Government Arts College, Melur , Tamilnadu, India. pons

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1 Volume 118 No , ISSN: (printed version); ISSN: (on-line version) url: doi: /ijpam.v118i10.53 ijpam.eu ODD VERTEX MAGIC TOTAL LABELING OF SOME GRAPHS CT. NAGARAJ Research Scholar, Department of Mathematics Research and Development centre Bharathiar University, Coimbatore , Tamilnadu, India. ct nagaraj@yahoo.co.in C.Y. PONNAPPAN 1, G. PRABAKARAN 1 Department of Mathematics Government Arts College, Melur , Tamilnadu, India. pons mdu1969@yahoo.com Department of Mathematics Thiagarajar College, Madurai , Tamilnadu, India. tcmaths@yahoo.in Abstract. Let G be Considered as a simple finite graph with n vertices and e edges. A vertex magic total labeling is a bijection f from V (G) E(G) to the consecutive integers 1,, 3,..., m + n with the property that for every u V (G), f(u) + f(uv) = k for some constant k. This paper, has introduced v N(u) a new concept of odd vertex magic total labeling of a graph. Some of the basic properties of labelings, to find some of the families of graphs that admit an odd vertex magic total labelings are studied here. It shows that other families of graphs do not, include it. Key words: Odd vertex magic labeling, odd vertex magic graph. 010 Mathematics Subject Classification: 05C Introduction All the graphs considered in this paper are just finite, simple and undirected. The graph G has a vertex set V = V (G) and edge set E = E(G) and let n = V and m = E.The degree of a vertex v is equal to the number of edges that have v as an end point and the set of neighbours of v are represented by N(v). Labeling of a graph G is a mapping that carries a set of graph elements, usually marked as vertices and/or edges into a set of numbers. Generally integers, 1 97

2 are called labels. Many kinds of labeling have been studied and an excellent survey of graph labeling can be found in [1]. MacDougall et al. [] introduced the concept of vertex magic total labeling. If type of graph G is a finite simple undirected graph with n vertices and m edges, then a vertex magic total labeling (V MT L) is a bijection f from V (G) E(G) the integers 1,, 3,..., m + n with the property that for every x in V (G), f(v) + f(uv) = k for some constant k.(which we call the magic constant for f). v N(u) MacDougall et al.[3] introduced the notion of super vertex magic total labeling. They call vertex magic total labeling as super if f(v (G)) = {1,, 3,..., n}.we could call it as V super vertex magic labeling. In this labeling, the smallest labels are assigned to the vertices. Swaminathan and Jeyanthi [6] introduced a concept with the same name of super vertex magic labeling. They call vertex magic total labeling as super if f(e(g)) = {1,, 3,..., m}. It may be noted that the smallest labels are assigned to the edges. MacDougall et al. [3] and Swaminathan and Jeyanthi [6] introduced different labelings with the same name of super vertex magic total labeling. To avoid a cofusion, Marimuthu and Balakrishan [4] called a vertex magic total labeling E-super if f(e(g)) = {1,, 3,..., m}. Note that the smallest labels are assigned to the edges. A graph G is called E-super vertex magic if it admits an E super vertex magic labeling. Nagaraj et al. [5] introduced the concept of an Even vertex magic total labeling. They call vertex magic total labeling as even if f(v (G)) = {, 4, 6,..., n}. In this labeling, the even integers are designated to the vertices. 98

3 3 Lemma 1.1. If G has a vertex magic total labeling then m n 3.. Definition and Main Results Definition.1. A vertex magic total labeling f is called an odd vertex magic total labeling if f(v (G)) = {1, 3, 5,..., n 1} A graph G could be called an odd vertex magic if there exists an odd verex magic total labeling of G. Example.. Figure 1: n=7, m=6, k=19 Theorem.3. Let G be a nontrivial graph.if G is an odd vertex magic, then the magic constant k is given by k = 1 + m + m n + m n. Proof. Let G be a nontrival graph. Let f be an odd vertex magic total labeling of a graph G with the magic number k. Then f(v ) = {1, 3, 5,..., n 1}. k = f(u) + f(uv) u V v N(u) nk = [ n 1] + [ m + n] [ n 1] k = 1 + m + m n + m n 99

