Cmjv02i02p0222

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J. Comp. & Math. Sci. Vol.2 (2), 222-233 (2011)

On the Dominating of a Graph and its Complement GANGADHARAPPA D. B1. and A. R. DESAI2 1

Sr. Lecturer in dept. of Mathematics Coorg Institute of Technology, Ponnampet South Kodagu, Karnataka, India. 2 Prof. Dept. of Maths, S.D.M. College of engg. Dharwad, Karnataka. India. e.mail-gangadharappadb.@ g.mail.com.

ABSTRACT A set D of vertices in a graph G=(V,E) is a dominating set of G, The domination number γ=γ(G) of a minimum cardinality of a dominating set . A dominating set G is a global dominating set (g.d.set) of G if D is also a dominating set of the compliment G of G.The global domination number γ୥ =γ୥ (G) of G is the minimum cardinality of a g.d.set similarly for total global domination numberγ୲୥ (G). Key words: domination number, total domination, global domination, and compliment of graph.

1 INTRODUCTION A set D of vertices in a graph G = (V, E) is a dominating set of G if every vertex in V-D is adjacent to some vertex in D. The domination number γ= γ(G) of G is the minimum cardinality of a dominating set. The upper domination number of G, Г(G) is the maximum number of vertices in a minimal dominating set. For details on γ (Fundamentals of domination in graphs and domination in graphs.advanced topics by Haynes, Hedetniemi and Slater). A dominating set D of G is a global dominating set (g.d.set)of G if D is also a of G. dominating set of the complement G The global domination numbe r γ = γ (G) of G is the minimum cardinality of a g.d. set.

Example, suppose G is a graph representing a network of roads linking various locations. Some essential goods are being supplied to these locations from supplying stations located at v , v and v . It may happen that these links (edges of G) may be broken for some reason or the other. So we have to think of maintaining the supply of goods to various locations uninterrupted through secret links (dotted lines: edges of the complement of G). What is the minimum number of supplying stations needed to accomplish this task? It is the global domination number of G (Fig. 1). The graphs G considered here have order p and size q (i.e. p verticesand q edges) and both G have no isolates.A and their complements G set D of vertices in a graph = (V, E) is a

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dominating set of G if every vertex in V -D is adjacent to some vertex in D. The domination number Îł (G) of G is the minimum cardinality of a dominating set (see Cockayne and Hedetniemi). A total dominating set T of G is a dominating set such that the induced subgraph (T) has no isolates. The total domination number Îł (G) of G is the minimum cardinality of a total dominating set. This concept was introduced in Cockayne. A dominating set D of G is a global dominating set (g.d.set) if d is alsoa . The global domination dominating set of G number Îł (G)of G is the minimum cardinality of a g.d.set (Sampthkumar). The purpose of this is to study the global aspect of total domination.A total dominating set T of G is a total global dominating set (t.g.d. . set) if T is also a total dominating set of G The total global domination number Îł (G) of G is the minimum cardinality of a t.g.d. set. This concept introducedby Kulli and Janakiram. We note that Îł (G) and Îł (G) for G with δ (G) ≼ 1 and Îł (G) is only ) ≼1, defined for G with δ (G) ≼ 1 and δ (G where δ(G) is the minimum degree of G.A Îł -set is a minimum total dominating set. Similarly a Îł -set and a Îł -set are defined. This concept was introduced by Sampathkumar. BASIC PROPERTIES Proposition 1: A dominating set D of G is a g.d.set if, and only if, for each v ∈ V-D, there exists a u ∈D such that u is not adjacent to v. ) and Îł = Îł (G ). Proposition 2 LetÎł = Îł(G follows directly from the definitions.

