Complementary Signed Dominating Functions in Graphs

Transcription

1 Int. J. Contemp. Math. Sciences, Vol. 6, 011, no. 38, Complementary Signed Dominating Functions in Graphs Y. S. Irine Sheela and R. Kala Department of Mathematics Manonmaniam Sundaranar University Tirunelveli, India Abstract Let G = (V, E) be a graph. A function f : V +1, 1} is called a complementary signed dominating function of G if u/ N[v f(u) 1 for every v V with d(v) n. The weight of f is defined as w(f) = v V f(v). The complementary signed domination number of G is defined as γ cs (G) = min w(f) : f is a minimal complementary signed dominating f unction of G}. In this paper, we determine the value of complementary signed domination number for some special class of graphs. We also determine bounds for this parameter and exhibit the sharpness of the bounds. We also characterize graphs attaining the bounds in some special classes. Mathematics Subject Classification: 05C69 Keywords: Dominating function, Signed dominating function, complementary signed dominating function 1 Introduction By a graph we mean a finite, undirected connected graph without loops or multiple edges. Terms not defined here are used in the sense of Haynes et. al. [3 and Harary [. Let G = (V,E) be a graph with n vertices and m edges. A subset S V is called a dominating set of G if every vertex in V -S is adjacent to at least one vertex in S. A function f : V 0, 1} is called a dominating function of G if u N[v f(u) 1 for every v V. Dominating function is a natural generalization of dominating set. If S is a dominating set, then the characteristic

2 186 Y. S. Irine Sheela and R. Kala function χ S (V ) is a dominating function. A Signed dominating function of G is a function f : V +1, 1} such that u N[v f(u) 1 for every v V. The signed domination number for a graph G is γ s (G) = min w(f) : f is signed dominating function of G}. Signed dominating function is defined in [1. Definition 1.1 A caterpillar is a tree T for which removal of all pendent vertices leave a path. Definition 1. The wheel W n is defined to be the graph K 1 +C n 1 for n 4. Main Results Definition.1 A function f : V +1, 1} is called a complementary signed dominating function of G if u/ N[v f(u) 1 for every v V with deg(v) n. Weight of a complementary signed dominating function f is defined as w(f) = v V f(v). The complementary signed domination number of G is defined as γ cs (G) = min w(f) : f is a minimal complementary signed dominating function of G}. Theorem. γ cs (G) = (n 4) if and only if G = K n e. Proof. Assume that G = K n e. Assigning the values 1 to all the full degree vertices and the value +1 to the remaining two vertices produces a complementary signed dominating function that is minimum. Therefore γ cs (G) = (n ) + = (n 4). Conversely assume that γ cs (G) = (n 4)and let f be the corresponding complementary signed dominating function of G. Now γ cs (G) = (n ) +. Hence there exists exactly two vertices with function value +1. We claim that G has exactly two nonadjacent vertices u and v. We first prove that f(u) = +1. If f(u) = 1 then there exists at least one vertex w which is nonadjacent to u with f(w) = +1.But now x/ N[w f(x) 0 which is impossible. Thus f(u) = +1. Hence there exists exactly one v such that v is nonadjacent to u and f(v) = +1. Also u should be adjacent to all the vertices with function value 1. Clearly < v : f(v) = 1} > is a complete graph. Therefore G = K n e. Theorem.3 If every vertex of G is non-adjacent to exactly one or two vertices, then γ cs (G) = n.

