Complementary nil vertex edge dominating sets

Dominating sets play a vital role in day-to-day life problems. For providing e ﬀ ective services in a location, central points are to be iden-ti ﬁ ed. This can easily be achieved by graph theoretic techniques. Such graphs and relevant parameters are introduced and extensively studied. One such parameter is complementary nil vertex edge dominating set(cnved-set). A vertex edge dominating set(ved-set) of a connected graph G with vertex set V is said to be a complementary nil vertex edge dominating set(cnved-Set) of G if and only if V − D is not a ved-set of G . A cnved-set of minimum cardinality is called a minimum cnved-set(mcnved-set)of G and this minimum cardinality is called the complementary nil vertex-edge domination number of G and is denoted by γ cnve ( G ) . We have given a characterization result for a ved-set to be a cnved-set and also bounds for this parameter are obtained.


Introduction & Preliminaries
Domination is an active topic in graph theory and has numerous applications to distributed computing, the web graph and adhoc networks.Haynes et al. [3] gave a comprehensive introduction to the theoretical and applied facets of domination in graphs.
A subset S of the vertex set V of G is said to be a dominating set of G if each vertex in V − D is adjacent to some vertex of D. The domination number γ(G) is the minimum cardinality of the dominating set of G [3].A subset E 0 of the edge set E of a graph G is said to be an edge dominating set of G if each edge in E − E 0 is adjacent to some edge in E 0 .The edge domination number γ 0 (G) is the minimum cardinality of the edge dominating set of G [3].A subset D of vertices is said to be a vertex edge dominating set of G if each edge in G has either one of its ends from D (or) one of its ends is adjacent to a vertex in D. The vertex edge domination number γ ve (G) is the minimum cardinality of the vertex edge dominating set of G [5].A subset D of vertices is said to be a com- The minimum cardinality of a complementary nil dominating set is called a complementary nil domination number of G and is denoted by γ cnd (G) [6].Many variants of vertex -vertex dominating sets have been studied.In the present paper, we introduce a new variant of vertex-edge dominating set namely complementary nil vertex edge dominating set.We have given the characterization result for vertex edge dominating set to be complementary nil vertex edge dominating set and characterized the graphs of order n having cnved number n, characterized trees of order n having cnved numbers n − 1, n − 2. Bounds for this parameter are also obtained.
All graphs considered in this paper are simple, finite, undirected and connected.For standard terminology and notation, we refer Bondy & Murthy [1].

Characterization and other relevant results
In this section, we mainly obtain characterization result for a proper subset of the vertex set of G to be a cnved-set of G.
Note: For a cnved-set there is an edge of G whose ends are in cnved-set.So cnved-set is not an independent set in G.
Corollary 2.2.Let D be a cnved-set of a (connected)graph G. Then D has atleast two enclaves in D.
Proof: By the above Characterization Result, there is atleast one edge , say not cnved-sets and this contradicts the hypothesis.
We now give the bounds for cnved number of connected graphs.
Proof: Let D be a minimum cnved-set of G.By the Characterization Result there is an edge Furthermore the lower bound is attained in the case of P 8 .Hence the bound is sharp.
A lower bound for γ cnve (G) is obtained in terms of the number of edges( ) and maximum degree ∆(G) of the vertices of G.

being the number of edges of G).
Proof: Clearly ∆(G) ≥ 2. Any vertex v can have atmost ∆(G) neighbours.Furthermore any neighbour u of v can dominate atmost ∆(G) − 1 edges(excluding the edge uv) Then by the above Theorem the inequality follows.
Note: The bound is sharp as it is attained in the case of C 4 .For any kregular graph G(6 = K 2 ) with n vertices , Proof: The proof follows from the above theorem and the fact that Theorem 2.7.If G is a connected graph of order n and having edges, then Proof: Let S be a minimum cnved-set of G. Since V − S is not a cnvedset of G, there is an edge uv such that N [uv] ⊆ S. Each of u, v are non adjacent to all the vertices in V − S.So Theorem 2.8 For a connected graph G with g(G) 6 = 3, 2δ(G) ≤ γ cnve (G) .
Proof: Let S be a minimum cnved-set of G. Since V − S is not a cnved-set of G, there is an edge e = uv such that N Note: The bound is sharp as it is attained in the case of C 5 .Corollary 2.9 For a connected Bipartite graph G , 2δ(G) ≤ γ cnve (G) .
Proof: The proof follows from the fact that G cannot have odd cycles.
Note The bound is sharp as it is attained in the case of C 4 .
Theorem 2.10.If G is a connected graph , then Proof: Let D be the minimum vertex edge dominating set for G. Then for any edge e in G, D

