Eccentric Domination in Boolean Graph BG 2 (G) of a Graph G

: Let G be a simple (p, q) graph with vertex set V(G) and edge set E(G). BG 2 (G) is a graph with vertex set V(G)  E(G) and two vertices are adjacent if and only if they correspond to two adjacent vertices of G, a vertex and an edge incident to it in G or two non-adjacent edges of G. In this paper, we studied eccentric domination number of Boolean graph BG 2 (G), obtained bounds of this parameter and determined its exact value for several classes of graphs.


1.Introduction
Let G be a finite simple, undirected graph on p vertices and q edges with vertex set V(G) and edge set E(G).For graph theoretic terminology refer to Harary [11], and Kulli [17].
The distance d(u, v) between two vertices u and v in G is the minimum length of a path joining them if any; otherwise d(u, v) = .Let G be a connected graph and u be a vertex of G.The eccentricity e(v) of v is the distance to a vertex farthest from v. Thus, e(v) = max{d(u, v): u  V}.The radius r(G) is the minimum eccentricity of the vertices, whereas the diameter diam(G) is the maximum eccentricity.For any connected graph G, r(G) ≤ diam(G) ≤ 2r(G).The vertex v is a central vertex if e(v) = r(G).The center C(G) is the set of all central vertices.The central sub graph C(G) of a graph G is the subgraph induced by the center.The vertex v is a peripheral vertex if e(v) = diam(G).The periphery P(G) is the set of all peripheral vertices.For a vertex v, each vertex at a distance e(v) from v is an eccentric vertex.Eccentric set of a vertex v is defined as E(v) = {u  V(G) : d(u, v) = e(v)}.A graph is self-centered if every vertex is in the center.Thus, in a self-centered graph G all vertices have the same eccentricity, so r(G) = diam(G).
A vertex and an edge are said to cover each other if they are incident.A set of vertices which covers all the edges of a graph G is called a point cover for G, while a set of edges which covers all the vertices is a line cover.The smallest number of vertices in any point cover for G is called its point covering number or simply covering number and is denoted by 0(G) or 0.Similarly, 1 is the smallest number of edges in any line cover of G and is called its line cover number.A set of vertices in G is independent if no two of them are adjacent.The largest number of vertices in such a set is called the point independence number of G and is denoted by 0(G) or 0.A set of edges in a graph is independent if no two edges in the set are adjacent.By a matching in a graph G, we mean an independent set of edges in G.The edge independence number 1(G) of a graph G is a maximum cardinality of an independent set of edges.A perfect matching is a matching with every vertex of the graph is incident to exactly one edge of the matching.The graph G + is obtained from the graph G by attaching a pendant edge to each of the vertices of G.
The open neighborhood N(v) of a vertex v is the set of all vertices adjacent to v in G. N[v] = N(v)  {v} is called the closed neighborhood of v.The second neighborhood N2(v) of a vertex v is the set of all vertices at distance two from v in G.
In 2007, Janakiraman, Bhanumathi and Muthammai defined the Boolean graph BG2(G) and studied its properties [12,14,15,16].Boolean graph BG2(G) is a graph with vertex set V(G)  E(G) and edge set ), where L(G) is the line graph of G and T(G) is the total graph of G.It is a graph with vertex set V(G)  E(G) and two vertices are adjacent if and only if they correspond to two adjacent vertices of G, a vertex and an edge incident to it in G or two non-adjacent edges of G.
The concept of domination in graphs was introduced by Ore [18].A set D  V(G) is said to be a dominating set of G, if every vertex in V(G)  D is adjacent to some vertex in D. D is said to be a minimal dominating set if D  {u} is not a dominating set for any u  D. The domination number (G) of G is the minimum cardinality of a dominating set [10].
