Distance between vertices of lattice polytopes

Transcription

1 Optmzaton Letters ORIGINAL PAPER Dstance between vertces of lattce polytopes Anna Deza 1 Antone Deza 2 Zhongyan Guan 2 Lonel Pournn 3 Receved: 18 January 2018 / Accepted: 1 October 2018 Sprnger-Verlag GmbH Germany, part of Sprnger Nature 2018 Abstract A lattce (d,k)-polytope s the convex hull of a set of ponts n dmenson d whose coordnates are ntegers rangng between 0 and k. We consder the largest possble dstance δ(d,k) between two vertces n the edge-graph of a lattce (d,k)-polytope. We show that δ(5,3) and δ(3,6) are equal to 10. Ths substantates the conjecture whereby δ(d,k) s acheved by a Mnkowsk sum of lattce vectors. Keywords Lattce polytope Dameter Mnkowsk sum 1 Introducton Boundng the maxmal possble dameter of the edge-graph of a polyhedron as a functon of ts dmenson d and the number n of ts facets s not only a natural queston of extremal dscrete geometry, but t s also hstorcally connected wth the theory of smplex methods. Larman [15] gave an upper bound on ths quantty that s lnear as a functon of n, but exponental as a functon of d, whch was subsequently refned by Barnette [3] and generalzed by Esenbrand et al. [9] and Labbé et al. [14]. Kala and Kletman [11] found an upper bound that s quas-polynomal as a functon of d and n, whch was subsequently refned by Todd [20] and Sukegawa [18]. Lower bounds B Antone Deza deza@mcmaster.ca Anna Deza anna.deza@mal.utoronto.ca Zhongyan Guan guanz@mcmaster.ca Lonel Pournn lonel.pournn@unv-pars13.fr 1 Unversty of Toronto, Toronto, ON, Canada 2 McMaster Unversty, Hamlton, ON, Canada 3 Unversté Pars 13, Vlletaneuse, France

2 A. Deza et al. have also been obtaned by Klee and Walkup [12] and by Santos [17], dsprovng the Hrsch conjecture for unbounded polyhedra and for polytopes, respectvely. In the case of a lattce polytope,.e. the convex hull of a set of ponts wth nteger coordnates, the range for the coordnates of the vertces can be used as an alternatve to n. A lattce (d,k)-polytope s the convex hull of a set of ponts n dmenson d whose coordnates are ntegers rangng between 0 and k.let δ(d,k) denote the largest possble dameter of a lattce (d,k)-polytope. The case when k = 1 was nvestgated by Naddef [16] who showed that δ(d,1) = d, and thus that lattce (d,1)-polytopes satsfy the Hrsch bound. Ths result was generalzed to δ(d,k) kd by Klenschmdt and Onn [13]. The case when d = 2 was studed ndependently by Thele [19], Balog and Bárány [2] and Acketa and Žunć [1]. It can also be found n Zegler s book [21] as Exercse In partcular, δ(2,k) s known for all k. Klenschmdt and Onn s upper bound was strengthened for k 2toδ(d,k) kd d/2 by Del Pa and Mchn [5] wth equalty for k = 2, and then to δ(d,k) kd 2d/3 (k 3) when k 3 by Deza and Pournn [7]. The quanttes δ(d,2) = 3d/2,δ(4, 3) = 8, δ(3,4) = 7, and δ(3,5) = 9 were determned n [4,5,7]. Investgatng the lower bound, Deza, Manoussaks, and Onn [6] bult lattce (d,k)-polytopes of dameter (k + 1)d/2. These polytopes are Mnkowsk sums of sets of the shortest possble lattce vectors, no two of whch are collnear. In ths paper, we nvestgate Conjecture 1 statng that δ(d,k) s acheved by such polytopes. Conjecture 1 [6] For any d and k, δ(d,k) s acheved, up to translaton, by a Mnkowsk sum of lattce vectors. In partcular, when k < 2d, δ(d,k) = (k + 1)d/2. Our man contrbuton s the determnaton of δ(5,3) and δ(3,6), reported n bold n Table 1 along wth the other known values of δ(d,k). The determnaton of δ(5,3) and δ(3,6) s detaled n Sect. 4. Theorem 1 δ(5,3) and δ(3,6) are equal to 10. The paper s organzed as follows. Structural propertes of lattce polytopes wth large dameter are presented n Sect. 2. Those propertes are used n Sect. 3 to generalze the computatonal framework ntroduced n [4] to determne smaller nstances, Table 1 The largest possble dameter δ(d,k) of a lattce (d,k)-polytope k d d. d. 32 d

