International J.Math. Combin. Vol.1 (2011), 59-72 Super Fibonacci Graceful Labeling of Some Special Class of Graphs R.Sridevi 1, S.Navaneethakrishnan 2 and K.Nagarajan 1 1. Department of Mathematics, Sri S.R.N.M.College, Sattur - 626 20, Tamil Nadu, India 2. Department of Mathematics, V.O.C.College, Tuticorin - 628 008, Tamil Nadu, India E-mail: r.sridevi 2010@yahoo.com, snk.voc@gmail.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 2), 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 2006. 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(2),..., 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(2),..., 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 show that some special class of graphs namely Fn, t Cn t and Sm,n t 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(2010): 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 [6] 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, 2,..., 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 [5]. We call a function f, a fibonacci graceful labeling of a graph G with q edges if f is an injection from the vertices of G to the set {0, 1, 2,..., F q }, where 1 Received November 1, 2010. Accepted February 25, 2011.
60 R.Sridevi, S.Navaneethakrishnan and K.Nagarajan F q is the q th fibonacci number of the fibonacci series F 1 = 1,F 2 = 2,F =, F 4 = 5,..., such that each edge uv is assigned the labels f(u) f(v), the resulting edge labels are F 1, F 2,...,...F q. An injective function f : V (G) {F 0, F 1,..., 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 1, F 2,..., F q }. In the labeling problems the induced labelings must be distinct. So to introduce fibonacci graceful labelings we assume F 1 = 1, F 2 = 2, F =, F 4 = 5,..., as the sequence of fibonacci numbers instead of 0, 1, 2, []. Generally, a Smarandache-Fibonacci Triple is a sequence S(n), n 0 such that S(n) = S(n 1) + S(n 2), where S(n) is the Smarandache function for integers n 0 ([2]). 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(2),..., S(q)} such that the induced edge labeling f (uv) = f(u) f(v) is a bijection onto the set {S(1), S(2),...,S(q)}. So a super Fibonacci graceful graph is a special type of Smarandache-Fibonacci graceful graph by definition. We have constructed some new types of graphs namely F n K + 1,m, C n P m, K 1,n K 1,2, F n P m and C n K 1,m and we proved that these graphs are super fibonacci graceful labeling in [7]. In this paper, we prove that F t n, C t n and S t m,n are super fibonacci graceful graphs. 2. Main Results In this section, we show that some special class of graphs namely F t n, C t n and S t m,n are super fibonacci graceful graphs. Definition 2.1 Let G be a (p, q) graph. An injective function f : V (G) {F 0, F 1, F 2,, 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 1, F 2,...,F q }. Definition 2.2 The one point union of t copies of fan F n is denoted by F t n. The following theorem shows that the graph Fn t is a super fibonacci graceful graph. Theorem 2. F t n is a super fibonacci graceful graph for all n 2. Proof Let u 0 be the center vertex of Fn t and u j i, where i = 1, 2,...,t, j = 1, 2,..., n be the other vertices of Fn t. Also, V (G) = nt + 1 and E(G) = 2nt t. Define f : V (F n t) {F 0, F 1,...,F q } by f(u 0 ) = F 0, f(u j 1 ) = F 2j 1, 1 j n. For i = 2,,..., t, f(u j i ) = F 2n(i 1)+2(j 1) (i 2), 1 j n. We claim that all these edge labels are distinct. Let E 1 = {f (u 0 u j 1 ) : 1 j n}. Then E 1 = { f(u 0 ) f(u j 1 ) : 1 j n} = { f(u 0 ) f(u 1 1 ), f(u 0) f(u 2 1 ),..., f(u 0) f(u n 1 1 ), f(u 0 ) f(u n 1 ) } = { F 0 F 1, F 0 F,..., F 0 F 2n, F 0 F 2n 1 } = {F 1, F,...,F 2n, F 2n 1 }.
