On Pairs of Disjoint Dominating Sets in a Graph

Transcription

1 International Journal of Mathematical Analysis Vol 10, 2016, no 13, HIKARI Ltd, wwwm-hikaricom On Pairs of Disjoint Dominating Sets in a Graph Edward M Kiunisala Mathematics Department College of Arts and Sciences Cebu Normal University Cebu City, Philippines Ferdinand P Jamil Department of Mathematics and Statistics College of Science and Mathematics MSU-Iligan Institute of Technology Iligan City, Philippines Copyright c 2016 Edward M Kiunisala and Ferdinand P Jamil This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Abstract In this paper, we investigate pairs of disjoint dominating sets A and B in a graph G, where B is either an independent or a total dominating set in G Mathematics Subject Classification: 05C69 Keywords: dominating set, independent dominating set, total dominating set, pair of disjoint dominating sets 1 Introduction In this study, we consider graphs G = (V (G), E(G)) which are finite, simple and undirected The basic terminologies used here are adapted from [4] For

2 624 Edward M Kiunisala and Ferdinand P Jamil S V (G), the symbol S denotes the cardinality of S In particular, V (G) is called the order of G The cardinality E(G) of the edge set E(G) of G is the size of G If E(G) = 0, then G is an empty graph An empty graph of order n is denoted by K n Given two vertices u and v of a graph G, a u-v geodesic is a shortest path in G joining u and v For a non-empty S V (G), S denotes that subgraph H of G for which E(H) is the maximum size of a subgraph of G with vertex set S Let G and H be any graphs The (disjoint) union of G and H is the graph G H with vertex set V (G H) = V (G) V (H) and edge set E(G H) = E(G) E(H), where V (G) and V (H) are considered disjoint The join of G and H is the graph G + H with vertex set V (G) V (H) and edge set E(G) E(H) {uv : u V (G), v V (H)} The corona G H of G and H is the graph obtained by taking one copy of G and V (G) copies of H, and then joining the i th vertex of G to every vertex in the i th copy of H We denote by H v that copy of H whose vertices are adjoined with the vertex v of G In effect, G H is composed of the subgraphs H v + v = H v + {v} joined together by the edges of G For any v V (G), G v is the resulting graph after removing from G the vertex v and all edges of G incident to v For any vertex v of a graph G, the closed neighborhood N G [v] of v consists of v and all vertices of G adjacent to v We also define N G (v) = N G [v] \ {v} For S V (G), we define N G [S] = v S N G [v] and N G (S) = v S N G (v) A dominating set in G is any S V (G) for which N G [S] = V (G) In this case, we also say S dominates V (G) If S = {u} dominates V (G), we simply say u dominates V (G) A dominating set S is an independent dominating set if uv / E(G) for all u, v S A dominating set S is a total dominating set if for each u S there is v S such that uv E(G) The minimum cardinality of a dominating (resp independent dominating, total dominating) set is called the domination number (resp independence domination number, total domination number) of G, denoted by γ(g) (resp γ i (G), γ t (G)) The symbols D(G), I (G) and T (G) are used to denote the collection of all dominating sets, the collection of all independent dominating sets, and the collection of all total dominating sets in G, respectively We also use the terminologies γ-set, γ i -set and γ t -set to mean any set in D(G), in I (G) and in T (G), respectively, of minimum cardinality Domination is one of the most well-studied concepts in graph theory The reader is referred to [1, 3, 5, 6, 7, 8, 10, 11, 13, 22, 24, 25] for the fundamental concepts and recent developments of the domination theory, and to [3, 12, 16, 25] for its various applications In 1962, O Ore gave the classical result which can be stated as follows: Theorem 11 [24] If a graph G has no isolated vertices and S is a minimum dominating set, then V (G) \ S is a dominating set in G

