The metric dimension of Cayley digraphs

Transcription

1 Discrete Mathematics 306 ( The metric dimesio of Cayley digraphs Melodie Fehr, Shoda Gosseli 1, Ortrud R. Oellerma 2 Departmet of Mathematics ad Statistics, The Uiversity of Wiipeg, 515 Portage Aveue, Wiipeg, Ma., Caada R3B 2E9 Received 9 July 2003; received i revised form 15 September 2005; accepted 19 September 2005 Available olie 4 Jauary 2006 Abstract A vertex x i a digraph D is said to resolve a pair u, v of vertices of D if the distace from u to x does ot equal the distace from v to x. A set S of vertices of D is a resolvig set for D if every pair of vertices of D is resolved by some vertex of S. The smallest cardiality of a resolvig set for D, deoted by dim(d, is called the metric dimesio for D. Sharp upper ad lower bouds for the metric dimesio of the Cayley digraphs Cay(Δ : Γ, where Γ is the group Z 1 Z 2 Z m ad Δ is the caoical set of geerators, are established. The exact value for the metric dimesio of Cay({(0, 1, (1, 0} :Z Z m is foud. Moreover, the metric dimesio of the Cayley digraph of the dihedral group D of order 2 with a miimum set of geerators is established. The metric dimesio of a (digraph is formulated as a iteger programme. The correspodig liear programmig formulatio aturally gives rise to a fractioal versio of the metric dimesio of a (digraph. The fractioal dual implies a iteger dual for the metric dimesio of a (digraph which is referred to as the metric idepedece of the (digraph. The metric idepedece of a (digraph is the maximum umber of pairs of vertices such that o two pairs are resolved by the same vertex. The metric idepedece of the -cube ad the Cayley digraph Cay(Δ : D, where Δ is a miimum set of geerators for D, are established Elsevier B.V. All rights reserved. Keywords: Metric dimesio; Cayley digraphs; Metric idepedece; Iteger programmig; Liear programmig 1. Itroductio Let G be a coected graph. A vertex x of G is said to resolve two vertices u ad v of G if the distace d(u, x from u to x does ot equal the distace d(v,xfrom v to x. A set S of vertices of G is said to be a resolvig set for G if, for every two distict vertices u ad v, there is a vertex x of S that resolves u ad v. Alteratively, suppose S ={x 1,x 2,...,x k } is a set whose vertices have bee assiged the give order. The k-vector r(v S = (d(v, x 1, d(v, x 2,...,d(v,x k is called the represetatio of v with respect to S. The S is a resolvig set for G if ad oly if o two vertices of G have the same represetatio with respect to S. Note that x i is the oly vertex of S for which the ith coordiate of its represetatio with respect to S is 0. Therefore, whe checkig if S is a resolvig set for G, oe eed oly check that the vertices of V (G S have distict represetatios with respect to S. The miimum cardiality of a resolvig set for G is called the metric dimesio of G ad is deoted by dim(g. A miimum resolvig set is called a metric basis for G. address: o.oellerma@uwiipeg.ca (O. Oellerma. 1 Research supported by NSERC USRA Caada. 2 Research supported by NSERC Grat Caada X/$ - see frot matter 2005 Elsevier B.V. All rights reserved. doi: /j.disc

2 32 M. Fehr et al. / Discrete Mathematics 306 ( Harary ad Melter [7] ad idepedetly Slater i [13,14] itroduced this cocept. Slater referred to the metric dimesio of a graph as its locatio umber ad motivated the study of this ivariat by its applicatio to the placemet of a miimum umber of soar/lora detectig devices i a etwork so that the positio of every vertex i the etwork ca be uiquely described i terms of its distaces to the devices i the set. It was oted i [6] that the problem of fidig the metric dimesio of a graph is NP-hard. Khuller et al. [8] gave a costructio that shows that the metric dimesio of a graph is NP-hard. Their iterest i this ivariat was motivated by the avigatio of robots i a graph space.a resolvig set for a graph correspods to the presece of distictively labelled ladmark odes i the graph. It is assumed that a robot avigatig a graph ca sese the distace to each of the ladmarks ad hece uiquely determie its