Super Fibonacci Graceful Labeling of Some Special Class of Graphs 61 Let E 2 = {f (u j 1 uj+1 1 ) : 1 j n 1}. Then E 2 = { f(u j 1 ) f(uj+1 1 ) : 1 j n 1} = { f(u 1 1 ) f(u2 1 ), f(u2 1 ) f(u 1 ),..., f(un 2 1 ) f(u n 1 1 ), f(u n 1 1 ) f(u n 1 ) } = { F 1 F, F F 5,..., F 2n 5 F 2n, F 2n F 2n 1 } = {F 2, F 4,..., F 2n 4, F 2n 2 }. For i = 2, we know that E = {f (u 0 u j 2 ) : 1 j n} = { f(u 0) f(u j 2 ) : 1 j n} = { f(u 0 ) f(u 1 2), f(u 0 ) f(u 2 2),..., f(u 0 ) f(u n 1 2 ), f(u 0 ) f(u n 2) } = { F 0 F 2n, F 0 F 2n+2,..., F 0 F 4n 4, F 0 F 4n 2 } = {F 2n, F 2n+2,..., F 4n 4, F 4n 2 }, E 4 = {f (u j 2 uj+1 2 ) : 1 j n 1} = { f(u j 2 ) f(uj+1 2 ) : 1 j n 1} = { f(u 1 2) f(u 2 2), f(u 2 2) f(u 2),..., f(u n 2 2 ) f(u n 1 2 ), f(u n 1 2 ) f(u n 2) } = { F 2n F 2n+2, F 2n+2 F 2n+4,..., F 4n 6 F 4n 4, F 4n 4 F 4n 2 } = {F 2n+1, F 2n+,..., F 4n 5, F 4n }. For i =, let E 5 = {f (u 0 u j ) : 1 j n}. Then E 5 = { f(u 0 ) f(u j ) : 1 j n} = { f(u 0 ) f(u 1 ), f(u 0) f(u 2 ),..., f(u 0) f(u n 1 ), f(u 0 ) f(u n ) } = { F 0 F 4n 1, F 0 F 4n+1,..., F 0 F 6n 5, F 0 F 6n } = {F 4n 1, F 4n+1,..., F 6n 5, F 6n }. Let E 6 = {f (u j uj+1 ) : 1 j n 1}. Then E 6 = { f(u j ) f(uj+1 ) : 1 j n 1} = { f(u 1 ) f(u2 ), f(u2 ) f(u ),..., f(un 2 ) f(u n 1 ), f(u n 1 ) f(u n ) } = { F 4n 1 F 4n+1, F 4n+1 F 4n+,..., F 6n 7 F 6n 5, F 6n 5 F 6n } = {F 4n, F 4n+2,..., F 6n 6, F 6n 4 }, Now, for i = t 1, let E t 1 = {f (u 0 u j t 1 ) : 1 j n}. Then E t 1 = { f(u 0 ) f(u j t 1 ) : 1 j n} = { f(u 0 ) f(u 1 t 1 ), f(u 0) f(u 2 t 1 ),..., f(u 0) f(u n 1 t 1 ), f(u 0) f(u n t 1 ) } = { F 0 F 2nt 4n t+, F 0 F 2nt 4n t+5,..., F 0 F 2nt 2n t 1, F 0 F 2nt 2n t+1 } = {F 2nt 4n t+, F 2nt 4n t+5,..., F 2nt 2n t 1, F 2nt 2n t+1 }.
62 R.Sridevi, S.Navaneethakrishnan and K.Nagarajan Let E t 1 = {f (u j t 1 uj+1 t 1 ) : 1 j n 1}. Then E t 1 = { f(u j t 1 ) f(uj+1 t 1 ) : 1 j n 1} = { f(u 1 t 1) f(u 2 t 1), f(u 2 t 1) f(u t 1),..., f(u n 2 t 1 ) f(un 1 t 1 ), f(un 1 t 1 ) f(un t 1 ) } = { F 2nt 4n t+ F 2nt 4n t+5, F 2nt 4n t+5 F 2nt 4n t+7,..., F 2nt 2n t F 2nt 2n t 1, F 2nt 2n t 1 F 2nt 2n t+1 } = {F 2nt 4n t+4, F 2nt 4n t+6,..., F 2nt 2n t 2, F 2nt 2n t }. F 4 4 : F 22 F 2 F 24 F 25 F 26 F 27 F 28 F 26 F 22 F 24 F 28 F 21 F 20 F 19 F 21 F 19 F 0 F 1 F F 1 F 2 F F 18 F 17 F 17 F 5 F 4 F 16 F 15 F 7 F 5 F 15 F 14 F 12 F 10 F 8 F 6 F 7 F 14 F 1 F 12 F 11 F 10 F 9 F 8 Fig.1 For i = t, let E t = {f (u 0 u j t ) : 1 j n}. Then E t = { f(u 0 ) f(u j t ) : 1 j n} = { f(u 0 ) f(u 1 t), f(u 0 ) f(u 2 t),..., f(u 0 ) f(u n 1 t ), f(u 0 ) f(u n t ) } = { F 0 F 2nt 2n t+2, F 0 F 2nt 2n t+4,..., F 0 F 2nt t 2, F 0 F 2nt t } = {F 2nt 2n t+2, F 2nt 2n t+4,..., F 2nt t 2, F 2nt t }.
