Secure Connected Domination in a Graph

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1 International Journal of Mathematical Analysis Vol. 8, 2014, no. 42, HIKARI Ltd, Secure Connected Domination in a Graph Amerkhan G. Cabaro Department of Mathematics College of Natural Sciences and Mathematics Mindanao State University Marawi City, Philippines Sergio S. Canoy, Jr., and Imelda S. Aniversario Department of Mathematics & Statistics College of Sciences and Mathematics Mindanao State University- Iligan Institute of Technology 9200 Iligan City, Philippines Copyright c 2014 Amerkhan G. Cabaro, Sergio S. Canoy, Jr. and Imelda S. Aniversario. This is an open access article 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 Let G = (V (G), E(G)) be a connected simple graph. A connected dominating set S of V (G) is a secure connected dominating set of G if for each u V (G) \ S, there exists v S such that uv E(G) and the set (S \ {v} {u}) is a connected dominating set of G. The minimum cardinality of a secure connected dominating set of G, denoted by γ sc (G), is called the secure connected domination number of G. We characterized secure connected dominating set in terms of the concept of external private neighborhood of a vertex. Also, we give necessary and sufficient conditions for connected graphs to have secure connected domination number equal to 1 or 2. The secure connected domination numbers of graphs resulting from some binary operations are also investigated. Mathematics Subject Classification: 05C69

2 2066 Amerkhan G. Cabaro, Sergio S. Canoy, Jr. and Imelda S. Aniversario Keywords: dominating set, connected dominating set, secure connected dominating set, external private neighbors 1 Introduction Let G = (V (G), E(G)) be a simple graph with vertex set V (G) of finite order and edge set E(G). The neighborhood of v is the set N G (v) = N(v) = {u V (G) : uv E(G)}. If X V (G), then the open neighborhood of X is the set N G (X) = N(X) = {N G (v) : v X}. The closed neighborhood of X is N G [X] = N[X] = X N(X). A vertex w V (G) \ X is an X-external private neighbor of v X if N G (w) X = {v}. The set of all external private neighbors of v X is denoted by epn(v, X). A subset S of V (G) is a dominating set(ds) in G if for every u V (G)\S, there exists v S such that uv E(G), i.e., N[S] = V (G). The domination number of G is the minimum cardinality of a dominating set in G and is denoted by γ(g). A set S is said to be a secure dominating set in G if for every u V (G) \ S there exists v S such that uv E(G) and (S \ {v}) {u} is a dominating set. The minimum cardinality of a secure dominating set in G is called the secure domination number of G and is denoted by γ s (G). This variant of domination was introduced in [5] and several studies is found in [1, 2, 4] and elsewhere. Another variant of domination was introduced by Sampathkumar and Walikar in [8] which they called connected domination. A dominating set S is said to be connected dominating set(cds), if the induced subgraph S is connected. Let S be a connected dominating set in G. A vertex v S is said to S-defend u, where u V (G) \ S, if uv E(G) and (S \ {v}) {u} is a connected dominating set in G. S is a secure connected dominating set (SCDS) in G if for each u V (G) \ S, there exists v S such that v S-defends u. The secure connected domination number γ sc (G) of G is the smallest cardinality of a secure connected dominating set in G. A dominating set ( resp. secure connected dominating set ) S in G with S = γ(g) ( resp. S = γ sc (G) ) is called a γ-set ( resp. γ sc -set). In this paper, we initiate a study of secure connected domination in graphs. Any undefined terms maybe found in [3] or [6]. 2 Results and Discussion Remark 2.1 Let G be a connected graph of order n. Then 1 γ sc (G) n. Theorem 2.2 Let S be a connected dominating set of G, v S and u V (G) \ S with uv E(G). v S-defends u if and only if epn(v, S) N G [u] and V (C) N G (u), for every component C of S \ {v}.

