Distance Three Labelings of Trees

Transcription

1 Dstance Three Labelngs of Trees Jří Fala Petr A. Golovach Jan Kratochvíl Bernard Ldcký Danël Paulusma Abstract An L(2, 1, 1)-labelng of a graph G assgns nonnegatve ntegers to the vertces of G n such a way that labels of adjacent vertces dffer by at least two, whle vertces that are at dstance at most three are assgned dfferent labels. The maxmum label used s called the span of the labelng, and the am s to mnmze ths value. We show that the mnmum span of an L(2, 1, 1)-labelng of a tree can be bounded by a lower and an upper bound wth dfference one. Moreover, we show that decdng whether the mnmum span attans the lower bound s an NP-complete problem. Ths answers a known open problem, whch was recently posed by Kng, Ras, and Zhou as well. We extend some of our results to general graphs and/or to more general dstance constrants on the labelng. 1 Introducton Classcal graph colorng nvolves the labelng of the vertces of some gven graph by ntegers usually called colors such that no two adjacent vertces receve the same color. We study a varant of ths problem that has been motvated by and fnds applcatons n wreless communcaton. In a wreless network, each transmtter s assgned a frequency channel for ts transmssons. However, two transmssons can nterfere f ther channels are too close. Whether ths happens depends on the physcal structure of the network; even f two transmtters use dfferent channels, there stll may be nterference f the two transmtters are located close to each other. Extended abstracts of results n ths paper appeared n the proceedngs of WG 2004 [7] and TAMC 2010 [17]. The paper s supported by Royal Socety Jont Project Grant JP Department of Appled Mathematcs, Charles Unversty, Prague. E-mals: {fala honza bernard}@kam.mff.cun.cz. Supported by the Mnstry of Educaton of the Czech Republc as project 1M0545 and GACR 201/09/0197 and by Charles Unversty as GAUK School of Engneerng and Computng Scences, Durham Unversty. E-mals: {petr.golovach danel.paulusma}@durham.ac.uk. Supported by EPSRC as project EP/G043434/1. 1

2 The rado spectrum gets more and more scarce, because the number of wreless networks s rapdly ncreasng. Thus the task s to mnmze the span of frequences whle avodng nterference. A wreless network can be modeled by an undrected graph G = (V, E) wth no loops and no multple edges. The transmtters are represented by vertces and the dstance dst G (u, v) between two transmtters u, v s the number of edges on a shortest path from u to v. A labelng of G s a mappng f : V {0, 1,...} that assgns each vertex of V a label f(v) representng a frequency channel (n ths settng, the conventon s to use the noton label nstead of color ). The dstance of two transmtters n a network mples certan requrements on the dfference of the channels assgned to them. We model ths by posng extra restrctons on the labelng. Ths approach s called dstance constraned labelng and t s done va a frequency graph H, whose vertces represent the avalable channels and are denoted by 0,..., V (H) 1. For postve ntegers p 1, p 2,..., p k, a labelng f of G wth f(v (G)) V (H) s called an H(p 1,..., p k )-labelng f dst H (f(u), f(v)) p for all u, v V G wth dst G (u, v) = holds for every = 1,..., k. The ntegers p 1,..., p k are called the dstance constrants mposed on the labelng. It s natural to assume that frequences must be farther apart f transmtters are closer to each other; so we restrct ourselves to dstance constrants p 1 p 2 p k. We can now formalze the aforementoned task as the followng decson problem: H(p 1,..., p k )-Labelng Parameters: Dstance constrants p 1,..., p k. Instance: Graphs G and H. Queston: Does G have an H(p 1,..., p k )-labelng? Not only for ts practcal applcatons but also because of ts many nterestng theoretcal propertes, dstance constraned labelng has receved much attenton n recent lterature, n partcular the cases n whch H s a path or a cycle. Below we dscuss these two cases; for a survey on known algorthmc results for other frequency graphs we refer to Fala, Golovach and Kratochvíl [9]. Lnear Metrc. Let H be the path P λ+1 on vertces 0,..., λ wth an edge between vertces and + 1 for = 0,..., λ 1. Then an H(p 1,..., p k )- labelng s called an L(p 1,..., p k )-labelng wth span λ, and H(p 1,..., p k )- Labelng s formulated as the problem: L(p 1,..., p k )-Labelng Parameters: Dstance constrants p 1,..., p k. Instance: A graph G and nteger λ. Queston: Does G have an L(p 1,..., p k )-labelng wth span λ? 2

3 The mnmum λ such that a graph G has an L(p 1,..., p k )-labelng s denoted by λ p1,...,p k (G). An L(1)-labelng of G s also called a colorng of G and λ 1 (G) + 1 s also called the chromatc number χ(g) of G. Cyclc Metrc. Let H be the cycle C λ on vertces 0,..., λ 1 wth an edge between vertces and + 1 for = 0,..., λ 1 (modulo λ). Then an H(p 1,..., p k )-labelng s called a C(p 1,..., p k )-labelng wth span λ, and the correspondng decson problem s denoted C(p 1,..., p k )-Labelng. We denote the mnmum λ such that G has a C(p 1,..., p k )-labelng of span λ by c p1,...,p k (G). Observe that whle for the lnear metrc, the span λ s the number of vertces of the frequency graph (path) mnus one, for the cyclc metrc, we follow Lu and Zhu [29] and defne the span as the number of vertces of the correspondng cycle. Known Results. Especally L(p 1, p 2 )-labelngs are well studed, see the surveys of Calamoner [2] and Yeh [33]. For a survey on a more general model we refer to Grggs and Král [18]. We start wth a number of algorthmc and complexty results for labelngs. Fala, Kloks and Kratochvíl [11] showed that L(2, 1)-Labelng s NPcomplete already for fxed λ 4. Král gave an exact exponental-tme algorthm for solvng the general channel assgnment problem [27]. Ths mples an O (4 n ) algorthm for L(2, 1)-Labelng (when λ s part of the nput). The latter was mproved to an O (3.885 n ) algorthm by Havet et al. [19] and further mproved to an O ( n ) algorthm by Junosza-Szanawsk and Rzazewsk [23]. Chang and Kuo [5] presented a nontrval dynamc programmng algorthm to show that L(2, 1)-Labelng can be solved n polynomal tme for trees. Hasunuma et al. [20] gave a sub-quadratc algorthm, and the same authors [21] found a lnear tme algorthm afterwards. For p 1 > 1, Chang et al. [4] showed that L(p 1, 1)-Labelng s polynomaltme solvable for trees even when p 1 s not fxed but part of the nput (see also Fala, Kratochvíl and Proskurowsk [13]). However, for any fxed p 1, p 2, the L(p 1, p 2 )-Labelng problem s NP-complete, even for trees, f p 2 2 and p 2 does not dvde p 1 [9]. It s also known that, for fxed p 1 2, L(p 1, 1)- Labelng s already NP-complete for graphs of treewdth two [8]. Ths s n contrast to the polynomal tme result of Zhou, Kanar and Nshzek [35] on L(1, 1)-Labelng for graphs of bounded treewdth (but L(1, 1)-Labelng s W[1]-hard when parameterzed by the treewdth of the nput graph [10]). Also L(p 1,..., p k )-labelngs wth k 3 have been studed. Zhou, Kanar and Nshzek [35] showed that L(1,..., 1)-Labelng can be solved n polynomal tme on graphs of bounded treewdth. Bertoss, Pnott and Rzz [1] showed the same for the class of nterval graphs. Golovach [15] proved that the prelabelng extenson of L(2, 1, 1)-Labelng s NP-complete for trees (n ths varant of the problem some vertces have preassgned labels). He also proved [16] that L(p 1, 1, 1)-Labelng s NP-complete for trees f p 1 s part of the nput. Calamoner et al. [3] presented lower and upper bounds 3

