Another Look at p-liar s Domination in Graphs

Transcription

1 International Journal of Mathematical Analysis Vol 10, 2016, no 5, HIKARI Ltd, wwwm-hikaricom Another Look at p-liar s Domination in Graphs Carlito B Balandra 1 Department of Arts and Sciences Southern Philippines Agri-Business Marine and Aquatic School of Technology Malita, Davao del Sur, 8200, Philippines Sergio R Canoy, Jr Department of Mathematics and Statistics College of Science and Mathematics MSU-Iligan Institute of Technology Tibanga, Iligan City, Philippines Copyright c 2015 Carlito B Balandra and Sergio R Canoy, Jr This is an 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 A set S V (G) is a p-liar s dominating set (plds) of graph G if N G (v) S 2 for every v V (G) \ S (that is, S is a 2- dominating set of G), and [N G (u) N G (v)] S 3 for any two distinct vertices u, v V (G) \ S The p-liar s domination number of G, denoted by γ plr (G), is the smallest cardinality of a p-liar s dominating set of G The 2-domination number γ ( G) of G is the smallest cardinality of a 2-dominating set of G It is shown that the difference γ plr (G) γ ( G) can be made arbitrarily large Further, the p-liar s dominating sets in the composition of graphs are characterized Keywords: p-liar s dominating set, graphs 1 Research is funded by the Commission on Higher Education (CHED), Philippines under Faculty Development Program Phase II

2 214 Carlito B Balandra and Sergio R Canoy, Jr 1 Introduction Let G = (V (G), E(G)) be a simple graph The neighborhood of v V (G) is the set N G (v) = {x V (G) : xv E(G)} Let S V (G) of a graph G A vertex w is called an internal private neighbor (abbreviated by ipn) of v S if w S and N(w) S = {v} A vertex w is an external private neighbor (abbreviated by epn) of v S if w V (G) \ S and N(w) S = {v} The set of all internal private neighbors (respectively, external private neighbors) of v is denoted by ipn(v; S)(respectively, epn(v; S)) A set S V (G) is a dominating set of G if for every v V (G) \ S, there exists u S such that uv E(G), that is, N G [S] = V (G) The domination number of G, denoted by γ(g), is the smallest cardinality of a dominating set of G Any dominating set in G of cardinality γ(g) is referred to as a γ-set of G A subset S of V (G) is called an almost dominating set of G if V (G) \ N[S] 1 The a-domination number of G, denoted by γ a (G), is the smallest cardinality of an almost dominating set of G An almost dominating set of G with cardinality γ a (G) is referred to as a γ a -set of G A set S V (G) is a total dominating set of G if for every vertex v V (G) there exists a vertex u S such that u and v are adjacent The total domination number of G, denoted by γ t (G), is the smallest cardinality of a total dominating set of G A total dominating set of G of minimum cardinality is called γ t -set of G A set S V (G) is a liar s dominating set (lds) of G if (i) N G [v] S 2 for every v V (G) (that is, S is a double dominating set of G), and (ii) (N G [u] N G [v]) S 3 for any two distinct vertices u, v V (G) The liar s domination number of G, denoted by γ LR (G), is the smallest cardinality of a liar s dominating set of G Any subset of V (G) with cardinality γ LR (G) is called a γ LR -set of G A set S V (G) is a p-liar s dominating set (plds) of G if (i) N G (v) S 2 for every v V (G) \ S (that is, S is a 2- dominating set of G), and (ii) [N G (u) N G (v)] S 3 for any two distinct vertices u, v V (G) \ S The p-liar s domination number of G, denoted by γ plr (G), is the smallest cardinality of a p-liar s dominating set of G Any subset of V (G) with cardinality γ plr (G) is called a γ plr -set of G Domination in graph as well some of its variations can be found in [2] The concept of liar s domination of a graph G was introduced by Slater and Roden in [4] and a similar study is introduced and investigated by Nikodem in [3] The p-liar s domination number of graph lies between the 2-domination number and the liar s domination number of the graph The concept of p-liar s domination in a graph introduced in [1] 2 Results The first result is taken from [1]