A set D  V(G) is an eccentric dominating set if D is a dominating set of G and for every v  V D, there exists at least one eccentric point of v in D. The eccentric domination number ed(G) of a graph G equals the minimum cardinality of an eccentric dominating set.Obviously, (G)  ed(G).Theorem 1.1 [15]: Let G be a connected graph.Then, (G)  (BG2(G))  (G) + 2. Theorem 1.2 [16]
Here, ed(G) = 6 and ed(BG2(G)) = 5.Theorem 2.1 Let G be a graph without isolated vertices.Set of all point vertices is an eccentric dominating set of BG2(G); and hence 1  ed(BG2(G))  p. Proof: Distance from a line vertex to point vertices is one or two.Also, distance from a point vertex to line vertices is also one or two.So if G has more than one edge, then V(G) is an eccentric dominating set of BG2(G).Hence, 1  ed(BG2(G))  p.  Proof: Let u and v be the central vertices of G.In BG2(G), NG(u) and NG(v) are dominating set of BG2(G).Let x, y be the any two peripheral vertices at distance atmost 3 in BG2(G).S = {x, y, u, v} is an eccentric dominating set of BG2(G).Hence, ed(BG2(G))  4.
If G has a vertex v of maximum degree which is a support of pendant vertices, then in BG2(G), let S = V(G)  NG(u)  {x, y}, where x, y are peripheral vertices of G.This S is an eccentric dominating set of BG2(G).Hence, ed(BG2(G))  p  (G) + 2. Theorem 2.5 Let G be a tree, then S dominates all the point vertices and u  S is eccentric to point vertices in V  S also.All the edges incident with u and elements of N(u) are dominated by u and vertices of N(u) in BG2(G).If an edge e1 is in N2(u) in G, then it is dominated by e in BG2(G).Also, all line vertices not in S is eccentric to some vertices of S in BG2(G).Therefore, S is an eccentric dominating set for BG2(G).Hence, ed(BG2(G))  2 + (G) and S is a connected eccentric dominating set for BG2(G).Theorem 2.7 If G  K1,n is of radius one with a unique central vertex u, then ed(BG2(G)) = 3. Proof: Let G be a graph with radius one with a unique central vertex u.In BG2(G), u dominates all point vertices and line vertices incident with u in G. Let e = uv  E(G).Now in BG2(G), Consider S = {u, v, e}  V(BG2(G)).S is a dominating set of BG2(G).BG2(G) is two self-centered by Theorem 1.4.In BG2(G), the line vertex e is eccentric to all point vertices except u and v; u is eccentric to all line vertices which are not incident with u in G; v is eccentric to all line vertices which are not incident with v in G. Therefore, S is a minimum eccentric dominating set of BG2(G).Hence, ed(BG2(G)) = 3. Theorem 2.8 If G is of radius two and diameter three and if G has a pendant vertex v of eccentricity three, then ed(BG2(G))  (G) + 2. Proof: If G has a pendant vertex v of eccentricity three, then its support vertex u is of eccentricity two.In BG2(G), NG(u) dominates all point and line vertices.Therefore, S = NG(u)  {v, e}, where uv = e is an eccentric dominating set of BG2(G).Hence, BG2(G)  (G) + 2. Theorem 2.9 If G is a graph with radius two, diameter three, then ed(BG2(G))  p  (G) + 2. Proof: Let u  V(G) with deg u = (G).Since radius of G is two and diameter three, all the point vertices in BG2(G) has their eccentric vertices atmost at distance three from u. Also, eccentricity of line vertices in BG2(G) is two by Theorem 1.4.All the edges incident with u are dominated by u in BG2(G) and are also eccentric to a point vertex w, where w N2(u).Suppose e1 is an edge in N(u), then e1 is not dominated by (V  N(u)).Hence the following cases arise: Case(i): If all the edges in N(u) are adjacent or incident at a vertex v, then (V  N(u))  {v} is an eccentric dominating set of BG2(G).Proof: G  K1,n.Consider a pendant vertex u  V(G) and let v  V(G) be its adjacent vertex in G, e = uv  E(G), v is a central vertex of G. Now in BG2(G), S = {u, v, e} is an eccentric dominating set.Thus, ed(BG2(G)) = 3 = c(BG2(G)).Theorem 2.12 Let G be a connected graph with p  3.Then, ed(BG2(G))  ed(G) + 2. Proof: Let D  V(G) be an eccentric dominating set of G with cardinality ed(G).Let u  D be such that u is adjacent to v  V(G), e = uv  E(G).Consider S = D  {v, e}  V(BG2(G)).The vertex v dominates incident edges in G and the edge e dominates non adjacent edges in G.All point vertices in V(BG2(G)) -S have their eccentric vertices in S. Also, the line vertices of V(BG2(G)) -S have u or v as eccentric vertices, since eccentricity of every line vertex is two in BG2(G).Therefore, S is an eccentric dominating set of BG2(G).Hence, ed(BG2(G))  ed(G) + 2. Remark 2.1 ed(G)  ed(BG2(G)) is not true.Refer Example 2.1.Theorem 2. 