Corporate Domination Number of the Cartesian Product of Cycle and Path

Domination in graphs is to dominate the graph G by a set of vertices , vertex set of G) when each vertex in G is either in D or adjoining to a vertex in D. D is called a perfect dominating set if for each vertex v is not in D, which is adjacent to exactly one vertex of D. We consider the subset C which consists of both vertices and edges. Let denote the set of all vertices V and the edges E of the graph G. Then is said to be a corporate dominating set if every vertex v not in is adjacent to exactly one vertex of , where the set P consists of all vertices in the vertex set of an edge induced sub graph , (E1 a subset of E) such that there should be maximum one vertex common to any two open neighborhood of different vertices in V(G[E1]) and Q, the set consists of all vertices in the vertex set V1, a subset of V such that there exists no vertex common to any two open neighborhood of different vertices in V1. The corporate domination number of G, denoted by , is the minimum cardinality of elements in C. In this paper, we intend to determine the exact value of corporate domination number for the Cartesian product of the Cycle and Path .

In this paper, all graphs G=(V,E) are considered as simple and undirected with vertex set V and edge set E. The order of G is the number of vertices in the vertex set V and the size of G is the number of edges in the edge set E.

Vertices and of are neighbors if
∈ . The open neighborhood of , denoted by ( ), is the set which consists of all the neighbors of v. The closed neighborhood of , denoted by [ ], consists of all the neighbors of including the vertex . The complement of the vertex set is denoted by . A detailed study of the dominating set and its algorithm of the Cartesian product of paths and cycles have been established by Polana Palvic, Janez Zerovnik [8]. The Cartesian product 1 □ 2 of graphs 1 and 2 is a graph with V ( 1 □ 2 ) =V ( 1 ) □ V ( 2 ) and (( 1 , 1 ),( 2 , 2 )) ∈ E( 1 □ 2 ) if and only if either 1 = 1 and 2 adjacent to 2 in 2 or 2 = 2 and 1 adjacent 1 in 1 .
In [6], the concepts of the variety of domination parameters such as total, perfect, mixed, and many more, have been studied by various authors. Domination with its variations is well studied in [7]. A set ⊆ is said to be a dominating set of if [ ] = . The domination number of , denoted by ( ), is the minimum number of vertices of any dominating set of . In [4] Gayathri A., Abdul. Muneera, Nageswara Rao T., Srinivas Rao T. have explained the significance of domination in various fields and expressed some real-life applications where dominations in a graph are used. In [9] Shobha Shukla, Vikas Singh Thakur have attempted to categorize domination concepts into Some categories. In [1] D. Bange, A. Barkauskas and P. Slater have obtained some results of the efficient dominating set.A set ⊆ is called an efficient and total efficient dominating set if | [ ] ∩ | = 1 and | ( ) ∩ | = 1 for every ∈ respectively. A dominating set S is perfect if each vertex is not in S , which is adjacent to exactly one vertex of S. The perfect domination number is the minimum number of vertices of any perfect dominating set and is denoted by ( ) .Analogous to perfect domination, perfect ev-domination has been introduced in [3] as follows.A set F of edges of a graph G is said to be a perfect ev-dominating set if every vertex of a graph is m-dominated by exactly one edge in F. The perfect ev-domination number of G, denoted by ( ) is the minimum number of edges of any perfect ev-dominating set. By incorporating both perfect and perfect evdomination, the corporate dominating set C is considered as a subset of ∪ and also by imposing certain conditions to dominate the graph G. Compared with Perfect domination number ( ) and Perfect ev-domination number ( ) , the corporate domination number ( ) is minimum. Corporate domination can be applied practically for the following problem. The CEO of the company has to get approval for a proposal from its board of directors.
Suppose there are 7 board of directors in that company setting at a table as a C 7 cycle. The vertices of C 7 represent the individual directors whereas the edges between any two vertices denote the trust between the respective two directors. It can be diagrammatically represented in Figure 1. The main aim of the CEO is to select the minimum number of directors to get approval for the proposal. On performing corporate domination, if the CEO has to select the individual director v 4 , so that v 4 can convince the directors v 3 and v 5 as they trust v 4 . Instead of selecting two other individual directors, the CEO can entrust the directors v 1 and v 7 who are both loyal to each other and can persuade the rest of the directors. Based on this corporate dominating set, graph C 7 is dominated and hence the proposal is approved with an optimum solution.
In  Figure 2.
A subset C is said to be a corporate dominating set if every vertex ∉ ∪ is adjacent to exactly one vertex of ∪ . The corporate domination number of G , denoted by γ ( ), is the minimum cardinality of elements in C.