4 on the mnmum span λ p,1,1 (G) for an outerplanar graph G n terms of the maxmum vertex degree of G. They also gave a lnear-tme approxmaton algorthm for obtanng the mnmum span λ p,1,1 for outerplanar graphs. Zhou [34] presented lower and upper bounds on the mnmum span of an L(p 1, p 2, p 3 )-labelng of a hypercube Q d extendng the work of Km, Du and Pardolos [25] and Ngo, Du and Graham [30] on L(1,, 1)-labelngs of hypercubes for the case k = 3, whereas Östergård[31] determned that λ 1,1,1 (Q d ) d converges to 2. Recently, Kng, Ras and Zhou [26] gave lower and upper bounds on the mnmum span of an L(p, 1, 1)-labelng of a tree. For the cyclc metrc, Fala and Kratochvíl [12] showed that C(2, 1)- Labelng s NP-complete already for fxed span λ 6. Smlarly to the lnear metrc, Fala, Golovach and Kratochvíl [8] showed that C(2, 1)- Labelng s already NP-complete for the class of graphs wth treewdth 2. On the postve sde, Lu and Zhu [29] presented a closed formula for the mnmum span of a C(p 1, p 2 )-labelng of a tree. Somewhat surprsngly the span only depends on the maxmum vertex degree n the tree. Ths mmedately mples that C(p 1, p 2 )-Labelng can be solved n polynomal tme for trees, even f p 1 and p 2 (and λ) are part of the nput. Our Results. In the frst part of our paper we show NP-hardness of the followng two problems: the L(2, 1, 1)-Labelng problem for general graphs for any fxed λ 5 (n Secton 3). the L(2, 1, 1)-Labelng problem for trees f λ s a part of the nput (n Secton 4). The remanng cases,.e., of λ 4 for general graphs and of fxed λ for trees, are shown to be polynomal-tme solvable. The latter case can be extended to general dstance constrants p 1,..., p k. In the second part (Secton 5) we prove an upper bound on the mnmum span c p1,p 2,p 3 (T ), whch s also an upper bound on the mnmum span λ p1,p 2,p 3 (T ), for a tree T. Because we gve an upper bound that s vald for the cyclc metrc, the upper bound on λ p,1,1 (T ) of Kng, Ras, and Zhou [26] s a better bound on λ p,1,1 (T ) than ours (after substtutng p 2 = p 3 = 1). Nevertheless, the bounds n our WG 2004 paper and ther 2010 paper concde for (p 1, p 2, p 3 ) = (2, 1, 1). The proof of our upper bound on λ p1,p 2,p 3 and c p1,p 2,p 3 for trees s constructve; just as the proof of Kng, Ras and Zhou [26] for ther upper bound on λ 2,1,1 for trees, t yelds a polynomal-tme algorthm for constructng a labelng that meets the upper bound. Both ther and our obtaned labelngs have the extra property that an nterval can be assgned to each vertex contanng all the labels of ts neghbors, such that the dstance constrant p 3 can be replaced by a correspondng dstance constrant on the ntervals assocated to two adjacent vertces. We call such labelngs elegant and 4

5 show how to fnd optmal elegant L(p, 1, 1)-labelngs and optmal elegant C(p, 1, 1)-labelngs of trees n polynomal tme for any p 1. For the case (p 1, p 2, p 3 ) = (2, 1, 1) the exstence of the above algorthms means that λ 2,1,1 (T ) and c 2,1,1 (T ) can be approxmated n polynomal tme wthn addtve factor 1 by determnng an optmal elegant L(2, 1, 1)-labelng or C(2, 1, 1)-labelng, respectvely. We observe that for the lnear metrc ths s n contrast wth the aforementoned NP-hardness of fndng optmal (but not necessarly elegant) L(2, 1, 1)-labelngs of trees, even though the dfference between the two spans s at most one. In Queston 10b of ther paper, Kng, Ras and Zhou ask whether there exsts a characterzaton of trees wth λ 2,1,1 equal to the sum of the maxmum total degree of two adjacent vertces. The +1 approxmaton algorthm for computng λ 2,1,1 and the NP-hardness of L(2, 1, 1)-Labelng for trees mply that the exstence of a good (.e., polynomal-tme verfable) characterzaton of such trees does not exst (unless P=NP). Our NP-hardness result also provdes a negatve answer (unless P=NP) to Queston 12 of ther paper, n whch they ask f L(2, 1, 1)-Labelng can be solved n polynomal tme for trees. 2 Prelmnares All graphs consdered n ths paper are smple,.e., wthout loops and multple edges. Let G be a graph. The vertex set of G s denoted by V (G) and ts edge set s denoted by E(G). For a vertex v, N G (v) = {uv u V (G)} s the (open) neghborhood of G, and deg G (v) = N G (v) denotes the degree of vertex v V (G). We may omt subscrpts f the graph under consderaton s clear from the context. The length of a cycle or a path s ts number of edges. A connected graph wthout a cycle as a subgraph s called a tree, ts vertces of degree one are called the leaves, and the other vertces are called the nner vertces. A star s a tree on at least two vertces that has at most one nner vertex, whch s called the center. We denote the star on k + 1 vertces by K 1,k for k 1. A double star s a tree wth exactly two nner vertces. A complete graph s a graph wth an edge between every par of vertces. We denote the complete graph on k vertces by K k for k 1. The vertex set of a complete graph s called a clque. The symbol ω(g) denotes the number of vertces of a largest clque n a graph G. The k-th dstance power G k of a graph G s the graph on the same vertex set V (G k ) = V (G) where edges of G k connect dstnct vertces that are at dstance at most k n G,.e., E(G k ) = {uv : u, v V (G k ), 1 dst G (u, v) k}. A tree decomposton of a graph G = (V, E) s a par (X, T ) where X = {X 1,..., X r } s a collecton of bags (sets of vertces) and T s a tree wth vertex set X such that the followng three propertes hold. Frst, 5

