π‘₯βˆˆπ‘‰βˆ’(𝑆βˆͺ𝐿), every stem in 𝑁(π‘₯) is adjacent to at least two leaves.

2. Graphs with 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1

Tamizh Chelvam and Grace Prema [8] characterized graphs for which 𝛾(𝐺)=𝛾′(𝐺)=(π‘›βˆ’1)/2. In this context, we attempt to characterize graphs 𝐺 with 𝛿(𝐺)β‰₯2 for which 𝛾(𝐺)+𝛾′(𝐺)=π‘›βˆ’1. 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 𝐺 be a connected graph on 𝑛 vertices with 𝛿(𝐺)β‰₯2. Then 𝛾(𝐺)+𝛾′(𝐺)=π‘›βˆ’1 implies that 𝛾(𝐺)=𝛾′(𝐺).

Proof. Let 𝐷 be a 𝛾-set of 𝐺 and π·ξ…ž be a π›Ύξ…ž-set of 𝐺 with respect to 𝐷. Note that 𝛾(𝐺)β‰€π›Ύξ…ž(𝐺). Assume that 𝛾(𝐺)+𝛾′(𝐺)=π‘›βˆ’1, and let 𝑉(𝐺)βˆ’{𝐷βˆͺπ·ξ…ž}={𝑀}. Let π‘†βŠ†π· be those vertices that are adjacent to more than one vertex in π·ξ…ž. Suppose that 𝛾(𝐺)<𝛾′(𝐺). Then |𝐷|<|π·ξ…ž| and so π‘†β‰ βˆ…. Let π‘†ξ…ž=𝑁(𝑆)βˆ©π·ξ…ž.
Claim  1. There is at most one vertex in π‘†ξ…ž which is adjacent to a vertex in π·βˆ’π‘†.
Suppose not, there are at least two vertices π‘‘ξ…ž, π‘Ÿξ…ž in π‘†ξ…ž and 𝑑,π‘Ÿβˆˆπ·βˆ’π‘† such that 𝑑′ is adjacent to 𝑑 and π‘Ÿβ€² is adjacent to π‘Ÿ. Then either both 𝑑,π‘Ÿβˆˆπ·βˆ’π‘† are adjacent to 𝑀 or at least one of 𝑑,π‘Ÿ is not adjacent to 𝑀.
Suppose that both 𝑑,π‘Ÿβˆˆπ·βˆ’π‘† are adjacent to 𝑀. Since π‘‘ξ…ž, π‘Ÿξ…ž are the only vertices in 𝑉(𝐺)βˆ’{𝐷βˆͺ{𝑀}} which are adjacent to 𝑑,π‘Ÿ, and 𝑑′,π‘Ÿβ€² are dominated by some vertices in 𝑆, 𝐷1=𝐷βˆͺ{𝑀}βˆ’{𝑑,π‘Ÿ}βŠ‚π· is a dominating set of 𝐺, which is a contradiction to the fact that 𝐷 is a 𝛾-set of 𝐺.
When at least one of them, say 𝑑, is not adjacent to 𝑀. Since π‘‘βˆˆπ·βˆ’π‘† and 𝛿(𝐺)β‰₯2, 𝑑 is adjacent to a vertex 𝑒 in 𝐷. Therefore 𝐷1=π·βˆ’{𝑑} is a dominating set of 𝐺, which is a contradiction to the fact that 𝐷 is a 𝛾-set of 𝐺. Hence, at most one vertex π‘‘ξ…žβˆˆπ‘†ξ…ž which is adjacent to a vertex in π·βˆ’π‘†. By similar argument as given above, one can prove that π‘‘ξ…ž is adjacent to exactly one vertex in π·βˆ’π‘†. Let us take π‘†ξ…ž1=ξ‚»π‘†ξ…žβˆ’{𝑑′}ifπ‘‘ξ…žπ‘†exists,ξ…žotherwise.(2.1)
Note that each vertex in 𝑆 has at least two neighbors in 𝑆′ and so π‘†ξ…ž1β‰ βˆ….
Claim  2. (π‘†ξ…ž1 is independent.)