locatio i the graph. They also gave approximatio algorithms for this ivariat ad established properties of graphs with metric dimesio 2. Motivated by a problem from Pharmaceutical Chemistry, this problem received reewed attetio i [1]. The metric dimesio of a coected digraph D has the expected defiitio, amely, the smallest cardiality of a set S of vertices with the property that, for every two vertices u, v of D, there is some x S such that d(u, x = d(v,x. Sice the distace betwee two vertices i a digraph eed ot be defied, the metric dimesio of a digraph may ot be defied. The metric dimesio of orieted graphs was first studied by Chartrad et al. i [2] ad further i [3]. It was poited out by these authors that it remais a ope problem to determie for which directed graphs the directed distace dimesio is defied. I this paper we study the metric dimesio of Cayley digraphs for which the metric dimesio is defied. These digraphs with their high degree of symmetry are of iterest i this cotext as the metric dimesio appears to be related to both local ad global symmetry i regular (digraphs. We establish sharp bouds for this ivariat ad coclude the paper with a iteger programmig formulatio of this problem as described i [4]. The liear programmig relaxatio yields a fractioal versio of the metric dimesio whose dual yields a dual for the metric dimesio of a graph called the metric idepedece of the graph. This ivariat is defied as the maximum umber of pairs of vertices i a coected graph G such that o vertex of G simultaeously resolves two distict pairs i such a set. I [4] a geometric proof was give to show that the metric idepedece of the -cube, Q,is2.However, the proof was foud to cotai a gap. We preset here a proof of this fact usig the fractioal versio of the metric dimesio ad a combiatorial argumet. The metric idepedece of the Cayley digraph for the dihedral group of order 2, with a miimum set of geerators, is also established. 2. The metric dimesio of Cayley digraphs I this sectio we focus o determiig the dimesio of Cayley digraphs. First, recall the defiitio of the Cayley digraph for a give group with a specified set of geerators (see [5]. Let Γ be a fiite group ad Δ a set of geerators for Γ. The Cayley digraph of Γ with geeratig set Δ, deoted by Cay(Δ : Γ, is defied as follows: (1 The vertices of Cay(Δ : Γ are precisely the elemets of Γ. (2 For u ad v i Γ, there is a arc from u to v if ad oly if ug = v for some geerator g Δ. Note that for a give fiite group Γ ad a specified set of geerators Δ of Γ, every elemet of the group ca be expressed as a product of geerators i Γ. Hece, i the graph G = Cay(Δ : Γ, there exists a path i G from ay vertex of G to every other vertex of G. Thus, ay Cayley digraph is strogly coected, ad the metric dimesio of ay Cayley digraph is therefore defied. We ow fid the metric dimesio of some specific Cayley digraphs. Let G be the Cayley digraph for the group Γ = Z 2 Z 4 with the caoical set of geerators Δ ={(1, 0, (0, 1}. The graph G is show i Fig. 1(a. The vertices of G are the elemets of the group Z 2 Z 4, ad the arcs betwee vertices correspod to the geerators i Δ. I Fig. 1(a the dashed arcs correspod to the geerator (1, 0, ad the solid arcs correspod to the geerator (0, 1. The two shaded vertices i Fig. 1(a costitute a resolvig set for G, ad oe ca verify that o sigle vertex of G resolves every pair of vertices of G. Thus, the dimesio of G is 2. Now cosider the group of symmetries of the regular -go, called the dihedral group of order 2, deoted by D. This group cosists of rotatios ad reflectios. For = 4, this is the group of symmetries of the square, cosistig of the four rotatios, deoted by R 0,R 90,R 180,R 270, ad the four reflectios, deoted by A, B, C, D. Let H be the