3 Another Look at p-liar s domination in graphs 215 Theorem 21 [1] Let n be a positive integer Then 1, if n = 1 (i) γ plr (K n ) = 2, if n = 2 or n = 3 3, if n 4; (ii) γ plr (P n ) = n+1 2 for n 1; and 3, if n = 4 n+1 (iii) γ plr (C n ) =, if n is odd 2, if n is even (n 4) n 2 Theorem 22 Let m and n be positive integers such that 2 m n Then 3, if m = 2 and n = 2 6, if n, m 6 γ plr (K m,n ) = n, if m = 2 and n 3 m, otherwise Proof : By Theorem 21 (iii), γ plr (K 2,2 ) = γ plr (C 4 ) = 3 Supppose that m, n 6 Pick a, b, c V (K m ) and x, y, z V (K n ) Then S = {a, b, c, x, y, z} is a p-liar s dominating set of K m,n Hence, γ plr (K m,n ) 6 Let S be a γ plr -set of K m,n and suppose that S 5 Let S 1 = S V (K m ) and S 2 = S V (K n ) Since S is a dominating set of K m,n, S 1 and S 2 This implies that there exist distinct vertices x and y in V (K m ) \ S 1 or V (K n ) \ S 2 which do not satisfy the second condition of Definition?? Hence, S is not a p-liar s dominating set, contrary to our assumption Thus, γ plr (K m,n ) = S 6 Therefore, γ plr (K m,n ) = 6 Next, suppose that m = 2 and n 3 Then D is a p-liar s dominating set of K m,n if and only if it is one of the following sets: V (K m,n ), V (K m ) [V (K n ) \ {v}] for some v V (K n ), {a} V (K n ) for some a V (K m ), and V (K n ) Hence, γ plr (K 2,n ) = n Finally, suppose that 3 m 5 Since V (K m ) is a plds of K m,n, it follows that γ plr (K m,n ) m Now let S be γ plr -set of K m,n and suppose that S m 1 Since S is a dominating set of K m,n, S 1 = S V (K m ) and S 2 = S V (K n ) From the additional assumption that S m 1, it follows that S V (K m ) m 2 and S V (K n ) m 2 Since m 2 3, it follows that there exist distinct vertices x, y in V (K m ) \ S 1 or V (K n ) \ S 2 that do not satisfy the second condition of definition of p-liar s dominating set This contradicts the fact that S is a p-liar s dominating set of K m,n Therefore S m, showing that γ plr (K m,n ) = m Theorem 23 Let a and b be positive integers such that 3 a b Then there exists a connected graph G such that γ 2 (G) = a and γ plr (G) = b

4 216 Carlito B Balandra and Sergio R Canoy, Jr Proof : Consider the following cases: Case 1 a = b = 3 Let G = K 3,3 in Figure 1 Then, by Theorem 22, γ plr (G) = 3 Also, γ 2 (G) = 3 G : u 1 u 2 u 3 v 1 v 2 v 3 Figure 1: A graph G with γ plr (G) = γ 2 (G) = 3 Case 2 3 = a < b Consider the graph G in Figure 2 Clearly, S = {y i : i = 1, 2,, b} is a plds of G Hence, γ plr (G) b Suppose that S is a γ plr -set of G Let A = {y 1, y 2,, y b 2 } and B = {y b 1, y b } Since S is a plds of G, S A and S B Moreover, S A b 3 If S A = b 3, then x 1, x 2 S since S is a 2-dominating set of G Thus, S b 3+2 = b 1 If S B = 1, then x 3 S Hence, S = (b 1) + 2 = b + 1 If S B = 2, then S = b + 1 In both cases, we obtain a contradiction Thus, S A = b 2 Hence, γ plr (G) b = b Therefore, γ 2 (G) = 3 = a and γ plr (G) = b y 1 y 2 x 1 G : x 2 y b 2 x 3 y b 1 y b Figure 2: A graph G with γ 2 (G) = 3 and γ plr (G) = b Case 3 3 < a = b Consider G = K 1,a in Figure 3 Then S = {x 1, x 2,, x a } is both a γ 2 -set and a γ plr -set of G Thus, γ 2 (G) = γ plr (G) = a Case 4 3 < a < b Consider the graph G in Figure 4 Let A 1 = {x 3, x 4, x a }, B 1 = {x 1, x 2 }, D 1 = {y 1, y 2,, y b a+2 }, and S 1 = {x 1, x 2,, x a } Clearly, S 1 is a 2-dominating