13 Let G be a graph with diam(G) = 2.If there exists a vertex v  V(G) such that N2(v) is totally disconnected, then ed(BG2(G))  (G) + 2. Proof: Let v  V(G) be such that N2(v) is totally disconnected Let S = N(v)  {u, w}, where u, w  N2(v).Since N2(v) is totally disconnected, all the edges of G are incident with vertices of S. Therefore, vertices of BG2(G)  S are adjacent to atleast one vertex in S. Also, the vertices of V(BG2(G))  S has u, w as eccentric vertices.Hence, ed(BG2(G))  S = N(v) + 2  (G) + 2. Theorem 2.14 Let G be a connected graph.Then line independent set of G is an eccentric dominating set for BG2(G) if and only if G is a graph with p  6 and G has a perfect matching with diam(G) ≤ 2. Proof: Let D be a line independent set of G.If D is an eccentric dominating set for BG2(G), it dominates every point vertices of BG2(G), that is D is a line cover of G. D is independent and cover all vertices of G implies that D is a perfect matching.If p  3 and 0(G)  3, then every edge in E(G) -D has atleast one edge in D, which is not adjacent to e in G. Thus D dominates all line vertices of BG2(G) also.Hence, D is a dominating set of Therefore, G must be a graph with even number of vertices and has a perfect matching.Also, eccentricity of every line vertex in BG2(G) is two and if diam(G)  3, then eccentricity of point vertex is three in BG2(G).Hence, D is an eccentric dominating set implies that G is a graph with p  6 and G has a perfect matching with diam(G) ≤ 2.
Conversely, let G has a perfect matching with diam(G) ≤ 2 and p  6.This implies, G cannot be K1,n.Let D be a perfect matching of G. D dominates all point and line vertices of BG2(G).Since diam(G) ≤ 2 and G  K1,n, line vertices of BG2(G) is of eccentricity two.Therefore, BG2(G) is a 2 self-centered graph.In BG2(G), every edge in E(G) -D is adjacent with some edge in D. Hence, in BG2(G), every line vertex has eccentric vertex in D. Every point vertices of V(G) is non incident with some edge of D in G. Therefore, point vertex in BG2(G) has eccentric vertex in D. Hence, D is an eccentric dominating set of BG2(G).Remark 2.1 If p = 4 and G has a perfect matching, then D cannot be a dominating set of BG2(G).Theorem 2. 15 Let G be a connected graph.Maximal independent set of G is an eccentric dominating set of BG2(G) if and only if G satisfies any one of the following (i) G = K1,n, n  3 (ii) G is bipartite and if v  V(G) -D such that eG(v) = 2 then v is not adjacent to atleast one element of D, if v V(G) -D such that eG(v)  3 then there exists w  S such that d(v, w)  3. Proof: Let G be a connected graph.Let D be the maximal independent set of G. So, D  V(G) such that D is independent.Since, D is maximal independent it is a dominating set of G. So, D dominates the point vertices in BG2(G).Now, to dominate the line vertices of BG2(G), D must be a point cover of G also.D is maximal independent implies that V(G) -D is a point cover of G. Also, D is a point cover of G implies that V(G) -D has no edges and so it is independent.Thus both D and V(G) -D are independent.Therefore, G is bipartite.When p  3 and G  Kn, every line vertex of BG2(G) has eccentric vertices in D. But point vertices which are not in D need not have eccentric vertices in D. D has eccentric vertices of other point vertices if D satisfies condition (ii) only.Hence the theorem is proved.On the otherhand, if all the conditions are satisfied, then any maximal independent set of G is an eccentric dominating set of BG2(G).Theorem 2.16 G is a connected (p, q) graph with p  4. Set of all line vertices is an eccentric dominating set of BG2(G) if and only if diameter of G is 1 or 2. Proof: Eccentricity of line vertices in BG2(G) is always two and eccentricity of point vertex is 1, 2 or 3. Hence, E(G) is an eccentric dominating set only when diam(G) ≤ 2 by Theorem 1. (ii) ed(BG2(Cn)) = n / 3 + 1, n = 3k + 1 or n = 3k +2.Proof: Let V(Cn) = {v1, v2, …, vn} and ei = vivi+1, 1  i  n  1 and en = vnv1.Let ui be the vertex corresponding to ei in BG2(Cn).Then v1, v2, v3, …, vn, u1, u2, u3, …, un  V(BG2 (Cn)).Thus V(BG2 (Cn)) = 2n.