Results on Corporate Domination
Inthis section, we establish γ (C 4k □ P n ) with illustrations.
Theorem 2.1Let C 4k (k ≥ 2) be any cycle and P 2n+1 (n ≥ 1) be any path. Then Proof:Let C 4k (k ≥ 2) be any cycle and P 2n+1 (n ≥ 1) be any path and let m = 4k. Consider the following cases.
For 0 ≤ ≤ 2 and2 ≤ t ≤ 2(n−1) where s ≡ 0(mod4) and t ≡ 2(mod4) and 0 ≤ i ≤ To prove C is minimum, let ′be any other corporate dominating set and ′ , ′ be the sets corresponding to ′ such that every vertex in ( ′ ∪ ′ ) is adjacent to exactly one vertex in ′ ∪ ′ . Furthermore, the set ′ will be in one of the following forms. i).
Then there exists at least one vertex ∈ ( ′ ∪ ′ ) which is not adjacent to any one of the vertices in ′ ∪ ′ . This is a contradiction. Hence Hence Case2: Let ≡ 1( 2)and = 4 .

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To prove C is minimum, let ′ be any other corporate dominating set and ′ , ′ be the sets corresponding to ′ such that every vertex in ( ′ ∪ ′ ) is adjacent to exactly one vertex in ′ ∪ ′ . Furthermore, the set ′ will be in one of the following forms. i).
Let C 4k (k ≥ 2) be any cycle and P 2n (n ≥ 1) be any path . Let m=4k. Consider the following cases.
.  . We shall prove that C is the minimum. Let C′ be any other corporate dominating set and P′, Q′be the sets corresponding to C′such that every vertex in (P′ ∪ Q′) is adjacent to exactly one vertex in ′ ∪ ′. Furthermore, the set C′will be in one of the following forms. i). . Then there exists at least one vertex ( ′ ∪ ′) which is not adjacent to any one of the vertices in ′ ∪ ′ . This is a contradiction. Hence | P′∪Q′| ≥ |P ∪Q| (=   . Here P = {v tm+1 , v tm+2 , v tm+3 , v tm+4 , ..., v (t+1)m−1 , v (t+1)m } and Q = φ Since every vertex not in P ∪Q is adjacent to exactly one vertex in P ∪Q , C is the corporate dominating set .
As every vertex in P ∪Q is adjacent to exactly two To prove C is minimum, let C′be any other corporate dominating set and P′, Q′be the sets corresponding to Q′such that, | ( ) ∩ ( ′ ∪ ′)| = 1 , ∀ ( ′ ∪ ′) . Furthermore, the set C′will be in one of the following forms. i).
Consider the following cases.

Conclusions
We have studied the new domination parameter namely, corporate domination,and explained the concepts with illustrations. The corporate domination number for some standard classes of graphs such as Path, Cycle, Wheel, Star, and Complete graphs has been computed. Also, we have found out that the exact value of the corporate domination number for the Cartesian product of C 4k and P n where ≥ 1 and ≥ 2 and illustrated the same with examples. Due to the minimum cardinality of the corporate dominating set, it can be beneficial when applied practically than perfect and perfect ev-domination. Further the general case for getting the corporate domination number for C m □P n where ≠ 4 and ≥ 2, □ ( , ≥ 3) , and P m □ P n ( , ≥ 2) can be investigated.