International J.Math. Combin. Vol. (010), -40 Super Fibonacci Graceful Labeling R. Sridevi 1, S.Navaneethakrishnan and K.Nagarajan 1 1. Department of Mathematics, Sri S.R.N.M.College, Sattur - 66 0, Tamil Nadu, India. Department of Mathematics, V.O.C.College, Tuticorin - 68 008, Tamil Nadu, India Email: r.sridevi 010@yahoo.com, k nagarajan srnmc@yahoo.co.in Abstract: A Smarandache-Fibonacci Triple is a sequence S(n), n 0 such that S(n) = S(n 1)+S(n ), where S(n) is the Smarandache function for integers n 0. Certainly, it is a generalization of Fibonacci sequence. A Fibonacci graceful labeling and a super Fibonacci graceful labeling on graphs were introduced by Kathiresan and Amutha in 006. Generally, let G be a (p, q)-graph and {S(n) n 0} a Smarandache-Fibonacci Triple. An bijection f : V (G) {S(0), S(1), S(),..., S(q)} is said to be a super Smarandache-Fibonacci graceful graph if the induced edge labeling f (uv) = f(u) f(v) is a bijection onto the set {S(1), S(),..., S(q)}. Particularly, if S(n), n 0 is just the Fibonacci sequence F i, i 0, such a graph is called a super Fibonacci graceful graph. In this paper, we construct new types of graphs namely F n K + 1,m, Cn Pm, K1,n K1,, Fn Pm and Cn K1,m and we prove that these graphs are super Fibonacci graceful graphs. Key Words: Smarandache-Fibonacci triple, graceful labeling, Fibonacci graceful labeling, super Smarandache-Fibonacci graceful graph, super Fibonacci graceful graph. AMS(000): 05C78 1. Introduction By a graph, we mean a finite undirected graph without loops or multiple edges. A path of length n is denoted by P n+1. A cycle of length n is denoted by C n. G + is a graph obtained from the graph G by attaching pendant vertex to each vertex of G. Graph labelings, where the vertices are assigned certain values subject to some conditions, have often motivated by practical problems. In the last five decades enormous work has been done on this subject [1]. The concept of graceful labeling was first introduced by Rosa [5] in 1967. A function f is a graceful labeling of a graph G with q edges if f is an injection from the vertices of G to the set {0, 1,,..., q} such that when each edge uv is assigned the label f(u) f(v), the resulting edge labels are distinct. The notion of Fibonacci graceful labeling and Super Fibonacci graceful labeling were introduced by Kathiresan and Amutha []. We call a function f, a Fibonacci graceful labeling of a graph G 1 Received June 0, 010. Accepted September 6, 010.
Super Fibonacci Graceful Labeling with q edges if f is an injection from the vertices of G to the set {0, 1,,..., F q }, where F q is the q th Fibonacci number of the Fibonacci series = 1, =,F =, F 4 = 5,, such that each edge uv is assigned the labels f(u) f(v), the resulting edge labels are,,...,f q. An injective function f : V (G) {F 0,,...,F q }, where F q is the q th Fibonacci number, is said to be a super Fibonacci graceful labeling if the induced edge labeling f(u) f(v) is a bijection onto the set {,,..., F q }. In the labeling problems the induced labelings must be distinct. So to introduce Fibonacci graceful labelings we assume = 1, =, F =, F 4 = 5,, as the sequence of Fibonacci numbers instead of 0, 1,,..., []. Generally, a Smarandache-Fibonacci Triple is a sequence S(n), n 0 such that S(n) = S(n 1) + S(n ), where S(n) is the Smarandache function for integers n 0 []. A (p, q)- graph G is a super Smarandache-Fibonacci graceful graph if there is an bijection f : V (G) {S(0), S(1), S(),..., S(q)} such that the induced edge labeling f (uv) = f(u) f(v) is a bijection onto the set {S(1), S(),...,S(q)}. So a super Fibonacci graceful graph is a special type of Smarandache-Fibonacci graceful graph by definition. In this paper, we prove that F n K + 1,m, C n P m, K 1,n K 1,, F n P m and C n K 1,m are super Fibonacci graceful graphs.. Main Results In this section, we show that some new types of graphs namely F n K + 1,m, C n P m, K 1,n K 1,, F n P m and C n K 1,m are super Fibonacci graceful graphs. Definition.1([4]) Let G be a (p, q) graph. An injective function f : V (G) {F 