Proposition 2: For any graph G (i) γ = γ (ii) γ ≤γ (iii)

γ γ ≤γ ≤γ+γ

For a real number r, let r be the smallest integer not less then r.. Proposition 3: (i) For a graph G with p vertices.γ (G)=p if and if, G=K or K . (ii) γ (K , ) = 2 for all p,q≼1

(iii) Îł (C ) = 2, Îł (C ) = 3 and Îł (Cp) = for all p ≼6. (iv) v (P = 2 for p = 2, 3 and Îł (Pp)= for p ≼ 4. Proof: We prove only (i), and (ii)-(iv) are ) = p. obvious. Clearly, Îł (Kp) = Îł (K Then G Suppose Îł (G) = P and G≠K ,K has at least one edge uv and vertex w not adjacent to, say v. Then V-{v} is a g.d.set and Îł (G) ≤p-1. For some graphs including trees,Îł is almost equal to Îł Proposition 4: Let D be a minimum dominating set of G. If there exists a vertex v in V-D adjacent to only vertices is D, thenÎł ≤γ+1 Proof: This follows since DâˆŞ v is a g.d.set. Corollary 1: Let G = V , V ,E) be a bipartite graph without isolates, where|V | q,|V | p. and q≤ p.thenÎł ≤q+1. proof: This follows from Îł ≤γ+1 sinceγ≤p.

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Corollary 2 : For any graph with a pendant vertex, Îł Îł 1, holds. In particular, Îł Îł holds for a tree. Corollary 3: If V-D is independent,then Îł Îł 1 holds. Let Îą and β respectively denote the covering and independence number of a graph. Proposition 5: For a (p,q) graph G without isolates. ≤γ ≤p-β +1 Proof: Let D be a minimum g.d.set. Then every vertex in V-D is not Adjacent to at least one vertex in D this implies q≤ -(p-Îł ) and the lower bound as follows. To eastablish the upper bound, let B be an independent set with β vertices,V-B is a dominating set of G. Clearly, for any v ĎľB,(V- B)âˆŞ v is a g. d. set of G, and the upper bound follows.SinceÎą +β =p for any graph of order p without isolates.

and Walikar,the total domination number of a graph). Let D be a minimum g. d. set. Then D induces a connected sub graph in G . Hence D is Connected dominating set or G .Thus, we have for any graph G, at of G or G least one of the following holds (i) γ ≤γ (ii) γ ≤γ where γ =γ (G). Corollary 6 : For any graph G, at least one of the following holds; (i) γ ≤ι +1, (ii)γ ≤ι +1 Proposition 6; For any graph G=(V,E), where χ(G) is the γ ≤max χ G , χ G

chromatic number of G. Proof : Let χ(G)=q, χ(G)=p and q ≤ p. Consider a≤(G)- partition V , V , ‌ , V and a≤(G)-partition V , V , V , ‌ , V of V. Clearly, no two vertices of any V can belong | to any V and conversely, we can select q | vertices V , V , ‌ , V such that (i) V Ͼ V , 1≤i≤q, and (ii) V , V , ‌ , V belong to

different sets in V , V , ‌ , V , say V ĎľV , 1≤j≤q. | Choose, V ĎľV , q+1≤j≤p. clearly, V , V , ‌ , V is a dominating set of G and V ,V ,. . . ,V . V ,. . . ,V is a dominating set of G and hence it is well-known that χ(G)≤∆+1, and if G is neither complete nor an odd cycle, then χ (G)≤∆, |

Corollary 4 : γ ≤ι 1 The independent domination number i(G) of G is the minimum cardinality of a dominating set which is also independent. It is well-known that γ≤i≤β Corollary 5: For any graph G of order p without isolates, (i) γ+γ ≤p+1 (ii) i+γ ≤p+1 The connected domination number γ of a connected graph G is the minimum cardinality of a dominating set D such that the sub graph <D> induced by D is connected. For detailson γ ,(Sampath kumar

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|

|

Corollary 7: For any graph G of order p +1 = max {p-δ , p-δ} and (i) Îł ≤max {∆+1,∆ (ii) If G is neither complete nor an odd cycle = max {p-1-δ , p-1-δ} Îł max {∆,∆ Corollary 8 : Let t = Îł orÎł . For any graph G

