ODD VERTEX MAGIC TOTAL LABELING OF TREES CT. NAGARAJ#1,C.Y. PONNAPPAN?1, G. PRABAKARAN#2 1

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1 ODD VERTEX MAGIC TOTAL LABELING OF TREES CT NAGARAJ#1,CY ONNAAN?1, G RABAKARAN# 1 Research Scholar, Department of Mathematics Research and Development centre Bharathiar University, Coimbatore , Tamilnadu, India ct nagaraj@yahoocoin Department of Mathematics Government Arts College, Melur , Tamilnadu, India pons mdu1969@yahoocom Department of Mathematics Thiagarajar College, Madurai , Tamilnadu, India tcmaths@yahooin Abstract Let G be a graph with vertex set V = V (G) and edge set E = E(G) and let m = E(G) and n = V (G) A one-to-one map f from V E onto the integers {1,,,, m + n} is called vertex magic total labeling there is a constant k so that for every vertex u, f (u) + f (uv) = k where the sum is over all vertices v adjacent to u Let us call the sum of labels at vertex u the weight wf (u) of the vertex under labeling f ; we require wf (u) = k for all u The constant k is called the magic constant for f Such a labeling is odd f (V (G)) = {1,, 5,, n 1} In this paper we present the odd vertex magic total labeling of trees Key words: Odd vertex magic labeling, Odd vertex magic graph, Tree 010 Mathematics Subject Classication: 05C78 age 74

2 1 Introduction will be the largest possible, and f assigns the largest integers m + 1, m +,, m + n All graphs considered in this paper are to the vertices, k will be the smallest finite, simple and undirected The graph G possible has vertex set V = V (G) and edge set E = E(G) and let n = V and MacDougall et al [4] introduced the m = E The degree of a vertex v is the notion of super vertex magic total labeling number of edges that have v as an end They call a vertex magic total labeling is point and the set of neighbours of v is super f (V (G)) = {1,,,, n} We call denoted by N (v) it as V Super vertex magic total labelingin this labeling, the smallest labels A one-to-one map are assigned to the vertices f (V E) = {1,,,, m + n} is a vertex Swaminathan and Jeyanthi [8] magic total labeling of G there is a introduced a concept with same name of constant k so that for every vertex super vertex magic labeling They call a u,wf (u) = f (u) + v N (u) f (uv) = kso vertex magic total labeling is super the magic requirement is the associated f (E(G)) = {1,,,, m} Note that the weight wf (u) = k for all u V (G) The smallest labels are assigned to the edges fixed integer k is called the magic constant They proved the following: n is super of f A graph is called vertex magic the vertex magic and only n is odd and graph has a vertex magic total labeling n ; Cn is super vertex magic and only Vertex magic total labeling first appeared n is odd; the star graph is super vertex in 00 in [] For more details of vertex magic and only it is ; and rcs is magic graphs see the book by Wallis [10], super vertex magic and only r and s and for other types of graphs labeling see are odd In [9], they proved the following: the dynamic survey by Gallian [] no super vertex magic total graph has two or more isolated vertices or an isolated Summing the weights of all vertices edge; a tree with s internal edges and st yields nk = u V wf (u) = Sn + Sm, leaves is not super vertex magic total where Sn is the sum of all vertex labels ; the graph obtained from a comb t > s+1 s and Sm is the sum of all edge labels This by attaching a pendant edge to each vertex observation always leads to upper and of degree is super vertex magic total; the lower bounds for k graph obtained by attaching a path with t edges to a vertex of an n cycle is super In articular, f assigns the smallest vertex magic total and only n + t is integers 1,,, n to the vertices, then k age 75

3 odd The use of the word super has introduce in [1] MacDougall et al [4] and Swaminathan and Jeyanthi [8] introduced dferent labelings with same name of super vertex magic total labeling To avoid cofusion, Marimuthu and Balakrishan [5] called a vertex magic total labeling is E-super f (E(G)) = {1,,,, m} Note that the smallest labels are assigned to the edges A graph G is called E-super vertex magic it admits an E super vertex magic labeling Theorem 1 A path n is an odd vertex magic total labeling and only n is odd Main Results Theorem 1 A star graph K1,r is odd vertex magic and only r = roof Let V (K1,r ) = {c, u1, u, ur } and E(K1,r ) = {cui : 1 i K} Nagaraj et al [6] introduced the Let f be an odd vertex magic labeling of concept of an Even vertex magic total K Then by 11, k = r +1 The minimum 1,r labeling They call a vertex magic total possible weight of c is r + r + 1 labeling is even ie wf (c) r + r + 1 f (V (G)) = {, 4, 6,, n} When r 6=, which is a contradiction to Nagaraj et al [7] introduced the wf (c) = k concept of an Odd vertex magic total When r =, Then by Theorem 1, labeling They call a vertex magic total K1, = is an odd vertex magic graph labeling is odd with a magic constant k = 7 f (V (G)) = {1,, 5,, n 1} Theorem If a tree T is odd vertex magic then n is odd The following results that will subsequently be very useful to prove some roof For a tree T, n = m + 1 They by Theorem 11,k = n theorems To prove that n is odd Theorem 11 Let G be a non trivial graph Suppose n is even If G is an odd vertex magic, then the magic Then k = n is even constant k is given by For any odd vertex magic labeling f, f (u) + v N (u) f (uv) = k, u V In m m k = 1 + m + + particular u is pendant vertex of T then, n n f (u) + f (uv) = k, age 76