3 Complementary signed dominating functions 1863 Proof. Let v be a vertex of G. If v is non-adjacent to exactly one vertex say w 1 and if f(w 1 ) = 1 then u/ N[v f(u) 0 and so f(w 1) = +1. Suppose v is non-adjacent to exactly two vertices say w 1 and w. If f(w 1 ) = +1 and f(w ) = 1 or f(w 1 ) = 1 and f(w ) = +1 then u/ N[v f(u) 0 and so f(w 1) = f(w ) = +1. In both cases, since v is nonadjacent to w 1, by the same argument f(v) = +1. Since v is arbitrary, f(v) = +1 for all v V (G). Hence γ cs (G) = n. Theorem.4 For any positive integer n 3, there exists a graph G with V (G) = n such that γ cs (G) = 0. Proof. We construct the required graph G as follows : Consider K n with V (K n ) = v 1,v,...,v n }. For every set S i of n vertices of K n, choose a vertex u i. Clearly 1 i n. For every i, draw an edge from u i to every vertex of S i. We claim that the resulting graph G is the required graph. Define f : V (G) +1, 1} by f(v i ) = 1 and f(u i ) = +1, 1 i n. By construction, f is a minimal complementary signed dominating function. Therefore γ cs (G) = n n = 0. Remark.5 For n, the graph in the above construction is the Hajo s graph. Remark.6 (1) For n, there is no graph G with γ cs (G) = 0. () For n =, by routine verification it is easy to observe that G = K 4 e is the only graph with γ cs (G) = 0. We now determine γ cs (G) for some special class of graphs. Theorem.7 For n, γ cs (K 1,n ) = 1 if n is even if n is odd Proof. Case (i) : n is even Let u be the central vertex of K 1,n and let v 1,v,...,v n be the other vertices. Define f : V (K 1,n ) +1, 1} by f(u) = 1 and f(v i ) = 1 if 1 i n +1 if n < i n

4 1864 Y. S. Irine Sheela and R. Kala Claim : f is a complementary signed dominating function. Let v be a pendent vertex with f(v) = 1. Then [ ( ) n n f(u) = ( 1) + n u/ N[v = (n ) n Let v be a pendent vertex with f(v) = +1. Then n n f(w) = ( 1) + n w/ N[v = (n ) + n Therefore f is a complementary signed dominating function. n n f(u) = 1 + ( 1) + n v V (K 1,n ) = 1 (n ) + n If the number of vertices with function value 1 is increased by 1, a vertex with function value +1 will not satisfy the condition necessary for a complementary signed dominating function. Therefore 1 is the minimum value of v V (K 1,n ) f(v). Hence γ cs (K 1,n ) if n is even. Case (ii) : n is odd Define f : V (K 1,n ) +1, 1} by f(u) = 1 and u/ N[v f(v i ) = 1 if 1 i n 3 +1 if n 3 < i n Claim : f is a complementary signed dominating function. Let v be a pendent vertex with f(v) = 1. Then f(u) = ( 1) + n = (n 3) n

5 Complementary signed dominating functions 1865 Let v be a pendent vertex with f(v) = +1. Then f(w) = ( 1) + n w/ N[v = (n 3) + n = Therefore f is a complementary signed dominating function. Now, f(v) = 1 + ( 1) + n v V (K 1,n ) = 1 (n 3) + n = If the number of vertices with function value 1 is increased by 1, a vertex with function value +1 will not satisfy the condition necessary for a complementary signed dominating function. Therefore is the minimum value of v V (K 1,n ) f(v). Hence γ cs (K 1,n ) = if n is odd. We observe that γ cs (P 3 ), γ cs (P 4 ) and γ cs (P 5 ). We now determine γ cs (P n ) for n 6. Theorem.8 For n 6, γ cs (P n ) = if n is even 3 if n is odd Proof. Let v 1,v,...,v n be the vertices of P n. Case (i) : n is even Define f : V (P n ) +1, 1} as follows : f(v 1 ) = f(v 3 ) = +1, f(v ) = 1 For 4 i n, +1 if i is even f(v i ) = 1 if i is odd We claim that f is a complementary signed dominating function. n n f(u) = ( ) + u/ N[v 1 =

6 1866 Y. S. Irine Sheela and R. Kala For i, 4, u/ N[v u/ N[v i u/ N[v i n n f(u) = ( ) + n n f(u) = ( ) + If i is odd and 5 i n, n n f(u) = + u/ N[v i = + 1 If i is even and 6 i n, n n f(u) = + = + Also u/ N[v n [( ) n f(u) = + = [( ) n Therefore f is a complementary signed dominating function. Since u/ N[v f(u), the labeling is minimum with respect to the vertices v 4,v 5,...,v n. If f(v n ) = 1 then u/ N[v 3 f(u) < 0. It is easy to observe that n n f(v) = + v V (P n) = is minimum. Hence γ cs (P n ) = if n is even. Case (ii) : n is odd. Define f : V (P n ) +1, 1} as follows : f(v 1 ) = f(v 3 ) = f(v n ) = +1 and f(v ) = 1. For 4 i n, +1 if i is even f(v i ) = 1 if i is odd