Suppose that there exists a vertex π‘₯ξ…žβˆˆπ‘†ξ…ž1 which is adjacent to π‘¦ξ…žβˆˆπ‘†ξ…ž1. Suppose that 𝑀 is not adjacent to both π‘₯ξ…ž and π‘¦ξ…ž. By the fact that each vertex in 𝑆 has at least two neighbors in π·ξ…ž and Claim 1, π·β€²βˆ’{π‘₯ξ…ž} is a π›Ύξ…ž-set of 𝐺, a contradiction. If 𝑀 is adjacent to one of π‘₯ξ…ž or π‘¦ξ…ž, say π‘₯ξ…ž, then π·ξ…žβˆ’{𝑦′} is a π›Ύξ…ž-set of 𝐺, which is a contradiction. Hence π‘†ξ…ž1 is independent.
Now we have the following three possibilities. (1)𝑀 is not adjacent to any of the vertices in π‘†ξ…ž1.(2)𝑀 is adjacent to exactly one vertex in π‘†ξ…ž1.(3)𝑀 is adjacent to more than one vertex in π‘†ξ…ž1.
Case  1. Suppose that 𝑀 is not adjacent to any of the vertices of π‘†ξ…ž1. If there exists π‘₯ξ…žβˆˆπ‘†ξ…ž1 which is adjacent to a vertex in π·ξ…žβˆ’π‘†ξ…ž1, then π·ξ…žβˆ’{π‘₯ξ…ž} is a π›Ύξ…ž-set with respect to 𝐷, a contradiction. Therefore, Claim 2 along with 𝛿(𝐺)β‰₯2 together implies that each vertex in π‘†ξ…ž1 has at least two neighbors in 𝑆. Suppose that there exists a vertex π‘₯βˆˆπ‘† which is adjacent to π‘¦βˆˆπ‘†, then, as in the proof of Claim 2, we get that either π·βˆ’{π‘₯} or Dβˆ’{𝑦} is a 𝛾-set of 𝐺, a contradiction. Thus 𝑆 is independent.
Case  1.1. Suppose that there exists a pair of vertices 𝑒,π‘£βˆˆπ‘† such that 𝑁𝑆′1(𝑒)βˆ©π‘π‘†β€²1(𝑣)={π‘’ξ…ž} for some π‘’ξ…žβˆˆπ‘†ξ…ž1.
Case  1.1.1.   If 𝑀 is adjacent to a vertex in π·βˆ’{𝑒,𝑣}, then, by the assumption in Case 1.1, the vertices in π‘†ξ…ž1βˆ’{π‘’ξ…ž}, dominated by either 𝑒 or 𝑣, are also adjacent to some vertex in π‘†βˆ’{𝑒,𝑣}, and so π·βˆ’{𝑒,𝑣}βˆͺ{π‘’ξ…ž} is a 𝛾-set of 𝐺, which is a contradiction.
Case 1.1.2. If 𝑀 is adjacent to one of 𝑒 or 𝑣, say 𝑒, and let π‘’ξ…žβ‰ π‘£ξ…žβˆˆπ‘π‘†β€²1(𝑣). Note that by assumption in Case 1, 𝑀 is not adjacent to π‘’ξ…ž as well as π‘£ξ…ž.
If there exists π‘₯βˆˆπ‘† such that 𝑁𝑆′1(π‘₯)={π‘’ξ…ž,π‘£ξ…ž}. Suppose that there exists no vertex in π‘†ξ…ž1βˆ’{𝑒′,π‘£ξ…ž} which is adjacent to only 𝑣 and π‘₯, and then π·βˆ’{𝑣,π‘₯}βˆͺ{π‘£ξ…ž} is a 𝛾-set of 𝐺, which is a contradiction. If there exists π‘₯ξ…žβˆˆπ‘†ξ…ž1βˆ’{𝑒′,π‘£ξ…ž} such that 𝑁𝑆(π‘₯ξ…ž)={𝑣,π‘₯}, then 𝐷1=π·βˆ’{𝑣}βˆͺ{π‘£ξ…ž} is a 𝛾-set of 𝐺 and π·ξ…ž1=π·β€²βˆ’{π‘£ξ…ž,π‘₯ξ…ž}βˆͺ{𝑣} is a π›Ύξ…ž-set of 𝐺 with respect to 𝐷1, a contradiction. If there exists π‘¦βˆˆπ‘†βˆ’{𝑒,𝑣,π‘₯} such that 𝑦 is adjacent to π‘£ξ…ž and π‘₯ξ…ž only, then by similar argument one can get a contradiction in all the cases.