Super Fibonacci Graceful Labeling of Some Special Class of Graphs 6 Let E t = {f (u j t uj+1 t ) : 1 j n 1}. Then E t = { f(u j t) f(u j+1 t ) : 1 j n 1} = { f(u 1 t ) f(u2 t ), f(u2 t ) f(u t ),..., f(un 2 t = { F 2nt 2n t+2 F 2nt 2n t+4, F 2nt 2n t+4 F 2nt 2n t+6,..., F 2nt t 4 F 2nt t 2, F 2nt t 2 F 2nt t } = {F 2nt 2n t+, F 2nt 2n t+5,..., F 2nt t, F 2nt t 1 }. ) f(u n 1 t ), f(u n 1 t ) f(u n t ) } Therefore, E = E 1 E 2,..., E t 1 E t = {F 1, F 2,..., F 2nt t } Thus, the edge labels are distinct. Therefore, Fn t admits super fibonacci graceful labeling. For example the super fibonacci graceful labeling of F 4 4 is shown in Fig.1. Definition 2.4 The one point union of t cycles of length n is denoted by C t n. Theorem 2.5 C t n is a super fibonacci graceful graph for n 0(mod). Proof Let u 0 be the one point union of t cycles and u 1, u 2,...,u t(n 1) be the other vertices of C t n. Also, V (G) = t(n 1) + 1, E(G) = nt. Define f : V (C t n) {F 0, F 1,...,F q } by f(u 0 ) = F 0. For i = 1, 2,..., t, f(u (n 1)(i 1)+(j 1)+1 ) = F nt n(i 1) 2(j 1), 1 j 2. For s = 1, 2,, n, i = 1, 2,..., t, f(u (n 1)(i 1)+j) = F nt 1 n(i 1) 2(j s 2)+(s 1), s j s + 2. Next, we claim that the edge labels are distinct. We find the edge labeling between the vertex u 0 and starting vertex of each copy of (u (n 1)(i 1)+1 ). Let E 1 = {f (u 0 u (n 1)(i 1)+1 ) : 1 i t}. Then E 1 = { f(u 0 ) f(u (n 1)(i 1)+1 ) : 1 i t} = { f(u 0 ) f(u 1 ), f(u 0 ) f(u n ),..., f(u 0 ) f(u nt 2n t+ ), f(u 0 ) f(u nt n t+2 ) } = { F 0 F nt, F 0 F nt n,..., F 0 F 2n, F 0 F n } = {F nt, F nt n,...,f 2n, F n } Now we determine the edge labelings between the vertex u (n 1)(i 1)+1 and the vertex u (n 1)(i 1)+2 of each copy. Let E 2 = {f (u (n 1)(i 1)+1 u (n 1)(i 1)+2 ) : 1 i t}. Then E 2 = { f(u (n 1)(i 1)+1 ) f(u (n 1)(i 1)+2 ) : 1 i t} = { f(u 1 ) f(u 2 ), f(u n ) f(u n+1 ),..., f(u nt 2n t+ ) f(u nt 2n t+4 ), f(u nt n t+2 ) f(u nt n t+ ) } = { F nt F nt 2, F nt n F nt n 2,..., F 2n F 2n 2, F n F n 2 } = {F nt 1, F nt n 1,..., F 2n 1, F n 1 } We calculate the edge labeling between the vertex u (n 1)(i 1)+2 and starting vertex u (n 1)(i 1)+
