Integrated Intelligent Research (IIR)

International Journal of Data Mining Techniques and Applications Volume: 01 Issue: 02 December 2012 Page No.26-28 ISSN: 2278-2419

A Note on Non Split Locating Equitable Domination P.Sumathi 1 G.Alarmelumangai 2 Head & Associate Professor in Mathematics, C.K.N college for Men,1 Anna Nagar, Chennai, India 2 Lecturer in Mathematics, E.M.G Yadava women’s college, Madurai, India. Email: 1 sumathipaul@yahoo.co.in2 alarmelu.mangai@yahoo.com 1

Abstract - Let G = (V,E) be a simple, undirected, finite nontrivial graph. A non empty set DV of vertices in a graph G is a dominating set if every vertex in V-D is adjacent to some vertex in D. The domination number (G) of G is the minimum cardinality of a dominating set. A dominating set D is called a non split locating equitable dominating set if for any two vertices u,wV-D, N(u)D N(w)D, N(u)D=N(w)D and the induced sub graph V-D is connected.The minimum cardinality of a non split locating equitable dominating set is called the non split locating equitable domination number of G and is denoted by nsle(G). In this paper, bounds for nsle(G) and exact values for some particular classes of graphs were found.

nsle (G). A set D is said to be a nsle – set if D is a minimum non split locating equitable dominating set.

Keywords-Domination number

A. Proposition

Number,non

Split

II. CHARACTERIZATION OF NON SPLIT LOCATING EQUITABLE DOMINATING SET. Observation 1. For any connected graph G, (G) nsle (G). 2. For any connected spanning subgraph H of G , nsle ( H ) nsle ( G ). Example: (i) nsle ( C4 ) = nsle ( P4 )=2. (ii) nsle ( K4 ) = 3, nsle ( C4 ) = 2.

domination

Pendant vertices are members of every nsle-set. Proof. Let v be vertex in G such that deg(v) =1 and let D be a nsle-set. If vV-D, then a vertex adjacent to v must be in D and hence V-D is disconnected, which is a contradiction.

I. INTRODUCTION For notation and graph theory terminology [2] is followed. Specifically, let G=(V,E) be a simple, undirected, finite nontrivial graph with vertex set V and edge set E. For a vertex vV, the open neighborhood of v is the set NG(v)={u /uv E}, NG(v) can be written as N(v) and the closed neighborhood of v is the set NG[v]= N(v){v} , NG[v] can be written as N[v] . The degree of a vertex v is the number of edges incident with v in G, i.e, d(v) =N(v). The maximum , minimum degree among the vertices of G is denoted by (G) ,(G) respectively. If deg v = 0,then v is called an isolated vertex of G. If deg v = 1 , then v is called a pendant vertex of G.As usual Kn, Cn, Pn and K1,n-1 denote the complete graph, the cycle, the path and the star on n vertices, respectively. The distance dG (u,v) or d (u,v) between two vertices u and v in a graph G, is the length of a shortest path connecting u and v. The diameter of a connected graph G is the maximum distance between two vertices of G it is denoted by diam(G) . A non empty set DV of vertices in a graph G is a dominating set if every vertex in V-D is adjacent to some vertex in D. The domination number (G) of G is the minimum cardinality of a dominating set. A dominating set D of a graph G = (V,E) is a non split dominating set if the induced subgraph V-D is connected . The non split domination number ns(G) of a graph G is the minimum cardinality of a non split dominating set. A set D is said be a ns-set if D is a minimum non split dominating set. A dominating set D is said to be non split locating equitable dominating set if for any two vertices u,wV-D, N(u)D N(w)D, N(u)D=N(w)D and the induced sub graph V-D is connected . The minimum cardinality of a non split locating equitable dominating set is called the non split locating equitable domination number of G and is denoted by

B.

Proposition

nsle (G) e where e is the number of pendant vertices. Proof. Since every pendant vertex is a member of each non split locating equitable dominating set. III. BOUNDS OF NON SPLIT LOCATING EQUITABLE DOMINATION NUMBER. n ber . Observation .- For any connected graph G with n 2, nsle (G) n -1. This bound is sharp for Kn. Theorem- For any connected (n,m) graph G with (G) 2, nsle (G) 3n- 2m-2. Proof. Let D be a nsle –set of G and let t be the number of edges in G having one vertex in D and the other in V-D. Number of vertices in <V-D> is n-nsle (G) and minimum number of edges in <V-D> is n-nsle (G) -1 . Hence viD deg (vi) + t. Since V-D= n-nsle (G) , there are atleast n-nsle (G) edges from VD to D. Also deg(vi ) (G) Therefore 26

Integrated Intelligent Research (IIR)