0,,,..., F q }, where F q is the q th Fibonacci number, is said to be a super Fibonacci graceful graphs if the induced edge labeling f (uv) = f(u) f(v) is a bijection onto the set {,,...,F q }. Definition. The graph G = F n P m consists of a fan F n and a Path P m which is attached with the maximum degree of the vertex of F n. The following theorem shows that the graph F n P m is a super Fibonacci graceful graph. Theorem. The graph G = F n P m is a super Fibonacci graceful graph. Proof Let {u 0 = v, u 1, u,..., u n } be the vertex set of F n and v 1, v,...,v m be the vertices of P m joined with the maximum degree of the vertex u 0 of F n. Also, V (G) = m + n + 1 and E(G) = n + m 1. Define f : V (G) {F 0,,...,F q } by f(u 0 ) = F 0, f(u i ) = n+m 1 (i 1), 1 i n, f(v i ) = F m (i 1), 1 i, f(v m ) = ifm 0(mod) ifm 1, (mod) f(v m 1 ) = F ifm 1(mod) ifm (mod) and f(v m ) = F 4 if m (mod). For l = 1,,..., m, or m 4, or m 5 according to m 0(mod) or m 1(mod)
4 R. Sridevi, S.Navaneethakrishnan and K.Nagarajan or m (mod), define f(v i+ ) = F m 1 (i 1)+(l 1), l i l. We claim that the edge labels are distinct. Let E 1 = {f (u i u i+1 ) : i = 1,,..., n 1}. Then E 1 = { f(u i ) f(u i+1 ) : i = 1,,...,n 1} = { f(u 1 ) f(u ), f(u ) f(u ),..., f(u n 1 ) f(u n ) } = { n+m 1 n+m, n+m n+m 5,..., F m+ F m+1 } = {n+m, n+m 4,..., F m+4, F m+ }, E = {f (u 0 u i ) : i = 1,,...,n} = { f(u 0 ) f(u i ) : i = 1,,..., n} = { f(u 0 ) f(u 1 ), f(u 0 ) f(u ),..., f(u 0 ) f(u n ) } = { F 0 n+m 1, F 0 n+m,..., F 0 F m+1 } = {n+m 1, n+m,...,f m+, F m+1 }, E = {f (u 0 v 1 ), f (v 1 v )} = { f(u 0 ) f(v 1 ), f(v 1 ) f(v ) } = { F 0 F m, F m F m } = {F m, F m 1 }. Let E 4 = {f (v v )}. The edge labeling between the vertex v and starting vertex v of the first loop is E 4 = { f(v ) f(v ) } = { F m F m 1 } = {F m }. For l = 1, let E 5 = {f (v i+ v i+ ) : 1 i }. Then E 5 = { f(v i+ ) f(v i+ ) : 1 i } = { f(v ) f(v 4 ), f(v 4 ) f(v 5 ) } = { F m 1 F m, F m F m 5 } = {F m, F m 4 }. Let E 1 5 = {f (v 5 v 6 )}. We find the edge labeling between the end vertex v 5 of the first loop and starting vertex v 6 of the second loop following. E 1 5 = { f(v 5 ) f(v 6 ) } = { F m 5 F m 4 } = {F m 6 }. For l =, let E 6 = {f (v i+ v i+ ) : 4 i 5}. Then E 6 = { f(v i+ ) f(v i+ ) : 4 i 5} = { f(v 6 ) f(v 7 ), f(v 7 ) f(v 8 ) } = { F m 4 F m 6, F m 6 F m 8 } = {F m 5, F m 7 }. For labeling between the end vertex v 8 of the second loop and starting vertex v 9 of the third loop, let E 1 6 = {f (v 8 v 9 )}. Then E 1 6 = { f(v 8) f(v 9 ) } = { F m 8 F m 7 } = {F m 9 },
Super Fibonacci Graceful Labeling 5 etc.. For l = m 5 1, let E m 5 1 = {f (v i+ v i+ ) : m 10 i m 9}. Then E m 5 1 = { f(v i+ ) f(v i+ ) : m 10 i m 9} = { f(v m 8 ) f(v m 7 ), f(v m 7 ) f(v m 6 ) } = { 0, } = {F 9, }. For the edge labeling between the end vertex v m 6 of the ( m 5 1) th loop and starting vertex v m 5 of the ( m 5 ) rd loop, let E 1 m 5 = {f (v 1 m 6 v m 5 )}. Then E 1 m 5 1 = { f(v m 6) f(v m 5 ) } = { } = { }, For l = m 4 E m 5 = {f (v i+ v i+ ) : m 7 i m 6} = { f(v i+ ) f(v i+ ) : m 7 i m 6} = { f(v m 5 ) f(v m 4 ), f(v m 4 ) f(v m ) } = {, F } = {, F 4 }. 1, let E m 4 1 = {f (v i+ v i+ ) : m 9 i m 8}. Then E m 4 1 = { f(v i+ ) f(v i+ ) : m 9 i m 8} = { f(v m 7 ) f(v m 6 ), f(v m 6 ) f(v m 5 ) } = { F 9, } = {, }. For the edge labeling between the end vertex v m 5 of the ( m 4 1) th loop and starting vertex v m 4 of the ( m 4 ) rd loop, let E 1 m 4 = {f (v 1 m 5 v m 4 )}. Then E 1 m 4 1 = { f(v m 5) f(v m 4 ) } = { } = {F 4 }. For l = m 4, let E m 4 = {f (v i+ v i+ ) : m 6 i m 5}. Calculation shows that E m 4 Now for l = m = { f(v i+ ) f(v i+ ) : m 6 i m 5} = { f(v m 4 ) f(v m ), f(v m ) f(v m ) } = { F 4, F 4 } = {, F }. 1, let E m 1 = {f (v i+ v i+ ) : m 8 i m 7}. Then E m 1 = { f(v i+ ) f(v i+ ) : m 8 i m 7} = { f(v m 6 ) f(v m 5 ), f(v m 5 ) f(v m 4 ) } = {, F 4 } = {, }. Similarly, for finding the edge labeling between the end vertex v m 4 of the ( m 1) th loop and starting vertex v m of the ( m ) rd loop, let E 1 m = {f (v 1 m 4 v m )}. Then E 1 m 1 = { f(v m 4) f(v m ) } = { F 4 } = {F }.