64 R.Sridevi, S.Navaneethakrishnan and K.Nagarajan of the first loop. Let E = {f (u (n 1)(i 1)+2 u (n 1)(i 1)+ ) : 1 i t}. Then E = { f(u (n 1)(i 1)+2 ) f(u (n 1)(i 1)+ ) : 1 i t} = { f(u 2 ) f(u ), f(u n+1 ) f(u n+2 ), f(u 2n ) f(u 2n+1 ),..., f(u nt 2n t+4 ) f(u nt 2n t+5 ), f(u nt n t+ ) f(u nt n t+4 ) } = { F nt 2 F nt 1, F nt n 2 F nt n 1, F nt 2n 2 F nt 2n 1,..., F 2n 2 F 2n 1, F n 2 F n 1 } = {F nt, F nt n, F nt 2n,...,F 2n, F n }. Now, for s = 1, let E 4 = t {f (u (n 1)(i 1)+j u (n 1)(i 1)+j+1 ) : j 4}. Then i=1 E 4 = t i=1{ f(u (n 1)(i 1)+j ) f(u (n 1)(i 1)+j+1 ) : j 4} = { f(u ) f(u 4 ), f(u 4 ) f(u 5 ) } { f(u n+2 ) f(u n+ ), f(u n+ ) f(u n+4 ) },..., { f(u nt 2n t+5 ) f(u nt 2n t+6 ), f(u nt 2n t+6 ) f(u nt 2n t+7 ) } { f(u nt n t+4 ) f(u nt n t+5 ), f(u nt n t+5 ) f(u nt n t+6 ) } = { F nt 1 F nt, F nt F nt 5 } { F nt n 1 F nt n, F nt n F nt n 5 },..., { F 2n 1 F 2n, F 2n F 2n 5 } { F n 1 F n, F n F n 5 } = {F nt 2, F nt 4 } {F nt n 2, F nt n 4 },..., {F 2n 2, F 2n 4 } {F n 2, F n 4 } For the edge labeling between the end vertex (u (n 1)(i 1)+5 ) of the first loop and starting vertex (u (n 1)(i 1)+6 ) of the second loop, calculation shows that E4 1 = { f(u (n 1)(i 1)+5 ) f(u (n 1)(i 1)+6 ) : 1 i t} = { f(u 5 ) f(u 6 ), f(u n+4 ) f(u n+5 ),..., f(u nt 2n t+7 ) f(u nt 2n t+8 ), f(u nt n t+6 ) f(u nt n t+7 ) } = { F nt 5 F nt 4, F nt n 5 F nt n 4,..., F 2n 5 F 2n 4, F n 5 F n 4 } = {F nt 6, F nt n 6,..., F 2n 6, F n 6 } For s = 2, let E 5 = t i=1 {f (u (n 1)(i 1)+j u (n 1)(i 1)+j+1 ) : 6 j 7}. Then
Super Fibonacci Graceful Labeling of Some Special Class of Graphs 65 E 5 = t i=1{ f(u (n 1)(i 1)+j ) f(u (n 1)(i 1)+j+1 ) : 6 j 7} = { f(u 6 ) f(u 7 ), f(u 7 ) f(u 8 ) } { f(u n+5 ) f(u n+6 ), f(u n+6 ) f(u n+7 ) },..., f(u nt 2n t+8 ) f(u nt 2n t+9 ), f(u nt 2n t+9 ) f(u nt 2n t+10 ) } { f(u nt n t+7 ) f(u nt n t+8 ), f(u nt n t+8 ) f(u nt n t+9 ) } = { F nt 4 F nt 6, F nt 6 F nt 8 } { F nt n 4 F nt n 6, F nt n 6 F nt n 8 },..., { F 2n 4 F 2n 6, F 2n 6 F 2n 8 } { F n 4 F n 6, F n 6 F n 8 } = {F nt 5, F nt 7 } {F nt n 5, F nt n 7 },..., {F 2n 5, F 2n 7 } {F n 5, F n 7 } Similarly, for finding the edge labeling between the end vertex (u (n 1)(i 1)+8 ) of the second loop and starting vertex (u (n 1)(i 1)+9 ) of the third loop, calculation shows that For s = n Then E 1 5 = { f(u (n 1)(i 1)+8 ) f(u (n 1)(i 1)+9 ) : 1 i t} = { f(u 8 ) f(u 9 ), f(u n+7 ) f(u n+8 ),..., f(u nt 2n t+10 ) f(u nt 2n t+11 ), f(u nt n t+9 ) f(u nt n t+10 ) } = { F nt 8 F nt 7, F nt n 8 F nt n 7,..., F 2n 8 F 2n 7, F n 8 F n 7 } = {F nt 9, F nt n 9,...,F 2n 9, F n 9 },, 1, let E n 1 = t i=1 {f (u (n 1)(i 1)+j u (n 1)(i 1)+j+1 ) : n 6 j n 5}. E n 1 = t i=1{ f(u (n 1)(i 1)+j ) f(u (n 1)(i 1)+j+1 ) : n 6 j n 5} = { f(u n 6 ) f(u n 5 ), f(u n 5 ) f(u n 4 ) } { f(u 2n 7 ) f(u 2n 6 ), f(u 2n 6 ) f(u 2n 5 ) },..., { f(u nt n t 4 ) f(u nt n t ), f(u nt n t ) f(u nt n t 2 ) } { f(u nt t 5 ) f(u nt t 4 ), f(u nt t 4 ) f(u nt t ) } = { F nt n+8 F nt n+6, F nt n+6 F nt n+4 } { F nt 2n+8 F nt 2n+6, F nt 2n+6 F nt 2n+4 },..., { F n+8 F n+6, F n+6 F n+4 } { F 8 F 6, F 6 F 4 } = {F nt n+7, F nt n+5 } {F nt 2n+7, F nt 2n+5 },..., {F n+7, F n+5 } {F 7, F 5 }
66 R.Sridevi, S.Navaneethakrishnan and K.Nagarajan We calculate the edge labeling between the end vertex (u (n 1)(i 1)+n 4 ) of the ( n 1) th loop and starting vertex (u (n 1)(i 1)+n ) of the ( n )rd loop as follows. E 1 n 1 = { f(u (n 1)(i 1)+n 4 ) f(u (n 1)(i 1)+n ) : 1 i t} = { f(u n 4 ) f(u n ), f(u 2n 5 ) f(u 2n 4 ),..., f(u nt n t+2 ) f(u nt n t 1 ), f(u nt t ) f(u nt t 2 ) } = { F nt n+6 F nt n+5, F nt 2n+4 F nt 2n+5,..., F n+4 F n+5, F 4 F 5 } = {F nt n+4, F nt 2n+,...,F n+, F } For s = n, let E n Then = t {f (u (n 1)(i 1)+j u (n 1)(i 1)+j+1 ) : n j n 2}. i=1 E n = t i=1{ f(u (n 1)(i 1)+j ) f(u (n 1)(i 1)+j+1 ) : n j n 2} = { f(u n ) f(u n 2 ), f(u n 2 ) f(u n 1 ) } { f(u 2n 4 ) f(u 2n ), f(u 2n ) f(u 2n 2 ) },..., f(u nt n t 1 ) f(u nt n t ), f(u nt n t ) f(u nt n t+1 ) } { f(u nt t 2 ) f(u nt t 1 ), f(u nt t 1 ) f(u nt t ) } = { F nt n+5 F nt n+, F nt n+ F nt n+1 } { F nt 2n+5 F nt 2n+, F nt 2n+ F nt 2n+1 },..., { F n+5 F n+, F n+ F n+1 } { F 5 F, F F 1 } = {F nt n+4, F nt n+2 } {F nt 2n+4, F nt 2n+2 },..., {F n+4, F n+2 } {F 4, F 2 } Calculation shows the edge labeling between the end vertex (u (n 1)(i 1)+n 1 ) of the ( n )rd loop and the vertex u 0 are E = { f(u (n 1)(i 1)+n 1 ) f(u 0 ) : 1 i t} = { f(u n 1 ) f(u 0 ), f(u 2n 2 ) f(u 0 ),..., f(u nt n t+1 ) f(u 0 ), f(u nt t ) f(u 0 ) } = { F nt n+1 F 0, F nt 2n+1 F 0,..., F n+1 F 0, F 1 F 0 } = {F nt n+1, F nt 2n+1,...,F n+1, F 1 }. Therefore, E = (E 1 E 2,..., E n ) (E4 1 E5, 1..., E 1 n 1) E = {F 1, F 2,...,F nt } Thus, the edge labels are distinct. Therefore, Cn t admits a super fibonacci graceful labeling.