3 On pairs of disjoint dominating sets in a graph 625 It has motivated the introduction of the concept of inverse domination (see [21]) as well as the concept of disjoint domination (see [14]) A subset S V (G) is an inverse dominating set in G if S is a dominating set in G and there is a minimum dominating set D in G such that S D = The minimum cardinality of an inverse dominating set in G is the inverse domination number of G, which is denoted by γ (G) For any graph G which has no isolated points, Theorem 11 guarantees the existence of a pair (S, D) of dominating sets in G with S D = Any such pair is called a dd-pair We denote by DD(G) the collection of all dd-pairs in G The minimum sum S + D among all dd-pairs in DD(G) is the disjoint domination number of G, which is denoted by γγ(g) That is, γγ(g) = min{ S + D : (S, D) DD(G)} A dd-pair (S, D) with S + D = γγ(g) is called a γγ-pair Inverse domination is studied further in [9, 19, 20] Disjoint domination is also investigated in [16, 17, 19, 23] In [26], an application of the concept in information retrieval system is cited: In an Information Retrieval System, we always have a set of primary nodes to pass on the information In case, the system fails, we have another set of secondary nodes to do the job in the complement Thus, the dominating sets and the elements in the inverse dominating sets can stand together to facilitate the communication process In what follows, we investigate disjoint domination which involves either a total dominating or an independent dominating set 2 A pair of disjoint dominating and total dominating sets The authors in [15, 17, 18] have determined certain conditions under which a graph G has a vertex set V (G) having a pair of subsets consisting of a dominating set and a total dominating set which are disjoint Theorem 21 [17] If G is a graph of minimum degree at least 2 such that no component of G is a chordless cycle of length 5, then V (G) can be partitioned into a dominating set D and a total dominating set T Theorem 22 [18] If G is a graph of minimum degree at least 3 with at least one component different from the Petersen graph, then G contains a dominating set D and a total dominating set T which are disjoint and satisfy D + T < V (G) A graph G is a (D, T )-graph if V (G) contains a pair of disjoint subsets where one is a dominating set and the other a total dominating set If G is

4 626 Edward M Kiunisala and Ferdinand P Jamil a nontrivial connected graph, then for any graph H, G + H and G H are (D, T )-graphs A dt-pair in G is any pair (S, T ) such that S is a dominating set and T is a total dominating set in G with S T = The symbol DT (G) denotes the collection of all dt-pairs in G, and γγ t (G) = min{ S + T : (S, T ) DT (G)} A dt-pair (S, T ) is called a γγ t -set if S + T = γγ t (G) For all (D, T )-graphs G, γ(g) + γ t (G) γγ t (G) V (G) (1) If G = K 3 K 1, then γ t (G) = 3 = γ(g) For this graph, γ(g) + γ t (G) = γγ t (G) = V (G) = 6 Theorem 23 Let G be a (D, T )-graph Then (i) γγ t (G) = 3 if and only if G = K 1 + H, where H is connected and γ t (H) = 2 (ii) γγ t (G) = 4 if and only if either (a) G = K 1 + H, where H is a connected graph with γ t (H) = 3, or (b) G has a pair of disjoint γ-sets {u, v} and {x, y} such that xy E(G) Proof : Statement (i) is clear (ii) Suppose that γγ t (G) = 4, and let (S, T ) be a γγ t -pair in G Then either S = 1 and T = 3 or S = 2 = T Suppose that S = 1 and T = 3, and put S = {v}, T = {a, b, c} and H = G v Then G = K 1 + H and T V (H) Moreover, T is a total dominating set in H so that H is connected and γ t (H) 3 In view of Theorem 23(i), γ t (H) = 3 Suppose that G = K 1 + H with H connected and γ t (H) = 3, and let T be a γ t -set in H Then (V (K 1 ), T ) is a dt-pair in G and 3 γγ t (G) = 4 In view of Theorem 23(i), γγ t (G) = 4 Suppose that S = T = 2 Clearly, T = K 2 In view of Statement (i), S and T are γ-sets of G Conversely, if γ(g) = 2, then γγ t (G) > 3 by Statement (i) Thus γγ t (G) = 4 The next theorem shows that γγ t (G) γ(g) γ t (G) and V (G) γγ t (G) can each be made arbitrarily large Theorem 24 For each positive integer n,