3 M. Fehr et al. / Discrete Mathematics 306 ( (0,0 (0,1 R0 R90 (1,0 (1,1 RA 0 R90A (1,3 (1,2 R270A R180A (0,3 (0,2 R270 R180 G = Cay ( {(1,0,(0,1} : Z2 Z 4 H = Cay ({R90, A} : D 4 (a (b Fig. 1. Cayley graphs for Z 2 Z 4 ad D 4. (1,4,4 W1 (1,2,4 (3,4,2 (2,3,5 (2,1,5 (3,4,4 (4,5,1 (4,3,3 (3,2,6 W2 (4,1,5 (6,3,3 (5,4,2 W3 (2,5,1 (4,3,5 (5,2,6 (6,5,3 (6,3,5 (5,4,6 (5,6,4 (4,5,3 (3,6,2 Fig. 2. Cay({(1, 2, (1, 2, 3, 4} :S 4. Cayley digraph for the group D 4 with geeratig set Δ ={R 90,A}. The graph H is show i Fig. 1(b. The structure of this graph is very similar to that of G i Fig. 1(a, except that the two directed 4-cycles i H are orieted i opposite directios, while those of G are orieted i the same directio. Both of these Cayley digraphs are costructed with two geerators, ad i both cases the miimum order of a geerator i Δ is 2. However, while the dimesio of G is 2, the dimesio of H is 4. The four shaded vertices i Fig. 1(b costitute a resolvig set for H, ad it ca be verified that o three vertices of H resolve every pair of vertices of H (see Theorem 4. Fig. 2 shows the Cayley digraph for the o-abelia symmetric group of degree 4, deoted by S 4, with geeratig set Δ={(1, 2, (1, 2, 3, 4}. The vertices w 1,w 2,w 3 form a resolvig set for this graph as the represetatios of the vertices of Cay({(1, 2, (1, 2, 3, 4} :S 4 with respect to the set {w 1,w 2,w 3 } are all distict. These distict represetatios are

4 34 M. Fehr et al. / Discrete Mathematics 306 ( W11 W12 W1j W1m W21 W22 W2j W2m Wt1 Wt2 Wtj Wtm W1 W2 Wj Wm H1 H2 Hj Hm Fig. 3. The graph H. displayed i Fig. 2. Furthermore, o two vertices of the graph costitute a resolvig set. Thus, the dimesio of this graph is 3. Some familiar graphs are Cayley digraphs. For example, the -cube is the Cayley digraph for the group Γ=Z 2 Z 2 Z 2 ( times, with the caoical set of geerators Δ ={(1, 0, 0,...,0, (0, 1, 0, 0,...,0,...,(0, 0,...,0, 1}. Later i this sectio, Corollary 3 shows that 2 dim(q. More geerally, we will establish bouds o the dimesio of the Cayley digraph for the direct product of ay umber of cyclic groups with the caoical set of geerators. To this ed, we first look at how the dimesio of the Cayley digraph for a group Γ chages whe we take the direct product of Γ ad the cyclic group of order m, where m is a positive iteger. Clearly, the Cayley digraph for the ew group Γ Z m ad its dimesio will deped o the set of geerators chose. The followig theorem bouds this dimesio subject to a specific choice of geerators. Theorem 1. Let Γ be a group of order ad let Δ ={g 1,g 2,...,g k } be a geeratig set for Γ. Let H = Cay(Δ : Γ. Let Δ ={(g 1, 0, (g 2, 0,...,(g k, 0, (e Γ, 1} be a geeratig set for the group Γ = Γ Z m, where m 2 ad e Γ is the idetity elemet of Γ. The for H = Cay(Δ : Γ, dim(h dim(h dim(h + m 1. Proof. The graph H cosists of m copies of the graph H. Label these copies H 1,H 2,...,H m. Let V(H j = {u 1j,u 2j,...,u j }, for 1 j m, where for each i (1 i, u ij is i the same positio i H j as u ik is i H k (for 1 j,k m. The arcs betwee the m copies of H are precisely the arcs o the directed cycles C i give by C i : u i1,u i2,...,u im,u i1 (for 1 i. That is, H is costructed by takig m copies of H, H 1,H 2,...,H m, ad placig arcs from H i to H i+1 (subscripts modulo m betwee correspodig vertices (for 1 i m. To establish the upper boud i the theorem we eed to fid a resolvig set for H of cardiality dim(h +m 1. Let W ={w 1,w 2,...,w t } be a basis for H. The dim(h =t, ad there exists a correspodig set W j ={w 1j,w 2j,...,w tj } of vertices of the graph H j which is a basis for H j (for 1 j m. Let W ={w 11,w 21,...,w t1,w 12,w 13,...,w 1m }. The W =t + m 1 = dim(h + m 1. We claim that W is a resolvig set for H. The graph H is show i Fig. 3. Note that, for simplicity, oly oe of the m-cycles of H is show i Fig. 3 ad that the shaded vertices i the figure are the vertices of W. To demostrate that W is a resolvig set for H, let u ad v be distict vertices of H. We show that u ad v are resolved by some vertex of W. We cosider two cases. Case 1: Both u ad v are vertices of H j for some j (1 j m. Sice W j is a basis for H j, it follows that d Hj (u, w ij = d Hj (v, w ij for some i (1 i t. For this i, the structure of H guaratees that d H (u, w i1 =d Hj (u, w ij +m j +1 = d Hj (v, w ij + m j + 1 = d H (v, w i1. Thus, d H (u, w i1 = d H (v, w i1, ad so w i1 resolves u ad v i this case.