4 Which is a contradiction Since f (u) is odd and f (uv) is even Therefore n is odd weights on the internals is Theorem If T has s internal vertices and st leaves then T does not admit an odd vertex magic labeling t > (s+1) s = ( s ) + (s + (s + ) + + (s + st )) + ( s 1) = s(t s + t(s 1) + s ) Since there are s internal vertices, roof If T has s internal vertices and st sk s(t s + t(s 1) + s ) leaves then k t s + t(s 1) + s n = (t + 1)s and m = ts + s 1 So the labels used for the veritices {1,, 5,, st + s 1} Therefore no labeling will be possible and for the edges are when t s + t(s 1) + s > st + 4s 1 {, 4, 6,, st + s } The maximum possible sum of weights on when st t (s + 1) > 0 p the leaves is s(s + 1) s+1 = t> {(s + 1) + (s + ) + + (s + st 1)} + s s {(s) + (s + ) + (s + 4) + + (s + st )} = st(st + 4s 1) Corollary 4 The odd vertex magic Since there are st leaves, we set labeling is impossible for a graph with s internal vertices and more than s + 1 stk st(st + 4s 1) leaves k (st + 4s 1) Theorem 5 If is the largest degree of any vertex in a tree T with n vertices and On the other hand, the minimum possible m edges then T doesnot admit an odd vertex sum of weights on the internal vertices n magic labeling whenever > occurs when the smallest labels {, 4, 6,, s } are assigned to internal roof Let c be the vertex of maximum edges (because they will be added twice), degree the labels {s, s +,, s + st } are The minimum possible weight of c is assigned to the remaining edges and the labels {1,, 5,, s 1} are assigned to Therefore, k 1 + ( + 1) Since there is an internal vertex of degree the vertices Hence the minimum possible sum of, there are atleast leaves in T age 77

5 so the maximum possible sum of weights there is a tree with 7 vertices and =, on the leaves is atmost the sum of the shown in Figure 1, which doesnot admit largest labels from f (E) and the largest any odd vertex magic labeling labels from f (v) Hence, k (n ) + (n + ) + + (n + ) + (n + 1) + (n + ) + + (n + 1) k (4n 1) k 4n 1 So a labeling whenever will be impossible 4n 1 < 1 + ( + 1) That is when + + ( 4n) > 0 > n Figure 1: The reason is as follows: The magic constant k = n = 19 The labels and 4 can be assigned only to the internal edges The only possible vertex label of d is 1 The labels 1, and 5 can be assigned only two internal vertices namely c and g Which is a contradiction Theorem 7 Let G be a graph obtained by joining a pendant vertex with a vertex of Remark 6 The following table shows degree of a comb graph Then G admits the maximum degree permitted by the an odd vertex magic labeling restriction given in 5 for some small roof Let the vertex V = {a, a,, a } 1 t values of n {a, a, a, a,, a } and the edge set 11 n f (ai ai+1 ) = t i Theorem and 5 donot provide sufficient condition, for a graph to be an odd vertex magic, since we can prove that t1 E = {a1 a11, a1 a1, a a1,, at at1 } {ai ai+1 / 1 i t 1} Here n = t + 1 and m = t Define f : E {1,,,, m} as follows 1 i t 1 1 i t f (a1 a1 ) = t f (ai ai1 ) = t + i age 78

6 The vertex labelings are as follows f (ai1 ) = 4t + 1 i 1 i t f (a1 ) = 4t + 1 f (ai ) = i 1 1 i t It can be easily veried that f is an odd vertex magic labeling with a vertex magic constant k = 6t + 1 Example 8 Example an odd vertex magic labeling of a graph G with k = 1 given in Figure [6] CTNagaraj, CYonnappan, Grabakaran, Even vertex magic total labeling, International Journal of ure and Applied Mathematics, Volume 115, No 9, (017),6-75 [7] CTNagaraj, CYonnappan, Grabakaran, Odd vertex magic total labeling of some graphs, Communicated [8] VSwaminathan, Jeyanthi, Super vertex magic labeling, Indian Jure ApplMath 4(6)(00) [9] VSwaminathan, Jeyanthi, On Super vertex magic labeling, JDiscrere MathSciCryptogr8(005) 17-4 [10] WDWallis, Magic Graphs, Birkhauser, Basel,001 Figure : References [1] HEnomoto, ASLlado, TNakamigawa GRingel, Super edge magic graphs, SUTJMath(1998) [] JA Gallian, A dynamic survey of graph labeling, Electron JCombinatorics 16 (009) DS6 [] JA MacDougall, MMiller, Slamin, WDWallis, Vertex magic total labeling of graphs, Utilitas Math 61 (00) -1 [4] JAMacDougall, MMiller,KASugeng,Super vertex magic total labelling of graphs, in: roceedings of the 15th Australian Workshop on Combinatorial Algorithms,004,pp-9 [5] GMarimuthu, MBalakrishnan, E-Super vertex magic labelings of graphs, Discrete Applied Mathematics, 160, , 01 age 79