6 R. Sridevi, S.Navaneethakrishnan and K.Nagarajan For l = m, let E m = {f (v i+ v i+ ) : m 5 i m 4}. Then Now let ( E (1) = ( E () = ( E () = E m = { f(v i+ ) f(v i+ ) : m 5 i m 4} = { f(v m ) f(v m ), f(v m ) f(v m 1 ) } = { F, F } = {F 4, }. E 1 E,..., E m E 1 E,..., E m 4 E 1 E,..., E m 5 ) ( ) E5 1 E 1 6,..., E 1 m, 1 ) ( ) E5 1 E 1 6,..., E 1 m 4 1 ) ( E5 1 E 1 6,..., E 1 m 5 1 If m 0(mod), let E 1 = {f (v m 1 v m )}, then E 1 = { f(v m 1 f(v m ) } = { } = { }. Thus, E = E 1 E (1) = {,,..., n+m 1 }. For example the super Fibonacci graceful labeling of F 4 P 6 is shown in Fig.1. F 4 P 6 : 1 0 F 9, ). 1 F9 F7 F F 4 F 0 F 4 F Fig.1 Thus, If m 1(mod), let E = {f (v m v m 1 ), f (v m 1 v m )}, then E = { f(v m f(v m 1, f(v m 1 f(v m ) } = { F F } = {, }. E = E E() = {,,..., n+m 1 }. For example the super Fibonacci graceful labeling of F 4 P 7 is shown in Fig.. F 4 P 7 : 4 0 1 F 9 4 0 F 4 F F 0 F 4 F Fig.
Super Fibonacci Graceful Labeling 7 If m (mod), let E = {f (v m v m ), f (v m v m 1 ), f (v m 1 v m )}, then E = { f(v m ) f(v m ), f(v m ) f(v m 1 ), f(v m 1 f(v m ) } = { F F 4, F 4, } = {, F, }. Thus, E = E E () = {,,..., n+m 1 }. For example the super Fibonacci graceful labeling of P 5 is shown in Fig.. P 5 : 4 1 0 F 9 4 0 F 4 F F 0 F F 4 Fig. Therefore, F n P m admits a super Fibonacci graceful labeling. Hence, F n P m is a super Fibonacci graceful graph. Definition.4 An (n, m)-kite consists of a cycle of length n with m-edge path attached to one vertex and it is denoted by C n P m. Theorem.5 The graph G = C n P m is a super Fibonacci graceful graph when n 0(mod). Proof Let {u 1, u,..., u n = v} be the vertex set of C n and {v = u n, v 1, v,..., v m } be the vertex set of P m joined with the vertex u n of C n. Also, V (G) = E(G) = m + n. Define f : V (G) {F 0,,...,F q } by f(u n ) = F 0, f(u 1 ) = F m+n, f(u ) = F m+n and for l = 1,,..., n, f(u i+ ) = F m+n 1 (i 1)+(l 1), and for l i l, f(v i ) = F m (i 1), and for 1 i, f(v m ) = if m 0(mod) if m 1, (mod), f(v m 1 ) = F if m 1(mod) if m (mod) and f(v m ) = F 4 when m (mod). For l = 1,,..., m or m 4 or m 5 according to m 0(mod) or m 1(mod) or m (mod), let f(v i+ ) = F m 1 (i 1)+(l 1) for l i l.