and uv = e  E(G).The vertex u dominates all point vertices of V2 and line vertices which are edges incident with u in G.The vertex v dominates all point vertices of V1 and line vertices which are edges incident with v in G.The line vertex e dominates all line vertices which are edges not incident with both u and v.The vertex u is an eccentric vertex of V1 and non incident edges of G and the vertex v is an eccentric vertex of V2 and non incident edges of G. Therefore, S is an minimum eccentric dominating set of BG2(Km,n).Hence, ed(BG2(Km,n)) = 3. Theorem 3.6 ed(BG2(Wn)) = 3, where Wn = K1 + Pn.Proof: Let S = {u, v, e}, where u and v are adjacent vertices and v is the central vertex.uv = e  E(G).u and v dominates all point vertices and incident edges in BG2(Wn) and e dominates non adjacent edges in BG2(Wn).The vertex u is a eccentric vertex of non adjacent point vertices and non incident line vertices and the vertex e is a eccentric vertex of non adjacent line vertices and non incident point vertices in G. Therefore, eccentricity of every point vertex and line vertex of BG2(Wn) is two.This implies, it is a self-centered graph.S is an eccentric dominating set of BG2(Wn).Also, S is a minimum eccentric dominating set of BG2(Wn).Hence, ed(BG2(Wn)) = 3. Theorem 3.7 ed(BG2(Fn)) = 3, where Fn = K1 + Pn.Proof: Let v1, v2, v3, …, vn, v (v is the central vertex of Fn) be the vertices of Fn and let ej = vvj, j = 1, 2, …, n, and vivj = eij (j = i + 1, i = 1, 2, 3, …, n) be the edges of Fn.Let v1, v2, …, vn, v, u1, u2,…, un, e12, e23, …, en1,n be the corresponding vertices of BG2(Fn).Thus V((BG2(Fn)) has 3n vertices.S = {v, v1, e1} is the eccentric dominating set of BG2(Fn).Eccentricity of every point vertex and line vertex of BG2(Fn) is two.Therefore it is a self-centered graph.The vertex v1 is an eccentric vertex of ej, j  1 in BG2(Fn).The vertex v is the eccentric vertex of the line vertex e12 in BG2(Fn).For other eij's v is the eccentric vertex in BG2(Fn).For point vertex vi, i  1, line vertex e is an eccentric point.Therefore, S is a minimum eccentric dominating set of BG2(Fn).Hence, ed(BG2(Fn)) = 3.

Corollary 2 . 1 Theorem 2 . 2
The bounds are sharp, since ed(BG2(G)) = 1 if and only if G = P2 and ed(BG2(G)) = p if and only if G = n K .If G is unicentral tree of radius 2, then ed(BG2(G))  p  degG(u), where u is a central vertex.Proof: If G is of radius two with unique central vertex u, then in, BG2(G), r(BG2(G)) = 2 and V  NG(u) dominates all point vertices and line vertices of BG2(G).Each vertex of V(BG2(G))  NG(u) has their eccentric vertices in V(G)  NG(u) only.Therefore, V(G)  NG(u) is an eccentric dominating set of BG2(G).Hence, ed(BG2(G))  p  degG(u).Theorem 2.3 For a bi-central tree T with radius 2, ed(BG2(G))  4.
3. Converse:Case(i): r(G) = d(G) = 1.That is G = Kn.In this case, E(G) is an eccentric dominating set of BG2(G).Case(ii): r(G) = 1, d(G) = 2.If G = K1,n, BG2(G) is of radius one and E(G) is an eccentric dominating set of BG2(G).When G  K1,n, BG2(G) is two self centered.For a point vertex u, a line vertex e which is not incident with u in G is an eccentric vertex in BG2(G).So, E(G) is an eccentric dominating set of BG2(G).Case(iii): r(G) = d(G) = 2.In this case also BG2(G) is 2 self-centered and E(G) is an eccentric dominating set of BG2(G).