If there is no vertex π‘₯βˆˆπ‘† such that 𝑁𝑆′1(π‘₯)={𝑒′,π‘£ξ…ž}, then, by Claim 2, 𝐷1=π·βˆ’{𝑣}βˆͺ{π‘£ξ…ž} is a 𝛾-set of 𝐺, and so π·ξ…ž1=π·βˆ’{𝑒′,π‘£ξ…ž}βˆͺ{𝑣} is a π›Ύξ…ž-set of 𝐺, which is a contradiction.
Case  1.2. Suppose that, for each pair of vertices π‘₯,π‘¦βˆˆπ‘†, there exist at least two vertices π‘₯ξ…ž,π‘¦ξ…žβˆˆπ‘†ξ…ž1 such that {π‘₯ξ…ž,π‘¦ξ…ž}βŠ†π‘π‘†β€²1(π‘₯)βˆ©π‘π‘†β€²1(𝑦).
Case  1.2.1. Suppose that, for some π‘’ξ…ž,π‘£ξ…žβˆˆπ‘†ξ…ž1, there exists at most one vertex π‘’βˆˆπ‘† such that 𝑁𝑆(π‘’ξ…ž)βˆ©π‘π‘†(π‘£ξ…ž)={𝑒}. If 𝑀 is adjacent to a vertex in π·βˆ’{𝑒}, then 𝐷1=π·βˆ’{𝑒}βˆͺ{π‘’ξ…ž} is a 𝛾-set, and so π·ξ…ž1=π·ξ…žβˆ’{π‘’ξ…ž,π‘£ξ…ž}βˆͺ{𝑒} is a π›Ύξ…ž-set of 𝐺, a contradiction. If 𝑁𝐷(𝑀)={𝑒}, then 𝐷1=π·βˆ’{𝑒}βˆͺ{𝑀} is a 𝛾-set and so π·ξ…ž1=π·β€²βˆ’{π‘’ξ…ž,π‘£ξ…ž}βˆͺ{𝑒} is a π›Ύξ…ž-set of 𝐺, a contradiction.
Case  1.2.2. Suppose that, for each pair of vertices π‘₯ξ…ž,π‘¦ξ…žβˆˆπ‘†ξ…ž1, there exist at least two vertices π‘₯,π‘¦βˆˆπ‘† such that {π‘₯,𝑦}βŠ†π‘π‘†(π‘₯ξ…ž)βˆ©π‘π‘†(π‘¦ξ…ž). Note that |𝑆|β‰₯2 and |π‘†ξ…ž1|β‰₯2. Assume that |𝑆|=π‘˜ and 𝑆={𝑒1,𝑒2,…,π‘’π‘˜}. If 𝑀 is adjacent to some vertex in 𝑆, say 𝑒1, then by the assumption in Case 1.2, there exist 𝑒2βˆˆπ‘† and π‘’ξ…ž1,π‘’ξ…ž2βˆˆπ‘†ξ…ž1 such that ⟨{𝑒1,𝑒2,π‘’ξ…ž1,π‘’ξ…ž2}⟩=𝐾2,2 as 𝑆, and π‘†ξ…ž1 are independent.
Assume that |𝑆|=π‘˜β‰₯3. Suppose that 𝑁𝑠′1(𝑒3)={π‘’ξ…ž1,π‘’ξ…ž2}. Since π‘’ξ…ž2 dominates both 𝑒2,𝑒3 and π‘’ξ…ž1,π‘’ξ…ž2 are the vertices dominated by 𝑒2 and 𝑒3, 𝐷1=π·βˆ’{𝑒3,𝑒2}βˆͺ{π‘’ξ…ž2} is a 𝛾-set of 𝐺, a contradiction. Thus 𝑒3 is adjacent to vertex in π‘†βˆ’{π‘’ξ…ž1,π‘’ξ…ž2}, say π‘’ξ…ž3. Suppose that 𝑁𝑆(π‘’ξ…ž3)βŠ†{𝑒1,𝑒2,𝑒3}. Since each pair of vertices in 𝑆 has at least two neighbors, we have 𝐷1=π·βˆ’{𝑒2}βˆͺ{π‘’ξ…ž2} as a 𝛾-set and π·ξ…ž1=π·ξ…žβˆ’{π‘’ξ…ž2,π‘’ξ…ž3}βˆͺ{𝑒2} as a π›Ύξ…ž-set of 𝐺, a contradiction. Thus π‘’ξ…ž3 is adjacent to some vertex in π‘†βˆ’{𝑒1,𝑒2,𝑒3}, take 𝑒4. Proceed like this up to π‘’π‘˜, and let π‘’ξ…žπ‘˜βˆˆπ‘π‘ β€²1(π‘’π‘˜). If 𝑁𝑆(π‘’ξ…žπ‘˜)βŠ†π‘†, then 𝐷1=π·βˆ’{π‘’π‘˜βˆ’1}βˆͺ{π‘’ξ…žπ‘˜βˆ’1} is a 𝛾-set, and so π·ξ…ž1=π·ξ…žβˆ’{π‘’ξ…žπ‘˜βˆ’1,π‘’ξ…žπ‘˜}βˆͺ{π‘’π‘˜βˆ’1} is a π›Ύξ…ž-set of 𝐺, a contradiction. Hence π‘’ξ…žπ‘˜ is adjacent to at least one vertex in π‘†βˆ’{𝑒1,…,π‘’π‘˜}=βˆ…, which is not possible.