8 R. Sridevi, S.Navaneethakrishnan and K.Nagarajan We claim that all these edge labels are distinct. Let E 1 = {f u n u 1 ), f (u 1 u )}. Then E 1 = { f(u n ) f(u 1 ), f(u 1 ) f(u ) } = { F 0 F m+n, F m+n F m+n } = {F m+n, F m+n 1 }. For the edge labeling between the vertex u and starting vertex u of the first loop, let E = {f (u u )}. Then E = { f(u ) f(u ) } = { F m+n F m+n 1 } = {F m+n }. For l = 1, let E = {f (u i+ u i+ ) : 1 i }. Then E = { f(u i+ ) f(u i+ ) : 1 i } = { f(u ) f(u 4 ), f(u 4 ) f(u 5 ) } = { F m+n 1 F m+n, F m+n F m+n 5 ) } = {F m+n, F m+n 4 }. For the edge labeling between the end vertex u 5 of the first loop and starting vertex u 6 of the second loop, let E (1) = {f (u 5 u 6 )}. Then E (1) = { f(u 5 ) f(u 6 ) } = { F m+n 5 F m+n 4 } = {F m+n 6 }. For l =, let E 4 = {f (u i+ u i+ ) : 4 i 5}. Then E 4 = { f(u i+ ) f(u i+ ) : 4 i 5} = { f(u 6 ) f(u 7 ), f(u 7 ) f(u 8 ) } = { F m+n 4 F m+n 6, F m+n 6 F m+n 8 } = {F m+n 5, F m+n 7 }. For the edge labeling between the end vertex u 8 of the second loop and starting vertex u 9 of the third loop, let E (1) 4 = {f (u 8 u 9 )}. Then etc.. For l = n E (1) 4 = { f(u 8 ) f(u 9 ) } = { F m+n 8 F m+n 7 } = {F m+n 9 }, 1, let E n 1 = {f (u i+ u i+ ) : n 8 i n 7}. Then E n 1 = { f(u i+ ) f(u i+ ) : n 8 i n 7} = { f(u n 6 ) f(u n 5 ), f(u n 5 ) f(u n 4 ) } = { F m+8 F m+6, F m+6 F m+4 ) } = {F m+7, F m+5 }. For finding the edge labeling between the end vertex u n 4 of the ( n starting vertex u n of the ( n ) rd loop, let E (1) n 1 = {f (u n 4 u n )}. Then E (1) n 1 = { f(u n 4) f(u n ) } = { F m+4 F m+5 } = {F m+ }. For l = n, let E n = {f (u i+ u i+ ) : n 5 i n 4}. Then E n = { f(u i+ ) f(u i+ ) : n 5 i n 4} = { f(u n ) f(u n ), f(u n ) f(u n 1 ) } = { F m+5 F m+, F m+ F m+1 } = {F m+4, F m+ }. 1) th loop and
Super Fibonacci Graceful Labeling 9 Let E 1 = {f (u n 1 u n )} and E = {f (u n v 1 ), f (v 1 v )}. Then E 1 = { f(u n 1) f(u n ) } = { F m+1 F 0 } = {F m+1 }, E = { f(u n ) f(v 1 ), f(v 1 ) f(v ) } = { F 0 F m, F m F m } = {F m, F m 1 }. For finding the edge labeling between the vertex v and starting vertex v of the first loop, let E = {f (v v )}. Then E = { f(v ) f(v ) } = { F m F m 1 } = {F m }. For l = 1, let E 4 = {f (v i+ v i+ ) : 1 i }. Then E 4 = { f(v i+ ) f(v i+ ) : 1 i } = { f(v ) f(v 4 ), f(v 4 ) f(v 5 ) } = { F m 1 F m, F m F m 5 } = {F m, F m 4 }. Now let E ( 1) 4 = {f (v 5 v 6 )}. Then E ( 1) 4 = { f(v 5 ) f(v 6 ) } = { F m 5 F m 4 } = {F m 6 }. For l =, let E 5 = {f (v i+ v i+ ) : 4 i 5}. Calculation shows that E 5 = { f(v i+ ) f(v i+ ) : 4 i 5} = { f(v 6 ) f(v 7 ), f(v 7 ) f(v 8 ) } = { F m 4 F m 6, F m 6 F m 8 } = {F m 5, F m 7 }. Let E ( 1) 5 = {f (v 8 v 9 )}. We find the edge labeling between the end vertex v 8 of the second loop and starting vertex v 9 of the third loop. In fact, etc.. For l = m 5 E ( 1) 5 = { f(v 8 ) f(v 9 ) } = { F m 8 F m 7 } = {F m 9 } 1, let E m 5 1 = {f (v i+ v i+ ) : m 10 i m 9}. Then E m 5 1 = { f(v i+ ) f(v i+ ) : m 10 i m 9} = { f(v m 8 ) f(v m 7 ), f(v m 7 ) f(v m 6 ) } = { 0, } = {F 9, }. Similarly, for finding the edge labeling between the end vertex v m 6 of the ( m 5 loop and starting vertex v m 5 of the ( m 5 1) th ) rd loop, let E ( 1) m 5 1 = {f (v m 6 v m 5 )}. Then E ( 1) m 5 1 = { f(v m 6) f(v m 5 ) } = { } = { }.