3 Dstance between vertces of lattce polytopes allowng to prove Theorem 1 n Sect. 4. Ths manuscrpt s dedcated to the memory of Mchel Deza who worked on a related queston: boundng the dameter of a polytope n terms of the lattce ponts t contans, see [8]. 2 Structural propertes of lattce polytopes wth large dameter Gven two vertces u and v of a polytope P, we call d(u,v) ther dstance n the edgegraph of P. IfF s a face of P, we further call d(u,f) = mn{d(u,v) : v F}. The dameter of the edge-graph of P s denoted by δ(p). The coordnates of a vector x R d are denoted by x 1 to x d, and ts scalar product wth a vector y R d by x y. We recall Lemma 1 ntroduced by Del Pa and Mchn. Lemma 1 [5] Consder a lattce (d,k)-polytope P. If u s a vertex of P and c R d s a vector wth nteger coordnates, then d(u,f) c u γ where γ = mn{c x : x P} and F ={x P : c x = γ }. We consder the followng 2d faces of P whch are key objects n our computatonal framework. Let γ (P) = mn{x : x P} and F (P) ={x P : x = γ (P)}. Smlarly, let γ + (P) = max{x : x P} and F + (P) ={x P : x = γ + (P)}. When there s no ambguty, F (P), and F + (P) wll be smply denoted by F and F +. Consderng paths from u to v gong though F (P) or F + (P), yelds: d(u,v) mn =1,...,d mn{δ(f ) + d(u, F ) + d(v, F ), δ(f + ) + d(u, F + ) + d(v, F + )}. (1) Usng nequalty (1) and settng c as a bass vector, or ts opposte, n Lemma 1 gve Corollary 1, whch s the key ngredent to show by nducton that δ(d,k) kd. Corollary 1 Let u and v be two vertces of a lattce (d,k)-polytope, then d(u,v) mn =1,...,d mn{δ(f ) + u + v,δ(f + ) + 2k u v }. Proposton 1 s borrowed from [10], see Corollary 12.2 and Proposton 12.4 theren. It s used to prove Lemma 2. Proposton 1 Let P 1 and P 2 be two polytopes n R d and P = P 1 + P 2 ther Mnkowsk sum. Let v = v 1 + v 2, such that v 1 P 1 and v 2 P 2. Then v s a vertex of P f and only f () v 1 and v 2 are vertces of P 1 and P 2, respectvely; and () there exsts an objectve functon c R d that s unquely mnmzed at v 1 n P 1 and at v 2 n P 2. Moreover, f u and v are adjacent vertces of P wth Mnkowsk decompostons u = u 1 + u 2 and v = v 1 + v 2, respectvely, then u and v are ether adjacent vertces of P, or they concde, for = 1, 2. Lemma 2 For any lattce (d,k)-polytope Q, there exsts a lattce (d,k)-polytope P of dameter at least δ(q) satsfyng γ (P) = 0 and γ + (P) = kfor= 1,...,d.

4 A. Deza et al. Proof Assume that, for some, γ + (Q) γ (Q) <k. Up to translaton, we can assume that γ (Q) = 0. Consder the segment σ = conv{0,(k γ + (Q))c } where c s the pont whose coordnates are all equal to 0 except for the -th coordnate that s equal to 1. By constructon, Q + σ s a lattce (d,k)-polytope such that γ (Q + σ ) = 0 and γ + (Q + σ ) = k.letu and v be two vertces of Q such that d(u,v) = δ(q).by Proposton 1, wth c = c, there exst two vertces u and v of Q + σ obtaned as the Mnkowsk sums of u and v, respectvely wth two (possbly dentcal) vertces of σ. Moreover, for any path of length l between u and v n the edge-graph of Q +σ, there exsts a path of length at most l between u and v n the edge-graph of Q. Consequently, the dstance of u and v n Q s at most the dstance of u and v n Q + σ. Thus, δ(q) δ(q + σ ).Ifγ j + (Q + σ ) γ j (Q + σ )<k for some j =, the above procedure can be repeated untl no such coordnate remans. Lemma 3 Assume that δ(d,k) = δ(d 1,k) + k g for an nteger g wth 0 g k. () If u and v are two vertces of a lattce (d,k)-polytope such that d(u,v) = δ(d,k), then u + v k gfor= 1,...,d. () There exsts a lattce (d,k)-polytope P of dameter δ(d,k) such that the ntersecton of P wth each facet of the hypercube [0,k] d s, up to an affne transformaton, a lattce (d 1,k)-polytope of dameter at least δ(d 1,k) 2g. Proof Settng d(u,v) = δ(d 1,k) + k g n Corollary 1 yelds: Thus, δ(d 1,k) + k g δ(f ) + (u + v ) for = 1,...,d, (2) δ(d 1,k) + k g δ(f + ) + 2k (u + v ) for = 1,...,d. (3) k g u + v + δ(f ) δ(d 1,k) for = 1,...,d, (4) k + g u + v + δ(d 1,k) δ(f + ) for = 1,...,d. (5) Hence, snce both δ(f ) and δ(f + ) are at most δ(d 1,k), the nequalty k g u +v k + g holds for = 1,...,d; that s, tem () holds. By Lemma 2 there exsts a lattce (d,k)-polytope P of dameter δ(d 1,k) + k g such that the ntersecton of P wth each facet of the hypercube [0,k] d s nonempty. Let u and v be two vertces of P such that d(u,v) = δ(p). Inequaltes (4) and (5) can be rewrtten as: δ(f ) δ(d 1,k) g + k (u + v ) for = 1,...,d, (6) δ(f + ) δ(d 1, k) g k + (u + v ) for = 1,...,d. (7) Thus, snce k g u + v k + g for = 1,...,d by tem (), δ(f ) and δ(f + ) are at least δ(d 1, k) 2g for = 1,...,d; that s, tem () holds. We recall that the bounds obtaned by Del Pa and Mchn [5] and Deza and Pournn [7] hold n general for lattce polytopes nscrbed n rectangular boxes.