Let |𝑆|=π‘˜=2. If |π‘†ξ…ž1|β‰₯3, then π‘’ξ…ž3 is adjacent to 𝑒1 and 𝑒2 only. Therefore 𝐷1=π·βˆ’{𝑒2}βˆͺ{π‘’ξ…ž2} is a 𝛾-set, and π·ξ…ž1=π·ξ…žβˆ’{π‘’ξ…ž2,π‘’ξ…ž3}βˆͺ{𝑒2} is a 𝛾′-set of 𝐺, a contradiction. If |π‘†ξ…ž1|=2, then 𝐷=π·ξ…ž, which is a contradiction.
Case  2. Suppose that 𝑀 is adjacent to exactly one vertex π‘₯ξ…žβˆˆπ‘†ξ…ž1. If π‘’ξ…žβˆˆπ‘†ξ…ž1βˆ’{π‘₯ξ…ž} is adjacent to a vertex in π·ξ…ž, then π·ξ…žβˆ’{π‘’ξ…ž} is a π›Ύξ…ž-set of 𝐺, a contradiction. Thus every vertex in π‘†ξ…ž1βˆ’{π‘₯β€²} has at least two neighbors in 𝑆.
If |π‘†ξ…ž1|β‰₯3, then, as in Case 1, replacing π‘†ξ…ž1 by π‘†ξ…ž1βˆ’{π‘₯ξ…ž}, we get contradiction in all the possibilities.
Let |π‘†ξ…ž1|=2 and π‘†ξ…ž1={π‘₯ξ…ž,π‘¦ξ…ž}. Since π‘¦ξ…ž has at least two neighbors in 𝑆, |𝑆|β‰₯2 and so |𝑆|β‰₯|π‘†ξ…ž1|, which is contradiction to |𝐷|<|π·ξ…ž|.
If |π‘†ξ…ž1|=1, then since |𝐷|<|𝐷′|, 𝑆=βˆ…, a contradiction to π‘†β‰ βˆ….
Case  3. Suppose 𝑀 is adjacent to more than one vertex in π‘†ξ…ž1, say π‘’ξ…ž,π‘£ξ…žβˆˆπ‘†ξ…ž1. If no vertex in 𝑆 is adjacent to only π‘’ξ…ž,π‘£ξ…ž, then π·β€²βˆ’{𝑒′,π‘£ξ…ž}βˆͺ{𝑀} is a 𝛾-set of 𝐺, a contradiction. Thus there exists a vertex π‘’βˆˆπ‘† such that 𝑁𝑆′1(𝑒)={π‘’ξ…ž,π‘£ξ…ž}. If 𝑀 is adjacent to 𝑒, then 𝐷1=π·βˆ’{𝑒}βˆͺ{𝑀} is a 𝛾-set and π·ξ…ž1=π·β€²βˆ’{π‘’ξ…ž,π‘£ξ…ž}βˆͺ{𝑒} is a 𝛾′-set of 𝐺, a contradiction. Now let 𝑒≠π‘₯βˆˆπ‘π·(𝑀) and let π‘₯β€²βˆˆπ‘π·β€²(π‘₯).
Suppose that there exists π‘¦βˆˆπ·βˆ’{π‘₯} such that π‘¦βˆˆπ‘(π‘₯ξ…ž). Suppose there exists π‘§βˆˆπ·βˆ’{𝑒,π‘₯} such that π‘π·ξ…ž(𝑧)βŠ†{π‘’ξ…ž,π‘£ξ…ž,π‘₯ξ…ž}. Then π·βˆ’{𝑒,𝑧}βˆͺ{π‘’ξ…ž} is a 𝛾-set of 𝐺, a contradiction. Otherwise, 𝐷1=π·βˆ’{𝑒,π‘₯}βˆͺ{π‘’ξ…ž,𝑀} is a 𝛾-set and π·ξ…ž1=π·ξ…žβˆ’{π‘’ξ…ž,π‘£ξ…ž,π‘₯ξ…ž}βˆͺ{π‘₯,𝑒} is a π›Ύξ…ž-set of 𝐺, a contradiction.
Suppose that 𝑁𝐷(π‘₯ξ…ž)={π‘₯}. If π‘₯ξ…ž is adjacent to 𝑀, then, π·ξ…žβˆ’{π‘’ξ…ž,π‘₯ξ…ž}βˆͺ{𝑀} is a π›Ύξ…ž-set of 𝐺. If π‘₯ξ…ž is not adjacent to 𝑀, then as 𝛿(𝐺)β‰₯2, π‘₯ξ…ž is adjacent to at least a vertex say π‘¦ξ…žβˆˆπ·ξ…ž. Since π‘₯ξ…žβˆˆπ‘(π‘¦ξ…ž), π‘₯,π‘’ξ…žβˆˆπ‘(𝑀) and 𝑁𝐷(π‘₯ξ…ž)={π‘₯}, we get π·ξ…žβˆ’{π‘’ξ…ž,π‘₯ξ…ž}βˆͺ{𝑀} is a π›Ύξ…ž-set of 𝐺, which is a contradiction.
Hence, 𝛾(𝐺)=π›Ύξ…ž(𝐺).