0 R. Sridevi, S.Navaneethakrishnan and K.Nagarajan For l = m 5, let E m 5 = {f (v i+ v i+ ) : m 7 i m 6}. Then For l = m 4 E m 5 = { f(v i+ ) f(v i+ ) : m 7 i m 6} = { f(v m 5 ) f(v m 4 ), f(v m 4 ) f(v m ) } = {, F } = {, F 4 }. 1, let E m 4 1 = {f (v i+ v i+ ) : m 9 i m 8}. We find that E m 4 1 = { f(v i+ ) f(v i+ ) : m 9 i m 8} = { f(v m 7 ) f(v m 6 ), f(v m 6 ) f(v m 5 ) } = { F 9, } = {, }. For getting the edge labeling between the end vertex v m 5 of the ( m 4 starting vertex v m 4 of the ( m 4 ) rd loop, let E ( 1) m 4 1 = {f (v m 5 v m 4 )}. Then E ( 1) m 4 1 = { f(v m 5) f(v m 4 ) } = { } = {F 4 }. For l = m 4, let E m 4 = {f (v i+ v i+ ) : m 6 i m 5}. Then For l = m E m 4 = { f(v i+ ) f(v i+ ) : m 6 i m 5} = { f(v m 4 ) f(v m ), f(v m ) f(v m ) } = { F 4, F 4 } = {, F }. 1, let E m 1 = {f (v i+ v i+ ) : m 8 i m 7}. Then E m 1 = { f(v i+ ) f(v i+ ) : m 8 i m 7} = { f(v m 5 ) f(v m 4 ), f(v m 4 ) f(v m ) } = {, F 4 } = {, }. For the edge labeling between the end vertex v m of the ( m vertex v m of the ( m ) rd loop, let E ( 1) m 1 = {f (v m v m )}. Then E ( 1) m 1 = { f(v m ) f(v m ) } = { F 4 } = {F }. Similarly, for l = m, let E m E m 1) th loop and 1) th loop and starting = {f (v i+ v i+ ) : m 5 i m 4}. Then = { f(v i+ ) f(v i+ ) : m 5 i m 4} = { f(v m ) f(v m ), f(v m ) f(v m 1 ) } = { F, F } = {F 4, }. Now let E (1) = ( ) ( ) E 1 E E n E 1 E 1 4 E 1 n ( ) ( ) E1 E E m E ( 1) ( 1) ( 1) 4 E 5 E m, 1
Super Fibonacci Graceful Labeling 1 and Thus, E () = E () = ( E 1 E E n ( E 1 E E m 4 )( ) E 1 E 1 4 E 1 n )( E ( 1) ( 1) 4 E 5 E ( 1) m 4 1 ( ) ( ) E 1 E E n E 1 E 1 4 E 1 n ( ) ( ) E1 E E m 5 E ( 1) ( 1) ( 1) 4 E 5 E m 5. 1 If m 0(mod), let E 1 = {f (v m 1 v m )}, then E 1 = { f(v m 1 f(v m ) } = { } = { }. E = E 1 E(1) = {,,..., F m+n }. For example the super Fibonacci graceful labeling of C 6 P 6 is shown in Fig.4. C 6 P 6 : F 9 0 1 F 9 ) 0 1 F 0 F F 4 F 4 F Fig.4 Thus, If m 1(mod), let E = {f (v m v m 1 ), f (v m 1 v m )}, then E = { f(v m f(v m 1, f(v m 1 f(v m ) } = { F F } = {, }. E = E E() = {,,..., F m+n }. For example the super Fibonacci graceful labeling of C 6 P 7 is shown in Fig.5. C 6 P 7 : 1 0 1 F10 F 0 F 9 F 4 F F 4 F Fig.5
R. Sridevi, S.Navaneethakrishnan and K.Nagarajan Thus, If m (mod), let E = {f (v m v m ), f (v m v m 1 ), f (v m 1 v m )}, then E = { f(v m ) f(v m ), f(v m ) f(v m 1 ), f(v m 1 f(v m ) } = { F F 4, F 4, } = {, F, }. E = E E() = {,,..., F m+n }. For example the super Fibonacci graceful labeling of C 6 P 5 is shown in Fig.6. C 6 P 5 : F 9 F 9 0 0 1 1 F 0 F 4 F F F 4 Fig.6 Therefore, C n P m admits a super Fibonacci graceful labeling. Hence, C n P m is a super Fibonacci graceful graph. Definition.6 The graph G = F n K + 1,m consists of a fan F n and the extension graph of K + 1,m which is attached with the maximum degree of the vertex of F n. Theorem.7 The graph G = F n K 1,m + is a super Fibonacci graceful graph. Proof Let V (G) = U V, where U = {u 0, u 1,...,u n } be the vertex set of F n and V = (V 1, V ) be the bipartion of K 1,m, where V 1 = {v = u 0 } and V = {v 1, v,...,v m } and w 1, w,..., w m be the pendant vertices joined with v 1, v,..., v m respectively. Also, V (G) = m + n + 1 and E(G) = m + n 1. Case 1 m, n is even. Define f : V (G) {F 0,,...,F q } by f(u 0 ) = F 0, f(u i ) = m+n 1 (i 1) if 1 i n; f(v i 1 ) = m 4(i 1) if 1 i m ; f(v i) = m 4(i 1) if 1 i m ; f(w i 1) = m 4(i 1) if 1 i m and f(w i) = m 1 4(i 1) if 1 i m. We claim that all these edge labels are distinct. Calculation shows that E 1 = {f (u i u i+1 ) : i = 1,,...,n 1} = { f(u i ) f(u i+1 ) : i = 1,,...,n 1} = { f(u 1 ) f(u ), f(u ) f(u ),..., f(u n 1 ) f(u n ) } = { n+m 1 n+m, n+m n+m 5,..., m+ m+1 } = {n+m, n+m 4,..., m+4, m+ },