5 On pairs of disjoint dominating sets in a graph 627 u 1 G : v 1 v 2 x y u 2 u n Figure 1: (i) there exists a connected graph G such that γγ t (G) γ(g) γ t (G) = n; (ii) there exists a connected graph G such that V (G) γγ t (G) = n Proof : (i) Write K 2 = [x, y] and obtain G as in Figure 1 by adding to K 2 the pendant edges v 1 x, v 2 x and u j y, j = 1, 2,, n Then {x, y} is a γ-set and at the same time a γ t -set in G On the other hand, there is a unique γγ t -pair in G, namely S = {v 1, v 2, u 1, u 2,, u n } and T = {x, y} For this graph G, γγ t (G) γ t (G) γ(g) = [(n + 2) + 2] 2 2 = n (ii) Consider the graph G = K 2 + K 1,n = K 1 + (K 1 + K 1,n ) Since γ t (K 1 + K 1,n ) = 2, γγ t (G) = 3 by Theorem 23 Thus, V (G) γγ t (G) = (n + 3) 3 = n 3 A pair of disjoint dominating and independent dominating sets Recall that by an independent set we mean any S V (G) such that uv / E(G) for all u, v S, and the symbol β(g) denotes the maximum cardinality of an independent set in G An independent set S with S = β(g) is called a β-set Lemma 31 let G be any graph with no isolated vertices If S V (G) is an independent set in G, then V (G) \ S is a dominating set in G Let G be a graph without an isolated vertex A di-pair in G is any pair (S, I) of subsets of V (G), where S is a dominating set and I is an independent dominating set in G such that S I = We use the symbol DI (G) to denote the collection of all di-pairs in G, and define γγ i (G) = min{ S + I : (S, I) DI (G)} A di-pair (S, I) with S + I = γγ i (G) is called a γγ i -pair It is worth noting that Lemma 31 guarantees existence of di-pairs (S, I), where I is a γ i -set in G An inverse independent dominating set in G is a dominating set S such that S V (G)\I for some γ i -set I in G The minimum cardinality of an inverse independent dominating set is denoted by γ i(g) Any

6 628 Edward M Kiunisala and Ferdinand P Jamil inverse independent dominating set with caridanilty γ i(g) is called a γ i-set in G Apparently, γ(g) γ i(g) V (G) γ i (G) (2) If G is the complete graph K m, where m 2, then γ(g) = 1 = γ i(g) If G is the complete bipartite graph K m,n, where 1 m n, then γ i (G) = m and γ i(g) = n so that γ i(g) = n = V (G) γ i (G) Hence, the above estimates are sharp The parameters γ i(g) and γ i (G) are not directly related Note, in particularly, that γ i (P 3 ) < γ i(p 3 ) On the other hand, γ i(g 1 ) < γ i (G 1 ), where G 1 is the graph as in Figure 2 Verify that γ i (G 1 ) = 3 and γ i(g 1 ) = 2, the latter being determined by the blackened vertices 1 x 2 x n+1 u v 1 y 2 y n+2 G 1 G 2 Figure 2: G 3 The parameters γ i(g) and γ (G) are not also directly related Note that γ (P 2 K 2 ) = 4 > 3 = γ i(p 2 K 2 ) On the other hand, we have γ (G 2 ) = 2 < 3 = γ i(g 2 ), where G 2 is the graph as in Figure 2 The blackened vertices comprise a γ i-set in G 2 In fact, γ (G) γ i(g) and γ i(g) γ (G) can each be made arbitrarily large Let n be a positive integer If G is the graph G 3 as in Figure 2, then the sets {x k, y j : k = 1, 2,, n + 1; j = 1, 2,, n + 2} and {u, y j : j = 1, 2,, n + 2} are a γ -set and a γ i-set in G, respectively In this case, γ (G) γ i(g) = (2n+3) (n+3) = n On the other hand, if G = K 2,n+2, then γ (G) = 2 and γ i(g) = n + 2 so that γ i(g) γ (G) = n Theorem 32 Let G be a connected graph of order n Then (i) γ i(g) = 1 if and only if either G = K 2 or G = K 2 + G, for some graph G of order n 2 (ii) For n 2, γ i(g) = n 1 if and only if G = K 1, n 1 ; (iii) For n 4, γ i(g) = n 2 if and only if either G = K 1 + (K n 3 K 2 ) or there exists u, v V (G) such that d G (u, v) = 2 and V (G) \ {u, v} = K n 2 Proof : Statement (i) is clear (ii) By Theorem 32(i), the result holds for n = 2 We proceed with n 3 Suppose that γ i(g) = n 1, and let S V (G) be a γ i-set in G Let