5 Another Look at p-liar s domination in graphs 217 x 1 G : v x 2 x 3 x a Figure 3: A graph G with γ 2 (G) = γ LR (G) = a set of G Thus γ 2 (G) a Let S be a γ 2 -set of G Since S is a 2-dominating set of G, it follows that A 1 S This implies that S a 2 Suppose that S B 1 = Then D 1 S since S is a 2-dominating set of G Thus, S (a 2) + (b a + 2) = b > a, contrary to an earlier inequality Hence, S B 1 1 If S B 1 = 1, then, again, D 1 S and we get a contradiction Thus, S B 1 = 2, that is, B 1 S This implies that γ 2 (G) = S a Therefore, γ 2 (G) = a Next, consider S 2 = A 1 D 1 Clearly, S 2 is a plds of G This implies that γ plr (G) (a 2) + (b a + 2) = b Let S be a γ plr -set of G Since S is a 2-dominating set of G, A 1 S Since S is a plds, S D 1 Moreover, S D 1 b a + 1 Now, if S D 1 = b a + 1, then x 1, x 2 S Thus, γ plr (G) = S (a 2)+(b a+1)+2 = b+1, contrary to an earlier inequality Therefore, S D 1 = b a + 2 and γ plr (G) = (a 2) + (b a + 2) = b G : v x 1 x 2 x 3 x a 1 x a y 1 y 2 y 3 y b a y b a+1 y b a+2 Figure 4: A graph G with γ 2 (G) = a and γ LR (G) = b This proves the assertion Corollary 24 For every n 4, there exists a connected graph G such that γ plr (G) γ 2 (G) = n, that is, the difference γ plr (G) γ 2 (G) can be made arbitrarily large

6 218 Carlito B Balandra and Sergio R Canoy, Jr 3 p-liar s Domination in the Composition of Graphs The composition of two graphs G and H is the graph G[H] with vertex-set V (G[H]) = V (G) V (H) and edge-set E(G[H]) satisfying the following conditions: (u 1, u 2 )(v 1, v 2 ) E(G[H]) if and only if either u 1 v 1 E(G) or u 1 = v 1 and u 2 v 2 E(H) Observe that a non-empty subset C of V (G[H]) = V (G) V (H) can be written as C = x S ({x} T x ),where S V (G) and T x V (H) for every x S Henceforth, we shall use this form to denote any non-empty subset C of V (G[H]) Theorem 31 Let G and H be non-trivial connected graphs of orders n 2 and m 3, respectively A non-empty subset C = x S ({x} T x ) of V (G[H]), where S V (G) and T x V (H) for each x S, is a p-liar s dominating set of G[H] if and only if S is a dominating set of G and satisfies each of the following: (i) T x is a p-liar s dominating set of H for each x S \ N G (S); (ii) For each x V (G) \ S, (a) T y 3 whenever N G (x) S = {y}; and (b) T y 2 or T z 2, whenever N G (x) S = {y, z} (iii) For each x S N G (S) such that N G (x) S = {y}, one of the following holds: (a) T y 3; (b) T x is an a-dominating set and T y = 2; (c) T y = 1 and T x is a dominating set such that epn(a; T x ) 1 for all a T x ; (iv) For each x S N G (S) such that N G (x) S = {y, z} and T y = T z = 1, T x is an almost dominating set of H Proof : Suppose that C is a plds of G[H] Let x V (G)\S and pick a V (H) Then (x, a) / C Since C is a dominating set of G[H], there exists (y, b) C such that (x, a)(y, b) E(G[H]) Thus, y S and xy E(G) Hence, S is a dominating set of G Let x S \ N G (S) and let q V (H) \ T x Then (x, q) / C and there exists distinct vertices (z, c), (w, p) C N G[H] ((x, q)) since C is a 2-dominating set of G[H] Since x / N G (S), it follows that z = w = x Thus, c, p T x (c p),

