SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH

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1 BULLETIN OF THE INTERNATIONAL MATHEMATICAL VIRTUAL INSTITUTE ISSN (p) 0-87, ISSN (o) /JOURNALS / BULLETIN Vol. 7(017), 5-6 Former BULLETIN OF THE SOCIETY OF MATHEMATICIANS BANJA LUKA ISSN (o), ISSN X (p) SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH B. Chaluvaraju and V. Chaitra Abstract. In this paper, we newly constructed the sign dominating outer (inner) switched graph µ o d (G) (µi d (G)) of a graph G = (V, E) and establish their properties. Also we determine number of edges and its relation between µ o d (G) and µi d (G) in some special classes of graphs are explored. 1. INTRODUCTION All the graphs considered in this paper are finite, nontrivial, simple and undirected. Let G = (V, E) be a simple graph with vertex set V (G) = V of order V = n, edge set E(G) = E of size E = m and let v be a vertex of V. The open neighborhood of v is N(v) = {u V/uv E(G)} and closed neighborhood of v is N[v] = {v} N(v). The graph G c is called complement of a graph G, if G and G c have the same vertex set and two vertices are adjacent in G if and only if they are not adjacent in G c. A subset S of V is called vertex independent set if no two vertices in S are adjacent in G. A clique in a graph is an induced complete subgraph. The maximum order of a clique in the graph G is called the clique number of G, denoted by ω(g). A collection of independent edges of a graph G is called a matching of G. If there is a matching consists of all vertices of G it is called a perfect matching. For standard terminology and notation in graph theory, we refer [6]. A sign dominating function of a graph G is a function f : V { 1, 1} such that f(n[v]) 1 for all v V. The sign domination number of a graph G is γ s (G) = min{w(f) : f is sign dominating function}. The concept of sign domination was initiated by Dunbar et al. [5]. For complete review on theory of domination and its related parameters, we refer [7], [8] and [1]. 010 Mathematics Subject Classification. 05C69, 05C70. Key words and phrases. Domination, sign domination, k-complement, k(i)-complement. Communicated by... 5

2 5 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH In a graph G, let P = {V 1, V,..., V k } be a partition of V of order k 1. The k-complement of a graph G denoted as G P k is defined as follows: for all V i and V j in P, i j remove the edges between V i and V j and add the edges which are not in G. The k(i)-complement of a graph G denoted as G P k(i) is defined as: for each set V r in the partition P, remove the edges of G inside V r and add the missing edges between them. The concept of generalized complement of a graph was studied by Sampathkumar et al. [9] and [10]. Analogously, the concept of -complement and (i)-complement of a graph is also known as switched graph, where switching of G assigns +1 or 1 to each vertex of a graph G. This type of switched graph was introduced by Van-Lint et al. [1]. Further, let V 1 and V 1 be set of vertices assigned 1 and 1 in G respectively. We denote ρ + (G) = ρ + = V 1 and ρ (G) = ρ = V 1. For more details on Generalized complements (switched invariants) and its related concept, we refer [1], [], [], [5] and [11].. Sign dominating outer switched graph The sign dominating outer switched graph of G denoted as µ o d (G) is defined as: for V 1 and V 1 of G remove the edges between V 1 and V 1 and add the edges which are not there in G. Theorem.1. Let G be a nontrivial graph. Then (i) µ o d (G) is a totally disconnected graph if and only if G is totally disconnected graph. (ii) V 1 is independent set if and only if µ o d (G) is totally disconnected graph. Proof. (i) Let G be totally disconnected graph. Every vertex of G belongs to V 1. In construction of µ o d (G), no new edge is added or any edge is deleted. Hence G = µ o d (G). Conversely, if µo d (G) is totally disconnected graph, then in G no two vertices of V 1 (or V 1 ) are connected. Also in G, vertices of V 1 and V 1 cannot be connected as a vertex v of V 1 which is adjacent to a vertex of V 1 should be adjacent to at least one vertex of V 1 such that f(n[v]) 1. Hence G is totally disconnected graph. (ii) If µ o d (G) is totally disconnected graph, then it is obvious that V 1 is independent set as every vertex of µ o d (G) is assigned 1. Conversely, suppose V 1 is independent set. To prove µ o d (G) is totally disconnected, we shall prove G is totally disconnected. Suppose G is not totally disconnected and a vertex v V 1 be adjacent to a vertex of V 1. But this vertex v should be adjacent to another vertex in V 1 as f(n[v]) 1 which is a contradiction to V 1 being independent. Hence G is totally disconnected. Theorem.. Let G be a nontrivial graph. Then, there is no perfect matching between vertices of V 1 and V 1. Proof. In a graph G, a vertex assigned 1 is adjacent to at least two vertices assigned 1, there cannot be a perfect matching between V 1 and V 1. Thus the required result follows.