7 On pairs of disjoint dominating sets in a graph 629 v V (G) \ S Then N G [v] = V (G), that is, vx E(G) for all x V (G) \ {v} We claim that xy / E(G) for all x, y V (G) \ {v} Suppose that there exist x, y V (G) \ {v} such that xy E(G) Then v, y N G [x] N G [S \ {y}] Thus, S \ {y} is an inverse independent dominating set in G, a contradiction Therefore, G = K 1, n 1 The converse is obvious (iii) Suppose that n 4 and γ i(g) = n 2, and let S be a γ i-set in G Let u, v V (G) \ S We consider two cases Case 1: Suppose that N G [v] = V (G) First, we claim that S = K n 2 Suppose that there exist x, y S such that xy E( S ) Either xu, yu / E(G) or, say, ux E(G) Since u, y, v N G [S\{y}], S\{y} is an inverse independent dominating set in G, a contradiction This shows that S = K n 2 Next, we claim that there is a unique x S such that xu E(G) For the existence part, note that if xu / E(G) for all x S, then G = K 1, n 1 and, by Theorem 32(ii), γ i(g) = n 1, a contradiction To show the uniqueness, suppose that there exists y S \ {x} with uy E(G) Then x, y N G [u] N G [S ], where S = (S \ {x, y}) {u} This means that S is an inverse independent dominating set in G and γ i(g) n 3, a contradiction The two claims above together imply that G is the graph obtained by adding to K 1,n 1 the edge xu That is, G = K 1 + (K n 3 K 2 ) Case 2: Suppose that {u, v} is a γ i -set in G Then uv / E(G) Let P = [u = x 1, x 2,, x k = v] be a u-v geodesic in G Since {u, v} is a γ-set in G, k 4 We claim that if k = 4, then G = P = P 4 Suppose that k = 4 and G P 4 Let y V (G) \ V (P 4 ) Then either uy E(G) or vy E(G) Assume that uy E(G) Note that u, v, x 2 N G [{y, x 3 }] N G [S \ {x 2 }]] This means that S \ {x 2 } is an inverse independent dominating set in G, a contradiction Thus, if k = 4, then G = P 4 In this case, d G (u, x 3 ) = 2 and V (G) \ {u, x 3 } = K 2 Suppose that k < 4 Since uv / E(G), k = 3 and P = [u = x 1, x 2, x 3 = v] This means that d G (u, v) = 2 Let x, y S such that xy E(G) Put S = S \ {y} whenever x = x 2, and put S = S \ {x} whenever x x 2 Then V (G) = N G [S ] Accordingly, S is an inverse independent dominating set in G, a contradiction Therefore, S = K n 2 Now, we prove the converse First, suppose that G = K 1 + (K n 3 K 2 ) Let V (K 1 ) = {v}, u V (K 2 ), and let S = V (G) \ {u, v} Then {v} and S are a γ i -set and a γ i-set, respectively, in G Thus, γ i(g) = S = n 2 Next, suppose that G has distinct vertices u and v such that d G (u, v) = 2 and S = K n 2, where S = V (G) \ {u, v} Since G is connected, for each x S, xu E(G) or xv E(G) Thus {u, v} is an independent dominating set in G Since u / N G [v], {v} is not a dominating set in G Similarly, {u} is not a dominating set in G Since n 4, {x} is not a dominating set in G for all x S Thus {u, v} is a γ i -set in G It also follows that S is a γ i-set in G, and γ i(g) = n 2