7 Another Look at p-liar s domination in graphs 219 and qc, qp E(H) This shows that T x is a 2-dominating set of H Next, let q, r V (H) \ T x such that r q Then (x, q), (x, r) / C and (x, q) (x, r) Since C is a p-liar s dominating set of G[H], [N G[H] ((x, q)) N G[H] ((x, r))] C = [N H (q) N H (r)] T x 3 Therefore, T x is a p-liar s dominating set of H This shows that (i) holds Let x V (G) \ S with N G (x) S = 1, say N G (x) S = {y} Let a, b V (H), where a b Then (x, a), (x, b) / C Since C is a p-liar s dominating set of G[H], there exist at least three distinct vertices (v, b), (w, c), (z, d) C (N G[H] ((x, a)) N G[H] ((x, b))) This implies that v = w = z = y, and b, c, d T y Hence, T y 3, showing that (a) holds Suppose further that N G (x) S = {y, z} Let m, n V (H), where m n Then (x, m), (x, n) / C Since C is a p-liar s dominating set of G[H], there exist at least three vertices (e, m), (f, l), (g, z) C (N G[H] ((x, m)) N G[H] ((x, n))) This implies that e = f = y, and m, l T y Thus, T y 2 Similarly, T z 2 if e = f = z, and m, l T z Hence, (b) holds Therefore, (ii) holds Let x S N G (S) with N G (x) S = 1, say N G (x) S = {y} Suppose T y 2 Suppose that T y = 2 and suppose further that T x is not an a-dominating set of H Then there exists two distinct vertices a, b V (H)\T x such that a, b / N H (T x ) Thus, (N G[H] ((x, a)) N G[H] ((x, b))) C = T y = 2, contrary to our assumption that C is a plds of G[H] Hence, T x is an almost dominating set of H, showing that (b) holds Next, suppose that T y = 1 Since C is a 2-dominating set of G[H], T x is a dominating set of H Let a T x and suppose that epn(a; T x 2, say l, m epn(a; T x ) (l m) Then (x, l), (x, m) / C and (N G[H] ((x, l)) N G[H] ((x, m))) C = {(x, a)} + T y = 2, a contrary to our assumption that C is a plds of G[H] Thus, epn(a; T x ) 1, showing that (c) holds Therefore, (iii) holds Let x S N G (S) with N G (x) S = {y, z} and T y = T z = 1 Let T y = {t} and T z = {s} Suppose that V (H) \ N[T x ] 2 Then there exist n, p V (H) \ T x such that N G[H] ((x, n)) C = {(y, t), (z, s)} = N G[H] ((x, p)) C This means that C is not a plds of G[H], contrary to our assumption Thus, T x is an almost dominating set of H This shows that (iv) holds For the converse, suppose that S is a dominating set of G and satisfies each of the given properties Let (x, a) V (G[H]) \ C Then consider the following cases: Case 1 x / S