3 Secure connected domination in a graph 2067 Proof : Suppose v S-defends u. Let w epn(v, S). Then w / S and N G (w) S = {v}. If w = u, then w N G [u]. If w u, then w / T, where T = (S \ {v}) {u}. Since T is a dominating set in G and N G (w) S = {v}, w N G (u). Hence, epn(v, S) N G [u]. Suppose that V (C) N G (u) = for some component C of S \ {v}. Then there is no path joining any a V (C) and u; hence T is not connected which is contrary to the assumption that v S-defends u. Now we suppose that epn(v, S) N G [u] and V (C) N G (u), for every component C of S \ {v}. Let w V (G) \ T, where T = (S \ {v}) {u}. If w = v, then uw E(G). If w v, then w / S. If w epn(v, S), then uw E(G). If w / epn(v, S), then N G (w) S {v}. Hence, there exists t T \ {v} T such that tw E(G). Thus, T is a dominating set in G. It remains to show that T is connected. Let a, b T, where a b and ab / E(G). Consider the following cases: Case 1: a = u and b V (D 1 ), where D 1 is a component of S \ {v} Let z V (D 1 ) N G (a). Then az E(G) and z V (D 1 ). Since D 1 is connected, there is a path, say P (z, b), joining z and b. Now, the path P (z, b) together with the edge az forms a path P (a, b) joining a and b. Case 2: a and b are in the same component D of S \ {v} Since D is connected, there is a path joining a and b. Case 3: a V (D 1 ) and b V (D 2 ), where D 1 and D 2 are different components of S \ {v} Let w V (D 1 ) N G (u) and x V (D 2 ) N G (u). Since D 1 is connected, there is a path P (a, w) joining a and w. Similarly, there is also a path P (x, b) joining x and b. The paths P (a, w) and P (x, b), together with the edges uw and ux form a path P (a, b) from a to b. Thus, in either case, T is connected. Corollary 2.3 Let S be a connected dominating set in G. Then S is a secure connected dominating set in G if and only if for every u V (G) \ S, v S N G (u) such that (i) epn(v, S) N G [u], and (ii) V (C) N G (u), for every component C of S \ {v}. Proof : Suppose S is an SCDS in G. Then for every u V (G) \ S, there exists v S N G (u) such that (S \ {v}) {u} is a CDS in G. This means that v S defends u. By Theorem 2.2, (i) and (ii) hold. The converse follows immediately. A leaf u of a graph G is a vertex of degree one and the support vertex of a leaf u is the unique vertex v such that uv E(G). We denote by L(G) and S(G) be the set of leaves and support vertices of G, respectively.

4 2068 Amerkhan G. Cabaro, Sergio S. Canoy, Jr. and Imelda S. Aniversario Theorem 2.4 Let G be a connected graph of order n 3 and let X be a secure connected dominating set of G. Then (i) L(G) X and S(G) X (ii) No vertex in L(G) S(G) is an X-defender. Proof : Suppose that x L(G). Let y S(G) such that xy E(G). Let S = (X \ {y}) {x}. Since S is not connected, x X. Suppose x S(G). Let z L(G) N G (x). Then z X. Since X is connected, x X. Let v L(G) S(G) and u V (G) \ X. Now v L(G) S(G) = v L(G) or v S(G). If v L(G), then uv / E(G). If v S(G), ( then ) there exists z V (G) such that N(z) = {v}. By (i), z X. Thus X \ {v} {u} is not connected. Thus, in either case, v does not X-defend u. Theorem 2.5 Let n be a positive integer. Then (i) γ sc (K n ) = 1 for all n 2 { 1, if n = 2 (ii) γ sc (P n ) = n, if n 3 { 1, if n = 3 (iii) γ sc (C n ) = n 1, if n > 3 Proof : First suppose that G = K n. Then S = {v} V (G) is a connected dominating set of G. Now x V (G), (S \ {v}) {x} = {x} which is a CDS of G. Thus S = {v} is an SCDS of G. Therefore, γ sc (G) 1. By Remark 2.1, it follows that γ sc (G) = 1. Next, let P n be a path of order n 2. For n = 2, P 2 = K2 and γ sc (P 2 ) = 1. Suppose n 3. Let S be a γ sc -set of P n = [v 1, v 2,..., v n ]. Then v 1, v n S by Theorem 2.4. Since S is connected, S = V (P n ). Thus γ sc (P n ) = S = n. Finally, let C n be a cycle of order n 3. Suppose n = 3, then C 3 = K3 and γ sc (C 3 ) = 1. For n > 3, let C n = [v 1, v 2,..., v n, v 1 ] and let S = {v 1, v 2,..., v n 1 }. Clearly, S is an SCDS in C n. Let S be a γ sc -set of C n. Since S is an SCDS in C n, S S = n 1. Suppose S = k < n 1. Since S is connected, V (C n ) \ S is connected, where V (C n ) \ S = m 2. Moreover, since S is a dominating set, m = 2. Let V (C n ) \ S = {v i, v i+1 }, i = 1, 2,..., n 1. Note that (S \ {v i 1 }) {v i } = {v 1, v 2,..., v i 2, v i, v i+1,..., v n } is not connected, which is contrary to the assumption that S = k < n 1. This shows that S = k = n 1. Consequently, γ sc (C n ) = n 1.