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+1} and (i) t ≤max{∆+1, ∆ (ii) If G is neither complete nor an odd cycle } . T ≤max{∆,∆ Let k and k respectively denote the . It is well known connectivity of G and G that k≤δ ,from corollary 7 of (ii) Corollary 9: For any graph G of order p, Îł ≤max {p-k-1, p-k-1} . For v ∈ V, let N(v) = {u ∈ V: u v ∈E} and N[v] = N(v) U {v}. A set D⊊V is full if N{v}∊D-V≠for all v ∈D. Also, D is g-full . The full if N(V)∊V-D≠both in G and G number f = f (G) ofG is the maximum cardinality of a full set G, and the g-full number f = f (G) of G is the maximum cardinality of a g-full set of G. Clearly, f (G) ). = f (G Proposition A : If G is of order p , Îł +f =p. The global domatic number of a graph A partition {V , V , ‌ , V of V is a domatic (global domatic) partition of G if | each V ,is a dominating set (g.d.set). The domatic number d = d(G) (global domatic number d = d (G) of G is the maximum order of a domatic (global domatic) partition of G. Clearly, for any graph G, d (G) = ). d (G Proposition 9: ) =1 (i) d (k ) = d (k

(ii) For any p≼1, d (C ) = 3, and d (C ) = d (C ) = 2. (iii) For any 2 ≤ q ≤p ,d (k , ) = p. Let d = = d G ), and d ). d(G Proposition 10: For any graph G of order p, (i) d≤δ(G)+ 1 (Cockayne and Hedetniemi ) (ii) d d δ G + 1 (iii) d = d g

iv d ≤ v d ≤min{δ+1, δ +1=min{δ+1,p-∆}

(vi) d ≤ (vii) For a tree T with p≼2 vertices (a) d(T)=2(Cockayne and Hedetniemi) (b) d T 2

Proof: The results (i) and (ii) follow from the definitions. Also, (i) and (ii) imply (iii) (iv). To establish (v), we have from (i) and δ δ

(ii). d ≤

=

δ ∆

≤

if G is not

regular , then δ-∆≤ -1, and d ≤ clearly,(b)follows from (i) and (a) Proposition 11 : If G is of order p, then (i) Îł+d≤ p+1 (Cockayne and Hedetniemi ) (ii) (ii) Îł + d ≤P+ 1 with equality in (i) or Kp. and (ii), if, and only if, G = K Proof-: Clearly, (ii) follows from Îł ≤ p-δ} and 1 max p-δ, max{∆+1, ∆ dŕ­Ľ min

δ

1 min 1, ∆ 1, δ

. Let G≠to equality holds when G = K or K K , or K . Suppose p-δ≼p-δ. Then from Îł 1 =max{p-δ ,p-δ}Îł p ( max ∆ 1,.∆

δ, and from proposition10 (v), d δ 1. We claim that one of these inequalities is

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strict. For suppose γ = P -δ = P -d + 1. Using the fact that

d ≤ , γ�

u

we have d (p-

d +1)≤or, (p-d )(d ( 1)≤0------(A)

w1 u11

wr

w2

u1p1 u21

u2p2

ur1

uqr

Figure 2. Tree T

V1

V2

V3

Figure. 1

Both factors in (A) are non negative, and because of (corollary10(vi)), we haved , there exists in G an =1 Since G ≠K , K edge uv and a vertex w not adjacent to, say v. Clearly, V -{v} is a g.d.set of G, and henceγ p-1. Thus, γ d p, a contradiction. The following results are due to D.F. Rall.It is immediate that Г G , where Г G of G is the upper global Г G domination number of G. 1. Minimum global domination in trees It is easy to see from the definition that γ (K , ) = 1 and γ (K , ) = 2 Consider the tree T in Figure 2 where r ≼ 2 and for each 1 ≤i ≤ r, n ≼2.γ (T)=r, γ T = r +1, with {u,w , ‌ , w ) a minimum global dominating set of T.