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 𝐺 be a connected graph on 𝑛 vertices with 𝛿(𝐺)β‰₯2. If 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1, then 𝐺 has two disjoint 𝛾-sets.

The following example shows that in general Theorem 2.1 is not true whenever 𝛿(𝐺)=1.

Example 2.3. (i) Consider the graph 𝑃6, the path on 6 vertices. Then 𝛾(𝑃6)=2 and π›Ύξ…ž(𝑃6)=3. Therefore, 𝛾(𝑃6)+π›Ύξ…ž(𝑃6)=5 but 𝛾(𝑃6)β‰ π›Ύξ…ž(𝑃6).
(ii) Consider the graph 𝐺 in Figure 1. Clearly, 𝛾(𝐺)=3 and π›Ύξ…ž(𝐺)=4. Therefore 𝛾(𝐺)+π›Ύξ…ž(𝐺)=7=π‘›βˆ’1 whereas 𝛾(𝐺)β‰ π›Ύξ…ž(𝐺).

831489.fig.001
Figure 1: Graph 𝐺.

Lemma 2.4. Let 𝐺 be a connected graph with 𝛿(𝐺)β‰₯2. Then 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1 if and only if 𝛾(𝐺)=π›Ύξ…ž(𝐺)=βŒŠπ‘›/2βŒ‹, and 𝑛 is odd.

Proof. If 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1, then, by Theorem 2.1, 𝛾(𝐺)=𝛾′(𝐺). Therefore 𝛾(𝐺)=π›Ύξ…ž(𝐺)=(π‘›βˆ’1)/2, and hence 𝑛 is odd. Conversely, assume that 𝛾(𝐺)=π›Ύξ…ž(𝐺)=βŒŠπ‘›/2βŒ‹ and 𝑛 is odd. Since 𝑛 is an odd integer, we get 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1.

Let π’ž and π’Ÿ be the families of graphs given in Figures 2 and 3, respectively.

831489.fig.002
Figure 2: Graphs in family π’ž.
831489.fig.003
Figure 3: Graphs in family π’Ÿ.

Note that the class π’ž is a subclass of the class π’œ, and the class π’Ÿ is same as the class ℬ where π’œ and ℬ are classes given in Theorem 2.6 [1, Page -45].

The next theorem characterizes all connected graphs 𝐺 with 𝛿(𝐺)β‰₯2 for which 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1. By Lemma 2.4 and Lemma 2.4 [1], we get the main theorem of this paper.

Theorem 2.5. Let 𝐺 be a connected graph with 𝛿(𝐺)β‰₯2. Then 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1 if and only if πΊβˆˆπ’žβˆͺπ’Ÿ.

It may be worth noting that, for any graph 𝐺, the disjoint domination number 𝛾𝛾(𝐺)≀𝛾(𝐺)+π›Ύξ…ž(𝐺). Due to this, we get an upper bound for 𝛾𝛾(𝐺) and 𝛾(𝐺)+π›Ύξ…ž(𝐺) which is better than the Ore’s observation for disjoint domination number and sum of domination number and inverse domination number [7].

Corollary 2.6. Let 𝐺 be a connected graph with 𝛿(𝐺)β‰₯2 and πΊβˆ‰{𝐻1,…,𝐻11}. Then 𝛾𝛾(𝐺)≀𝛾(𝐺)+π›Ύξ…ž(𝐺)β‰€π‘›βˆ’2.

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

Open Problems
(1) Find a necessary and sufficient condition for a graph 𝐺 with 𝛿(𝐺)=1 and 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’1.
(2) Characterize all connected graphs 𝐺 with 𝛿(𝐺)β‰₯2 for which 𝛾(𝐺)+π›Ύξ…ž(𝐺)=π‘›βˆ’2.

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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