3 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH 55 Theorem.. For any nontrivial graph G, (µ o d(g)) c = µ o d (G c ). Proof. Let u and v be two non adjacent vertices of G. Then they are adjacent in G c. We prove the result in following cases: Case 1. If u and v belongs to same set V 1 or V 1, then they are non adjacent in µ o d (G), implies they are adjacent in (µo d (G))c. Also they are adjacent in µ o d (Gc ). Case. If u and v belongs to different sets, then they are adjacent in µ o d (G), implies they non adjacent in (µ o d (G))c. Also they are non adjacent in µ o d (Gc ). From above two cases, the required result follows. Theorem.. Let G = K p,q be a complete bipartite graph with bipartition P 1 q and P such that P 1 = p and P = q with p q. If = r, then m(µ o d(g)) = (q r)(p + r). Proof. Let G = K p,q. Since p q, degree of every vertex of P 1 is greater than or equal to degree of every vertex of P. Let every vertex of P 1 be assigned 1. Since a vertex assigned 1 should be adjacent to at least two vertices assigned q 1, number of vertices assigned 1 in P should be = r. p vertices of P 1 and (q r) vertices of P which are assigned 1 forms an induced bipartite graph. Now in µ o d (G), (q r) vertices assigned 1 and r vertices assigned 1 are adjacent. These (q r) vertices are adjacent to p vertices in µ o d (G). Hence µo d (G) is complete bipartite graph K r1,r, where r 1 = q r and r = r + p. To prove our next result we make use of the following result due to Bohdan Zelinka []. Theorem.5. Let G = K p,q be a complete bipartite graph with bipartition P 1 and P such that P 1 = p and P = q, with p q. Then (i) for p = 1, γ s (G) = q + 1. { p if q is even, (ii) for p, γ s (G) = p + 1 if q is odd. if both p and q are even, (iii) for p, γ s (G) = 6 if both p and q are odd, 5 if one out of p or q is even. q Theorem.6. Let G = K p,q be a complete graph with p q. If = r, r 1 = q r and r = p + r, then (i) for r = 1, γ s (µ o d (G)) = r

4 56 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH { (ii) for r, γ s (µ o d (G)) = r if r 1 is even, r + 1 if r 1 is odd. if both r 1 and are even, (iii) for r, γ s (µ o d (G)) = 6 if both r 1 and r are odd, 5 if one out of r 1 or r is even. Proof. From Theorem., if G is complete bipartite graph, then µ o d (G) is also complete bipartite graph isomorphic to K r1,r, where r 1 = q r and r = p + r. From Theorem.5, the desired results follows. Theorem.7. Let G be a nontrivial graph. If G = K n with n vertices, then (i) µ o d (G) K n. n (ii) ρ (G) = n 1 if n is even, if n is odd. (iii) µ o d (G) = G ρ + + G ρ, where G ρ + and G ρ are clique graphs of G with m(µ o d (G)) = ρ + (G) = n + and ρ (G) = n if n is even, ρ +(G) = n + 1 and ρ (G) = n 1 if n is odd. n n + (iv) m(µ o d (G)) = if n is even, n n + 1 if n is odd. if both ρ + (G) and ρ (G) are odd, (v) γ s (µ o d (G)) = if both ρ + (G) and ρ (G) are even, if one is odd and other is even. Proof. (i) If possible let µ o d (G) = K n with n vertices, then we consider following two cases: Case 1. Let v be a vertex of V 1 (G) and w be a vertex of V 1 (G). In µ o d (G) adjacency of vertices within V 1 and within V 1 are retained as it is in G. Remove edge connecting vertices v and w and connect v to vertices other than w in V 1. Hence in µ o d (G), v is not connected to w which is a contradiction to µo d (G) being complete graph. Case. Suppose V = V 1 or V 1. As V V 1 implies V = V 1. Hence G = µ o d (G). But G is not a complete graph with V = V 1 for n. Hence µ o d (G) K n for n. (ii) Let v be a vertex of a graph G. Then consider the following two cases:

5 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH 57 Case. If n is even, then v is adjacent to odd number of vertices. Out of which n vertices are assigned 1 and n vertices are assigned 1. The remaining (n 1) th vertex cannot be assigned 1 as it makes the weight of every vertex either 0 or depending on v being assigned 1 or 1. So the (n 1) th vertex is assigned 1 and v is also assigned 1. Hence ρ (G) = n. Case. If n is odd, then v is adjacent to even number of vertices. Out of which n 1 vertices are assigned 1, remaining n 1 vertices are assigned 1 and v is assigned 1 so as f(n[v]) 1 for all v V. Hence ρ (G) = n 1. (iii) Any vertex v in a graph G is adjacent to n 1 vertices, in µ o d (G) we remove edges between vertices of V 1 and V 1 and no new edge is added. Hence µ o d (G) = G ρ + +G ρ, where G ρ + is graph induced by vertices of V 1 and G ρ is graph induced by vertices of V 1 in G. Also in µ o d (G), each G ρ + and G ρ is complete. From (ii), if n is even, then ρ + (G) = n + then ρ + (G) = n + 1 and ρ (G) = n 1. (iv) when n is even: when n is odd: and ρ (G) = n. And if n is odd, ρ + (G) = n + and ρ (G) = n. m(µ o d(g)) = 1 [ ( ) n + n n ( n m(µ o d(g)) = 1 ( n n + ). ρ + (G) = n + 1 and ρ (G) = n 1. m(µ o d(g)) = 1 [ ( ) n + 1 n n 1 ( n 1 m(µ o d(g)) = 1 ( n n + 1 ). )] 1. )] 1. (v) Since γ s (K n ) = { 1 if n is odd, if n is even. Case 5. If both ρ + (G) and ρ (G) are odd, then γ s (µ o d(g)) = γ s (G ρ +) + γ s (G ρ ) =. Case 6. If both ρ + (G) and ρ (G) are even, then γ s (µ o d(g)) = γ s (G ρ +) + γ s (G ρ ) =.

6 58 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH Case 7. If one of ρ + (G) or ρ (G) is odd and the other is even, then γ s (µ o d(g)) = γ s (G ρ +) + γ s (G ρ ) =. From all the above cases, the required results follows. Theorem.8. For a cycle C n with n vertices, n where = r. m(µ o d(g)) = r + r(n ) + n, Proof. Let G = C n with n vertices. In a cycle C n, number of vertices n assigned 1 are = r and number of vertices assigned 1 are n r. Construction of µ o d (G) is as follows: In G, since degree of vertex is, no two vertices assigned 1 are adjacent as f(n[v]) 1 for every vertex v V. Step 1. A vertex assigned 1 is adjacent to exactly two vertices assigned 1. In µ o d (G), remove two edges for one vertex assigned 1. Since there are r such vertices, number of edges removed are r and number of edges remaining in µ o d (G) are n r. Step. In µ o d (G) a vertex assigned 1 is adjacent to n r vertices so we connect these by n r edges. Repeat this process for all r vertices assigned 1. Therefore number of edges connecting vertices assigned 1 and vertices assigned 1 in µ o d (G) are r(n r ) edges. From above two steps, m(µ o d (G)) = n r + r(n r ) = r + r(n ) + n follows. Theorem.9. For a path P n with n vertices, m(µ o d(g)) = r + n(r + 1) r 1, n where = r. Proof. Let G = P n with n vertices, then end vertices and support vertices of a graph G cannot be assigned 1 as f(n[v]) 1. Hence such vertices belongs to V 1. After this assignment number of vertices left for assignment are n. In a path, since degree of any vertex other than end vertex is, a vertex assigned 1 should have exactly two neighbours assigned 1. So for a minimum of vertices, only one vertex as 1 provided f(n[v]) 1 for every v V (G). Hence ρ (G) = n = r(say) and ρ + (G) = n r. Construction of µ o d (G) is as follows: Step 1. Remove the edges between vertices of V 1 and V 1 : A vertex assigned 1 is adjacent to exactly two vertices assigned 1, so remove two edges connecting them. This process is repeated for r vertices of V 1. Then total number of edges deleted are r and edges left in µ o d (G) are n 1 r. Step. Add edges between vertices of V 1 and V 1 which are non adjacent in G:

7 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH 59 A vertex of V 1 is adjacent to exactly two vertices of V 1 and no vertex assigned 1. Hence in µ o d (G) a vertex of V 1 is adjacent to n r vertices of V 1. This process is repeated for r vertices of V 1. Hence total edges added are r(n r ). From above two steps, m(µ o d (G)) = r + n(r + 1) r 1 follows. Theorem.10. Let G be a nontrivial graph with m(µ o d (G)) = m(g). Then one of the following condition holds: (i) G = µ o d (G). (ii) ρ (G) = 0. (iii) For every v V 1, number of vertices adjacent to v is same as number of vertices of V 1 non adjacent to v. Proof. For a graph G, (i) If G = µ o d (G), then m(g) = m(µo d (G)). (ii) If ρ (G) = 0, then every vertex of G is assigned 1. Hence in µ o d (G) no new edges are added or deleted. (iii) If any two vertices of V 1 (or V 1 ) are adjacent in G, then they are adjacent in µ o d (G). Now if a vertex v of V 1 is adjacent to k vertices of V 1 in G, then in µ o d (G), k edges are removed and if v is non adjacent to k vertices of V 1, then k edges are added in µ o d (G). This holds for every vertex of V 1. Hence total number of edges added and deleted are same in µ o d (G). Therefore m(µo d (G)) = m(g). Theorem.11. For any nontrivial graph G, γ s (µ o d(g)) n. Further equality is obtained if every vertex of G is an end vertex or a support vertex. Proof. If G is a graph with n vertices, then µ o d (G) is also a graph with n vertices for which γ s (µ o d (G)) n is obvious. If every vertex of G is either an end vertex or support vertex, then these vertices belong to V 1. The equality follows.. Sign dominating inner switched graph The Sign dominating inner switched graph of G denoted as µ i d (G) is defined as: remove the edges of G inside V 1, V 1 and add the missing edges joining vertices inside V 1 and V 1. Theorem.1. Let G be a nontrivial graph. Then (i) µ i d (G) = K n if G = K c n. (ii) µ i d (G) Kc n. Proof. (i) Let G be totally disconnected graph, then every vertex of G belongs to V 1. In µ i d (G), every vertex of G is connected to remaining n 1 vertices of G. Hence µ i d (G) is complete graph. (ii) On the contrary, if µ i d (G) = Kn, c then following cases arise Case 1. In G, there are no edges between vertices of V 1 and between vertices of V 1.

8 60 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH Case. In G, V 1 is complete. Case. In G, V 1 is complete. Since a vertex of V 1 should be adjacent to atleast two vertices of V 1. Hence, Case 1 and Case are not possible. If V 1 is complete, then γ s (G) is not minimum, which is a contradiction of our assumption. Thus the results follows. Theorem.. For any nontrivial graph G, (i) µ i d (G))c = µ i d (Gc ). (ii) (µ o d (G))c = µ i d (G). (iii) m(µ o d (G)) + m(µi d (G)) = ( n ). Proof. (i) Let u and v be two non adjacent vertices of a graph G. Then they are adjacent in G c. We prove the result in following cases: Case 1. If vertices u and v belongs to same set V 1 or V 1, then they are adjacent in µ i d (G), implying that they are non adjacent in (µi d (G))c. Also they are non adjacent in µ i d (Gc ). Case. If u and v belongs to different sets, then they are non adjacent in µ i d (G), implying they are adjacent in (µi d (G))c. Also they are adjacent in µ i d (Gc ). From above two cases (i) follows. (ii) Let u and v be two non adjacent vertices in µ o d (G). u and v are adjacent in (µ o d (G))c. If both u and v belongs to V 1 or V 1, then they are non adjacent in G, implies they are adjacent in µ i d (G). If u and v belongs to different sets, then they are adjacent in G implies they are adjacent in µ i d (G). Thus (ii) follows. (iii) From (ii), as graph µ i d (G) is complement of µo d (G), sum of their edges should be equal to nc. Theorem.. Let G = K p,q be a complete bipartite graph with bipartition P 1 q and P such that P 1 = p and P = q with p q. If r =, then m(µ i d(g)) = 1 [p(p 1) + q(q 1) + r(p q + r)]. Proof. From Theorems. and., m(µ i (p + q)(p + q 1) d(g)) = m(µ o d(g)). m(µ i (p + q)(p + q 1) d(g)) = (p + r)(q r). m(µ i p(p 1) + q(q 1) + r(p q + r) d(g)) =.