The main result of this section is that the stars and trees of diameter four of the type above are the only trees for whichγ(T)<γ T . Theorem 1: Let T be a tree. γ (T) = y(T) + 1 if and only if T is a star or T is a tree of diameter four which is constructed from two or more stars, each having at least two leaves, by connecting the centers of these stars to a common vertex. Proof: If T is a star or of the type described in the theorem (also see Figure 2), then it can easily be verified that γ (T) = γ(T) + 1. Conversely, assume γ (T) = γ(T) + 1 and that T is not a star. It follows that the diameter of T is Atleast three. Assume that T has diameter three and let u , u , u , u be a longest pat in T. Deg (u ) = 1 = deg (u ), and if U1 has any other neighbours in T besides u and u , they must be leaves. A similar statement is true for u . Therefore, D = {u ,u } is a minimum dominating set of T and is also a minimum global dominating set of T. This contradicts ourassumption that γ (T) >γ(T) and so the diameter of T must be atleast four. Suppose T has diameter p≼ 5 and let u ,u , ...,u be a longest pth in T. u and u are leaves of T and there is a minimum dominating set D for T which

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contains both u andu . Let C be the component of T -u u containing u , and let C be the component of T -u u containing u . In T every vertex of C is adjacent to u , and every vertex of C is adjacent to u in T. That is, {u ,u } dominates T and so D is a global dominating set of T. This implies γ (T) = γ (T), a contradiction.Hence the diameter of T must be four. Let u ,u ,u ,u ,u be a longest path in T, let u , v ,...,v be the leaves of T adjacent to u and let u , w ,...,w be the leaves of T adjacent to u . If u has a leaf as a neighbor, then there is a minimum dominating set A of T which contains u and u . Since such an A dominates T as well, γ (T) = γ (T) a contradiction. If u , a , a is a P attached at u and deg(a ) = 2, then there is a minimum dominating set A of T whichincludes u and a . But then again A dominates T so γ (T) = γ(T), also a contradiction. Therefore, if v is any neighbor of u in T, it follows that deg(v) ≥3 and v has atleast two leaves as neighbors. Hence T has the structure asclaimed in the statement of the theorem. 2. GLOBAL DOMINATION IN GRAPHS OF DIAMETER AT LEAST FIVE The proof of the following proposition follows directly from the definitions and is omitted. Proposition 12: Suppose A is a minimal dominating set of a graph G. If A also , then A is a minimal global dominates G dominating set of G.In general, a graph G will contain minimal global dominating sets which are not minimal dominating sets of G and also minimal dominating sets which do not dominate the complement of G. As an

example consider the graph H of Figure 3. {x, y, z, w} is a minimal global dominating set of H containing {y, z, w}, a minimal dominating set of H. However, we shall now show that for graphs having diameter at least five this cannot occur. X

Y

W

Z

Figure 3.H

Theorem 2: Let G be a graph diameter at least five, and let A be a subset of vertices in G. A is a minimal dominating set of G if and only if A is a minimal global dominating set of G. Proof: For vertices u and v in G we will let d(u,v) denote the length of a shortest u-v path in G if one exists. If G is disconnected and u and v belong to different components of G let d (u ,v) = ∞ with the assumption that ∞is large than any positive integer. Assume that A is a minimal dominating set of G and choose u, v ∈V(G) so that d(u, v,) ≥5. A must have a nonempty intersection with every closed neighborhood of G, so choose u ∈A∩ N[u] and v ∈A∩N[v]. For any x ∈ V -{u , v }, either xu ∉E(G) or xv ∉E(G), for otherwise G has a uv-path of length at most 4. Thus, {u , v } (and so also A) . By usingproposition 2 it dominates G follows that A is a minimal global dominating set in G. Conversely, assume