International Journal of Data Mining Techniques and Applications Volume: 01 Issue: 02 December 2012 Page No.26-28 ISSN: 2278-2419 equitable dominating set. Therefore G is the graph obtained from a complete graph by attaching pentant edges at atleast one of the vertices. Case (ii) (G) = 2 . V-D ={w1,w2 }. Let w be vertex of degree 3 in G and w V – D and w = w’. Let each vertex of D be adjacent to both w1 and w2. If <D> is complete, then G is complete. Assume < D > is not complete. Then there exists atleast one pair of non-adjacent vertices in D, say u,vD and V – { u, v, w1 } is a non split locating equitable dominating set of G containg (n-3 ) vertices, which is a contradiction.Therefore there exists a vertex in D which is adjacent to exactly one of w1 and w2 and again we get a non split locating equitable dominating set having (n-3) vertices and hence wD. Since deg (w )3, there exists atmost one vertex, say vD, adjacent to w. Then either V – { v,w,w1 } or V- {v,w, w2 } will be a non split locating equitable dominating set of G. Therefore there exists no vertex of degree 3 in G and hence each vertex in G of degree 2 and G is a cycle. Case (iii) (G) = 1, Let u,v be non adjacent vertices in < D >. Then either V – {u,v w1 } or V- { u,v,w2} will be a non split locating equitable dominating set, which is a contradiction. Therefore , < D {w1,w2}> is path. Hence G Pn..

(G) nsle (G) + n-nsle (G) 2 [ m –(n--nsle (G) -1]. But (G) 2, Hence, 2 nsle (G) + n-nsle (G) 2 [ m –n + nsle (G) +1]. Thus, 3n- 2m -2 nsle (G) Therefore nsle (G) 3n- 2m-2. Remark. This bound is attained if G Cn, n 3. Theorem 3.3. Let G be a connected graph and (G) =1. Then, nsle (G) 3n- 2m-e-2, where e is the number of pendant vertices. Proof. Let D be a nsle - set of G , such that D= nsle (G) and let t be the number of edges in G having one vertex in D and the other in V-D. As in Theorem 3.2, 2 [ m – ( n - nsle (G) – 1] = viD deg ( vi ) + t e + 2 ( nsle (G) – e ) + n-nsle (G) Hence , nsle (G) 3n- 2m-e-2. Remark. This bound is attained if G Pn, n 3. Corollary 3.4 . If G is a connected k-regular graph ( k > 3 ) with n vertices, then nsle (G) (n ( k -3) +2 ) / ( k – 3 ).

Theorem 3.8. For a connected graph G, nsle (G) = e if and only if each vertex of degree atleast 2 is a support, where e is the number of pendant vertices in G. Proof. Assume each vertex of degree atleast 2 is a support. If S is the set of all pendant vertices in G , then S is a dominating set in G and since <V-S> is connected, S is a nsle – set of G. Therefore nsle (G) e By proposition 2.3 theorem follows. Conversely, let u be a vertex in G such that deg (u) 2 and Let D be a nsle – set of G. If u is not a support of G, then u is not adjacent to any of the vertices in D, which is a contradiction.

Corollary 3.5. If (G) > 3, then nsle (G) ( 2m- 3n + 2) / ((G) – 3). Theorem 3.6. Let G be a connected graph with n 2, nsle (G) = n-1 if and only if G is a star and G is a complete graph on n vertices. Proof. If G K1, n-1 then the set of all pendant vertices of ,. K1, n-1 forms a minimal non split locating equitable dominating set for G. Hence nsle (G) = n-1. Conversely assume nsle (G) = n-1. Then there exists a non split locating equitable dominating set D containing n-1 vertices. Let V- D = {v}. Since D is a dominating set of G, v is adjacent to atleast one of the vertices in D,say u . If u is adjacent to any of the vertices in D, then the vertex u must be in V-D. Since D is minimal, u is adjacent to none of the vertices in D. Hence G K1, n-1 and Kn.

Theorem 3.9. If G is a connected graph which is not a star, then nsle (G) n-2. Proof. Since G is not a star, there exists two adjacent cut vertices u and v with deg (u), deg(v) 2. Then V- {u,v} is a non split locating equitable dominating set of G. Hence nsle (G) n-2.

Theorem 3.7. Let G bea connected graph nsle (G) = n-2 if and only if G is isomorphic to one of the following graphs. Cn, Pn or G is the graph obtained from a complete graph by attaching pendant edges at atmost one of the vertices of the complete graph and atmost at n-1 vertices. Proof. For all graphs given in the theorem, nsle (G) = n-2. Conversely, let G be a connected graph for which nsle (G) = n-2. and let D be a non split locating equitable dominating set of G such that D= n – 2 and V- D = {w1,w2 } and < V – D > K2. Case (i) By Proposition on 2.2, all vertices of degree 1 are in D and any vertex of degree 1 in D are adjacent to atmost one vertex in V-D since < V – D > K2. Also each vertex in V- D is adjacent to atmost on vertex in D. Let D’ = D – { pendant vertices }. Then {w1,w2 } D’ will be a complete graph. Otherwise, there exists a vertex u D’ , such that u is not adjacent to atleast one of the vertices of D’– {u} and hence D-{u} is a non split locating

