Applied Mathematics Letters 20 (2007) 59 64 www.elsevier.com/locate/aml The L(2, 1)-labeling on the skew and converse skew products of graphs Zhendong Shao a,,rogerk.yeh b, David Zhang c a Department of Computer Science and Technology, Shenzhen Graduate School, Harbin Institute of Technology, Shenzhen, 518055, PR China b Department of Applied Mathematics, Feng Chia University, Taichung 40724, Taiwan c Department of Computing, Hong Kong Polytechnic University, Kowloon, Hong Kong Received 10 February 2006; accepted 13 February 2006 Abstract An L(2, 1)-labeling of a graph G is a function f from the vertex set V (G) to the set of all nonnegative integers such that f (x) f (y) 2 if d(x, y) = 1 and f (x) f (y) 1 if d(x, y) = 2, where d(x, y) denotes the distance between x and y in G. TheL(2, 1)-labeling number λ(g) of G is the smallest number k such that G has an L(2, 1)-labeling with max{ f (v) : v V (G)} =k. Griggs and Yeh conjecture that λ(g) Δ 2 for any simple graph with maximum degree Δ 2. This work considers the graph formed by the skew product and the converse skew product of two graphs. As corollaries, the new graph satisfies the above conjecture (with minor exceptions). c 2006 Elsevier Ltd. All rights reserved. Keywords: Channel assignment; L(2, 1)-labeling; Graph skew product; Graph converse skew product 1. Introduction The frequency assignment problem is to assign a frequency to each radio transmitter so that interfering transmitters are assigned frequencies whose separation is not in a set of disallowed separations. Hale [10] formulated this into a graph vertex coloring problem. In a private communication with Griggs, Roberts proposed a variation of the channel assignment problem in which close transmitters must receive different channels and very close transmitters must receive channels that are at least two channels apart. To translate the problem into the language of graph theory, the transmitters are represented by the vertices of a graph; two vertices are very close if they are adjacent and close if they are of distance two in the graph. Motivated by this problem, Yeh [19] and then Griggs and Yeh [9] proposed the following labeling on a simple graph. An L(2, 1)-labeling of a graph G is a function f fromthe vertexset V (G) to the set ofall nonnegative integers such that f (x) f (y) 2ifd(x, y) = 1and f (x) f (y) 1ifd(x, y) = 2, where d(x, y) denotes the distance between x and y in G. A k-l(2, 1)-labeling is an L(2, 1)-labeling such that no label is greater than k. The L(2, 1)-labeling number of G, denoted by λ(g), is the smallest number k such that G has a k-l(2, 1)-labeling. Corresponding author. E-mail addresses: zhdshao0026@126.com (Z. Shao), rkyeh@math.fcu.edu.tw (R.K. Yeh), csdzhang@comp.polyu.edu.hk (D. Zhang). 0893-9659/$ - see front matter c 2006 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2006.02.032