Super Fibonacci Graceful Labeling of Some Special Class of Graphs 67 C 2 9 : F 6 F 7 F 8 F 6 F 7 F 8 F 9 F 10 F 11 F 12 F 1 F 14 F 12 F 1 10 F9F F 5 F 0 F 14 F 18 F 1 F 4 F F 5 F 4 F F 2 F 1 F 18 F 17 F 16 F 17 F 15 F 16 F 15 Fig.2 For example the super fibonacci graceful labeling of C9 2 is shown in Fig.2. Definition 2.6([4]) Let S m,n stand for a star with n spokes in which each spoke is a path of length m. Definition 2.7 The one point union of t copies of S m,n is denoted by S t m,n. Next theorem shows that the graph Sm,n t is a super Fibonacci graceful graph. Theorem 2.8 Sm,n t is a super fibonacci graceful graph for all m, n, when n 1(mod). Proof Let v 0 be the center of the star and vj i, i = 1, 2,...,mt, j = 1, 2,...,n be the other vertices of Sm,n t. Also, V (G) = mnt + 1 and E(G) = mnt. Define f : V (St m,n ) {F 0, F 1,...,F q } by f(v 0 ) = F 0. For i = 1, 2,...,mt, f(vj i) = F mnt n(i 1) 2(j 1), 1 j 2. For i = 1, 2,..., mt, F(vj i) = F mnt (2j n 1) n(i 1), n 1 j n. For s = 1, 2,..., n 4, i = 1, 2,...,mt f(vj i) = F mnt 1 n(i 1) 2(j s 2)+(s 1), s j s+2. We claim that all these edge labels are distinct. Let E 1 = {f (v 0 v1 i ) : 1 i mt}. Calculation shows that E 1 = { f(v 0 ) f(v1 i ) : 1 i mt} = { f(v 0 ) f(v 1 1), f(v 0 ) f(v 2 1),..., f(v 0 ) f(v mt 1 1 ), f(v 0 ) f(v mt 1 ) } = { F 0 F mnt, F 0 F mnt n,..., F 0 F 2n, F 0 F n } = {F mnt, F mnt n,..., F 2n, F n }. Let E 2 = {f (v i 1v i 2) : 1 i mt}. Then E 2 = { f(v1 i ) f(vi 2 ) : 1 i mt} = { f(v 1 1) f(v 1 2), f(v 2 1) f(v 2 2),..., f(v mt 1 1 ) f(v mt 1 2 ), f(v mt 1 ) f(vmt 2 ) } = { F mnt F mnt 2, F mnt n F mnt n 2,..., F 2n F 2n 2, F n F n 2 } = {F mnt 1, F mnt n 1,..., F 2n 1, F n 1 }
68 R.Sridevi, S.Navaneethakrishnan and K.Nagarajan For the edge labeling between the vertex v2 i and starting vertex vi E = {f (v2v i ) i : 1 i mt}. Calculation shows that of the first loop, let E = { f(v2 i ) f(vi ) : 1 i mt} = { f(v2 1 ) f(v1 ), f(v2 2 ) f(v2 ),..., f(vmt 1 2 ) f(v mt 1 ), f(v mt 2 ) f(v mt ) } = { F mnt 2 F mnt 1, F mnt n 2 F mnt n 1,..., F 2n 2 F 2n 1, F n 2 F n 1 } = {F mnt, F mnt n,...,f 2n, F n }. For s = 1, let E 4 = mt i=1 {f (vj ivi j+1 ) : j 4}. Then E 4 = mt i=1 { f(vi j ) f(vi j+1 ) : j 4} = { f(v 1 ) f(v 1 4), f(v 1 4) f(v 1 5) } { f(v 2 ) f(v 2 4), f(v 2 4) f(v 2 5) },..., { f(v mt 1 ) f(v mt 1 4 ), f(v mt 1 4 ) f(v mt 1 5 ) } { f(v mt ) f(v mt 4 ), f(v mt 4 ) f(v mt 5 ) } = { F mnt 1 F mnt, F mnt F mnt 5 } { F mnt n 1 F mnt n, F mnt n F mnt n 5 },..., { F 2n 1 F 2n, F 2n F 2n 5 } { F n 1 F n, F n F n 5 } = {F mnt 2, F mnt 4 } {F mnt n 2, F mnt n 4 },..., {F 2n 2, F 2n 4 } {F n 2, F n 4 }. We find