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that A is a minimal global dominating set of G but that A is not a minimal dominating set of G. There must be a vertex x∈ A so that A -{x} is a dominating set of G. Choosing u, v ∈ V as before as before with d(u, v) ≥ 5 it follows that there are vertices u and v with u ∈ (A -{x})∩N[u] and v ∈(A- {x})∩N[v]. Similar to before {u ,v } dominates G and so A- {x} is a global dominating set for G. This contradicts the minimallty assumption on A. Therefore, A is also minimal dominating set of G. Corollary 4.9 : If G is graph with diameter at least five, then γ(G) = γ (G) and Г (G) = Гg(G). 3. GLOBAL DOMINATION FOR SMALL DIAMETER As was shown in Theorem 2 the connection between domination and global domination in a graph of diameter five or larger is stronger than thei equality of the respective parameters. In this section we show that for graphs of smaller diameter the situation is very different, although for a tree T we have γ (T) = γ (T) with the two exception as given in Theorem 1 where γ and γ differ. Stars are the only trees for which Г and Г assume different values.In order to effectively discuss the case of trees of diameter four we introduce the following notation. Let T be a tree of diameter four having unique center vertex v. T will be said to have type [r, s, t] for nonnegative integers r, sand t if and only if v is adjacent to r leaves, v is adjacent to svertices of degree two and v is adjacent to t vertices of degree at least three. Note that s + t ≥2.There are thus six different types of trees of diameter four-those given in the following list:

s ≥2; r ≥1, s ≥ 2; s ≥1, t ≥1; r≥1, s≥1, t≥1: t≥2: r≥1, t≥2

[0, s, 0] [ r, s, 0] [0, s, t] [r,s,t] [0,0,t] [r, 0, t]

For example, the tree in Figure 2 is of type [0, 0, t]. It T is the star k1,p then it is clear thatγ(T) = 1 < 2 + γg (T) = Гg (T) ≤p = Г(T). Diameter three and four trees are covered in the following theorem. Theorem3: Let T be a tree of diameter three or four. Г(T) = Гg (T). However, unless T is the path P4 or of type [r, s, 0], there are minimal global dominating sets which fail to be minimal dominating. Proof: We do not present a complete proof of the theorem there but consider only one of the six types of trees of diameter four. The proofs in the remaining cases (of diameter three and four) are similar. In particular let T be the three of type [0, s, 0] shown in V

u1

v1

u2

us

v2

vs

Figure 4. type {0, s, 0} tree, T .

Let U = {u1 ,. . ., us } and V = {v1 , ..., vs }. It is clear that both U ∪ {v} and V ∪ {v} are minimal global dominating sets of T.

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Suppose A is an arbitrary minimal global dominating set of T. Assume first that v ∈A. If U is not a subset of A, then there exists uj ∉ A. If uk ∈ A for some k ≠ j, then {uk , vj } dominates T and so A –{v} is a global dominating set of T, a contradiction. Therefore, if v ∈ A either A = U ∪ {v} or A = V ∪ {v}. If v ∉ A, let X = {i / uj ∈A}. Since A dominates T, it follows that if j ∉ X, then vj ∈ A. Because A ≠ U and A ≠ V, there exist 1 ≤ q < p ≤ s with q ∈ X and p ≠ X. This yields uq ∈ A and vp ∈ A. {uq , vp } dominates T, and so it follows that for each i, 1 ≤ i ≤ s, exactly one of ui , vi is in A. Therefore A = {ui / i∈ X} U { vi / i ∉X }, both of these subsets are nonempty and soIAI = s. it now follows that Г(T) = s + 1 = Гg (T). If G is any graph and D is any global dominating set of G, then.|D|≥γ (G) and ) and so the following theorem |D|≥γ G holds . Theorem 4 : For any graph G, γg (G) ≥ max )}.One might expect to be able to {γ (G), γ(G establish a similar type of bound for Гg (G) ). However, which involves Г(G) and Г (G the following four examples by necessity each has diameter less than five-show that one of the four possible bounds for Гg (G) holds in general. Example 1 : Let q ≥ 2 and p≥ 2 be positive integers. Let H be the graph in Figure 5 where A = {u1 , ..., uq 1} induces a complete subgraph, B={v1 , ...,vp 1 }induces an independent set, N(x) = A ∪ Band N (y) = B. It is straightforward to verity that if D is a minimal global dominating set of H then D ={x,vi } or D = {y, ui } for some i, and so γg (H) = Гg (H) = 2. Yet Г (H) ≥ p and Г (H)