4 4 Theorem.4. If a graph G has an odd vertex magic total labeling with constant k, then k 16n Proof. This result directly follows from Lemma 1.1 and Theorem.3. Theorem.5. Let G be any (n, m) graph. If G permits an odd vertex magic total labeling, then m n 1. Proof. This result directly follows from Definition.1. Theorem.6. A path P n is an odd vertax magic total labeling if and only if n is odd. Proof. Suppose an odd vertex magic labeling f of P n exists with the magic constant k. Then by Theorem.3, k = 3n To prove that n is odd. If n is even. Then k = 3n is considered as even. For any odd vertex magic labeling f, f(u) + f(uv) = k u V. v N(u) In particular if u is a pendent vertex of P n then,f(u) + f(uv) = k, which is a contradiction. Since f(u) is odd and f(uv) is even. Therefore n is odd. Converse: Let n be an odd integer. Let V (P n ) = {v 1, v, v 3,..., v n } and E(P n ) = {e i = v i v i+1 /1 i n 1}. Define f(v E) = {1,,..., n 1} as follows. n 1, if i = 1 f(v i ) = i 3, if i n n i if i is odd f(e i ) = n i if i is even and 100

5 It is clearly seen that f is an odd vertex magic total labeling with the magic constant 3n. Theorem.7. A cycle C n is an odd vertex magic iff n is odd. Proof. Suppose an odd vertex magic total labeling f of C n exists with the magic constant k. Then by theorem.3, k = 3n +. To prove that n is odd. Suppose n is even. Then k = 3n + is even. For any odd vertex magic total labeling f,f(u) + f(uv) = k, u V. v N(u) In particular f(v 1 ) + f(v 1 v n ) + f(v 1 v ) = k, which is a contradiction. Since f(v 1 ) is odd and f(v 1 v n ) and f(v 1 v ) are even. Therefore n is odd. Converse: Let n be an odd integer, V (C n ) = {v 1, v,..., v n } and E(C n ) = {v n v 1 } {v i v i+1 : 1 i n 1} Define f : V E {1,,..., n} as follows: 5 f(v i ) = n i + 1 if 1 i n, f(v i v i+1 ) = i + 1 if i is odd = n i if i is even f(v n v 1 ) = n + 1 It is easily seen that f is an odd vertex magic total labeling with the magic constant 3n+. Example.8. An odd vertex magic labeling of C 9 is given in figure. Figure : n = 9, m = 9, k = 9 101

6 6 3. ODD VERTEX MAGIC TOTAL LABELING ON A DISCONNECTED GRAPH This section an odd vertex magic total labeling for the disconnected graph rc s is given that is, the incoherent union of r copies of cycles of length s, where r and s are odd. Theorem 3.1. rc s is an odd vertex magic iff r and s are odd. Proof. Assume that rc s is odd vertex magic. Number of vertices of rc s is rs. Number of edges of rc s is rs. Then by Theorem.3, k = 3rs +. For any odd vertex magic total labeling f, k = f(u) + v N(u) f(uv) u V. Since any vertex of rc s is adjacent to only two edges, f(u) + f(e 1 ) + f(e ) = k, where e 1 and e are the edges adjacent to u. Here f(u) is an odd number and f(e 1 ) and f(e ) are even numbers. Therefore k = 3rs + is odd iff r and s are odd. Let r and s be odd integers. Assume that the graph rc s has the vertex set V = V 1 V V 3,..., V r, where V i = {Vi 1, Vi, Vi 3,..., Vi s, } and the edge set E = E 1 E E 3... E r where E i = {e 1 i, e i, e 3 i,..., e s i }, and e j i = vj i v(j+1) i for 1 i r,1 j s 1, e s i = vi s vi 1. 10

7 7 Define f(v E) = {1,, 3,..., rs} as follows: f(vi 1 ) = (rs i) + 1, i = 1,,..., r (rs r + i) 1, f(vi ) = (rs 3r + i) 1, For j = 3, 4, 5,..., s f(v j i ) = rs (j 1)r i, rs (j 3)r i, r i + 1, f(e 1 i ) = 3r i + 1, j =, 4, 6,..., s 1 1 i (r 1) (r+1) i r. 1 i (r 1) (r+1) i r. 1 i (r 1) (r+1) i r. f(e s i ) = rs i + + (j + 1)r, 1 i r. j = 3, 5, 7,..., s f(e j i ) = 4i + (j 1)r, 4i + (j 3)r, 1 i (r 1) (r+1) i r It is clearly verified that f is odd vertex magic labeling of rc s constant k = 3rs +. with the magic Example 3.. An odd vertex magic labeling of 5C 3 is given in figure. Figure 3: n = 15, m = 15, k =