the edge labeling between the vertex v5 i of the first loop and starting vertex v6 i of the second loop. Let E4 1 = {f (v5 ivi 6 ) : 1 i mt}. Then E4 1 = { f(v5 i ) f(vi 6 ) : 1 i mt} = { f(v 1 5) f(v 1 6), f(v 2 5) f(v 2 6),..., f(v mt 1 5 ) f(v mt 1 6 ), f(v mt 5 ) f(vmt 6 ) } = { F mnt 5 F mnt 4, F mnt n 5 F mnt n 4,..., F 2n 5 F 2n 4, F n 5 F n 4 } = {F mnt 6, F mnt n 6,..., F 2n 6, F n 6 }. For s = 2, let E 5 = mt i=1 {f (vj ivi j+1 ) : 6 j 7}. Then
Super Fibonacci Graceful Labeling of Some Special Class of Graphs 69 E 5 = mt i=1{ f(v i j) f(v i j+1) : 6 j 7} = { f(v 1 6 ) f(v1 7 ), f(v1 7 ) f(v1 8 ) } { f(v2 6 ) f(v2 7 ), f(v2 7 ) f(v2 8 ) },...,...,{ f(v mt 1 6 ) f(v mt 1 7 ), f(v mt 1 7 ) f(v mt 1 8 ) } { f(v mt 6 ) f(v mt 7 ), f(v mt 7 ) f(v mt 8 ) } = { F mnt 4 F mnt 6, F mnt 6 F mnt 8 } { F mnt n 4 F mnt n 6, F mnt n 6 F mnt n 8 },..., { F 2n 4 F 2n 6, F 2n 6 F 2n 8 } { F n 4 F n 6, F n 6 F n 8 } = {F mnt 5, F mnt 7 } {F mnt n 5, F mnt n 7 },..., {F 2n 5, F 2n 7 } {F n 5, F n 7 }. Let E5 1 = {f (v8 ivi 9 ) : 1 i mt}. Calculation shows that the edge labeling between the vertex v8 i of the second loop and starting vertex vi 9 of the third loop are For s = n 4 E5 1 = { f(v8 i ) f(vi 9 ) : 1 i mt} = { f(v 1 8) f(v 1 9), f(v 2 8) f(v 2 9),..., f(v mt 1 8 ) f(v mt 1 9 ), f(v mt 8 ) f(vmt 9 ) } = { F mnt 8 F mnt 7, F mnt n 8 F mnt n 7,..., F 2n 8 F 2n 7, F n 8 F n 7 } = {F mnt 9, F mnt n 9,..., F 2n 9, F n 9 }. 1, let E n 4 1 = mt i=1 {f (vj ivi j+1 ) : n 7 j n 6}. Then E n 4 1 = mt i=1{ f(vj) i f(vj+1) i : n 7 j n 6} = { f(vn 7 1 ) f(v1 n 6 ), f(v1 n 6 ) f(v1 n 5 ) } { f(v 2 n 7) f(v 2 n 6), f(v 2 n 6) f(v 2 n 5) },..., { f(vn 7 mt 1 ) f(vmt 1 n 6 ), f(vmt 1 { f(vn 7 mt ) f(vmt n 6 ), f(vmt n 6 ) f(vmt n 5 ) } n 6 ) f(vmt 1 = { F mnt n+9 F mnt n+7, F mnt n+7 F mnt n+5 } n 5 ) } { F mnt 2n+9 F mnt 2n+7, F mnt 2n+7 F mnt 2n+5 },..., { F n+9 F n+7, F n+7 F n+5 } { F 9 F 7, F 7 F 5 } = {F mnt n+8, F mnt n+6 } {F mnt 2n+8, F mnt 2n+6 },..., {F n+8, F n+6 } {F 8, F 6 }. Similarly, for the edge labeling between the end vertex vn 5 i of the (n 4 1) th loop and starting vertex vn 4 i of the (n 4 )rd loop, let E 1 = {f n 4 (v i 1 n 5 vi n 4 ) : 1 i mt}. Calcula-
70 R.Sridevi, S.Navaneethakrishnan and K.Nagarajan tion shows that E 1 n 4 1 = { f(vn 5 i ) f(vi n 4 ) : 1 i mt} = { f(vn 5) 1 f(vn 4), 1 f(vn 5) 2 f(vn 4), 2..., f(v mt 1 Now for s = n 4, let E n 4 E n 4 n 5 ) f(vmt 1 n 4 ), f(vmt n 5 ) f(vmt n 4 ) } = { F mnt n+5 F mnt n+6, F mnt 2n+5 F mnt 2n+6,..., F n+5 F n+6, F 5 F 6 } = {F mnt n+4, F mnt 2n+4,..., F n+4, F 4 }. {f (vj ivi j+1 ) : n 4 j n }. Then = mt i=1 = mt i=1 { f(vi