ON DISJUNCTIVE DOMINATION NUMBER OF CORONA RELATED GRAPHS

unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract. A disjunctive dominating set in a graph G is a set S⊂V (G) such that every vertex v ∈V (G)\S is either adjacent to a vertex in S or has at least 2 vertices in S at a distance 2 from it in G. The disjunctive domination number of G, denoted by γd 2 (G), is the minimum cardinality of a disjunctive dominating set in G. In this paper we investigate disjunctive domination number of some corona related graphs.


INTRODUCTION
Domination in graphs is one of the best tools for understanding situations that can be modeled by networks. A dominating set can be interpreted as a set of processors from which information can be passed on to all the other processors in the network. Hence the determination of the domination parameters is a significant problem. There are many variations of domination. A vertex can exert influence on, or dominate, all vertices not only in its immediate neighborhood, but in a specified distance not too far from it. Such a situation is handled in distance domination introduced by Meir and Moon [11] in 1975. For an integer k ≥ 1, a distance k-dominating set of a connected graph G is a set S of vertices of G such that every vertex of V (G) is at distance at most k from some vertex of S. The distance k-domination number γ k (G) of G is the minimum cardinality of a distance k-dominating set of G. A similar situation also occurs in exponential domination proposed by Dankelmann et al. [2] in 2009. Motivated by distance domination and exponential domination, Goddard et al. [5] introduced disjunctive domination. A subset S of vertices of G is called a disjunctive dominating set if every vertex in V \ S is either adjacent to a vertex in S or has at least 2 vertices in S at a distance 2 from it in G. The disjunctive domination number of G, denoted by γ d 2 (G), is the minimum cardinality of a disjunctive dominating set in G. Some properties of disjunctive domination are studied in [8].
The corona of two graphs is defined in [7] and some results on the corona of two graphs are given in [3]. Domination in corona of two graphs is studied in [4] and disjunctive domination in corona of two graphs is studied in [5]. Two variants of corona of graphs are defined in [10] and [6]. Exponential domination of these corona graphs are studied in [1]. In this paper we study disjunctive domination in neighborhood and edge corona of graphs.

Terminology and Notation.
From the definition of disjunctive domination we note that if S is a disjunctive dominating set and v ∈ V is not dominated by S, then it has at least 2 vertices in S at a distance 2 from it. In this case we say that v ∈ V is disjunctively dominated by S.
A universal vertex is a vertex that is adjacent to every other vertex in the graph.
A subdivision of an edge e = uv of a graph G is the replacement of the edge e by a path (u, w, v). The graph obtained from a graph G by subdividing each edge of G exactly once is called the subdivision graph of G and is denoted by S(G).
For all standard terminology and notation we follow [7]. The terms related to the theory of domination in graphs are used as in the sense of Haynes et al. [9]. The graph G considered in this paper is connected and nontrivial and H is an arbitrary graph, unless otherwise specified.