5 M. Fehr et al. / Discrete Mathematics 306 ( Case 2: u V(H i ad v V(H j for some i ad j (where 1 i<j m. We cosider two subcases. Subcase 2.1: u ad v are i correspodig positios i H i ad H j, respectively. That is, u = u qi ad v = u qj for some q (1 q. I this case, w 11 resolves u ad v. To see this ote d Hi (u, w 1i = d Hj (v, w 1j,soifi = 1wehave that d H (u, w 11 = d Hi (u, w 1i <d Hj (v, w 1j + m j + 1 = d H (v, w 11, ad if i = 1, sice i<j, d H (u, w 11 = d Hi (u, w 1i + m i + 1 >d Hj (v, w 1j + m j + 1 = d H (v, w 11. I either case, d H (u, w 11 = d H (v, w 11. Subcase 2.2: u ad v are i differet positios i H i ad H j, respectively. That is, u = u si ad v = u rj for some s, r where 1 s = r. Ifu ad v are resolved by w 11, the u ad v are resolved by a vertex of W,sowemay assume that u ad v are ot resolved by w 11. The d H (u, w 11 = d H (v, w 11.Nowifi = 1 ad u V(H 1, the d H (u, w 1j = d H (u, w 11 + j 1 = d H (v, w 11 (m j + 1 = d H (v, w 1j, ad so d H (u, w 1j = d H (v, w 1j. (These distaces differ by m. Thus if i = 1, w 1j resolves u ad v. O the other had, if i = 1 ad u V(H i for some i {2, 3,...,m}, the d H (u, w 1i = d H (u, w 11 (m i + 1 = d H (v, w 11 + (i 1 = d H (v, w 1i, ad so d H (u, w 1i = d H (v, w 1i. (Agai these distaces differ by m. Thus if i = 1, w 1i resolves u ad v. I either case, u ad v are resolved by some vertex of W. Thus W is a resolvig set for the graph H, as claimed, ad so dim(h W =t + m 1 = dim(h + m 1. To establish the lower boud i the theorem, let H 1,H 2,...,H m be the m copies of H i H. Let W be a basis for H. Let W i = W V(H i (for 1 i m. Let W i be the vertices of H 1 that correspod to the vertices of W i i H i (2 i m. Let U 1 V(H 1 be the uio of W 1 ad the sets W 2,W 3,...,W m. Thus, ( U 1 = W m 1 2 W i W 1 + W 2 + W W m = W 1 + W 2 + W W m = W. We claim that U 1 is a resolvig set for H 1. Let u ad v be distict vertices of H 1. We show that u ad v are resolved by some vertex of U 1. Sice W is a basis for H, ad u ad v are vertices of H, there exists a vertex w W such that d H (u, w = d H (v, w. Recall that W = W 1 W 2 W m. Thus either w W 1 or w W i for some i {2, 3,...,m}. If w W 1, the w U 1 sice W 1 U 1, ad w resolves u ad v i H 1. This follows from the fact that d H1 (u, w = d H (u, w = d H (v, w = d H1 (v, w. If w W i for some i {2, 3,...,m}, the let w be the vertex i W i U 1 correspodig to w. The w resolves u ad v. This follows from the fact that d H1 (u, w = d H (u, w (i 1 = d H (v, w (i 1 = d H1 (v, w. I either case, u ad v are resolved by some vertex of U 1.SoU 1 is a resolvig set for H 1. This implies that dim(h = dim(h 1 U 1 W =dim(h, from which the lower boud i the theorem follows. Recall the Cayley digraph i Fig. 1(a for the group Z 2 Z 4 with the caoical set of geerators. This graph has dimesio 2. I the followig theorem we geeralize this result ad show that, for positive itegers m ad, the dimesio of the Cayley digraph for the group Z m Z with the caoical set of geerators {(1, 0, (0, 1} is the miimum of m ad. Theorem 2. Let m ad be positive itegers. Let H be the Cayley digraph for the group Z Z m with geeratig set {(1, 0, (0, 1}. The dim(h = mi(m,. Proof. Suppose that m. First we show that dim(h mi(m, = m. Let H be the Cayley digraph for the group Γ = Z with geeratig set Δ ={1}. The H is the directed -cycle, which clearly has dimesio 1. Let Δ ={(1, 0, (e Γ, 1}={(1, 0, (0, 1}. The Δ is a geeratig set for the group H = H Z m = Z Z m. By Theorem 1, dim(h dim(h + m 1 = 1 + m 1 = m. It remais to show that dim(h m. Suppose, to the cotrary, that there exists a basis B for H such that B <m. As i Theorem 1, H is costructed from m copies of the directed -cycle, label them H 1,H 2,...,H m, by placig arcs from H i to H i+1 (subscripts modulo m betwee correspodig vertices (for 1 i m. Thus, there are vertex disjoit directed m-cycles i the graph H, as well as m vertex disjoit directed -cycles. A vertex i the ith -cycle has first coordiate i (for 0 i m 1, ad a vertex i the jth m-cycle has secod coordiate j (for 0 j 1.

6 36 M. Fehr et al. / Discrete Mathematics 306 ( Table 1 Table of dimesios of the graph Cay({(1, 0, 0, (0, 1, 0, (0, 0, 1} :Z m Z Z k for m,, k 5 m k Lower boud Upper boud Dimesio Sice there are less tha m vertices i the basis B, ad there are m-cycles, there must be at least oe directed -cycle which cotais o vertex of B. Due to the symmetry of the graph H we ca assume, without loss of geerality, that the 0th -cycle cotais o vertex of B. Also, sice B <m, ad there are m-cycles, there is at least oe directed m-cycle which cotais o vertex of B. Agai, by the symmetry of the graph H, we ca assume that the 0th m-cycle cotais o vertex of B. Now cosider the vertices (1, 0 ad (0, 1 ad ay vertex w B. Sice o vertex of B lies o either the 0th m-cycle or the 0th -cycle, there exists a shortest path from (0, 1 to w, ad also oe from (1, 0 to w, which cotais (1, 1. However, both vertices (0, 1 ad (1, 0 are adjacet to (1, 1. Thus, for ay vertex w of B, d((0, 1, w = d((1, 0, w, ad so the vertices (0, 1 ad (1, 0 are ot resolved by ay vertex of B, which cotradicts the fact that B is a basis for H. Hece dim(h m. Theorem 2 illustrates that the upper boud of Theorem 1 is attaied for Cay({(1, 0, (0, 1} :Z Z m. Usig the iteger programmig formulatio for (digraphs, as described i the ext sectio, values for the metric dimesio of the Cayley digraphs for the direct product of three cyclic groups with the caoical set of geerators are obtaied (see Table 1. The upper ad lower bouds of Theorem 1 are also icluded i Table 1. From these values we coclude that it is possible to have equality for either boud i Theorem 1 ad that itermediate values ca also be attaied. Fidig exact values for the metric dimesio of Cayley digraphs of abelia groups of the form Z 1 Z 2 Z k (for k 3 with the caoical set of geerators Δ ={(1, 0, 0,...,0, (0, 1, 0, 0,...,0,...,(0, 0,...,0, 1} is a ope problem. However, the previous two theorems ca be used to boud the dimesio of these Cayley digraphs. These bouds are give i the followig corollary. Corollary 3. Let k, 1, 2,..., k be positive itegers where k 2 ad k. Let Γ = Z 1 Z 2 Z k ad Δ ={(1, 0, 0,...,0, (0, 1, 0, 0,...,0,...,(0, 0,...,0, 1}. If G = Cay(Δ : Γ, the k 2 dim(g 2 + ( i 1. i=3 Proof. This result follows immediately from repeated applicatios of Theorem 1 ad from Theorem 2. The values give i Table 1 support our ituitio that there appears to be a correlatio betwee higher degrees of symmetry i a graph ad the metric dimesio. I particular, if m, ad k are all distict (so that there is less symmetry the lower boud of the previous corollary is always achieved.