j ) f(vi j+1 ) : n 4 j n } = { f(v 1 n 4 ) f(v1 n ), f(v1 n ) f(v1 n 2 ) } { f(v 2 n 4) f(v 2 n ), f(v 2 n ) f(v 2 n 2) },..., { f(vn 4 mt 1 ) f(vmt 1 n ), f(vmt 1 n 4 ) f(vmt 1 n ) } { f(v mt n 4) f(v mt n ), f(v mt n ) f(v mt n 2) } = { F mnt n+6 F mnt n+4, F mnt n+4 F mnt n+2 } { F mnt 2n+6 F mnt 2n+4, F mnt 2n+4 F mnt 2n+2 },..., { F n+6 F n+4, F n+4 F n+2 } { F 6 F 4, F 4 F 2 } = {F mnt n+5, F mnt n+ } {F mnt 2n+5, F mnt 2n+ },..., {F n+5, F n+ } {F 5, F }. We find the edge labeling between the end vertex vn 2 i of the ( n 4 )rd loop and the vertex vn 1 i. Let E 1 = {f (vn 2 i vi n 1 ) : 1 i mt}. Then E1 = { f(vn 2 i ) f(vi n 1 ) : 1 i mt} = { f(v 1 n 2) f(v 1 n 1), f(v 2 n 2) f(v 2 n 1),..., f(v mt 1 n 2 ) f(vmt 1 n 1 ), f(v mt n 2 ) f(vmt n 1 ) } = { F mnt n+2 F mnt n+, F mnt 2n+2 F mnt 2n+,..., F n+2 F n+, F 2 F } = {F mnt n+1, F mnt 2n+1,...,F n+1, F 1 }. Let E 2 = {f (v i n 1v i n) : 1 i mt}. Then E 2 = { f(v i n 1) f(v i n) : 1 i mt} = { f(v 1 n 1 ) f(v1 n ), f(v2 n 1 ) f(v2 n ),..., f(v mt 1 n 1 ) f(vmt 1 n ), f(vn 1) mt f(vn mt ) } = { F mnt n+ F mnt n+1, F mnt 2n+ F mnt 2n+1,..., F n+ F n+1, F F 1 } = {F mnt n+2, F mnt 2n+2,..., F n+2, F 2 }.
Super Fibonacci Graceful Labeling of Some Special Class of Graphs 71 Therefore, E = (E 1 E 2,..., E n 4 ) (E4 1 E5, 1..., E 1 n 4 1) E 1 E2 = {F 1, F 2,...,F mnt } Thus, S t m,n admits a super fibonacci graceful labeling. For example the super fibonacci graceful labeling of S 2,7 is shown in Fig.. F 42 F 41 F 40 F 9 F 41 F 40 F 9 F 8 F 7 F 6 F 8 F 7 F 6 F 42 F 5 F 4 F F 2 F 4 F F 2 F 1 F 0 F 29 F 1 F 0 F 29 F 5 F 0 F 7 F 14 F 28 F 28 F 27 F 26 F 25 F 27 F 26 F 25 F 24 F 2 F 22 F 24 F 2 F 22 F 21 F 7 F 6 F 5 F 4 F 6 F 5 F 4 F F 2 F 1 F F 2 F 1 F 14 F 1 F 12 F 11 F 1 F 12 F 11 F 10 F 9 F 8 F 10 F 9 F 8 F 21 F 20 F 19 F 18 F 20 F 19 F 18 F 17 F 16 F 15 F 17 F 16 F 15 Fig. Theorem 2.9 The complete graph K n is a super fibonacci graceful graph if n. Proof Let {v 0, v 1,..., v n 1 } be the vertex set of K n. Then v i (0 i n 1) is adjacent to all other vertices of K n. Let v 0 and v 1 be labeled as F 0 and F q respectively. Then v 2 must be given F q 1 or F q 2 so that the edge v 1 v 2 will receive a fibonacci number F q 2 or F q 1. Therefore, the edges will receive the distinct labeling. Suppose not, Let v 0 and v 1 be labeled as F 1 and F q or F 0 and F q 2 respectively. Then v 2 must be given F q 1 or F q 2 so that the edges v 0 v 2 and v 1 v 2 will receive the same edge label F q 2, which is a contradiction by our definition. Hence, K n is super fibonacci graceful graph if n.
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