Super Fibonacci Graceful Labeling E = {f (u 0 u i ) : i = 1,,...,n} = { f(u 0 ) f(u i ) : i = 1,,...,n} = { f(u 0 ) f(u 1 ), f(u 0 ) f(u ),..., f(u 0 ) f(u n 1 ), f(u 0 ) f(u n ) } = { F 0 n+m 1, F 0 n+m,..., F 0 m+, F 0 m+1 } = {n+m 1, n+m,...,m+, m+1 }, E = {f (u 0 v i 1 ) : 1 i m } = { f(u 0 ) f(v i 1 ) : 1 i m } = { f(u 0 ) f(v 1 ), f(u 0 ) f(v ),..., f(u o ) f(v m ), f(u 0 ) f(v m 1 ) } = { F 0 m, F 0 m 4,..., F 0, F 0 F 4 } = {m, m 4,...,, F 4 }, E 4 = {f (u 0 v i ) : 1 i m } = { f(u 0 ) f(v i ) : 1 i m } = { f(u 0 ) f(v ), f(u 0 ) f(v 4 ),..., f(u 0 ) f(v m ), f(u 0 ) f(v m ) } = { F 0 m, F 0 m 7,..., F 0, F 0 } = {m, m 7,...,, }, E 5 = {f (v i 1 w i 1 ) : 1 i m } = { f(v i 1 ) f(w i 1 ) : 1 i m } = { f(v 1 ) f(w 1 ), f(v ) f(w ),..., f(v m ) f(w m ), f(v m 1 ) f(w m 1 ) } = { m m, m 4 m 6,...,, F 4 } = {m 1, m 5....,, F }, E 6 = {f (v i w i ) : 1 i m } = { f(v i ) f(w i ) : 1 i m } = { f(v ) f(w ), f(v 4 ) f(w 4 ),..., f(v m ) f(w m ), f(v m ) f(w m ) = { m m 1, m 7 m 5,...,, F } = {m, m 6....,, }. Therefore, E = E 1 E E6 = {,,..., m+n 1 }. Thus, the edge labels are distinct. Therefore, F n K 1,m + admits super Fibonacci graceful labeling. Hence, F n K 1,m + is a super Fibonacci graceful graph. For example the super Fibonacci graceful labeling of F 4 K 1,4 + is shown in Fig.7.
4 R. Sridevi, S.Navaneethakrishnan and K.Nagarajan 5 4 1 0 F 9 5 1 F 9 F 4 K + 1,4 : F 0 F 4 F 4 F F Fig.7 Case m even, n odd. Proof of this case is analogous to case(i). For example the super Fibonacci graceful labeling of K 1,4 + is shown in Fig.8. 7 5 1 F 9 6 4 0 K + 1,4 : 7 5 1 F9 F 0 F F 4 8 F 4 F 7 F F Fig.8 Case m, n is odd. Define f : V (G) {F 0,,...,F q } by f(u 0 ) = F 0 ; f(u i ) = m+n 1 (i 1) if 1 i n; f(w m ) = ; f(v i 1 ) = m 4(i 1) if 1 i m + 1 ; f(v i ) = m 4(i 1) if 1 i m 1 ; f(w i 1 ) = m 4(i 1) if 1 i m 1 and f(w i ) = m 1 4(i 1) if 1 i m 1. We claim that the edge labels are distinct. Calculation shows that E 1 = {f (u i u i+1 ) : i = 1,,..., n 1} = { f(u i ) f(u i+1 ) : i = 1,,..., n 1} = { f(u 1 ) f(u ), f(u ) f(u ),..., f(u n ) f(u n 1 ), f(u n 1 ) f(u n ) } = { n+m 1 n+m, n+m n+m 5,..., m+5 m+, m+ m+1 } = {n+m, n+m 4,..., m+4, m+ },
Super Fibonacci Graceful Labeling 5 E = {f (u 0 u i ) : i = 1,,...,n} = { f(u 0 ) f(u i ) : i = 1,,...,n} = { f(u 0 ) f(u 1 ), f(u 0 ) f(u ),..., f(u 0 ) f(u n 1 ), f(u 0 ) f(u n ) } = { F 0 n+m 1, F 0 n+m,..., F 0 m+, F 0 m+1 } = {n+m 1, n+m,...,m+, m+1 }, E = {f (u 0 v i 1 ) : 1 i m + 1 } = { f(u 0 ) f(v i 1 ) : 1 i m + 1 } = { f(u 0 ) f(v 1 ), f(u 0 ) f(v ),..., f(u o ) f(v m ), f(u 0 ) f(v m ) } = { F 0 m, F 0 m 4,..., F 0, F 0 } = {m, m 4,...,, }, E 4 = {f (u 0 v i ) : 1 i m 1 } = { f(u 0 ) f(v i ) : 1 i m 1 } = { f(u 0 ) f(v ), f(u 0 ) f(v 4 ),..., f(u o ) f(v m ), f(u 0 ) f(v m 1 ) } = { F 0 m, F 0 m 7,..., F 0, F 0 F } = {m, m 7....,, F }, E 5 = {f (v m w m )} = { f(v m ) f(w m ) } = { } = { }, E 6 = {f (v i 1 w i 1 ) : 1 i m 1 } = { f(v i 1 ) f(w i 1 ) : 1 i m 1 } = { f(v 1 ) f(w 1 ), f(v ) f(w ),..., f(v m 4 ) f(w m 4 ), f(v m ) f(w m ) } = { m m, m 4 m 6,...,, F 4 } = {m 1, m 5...., F 9, }, E 7 = {f (v i w i ) : 1 i m 1 } = { f(v i ) f(w i ) : 1 i m 1 } = { f(v ) f(w ), f(v 4 ) f(w 4 ),..., f(v m ) f(w m ), f(v m 1 ) f(w m 1 ) } = { m m 1, m 7 m 5,..., F 9, F } = {m, m 6....,, F 4 }. Therefore, E = E 1 E E7 = {,,..., m+n 1 }.