)} nor max ≥ q. Henceneither min [Г(G), Г(G )} can, in general, be a lower [Г(G), Г(G bound for Гg (G). u1

A

u2 u q-1

X

V1

B

Vq-1

V2

y Figure 5. H

Example 2 : Let 2 ≤q < p, let A = {v1 , ..., vp } and B = {u1 , ..., up }. Let H be the graph of order 2p where each of A and B induces a complete subgraph of H. In addition, let H contain all edges vi uj where 1 ≤ i ≤ q, 1 ≤ j ) = p ≤q and i ≠j. See Figure 6. Г(H) = 2, Г(H and Гg (H) = q+1. Therefore, for an arbitrary graph G, Гg (G) is not bounded above by min )}. {Г(G), Г(G V1

V2

Vq

Vq+1

Vp

A

B U1

U2

Uq

Uq+1

Up

Figure 6. H

)} is The 5-cycle shows that max {Г(G), Г (G not an upper bound for Гg G . We initially suspected that Гg (G) could )} + 1, never be larger than max {Г(G), Г(G

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but a computer search has produced a graph G of order 12 with Гg (G) = 6 and Г (G) = 4 ). We thus end with the following = Г(G open question. Given positive integers r, s, and t does there exist a graph G with Г(G) = ) = s and Гg (G) = t? r, Г(G Total domination global set of a graph RESULTS Theorem 1 -A total dominating set T of G is a t.g.d.set if and only if for each vertex v ∈V there exists a vertex u ∈ T such that v is not adjacent to u. Theorem 2 -Let G be a graph such that have an isolated vertex. neither G nor G Then, ); (i) γ (G)=γ (G (ii) γt (G)≤γtg (G); iii) γg (G) ≤γtg (G); )/2≤γ (G) ≤γ (G) + γ (G ). (iv) (γ (G) +γ (G t

t

tg

t

t

Next we characterize graphs G which have total global domination number equal to the order p of G. Theorem 3 -Let G be a graph such that have an isolated vertex. neither G nor G Then γtg (G) = P , if and only if G = P4 (a path on 4 vertices) or q k2 or q k 2 :q ≥ 2. Proof: Suppose γtg (G) =p holds, On the 2: q≥2. contrary; suppose G ≠P4 , q K2, q K Then we consider the following cases: ) are ≤ p -3, where Case 1 : If ∆ (G) and ∆(G ∆(G) is the maximum degree of-G, then both have no vertices of degree 1 and G and G hence for any vertex v∈V, V -{v} is a t.g.d. set of G, a contradiction.