DISJUNCTIVE DOMINATION IN NEIGHBORHOOD CORONA OF GRAPHS
Definition 2.1. [6] Let G and H be two graphs on n and m vertices respectively. Then the neighborhood corona, G H is the graph obtained by taking n copies of H and for each i, making all vertices in the i th copy of H adjacent with the neighbors of v i ∈ G, i = 1, 2, ..., n.
Notation: H v denotes the copy of H in G H corresponding to v ∈ G.
In neighborhood corona G H if we consider H = K 1 then G H becomes a splitting graph.
The splitting graph was introduced by Sampathkumar and Walikar [12].   Theorem 2.6. For any two graphs G and H, where γ 2 (G) is the distance -2 domination number of G.
. Now let S be a distance-2 dominating set of G and let S be a set of vertices formed by taking exactly one vertex from each H v , v ∈ S.
The bounds given in the above theorem are sharp. For example the lower bound is achieved by the family of graphs G given in figure 1. The upper bound is achieved by the family of graphs G = S(K 1,n ) obtained from K 1,n by subdividing each edge once. The case when n = 2 and H = K 1 is illustrated in figure 2.
Observation 2.7. Corresponding to each positive integer k ≥ 2, there exists a graph G for which γ 2 (G) = k and γ d 2 (G H) = k + i where 0 ≤ i ≤ k. This is illustrated in figure 3. If there are (k − 1) copies of C 6 in each of the graphs G 0 , G 1 ,..., G k as shown in figure 3, then In each graph, there are k − 1 copies of C 6 where k ≥ 2.
Theorem 2.8. For any two graphs G and H, Equality is attained if and only if G has a universal vertex.
Proof. Let S be a γ d 2 -set of G. It can be observed that all vertices in G H, except the vertices in the copy H v corresponding to v ∈ S are dominated or disjunctively dominated by S. Let S be a set formed by taking exactly one vertex from H v corresponding to each v ∈ S. Then S ∪ S is a disjunctive dominating set of G H and |S ∪ S | = 2|S|. Hence γ d 2 (G H) ≤ 2γ d 2 (G). If G has a universal vertex, it follows from theorem 2.5 that If G has no universal vertex then, every γ d 2 -set of G must contain at least two vertices. Let S be a γ d 2 -set of G and u ∈ S. Then there exist at least one vertex v ∈ S such that d(u, v) ≤ 4.
Let S be a γ d 2 -set of G and S be a set formed by taking exactly one vertex from each copy of a vertex in S \ {u, v}. Then S ∪ S ia a disjunctive dominating set of G H and are both less than or equal to 2. If S and S are sets as in case.1, because the centre vertex together with one vertex in its copy is a disjunctive dominating set of G H. Hence in this case, Theorem 2.10. For any two graphs G and H, γ d Proof. Let S be a γ  and if n ≥ 3, Proof. Let v i be vertices of P n and H v i be the copy of H corresponding to v i ∈ P n , where i ∈ {1, 2, ..., n}.
set if k is odd. Number of vertices in these sets are 2k. Hence γ d 2 (P n H) ≤ 2k. The reverse inequality can be seen as follows. Let S be a disjunctive dominating set. In order to dominate or disjunctively dominate the vertex v 1 , the set S must contain a vertex from the set {v 1 , v 2 } ∪ H v 2 or two vertices from H v 1 ∪ {v 3 } ∪ H v 3 respectively. If S dominates the vertex v 1 , then either the vertex v 2 or vertices in H v 1 or Hv 2 is at a distance 2 from this vertex. In order to dominate or disjunctively dominate this vertex or vertices another vertex is required in its first or second neighborhood. If S disjunctively dominates v 1 , then it must contain at least 2 vertices from the set H v 1 ∪ {v 3 } ∪ H v 3 . Thus in order to dominate or disjunctively dominate vertices v 1 , v 2 and their copies of H at least two vertices are required from their first or second neighborhood. But these vertices can dominate or disjunctively dominate at the most vertices up to v 5 and their copies. Thus S must contain at least 2 vertices from the first five vertices on P n or their copies.
A similar argument shows that at least 2 vertices are required from every set of five consecutive vertices on P n and their copies.
case (ii) n ≡ 1, 2 (mod 5). Let n = 5k + 1 or 5k + 2. As in case (i) it can be seen that every disjunctive dominating set must contain at least 2 vertices to dominate or disjunctively dominate any set of 5 consecutive vertices and their copies. Thus 2k vertices are needed to dominate or disjunctively dominate any set of 5k consecutive vertices and their copies. In order to dominate or disjunctively dominate the remaining one or two vertices and their copies one more vertex is required. Hence γ d 2 (P n H) ≥ 2k + 1. On the other hand the set {v 2 , v 4 , v 8 , v 10 , v 14 , ..., v 5k , v 5k+1 } is a disjunctive dominating set of cardinality 2k + 1. Thus γ d 2 (P n H) ≤ 2k + 1. Thus when n = 5k + 1 or 5k + 2, γ d 2 (P n H) = 2k + 1 = 2 n 5 − 1.