6 R. Sridevi, S.Navaneethakrishnan and K.Nagarajan Thus, the edge labels are distinct.therefore, F n K 1,m + admits super Fibonacci graceful labeling. Whence, F n K 1,m + is a super Fibonacci graceful graph. For example the super Fibonacci graceful labeling of K 1, + is shown in Fig.9. 5 1 F 9 4 0 K + 1, : 5 1 F 9 F 0 F F F 4 F 4 Fig.9 Case 4 m odd, n even. Proof of this case is analogous to Case 4. For example the super Fibonacci graceful labeling of F 4 K 1, + is shown in Fig.10. 1 F 9 0 1 F 9 F 4 K + 1, : F 0 F F F 4 F 4 Fig.10 Definition.8 The graph G = C n K 1,m consists of a cycle C n of length n and a star K 1,m is attached with the vertex u n of C n. Theorem.9 The graph G = C n K 1,m is a super Fibonacci graceful graph when n 0(mod). Proof Let V (G) = V 1 V, where V 1 = {u 1, u,..., u n } be the vertex set of C n and V = {v = u n, v 1, v,..., v m } be the vertex set of K 1,m. Also, V (G) = E(G) = m+n. Define f : V (G) {F 0,,,..., F q } by f(u n ) = F 0 ; f(u i ) = F m+n (i 1) if 1 i ; f(v i ) = F i if 1 i m and for l = 1,,..., n, f(u i+ ) = F m+n 1 (i 1)+(l 1) if l i l.
Super Fibonacci Graceful Labeling 7 We claim that the edge labels are distinct. Calculation shows that E 1 = {f (u n v i ) : 1 i m} = { f(u n ) f(v i ) : 1 i m} = { f(u n ) f(v 1 ), f(u n ) f(v ),..., f(u n ) f(v m 1 ), f(u n ) f(v m ) } = { F 0, F 0,..., F 0 F m 1, F 0 F m } = {,,...,F m 1, F m }, E = {f (u n u 1 ), f (u 1 u )} = { f(u n ) f(u 1 ), f(u 1 ) f(u ) } = { F 0 F m+n, F m+n F m+n } = {F m+n, F m+n 1 }. For the edge labeling between the vertex u and starting vertex u of the first loop, let E = {f (u u )}. Then E = { f(u ) f(u ) } = { F m+n F m+n 1 } = {F m+n }. For l = 1, let E 4 = {f (u i+ u i+ ) : 1 i }. Then Let E (1) 4 = {f (u 5 u 6 )}. Then E 4 = { f(u i+ ) f(u i+ ) : 1 i } = { f(u ) f(u 4 ), f(u 4 ) f(u 5 ) } = { F m+n 1 F m+n, F m+n F m+n 5 ) } = {F m+n, F m+n 4 }. E (1) 4 = { f(u 5 ) f(u 6 ) } = { F m+n 5 F m+n 4 } = {F m+n 6 }. For l =, let E 5 = {f (u i+ u i+ ) : 4 i 5}. Then E 5 = { f(u i+ ) f(u i+ ) : 4 i 5} = { f(u 6 ) f(u 7 ), f(u 7 ) f(u 8 ) } = { F m+n 4 F m+n 6, F m+n 6 F m+n 8 ) } = {F m+n 5, F m+n 7 }. For finding the edge labeling between the end vertex u 8 of the second loop and starting vertex u 9 of the third loop, let E (1) 5 = {f (u 8 u 9 )}. Then E (1) 5 = { f(u 8 ) f(u 9 ) } = { F m+n 8 F m+n 7 } = {F m+n 9 } etc.. Similarly, for l = n 1, let E n 1 = {f (u i+ u i+ ) : n 8 i n 7}. Then E n 1 = { f(u i+ ) f(u i+ ) : n 8 i n 7} = { f(u n 6 ) f(u n 5 ), f(u n 5 ) f(u n 4 ) } = { F m+8 F m+6, F m+6 F m+4 ) } = {F m+7, F m+5 }.