5 Secure connected domination in a graph 2069 Theorem 2.6 Let G be a connected graph of order n. Then γ sc (G) = 1 if and only if G = K n. Proof : Suppose γ sc (G) = 1 and let S = {v} be an SCDS of G. Suppose G K n, then there exists x, y V (K n ) such that d(x, y) = 2. Then (S \ {v}) {x} = {x}, which is not a dominating set of G, since xy / E(K n ). Therefore, G = K n. The converse is true by Theorem 2.5. The following Theorem shows that it is not possible to find a non-complete connected graph G such that γ(g) = γ sc (G). Theorem 2.7 Let G be a non-complete connected graph and let S be a secure connected dominating set in G. Then the set S \ {v} is a dominating set for every v S. In particular, 1 + γ(g) γ sc (G). Proof : Let v S and T = S \ {v}. We will show that T is a dominating set in G. Let w V (G) \ T. If w = v, then u S such that uw E(G) because S is connected. If w v, then w / S. Since S is a secure dominating set in G, there exists x S N G (w) such that x S defends w. If x v, then x T. If x = v, then (S \ {v}) {w} is a CDS. This implies that u T such that uw E(G). Therefore, in any case, T = S \ {v} is a DS. Thus, 1 + γ(g) S. If, in particular, S is a γ sc -set, then 1 + γ(g) γ sc (G). Theorem 2.8 Let G be a connected graph of order n 4. Then γ sc (G) = 2 if and only if there exists a non-complete graph H such that G = K 2 + H. Proof : Suppose G = K 2 + H, for some non-complete graph H. Then G is not complete and by Theorem 2.6, it follows that γ sc (G) 2. Let S = {x, y}, where x, y V (K 2 ). Then S is a connected dominating set in G. Let z V (G)\S. Then z V (H), xz E(G) and (S \ {x}) {z} = {y, z}a connected dominating set of G. Thus S is an SCDS in G. Therefore, γ sc (G) = 2. For the converse, suppose that γ sc (G) = 2. Let S = {x, y} be a γ sc -set of G. Let z V (G) \ S. Suppose xz / E(G). Since S is an SCDS of G, yz E(G) and S = {x, z} is a connected dominating set of G. This is not possible because xz / E(G). Therefore, xz E(G). Similarly, yz E(G). Let H = V (G) \ S and let K 2 = S. Then G = K 2 + H. Moreover, since G is not complete, H is non-complete. Corollary 2.9 Let G be a non-complete connected graph and n 2. Then γ sc (G + K n ) = 2. Corollary 2.10 Let G be a non-complete connected graph. Then γ sc (K 1 + G) = 2 if and only if one of the following is true: (i) γ(g) = 1

6 2070 Amerkhan G. Cabaro, Sergio S. Canoy, Jr. and Imelda S. Aniversario (ii) γ sc (G) = 2 Proof : Suppose γ sc (K 1 + G) = 2 and let V (K 1 ) = {a}. Let S = {x, b} be an SCDS in K 1 + G, where b V (G). Consider the following cases: Case 1. Suppose x = a. If {b} were not a dominating set of G, then y V (G) \ {b} such that by / E(G). Since S is an SCDS of K 1 + G, S \ {x} {y} = {y, b} is a connected dominating set of K 1 + G. This, however, s impossible to happen since by / E(G). Thus, {b} is a dominating set of G. This shows that γ(g) = 1. Case 2. Suppose x a. Then x G. Hence, S is an SCDS in G. Since K 1 + G is not complete, it follows that γ sc (G) = S = 2. For the converse, suppose first that γ(g) = 1, say {b} is a dominating set of G. Let H = V (G) \ {b}. Then H is non-complete and K 1 +G {a, b} +H. By Theorem 2.8, γ sc (K 1 + G) = 2. Suppose now that γ sc (G) = 2. Then by Theorem 2.8, G = K 2 + H, where H is a non-complete graph. Let H = a + H. Then H is non-complete and K 1 + G K 2 + H. Therefore, by Theorem 2.8, γ sc (K 1 + G) = 2. Theorem 2.11 Let G be a non-complete connected graph and K 1 = v. Then S V (K 1 + G) is a SCDS of K 1 + G iff one of the following holds: (i) S V (G) and S is a SCDS of G. (ii) v S and S \ {v} is a dominating set in G. Proof : Suppose S V (K 1 + G) is an SCDS in K 1 + G. If S V (G), then S is an SCDS of G. Suppose v S and let x V (G) \ (S \ {v}). Suppose xy E(K 1 + G) for all y V (G) \ (S \ {v}). Then (S \ {v}) {x} is not connected, contrary to our assumption that S is an SCDS in K 1 + G. Thus, S \ {v} is a dominating set in G. For the converse suppose first that S is an SCDS in G. Then S is an SCDS in K 1 + G. Next suppose that v S and S \ {v} is a dominating set in G. Then S is a connected dominating set in K 1 + G. Let z V (K 1 + G) \ S. Then z V (G) \ (S \ {v}). Since S \ {v} is a DS in G, there exists x S \ {v} such that xz E(G) and (S \ {x}) {z} is a CDS in K 1 + G. Thus, S is an SCDS in K 1 + G. Corollary 2.12 Let G be a non-complete connected graph. Then γ sc (K 1 + G) = min{γ sc (G), γ(g) + 1}.