230

) = P -2 say Case 2 : If either ∆(G) or ∆(G ∆(G) = P -2 and u is a vertex of degree p -2, then there exists exactly one vertex v such that v is not adjacent to u. If v is also of 2; q ≥ 2, there degree p - 2, then, as G = q K exists a vertex w such that w is not adjacent to at least two vertices. If some nonneighbour x of w has degree p -2, then V -x is a t.g.d.set. Otherwise each non-neighbour of w has at least two non-neighbours and V w is a t.g.d.set. Suppose degG (v) < p-2 If u has no neighbour of degree 1, then V -v is a t .g.d.set of G, otherwise let u be adjacent to w, degG (w) = 1, and let x be adjacent to v (and necessarily adjacent to u non-adjacent to w) then {v, x, u, w} is a t.g.doset (and G = ) = P -2, then we can apply the P4 ). If ∆ (G (and obtain than G = same argument to G P4 , so that G = P 4 = P4 ). This proves the necessity. Sufficiency is obvious. Theorem 4 -Let G be a graph such that have an isolated vertex and neither G nor G T be a γt -set of G with each x in T has nonneighbour in T.,if there exists a vertex u ∈V -T which is adjacent only to vertices in T, then, γtg (G)≤γt (G)+2. Proof: We consider the following cases: Case 1 : If V -T = {u}, then there exists a vertex v ∈T such that v is not adjacent to u and hence T is a t.g.d. set. Thus γtg (G)≤γt (G)+2, holds. Case 2 : If V- T ≠ {u}, then there exists a vertex v ∈ V- T and hence T ∪ {u, v} is a t.g.d.set. Thus γtg (G) ≤γt (G)+2, follows. Now we obtain a lower bound on γtg (G). Theorem 5 -Let G be a graph such that have an isolated vertex. neither G nor G Then, 2q - p (p-3) ≤γtg (G).

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Theorem 6 -Let G be a graph such that have an isolated vertex. neither G nor G Then,γtg (G) ≤2α0 (G).

Theorem 7 : Let G be a graph with diam(G) ≥ 5. Then T ⊊V is a total dominating set of G if and only if T is a t.g.d.set. Proof: Suppose T is a total dominating set of G. Let u, v E V such that d(u, v)≥5. Then T∩ N(u) ≠∅ and T∩ N(v) ≠∅. Let u1 ∈ T∩ N(u) and v1 ∈T∩N(v). Then u1 and v1 are not adjacent and further every vertex in V -{u1 , v1 } is adjacent to at most one of u1 and v1 . This implies that {u1 , v1 } is a total and hence T is a dominating set of G t.g.d.set.The converse is obvious.For a set D ⊊ V, the minimum degree of the induced subgraph <D> is denoted by δ( <D>).

Proof: Let S be a vertex cover of G with |S| = α0 (G). Let S= {u1 , u2 , ..., us }, S = {u1 } is impossible since u1 has a non-neighbour x, and x has a neighbour which can only be u1 , so s≥ 2. Each u ∈ S has a neighbour, a nonneighbour, or both, in S -u. If u has no neighbour in D, then choose v ∈V –S a neighbour of u. If u has no non-neighbour in S, then choosen in V -S a non- neighbour v of u. Thus construct D= {v1 , v2 , ..., vs }, s'≤ s, not all vertices vi , vj need be distinct, so |D|≤|S|. If D= {v1 } and if v1 is adjacent to each vertex of s then add a non-neighbour v1 to D so that |D|≥2. S∪D is a t.g.d.set in G because x ϵ V -{S∪D} has at least one neighbour which necessarily belongs to S and each vertex of D is a non- neighbour of x. x ϵS has by construction both a neighbour and a non- neighbour in S∪D. x ϵD has a neighbour in S and as |D|≥2, x has a non-neighbour in D. Thus γtg (G) ≤|S . D|= |S| + |D|≤ 2 ISI = 2α0 (G). For a vertex v ∈ V, the eccentricity is defined as e (v) = max {d(u, v)/uϵV}. The maximum of the eccentricities is defined to be the diameter of G, denoted diam (G).