8 R. Sridevi, S.Navaneethakrishnan and K.Nagarajan For finding the edge labeling between the end vertex u n 4 of the ( n starting vertex u n of the ( n ) rd loop, let E (1) n 1 = {f (u n 4 u n )}. Then E (1) n 1 = { f(u n 4) f(u n ) } = { F m+4 F m+5 } = {F m+ }. For l = n, let E n = {f (u i+ u i+ ) : n 5 i n 4}. Then E n Let E 1 = {f (u n 1 u n )}. Then = { f(u i+ ) f(u i+ ) : n 5 i n 4} = { f(u n ) f(u n ), f(u n ) f(u n 1 ) } = { F m+5 F m+, F m+ F m+1 ) } = {F m+4, F m+ }. E 1 = { f(u n 1) f(u n ) } = { F m+1 F 0 } = {F m+1 }. Therefore, ( ) ( E = E 1 E E n = {,,..., F m+n }. E (1) 4 ) (1) (1) E 5 E n E 1 1 1) th loop and Thus, all edge labels are distinct. Therefore, the graph G = C n K 1,m admits super Fibonacci graceful labeling. Whence, it is a super Fibonacci graceful graph. Example.10 This example shows that the graph C 6 K 1,4 is a super Fibonacci graceful graph. F 9 C 6 K 1,4 : F 9 0 F 4 F 0 F 0 F 4 F Fig.11 Definition.11 G = K 1,n K 1, is a graph in which K 1, is joined with each pendant vertex of K 1,n. Theorem.1 The graph G = K 1,n K 1, is a super Fibonacci graceful graph.
Super Fibonacci Graceful Labeling 9 Proof Let {u 0, u 1, u,...,u n } be the vertex set of K 1,n and v 1, v,..., v n and w 1, w,..., w n be the vertices joined with the pendant vertices u 1, u,...,u n of K 1,n respectively. Also, V (G) = n + 1 and E(G) = n. Define f : V (G) {F 0,,,..., F q } by f(u 0 ) = F 0, f(u i ) = F n (i 1), 1 i n, f(v i ) = F n 1 (i 1), 1 i n, f(w i ) = F n (i 1), 1 i n. We claim that the edge labels are distinct. Calculation shows that E 1 = {f (u 0 u i ) : i = 1,,...,n} = { f(u 0 ) f(u i ) : i = 1,,...n} = { f(u 0 ) f(u 1 ), f(u 0 ) f(u ),..., f(u 0 ) f(u n 1 ), f(u 0 ) f(u n ) } = { F 0 F n, F 0 F n,..., F 0, F 0 F } = {F n, F n,...,, F }, E = {f (u i v i ) : i = 1,,..., n} = { f(u i ) f(v i ) : i = 1,,...,n} = { f(u 1 ) f(v 1 ), f(u ) f(v ),..., f(u n 1 ) f(v n 1 ), = f(u n ) f(v n ) } = { F n F n 1, F n F n 4,...,, F } = {F n, F n 5,..., F 4, }, E = {f (u i w i ) : i = 1,,..., n} Therefore, = { f(u i ) f(w i ) : i = 1,,...,n} = 1{ f(u 1 ) f(w 1 ), f(u ) f(w ),..., f(u n 1 ) f(w n 1 ), f(u n ) f(w n ) } = { F n F n, F n F n 5,..., F 4, F } = {F n 1, F n 4,...,, }. E = E 1 E E = {,,...,F n }. Thus, all edge labels are distinct. Therefore, K 1,n K 1, admits super Fibonacci graceful labeling. Whence, it is a super Fibonacci graceful graph. Example.1 This example shows that the graph K 1, K 1, is a super Fibonacci graceful graph. F 0 F 9 F K 1, K 1, : F 9 F F 4 F 4 Fig.1
40 R. Sridevi, S.Navaneethakrishnan and K.Nagarajan References [1] G.J.Gallian, A Dynamic survey of graph labeling, The electronic Journal of Combinotorics, 16(009), #DS6, PP 19. [] Henry Ibstedt, Surfing on the Ocean of Numbers a Few Smarandache Notions and Similar Topics, Ethus University Press, Vail 1997. [] Ian Anderson, A First Course in Combinatorial Mathematics, Claridon Press-Oxford, 8(1989) -1. [4] K.M.Kathiresan and S.Amutha, The Existence and Construction of Certain Types of Labelings for Graphs, PhD. thesis, Madurai Kamaraj University, October 006. [5] A.Rosa, On certain valuations of the vertices of a graph, Theory of Graphs (International Symposium, Rome), July (1966).