8 630 Edward M Kiunisala and Ferdinand P Jamil For all graphs G with no isolated vertices, γ i (G) + γ(g) γγ i (G) γ i (G) + γ i(g) (3) Theorem 33 Let G be a nontrivial connected graph of order n 2 Then (i) γγ i (G) = 2 if and only if G has two distinct vertices each of which dominates V (G) (ii) γγ i (G) = 3 if and only if G = K 1 + H, where γ(h) = 2 (iii) γγ i (G) = 4 if and only if either G = K 1 + H where γ(h) = 3 or γ i (G) = 2 = γ i(g) Proof : (i) Clearly, γγ i (G) = 2 if and only if γ(g) = 1 and γ (G) = 1 The latter holds if and only if G has two distinct vertices each of which dominates V (G) (ii) Suppose that γγ i (G) = 3 and (D, I) is a γγ i -pair in G Either D = 1 and I = 2 or I = 1 and D = 2 In either case, G has a vertex v such that N G [v] = V (G) Thus, G = K 1 + H for some subgraph H of G By Theorem 33(i), it is necessary that γ(h) = 2 Conversely, suppose that G = K 1 + H, where γ(h) = 2 Then γ i (G) = 1 and γ i(g) = 2 Inequality 3 implies that γγ i (G) 3 In view of Theorem 33(i), γγ i (G) = 3 (iii) Suppose that γγ i (G) = 4, and let (D, I) be a γγ i -pair in G Then exactly one of the following holds: (1) D = 1 and I = 3; (2) I = 1 and D = 3; (3) D = 2 = I If (1) or (2) holds, then G has a vertex that dominates V (G) so that G = K 1 +H for some subgraph H of G Moreover, by Theorem 33(i) and Theorem 33(ii), γ(h) = 3 Suppose that I = 2 = D, and let I = {u, v} Since I is an independent dominating set in G, γ i (G) 2 Suppose that γ i (G) = 1, and let w V (G) for which N G [w] = V (G) Since uv / E(G), w u and w v Thus {u, v} and {w} constitute a di-pair in G so that γγ i (G) 3, a contradiction This means that γ i (G) = 2 and {u, v} is a γ i -set in G Consequently, D is an inverse independent dominating set in G Hence, γ i(g) D = 2 If γ i(g) = 1, then, by Theorem 32(i), either G = K 2 or G = K 2 + G for some subgraph G of G In either case, γγ i (G) = 2, a contradiction Therefore, γ i(g) = 2 Conversely, suppose that G = K 1 + H where γ(h) = 3 Then γγ i (G) 4 In view of the preceeding results, γγ i (G) 4 Finally suppose that γ i (G) = 2 = γ i(g) Then 3 γγ i (G) 4 Theorem 33(ii) above implies that γγ i (G) = 4 4 Join of graphs Proposition 41 Let G and H be graphs with γ i (G) < γ i (H)