7 M. Fehr et al. / Discrete Mathematics 306 ( (1, 0 ( 1, -1 (1,1 (0,0 ( 0, -1 (0,1 (0,i-1 (0,i ( 1, i-1 (1,i Fig. 4. Cay({R 360/,A}:D. Recall that the Cayley digraph for the dihedral group D 4 with geeratig set Δ ={R 90,A} has dimesio 4. I the ext theorem we geeralize this result to the dihedral group D of order 2. Theorem 4. Let be a positive iteger, 3. Let G be the Cayley digraph for the group D with geeratig set {R 360/,A}, where A is ay reflectio i the group D. The dim(g =. Proof. Label each vertex of G with a ordered pair, where the first coordiate is 0 (or 1 if the vertex is o the ier (or outer -cycle, respectively. The outer -cycle is directed couter-clockwise, ad the ier -cycle is directed clockwise. The secod coordiate deotes the positio of the vertex o the -cycle, from 0 to 1 i the clockwise directio. The resultig Cayley digraph is show i Fig. 4. To show that the dimesio is at most, let W ={(1, 0, (1, 1,...,(1, 1}. All pairs of vertices i V (G W (i.e. pairs of vertices o the ier cycle are resolved sice (0,iis the uique vertex i V (G W that is adjacet to (1,iad is thus the oly vertex whose represetatio has ith coordiate 1. Hece dim(g. To establish the lower boud, observe that the oly vertices that resolve the pair {(0,i 1, (1,i} (for 1 i are the two vertices i the pair (see Fig. 4. Hece, ay resolvig set cotais at least vertices. Thus, dim(g. 3. A fractioal versio of the metric dimesio problem ad its dual Currie ad Oellerma i [4] formulated the problem of fidig the metric dimesio of a graph as a iteger programme. This formulatio aturally gives rise to a fractioal versio of the metric dimesio of a graph, ad its fractioal dual implies a iteger dual for the metric dimesio of a graph. Fractioal versios of other graph ivariats are discussed i [11]. Let G be a coected graph of order. Suppose V is the vertex set of G ad V p the collectio of all ( 2 pairs of vertices of G. Let R(G deote the bipartite graph with partite sets V ad V p such that x i V is joied to a pair {u, v} i V p if ad oly if x resolves u ad v i G. We call R(G the resolvig graph of G. The smallest cardiality of a subset S of V such that the eighborhood N(S of S i R(G is V p is thus the metric dimesio of G. Suppose V ={v 1,v 2,...,v } ad V p ={s 1,s 2,...,s ( 2 }. Let A = (a ij be the ( 2

8 38 M. Fehr et al. / Discrete Mathematics 306 ( matrix with { 1 if si v a ij = j E(R(G, 0 otherwise for 1 i ( 2 ad 1 j. The iteger programmig formulatio of the metric dimesio is give by: miimize f(x 1,x 2,...,x = x 1 + x 2 + +x subject to the costraits ad Ax [1] ( 2 x [0], where x =[x 1,x 2,...,x ] T, [1] k is the k 1 matrix all of whose etries are 1, [0] is the 1 matrix all of whose etries are 0 ad x i {0, 1} for 1 i. If we relax the coditio that x i {0, 1} for every i ad require oly that x i 0 for all i, the we obtai the followig liear programmig problem: miimize f(x 1,x 2,...,x = x 1 + x 2 + +x subject to the costraits ad Ax [1] ( 2 x [0]. I terms of the resolvig graph R(G of G, solvig this liear programmig problem amouts to assigig oegative weights to the vertices i V so that for each vertex i V p the sum of the weights i its eighborhood is at least 1 ad such that the sum of the weights of the vertices i V is as small as possible. The smallest value for f is called the fractioal dimesio of G ad is deoted by frdim(g. The dual of this liear programmig problem is give by: maximize f(y 1,y 2,...,y ( 2 = y 1 + y 2 + +y ( 2 subject to the costraits ad A T y [1] y [0] ( 2, where y =[y 1,y 2,...,y ( 2 ]T. For the resolvig graph R(G of G this amouts to assigig oegative weights to the vertices of V p so that for each vertex i V the sum of the weights i its eighborhood is at most 1 ad subject to this such that the sum of the weights of the vertices i V p is as large as