6 r =1 X = V. Second, for each uv E, there exsts a bag X such that {u, v} X. Thrd, f v X and v X j then v s n every bag on the (unque) path n T between X and X j. The wdth of (X, T ) s max 1 r X 1 and the treewdth of G s the mnmum wdth over all possble tree decompostons of G. For nonnegatve ntegers j, we defne the (dscrete) nterval [, j] = {, +1,..., j}. Let µ be a postve nteger. For ntegers, j {0,..., µ}, we defne the nterval modulo µ+1 denoted by [, j] µ+1 as [, j] µ+1 = {, +1, + 2,..., j} f j, and [, j] µ+1 = {,..., µ, 0,..., j} f > j. For any par of ntegers and j, we defne [, j] µ+1 = [ mod (µ + 1), j mod (µ + 1)] µ+1. Here x mod (µ + 1) = y [0, µ] such that µ + 1 dvdes x y. By [, j] 2 we denote the set of all even ntegers n the nterval [, j]. Let G be a graph. Then the vertces of every clque n G k must get labels parwse at least p k apart n any L(p 1,..., p k )-labelng of G. Furthermore, a colorng of G k can be transformed to an L(p 1,..., p k )-labelng by usng labels that form an arthmetc progresson of dfference p 1 as labels. Hence, we can make the followng observaton. Observaton 1. For any p 1 p 2 p k 1 and any graph G t holds that p k (ω(g k ) 1) λ p1,...,p k (G) p 1 (χ(g k ) 1). 3 Complexty of L(2, 1, 1)-Labelng wth fxed span Note that for fxed λ, we can descrbe the L(p 1,..., p k )-Labelng problem n Monadc Second-Order Logc. Then by the well-known theorem of Courcelle [6] we mmedately have the followng clam. Proposton 1. For any p 1... p k 1 and any fxed λ, the L(p 1,..., p k )-Labelng problem can be solved n lnear tme for graphs of bounded treewdth. For general graphs the stuaton s dfferent. To show ths we present a complete computatonal complexty characterzaton of the L(2, 1, 1)- Labelng problem for general graphs for fxed values of the parameter λ. Theorem 1. The L(2, 1, 1)-Labelng problem s NP-complete for every fxed λ 5 and t s solvable n lnear tme for all λ 4. Proof. We start wth the second part of the theorem and prove that the labelng problem s tractable for λ 4. Let G be a graph. We may assume that G s connected, as otherwse we consder each component of G separately. We frst observe that G allows an L(2, 1, 1)-labelng of span at most 3 f and only f G s a path on at most four vertces. (The labels along the path P 4 are 1, 3, 0, 2.) Hence, we are left to consder the case λ = 4. 6

7 F 1 F 2 F 3 F 4 F 5 F 6 F 7 F 8 F 9 Fgure 1: The graphs F 1,..., F 9 wth λ 2,1,1 (F ) > 4 for = 1,..., 9. We clam that none of the graphs F (1 9) depcted n Fgure 1 allows an L(2, 1, 1)-labelng of span 4 ths can be verfed by a straghtforward case analyss. Ths means that our nput graph G has no L(2, 1, 1)- labelng of span 4 f t contans one of these nne graphs as a subgraph. We test ths as follows. Frst, we check n lnear tme f G has maxmum degree 3. If not, then G contans F 1 as a subgraph. In the other case,.e., f G has maxmum degree 3, we can check n lnear tme f G contans a graph F (2 9) as a subgraph. In any such case we output No. From now on, assume that G contans no graph F (1 9) as a subgraph. Assume frst that G contans a cycle of length at least four, and let us fx a longest one. Observe that every edge of G s ncdent wth a vertex of ths cycle otherwse we would get F 5 or F 9. If two vertces of the cycle that share no common neghbor along the cycle were connected by an edge, a so-called shortcut, then we would get F 2. Hence all shortcuts produce trangles. These trangles must be edge-dsjont as otherwse we would get F 4. Fnally, f a vertex outsde the cycle were adjacent to two vertces of the cycle then ether the cycle could be extended (f the two neghbors were adjacent) or we would get F 7 or F 8 (f they shared another common neghbor) or get F 3 (f they were farther apart). Hence, any vertex outsde the cycle s a leaf. Moreover vertces of the cycle that are adjacent to the leaves must be parwse at dstance at least three (due to F 2 and F 3 ) as well as they should belong nether to a trangle nor to the neghborhood of one (ths would yeld F 2 or F 6 ). Some specfc cases are also excluded f the longest cycle s of length four, namely the forbdden graphs F 4, F 7 and F 8. By analogous arguments we get that f G has no cycle of length at least four, then t s formed from a longest path wth possbly some shortcuts formng trangles and/or possbly some pendant leaves, whereas these trangles and leaves are suffcently separated as n the prevous case. It s not dffcult to show that n both cases G has treewdth at most 3, and hence the exstence of an L(2, 1, 1)-labelng of span 4 can be tested n lnear tme by Proposton 1. 7

8 λ odd E λ 4 λ even K p w 1 w 2 u 1 w 1 u 2 w 1... w w 2 K p M p 1. E p 2 u 1 r 1 u 2 r 2. v 1 v 2 K p w 1 w 2 v 1 v 2 Fgure 2: Varable gadgets. To prove NP-hardness for λ 5, we reduce from the Monotone Not- All-Equal p-satsfablty problem for p = λ 2. An nstance of Monotone Not-All-Equal p-satsfablty s a formula Φ n the conjunctve normal form wth p postve lterals n each clause,.e., no negatons are allowed. The queston s whether Φ has a truth assgnment such that each clause contans at least one postvely valued lteral and at least one negatvely valued lteral. Schäfer [32] showed that Monotone Not-All-Equal 3-Satsfablty s NP-complete. Ths also holds for any fxed p 4, the proof of whch s straghtforward and folklore. Let Φ be a formula that s an nstance of the Monotone Not-All- Equal λ 2 -Satsfablty problem. Note that p = λ 2. For each varable x we construct a gadget consstng of a chan of copes of the graph depcted n Fg 2. The length of the chan correspondng to the varable x s the number of occurrences of x n Φ. The symbols E n and M n n Fgure 2 denote an ndependent set of n vertces, and a matchng on n edges, respectvely; recall that K n denotes a complete graph on n vertces. We argue that any L(2, 1, 1)-labelng of span λ of the constructed varable gadget satsfes: All vertces u are labeled by the same label, ether by 0 or by λ. If u s labeled by λ, then the vertex v s gven a label from the set L = [0, λ 4 + (λ mod 2)] 2, and analogously f u s labeled by 0, then the label of v belongs to L = {λ l : l L}. In both cases, vertces u are of degree λ 1, hence t would be mpossble to gve these vertces labels dfferent from 0 or λ. If u s gven λ, then the 8

9 label λ 1 must be used on the vertces w 1 and w +1, hence u 1 and u +1 must be also gven label λ. As a mrror argument holds for the label 0, the frst clam follows. For the case of and odd λ, observe that the subgraph consstng of the two complete subgraphs K p contans exactly λ+1 vertces and s of dameter three. Hence all labels from [0, λ] must be used, each on exactly one vertex of ths subgraph. In partcular, one K p wll only host even labels, whle the other one hosts all odd labels. If u s labeled by λ, then w s labeled by λ 1 by the same argument as for w. Then the upper K p uses even labels, the bottom all odd labels, and only the label λ 1 remans for w. As the vertex v s at dstance at most three from all vertces from the bottom K p, t may only be labeled by a label from the set L = [0, λ 3] 2, as clamed above. When λ s even then u together wth K p forms a clque on p+1 vertces, hence all even labels,.e., the set [0, λ] 2, are used to label ths subgraph. If a vertex u s labeled by λ, then ts remanng neghbors are gven odd labels from the set [1, λ 3]. (Recall that w s labeled by λ 1 n ths case.) In partcular, the same label s used for all copes of r and the remanng labels n [1, λ 3] for all copes of E p 2. Hence, all possble labels of v fall n the set [0, λ 4] 2, as clamed. In both cases when u s labeled by 0 the clam s obtaned by the symmetry of the labelng. We fnalze the constructon of the graph G by jonng varable gadgets through clause vertces as follows. For each clause C of the formula Φ we nsert an extra new vertex z C. For each varable x that appears n C we lnk z C by an edge wth a unque vertex v of the varable gadget assocated wth x. Hence, each clause vertex s of degree p. The propertes of the varable gadgets assure that G allows an L(2, 1, 1)- labelng of span λ f and only f Φ has a requred assgnment. These labelngs are related to assgnments e.g. by lettng x = true whenever the vertces u of the gadget for x are all labeled by λ, and x = false f u gets 0. Observe that for any clause vertex z C t holds that deg(z C ) = p > L = L. Hence labels both from L \ L and from L \ L must be present n the neghborhood of z C. Consequently, these labelngs ndcate only vald assgnments,.e., at least one of the adjonng gadgets represents a postvely valued varable and at least one stands for a negatvely valued one. In the opposte drecton, each assgnment for Φ can be converted nto an L(2, 1, 1)-labelng of G n a straghtforward way (by usng labelngs of the gadgets wth propertes dscussed above). Observe n partcular that n the case of even λ, each vertex w C together wth ts p neghbors wll requre p + 1 even labels, whch s just the number of even labels n the nterval [0, λ]. 9