8 220 Carlito B Balandra and Sergio R Canoy, Jr Since S is a dominating set of G, it follows that N G (x) S 1 If N G (x) S = 1, N G (x) S = {y}, then T y 3 by (ii) Hence, N G[H] ((x, a)) C 3 If N G (x) S 2, then N G[H] ((x, a)) C 2 Case 2 x S Then a / T x Suppose that x S \ N G (S) Then T x is a plds of H by (i) Hence, N G[H] ((x, a)) C = N H (a) T x 2 Next, suppose that x S N G (S) Suppose first that N G (x) S = 1 Then by (a), (b), and (c) of (iii), we have N G[H] ((x, a)) C 2 Suppose that N G (x) S = 2 Then N G[H] ((x, a)) C 2 by (iv) Therefore, C is a 2-dominating set of G[H] Let (x, a), (y, b) V (G[H])\C, where (x, a) (y, b) Consider the following cases: Case 1 x = y Let x S \ N(S) Then T x is p-liar s dominating set of H by (i) Since a, b / T x, it follows that [N G[H] ((x, a)) N G[H] ((y, b))] C = (N H (a) N H (b)) T x 3 Suppose that x S N G (S) Then [N G[H] ((x, a)) N G[H] ((y, b))] C 3 by (iii) and (iv) If x / S, then [N G[H] ((x, a)) N G[H] ((y, b))] C 3 by (ii) Case 2 x y Suppose that x, y S If xy / E(G), then [N G[H] ((x, a)) N G[H] ((y, b))] C 3 by (i), (iii) and (iv) Suppose xy E(G) If N G (x) S = 1, then N G (x) S = {y} Thus, [N G[H] ((x, a)) N G[H] ((y, b))] C 3 by (iv) Suppose x / S or y / S Then by (ii), [N G[H] ((x, a)) N G[H] ((y, b))] C 3 Accordingly, C is a p-liar s dominating set of G[H] Corollary 32 Let G and H be connected graphs of orders n 2 and m 4, respectively Then γ plr (G[H]) min{γ(g)γ plr (H), 3γ t (G)} Proof : Let S 1 and S 2 be γ-set and γ t -set of G, respectively, and let D be a γ plr -set of H By Theorem 31, C 1 = x S1 [{x} T x ] and C 2 = y S2 [{y} E y ], where T x = D for each x S 1 and E y = {a, b, c} V (H) for each y S 2, are plds of G[H] Thus, γ plr (G[H]) min{ C 1, C 2 } = min{γ(g)γ plr (H), 3γ t (G)} Remark 33 The bounds in Corollary 32 are sharp Also, the strict inequality can be obtained To see this, consider P 2 [P 4 ], P 4 [P 5 ], and P 5 [P 6 ] in Figure 5 Now, γ plr (P 4 [P 5 ]) = 6 = 3γ t (P 4 ) = γ(p 4 )γ plr (P 5 ), and γ plr (P 5 [P 6 ]) = 7 < 8 = min{γ(p 5 )γ plr (P 6 ), 3γ t (P 5 )} By Theorem 31, γ plr (P 4 ) = γ plr (P 5 ) = 3 Hence, γ plr (P 2 [P 4 ]) = 3 = γ(p 2 )γ plr (P 4 ) < 6 = 3γ t (P 2 )

9 Another Look at p-liar s domination in graphs 221 P 2 [P 4 ] P 5 [P 6 ] P 4 [P 5 ] References Figure 5: The graphs P 2 [P 4 ], P 5 [P 6 ], and P 4 [P 5 ] [1] CB Balandra, SR Canoy Jr and IS Aniversario, p-liar s Dominating Sets in Graphs, Applied Mathematical Sciences, 9 (2015), no 107, [2] TW Haynes, ST Hedetmiemi and PJ Slater, Fundamentals of Domination in Graphs, Vol 208 of Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, USA, 1998 [3] M Nikodem, False Alarm in Fault-Tolerant Dominating Sets in Graphs, Opuscula Mathematica, 32 (2012), [4] PJ Slater and ML Roden, Liar s Domination in Graphs, Discrete Mathematics, 309 (2009), [5] C Sterling, Liar s Domination in Grid Graphs, Thesis, Master of Science in Mathematical Sciences, East Tennesse State University, 2012 Received: November 29, 2015; Published: February 5, 2016