60 Z. Shao et al. / Applied Mathematics Letters 20 (2007) 59 64 There are a considerable articles studying the L(2, 1)-labelings (see [1 9,11 16,18,19]). Most of the papers consider the values of λ on particular classes of graphs. Griggs and Yeh [9] provided an upper bound Δ 2 + 2Δ for a general graph with the maximum degree Δ. Later, Chang and Kuo [4] improved the bound to Δ 2 + Δ. Recently, Král and Skrekovski [12] reduced the bound to Δ 2 + Δ 1. If G is a diameter 2 graph, then λ(g) Δ 2. The upper bound is attainable by Moore graphs (diameter 2 graph with order Δ 2 + 1). (Such graphs exist only if Δ = 2, 3, 7, and possibly 57; cf. [9].) Thus Griggs and Yeh [9] conjectured that the best bound is Δ 2 for any graph G with the maximum degree Δ 2(cf.[9]). (This is not true for Δ = 1. For example, Δ(K 2 ) = 1butλ(K 2 ) = 2.) Determining the value of λ is proved to be NP-complete (cf. [9]). Graph products play an important role in connecting many useful networks. In [16], we consider the graph formed by the cartesian product and the composition of graphs and prove that the L(2, 1)-labeling number of the graph is bounded by the square of its maximum degree and hence Griggs and Yeh s conjecture holds in both cases (with minor exceptions). In this work, we study the graph formed by the skew product and the converse skew product. As corollaries, the L(2, 1)-labeling number of the graph is bounded by the square of its maximum degree. Griggs and Yeh s conjecture holds in both cases (with minor exceptions). 2. A labeling algorithm A subset X of V (G) is called an i-stable set (or i-independent set) if the distance between any two vertices in X is greater than i. A 1-stable (independent) set is a usual independent set. A maximal 2-stable subset X of a set Y is a 2-stable subset of Y such that X is not a proper subset of any 2-stable subset of X. Chang and Kuo[4] proposedthe following algorithmfor obtaining an L(2, 1)-labeling and the maximum value of that labeling on a given graph. Algorithm 2.1. Input: AgraphG = (V, E). Output: The value k is the maximum label. Idea: In each step, find a maximal 2-stable set from these unlabeled vertices that are distance 2 away from those vertices labeled in the previous step. Then label all vertices in that 2-stable subset with the index i in the current stage. The index i starts from 0 and then increases by 1 in each step. The maximum label k is the final value of i. Initialization: Set X 1 = ; V = V (G); i = 0. Iteration: 1. Determine Y i and X i. Y i ={x V : x is unlabeled and d(x, y) 2forally X i 1 }. X i a maximal 2-stable subset of Y i. If Y i = then set X i =. 2. Label those vertices in X i (if there are any) with i. 3. V V \ X i. 4. V,theni i + 1; go to Step 1. 5. Record the currenti as k (which is the maximum label). Stop. Thus k is an upper bound on λ(g). We would like to find a bound in terms of the maximum degree Δ(G) of G analogous to the bound in terms of the chromatic number χ(g). Let x be a vertex with the largest label k obtained by Algorithm 2.1. Define I 1 ={i : 0 i k 1andd(x, y) = 1forsomey X i }, I 2 ={i : 0 i k 1andd(x, y) 2forsomey X i }, I 3 ={i : 0 i k 1andd(x, y) 3forally X i }. It is clear that I 2 + I 3 =k. For any i I 3, x Y i ;otherwisex i {x} is a 2-stable subset of Y i, which contradicts the choice of X i.thatis, d(x, y) = 1forsomevertexy in X i 1 ; i.e., i 1 I 1.So, I 3 I 1. Hence k I 2 + I 3 I 2 + I 1.