9 On pairs of disjoint dominating sets in a graph 631 (i) If G is an empty graph, then γ i(g + H) = γ(h) (ii) If G is a nonempty graph, then { γ i(g 1, if G = K 2, K 2 + G for some G < G, + H) = 2, otherwise Proof : Let S V (G + H) be a γ i-set in G + H, and let I V (G + H) be a γ i -set in G + H such that S V (G + H) \ I Since γ i (G) < γ i (H), I V (G) If G is an empty graph, then I = V (G) and, consequently, S V (H) and S is a γ-set in H That is, γ i(g + H) = S = γ(h) This proves (i) Suppose that G is a nonempty graph Then V (G) I Take u V (G) \ I and v V (H) Then S = {u, v} is a dominating set in G + H and S V (G + H) \ I Thus, γ i(g + H) S = 2 Consequently, S S Suppose that S = 1 Then I and S and are both singleton subsets of V (G) Thus G has two distinct vertices u and v such that both u and v dominate V (G) Accordingly, either G = K 2 or G = K 2 + G for some subgraph G of G On the other hand, it can be readily verified that if G = K 2 or G = K 2 + G for any graph G, then γ(g + H) = 1 = γ i(g + H) This proves (ii) Proposition 42 Let G and H be graphs with γ i (G) = γ i (H) (i) If G is an empty graph, then { γ i(g V (G), if γ(h) = 1 or H is an empty graph, + H) = 2, otherwise (ii) If G is a nonempty graph, then γ i(g + H) = { 1, if γ(g) = 1, 2, otherwise Proof : Suppose that G is an empty graph Then V (G) = γ i (H) V (H) If V (G) = V (H), then H is an empty graph, and γ i(g+h) = V (G) Suppose that V (H) > V (G), and I V (H) is a γ i -set in H Then I V (H) Pick u V (H) \ I For any v V (G), {u, v} is an inverse independent dominating set in G + H This implies that γ i(g + H) 2 Let S be a γ i-set in G + H Then, in particular, S = 1 if and only if V (G) = 1 and γ(h) = 1 This proves (i) Suppose that G is a nonempty graph Then γ i (G) < V (G) Let I V (G) be a γ i -set in G Choose u V (G) \ I and v V (H) Then {u, v} is an inverse independent dominating set in G + H Thus γ i(g + H) 2 Clearly, γ i(g + H) = 1 if and only if γ(g) = 1 This proves (ii)

10 632 Edward M Kiunisala and Ferdinand P Jamil Corollary 43 Let G be a graph without isolated vertices Then γ i(g) = 1 if and only if G = K 2 + H for some graph H of order V (G) 2 Proof : Suppose that γ i(g) = 1 Then there exists distinct(vertices) u and v V (G) such that {u} and {v} are γ-sets in G Since uv E(G), G = {u, v} + H = K 2 + H The converse immediately follows from Proposition 41 and Proposition 42 Proposition 44 Let G and H be graphs with γ i (G) γ i (H) (i) If G and H are nonempty, then either γγ i (G + H) = 2 or γγ i (G + H) = 2 + γ i (G) Further, γγ i (G + H) = 2 if and only if γ(g) = γ(h) = 1 or G contains two vertices each of which dominates V (G) (ii) If G is an empty graph, then { 2 + V (G), if γ i (G) = γ i (H) < V (H), γγ i (G + H) = V (G) + γ(h), otherwise Proof : To prove (i), let G and H be nonempty graphs Let (S, I) be a γγ i - pair in G + H Then either I V (G) or I V (H) Note that since G is nonempty, if I V (G), then I V (G) Similarly, if I V (H), then I V (H) In any case, we may choose u V (G) \ I and v V (H) \ I Since {u, v} is a dominating set in G + H, the definition of S implies that S 2 Suppose that S = 1 If S V (H), then γ i (H) = 1 so that γ i (G) = 1 This means that γ(g) = γ(h) = 1 and γγ i (G + H) = 2 Suppose that S V (G) Pick x V (G) \ S and y V (H) Then {x, y} and S constitute a di-pair in G + H It follows that 2 γγ i (G + H) 3 If γγ i (G + H) 2, then γγ i (G + H) = 3 = 2 + S = 2 + γ i (G) On the other hand, γγ i (G + H) = 2 if and only if I = S = 1 Note that S = I = 1 if and only if either γ(g) = γ(h) = 1 or G contains two vertices each of which dominates V (G) Finally, suppose that S = 2 Since γ i (G+H) = γ i (G) and (S, I) is a γγ i -pair, γγ i (G + H) = 2 + γ i (G) Now, we prove (ii) If G is the trivial graph, then γγ i (G + H) = 1 + γ(h) Let G be an empty graph Then γ(g) 2 Suppose that γ i (G) = γ i (H) < V (H) Then γ i (G + H) = γ i (H) Take a γ i -set I in H and take u V (G) and v V (H) \ I Then I and {u, v} constitute a γγ i -pair in G + H so that γγ i (G+H) = 2+ V (G) Next, if H is empty, then G+H = K m,n, where m = V (G) and n = V (H) In this case, γγ i (G + H) = m + n = V (G) + γ(h) Finally, suppose that H is nonempty but γ(g) < γ(h) Then V (G) and any γ-set in H constitute a γγ i -pair in G+H Thus, γγ i (G+H) = V (G) +γ(h)