Theorem 8 -Let G be a graph such that have an isolated vertex and neither G nor G diam(G) ≥ 5. A set D ⊊ V with δ(<D>) ≥ 1, is a g.d.set of G if and only if D is a t.g.d.set. Proof: Let D be a g.d.set of G. Suppose there exists a vertex u∈D such that u is adjacent to every vertex in D. Then, diam(G) ≤ 4, a contradiction this implies that D is a . Since (<D>) has total dominating set of G no isolates, D is a t.g.d.set.The converse is immediate. Corollary 1 : Let G be a graph such that have an isolated vertex and neither G nor G D be a γg -set of G with δ(<D>) ≥ 1. If diam(G) ≥ 5, then,γr (G) = γtg (G), γg (G) = γtg (G). Proposition A:(i) For any complete bipartite graph Kq,p with 2 ≤ q ≤p, γt (Kq,p) = 2. (ii) For any cycle Cp with p ≥r, vertices, γt (Cp ) = (p/2) + 1 if P ≡ 2 (mod 4); . (iii)For any path PP with p≥4, vertices. p γt (Pp )=(2)+1 if p≡2(mod4); Now we list the exact values of γtg (G) for some standard graphs.

Proof: Let T be a γtg -set of G. Then by Theorem 1, each vertex v ∈ V is not adjacent to at least one vertex in T. This |T| p implies q≤ –|V ( T| + thus, 2q-p (p2 2 3)≤γtg (G).In a graph G, a vertex and an edge incident with it are said to cover each other. The vertex covering number α0 (G) equals the minimum number of vertices in a set D which covers every edge of G.

Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)

Gangadharappa, et al., J. Comp. & Math. Sci. Vol.2 (2), 222-233 (2011)

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Proposition B : (i) For any complete bipartite graph Kq,p with 2≤ q ≤ p, γtg (Kq,p) =4.(ii) For any cycle Cp with p ≥ 4 vertices γtg (Cp ) = (p/2) + 1 if P ≡ 2 (mod 4); (iii) For any path Pp with p≥4, vertices γtg (Pp ) = (p/2) + 1 if P ≡ 2 (mod 4);

Theorem 11: Let diam(G) and diam (G) ≥ 3.If G is connected, then γtg (G)≤ min {p∆(G) + 2, δ(G) + 4}. If G is disconnected, then γtg (G) ≤min {p-∆(G) + 3, δ(G) + 3}. Proof: By the result, γtg (G)≤γt (G)+2 Since ), γ (G) ≤ min {γ (G) + 2, γtg (G) = γtg (G tg t ) + 2}. Suppose G is connected. Since γ (G

Theorem 9: Let diam (G) = k.

∆(G) < p-1, by Cockayne. γtg (G) ≤ P -∆(G). γt (G) ≤ P -∆(G) + 1 = 2 + δ(G). Thus the result (i) of theorem11 holds. Similarly, we can prove (ii) of theorem11, when G is disconnected.

(i) If k = 4, then γ (G) ≤γ (G) + 1. (ii) If k =3, then, γtg (G)≤γt (G)+2, holds. Proof: Let T be a γt -set of G. Suppose k = 4 and u, v be two vertices with d(u,v) = 4. Then T∩N[u] ≠∅. Let u1 ∈T∩N[u]. Then no vertex in G is adjacentto both u1 and v and . hence {u1 , v} is a total dominating set of G Thus T ∪ {v}is a t.g.d.set. Hence γtg (G)≤γt (G)+1,holds. If k = 3 and u, v be two vertices with d(u, v) = 3. then no vertex in G is adjacent to both u and v and hence . Thus {u, v} is a total dominating set of G T∪ {u, v} a t.g.dset. Hence γtg (G) ≤γt (G) +2,holds.Similarly, we can prove Theorem 10:Let D be a γg -set G such that (D) has no isolates and diam (G) = k. (i) If k = 4, then, γtg (G) ≤γg (G) + 1. (ii) If k = 4, then, γtg (G) ≤γg (G) + 2. When diam(G) = 2, then the difference between γtg (G) andγt (G) as well as between γtg (G) and γg (G) may be p with P ≥ very large. For example, let G = C 19 vertices. Then any two vertices u and v at a distance at least four in Cp form a γt -set for G and hence γr (G) = 2. Also, by proposition B, γtg (G) = γtg (Cp ) ≥┌ p/2┐. Further, γg (G) = γg (Cp ) = Гp/3┐.

t

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Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)

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