possible. The correspodig iteger programmig problem asks for a assigmet of 0 s ad 1 s to the vertices i V p such that the sum of the weights of the eighbors of every vertex i V is at most 1 ad such that the sum of the weights of the vertices i V p is as large as possible. This iteger programmig problem, which correspods to the dual of the fractioal form of the metric dimesio problem, is equivalet to fidig the largest collectio of pairs of vertices of G o two of which are resolved by the same vertex. This maximum is called the metric idepedece umber of G, deoted by mi(g. A collectio of pairs of vertices of G, o two of which are resolved by the same vertex, is called a

9 M. Fehr et al. / Discrete Mathematics 306 ( idepedetly resolved collectio of pairs. The fractioal metric idepedece umber of G is defied i the expected maer ad is deoted by frmi(g. Clearly, dim(g frdim(g ad frmi(g mi(g. It follows from the Duality Theorem for liear programmig that frdim(g = frmi(g. We thus obtai the followig strig of iequalities: dim(g frdim(g = frmi(g mi(g. Note that, for ay coected graph G, mi(g frdim(g. We ow use this fact to show that the metric idepedece of the -cube, deoted by Q, is 2 for all positive itegers 2. To this ed, the followig two lemmas, which ca be established i a straightforward maer usig iductio, are useful. Lemma 5. For all positive itegers k, ( 2k 2 2k 1. k Lemma 6. For all positive itegers k, ( 2k 1 2 2k 2. k 1 We are ow ready to prove the followig theorem. Theorem 7. For all 2, mi(q = 2. Proof. To see that mi(q 2, cosider the two pairs of vertices that are diametrically opposite to oe aother o ay 4-cycle i Q. These two pairs are ot resolved by the same vertex ad are thus metrically idepedet. It follows that mi(q 2. It remais to show that mi(q 2. We recall here two differet ways of describig the graph Q. (i Q is the graph whose vertex set cosists of all 2 -tuples of 0 s ad 1 s, ad where two -tuples are joied by a edge if ad oly if they differ i exactly oe positio. (ii Q ca be obtaied from two copies of the ( 1-cube, Q 1, by joiig correspodig vertices. Let Q 1 ad Q 1 deote two vertex disjoit copies of the ( 1-cube i the graph Q. We may assume that all of the vertices of Q 1 havea0ithefirst positio ad those i Q 1 havea1ithefirst positio of their -tuples. Assig each vertex of Q 1 a value of 1/2 2 ad each vertex of Q 1 a value of 0. Let R(Q be the resolvig graph of Q defied above. Let V p be the collectio of all pairs of vertices of Q. If we ca show, with this assigmet of fractioal values to the vertices of Q, that the sum of the values of the eighbors of the vertices i V p is at least 1, the we have show that ( 1 frdim(q (0 = 2. Sice mi(q frdim(q, the result will follow. Sice Q is bipartite, vertices from distict partite sets are resolved by every vertex of Q ad hece every vertex of Q 1. So, for such a pair, the sum of the values of its eighbors i R(Q is at least 2 1 (1/2 2 = 2. Suppose ow that u ad v are distict vertices that belog to the same partite set ad suppose that d(u, v = d. The d is ecessarily eve. Let P be the collectio of all positios for which the -tuples of u ad v agree. There are d such positios. Let P be the collectio of all positios where the -tuples for u ad v disagree. The P =d. Ifa vertex z of Q does ot resolve u ad v, the it is the same distace from u ad v. Note that the umber of positios i P where the -tuple for z differs from the oe for u is the same as for v. Suppose that the umber of positios i P where the -tuple for z differs from the -tuple for u is k. The the umber of positios i P where the -tuple for z differs from the -tuple for v is d k. Sice d(u, z = d(v,z we have that k = d k.sothe-tuple for z differs from the oe for u ad the oe for v i exactly k positios belogig to P. We cosider two cases.

