Short Paper Proc. of Int. Conf. on Advances in Computer Science 2012

On the Radio Number of Graphs P.Venkata Subba Reddy National Institute of Technology, Department of Computer Science and Engineering, Tiruchirapalli, India Email: venkatpalagiri@gmail.com

Proposition 1. Let Tn be a tree of order n. (i) If En be an even symmetric tree of order n with n 1and D=diam(En). Then rn(En) 2rcD(Tn)+D. (ii) If On be an odd symmetric tree of order n with n 1and D=diam(On). Then rn(On) 2rcD(Tn)+2D. (i) We denote the two copies of Tn ‘s of En repectively by T’ and T’’ and assume |V(T’)|=|V(T’’)|=n1. The vertices of T’ are labeled level by level from x1 to xn1 sequentially from left to right starting with the root. Similarly the vertices of T’’ are labeled from xn1+1 to x2n1. Let f : V(T’) {0,1,2, …,rcD(Tn)} be an optimal radio D-labeling of T’. Define a labeling g : V(En) {0,1,2, …, } recursively by g(xi)=f(xi), if 1 i n1. rcD(Tn)+f(xi-n1)+D, otherwise. It is easy to see that the maximum label used is 2rcD(Tn)+D. Now we show that g is a radio labeling of En i.e. for any two distinct vertices xi and xj of En the radio condition i.e. inequality (1) is satisfied. We distinguish three cases : Case 1: 1 i,j n1. It is easy to see that xi and xj satisfy the radio condition from the definition of g. Case 2: n1+1 i,j 2n1. From the labeling g we have |g(xi)-g(xj)| = |g(xi-n1)-g(xj-n1)| e” D+1-d(xi-n1,xj-n1) = D+1-d(xi,xj) Case 3: 1 min{i ,j} n1 and n1+1 max {i,j } 2n1. Clearly |g(x_i)-g(x_j)| D D+1-d(xi,xj), so the radio condition is satisfied. Therefore g is a radio labeling of En. Hence rn(En) 2rcD(Tn)+D. (ii) We denote the left and right subtrees of root of On by Tl and Tr respectively. Let |V(Tl)|=|V(Tr)|= n1 then |V(On)|=2n1+1. The vertices of Tl are labeled level by level from x1 to xn1 sequentially from left to right starting with the root. Similarly the verices of Tr are labeled from xn1+2 to x2n1+1 and the root is labeled as xn1+1. Let f : V(Tl) {0,1,2, …,rcD(Tn)} be an optimal radio D-labeling of Tl. Define a labeling g : V(On) {0,1,2, …, } recursively by g(xi)= f(xi), if 1 i n1, rcD(Tn)+D, if i=n1+1 rcD(Tn)+f(xi-n1-1)+2D, otherwise. Clearly the maximum label used by g is 2rcD(Tn)+2D. Now we show that for any two distinct vertices xi and xj of On the radio condition is satisfied. We have the following cases: Case 1: 1 i,j n1 or n1+2 i,j 2n1+1. It is easy to see that xi and xj satisfy the radio condition from the definition of g. Case 2: i=n1+1 or j=n1+1. Then we have |g(xi)-g(xj)| e =D+1d(xi,xj), hence the radio condition is satisfied.

Abstract— Let G be a simple, connected and undirected graph with diameter d. For a positive integer k, a radio k-labeling f of G is an assignment of non-negative integers, called labels to the vertices of G such that if u,v V(G) are distinct then |f(u) - f(v)| k + 1-d(u, v), where d(u,v) is the distance between u and v. The maximum label (positive integer) assigned by f to some vertex of G is called the span of f. The radio number of G denoted by rn(G) is the minimum span over all radio dlabelings of G. In this paper, we obtain an upper bound for the radio number of even symmetric tree, odd symmetric tree and the strong product of paths. Index Terms— Radio labeling, radio number, even symmetric tree, odd symmetric tree, graph products

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Short Paper Proc. of Int. Conf. on Advances in Computer Science 2012 Case 3: 1 min{i ,j} n1 and n1+2 max {i,j } 2n1+1. Clearly |g(x_i)-g(x_j)| 2D e +1- d(xi,xj), hence the radio condition is satisfied. Therefore g is a radio labeling of On. Hence rn(On) 2rcD(Tn)+2D.

diam(G)+1-dG(vij, vkl). So inequality (1) is satisfied. Therefore h is a radio labeling of G. So rn(G) d” max and hence the result.