Z. Shao et al. / Applied Mathematics Letters 20 (2007) 59 64 61 Fig. 1. (Skew) product of two graphs. Fig. 2. (Converse skew) product of two graphs. In order to find k, it suffices to estimate B = I 1 + I 2 in terms of Δ(G). We will investigate the value of B with respect to a particular graph. For convenience, the notation which has been introduced in this section will also be used in the following section. 3. The skew and converse skew products of graphs Graph products have an important role in constructing many useful networks. Shibata and Kikuchi [17] put forward two interesting graph products called the skew product and converse skew product, respectively, based on the distance in graphs. The two graph products are defined as follows: each product of graphs G and H has vertex set V (G) V (H ). The skew product: G H has edge set E(G) ={((u 1, u 2 ), (v 1,v 2 )) [u 1 = v 1 and u 2 v 2 E(G 2 )] or [u 1 v 1 E(G 1 ) and u 2 v 2 E(G 2 )]}.(SeeFig. 1 for an example.) The converse skew product: G H has edge set E(G) ={((u 1, u 2 ), (v 1,v 2 )) [u 2 = v 2 and u 1 v 1 E(G 1 )] or [u 1 v 1 E(G 1 ) and u 2 v 2 E(G 2 )]}.(SeeFig. 2 for an example.) By the definition of the skew product G H of two graphs G and H,ifΔ(G) = 0orΔ(H ) = 0, then G H is disjoint copies of H or empty. Therefore we assume Δ(G) 1andΔ(H ) 1. In this section, we obtain an upper bound in terms of the maximum degree of G H for any two graphs G and H. Theorem 3.1. Let Δ, Δ 1, Δ 2 be the maximum degree of G H, G, H respectively. Then λ(g H ) Δ 2 + Δ 1 Δ 2 + Δ 2 Δ 1 (Δ 1 1)(Δ 2 1) Δ 2 (Δ 2 1)Δ 1. Proof. Let x = (u,v) in V (G) V (H ). Thendeg G H (x) = deg G (u)deg H (v) + deg H (v). Defined = deg G H, d 1 = deg G (u), d 2 = deg H (v), Δ 1 = Δ(G) and Δ 2 = Δ(H ). Hence d = d 1 d 2 + d 2 and Δ = Δ(G H ) = Δ 1 Δ 2 + Δ 2. (See Fig. 3 for this paragraph. In Fig. 3, d(v) denotes the degree of v in H,i.e.,d(v) = d 2.) For any vertex u in G with distance 2 from u, there must be a path u u u of length 2 between u and u in G; butthedegreeofv in H is d 2,i.e.,v has d 2 adjacent vertices in H, and so by the definition of a strong product G H, there must be d 2 internally disjoint paths of length 2 between (u,v)and (u,v). Hence for any vertex in G with distance 2 from u, there must be corresponding d 2 vertices with distance 2 from x = (u,v)which coincide in G H ;butthereared 1 (d 1 1) vertices
62 Z. Shao et al. / Applied Mathematics Letters 20 (2007) 59 64 Fig. 3. Fig. 4. with distance 2 from u in G, and hence in this sense the number of vertices with distance 2 from x = (u,v)in G H will decrease d 1 (d 1 1)(d 2 1) from the value d(δ 1) (the number d(δ 1) is the best possible). (See Fig. 4 for the following proof. In Fig. 4, d(u) denotes the degree of u in G, i.e., d(u) = d 1.) For any vertex v in H with distance 2 from v, there must be a path v v v of length 2 between v and v in H ;butthedegreeofu in G is d 1, i.e., u has d 1 adjacent vertices in G, and so by the definition of a skew product G H, there must be d 1 + 1 internally disjoint paths of length 2 between (u,v ) and (u,v) in G H. Hence for any vertex in H with distance 2 from v, there must be corresponding d 1 + 1 vertices in G H with distance 2 from x = (u,v) which coincide in G H ;butthereared 2 (d 2 1) vertices with distance 2 from v in H, and hence in this sense the number of vertices with distance 2 from x = (u,v)in G H will decrease d 2 (d 2 1)(d 1 +1 1) = d 2 (d 2 1)d 1 from the value d(δ 1) (the number d(δ 1) is the best possible). Hence the number of vertices with distance 2 from x = (u,v)in G H will decrease d 1 (d 1 1)(d 2 1) + d 2 (d 2 1)d 1 from the value d(δ 1) altogether.