10 u v w x 1 v 1 w 1 T 1 u x 2 v 2 w 2 u v w T 3 T 2 Fgure 3: Gadgets T 1, T 2 and T 3 for λ = NP-completeness of L(2, 1, 1)-Labelng for trees By Proposton 1, the L(2, 1, 1)-Labelng problem can be solved n polynomal tme for trees f the span λ s fxed,.e., not part of the nput. If λ s consdered to be part of the nput, then the problem s dffcult. Theorem 2. The L(2, 1, 1)-Labelng problem s NP-complete for the class of trees. The remanng part of ths secton contans the proof of ths theorem. 4.1 Auxlary constructons We frst construct gadgets where some vertces are forced predetermned labels n an arbtrary L(2, 1, 1)-labelng. A set of ntegers S [0, λ] s called symmetrc f for each S, λ S. Note that for any L(p 1,..., p k )- labelng l of a graph G of span λ, the mappng l : V (G) [0, λ], such that l(v) = λ l(v) for v V (G), s an L(p 1,..., p k )-labelng of G of span λ too. Hence our gadgets force symmetrc sets of labels. From now on we assume that λ s an even postve nteger and that λ 16. We consder a star K 1,λ 1 wth the center u. Then a new vertex w s added and joned by an edge wth a leaf v of the star. Denote the obtaned tree by T 1. We say that w s the root of T 1. An example of T 1 s shown n Fgure 6. We need the followng propertes of T 1. Lemma 1. For any L(2, 1, 1)-labelng of T 1 wth span λ, 10

11 the vertex u s labeled by an nteger from the set {0, λ}; f u s labeled by 0 then the root w s labeled by 1 and f u s labeled by λ then w s labeled by λ 1. For any {1, λ 1} and any nteger j [3, λ 3], there s an L(2, 1, 1)- labelng l of T 1 wth span λ such that l(w) =, l(v) = j. Proof. Snce all vertces of N T1 (u) should be labeled by dfferent labels whch are 2-dstant from the label of u and snce deg T1 (u) = λ 1, for any L(2, 1, 1)- labelng of T 1 wth span λ, the vertex u can only be labeled ether by 0 or λ. Assume that u s labeled by 0. Then vertces of N T1 (u) are labeled by all ntegers from [2, λ]. Hence, w should be labeled by 1. Symmetrcally, f u s labeled by λ, then w s labeled by λ 1. The second clam of the lemma can be verfed drectly. The next gadget s denoted by T 2 and s constructed as follows (see Fgure 3). We ntroduce a star K 1,λ 3 wth center u and add a copy of T 1 rooted n u. Then a new vertex w s added and joned by an edge wth a leaf v of the tree adjacent to u. The vertex w s the root of T 2. The propertes of T 2 are gven n the followng lemma. Lemma 2. For any L(2, 1, 1)-labelng of T 2 wth span λ, the vertex u s labeled by an nteger from the set {1, λ 1}; f u s labeled by 1 then the root w s labeled by an nteger from {0, 2}, and f u s labeled by λ 1 then w s labeled by a label from {λ 2, λ}. For any {0, 2, λ 2, λ} and any nteger j [5, λ 5], there s an L(2, 1, 1)- labelng l of T 2 wth span λ such that l(w) =, l(v) = j. Proof. By Lemma 1 the vertex u s labeled ether by 1 or λ 1. Assume that u s labeled by 1. Snce deg T1 (u) = λ 2, for any L(2, 1, 1)-labelng of T 2 wth span λ, the vertces N T2 (u) are labeled by all ntegers from [3, λ]. Therefore, w should be labeled by 0 or 2. Symmetrcally, f u s labeled by λ 1, then w s labeled by λ 2 or λ. The second clam of the lemma can be verfed drectly. Now we construct the gadget T 3 (see Fgure 3). We consder a star K 1,λ 2 wth center u. Then two copes of T 1 rooted n two dfferent leaves x 1, x 2 of the star are added. Fnally we add two vertces w 1, w 2 and jon them by edges wth two dfferent leaves (v 1 and v 2 respectvely) of the constructed tree adjacent to u. We call w 1 and w 2 the roots of T 3. The propertes of T 3 are summarzed n the next lemma. Lemma 3. For any L(2, 1, 1)-labelng of T 3 wth span λ, 11

12 the vertex u s labeled by an nteger from [3, λ 3]; f u s labeled by, then w 1, w 2 are labeled by labels from { 1, + 1}. For any nteger [3, λ 3], any par of ntegers j 1, j 2 { 1, + 1} and any par of dfferent ntegers r 1, r 2 [+3, λ (+3)], there s an L(2, 1, 1)- labelng l of T 3 wth span λ such that l(u) =, l(w 1 ) = j 1, l(w 2 ) = j 2, l(v 1 ) = r 1 and l(v 2 ) = r 2. Proof. By Lemma 1, the vertces x 1 and x 2 can be labeled ether 1 or λ 1. Snce they must have dfferent labels, one of them s labeled by 1 and the other one s labeled by λ 1. Hence, u can only be labeled by an nteger from [3, λ 3]. Assume that u s labeled by. For any L(2, 1, 1)-labelng of T 3 wth span λ, the vertces n N T3 (u) are labeled by all ntegers from [0, λ] \ [ 1, + 1]. Therefore, w 1 and w 2 can only be labeled by ntegers from { 1, + 1} As before, the second clam of the lemma can be verfed drectly. Note that neghbors of x 1 and x 2 dfferent from u can always be labeled by 1 and + 1. For our gadgets constructed below we assume that k s a postve nteger and 2 k λ/4 2; the latter s a vald assumpton because λ 16 holds. We construct a rooted tree T (k) such that the root can only be labeled by ntegers from [2, 2k] 2 [λ 2k, λ 2] 2. To do t we ntroduce k 1 copes of trees T 3. For {1,..., k 1}, denote by u (), v () 1, v() 2, w() 1, w() 2 the vertces u, v 1, v 2, w 1, w 2 of the -th copy of T 3. Then vertces w ( 1) 2 and w () 1 are dentfed for {2,..., k 1}. Fnally, a copy of T 2 rooted n w (1) 1 s added. Let u (0) and v (0) be the vertces u and v of T 2, respectvely. The vertex w (k 1) 2 s the root of T (k). The constructon of T (k) s shown n Fgure 4. Lemma 4. For any L(2, 1, 1)-labelng of T (k) wth span λ, the root w (k 1) 2 s labeled by an nteger from [2, 2k] 2 [λ 2k, λ 2] 2 ; f w (k 1) 2 s labeled by, then u (k 1) s labeled ether 1 or + 1 f < 2k and u (k 1) s labeled by 1 f = 2k. For any nteger [2, 2k] 2 [λ 2k, λ 2] 2 and any nteger r [2k + 2, λ (2k + 2)] there s an L(2, 1, 1)-labelng l of T (k) wth span λ such that l(w (k 1) 2 ) = and l(v (k 1) 2 ) = r. Proof. Note that by Lemma 2 the vertex w (1) 1 s labeled by an nteger from the set {0, 2, λ 2, λ}. Snce by Lemma 3 t cannot be labeled by 0 or λ, ths vertex s labeled ether by 2 or λ 2. Then the frst clam of the lemma s proved by nductve applcatons of Lemma 3. We use the fact that f w (j) 1 12