case (iii) n ≡ 3, 4 (mod 5). Let n = 5k + 3 or 5k + 4. As in the above cases it can be proved that 2k + 2 vertices are necessary and sufficient to dominate or disjunctively dominate all the vertices in P n H, giving the result γ d 2 (P n H) = 2k + 2 = 2 n 5 . By summing up the cases (i), (ii) and (iii) we get the theorem. case (i) n ≡ 0 (mod 6). Let n = 6k. The set is a disjunctive dominating set of order 2k. Hence γ d 2 (C n H) ≤ 2k. The reverse inequality can be seen as follows. Let S be a disjunctive dominating set. Consider an arbitrary vertex v i on C n . Then S must contain v i or one of its neighbors to dominate it or two vertices in its second neighborhood to dominate it disjunctively. If v i is dominated by a vertex in S, then there is at least one vertex in the neighborhood of v i or copy of v i or its neighbor which is at a distance 2 from this vertex in S. S must contain at least one more vertex from its first or second neighborhood for the disjunctive domination of this vertex . But these two vertices in S together can dominate or disjunctively dominate at most 6 consecutive vertices on C n and their copies.
Thus S must contain at least 2 vertices from every set of 6 consecutive vertices on C n and their copies. Hence γ d 2 (C n H) ≥ 2k. Thus when n = 6k, γ d 2 (C n H) = 2k = 2 n 6 .
case (ii) n ≡ 1, 2 (mod 6). Let n = 6k + 1 or 6k + 2. As in case (i) it can be seen that every disjunctive dominating set must contain at least 2 vertices to dominate or disjunctively dominate any set of 6 consecutive vertices and their copies. Thus 2k vertices are needed to dominate or disjunctively dominate any set of 6k consecutive vertices and their copies.
In order to dominate or disjunctively dominate the remaining one or two vertices and their copies one more vertex is required. Hence γ d 2 (C n H) ≥ 2k + 1. On the other hand the set case (iii) n ≡ 3, 4, 5 (mod 6). Let n = 6k + 3 or 6k + 4 or 6k + 5. As in the above cases it can be proved that 2k + 2 vertices are necessary and sufficient to dominate or disjunctively dominate all the vertices in C n H, giving the result γ d 2 (C n H) = 2k + 2 = 2 n 6 . By summing up the cases (i), (ii) and (iii) we get the theorem.
Theorem 2.13. γ d 2 (G H) = 2 for G ∼ = K n , K 1,n , W 1,n where n is a positive integer greater than 3 for W 1,n .
Proof. Since these graphs have a universal vertex it follows from theorem 2.5.
Note. It may be noted that γ d 2 (G H) can be much larger than γ d 2 (G). For example if G is the friendship graph F n , which is constructed by joining n copies of C 3 with a common vertex, then  Hence γ d 2 (P n H) ≤ k. For the reverse inequality let S be a disjunctive dominating P n H . In order to dominate or disjunctively dominate the vertex v 1 , the set S must contain at least one vertex from {v 1 , v 2 }∪H 1 or two vertices from {v 3 } ∪ H 2 respectively. Thus S must contain at least one vertex from first three vertices or from the copies of H corresponding to the edges between these vertices. If v 1 is dominated by a vertex in S, this vertex can dominate at most vertices up to v 3 . Hence to dominate or disjunctively dominate vertices in H 3 , S must contain at least one more vertex from If v 1 is disjunctively dominated by two vertices in S, then those two vertices can dominate or disjunctively dominate vertices upto v 4 . Hence to dominate or disjunctively dominate H 4 , another vertex is required in its first or second neighborhood. It can be observed that at least one vertex is required from every set of three consecutive vertices or from the copies of H corresponding to the edges between these vertices. Hence, γ d 2 (P n H) ≥ k Thus when n = 3k, γ d 2 (P n H) = k = n 3 .
Hence γ d 2 (P n H) ≤ k + 1. The reverse inequality can be seen as follows. Let S be an arbitrary disjunctive dominating set. As before it can be seen that at least one vertex is required from every set of three consecutive vertices or from the copies of H corresponding to the edges between these vertices. Hence, γ d 2 (P n H) ≥ k + 1 Thus when n = 3k + 1 or 3k + 2, γ d 2 (P n H) = k + 1 = n 3 . By summing up the cases (i) and (ii) we get the theorem.  Proof. Let u be an arbitrary vertex in K n . It dominates all the vertices in K n and the copies of H  V needs at least one more vertex at a distance 2 from it. Since at least n 4 + 1 vertices are required for this γ d 2 (W 1,n H) ≥ n 4 + 1. Since S = {v 1 , v 5 , v 9 , ..., v 4k+1 } of the vertices on the rim of the wheel is such a set, we get {u} ∪ S is a disjunctive dominating set of W 1,n H. Hence

CONCLUSION
In this paper we have made an attempt to study the properties of disjunctive domination of neighborhood and edge corona of graphs. We also found the disjunctive domination number of neighborhood and edge corona of some standard classes of graphs. Such studies are significant because if a larger network can be broken into smaller networks, the solution of smaller ones can be used to find that of larger one. It will be interesting to study the impact of other graph operations on disjunctive domination.

CONFLICT OF INTERESTS
The author(s) declare that there is no conflict of interests.