S
N [e] is a cnved-set of G. Hence Note: The bound is sharp as it is attained in the case of C 6 .
Proof: Suppose that D is a minimum vertex edge dominating set for G.
Let v be adjacent to pendant vertex u(say Note: The bound is sharp as it is attained in the case of P 5 . Corollary 2.12.If G = P n , then γ cnve (G) ≤ γ ve (G) + 2.
Proof: Since ∆(G) = 2, the proof follows from the above theorem.Theorem 2.13.If G is a connected graph, then Proof: Let E 0 = {e 1 = x 1 y 1 , e 2 = x 2 y 2 , ..., e γ 0 (G) = x γ 0 (G) y γ 0 (G) } be a minimum edge dominating set for G. Let e i (1 ≤ i ≤ γ 0 (G)) be any edge in Note: The bound is sharp as it is attained in the case of K 2 .
Proof: Clearly the non pendant vertices of G along with all the neighbours of a support vertex v(say) forms a cnved -set for G, whose cardinality is n − ν 0 + d(v).Hence the inequality follows.
Proof: By the Remark in Theorem.2.10, Proof: By the Remark in Theorem.2.10 and [2] the proof follows.
Proof: For any graph G, So, By Theorem.2.19.there is an edge Supposing that all the vertices in V − V (T ) are adjacent to atleast two vertices in T we get a contradiction to that N Hence there is atleast one vertex in V − V (T ) which is adjacent to exactly one vertex in T .
The converse part is clear.
Proof: By Theorem.2.19, observe that for any graph G, Now we construct the proof as in the case of Theorem.2.18.
Proof: Assume that γ cnve (G) = n.Then by the Characterization Result for each edge which implies one of u, v is adjacent to all the vertices in G. W.l.g assume that v is the vertex adjacent to all the vertices in G(i.e u is a pendant vertex).Since G is a tree there is no edge between v 1 , v 2 ∈ V −{v}(i.e.all the vertices in V −{v} are pendant).Hence G ∼ = S n .
For the converse part, any edge uv in S n has the property that N (u) S N (v) = V .Hence by the Characterization Result the claimant holds.Now we characterize the graphs for which γ cnve (G) = 3.
Theorem 2.26.G be a tree, then γ cnve (G) = 3 iff G = P 3 or G is obtained by adding zero or more leaves to exactly one pendant vertex of P 4 .
Clearly atleast one of v 1 or v 3 is a pendant vertex.
Suppose both v 1 , v 3 are pendant vertices.
If there is any vertex in G adjacent to v 2 then it should be a member of D, which is a contradiction to our assumption.Hence G = P 3 .Suppose exactly one of v 1 , v 3 is a pendant vertex.W.l.g assume that v 1 is the pendant vertex.Clearly no vertex other than v 1 , v 3 can be adjacent with v 2 .Hence any vertex or edge in < V − {v 1 , v 2 , v 3 } > is adjacent to v 3 .Then G = P 4 or G is obtained by adding zero or more leaves to exactly one pendant vertex of P 4 .
The converse part is clear.
Theorem 2.27.G be a connected graph with δ(G) ≥ 2. Then γ cnve (G) = 3 iff (i) there is a C 3 edge disjoint with the other cycles in G.
(ii) each edge in G lies on a cycle of length atmost four that has a common vertex with C 3 .
Proof: Assume that γ cnve (G) = 3.Let D be a minimum cnved-set of G.By the Characterization Result and by the hypothesis, < D > = C 3 = < uvw > (say).By our assumption none of the edges of C 3 are common to two cycles.Hence C 3 is edge disjoint with the remaining cycles in G.If any edge in G lies on a cycle of length greater than four then there is atleast one edge not dominated by the vertices of D. Hence (ii) holds.
Conversely let C 3 be the cycle through u, v, w satisfying conditions (i)&(ii).Denote D = {u, v, w}.Let v be the vertex in C 3 through which the vertices in V − V (C 3 ) reach the vertices {u, w}.Let v 1 v 2 be an any edge in G. Then by (ii) of our assumption, v 1 v 2 lies on a cycle of length three or four for which v is one of the vertices.In either case v 1 v 2 is ve dominated by v. Hence D is a cnved-set.Since δ(G) ≥ 2, D is the γ cnve (G) − set of cardinality 3.
This proves the result.
G is a union of edge disjoint triangles, where all the triangles having a common vertex, then γ cnve (G) = 3.