11 On pairs of disjoint dominating sets in a graph 633 Proposition 45 Let G be a graph with no isolated vertex (i) If γ(g) = 1, then γγ t (G + K 1 ) = 3 (ii) If γ(g) > 1, then γγ t (G + K 1 ) = min{1 + γ t (G), 2 + γ(g)} Proof : (i) If γ(g) = 1, then G is connected and γ t (G) = 2 By Theorem 23, γγ t (G + K 1 ) = 3 (ii) Suppose that γ(g) > 1, and let D = V (K 1 ) = {v} Let T V (G) be a γ t -set in G Then (D, T ) is a dt-pair in G+K 1 Thus, γγ t (G+K 1 ) 1+γ t (G) Let D V (G) be a γ-set in G Since G has no isolated vertex, D V (G) Let u V (G) \ D, and put T = {u, v} Then T is a γ t -set in G + K 1 and (D, T ) is a dt-pair in G+K 1 Hence, γγ t (G+K 1 ) 2+γ(G) Combining with the above result, γγ t (G) min{1 + γ t (G), 2 + γ(g)} Now, let (S, T ) be a dtpair in G + K 1 Either v S or v T Suppose that v S Then T V (G), and is a total dominating set in G Thus, S + T 1 + γ t (G) Suppose that v T Then T V (G) and S V (G) is a dominating set in G This yields S + T 2 + γ(g) This means that S + T min{1 + γ t (G), 2 + γ(g)} Since (S, T ) is arbitrary, γγ t (G + K 1 ) min{1 + γ t (G), 2 + γ(g)}, and the conclusion follows Proposition 46 Let G and H be nontrivial graphs Then { 3, if γ(g) = 1 or γ(h) = 1, γγ t (G + H) = 4, otherwise Proof : Clearly 3 γγ t (G + H) 4 Suppose that γγ t (G + H) = 3 By Theorem 23, G + H = K 1 + G, where G is connected and γ t (G ) = 2 Let V (K 1 ) = {v} If v G, then, in particular, vx E(G) for all x V (G) \ {v} so that γ(g) = 1 Similarly, if v V (H), then γ(h) = 1 Conversely, suppose that γ(g) = 1 and let v V (G) such that N G [v] = V (G) Then N G+H [v] = V (G + H) Since G is nontrivial, we can pick u V (G) \ {v} For any w V (H), {u, w} is a total dominating set in G = (G + H) v Thus γ t (G ) = 2 and, by Theorem 23, γγ t (G + H) = 3 5 Corona of graphs Canoy et al provided the following four results for the corona of graphs Theorem 51 [11] Let G be a connected graph and H any graph Then C V (G H) is a dominating set in G H if and only if C V (H v + v) is a dominating set in H v + v for every v V (G)