5 Dstance between vertces of lattce polytopes Corollary 2 (Remark 4.1 n [7]) Let δ(k 1,...,k d ) denote the largest possble dameter of a polytope whose vertces have ther -th coordnate n {0,...,k } for = 1,...,d and, up to relabelng, k 1 k 2... k d. The followng nequaltes hold: () δ(k 1,...,k d ) k 2 + k 3 + +k d d/2 +2 when k 1 2, () δ(k 1,...,k d ) k 2 + k 3 + +k d 2d/3 +3 when k 1 3. Observe that the statement of Remark 4.1 n [7] contans a typographcal ncorrectness as k 1 and k d were nterchanged n () and n (). Conjecture 1 can also be stated for lattce polytopes nscrbed n rectangular boxes; that s, δ(k 1,...,k d ) s at most (k 1 + k k d + d)/2, and s acheved, up to translaton, by a Mnkowsk sum of lattce vectors. Note that ths generalzaton of Conjecture 1 holds for d = 2 and for (k 1,k 2,k 3 ) = (2,2,3) and (2,3,3). Moreover, δ(k 1,k 2 ) = δ(k 1,k 1 ), and δ(2,2,3) = δ(2,3,3) = 5. 3 Computatonal determnaton of ı(d,k) 3.1 Computatonal framework The computatonal framework ntroduced n [4] can only determne whether δ(d,k) s equal to δ(d 1,k) + k.inthetermsoflemma3, ths amounts to assume that g = 0. Ths case s sgnfcantly easer than when g > 0 snce t can then be assumed that both δ(f ) and δ(f + ) are equal to δ(d 1,k) and that the vertces u and v such that d(u,v) = δ(d,k) satsfy u + v = k for = 1,...,d. In addton, the computatons were performed for d = 3; that s, for nstances such that the determnaton of all lattce (d 1,k)-polytopes of dameter δ(d 1,k) s computatonally nexpensve compared to hgher dmensons. To handle the case g > 0 and be able to determne all lattce (d 1,k)-polytopes of dameter δ(d 1,k) for d > 3, we ntroduce an enhanced algorthm explotng the structural propertes presented n Sect. 2. Weare able to recompute prevously determned values of δ(d,k) n a few seconds and obtan prevously ntractable values. In addton, the enhanced algorthm can be used to generate all the lattce (d,k)-polytopes maxmzng the dameter. The enhanced algorthm s presented n Sect. 3.2 and llustrated for small nstances of (d,k). The determnaton of δ(d,k) s performed va the determnaton of g(d,k) = δ(d 1,k) + k δ(d,k). For example, g(d,1) = 0ford 2, g(d,2) = d mod 2 for d 2, and g(3,3) = 1. Assumng that g(d 1,k) s known, the determnaton of g(d,k) requres to frst determne all the lattce (d 1,k)-polytopes achevng a dameter of δ(d 1,k). The algorthm mplctly enumerates all possble lattce (d,k)-polytopes P of dameter at least some target value. The search space s reduced by explotng ntegralty and convexty. In partcular, a dvde and conquer strategy s adopted by mplctly enumeratng all pars (u,v) of vertces such that d(u,v) = δ(p). The algorthm crucally explots the knowledge about all lattce (d 1,k)- polytopes wth suffcently large dameters, and the symmetres of the hypercube. The constructon conssts of two steps: the shellng step and the nner step. In the shellng step, for a gven par (u,v), the algorthm tres to embed lattce (d 1,k)-polytopes wth suffcently large dameters onto the 2d ntersectons of P wth the facets of the

6 A. Deza et al. hypercube [0,k] d. These embeddngs must be consstent; that s, gven two embeddngs E 1 and E 2, the ntersecton of E 1 wth the facet of [0,k] d contanng E 2 should be equal to the ntersecton of E 2 wth the facet of [0,k] d contanng E 1. The algorthm also upper bounds the dstance of u and v from each ntersecton of P wth the facets of the hypercube [0,k] d, and t aborts f t detects () a shortcut between u and v, () that u or v become not extremal, or () that no shellng would acheve the target dameter. If none of the stoppng crtera s met, the shellng step returns a lst of shellngs; that s, a lst of choces for the 2d ntersectons of P wth the facets of the hypercube [0,k] d, for whch t mght be possble to construct a lattce (d,k)-polytope achevng the target dameter. Consequently, for each obtaned shellng, the nner step s performed where all possble ponts n {1, 2,...,k 1} d are consdered to be added as vertces of P. Fnally, the lattce (d,k)-polytopes wth an empty ntersecton wth at least one facet of [0,k] d are derved from the current output to complete the enumeraton. If the output s empty we can conclude that no lattce (d,k)-polytope wth the target dameter exsts; the target dameter s lowered by one, and so forth. If the output s nonempty, we obtan all lattce (d,k)-polytopes wth the target dameter and the algorthm can run agan for d + 1. In the remander of ths secton, a more detaled descrpton s gven. 3.2 Algorthm to determne whether ı(d,k) = ı(d 1,k) + k g Assumng that the value for δ(d 1,k) s known, the ntal lower bound for g(d,k) s g = 0; that s, the ntal upper bound used for δ(d,k) s δ(d 1,k) + k. Usng the necessary condtons derved from the structural propertes presented n Lemma 2, the algorthm checks whether there exsts a lattce (d,k)-polytope of dameter δ(d 1,k) + k. If no such polytope exsts, g s update to 1; that s, the upper bound s updated to δ(d 1,k)+k 1, and the computatonal framework checks whether there exsts a lattce (d,k)-polytope of dameter δ(d 1,k) + k 1, and so on. The lower bound for δ(d,k), and thus the correspondng upper bound for g(d,k), s provded by the Mnkowsk sum of prmtve lattce vectors proposed by Deza, Manoussaks, and Onn [6]. For nstance, the ntal upper bound for (d,k) = (3,6) s δ(3,6) δ(2,6) + 6 = 12 whle the lower bound s δ(3,6) 10; that s, 0 g(3,6) 2. The algorthm frst assumes that δ(3,6) s equal to 12 and determnes that no lattce (3,6)-polytope has dameter 12; that s, g(3,6) 1. Then, assumng that δ(3,6) s equal to 11, the algorthm determnes that no lattce (3,6)-polytope has dameter 11; that s, g(3,6) 2. Thus, we can conclude that δ(3,6) s equal to 10; that s, g(3,6) s equal to Input and output The nput of the algorthm conssts of a trple (d,k,g) where: () d s the dmenson, () k s the largest possble entry of a coordnate of a vertex; that s, x {0, 1,...,k}, () g s a parameter determnng the dameter we wsh to rule out (or acheve);.e. we assume that the dameter s at least δ(d 1,k) + k g.