9 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH 61 Theorem.. For any graph G = K n with n vertices, n if n is even, m(µ i d(g)) = n 1 if n is odd. Proof. From Theorem., m(µ i n(n 1) d (G)) = m(µ o d (G)). From Theorem.7, for n being even m(µ i d(g)) = n(n 1) n n + = n, and for n being odd m(µ i d(g)) = n(n 1) n n + 1 Theorem.5. For a cycle C n with n vertices, = n 1. m(µ i d(g)) = 1 n where r =. [ n n + r r(n ) ], Proof. From Theorem. and.8, m(µ i n(n 1) d(g)) = m(µ o d(g)). m(µ i n(n 1) d(g)) = [r + r(n ) + n]. m(µ i d(g)) = 1 [ n n + r r(n ) ]. Theorem.6. For a path P n with n vertices, m(µ i d(g)) = 1 [ n + r n nr + (r + 1) ], n where r =. Proof. From Theorem. and.9, m(µ i n(n 1) d(g)) = m(µ o d). m(µ i n(n 1) d(g)) = [r + n(r + 1) r 1]. m(µ i d(g)) = 1 [ n + r n nr + (r + 1) ].

10 6 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH Theorem.7. Let G be a nontrivial graph. Then m(µ o d (G)) = m(µi d (G)) if and only if (n 1) 1 is a multiple of 16. Proof. If m(µ o d (G)) = m(µi d (G)) = k, then from Theorem., n(n 1) k = n n k = 0 n = 1 ± k On simplifying, n is a positive integer for (n 1) 1 a multiple of 16. Conversely, if (n 1) 1 is a multiple of 16k, then n = k + 1. For n =, 5,..., we can generate the graph with m(µ o d (G)) = m(µi d (G)). Open problem: Characterize the graphs for which m(g) = m(µ i d(g)). Acknowledgement: The authors wish to thank Professor E. Sampathkumar for his valuable suggestions. References [1] B. D. Acharya. On characterizing graphs switching equivalent to acyclic graphs. Indian J. pure appl. Math., 1(10)(1981), [] B. Zelinka. Signed and Minus domination in bipartite graphs. Czechoslovak Math. J., 56()(006), [] B. Chaluvaraju. -global and (i)-global dominating sets in graphs. Ultra scientist of Phy. Sc. - Section A: Mathematics, 0()(008), [] B. Chaluvaraju, C. Nandeesh Kumar and V. Chaitra, Special kind of colorable complements in Graphs, International J. Math. Combin., (01), 5-. [5] J.E. Dunbar, S.T. Hedetniemi, M.A. Henning, P.J. Slater. Signed domination in graphs. In: Y. Alavi, A. Schwenk (Eds.), Graph Theory, Combinatorics, and Algorithms, Proceedings of the Seventh International Conference in Graph Theory, Combinatorics, Algorithms, and Applications, Kalamazoo, MI, 199, (pp. 11-1). WileyInterscience, New York [6] F. Harary. Graph theory. Addison-Wesley Publishing Company, MA, [7] T. W. Haynes, S. T. Hedetniemi and P. J. Slater. Fundamental of Domination in Graphs. Marcel Dekker, Inc., New York [8] V. R. Kulli, Theory of domination in graphs, Vishwa Internat. Publ., Gulbarga 010. [9] E. Sampathkumar and L. Pushpalatha. Complement of a graph: A generalization. Graphs and Combinatorics, 1()(1998), [10] E. Sampathkumar, L. Pushpalatha, C. V. Venkatachalam and Pradeep Bhat. Generalized Complements of a graph. Indian J. Pure Appl. Math., 9(6)(1998) [11] N. D. Soner, B. Janakiram and B. Chaluvaraju. Domination in -complement ((i)- complement) of a graph. Adv. Stud. Contemp. Math., Kyungshang, 7()(00), [1] J. H. Van-Lint and J. J. Seidel. Equilateral point sets in elliptic geometry. Proceedings of the Koninklijke Nederlandse Akademie van Wetenschappen: Series A: Mathematical Sciences, 69()(1966), 5-8. [1] L. Volmann and B. Zelinka. Signed domatic number of a graph. Discrete Appl. Math. 150(1- )(005), Received by editors ; Available online

11 SIGN DOMINATING SWITCHED INVARIANTS OF A GRAPH 6 Department of Mathematics, Bangalore University, Jnana Bharathi Campus, Bangalore , India address: bchaluvaraju@gmail.com Department of Mathematics, B. M. S. College of Engineering, Basavangudi, Bangalore , India address: chaitrashok@gmail.com