B. Strong Product Of Paths Definition 4. Strong product: The strong product G H of graphs G and H is defined as follows: V(G H)= Cartesian product of V(G) and V(H) and E(G \otimes H)=Ec U Ed, where Ec={((x1,x2),(y1,y2)): (x1,y1) ª E(G) and x2=y2, or (x2,y2) ª E(H) and x1=y1} and Ed={((x1,x2),(y1,y2)): (x1,y1) ª E(G) and (x2,y2) ª E(H)} We have the following theorem from [7]. Theorem 1. For any n 3, rn(Pn)=2k2+2, if n=2k+1 2(k2-k)+1, if n=2k. Proposition 2. For n 3, let G=Pn Pn be the strong product of graphs Pn and Pn. Then rn(G) 4(k2+1)(k+1), if n=2k+1. 2(2k3-k2+k+1), if n=2k. Clearly diam(G)=n-1=diam(Pn). We will identify each vertex v of G by its coordinates (xi,yj), for all 1 i,j n and let vij=(xi,yj). Hence V(G)={v11, … ,v1n,v21, … ,v2n, … ,vn1, … ,vnn}. We adapt the path radio labeling algorithm in [7] to G and define the functions f,φ and g based on it. We distinguish two cases: Case 1: n=2k+1. Define a mapping g : V(Pn) {z1,z2, …,zn}as g(v1)=z3,g(v2)=z2k,g(vk)=z1,g(vk+1)=z4,g(vk+2)=z2k+1,g(v2k)=z2,g(v2k+1)=z5 and g(vi)=z2i if i<k and z2(i-k)+1 otherwise. A mapping φ : { v1,v2, …,vn } ’! {1,2,…,n} is defined as φ(vi)=j, such that g(vi)=zj. Another mapping f : {z1,z2, …,zn} {0,1,…,} is defined as follows: f(zi)= 0, if i=1 f(zi-1)+n-dPn(zi,zi-1), otherwise We now define a labeling h : V(G) {0,1,2,…,} based on the functions f,g and φ as follows: h(vij)=(φ(vi)-1)rn(Pn)+f(g(vj))+ Qi, where Qi= f(zi+1)-f(zi)} Case 2: n=2k. The function g is redefined in this case as follows. The mapping g : V(Pn) {z1,z2, …,zn} is defined as g(v1)=z2k-1,g(vk)=z1, g(vk+1)=z2k, g(v2k)=z2 and g(vi)=z2i-1 if i<k and z2(i-k) otherwise. It is easy to verify that the maximum label in the labeling h is max=(n+1)rn(Pn)=4(k2+1)(k+1) if n=2k+1 and 2(2k3-k2+k+1) otherwise. Fig. 1. shows an example of this labeling for n=7. Next we show that h is a radio labeling of G i.e. for any two distinct vertices vij and vkl of G inequality (1) holds. We now distinguish three cases Case i) i=k. Is obvious. Case ii) |i-k|=1. From the labeling h we have |h(vij)-h(vkl)| |f(vi)-f(vk)| n-d(vi,vk) n-d(vij , vkl). Hence inequality (1) holds. Case iii)|i-k|>1. Then clearly |h(vij)-h(vkl)|>rn(Pn)>n> 45 © 2012 ACEEE DOI: 02.AETACS.2012.3.535

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CONCLUSIONS In this paper, we obtained an upper bound for the radio number of even symmetric tree, odd symmetric tree and the strong product of paths. Bound on radio number of strong product of paths is calculated based on the radio number of paths. One can improve the bounds obtained in this paper by cleverly choosing the labeling and proving that this labeling is radio labeling. ACKNOWLEDGMENT I would like to thank Prof. K. Viswanathan Iyer, for getting me introduced to this topic, for his valuable suggestions, comments and encouragement. REFERENCES [1] J. A. Bondy, U. S. R. Murthy, Graph Theory with Applications, North-Holland, Elsevier, 1982. [2] G. Chartrand, P. Zhang, Introduction to Graph Theory, McGraw-Hill, 2006. [3] J. Georges, D. Mauro, M. Stein, Labeling products of complete graphs with a condition at distance two, SIAM J. Discrete Math., 14 (2001), pp. 28-35. [4] N. Jia, K. W. McLaughlin, Fibonacci Trees: A Study of the Asymptotic Behavior of Balaban’s Index, MATCH Commun. Math. Comput. Chem. 51 (2004), pp. 79-95. [5] W. K. Hale, Frequency Assignment: Theory and Applications, IEEE Proc. 68(12), 1980.

Short Paper Proc. of Int. Conf. on Advances in Computer Science 2012 [8] D. D. F. Liu, M. Xie, Radio number for square cycles, Cong. Numer., 169(2004), pp. 101-125. [9] D. D. F. Liu, M. Xie, Radio number for trees, Discrete Math. Vol. 308 (2008), pp. 1153-1164. [10] D. D. F. Liu, M. Xie, Radio number for square paths, Ars Combin. 90, pp. 307-319.

[6] M. Kchikech, R. Khennoufa, O. Togni, Linear and cyclic radio k-labelings of Trees, Discussiones Mathematicae Graph Theory, 27(1) pp. 105-123, 2007. [7] D. D. Liu, X. Zhu, Multilevel distance labelings for paths and cycles, SIAM J. Discrete Math. 19 (2005), pp. 610621.

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