7 Dstance between vertces of lattce polytopes The output conssts of, up to the symmetres of the hypercube [0, k] d, all the lattce (d,k)-polytopes of dameter at least δ(d 1,k) + k g. An empty output provdes a certfcate that δ(d,k) <δ(d 1,k)+k g, and a nonempty output shows that δ(d,k) δ(d 1,k) + k g. Note that the algorthm requres the determnaton of all lattce (d 1,k)-polytopes of dameter δ(d 1,k).Ford = 3, the enumeraton of all lattce (2,k)-polygons of dameter δ(2,k) s straghtforward and nexpensve compared to the overall computatonal cost. For d 4, the enumeraton of all, up to the symmetres of [0,k] d, lattce (d 1,k)-polytopes of dameter δ(d 1,k) s obtaned as an output of the algorthm run wth the nput (d 1, k, g) where δ(d 1,k) = δ(d 2,k) + k g assumng that all lattce (d 2,k)-polytopes of the requred dameter are known. For example, n order to determne whether δ(4,3) = 9 we need to run the algorthm wth (d,k,g) = (4,3,0). Thus, we need to enumerate all lattce (3,3)-polytopes of dameter 6; that s, we need to run the algorthm wth (d, k, g) =(3,3,1). The algorthm output the 9, up to the symmetres of [0,3] 3, lattce (3,3)-polytopes of dameter 6. Then, runnng the algorthm wth (d, k, g) =(4,3,0) amounts to fndng a consstent combnaton of embeddngs of lattce (3,3)-polytopes of dameter 6 onto the 8 facets of the hypercube [0,3] 4. As any such obtaned lattce (4,3)-polytope s of dameter at most 8, the output s the empty set, and thus δ(4,3) <9. Smlarly, f the enumeraton of all lattce (d 1,k)-polytopes of dameter at least δ(d 1,k) 1 s requred, the lst, up to the symmetres of [0,k] d 1, s obtaned as an output of the algorthm run wth the nput (d 2,k,g) where δ(d 1,k) 1 = δ(d 2,k) + k g Generatng all potental pars (u,v) of vertces of a lattce (d,k)-polytope such that d(u,v) = ı(d 1,k) + k g A crtcal ngredent of the method conssts n reducng, as much as possble, the number of pars {u,v} that must be consdered. Ths s acheved by notcng that some restrctng condtons can be assumed wthout loss of generalty. Frst, by the symmetres of the hypercube [0, k] d, we can assume that the coordnates of u satsfy: u u +1 k/2 for = 1,...,d 1. In addton, by tem () of Lemma 3, we can assume that the coordnates of u and v satsfy: k g u + v k + g for = 1,...,d. Further, by the symmetres of the hypercube [0,k] d actng on the par {u,v} and assumng that all u are generated n the lexcographc order (denoted by n the followng), we can assume that the coordnates of u and v satsfy the followng condtons where ṽ s the pont consstng of the coordnates of v reordered lexcographcally: {v v +1 f u = u +1 } for = 1,...,d 1, ṽ (k,...,k) u f {v k/2 for = 1,...,d}.

8 A. Deza et al. Fnally, we can explot the fact that the ntersectons wth the facets of [0,k] d must be of suffcently large dameter. Let V d,k,g denote the set formed by all the vertces of all the lattce (d,k)-polytopes of dameter at least δ(d,k) g.let v denote the pont n R d 1 consstng of all coordnates of v except v, and let g = g + u + v k and g + = g + k (u + v ). The followng condtons are necessary for u and v to be vertces of a lattce (d, k)-polytope such that d(u,v) s at least δ(d 1,k) + k g: {ū V d 1,k,g f u = 0} for = 1,...,d, { v V d 1,k,g + f v = k} for = 1,...,d. Let P d,k,g denote the set of all the ponts wth nteger coordnates that belong to the ntersecton of all the lattce (d, k)-polytopes of dameter at least δ(d,k) g.letc u,v d,k,g denote the convex hull of u, v, and the followng set of ponts: [ d ] [ d ] {x R d : x = 0and x P d 1,k,g } {x R d : x = k and x P d 1,k,g + }. =1 The followng condton s necessary for u and v to be vertces of a lattce (d,k)- polytope such that d(u,v) s at least δ(d 1,k) + k g: =1 u and v are vertces of C u,v d,k,g. Let us llustrate the condtons that can be assumed for the par {u,v} by consderng the case (d, k, g) =(3,6,0). In other words, we assume that u and v are vertces of a lattce (3,6)-polytope of dameter 12. Snce g = 0, we can assume that: u 1 u 2 u 3 3, u + v = (6,6,6), {ū V 2,6,0 f u = 0} for = 1, 2, 3, u s a vertex of C u,v 3,6,0 f u 1 = 0. Among the 20 ponts u satsfyng u 1 u 2 u 3 3, the only ones such that u 1 = 0 and (u 2, u 3 ) V 2,6,0 are (0,0,1), (0,0,2), (0,0,3), (0,1,1), and (0,1,2). In addton, no pont such that u 1 = 0 s a vertex of C u,v 3,6,0. Thus, snce u + v = (6, 6, 6), we need to consder only 5 pars of vertces {u,v}. ThesetsV 2,6,0 and P 2,6,0 can be easly computed and are llustrated n Fg. 1. Note that both V d,k,g and P d,k,g are nvarant under the symmetres of [0, k] d Man subroutne to determne whether there exsts a lattce (d, k)-polytope wth vertces u and v such that d(u,v) = ı(d 1,k) + k g For each par {u,v} generated by the steps descrbed n Sect , we run the man subroutne to determne whether there exsts a lattce (d,k)-polytope wth vertces u and v such that d(u,v) = δ(d 1,k) + k g. Such polytopes are generated by