Theorem 3.10. Let T be a tree with n vertices which is not a star. Then nsle (G) = n-2 if and only if T is a path or T is obtained from a path by attaching pendant edges at atleast one of the end vertices. Proof. Let T be a tree which is not a star. It can be easily verified that for all trees stated in the theorem nsle (G) = n-2. Conversely, Assume nsle (G) = n-2. Let D be a nsle- set containing n-2 vertices and let V-D= {w1,w2} and <V-D> K2. Since T is a tree, each vertex in D is adjacent to atmost one vertex in V-D. Since D is a dominating set, each vertex in V-D is adjacent to atleast one vertex in D. (i) If <D> is independent, then T double star. (ii) Assume <D> is not independent. Then there exists a vertex u <D> such that deg(u) 1 in <D>. Also either Nj(u)= 1, 1 j diam ( T ) – 3 or if Nj(u) 2 for some j , 27

Integrated Intelligent Research (IIR)

International Journal of Data Mining Techniques and Applications Volume: 01 Issue: 02 December 2012 Page No.26-28 ISSN: 2278-2419

j diam ( T ) – 4, Then < Nj(u)> in D is independent, since otherwise D- {u} is a non split locating equitable dominating set of T. In the first case T is a path. In the second case, T is a tree obtained from a path by attaching pendant edges at atleast one of the end vertices of the path.

(G) = n-2 and nsle (G) + (G) < 2n-3 implies n=4.Hence, G C4. Let G be the graph obtained from a cycle by attaching pendant edges at atleast one of the vertices of the cycle. Let n be the number of vertices in the cycle and e be the maximum number of pendant edges attached. Hence (G) = n – 5 +e, Therefore nsle (G) + (G) < 2n-3. Let G Pn, nsle (G) + (G) < 2n-3 implies n =4 . Hence G P4. Let G be a graph obtained from a path by attaching pendant edges at atleast one of the end vertices. Let e be the number of pendant edges attacted , e 2. Hence, nsle (G) + (G) < 2n-3 .

IV. RELATION BETWEEN NON SPLIT LOCATING EQUITABLE DOMINATION NUMBERAND OTHER PARAMETER Theorem 4.1. For any tree with n vertices, c ( T) + nsle ( T ) n. Proof. If D1 is the set of all cut vertices of T with D1= n1, then c ( T) = n1. If D2 is the set of all pendant vertices of T with D2= n2, then nsle ( T ) n2. But V ( T1 )= n1 + n2 implies that c ( T) + nsle ( T ) n. By theorem 3.6 , equality holds if and only if each vertex of degree atleast 2 is a support. Proposition 4.5. For any connected (n,m) graph G, nsle ( G ) + (G) 2n-2.

REFERENCS [1] J. Cyman, The outer-connected domination number of a graph. Australasian J. Comb. 38 (2007), 35-46. [2] T.W. Haynes, S.T.Hedetniemi, and P.J. Slater, Fundamentals of domination Graphs, Marcel Dekker, New York, 1998. [3] V.R. Kulli and B. Janakiram, The non split domination number of a graph. India J. Pure Appl.math; 31(2000)545-550. [4] P. Sumathi , G.Alarmelumangai, Non split locating equitable domination, Proceedings.Of the International Conference on Mathematics in Engineering & Business Management, March 9-10,2012.ISBN: 978-81-8286-015-5. [5] Hongxing Jiang, Erfang shan, outer-connected domination in graphs, Utilitas Mathematics 81(2010).

Proof . For any graph with n vertices, (G) n-1. By observation 3.1, the proposistion follows. Proposition 4.6 For any connected (n,m) graph G, nsle ( G ) + (G) = 2n-2 if and only if G K1,n-1 or Kn. Proof. When G K1,n-1 or Kn., nsle (G) + (G) = 2n-2. Conversely, nsle (G) + (G) = 2n-2. is possible if nsle (G) = n-1 and (G) = n-1. But, nsle ( G ) = n-1 is possible if and only if G is a star and G is a complete graph. Theorem 4.2. For any connected (n,m) graph G , nsle (G) + (G) > 2n-3 and nsle (G) + (G) < 2n-3 if and only if G is one of the following. (i) C3,P3,Kn, or G is the graph obtained from a complete graph by attaching pendant edges at exactly one of the vertices of complete graph. (ii) C4, P4 or G is the graph obtained from a cycle or path by attaching pendant edges at exactly one of the vertices of cycle or path. Proof. For the graphs given in the theorem, nsle (G) + (G) > 2n-3 and nsle (G) + (G) < 2n-3 . Conversely, nsle (G) + (G) > 2n-3 is possible if nsle (G) = n-1 and (G) = n-1 and nsle (G) + (G) < 2n-3 possible if . nsle (G) = n-2 and (G) = n-2. In the first case, nsle (G) = n-1 if and only if G is a star on n vertices .But for a star (G) = n-1 and hence this case is possible.In the second case, nsle (G) = n-2 if and only if G is isomorphic to one of the following graphs. (a) Cn , Pn or G is obtained from a path by attaching pendant edges at atleast one of the end vertices.Let G Cn, then . nsle (G) = n-2 and 28

- Let G = (V,E) be a simple, undirected, finite nontrivial graph. A non empty set DV of vertices in a graph G is a dominating set if every...

Published on Apr 16, 2019

- Let G = (V,E) be a simple, undirected, finite nontrivial graph. A non empty set DV of vertices in a graph G is a dominating set if every...

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