13 w (1) 1 w (1) 2 = w (2) 1 w (2) 2 = w (3) 1 w (k 2) 2 = w (k 1) 1 w (k 1) 2 v (0) v (1) 1 v (1) 2 v (2) 1 v (2) 2 v (k 1) 1 v (k 1) 2 u (0) u (1) u (2) u (k 1) T 2 T 3 T 3 T 3 T (k) Fgure 4: Gadget T (k). s labeled by then u (j) s labeled by 1 or + 1 and w (j) 2 s labeled by an nteger from { 2,, + 2}. The second clam mmedately follows from Lemmas 2 and 3. It s suffcent to note that for j {1,..., k 1}, vertex v (j) 1 can be labeled by r + 1 or r 1, whereas v (j) 2 and v (0) can be labeled by r. Usng gadgets T (k) t s possble to construct a rooted tree F (k) (see Fgure 5) such that the root can only be labeled by an nteger 2k or λ 2k. We construct a star K 1,2k+1 wth the center v and leaves w 0,..., w 2k. Then four copes of T 2 rooted n w 1, w 2, w 3 and w 4 respectvely are ntroduced, and for each {2,..., k 1}, two copes of T () rooted n w 2+1 and w 2+2 are added. Fnally, a copy of T (k) rooted n w 0 s constructed. The vertex w 0 s declared the root of F (k). Lemma 5. For any L(2, 1, 1)-labelng of F (k) wth span λ, the root w 0 s labeled ether by 2k or λ 2k; the vertces at dstance two from the root are labeled by all ntegers from [0, 2k 2] 2 [λ (2k 2), λ] 2 and one vertex s labeled by 2k 1 or λ (2k 1). For any par of dfferent ntegers r 1, r 2 [2k + 2, λ (2k + 2)] there s an L(2, 1, 1)-labelng l of F (k) wth span λ such that the vertces adjacent to the root are labeled by r 1 and r 2. Proof. By Lemma 2 vertces w 1, w 2, w 3, w 4 have to be labeled by 0, 2, λ 2, λ. By nductve applcaton of Lemma 4 and the fact that all labels of w 5,..., w 2k have to be dfferent, we conclude that w 5,..., w 2k are labeled by all even ntegers from [4, 2k 2] 2 [λ (2k 2), λ 4] 2. Then agan 13

14 T 2 T 2 T 2 T 2 T 2 T 2 T 2 T 2 T (k) w 1 w 2 w 3 w 4 T (k) v 1 v 2 v 3 v 4 w 0 v v 0 u w 2k w 2k 1 w 6 w 5 T (k 1) T (2) v 2s+4 v 2s+3 v 6 v 5 F (p s /2) F (p 1 /2) F (k) R(S) Fgure 5: Gadgets F (k) and R(S). by Lemma 4 the vertex w 0 s labeled ether by 2k or λ 2k and the vertex at dstance two from w 0 n the copy of T (k) s labeled ether by 2k 1 or λ (2k 1). The second clam follows from Lemmas 2 and 4, snce v can be labeled by r 1 and the other vertces adjacent to w 0,..., w 2k can be labeled by r 2. We proceed by constructng a rooted tree R(S) such that the root can only be labeled by ntegers from the set of labels S (see Fgure 5). Let S [4, 2k] 2 [λ 2k, λ 4] 2 be a symmetrc set of even ntegers. Denote by X the set of all ntegers from [4, 2k] 2 \ S, and let X = {p 1,..., p s }. We construct a star K 1,2s+5 wth the center u and leaves v 0,..., v 2s+4. Then four copes of T 2 rooted n v 1, v 2, v 3, v 4, respectvely, are ntroduced, and for each {1,..., s}, two copes of F (p /2) rooted n v 2+3 and v 2+4 are added. Fnally, a copy of T (k) rooted n v 0 s constructed. The vertex v 0 s declared the root of R(S). Lemma 6. For any L(2, 1, 1)-labelng of R(S) wth span λ, the root v 0 s labeled by an nteger from S; the vertces at dstance two from the root are labeled by ntegers from [0, 2k] [λ 2k, λ]. For any nteger t S and any par of dfferent ntegers r 1, r 2 [2k + 2, λ (2k + 2)] there s an L(2, 1, 1)-labelng l of R(S) wth span λ such l(v 0 ) = t and the vertces adjacent to the root are labeled by r 1 and r 2. 14

15 Proof. By Lemma 2, vertces v 1, v 2, v 3, v 4 have to be labeled by 0, 2, λ 2, λ. By Lemma 5, vertces v 5,..., v 2p+4 are labeled by all ntegers from [4, 2k] 2 [λ 2k, λ 4] 2 \ S. By Lemma 4, vertex v 0 s labeled by an even nteger from [4, 2k] 2 [λ 2k, λ 4], and ths vertex s 2-dstant from the vertces v 1,..., v 2p+4. Therefore t can only be labeled by ntegers from S. The fact that the vertces at dstance two from the root are labeled by ntegers from [0, 2k] [λ 2k, λ] mmedately follows from Lemmas 2, 3 and 5. To prove the second clam, let us note that by Lemmas 2 and 5 there are labelngs of all copes of T 2 and F (p /2) such that the vertces adjacent to the roots of these trees are labeled by r 1. Usng Lemma 4 we observe that there s a labelng of T (k), such that the root s labeled by t and the vertex adjacent to the root s labeled by r 1. It remans to label u by r 2 to receve the L(2, 1, 1)-labelng of R(S) from these labelngs of these auxlary gadgets. We conclude ths part of the proof by the followng easy observaton. Lemma 7. The tree R(S) has O(λ 4 ) vertces. 4.2 Polynomal reducton We proceed wth reducton of the well-known NP-complete 3-Satsfablty problem [14, problem L02, page 259] to our L(2, 1, 1)-Labelng problem for trees. Let Φ be a boolean formula n conjunctve normal form wth varables x 1, x 2,..., x n and clauses C 1, C 2,..., C m. Each clause conssts of three lterals. We choose λ = 8n + m + 9 f m s odd and λ = 8n + m + 10 otherwse. For each varable x we defne the set of ntegers X = {4, 4 + 2, λ (4 + 2), λ 4} and construct three copes of trees R(X ) wth roots x (1), x (2) and x (3). For each clause C j we defne the set of sx ntegers Y j as follows. For each lteral z n C j, ntegers 4, λ 4 are ncluded n Y j f z = x and ntegers 4 + 2, λ (4 + 2) are ncluded n Y j f z s a negaton of the varable x for some {1,..., n}. Then a copy of R(Y j ) wth a root y j s constructed. Fnally, we add a vertex u and jon t wth all vertces x (1), x (2), x (3) by edges and wth all vertces y j by paths of length two wth mddle vertces v 1,..., v m. Denote the obtaned tree by T (see Fgure 6). Lemma 8. The tree T has an L(2, 1, 1)-labelng of span λ f and only f the formula Φ can be satsfed. Proof. Suppose that there s an L(2, 1, 1)-labelng of T wth span λ. By Lemma 6 for each {1,..., n}, vertces x (1), x (2), x (3) are labeled by ntegers from X. Snce these vertces are 2-dstant n T, the labels have to be dfferent. Hence exactly one label from X s not used for the labelng 15