9 Dstance between vertces of lattce polytopes Fg. 1 The sets V 2,6,0 and P 2,6,0 consderng all possble choces for the 2d ntersectons wth the facets of the hypercube [0,k] d. Ths step s called the shellng step. The order n whch the ntersectons are consdered s crtcal to reduce the search space. It s equally crtcal to dentfy, as early as possble, paths between u and v possbly nduced by the shellng process. A set of choces for the 2d ntersectons wth the facets of the hypercube [0,k] d obtaned by the shellng step s called a shellng. The output of the shellng step conssts n the lst of all the shellngs generated by performng the shellng step for all possble pars {u,v}. Shellngs that are duplcate, up to the symmetres of the hypercube [0,k] d, are removed. If the output of the shellng step s empty, the process stops and we can conclude that δ(d,k) <δ(d 1,k) + k g. If the output of the shellng step s nonempty, all possble ponts wth nteger coordnates rangng from 1 to k 1 are consdered as potental addtonal vertces for each generated shellng. Ths step s called the nner step. Agan, structural propertes are crtcal to prune the search space. The output of the nner step conssts n the lst of all the generated lattce (d,k)-polytopes after duplcates, up to the symmetres of the hypercube [0,k] d, and lattce (d,k)-polytopes of dameter at most δ(d 1,k)+k g 1 have been removed. If the output of the nner step s empty, we can conclude that δ(d,k) <δ(d 1,k) + k g Two certfcates that no lattce (d,k)-polytopes wth vertces u and v such that d(u,v) = ı(d 1,k) + k g exst Let Γ denote the graph defned by the currently known edges and vertces of a lattce (d,k)-polytope. Γ s ntally set to {u,v}. Letd Γ (x,y) denote the dstance n Γ between two vertces x and y. We consder the followng upper bounds for the dstance between u or v and the ntersecton wth a facet of the [0,k] d, and two upper bounds for d(u,v). d(u, F ) = mn {d Γ (u,w)+ w } w Γ d(u, F + ) = mn {d Γ (u,w)+ k w } w Γ d(v, F ) = mn {d Γ (v, w) + w } w Γ for = 1,...,d, for = 1,...,d, for = 1,...,d,

10 A. Deza et al. d(v, F + ) = mn {d Γ (v, w) + k w } w Γ for = 1,...,d. Note that settng w to u or v gves d(u, F ) u, d(v, F ) v, d(u, F + ) k u, and d(v, F + ) k v. The followng quantty d (u,v), where both δ(f ) and δ(f + ) are bounded from above by δ(d 1, k), s an upper bound for d(u,v) by nequalty (1): d (u,v)= mn {mn{ d(u, F =1,...,d ) + d(v, F ) + δ(f ), d(u, F + ) + d(v, F + ) + δ(f + )}}. Each tme a choce for the ntersecton wth a facet of the hypercube [0, k] d s consdered, the value of d (u,v) s updated. Smlarly, snce Γ s a subgraph of the edge-graph, d Γ (u,v)s another upper bound for d(u,v). Thus, we consder the followng nonnegatve parameter γ defned as: γ = δ(d 1, k) + k g mn{d Γ (u,v),d (u,v)}. Consequently, γ>0 s a certfcate that there does not exst a lattce (d, k)-polytope wth vertces u and v such that d(u,v)= δ(d 1, k) + k g. Another estmate that s updated along wth Γ s the convex hull Cd,k,g Γ of the vertex set of Γ and the followng set of ponts: [ d ] [ d ] {x R d : x = 0and x P d 1,k,g } {x R d : x = k and x P d 1,k,g + }. =1 The followng condton s a certfcate that there does not exst a lattce (d, k)-polytope wth vertces u and v such that d(u,v)= δ(d 1, k) + k g: Shellng step =1 u or v s not a vertex of C Γ d,k,g. Gven a trple (d, k, g) and a par {u,v},thescores of F and F + are g = g + u + v k and g + = g + k (u + v ), respectvely. The ntersectons wth the facets of [0, k] d are ordered by ther score, n a non-decreasng order. If two ntersectons or more have the same score, the number of currently known vertces of P belongng to the ntersecton s used as a te-breaker, startng wth the one contanng the largest number of such vertces. As a secondary te-breaker, an ntersecton contanng u or v s consdered before one contanng nether, wth contanng u gven prorty over contanng v as a further te-breaker. If none of those rules apply, the default order s F1,...,F d, F+ 1,...,F+ d. Every tme an ntersecton F or F + s consdered, we generate the set of lattce (d 1, k)-polytopes of dameter at least δ(d 1, k) g or at least δ(d 1, k) g +, respectvely, and havng x as a vertex when x s a vertex from Γ such that x = 0or