12 634 Edward M Kiunisala and Ferdinand P Jamil Theorem 52 [5] Let G be a connected graph and H any graph Then C V (G H) is an independent dominating set in G H if and only if C V (G) is an independent set in G and C V (H v + v) is an independent dominating set in H v + v for every v V (G) Theorem 53 [5] Let G be a connected graph of order n and H any graph with γ(h) 1 If C V (G H) is a γ i -set in G H, then C V (G) is a β-set in G Theorem 54 [11] Let G be a connected graph and let H be any graph Then C V (G H) is a total dominating set in G H if and only if for every v V (G), either (i) V (v + H v ) C is a total dominating set in H v + v; or (ii) v C and N G (v) C In particular, Theorem 52 and Theorem 53 yield the following corollary Corollary 55 Let G and H be connected graphs with γ(h) > 1, and let S V (G H) Then S is a γ i -set in G H if and only if S V (G) is a β-set in G and S V (H v ) is a γ i -set in H v for all v V (G) \ S Proposition 56 Let G and H be any graphs Then (i) γ i(g H) = V (G) + (γ(h) 1)β(G); and (ii) γγ i (G H) = V (G) (1 + γ i (H)) (γ i (H) γ(h))β(g) Proof : Verify that both claims are true for V (G) = 1 or γ(h) = 1 In what follows, we assume that V (G) > 1 and γ(h) > 1 Let S V (G H) be a γ i-set in G H, and let I V (G H) \ S be a γ i -set in G H By Theorem 53, I V (G) is a β-set in G and for each v V (G H) \ I, I V (H v ) is a γ i -set in H v For each v I V (G), choose a γ-set S v in H v, and define S = ( v I V (G) S v ) (V (G) \ I) By Theorem 51, S is a dominating set in G H Since S I =, S is an inverse independent dominating set in G H Thus, S S On the other hand, for each v I V (G), v / S so that S V (H v + v) = S V (H v ) By Theorem 52, S V (H v ) is a dominating set in H v, and thus, S v S V (H v + v) for all v I V (G) This yields S v I V (G) S v + V (G) \ I = S Therefore, S = γ(h)β(g) + V (G) β(g) = V (G) + (γ(h) 1)β(G)

13 On pairs of disjoint dominating sets in a graph 635 Let (S, I) be a γγ i -pair in G H, and for each v V (G), let I v = I V (H v + v) and S v = S V (H v + v) By Theorem 52, I V (G) is an independent set in G and I v = I V (H v ) is an independent dominating set in H v for each v V (G) \ I Also, by Theorem 51, S v is a dominating set in H v + v for each v V (G) By the choice of (S, I), S v + I v = γ(h)+1 for all v I V (G), and S v + I v = 1 + γ i (H) for all v V (G) \ I Since γ(h) γ i (H), I V (G) = β(g) Thus, S + I = β(g)(1 + γ(h)) + ( V (G) β(g))(1 + γ i (H)) Or, equivalently, γγ i (G H) = V (G) (1 + γ i (H)) β(g)(γ i (H) γ(h)) Corollary 57 If γ i (H) = γ(h), then γγ i (G H) = γγ(g H) for any graph G Theorem 58 Let G be a connected nontrivial grah and H any graph Then γγ t (G H) = V (G) (1 + γ(h)) Proof : For each v V (G), let D v V (H v ) be a γ-set in H v Then D = v V (G) D v is a dominating set in G H Thus, (D, V (G)) is a dt-pair in G H This means that γγ t (G H) D + V (G) = V (G) + γ(h) = V (G) (1 + γ(h)) v V (G) Conversely, let (D, T ) be a γγ t -pair in G H, and let v V (G) By Theorem 51, D v = D V (H v + v) is a dominating set in H v + v In view of Theorem 54, we consider two cases: Case 1: Suppose that T v = V (H v + v) T is a total dominating set in H v + v Since (D, T ) is a γγ t -pair and {v} dominates V (H v +v), if v / T v, then D v = 1 and T v V (H v ) and T v = γ t (H) In this case, D v + T v = 1 + γ t (H) On the other hand, if v T v, then D v + T v = γ(h) + 2 Case 2: Suppose that v T and N G (v) T By the same reason, T v = {v} and D v V (H v ) with D v = γ(h) Thus, D v + T v = 1 + γ(h) Combining all the above cases, D v + T v 1 + γ(h) Therefore, γγ t (G H) (1 + γ(h)) = V (G) (1 + γ(h)) v V (G) This completes the proof of the theorem References [1] RB Allan and R Laskar, On domination and independent domination numbers of a graph, Discrete Mathematics, 23 (1978), no 2,

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