16 R(X 1 ) R(X 2 ) x (1) 1 x (2) 1 x (3) 1 v 1 y 1 v 2 y 2 R(Y 1 ) R(Y 2 ). u. R(X n 1 ) R(X n ) x (1) n x (2) n x (3) n v m 1 v m y m 1 y m R(Y m 1 ) R(Y m ) Fgure 6: The tree T. of x (1), x (2), x (3). Denote ths label by p. If p = 4 or p = λ 4 then we assume that x = true and x = false otherwse. We prove that these values gve a truth assgnment whch satsfes Φ. By Lemma 6 the vertex y j s labeled by an nteger from the Y j. Assume that y j s labeled by 4 or λ 4 for some {1,..., n}. Ths label should be dfferent from the labels of vertces x (1), x (2), x (3). Therefore C j contans the lteral x and x = true. Smlarly, f y j s labeled by or λ (4 + 2) for some {1,..., n}, then ths label s not used for the labelng of x (1), x (2), x (3),.e., C j contans the lteral x and x = false. Assume now that the formula Φ has a satsfyng truth assgnment and varables x 1,..., x n have correspondng values. Note that sets X 1,..., X n do not ntersect. We label x (2) by λ (4+2) and x (3) by λ 4 for {1,..., n}. The vertex x (1) s labeled by f x = true, and x (1) s labeled by 4 f x = false. Each clause C j contans a lteral z = true. If z = x for some {1,..., n} then Y j contans the nteger 4 and ths label was not used for the labelng of x (1), x (2), x (3). We use 4 to label y j. Smlarly, f z = x for some {1,..., n} then Y j contans the nteger and snce ths label was not used for the labelng of x (1), x (2), x (3), we label y j by By lemma 6 these labelng of roots of trees R(S) can be extended to the labelngs of all vertces of these trees such that the vertces at dstance two from the root are labeled by ntegers from [0, 4n + 2] [λ (4n + 2), λ] and the vertces adjacent to the roots are labeled by 4n + 4 and 4n + 6. We extend ths labelng to the L(2, 1, 1)-labelng of T by labelng u by 4n + 5 and v 1,..., v m by 4n + 7,..., 4n + m + 6. To conclude the proof of Theorem 2 t remans to note that t follows from Lemma 7 that T has O((n + m) 5 ) vertces. 16

17 Fgure 7: An example of a tree T wth c 2,2,1 (T ) = 9 < 10 = c 2,2,1 (T ). 5 Elegant labelngs Let f be an L(p 1,..., p k )-labelng or a C(p 1,..., p k )-labelng of a graph G wth span λ. Then f s called elegant f for every vertex u, there exsts an nterval I u modulo λ + 1 or modulo λ, respectvely, such that f(n(u)) I u and for every edge uv E(G), I u I v =. Observe that only trangle-free graphs may admt elegant labelngs. On the other hand, t s not hard to deduce that every tree allows an elegant labelng for an arbtrary collecton of dstance constrants. An example of a C(2, 2, 1)-labelng and of an elegant C(2, 2, 1)-labelng of a tree T s depcted n Fgure 7. We note that the C(2, 2, 1)-labelng n ths fgure has mnmum span. Ths can be seen as follows. Because the maxmum dstance n T s at most three, every vertex of T must receve a dfferent label. We may wthout loss of generalty assume that the rght nner vertex of T gets label 0. Then the remanng fve vertces must get label at least 2. However, f labels 2,..., 6 are used, then the label of the left nner vertex of T s of dstance one to a label of some other vertex. Ths means that a label l 7 must be used. Hence, the C(2, 2, 1)-labelng n Fgure 7 has mnmum span c 2,2,1 (T ). Note that the span of ths labelng s c 2,2,1 (T ) = 9; otherwse label 0 of the rght nner vertex s of dstance one to the vertex wth label 7. The mnmum λ for whch a graph G allows an elegant L(p 1,..., p k )- labelng, and C(p 1,..., p k )-labelng respectvely, of span λ s denoted by λ p 1,...,p k (G) and c p 1,...,p k (G), respectvely (these parameters are set to + f no elegant labelng exsts). The elegant C(2, 2, 1)-labelng n Fgure 7 has span equal to 10 = c 2,2,1 (T ). The latter equalty follows from Proposton 4. We observe that every C(p 1,..., p k )-labelng wth span λ + 1 s an L(p 1,..., p k )-labelng wth span λ and that every elegant labelng s a vald labelng. Ths leads to the followng nequaltes. Proposton 2. For any p 1 p k 1 and any graph G t holds that λ p1,...,p k (G) + 1 c p1,...,p k (G) c p 1,...,p k (G), λ p1,...,p k (G) + 1 λ p 1,...,p k (G) + 1 c p 1,...,p k (G). Elegant labels are useful already for dstance constrants p 1, p 2, p 3, because we only need to mantan a separaton of dstance p 3 between the ntervals assocated to adjacent vertces nstead of checkng every par of vertces at dstance three. We explan ths n detal n Secton

18 5.1 An upper bound for elegant C(p 1, p 2, p 3 )-labelngs of trees We present an upper bound on the mnmum span of an elegant C(p 1, p 2, p 3 )- labelng of a tree. We frst present closed formulas for stars and double stars. In the proof of Proposton 3, we show that every L(p 1,..., p k )-labelng and every C(p 1,..., p k )-labelng of a star s elegant. However, for double stars ths s already not true anymore, as can be seen from Fgure 7. Proposton 3. For any p 1 p k 1 and any n-vertex star T t holds that λ p1,...,p k (T ) = λ p 1,...,p k (T ) = p 1 + (n 2)p 2, c p1,...,p k (T ) = c p 1,...,p k (T ) = 2p 1 + (n 2)p 2. Proof. Let T be a star on vertces u, v 1,..., v n 1, where u s the center vertex. We assgn label 0 to u and label p 1 + ( 1)p 2 to each v. Ths yelds λ p1,...,p k (T ) = p 1 + (n 2)p 2 and c p1,...,p k (T ) = 2p 1 + (n 2)p 2. We now show that every L(p 1,..., p k )-labelng and every C(p 1,..., p k )- labelng of T s elegant. Let f be an L(p 1,..., p k )-labelng or C(p 1,..., p k )- labelng wth span λ. We defne I u = [f(u)+1, f(u) 1] λ+1 n the frst case, and I u = [f(u) + 1, f(u) 1] λ n the second case. For = 1,..., n 1, we defne I v = [f(u), f(u)]. Ths completes the proof of Proposton 3. Proposton 4. For any p 1 p k 1 and any double star T wth nner vertces of degree d and d, resp., wth d d t holds that λ p 1,...,p k (T ) = (d + d 2)p 2 + max{p 1 (d 1)p 2, p 3 }, c p 1,...,p k (T ) = (d+d 2)p 2 +max{p 1 d 1 2 p 2, p 3 }+max{p 1 d 1 2 p 2, p 3 }. Proof. Let T be a double star wth nner vertces u and u of degree d and d, respectvely, such that d d ; see Fgure 8a. We start wth the lnear metrc; ths case s llustrated n Fgure 8b. The mnmum length of a possble nterval I u s (d 1)p 2. Analogously, we have that I u s of length at least (d 1)p 2. In addton, every label of I u should be at least p 3 apart from every label of I u, because the dameter of T s three. Ths means that λ (d + d 2)p 2 + p 3. For any elegant L(p 1,..., p k )-labelng of T wth span λ, we also have that λ p 1 + (d 1)p 2 = (d + d 2)p 2 + p 1 (d 1)p 2, because the label of u should be at least p 1 apart from the nterval I u. Combnng ths bound wth the prevous bound yelds λ (d + d 2)p 2 + max{p 1 (d 1)p 2, p 3 }. An elegant L(p 1,..., p k )-labelng f of T wth span λ = (d + d 2)p 2 + max{p 1 (d 1)p 2, p 3 } can be obtaned by usng the arthmetc progresson 0, p 2,..., (d 1)p 2 on N(u) wth f(u ) = 0 and arthmetc progresson r, r + p 2,..., r + (d 1)p 2 on N(u ) wth f(u) = r + (d 1)p 2, where r = (d 1)p 2 + max{p 1 (d 1)p 2, p 3 }. Ths proves the frst statement of Proposton 4. 18