11 Dstance between vertces of lattce polytopes Fg. 2 The unque shellng generated for (d, k, g) = (3,4,0) x = k, respectvely. After one such lattce (d 1, k)-polytope s assgned to form, up to an affne transformaton, the chosen ntersecton wth [0, k] d, ts vertces and edges are added to Γ. Consequently, the values of γ and Cd,k,g Γ are updated. If u or v s not a vertex of Cd,k,g Γ,orfγ>0, the search can be pruned at ths node. Typcally, the very frst chosen ntersecton wth [0, k] d may yeld a certfcate of non-exstence. Note that all scores g and g + are updated,.e. monotoncally nonncreased, durng the shellng process. Namely, each tme a choce for the ntersecton wth a facet of the hypercube [0, k] d s consdered, the scores g and g + of not yet consdered ntersectons are updated, f lowered, to g = g + d(u, F ) + d(v, F ) k and to g + = g + d(u, F + ) + d(v, F + ) k. In order to llustrate the shellng step, we frst consder the case (d,k,g) = (3,6,0). As dscussed n Sect , there are 5 pars {u,v} to consder. Snce g = 0, the score of any ntersecton wth [0,k] d s zero. The only currently known vertces are u and v and u 1 = 0 for all the 5 pars. Thus, the frst consdered ntersecton s F1. Consequently, for each {u,v}, we generate the set of lattce (2,6)-polytopes of dameter 6 havng (u 2, u 3 ) as a vertex. One can easly check that γ>0for each such choce. Thus, the shellng step termnates after consderng F1 for all possble pars {u,v} and we can conclude that δ(3,6) <12 n a matter of seconds. Another smple example s the case (d,k,g) = (3,4,0) for whch the output of the shellng step conssts n a unque shellng where the 6 ntersectons wth the facets of [0,4] 3 are, up to an affne transformaton, the octagonal lattce (2,3)-polytope. See Fg. 2 for an llustraton where the edges of the shellng are shown n blue. As no pont whose coordnates are {1,2,3}-valued can be added to ths unque shellng as a potental vertex, the nner step s reduced to check whether the dameter of the convex hull assocated to the unque shellng acheves δ(2,4) = 8. As the dameter s equal to 7, the output of the nner step s empty. Thus, we can conclude that δ(3,4) <8; that s, δ(3,4) = Inner step The nput for the nner step s the output of the shellng step, assumng t s nonempty. For each shellng, the nner step consders all the nner ponts p; that s, all the ponts

12 A. Deza et al. Fg. 3 A polytope consdered by the nner step for (d, k, g) = (3,4,2) and {u,v}={(0,0,0), (4,4,4)} p such that p {1,...,k 1} for = 1,...,d.LetC Γ p d,k,g denote the convex hull of p and Cd,k,g Γ. A necessary condton for p to be a vertex of a lattce (d, k)-polytope of dameter δ(d 1, k) + k g s: p s a vertex of C Γ p d,k,g. The generated lattce (d, k)-polytopes whose dameter s at most δ(d 1, k)+k g 1 are removed. Smlarly the duplcates, up to the symmetres of the hypercube [0, k] d, are removed. If the output of the nner step s empty, we can conclude that δ(d, k) s strctly less than δ(d 1, k) + k g. Otherwse, we can conclude that δ(d, k) = δ(d 1, k) + k g, and the output of the nner step provdes, up to the symmetres of the hypercube [0, k] d, all lattce (d, k)-polytopes of dameter δ(d 1, k)+k g whose ntersecton wth each facet of the hypercube [0, k] d s nonempty. Further computatons allow to determne all lattce (d, k)-polytopes of dameter δ(d 1, k) + k g wth an empty ntersecton wth at least one facet of the hypercube [0, k] d, as detaled n Sect In order to llustrate the nner step, we consder the case (d, k, g) =(3,4,2) and the par {u,v}={(0, 0, 0), (4, 4, 4)}. In other words, we assume that u = (0,0,0) and v = (4,4,4) are vertces of a lattce (3,4)-polytope such that d(u,v) = 5. Consderng ths par, we frst perform the shellng step. Snce the scores satsfy g 1 = g 2 = g 3 = g + 1 = g+ 2 = g+ 3 = 2, the 6 ntersectons wth the facets of [0, 4]3 are of dameter at least 2. One can check that the shellng step output ncludes the shellng consstng of the 6 dentcal facets depcted n Fg. 3 usng blue lnes. Each of these facets s, up to an affne transformaton, equal to the square [0,1] 2. Out of the 27 ponts whose coordnates are {1,2,3}-valued, 15 are contaned n the convex hull of ths shellng. Thus, the nner step must consder 12 nner ponts as possble vertces to be added to ths shellng. One can check that, up to the symmetres of [0,4] 3, 214 lattce (3,4)-polytopes are generated and that all of them have dameter at most 5. There are exactly 8 polytopes among them whose dameter s equal to 5. One of these lattce (3,4)-polytopes of dameter 5 s represented n Fg. 3 where the 6 added vertces are show n green and the edges of the ntersectons wth the facets of [0,4] 3 are shown n blue. Note that snce polytopes of dameter at most δ(d, k 1) + k g 1 = = 5are