7 Secure connected domination in a graph 2071 Proof : Let S 1 and S 2 be, respectively, γ sc -set and γ-set of G. By Theorem 2.11, S 1 and S 2 V (K 1 ) are SCDS in K 1 + G. Thus, γ sc (K 1 + G) min{ S 1, S 2 V (K 1 ) } = min{γ sc (G), γ(g) + 1}. Let r = min{γ sc (G), γ(g) + 1}. Let S be a γ sc -set of K 1 + G. By Theorem 2.11, γ sc (K 1 + G) = S r. This proves the desired equality. From this result, we can readily solve for the secure connected domination number of the graphs: F n and W n. Remark 2.13 Let F n, and W n be the fan and wheel of order n + 1, respectively. Then { 1, if n = 1 (i) γ sc (F n ) = n , if n 2 (ii) γ sc (W n ) = { 1, if n = 3 n 3 + 1, if n 4. Remark 2.14 Corollary 2.10 also follows from Corollary Let G and H be non-complete connected graphs. Then G + H is noncomplete. Thus γ sc (G + H) 2. Now, choose a, b V (G) and x, y V (H). Then S = {a, b, x, y} is a SCDS in G + H. It follows that γ sc (G + H) 4. This result is stated formally in the following remark. Remark 2.15 Let G and H be non-complete connected graphs. γ sc (G + H) 4. Then 2 Theorem 2.16 Let G and H be non-complete connected graphs. Then γ sc (G+ H) = 2 if and only if one of the following holds: (i) γ sc (G) = 2. (ii) γ sc (H) = 2. (iii) γ(g) = 1 and γ(h) = 1. Proof : Suppose γ sc (G + H) = 2. Then by Theorem 2.8, G + H = K 2 + H 1, where H 1 is a non-complete graph. Let V (K 2 ) = {x, y} = S, then S is a SCDS in G + H. Consider the following cases: Case 1: Suppose x, y V (G) Then G = K 2 + J, where J is a non-complete subgraph of G. Hence by Theorem 2.8, γ sc (G) = 2. Case 2: Suppose x, y V (H) As in case 1, γ sc (H) = 2.

8 2072 Amerkhan G. Cabaro, Sergio S. Canoy, Jr. and Imelda S. Aniversario Case 3: Suppose x V (G) and y V (H) Let z V (G) \ {x}. Since G + H = K 2 + H 1, xz E(G). Hence {x} is a dominating set in G. It follows that γ(g) = 1. Similarly, γ(h) = 1. For the converse, suppose that γ sc (G) = 2. Then by Theorem 2.8, G = K 2 + J, where J is a non-complete graph. Thus, G + H = (K 2 + J) + H = K 2 + H 1, where H 1 = J + H is non-complete. It follows from Theorem 2.8 that γ sc (G + H) = 2. Similarly, γ sc (G + H) = 2 if γ sc (H) = 2 holds. Suppose (iii) holds. Let {x} and {y} be a dominating set of G and H, respectively. Then G = x + J and H = y + K where J and K are non-complete graphs. Then, G+H = ( x +J ) + ( y +K ) = K 2 +H 1, where V (K 2 ) = {x, y} and H 1 = J + K is a non-complete graph. Hence by Theorem 2.8, γ sc (G + H) = 2. Theorem 2.17 Let G and H be non-complete connected graph such that γ sc (G+ H) 2. Then γ sc (G + H) = 3 if and only if one of the following holds: (i) γ sc (G) = 3 or γ sc (H) = 3 (ii) γ(g) = 2 or γ(h) = 2. Proof : Suppose that γ sc (G + H) = 3. Let S = {x, y, z} be a SCDS in G + H. Consider the following cases: Case 1: Suppose S V (G) Then S is an SCDS in G. It follows that γ sc (G) = 3 Case 2: If S H Then S is an SCDS in H and γ sc (H) = 3. Case 3: Suppose V (G) S and V (H) S Let w( V (G) )\ {x, y}. Suppose that w / N G (X). Then wz E(G + H). Hence, S \ {z} {w} = {x, y, w} is not a CDS, which is contrary to our assumption about S. Therefore, N G [X] = V (G), that is, X is a DS in G. Since G is non-complete, γ(g) = 2. For the converse, suppose (ii) holds, say γ(g) = 2. Let X = {a, b} be a DS in G. Pick c V (H) and set S = {a, b, c}. Let z V (G + H) \ S. Then z V (G) or z V (H). If z V (G), then z N G (X), say az E(G + H), and S = {b, z, c} is a CDS in G + H. If z V (H), then az E(G + H) and S = {b, z, c} is a CDS in G + H. Thus, S is an SCDS in G + H. Therefore γ sc (G + H) = 3. We obtain the same conclusion if we assume that γ(h) = 2. Suppose (i) holds, say γ sc (H) = 3. Let S = {a, b, c} be a SCDS in H. Then, clearly S is a SCDS in G + H. Therefore, γ sc (G + H) = 3. Theorem 2.18 Let G be a non-trivial connected graph and let H be any graph. Then C V (G H) is a secure connected dominating set in G H if and only if C = V (G) ( v V (G) S v ), where each S v is a dominating set in H v.