19 u 1 u 1 p 1 u u I u I u f(u) f(u ) p 3 p 2 u d 1 a) u d 1 p 1 b) p 3 p 2 f(u) p 1 f(u ) I u p 3 I u p 1 c) Fgure 8: The double star T and the two assocated metrcs. We llustrate the case of the cyclc metrc n Fgure 8c. Here, the lower bound (d + d 2)p 2 + 2p 3 comes from the separaton of I u and I u on both sdes. We also fnd that λ 2p 1 +(d 1)p 2 = (d+d 2)p 2 +p 1 d 1 2 p 2 + p 1 d 1 2 p 2, because the label of u should be at least p 1 apart from both ends of the cyclc nterval I u. Suppose d s odd. Then the above two bounds combne nto the value specfed n the second statement of Proposton 4; observe that both maxma attan the same value, because d 1 2 = d 1 2 n ths case. Suppose d s even. The label of u dvdes the nterval I u nto two parts. Assume that one part contans t labels of vertces from N(u). Then the other part contans d t 1 of them (we do not count the label of u n none of these two parts). Ths means that λ max{p 1, p 3 + tp 2 } + max{p 1, p 3 + (d t 1)p 2 } + (d 1)p 2. Ths expresson s mnmzed when we choose t and d t 1 as close as possble,.e., when t = d 1 2. By ths choce we agan get the bound gven n the second statement of Proposton 4. To construct an optmal elegant C(p 1,..., p k )-labelng f of T we use analogous arthmetc progressons for f as n the case of lnear metrc. To 19

20 be more precse, we defne ntervals I u and I u of length (d 1)p 2 and (d 1)p 2, respectvely, and we place the d labels of N(u) at dstance p 2 from each other n I u, and the d labels of N(u ) at dstance p 2 from each other n I u. In ths way, the dstance constrant p 2 s respected. Let the labels of u 1 and u d 1 be the two endponts of I u, and let the labels of u 1 and u d 1 be the two endponts of I u. Then we set f(u 1 ) = 0 f(u ) = d 1 2 p 2 f(u d 1 ) = (d 1)p 2 f(u 1 ) = (d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } f(u) = (d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } + d 1 2 p 2 f(u d 1 ) = (d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } + (d 1)p 2. Recall that f has span (d + d 2)p 2 + max{p 1 d 1 2 p 2, p 3 } + max{p 1 d 1 2 p 2, p 3 } n order to be an optmal elegant C(p 1,..., p k )-labelng of T. Ths means that we may wrte f(u 1 ) = 0 = (d + d 2)p 2 + max{p 1 d 1 2 p 2, p 3 } + max{p 1 d 1 2 p 2, p 3 }. In order to show that the dstance constrant p 3 s respected, t suffces to consder the extreme cases, whch are as follows. Frst, the dstance between f(u 1 ) and f(u d 1 ) s (d + d 2)p 2 + max{p 1 d 1 2 p 2, p 3 } + max{p 1 d 1 2 p 2, p 3 } ((d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } + (d 1)p 2 ) = max{p 1 d 1 2 p 2, p 3 } p 3. Second, the dstance between f(u 1 ) and f(u d 1) s (d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } (d 1)p 2 = max{p 1 d 1 2 p 2, p 3 } p 3. We are left to show that the dstance constrant p 1 s respected. Agan, we only consder the extreme cases, whch are four n total. Frst, the dstance between f(u 1 ) and f(u) s (d + d 2)p 2 + max{p 1 d 1 2 p 2, p 3 } + max{p 1 d 1 2 p 2, p 3 } ((d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } + (d 1) 2 p 2 ) = (d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } (d 1) 2 p 2 p 1 + (d 1)p 2 d 1 2 p 2 (d 1) 2 p 2 p 1, where the last nequalty follows from our assumpton that d d. Second, the dstance between f(u) and f(u d 1 ) s (d 1)p 2 +max{p 1 d 1 2 p 2, p 3 }+ d 1 2 p 2 (d 1)p 2 p 1 d 1 2 p 2 + d 1 2 p 2 p 1. Thrd, the dstance between f(u 1 ) and f(u ) s (d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } d 1 2 p 2 p 1 + (d 1)p 2 d 1 2 p 2 d 1 2 p 2 = p 1. Fourth, we may wrte f(u ) = 20

21 (d + d 2)p 2 + max{p 1 d 1 2 p 2, p 3 } + max{p 1 d 1 2 p 2, p 3 } + d 1 2 p 2. We then fnd that the dstance between f(u ) and f(u d 1 ) s (d + d 2)p 2 + max{p 1 d 1 2 p 2, p 3 } + max{p 1 d 1 2 p 2, p 3 } + d 1 2 p 2 ((d 1)p 2 + max{p 1 d 1 2 p 2, p 3 } + (d 1)p 2 ) = max{p 1 d 1 2 p 2, p 3 } + d 1 2 p 2 p 1. Ths completes the proof of Proposton 4. In Theorem 3, we consder trees that are not stars. We note that the gven upper bound holds for double stars and use Proposton 4 as the base case n our nducton proof. We also make the followng observatons. It s well known (see [24, 28]) that every power T k of a tree T s a chordal graph, and consequently, χ(t k ) = ω(t k ). Ths property enables us to compare the general upper bound of Observaton 1 wth the upper bound n Theorem 3. We note that the coeffcent n the man term ω(t 3 ) = χ(t k ) becomes p 2 nstead of p 1. Hence, the upper bound n Theorem 3 s an essental mprovement f p 2 p 1 and ω(t 3 ) s suffcently large. For the case (p 1, p 2, p 3 ) = (2, 1, 1) the upper bound become almost tght; we explan ths n Secton 5.3. Fnally, we note that Kng, Ras and Zhou [26] proved that λ p,1,1 (T ) ω(t 3 ) + p 1 for any tree T that s nether a star nor a double star. However, ther bound s not vald for c p,1,1 (T ). Theorem 3. For any p 1 p 2 p 3 1 and any tree T dfferent from a star, t holds that c p 1,p 2,p 3 (T ) p 2 ω(t 3 ) + p 1 p 2 + max{p 1 p 2, p 3 } 1. Proof. Let T be a tree that s not a star. Clam 1. T has an elegant labelng f such that for each nner vertex u, f(n(u)) s an arthmetc progresson (modulo λ) of length deg(u) and dfference p 2. We prove Clam 1 by nducton on the number of nner vertces of T. Let = 2. Then T s a double star. Let d and d wth d d denote the degrees of the two nner vertces u and u of T, respectvely. Because T s a double star, ω(t 3 ) = d + d. Then, by Proposton 4 and the fact that p 1 p 3, we obtan that c p 1,p 2,p 3 (T ) = (d + d 2)p 2 + max{p 1 d 1 2 p 2, p 3 } + max{p 1 d 1 2 p 2, p 3 } = p 2 ω(t 3 ) p 2 p 2 + max{p 1 d 1 2 p 2, p 3 } + max{p 1 d 1 2 p 2, p 3 } p 2 ω(t 3 ) p p 1 + max{p 1 p 2, p 3 }. Hence, Clam 1 holds for = 2. 21