13 Dstance between vertces of lattce polytopes removed, none of the 214 lattce (3,4)-polytopes generated by ths shellng are part of the output of the nner step for (d, k, g) = (3,4,2). 3.3 Generaton of all the lattce (d,k)-polytopes of dameter at least ı(d,k 1) + k g Runnng the algorthm for (d,k,g) allows to determne, up to the symmetres of [0,k] d, the set of all the lattce (d,k)-polytopes wth dameter at least δ(d,k 1)+k g whose ntersecton wth each facet of [0,k] d s nonempty. In ths secton we outlne how the ones wth an empty ntersecton wth at least one facet of [0,k] d can be derved from ths set. Usng the notatons of Lemma 2, leti (Q) denote the set of the coordnates such that γ + (Q) γ (Q) <k. Consder a lattce (d, k)-polytope Q of dameter at least δ(d, k 1) + k g such that I (Q) =. For all I (Q), we can assume, up to translaton, that γ (Q) = 0 and consder the segment σ = conv{0,(k γ + (Q))c }. Let S denote the Mnkowsk sum of all σ for I (Q). As shown n the proof of Lemma 2, Q + S s a lattce (d, k)-polytope of dameter at least δ(q) satsfyng I (Q + S) =. In other words, Q + S s, up to the symmetres of [0, k] d, n the output of the algorthm ran for (d, k, g). Note that settng P 1 = Q and P 2 =[0, s] where s 0 for all n Proposton 1 gves Remark 1. Remark 1 Consder a segment σ =[0, s]; a pont v s a vertex of Q + σ f and only f there exsts an objectve functon c R d that s unquely mnmzed at v n Q and () v = v and c s unquely mnmzed at 0 n σ, or () v = v + s and c s unquely mnmzed at s n σ. Moreover, f u and v are adjacent vertces of Q + σ, then ether (u,v ) s equal to (u,v)or to (u + s,v+ s) where u and v are adjacent vertces of Q, or t s equal to (u, u + s) where u s a vertex of Q. Consequently, up to translaton and up to the symmetres of the hypercube [0, k] d, the set of the lattce (d, k)-polytopes Q of dameter at least δ(d, k 1) + k g such that I (Q) = can be generated as follows: () for each lattce (d, k)-polytope P n the output of the algorthm ran for (d, k, g), check whether P = Q + σ where Q s a lattce (d, k)-polytope and σ a lattce segment. By Remark 1, ths can be done by checkng whether P and P + σ have the same number of vertces, () for each P such that P = Q+σ found at step (), determne Q and check whether δ(q) δ(d, k 1) + k g. As for the shellng and nner steps, the symmetres of the hypercube [0, k] d are used to remove duplcates generated wthn steps () and (). The set of lattce segments σ consdered n step () can be lmted to a few segments whose coordnates are relatvely prme and used teratvely. For an llustraton, we consder the case (d, k, g) =(3,3,1). As dscussed n Sect. 4.2, the output of the algorthm conssts n 9 lattce (3,3)- polytopes of dameter 6 whose ntersecton wth each facet of [0, 3] 3 s nonempty. One can check that, n order to perform step (), t s enough to consder for σ, teratvely, the vectors v such that v 2 2. All the 9 consdered lattce (3, 3)-polytopes

14 A. Deza et al Fg. 4 All lattce (2,6)-polytopes of dameter 6 of dameter 6 can be wrtten as Q + σ. Performng step (), one can check that δ(q) = 5 for each such Q. Thus, there s no lattce (3,3)-polytope Q of dameter 6 such that I (Q) =. 4 Proof of Theorem 1 Theorem 1 s obtaned by computatonally verfyng that the output of the nner step s empty for (d, k, g) = (3,6,1) and (5,3,0). Thus, δ(3,6) <11 and δ(5,3) <11; that s, δ(3,6) = δ(5,3) = 10. Runnng the algorthm for (5, 3, 0) requres the determnaton of all lattce (3,3)-polytopes of dameter 5 or 6 and all lattce (4,3)-polytopes of dameter 8. The algorthm s mplemented n C# and experments were carred out on a MacBook Pro wth a 2.8 GHz 7 processor and 16GB of RAM. 4.1 Determnaton of ı(3, 6) As mentoned n Sect , the output of the shellng step s empty for (d, k, g)= (3,6,0) and thus we can conclude that δ(3,6) <12. Runnng the algorthm for (d, k, g)= (3,6,1) s computatonally effcent because of two key propertes. Frst, there are only 4 lattce (2,6)-polytopes of dameter 6, see Fg. 4 for an llustraton. Second, for d = 2, there are only 8 lattce vectors v such that v 1. Thus, any lattce (2,6)-polytope of dameter 5 or 6 ncludes at least 2 edges e such that e 2.

15 Dstance between vertces of lattce polytopes Fg. 5 Intal teraton of the shellng step for (d, k, g) = (3,6,1) and {u,v} ={(0,0,0), (6,6,6)} v u u Consequently, unless both u and v are nner ponts, the update of the scores g = g + d(u, F ) + d(v, F ) k, respectvely of g + = g + d(u, F + ) + d(v, F + ) k, mples that g or g + s updated to zero for some after the frst ntersecton wth a facet of [0, 6] 3 s consdered n the shellng step. As g = 0org + = 0 mples that δ(f ) = 6orδ(F + ) = 6, respectvely, there are at most 4 lattce (2,6)-polytopes to consder for the next ntersecton wth a facet of [0,6] 3, and so forth. As an llustraton, consder the par {u,v}={(0,0,0), (6,6,6)}. Intally, the scores satsfy g1 = g 2 = g3 = g 1 + = g 2 + = g 3 + = 1 and the shellng step starts by consderng a lattce (2,6)-polytope of dameter at least 5 for F1. For example, assume that F 1 s, up to an affne transformaton, the lattce (2, 5)-polytope obtaned as the Mnkowsk sum of (1,0),(2,1),(1,1),(1,2), and (0,1). Before the next ntersecton wth a facet of [0,6] 3 s consdered, d(u, F 2 + ) s updated to 5 as d(u,u ) = 2 and u 2 = 3, see Fg. 5 where the vertex u s coloured black whle u and v are coloured red. The second edge on the path from u to u satsfes e 2 2. Consequently, g 2 + s updated to g + d(u, F 2 + ) + d(v, F 2 + ) 6 = = 0. Thus, δ(f+ 2 ) = 6 whch s mpossble snce v 2 = (6,6) / V 2,6,0 ; that s, there s no shellng wth the current choce of F1. The same holds for any choce of F 1 snce any lattce (2,6)-polytope of dameter at least 5 ncludes at least one edge e such that e 2. Consequently, there s no shellng for {u,v}={(0,0,0), (6,6,6)}. Table 2 lsts the 69 consdered pars {u,v} of vertces of a lattce (3,6)-polytope P such that d(u,v) = 11 where P s assumed to have a nonempty ntersecton wth each facet of [0,6] Determnaton of ı(5,3) The determnaton of δ(5,3) requres the lst of all lattce (4,3)-polytopes of dameter 8uptothesymmetresof[0,3] 4. In order to obtan all lattce (4,3)-polytopes of dameter 8, we frst determne all lattce (4,3)-polytopes of dameter 8 wth a nonempty ntersecton wth each facet of [0, 3] 4 by runnng the algorthm for (d, k, g) = (4, 3, 1).