8 8 4. KITE (s,t)-kite consists of a cycle of length s with a t edge path (the tail) attached to one vertex. Theorem 4.1. (s, t) kite graph admits an odd vertex magic labeling iff s + t is odd. Proof. Let G be a (s, t) kite graph. Let the vertex set V = {v 1, v, v 3,..., v s } {u 1, u, u 3,..., u t } and the edge set E = {e i = v i v i+1, e n = v s v 1, 1 i s 1} {x i = u i v i+1, x t = u t v 1 : 1 i t 1}. Hence n = m = s + t. Suppose G admits an odd vertex magic total labeling f with a vertex magic constant k. Then nk = f(u) + f(e). Hence k = 3(s + t) +. u V e E To prove that s + t is odd. Suppose s + t is even. Therefore k is even. If u is a pendent vertex of (s, t) kite, then f(u)+f(uv) = k, which is a contradiction. Since f(u) is odd and f(uv) is even.therefore s + t is odd. Conversely: Suppose s + t is odd. Hence either s or t is odd. We consider two cases. case(i): s is odd and t is even. Define f : V E {1,, 3,..., s + t} as follows, For 1 i s, t + i + 1, if i is odd f(e i ) = t + s i, if i is even F or1 i t, t + s + i +, if i is odd f(x i ) = i, if i is even 104

9 9 The vertex labeling as follows: f(v i ) = s i + 1, if 1 i s f(u i ) = s + t + 1 i, if 1 i t It could be easily verified that f k = 3(s + t) +. is an vertex magic labeling of G with Case(ii): S is even and t is odd. We consider two subcases. Subcase(i): t > s. For 1 i s s i + 1, if i is odd f(e i ) = 3s + t i + 1, if i is even and for 1 i t, 3s + t + i, if i = 1, 3, 5,..., t s f(x i ) = i (t s), if i = t s +, t s + 4,..., t s + i, if i is even The vertex labelings are as follows, f(u 1 ) = t + 1 t i s + 3, if i t s + 1 f(u i ) = 4t i + 3, if t s + i t f(v i ) = t + i s 1 for 1 i s It could be verified that f is an odd vertex magic total labeling of G with k = 3(s + t)

10 10 Subcase(ii): t s. For 1 i s t i, if i = 1, 3,..., t f(e i ) = 3t + s i, if i = t, t +,..., s 1 t + s i, if i =, 4,..., s t i, if i =, 4,..., t 1, 1 i t f(x i ) = 3t + s i, if i = 1, 3,..., t 1 i t The vertex labelings are as follows: f(v 1 ) = s t + 3 s + i + 1, if i t 1 f(v i ) = i t + 1, if t i s f(u 1 ) = s + 3 f(u i ) = s t + i + 1, if i t, It can be clearly proved that f is an odd vertex magic total labeling with k = 3(s + t) + Example 4.. Example of an odd vertex magic labeling of a kite graph with s = 5 and t = 8 is given in figure. Figure 4: n = 13, m = 13, k = 41 Example 4.3. An odd vertex magic total labeling of a kite graph with s = 4 and t = 9(t > s) and s = 8, t = 5(t < s) is given in figure. 106

11 11 Figure 5: n = 13, m = 13, k = 41 Figure 6: n = 13, m = 13, k = SUNS An n suns is a cycle c n with an edge finite in a vertex of degree 1 which attached to each vertex. Theorem 5.1. All n suns are not an odd vertex magic labeling. Proof. All n suns n vertices and n edges. For any odd vertex magic labeling f, f(u) + Σf(uv) = k, u V.f(u) is odd and each f(uv) is even. Therefore k is odd. Then by theorem.3, k = 6n + is even for any n which is a incompatibility to k is odd. Therefore all n suns are not an odd vertex magic labeling. 107

12 1 References [1] J.A. Gallian, A dynamic survey of graph labeling, Electron J.Combin. 16 (009) DS6. [] J.A. MacDougall, M.Miller, Slamin, W.D.Wallis, Vertex magic total labeling of graphs, Util.Math 61 (00) 3-1. [3] J.A.MacDougall, M.Miller,K.A.Sugeng,Super vertex magic total labelling of graphs, in: Proc.of the 15 t h Australian Workshop on Combinatorial Algorithms,004,pp.-9. [4] G.Marimuthu, M.Balakrishnan, E-Super vertex magic labelings of graphs, Discrete Applied Mathematics, 160, , 01. [5] CT.Nagaraj, C.Y.Ponnappan, G.Prabakaran, Even vertex magic total labeling, International Journal of Pure and Applied Mathematics, Volume 115,No.9,(017), [6] V.Swaminathan, P.Jeyanthi, Super vertex magic labeling, Indian J.Pure Appl.Math 34(6)(003)

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