22 Let 3, so T has at least three nner vertces. The subtree nduced by the nner vertces of T s called the nner tree of T. Let u and v be two adjacent nner vertces such that v s a leaf n the nner tree of T. Here, we choose u and v such that the sum deg T (u) + deg T (v) s mnmum over all pars of adjacent nner vertces wth the property that one of the vertces s a leaf of the nner tree of T. Let T be the tree obtaned from T after removng all neghbors of v except u. Note that these neghbors are all leaves of T. By defnton of T 3, every maxmal clque n T s obtaned by addng all possble edges n the subgraph of T nduced by two adjacent vertces and all ther neghbors. Let s and t be two adjacent vertces such that ω(t 3 ) = deg T (s) + deg T (t). Because 3, we may assume that s and t are nner vertces. Then, by our choce of u and v, we fnd that v / {s, t}. Ths means that deg T (s) + deg T (t) = deg T (s) + deg T (t). Hence, ω((t ) 3 ) deg T (s) + deg T (t) = deg T (s) + deg T (t) = ω(t 3 ). Because T s a subgraph of T, we also have ω((t ) 3 ) ω(t 3 ). We conclude that ω((t ) 3 ) = ω(t 3 ). We apply the nducton hypothess and fnd that T allows an elegant labelng f of span λ = ω((t ) 3 )p 2 + p 1 p 2 + max{p 1 p 2, p 3 } 1 = ω(t 3 )p 2 + p 1 p 2 + max{p 1 p 2, p 3 } 1 such that f (N(u)) s an arthmetc progresson (modulo λ) of length deg T (u) = deg T (u) and dfference p 2, say the arthmetc progresson on f (N(u)) s of the form a, a + p 2,..., a + (deg T (u) 1)p 2 (wth elements taken modulo λ). Then the vertces of N(v) should avod nterval I 1 = [a p 3 + 1, a + (deg(u) 1)p 2 + p 3 1] λ due to the dstance three constrant p 3. Also, they should avod nterval I 2 = [f (v) p 1 + 1, f (v) + p 1 1] λ due to the dstance one constrant p 1. Because f (v) s of dstance at least p 3 1 from the boundary of I 1, and of dstance at least p 1 1 from the boundary of I 2, we fnd I 1 I 2 p 3 +max{p 1, (deg T (u) 1)p 2 +p 3 } 1 p 3 +max{p 1, p 2 +p 3 } 1. Then I = [0, λ 1] \ (I 1 I 2 ) s an nterval of sze I = λ I 1 I 2 + I 1 I 2 ω(t 3 )p 2 + p 1 p 2 + max{p 1 p 2, p 3 } 1 (a + (deg T (u) 1)p 2 + p 3 1 a + p ) (f (v) + p 1 1 f (v) + p ) + p 3 + max{p 1, p 2 + p 3 } 1 = ω(t 3 )p 2 + p 1 p 2 + max{p 1 p 2, p 3 } 1 ((deg T (u) 1)p 2 + 2p 3 1) (2p 1 1) + p 3 + max{p 1, p 2 + p 3 } 1 = (ω(t 3 ) deg T (u))p 2 + max{p 1 p 2, p 3 } p 3 + max{p 1, p 2 + p 3 } p 1 deg T (v)p 2. Hence, I can accommodate an arthmetc progresson A of length deg(v) and dfference p 2 that contans f (u) as one of ts elements. We extend f nto a 22

23 labelng f of T by usng the elements of A \ f (u) as the labels for the (leaf) vertces adjacent to v n T. Ths concludes the proof of Theorem Optmal elegant L(p, 1, 1)- and C(p, 1, 1)-labelngs of trees The proof of Theorem 3 s constructve and can be straghtforwardly converted nto a polynomal-tme algorthm that fnds a C(p 1, p 2, p 3 )-labelng wthn the clamed upper bound. Here, we consder dstance constrants (p, 1, 1) wth p 1. For these constrants we show a stronger result, namely that λ p,1,1 (T ) and c p,1,1 (T ) can be computed n polynomal tme for any p 1. We use a dynamc programmng approach, smlarly to the approach used n the algorthm that computes λ 2,1 (T ) (see [5, 13]). Theorem 4. For any p 1 and any tree T, λ p,1,1 (T ) and c p,1,1 (T ) can be computed n polynomal tme. Proof. Let T be an n-vertex tree and λ be a postve nteger. We descrbe an algorthm that decdes whether c p,1,1 (T ) λ. The algorthm for the lnear metrc dffers only n some mnor detals. If T s a star or double star then we can apply Proposton 3 or Proposton 4, respectvely. Hence, we may assume that T s nether a star nor a double star. We may assume that λ n + 2p 4. Ths can be seen as follows. By Theorem 3, we know that T has an elegant C(p, 1, 1)-labelng wth span λ f λ ω(t 3 ) + p 2 + max{p 1, 1}. As mentoned at the start of Secton 5.2, we can construct such a labelng n polynomal tme. Suppose p = 1. Then, by Theorem 3, tree T has an elegant C(1, 1, 1)- labelng wth span λ f λ ω(t 3 ). Because T s not a double star, ω(t 3 ) n 1. Hence, T has an elegant C(1, 1, 1)-labelng f λ n 1 = n + 2p 3. Suppose p 2. We apply Theorem 3 and use ω(t 3 ) n 1 to fnd that T has an elegant C(1, 1, 1)-labelng wth span λ f λ n 1+p 2+p 1 = n + 2p 4. The dstncton n the two cases above shows that from now on we may assume that λ n + 2p 4. We frst choose a leaf r as the root of T, whch defnes the parent-chld relaton between every par of adjacent vertces. For any edge uv such that u s a chld of v, we denote by T uv the subtree of T that s rooted n v and that contans u and all descendants of u. For every such edge and for every par of ntegers, j [0, λ 1] and for every nterval I modulo λ wth j I, we ntroduce a boolean functon φ(u, v,, j, I). Ths functon s evaluated true f and only f T uv has an elegant C(p, 1, 1)-labelng f wth f(u) =, f(v) = j and I u = I. It can be calculated as follows: 1. Set an ntal value φ(u, v,, j, I) = false for all edges uv, ntegers, j [0, λ 1] and ntervals I (j I). 23