16 A. Deza et al. Table 2 All consdered pars {u,v} for (d,k,g) = (3,6,1) u v (0,0,0) (6,6,6) (0,0,1) (5,5,4), (5,5,5), (5,5,6), (5,6,4), (5,6,5), (5,6,6), (6,6,4), (6,6,5) (0,0,2) (5,5,3), (5,5,4), (5,5,5), (5,6,3), (5,6,4), (5,6,5), (6,6,3), (6,6,4) (0,0,3) (5,5,2), (5,5,3), (5,5,4), (5,6,2), (5,6,3), (5,6,4), (6,6,2), (6,6,3) (0,1,1) (5,4,4), (5,4,5), (5,4,6), (5,5,5), (5,5,6), (6,4,4), (6,4,5), (6,5,5) (0,1,2) (5,4,3), (5,4,4), (5,4,5), (5,5,3), (5,5,4), (5,5,5), (5,6,4), (6,4,3), (6,4,4), (6,4,5), (6,5,3), (6,5,4) (0,1,3) (6,4,2), (6,4,3), (6,4,4), (6,5,2), (6,5,3), (6,6,2) (0,2,2) (6,3,4), (6,4,4) (0,2,3) (6,3,2), (6,3,4), (6,4,2), (6,4,3), (6,5,2) (1,1,1) (4,5,5), (5,5,5) (1,1,2) (5,5,3), (5,5,4) (1,1,3) (5,5,2), (5,5,3), (5,6,2), (6,6,2) (1,2,2) (5,4,4) (1,2,3) (6,5,2) (2,2,3) (4,5,2) Table 3 All consdered pars {u,v} for (d, k, g) = (3,3,1) u v (0,0,0) (3,3,3) (0,0,1) (2,3,2), (2,3,3), (3,3,1), (3,3,2) (0,1,1) (3,2,2) Then, usng the procedure descrbed n Sect. 3.3, we can use the output of the algorthm for (d, k, g) =(4,3,1) to determne all the lattce (4,3)-polytopes of dameter 8 wth an empty ntersecton wth at least one facet of [0,3] 4. Note that runnng the algorthm for (d, k, g) =(4,3,1) requres the lst of all the lattce (3,3)-polytopes of dameter 5 or 6. Ths s acheved by runnng the algorthm for (d, k, g) =(3,3,2) and usng the procedure descrbed n Sect Table 3 lsts the 6 consdered pars {u,v} of vertces of a lattce (3,3)-polytope P such that d(u,v) = 6 where P s assumed to have a non-empty ntersecton wth each facet of [0,3] 3. The output of the algorthm s made up of the 9 lattce (3,3)-polytopes of dameter 6, shown n Fg. 6 up to the symmetres of [0,3] 3. In ths fgure, the edges of the ntersectons wth the facets of [0, 3] 3 are shown n blue. Usng the procedure descrbed n Sect. 3.3, one can check that there s no lattce (3,3)-polytope of dameter 6 wth an empty ntersecton wth at least one facet of [0,3] 3. In other words, any lattce (3,3)-polytope of dameter 6 s, up to the symmetres of [0,3] 3, one of the 9 polytopes depcted n Fg. 6. Table 4 provdes the numbers f 0 (P) and f 2 (P) of vertces and facets of the 9 polytopes represented n Fg. 6. The breakdown by ncdence s also

17 Dstance between vertces of lattce polytopes Fg. 6 All, up to the symmetres of [0, 3] 3, lattce (3,3)-polytopes of dameter 6 Table 4 Some combnatoral propertes of the lattce (3,3)-polytopes wth maxmal dameter Polytope f 0 (P) Vertex ncdence f 2 (P) Facet ncdence P {3} + 6{4} 18 12{4} + 6{6} P {3} + 3{4} 14 9{4} + 5{6} P {3} 12 6{4} + 6{6} P {3} 14 8{3} + 6{8} P {3} 14 4{3} + 3{4} + 4{6} + 3{8} P {3} + 1{4} 14 4{3} + 3{4} + 4{6} + 2{7} + 1{8} P {3} + 1{4} 14 4{3} + 3{4} + 4{6} + 2{7} + 1{8} P {3} 13 2{3} + 4{4} + 6{6} + 1{8} P {3} 14 6{4} + 8{6}

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