9 Secure connected domination in a graph 2073 Proof : Suppose that C V (G H) is a secure connected dominating set in G H and let v V (G). Choose w V (G) \ {v}. Since C is a DS, V (v + H v ) C and V (w + H w ) C. Since C is connected, v C. Let S v = V (H v ) C and x V (H v ) \ S v. Since C is an SCDS of G H, there exists y C such that xy E(G H) and (C \ {y}) {x} is a CDS. Note that since (C \ {v}) {x} is not connected in G H, it follows that y v. Thus, y S v, showing that S v is a DS in H v. Conversely, suppose that C = V (G) ( v V (G) S v ). Then, clearly, C is a CDS in G H. Let x V (G H)\C and let v V (G) such that x V (H v )\S v. Since S v is a DS in H v, there exists z S v such that xz E(G H). Also, (C \ {x}) {z} is a CDS in G H. Therefore, C is an SCDS in G H. Corollary 2.19 Let G be a non-trivial connected graph of order m and let H be any graph of order m. Then γ sc (G H) = m(1 + γ(h)). Proof : Let S be a γ-set of H. For each v V (H), choose S v V (H v ) such that S v = S. Then, by Theorem 2.18, C = V (G) ( Sv ) is an SCDS in G H. Hence, γ sc (G H) m + S v = m + mγ(h) = m(1 + γ(h)). On the other hand, if C is a γ sc -set in G H, then C = V (G) ( Sv), where each Sv is a DS in H v, by Theorem Thus, γ sc (G H) = C = m + Sv m + mγ(h) = m(1 + γ(h)). This proves the desired equality. Acknowledgements. The researcher would like to thank the Commission of Higher Education of the Republic of the Philippines for the partial financial support extended through its Faculty Development Program Phase II. References [1] A.P. Burger, M.A. Henning, and J. H. Van Vuuren, Vertex Cover and Secure Domination in Graph,Quaestiones Mathematicae, 31:2 (2008), [2] E. Castillano, R.A. Ugbinada and S. Canoy, Jr., Secure Domination in the Join of Graphs, Applied Mathematical Sciences, submitted. [3] G. Chartrand and L. Lesniak, Graphs and Digraphs, Third Edition, Chapman and Hall, [4] E.J. Cockayne, O. Favaron, and C.M. Mynhardt, Secure Domination, Weak Roman Domination and Forbidden Subgraphs, Bull. Inst. Combin. Appl., 39 (2003), [5] E.J. Cockayne, P.J.P. Grobler, W.R. Grundlingh, J. Munganga, and J.H. Van Vuuren, Protection of a Graph, Util. Math., 67 (2005),

10 2074 Amerkhan G. Cabaro, Sergio S. Canoy, Jr. and Imelda S. Aniversario [6] T.W. Haynes, S.T. Hedetniemi, and P.J. Slater(eds), Fundamentals of Domination in Graphs, Marcel Dekker, Inc. New York, [7] S.T. Hedetniemi, R.C. and Laskar, Connected Domination in Graphs, In B. Bollobás, editor, Graph Theory and Combinatorics, Academic Press, London, [8] E. Sampathkumar, and H.B. Walikar, The Connected Domination of a Graph, Math. Phys. Sci. 13 (1979), Received: July 1, 2014