10 40 M. Fehr et al. / Discrete Mathematics 306 ( ( Case 3: Suppose positio 1 belogs to P. The u ad v either both belog to Q 1 or to Q 1. There are 2 2k 1 2k ( k vertices i Q 1 that are the same distace from u ad v. By Lemma 5, 2 2k 1 2k k 2 2k 1 (2 2k 1 = 2 2.So there are at least =2 2 vertices of Q 1 that resolve u ad v. Hece, the sum of the values of the eighbors of {u, v} i R(Q is at least 2 2 (1/2 2 = 1. Case 4: Suppose positio 1 belogs to P. The oe of u ad v belogs to Q 1, ad the other to Q 1. Suppose u V(Q 1 ad v V(Q 1. Sice v ecessarily differs i ( positio 1 from all vertices of Q 1 ad as it differs i exactly k positios from z that belog to P, there are 2 2k 2k 1 k 1 vertices z i Q ( 1 that do ot resolve u ad v. By Lemma 6, 2 2k 2k 1 k 1 (2 2k (2 2k 2 = 2 2. So there are at least = 2 2 vertices i Q 1 that resolve the pair {u, v}. Thus the sum of the values of the eighbors of {u, v} i R(Q is at least 2 2 (1/2 2 = 1. Hece frdim(q 2, ad the result follows. Note that for ay coected graph G of order, ay set S of idepedetly resolved pairs of vertices of G must be pairwise disjoit; otherwise, ay vertex commo to two pairs i the set resolves both pairs, cotradictig the fact that the pairs of vertices i S are resolved idepedetly. Thus, mi(g /2. This fact ad the proof of Theorem 4 lead to the followig result. Theorem 8. Let be a positive iteger, 3. Let G be the Cayley digraph for the dihedral group D with geeratig set {R 360/,A}, where A is ay reflectio of D. The mi(g =. Proof. The proof of Theorem 4 demostrates that the Cayley digraph G cotais idepedetly resolved pairs of vertices. Thus mi(g. O the other had, sice G has order 2, mi(g 2/2 =. 4. Closig remarks This paper studies the metric dimesio of Cayley digraphs (with miimal geeratig sets. Bouds for the metric dimesio of the Cayley digraphs Cay(Δ : Γ, where Γ is the group Z 1 Z 2 Z m ad Δ is the caoical set of geerators, are established ad it is show that these bouds are sharp if m = 2. The case where m = 3 has bee studied i more depth i [9] but is still partially uresolved. The case where m 4 remais ope. The udirected versio was ivestigated i [10] ad was show to be urelated to the size of the cyclic groups. More specifically it was show, for the case m = 2 ad if at least oe of the cyclic groups has order at least 3, that the metric dimesio is 3 or 4 ad depeds o the parity of the cyclic groups. A iteger programmig formulatio of the metric dimesio of (digraphs ad its correspodig dual, the metric idepedece, was studied for Cayley digraphs. It was show that the metric dimesio ad the metric idepedece for the Cayley digraph of the dihedral group D, with a miimal set of geerators, are both equal to. O the other had, it is show that the metric dimesio of the -cube is 2. The asymptotically exact value for the metric dimesio of the -cube is 2/ log (see [12]. Thus the ratio of the metric dimesio to the metric idepedece of Cayley (digraphs maybe arbitrarily large. The metric dimesio of the Cayley digraph for the o-abelia symmetric group S 4, with geeratig set Δ = {(1, 2, (1, 2, 3, 4}, is show to be 3. However, it remais a ope problem to determie the metric dimesio ad metric idepedece of these Cayley digraphs for >4. Refereces [1] G. Chartrad, L. Eroh, M. Johso, O.R. Oellerma, Resolvability i graphs ad the metric dimesio of a graph, Discrete Appl. Math. 105 ( [2] G. Chartrad, M. Raies, P. Zhag, The directed distace dimesio of orieted graphs, Math. Bohemica 125 ( [3] G. Chartrad, M. Raies, P. Zhag, O the dimesio of orieted graphs, Utilitas Math. 60 ( [4] J. Currie, O.R. Oellerma, The metric dimesio ad metric idepedece of a graph, J. Combi. Math. Combi. Comput. 39 (

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