Z. Shao et al. / Applied Mathematics Letters 20 (2007) 59 64 63 Hence for the vertex x, the number of vertices with distance 1 from x is no greater than Δ. The number of vertices with distance 2 from x is no greater than d(δ 1) d 1 (d 1 1)(d 2 1) d 2 (d 2 1)d 1. Hence I 1 d, I 2 d + d(δ 1) d 1 (d 1 1)(d 2 1) d 2 (d 2 1)d 1 = dδ d 1 (d 1 1)(d 2 1) d 2 (d 2 1)d 1. Then B = I 1 + I 2 d + dδ d 1 (d 1 1)(d 2 1) d 2 (d 2 1)d 1 = d(δ + 1) d 1 (d 1 1)(d 2 1) d 2 (d 2 1) d 1 = (d 1 d 2 + d 2 )(Δ 1 Δ 2 + Δ 2 + 1) d 1 (d 1 1)(d 2 1) d 2 (d 2 1)d 1.Define f (s, t) = (st + t)(δ 1 Δ 2 + Δ 2 + 1) s(s 1)(t 1) t(t 1)s. Then f (s, t) has the absolute maximum at (Δ 1, Δ 2 ) on [0, Δ 1 ] [0, Δ 2 ]; f (Δ 1, Δ 2 ) = (Δ 1 Δ 2 + Δ 2 )(Δ 1 Δ 2 + Δ 2 + 1) Δ 1 (Δ 1 1)(Δ 2 1) Δ 2 (Δ 2 1)Δ 1 = Δ(Δ + 1) Δ 1 (Δ 1 1)(Δ 2 1) Δ 2 (Δ 2 1)Δ 1 = Δ 2 + Δ Δ 1 (Δ 1 1)(Δ 2 1) Δ 2 (Δ 2 1)Δ 1. Then λ(g H ) k B Δ 2 + Δ Δ 1 (Δ 1 1)(Δ 2 1) Δ 2 (Δ 2 1)Δ 1 = Δ 2 + Δ 1 Δ 2 + Δ 2 Δ 1 (Δ 1 1) (Δ 2 1) Δ 2 (Δ 2 1)Δ 1. Corollary 3.2. Let Δ be the maximum degree of G H. Then λ(g H ) Δ 2 except for when one of Δ(G) and Δ(H ) is 1. Proof. For Δ 1 (Δ 1 1)(Δ 2 1) + Δ 2 (Δ 2 1)Δ 1 Δ 1 Δ 2 Δ 2 = Δ 1 (Δ 1 1)(Δ 2 1) + Δ 2 (Δ 2 1)(Δ 1 1 + 1) Δ 1 Δ 2 Δ 2 = Δ 1 (Δ 1 1)(Δ 2 1) + Δ 2 (Δ 2 1)(Δ 1 1) + Δ 2 (Δ 2 1) Δ 1 Δ 2 Δ 2 = (Δ 1 + Δ 2 )(Δ 2 1)(Δ 1 1) + Δ 2 (Δ 2 1) Δ 1 Δ 2 Δ 2. But Δ 1 Δ 2 = (Δ 1 1)(Δ 2 1) + Δ 1 + Δ 2 1. Hence (Δ 1 + Δ 2 )(Δ 2 1)(Δ 1 1) + Δ 2 (Δ 2 1) Δ 1 Δ 2 Δ 2 = (Δ 1 + Δ 2 )(Δ 2 1)(Δ 1 1) + Δ 2 (Δ 2 1) ((Δ 1 1)(Δ 2 1) + Δ 1 + Δ 2 1) Δ 2 = (Δ 1 + Δ 2 1)(Δ 2 1)(Δ 1 1) + Δ 2 (Δ 2 1) (Δ 1 + Δ 2 1) Δ 2 = (Δ 1 + Δ 2 1)((Δ 2 1)(Δ 1 1) 1) + Δ 2 (Δ 2 2). If one of Δ 1 or Δ 2 is 1 then G H is still a general graph. Hence we can suppose Δ 1 2andΔ 2 2. Then Δ 1 (Δ 1 1)(Δ 2 1) + Δ 2 (Δ 2 1)Δ 1 Δ 1 Δ 2 Δ 2 = (Δ 1 + Δ 2 1)((Δ 2 1)(Δ 1 1) 1) + Δ 2 (Δ 2 2) 0. This implies λ(g H ) Δ 2 + Δ 1 Δ 2 + Δ 2 Δ 1 (Δ 1 1)(Δ 2 1) Δ 2 (Δ 2 1)Δ 1 Δ 2. Therefore the results follow. By the definitions of the skew product G H and the converse skew product G H, the two products have symmetric structures. Hence we can obtain an upper bound in terms of the maximum degree of G H for any two graphs G and H similarly. Theorem 3.3. Let Δ, Δ 1, Δ 2 be the maximum degrees of G H, G, H respectively. Then λ(g H ) Δ 2 + Δ 1 Δ 2 + Δ 1 Δ 2 (Δ 2 1)(Δ 1 1) Δ 1 (Δ 1 1)Δ 2. Corollary 3.4. Let Δ be the maximum degree of G H. Then λ(g H ) Δ 2 except for when one of Δ(G) and Δ(H ) is 1. Acknowledgment The second author was supported in part by the National Science Council under grant NSC-92-2115-M-035-001. References [1] H.L. Bodlaender, T. Kloks, R.B. Tan, J.v. Leeuwen, λ-coloring of Graphs, in: Lecture Notes in Computer Science, vol. 1770, Springer-Verlag, Berlin, Heidelberg, 2000, pp. 395 406.
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