7 Complementary signed dominating functions 1867 For i, 4, u/ N[v i u/ N[v 1 u/ N[v f(u) = (3 ) + f(u) = (3 ) + = f(u) = (3 ) + = If i is odd and 5 i n then f(u) + u/ N[v i + 1 = If i is even and 6 i n 3 then f(u) + u/ N[v i u/ N[v n 1 + f(u) = (3 ) + u/ N[v n = f(u) = (3 ) = Therefore f is a complementary signed dominating function. Since u/ N[v n f(u), the labeling is minimum with respect to the vertices v 1,v,...,v n. If f(v 1 ) = 1 then u/ N[v 3 f(u) = 0. It is easy to observe that f(v) + is minimum. v V (P n) Hence γ cs (P n ) if n is odd.

8 1868 Y. S. Irine Sheela and R. Kala We observe that γ cs (C 4 ), γ cs (C 5 ) = 5, γ cs (C 6 ) =, γ cs (C 7 ) = 5, γ cs (C 8 ),γ cs (C 9 ). We now determine γ cs (C n ) for n 10. Theorem.9 For n 10, γ cs (C n ) = 4 if n is even 5 if n is odd Proof. Let u 1,u,...,u n be the vertices of C n. Case(i) : n is even Define f : V (C n ) +1, } as follows : f(u 1 ) = f(u ) = f(u 3 ) = f(u 4 ) = +1 1 if i is even For 5 i n, f(u i ) = +1 if i is odd We claim that f is a complementary signed dominating function. For i =, 3 u/ N[u 1 u/ N[u 5 f(u) + u/ N[u i u/ N[u 4 f(u) 3 + f(u) + f(u) + If i is even and 6 i n then f(u) + u/ N[u i

9 Complementary signed dominating functions 1869 If i is odd and 5 < i < n then f(u) + u/ N[u i u/ N[u n = 5 f(u) + Therefore f is a complementary signed dominating function. Since f(u), the labeling is minimum with respect to the vertices u/ N[u u 4, u 5,..., u n 1, u n. If f(u 5 ) = 1, then f(u) < 0. It is easy to observe u/ N[u 3 that f(u) is minimum. Hence γ cs (C n ) if n is even. u V (C n) Case(ii) : n is odd Define f : V (C n ) +1, } as follows : f(u 1 ) = f(u ) = f(u 3 ) = f(u 4 ) = f(u 5 ) = +1 1 if i is even For 6 i n, f(u i ) = +1 if i is odd. We claim that f is a complementary signed dominating function. [( ) ( ) f(u) = 5 + u/ N[u 1 For i =, 3, 4 u/ N[u 5 u/ N[u 6 u/ N[u i = f(u) = = f(u) = 5 + f(u) = 5 +

10 1870 Y. S. Irine Sheela and R. Kala If i is odd and 7 i n, then f(u) = 5 + u/ N[u i = 5 + = 6 If i is even and 8 i < n, then f(u) = 5 + Also u/ N[u i u/ N[u n f(u) = 5 + Therefore f is a complementary signed dominating function. Since f(u) =, the labeling is minimum with respect to the vertices u/ N[u 4 u 1, u, u 6,..., u n. If f(u 1 ) = 1, then f(u) = 0. It is easy to observe that u V (C n) Hence γ cs (C n ) = 5 if n is odd. References u/ N[u 3 f(u) = 5 + = 5 is minimum. [1 J. E. Dunbar, S. T. Hedetniemi, M.A. Henning and P. J. Slater, Signed domination in graphs, In Y. Alavi and A. J. Schwenk, editors, Graph theory, Combinatorics and applications, Proc. 7th International Conf. Combinatorics, Graph theory, Applications, Volume 1, pages 311-3, John Wiley and Sons, Inc.,1995 [ Harary. F, Graph theory, Reading mass, [3 Teresa W. Haynes, S.T. Hedetniemi and Peter J. Slater, Fundamentals of domination in graphs, Marcel Dekker, Received: April, 011