The On-Line Heilbronn s Triangle Problem in d Dimensions

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1 Discrete Comput Geom 38: DI: 0.007/s x Discrete & Computatioal Geometry 2007 Spriger Sciece+Busiess Media, Ic. The -Lie Heilbro s Triagle Problem i d Dimesios Gill Barequet ad Alia Shaikhet Departmet of Computer Sciece, The Techio Israel Istitute of Techology, Haifa 32000, Israel {barequet dalia}@cs.techio.ac.il Abstract. I this paper we show a lower boud for the o-lie versio of Heilbro s triagle problem i d dimesios. Specifically, we provide a icremetal costructio for positioig poits i the d-dimesioal uit cube, for which every simplex defied by d + of these poits has volume / d+ l d d for d 5.. Itroductio The off-lie versio of the ow famous triagle problem was posed by Heilbro [R] more tha 50 years ago. It is formulated as follows: Give poits i the uit square, what is H2 off-lie, the maximum possible area of the smallest triagle defied by some three of these poits? There is a large gap betwee the best curretly kow lower ad upper bouds o H2 off-lie, log / 2 [KPS2] ad / 8/7 ε for ay ε>0 [KPS]. Jiag et al. [JLV] showed that the expected area of the smallest triagle, whe the poits are put uiformly at radom i the uit square, is / 3. Barequet [B] geeralized the off-lie problem to d dimesios: Give poits i the d-dimesioal uit cube, what is Hd off-lie, the maximum possible volume of the smallest simplex defied by some d + of these poits? The best curretly kow lower boud o Hd off-lie is log / d [L]. ther versios, i which the dimesio of the optimized simplex is lower tha that of the cube, were ivestigated i [L2], [L3], ad [BN]. Work o this paper has bee supported i part by the Fud for the Promotio of Research at the Techio. A prelimiary versio of this paper appeared i CCN 06.

2 52 G. Barequet ad A. Shaikhet The o-lie versio of the triagle problem is harder tha the off-lie versio because the value of is ot specified i advace. I other words, the poits are positioed oe after the other i a d-dimesioal uit cube, while is icremeted by oe after every poitpositioig step. The procedure ca be stopped at ay time, ad the already-positioed poits must have the property that every subset of d + poits defies a polytope whose volume is at least some quatity Hd o-lie, where the goal is to maximize this quatity. Schmidt [S] showed that H2 o-lie = / 2. Barequet [B2] used ested packig argumets to demostrate that H3 o-lie = / 0/3 = / ad H4 o-lie = / 27/24 = / I this paper we preset a otrivial geeralizatio of the latter method to d dimesios, showig that for a fixed value of d 5 we have Hd o-lie = / d+ l d d We provide a icremetal procedure for positioig poits oe by oe i a d-dimesioal uit cube so that o subset of up to d + poits is too dese. Specifically, the distace betwee ay two poits is at least a / /d for some costat a > 0, o three poits defie a triagle whose area is less tha a 2 / 2/d for some costat a 2 > 0, ad so o. The values of the costats are tued at the ed of the costructio. It is the prove that all the d-dimesioal simplices defied by d + -tuples of the poits have volume / d+ l d d The Costructio 2.. Notatio ad Pla We use the followig otatio. Let p i, p i2,...,p iq be ay q poits i R d. The p i p i2 deotes the distace betwee two poits p i, p i2 ; p i p i2 p i3 deotes the area of the triagle p i p i2 p i3 ; p i p i2 p i3 p i4 deotes the three-dimesioal volume of the tetrahedro p i p i2 p i3 p i4 ; ad, i geeral, p i p i2 p iq deotes the volume of the q - dimesioal simplex p i p i2 p iq. We deote by C d the d-dimesioal uit cube, ad by B d r a d-dimesioal ball of radius r. The lie defied by the pair of poits p i, p i2 is deoted by l i i 2. Throughout the costructio we refer to d as a fixed costat. Therefore, we omit factors that deped solely o d, except whe they appear i powers of. We wat to costruct a set S of poits i C d such that: [] p i p i2 V = a / /d, for ay pair of distict poits p i, p i2 S ad for some costat a > 0. [2] p i p i2 p i3 V 2 = a 2 / 2/d, for ay triple of distict poits p i, p i2, p i3 S ad for some costat a 2 > 0. [3] p i p i2 p i3 p i4 V 3 = a 3 / 4d2 5d /dd d 2, for ay quadruple of distict poits p i, p i2, p i3, p i4 S ad for some costat a 3 > 0.. [q ] p i p i2 p iq V q = a q V q 2 /a q 2 dq 2+q 3/dd q+2, for ay q-tuple 4 q d + of distict poits p i, p i2,...,p iq S ad for some costat a q > 0. Note that coditio [q ] holds for q = 4, i which case it coicides with [3].

3 The -Lie Heilbro s Triagle Problem i d Dimesios 53 The goal is to costruct S icremetally. That is, assume that we have already costructed a subset S v of v poits, for v<, which satisfies coditios [] [q ] above. We wat to show that there exists a ew poit p C d that satisfies: [ ] pp i V = a / /d, for each poit p i S. [2 ] pp i p i2 V 2 = a 2 / 2/d, for ay pair of distict poits p i, p i2 S. [3 ] pp i p i2 p i3 V 3 = a 3 / 4d2 5d /dd d 2, for ay triple of distict poits p i, p i2, p i3 S.. [q ] pp i p i2 p iq V q = a q V q 2 /a q 2 dq 2+q 3/dd q+2, for ay q -tuple 4 q d + of distict poits p i, p i2,...,p iq S. We will show this by summig up the volumes of the forbidde portios of C d where oe of the iequalities [ ] [q ] is violated, ad by showig that the sum of these volumes is less tha. This implies the existece of the desired poit p, which we the add to S v to form S v+. We cotiue i this maer util the etire set S is costructed Forbidde Balls The forbidde regios where oe of the iequalities [ ] is violated are v d-dimesioal balls of radius r = a / /d. Their total volume is at most v v Br d = = Forbidde Cyliders The forbidde regios where oe of the iequalities [2 ] is violated are v 2 d-dimesioal cyliders G ij, for i < j v. The cylider G ij is cetered at l ij, its legth is at most d, ad its cross-sectio perpedicular to l ij is a d -dimesioal sphere of radius r i, j = i< j v 2a 2 2/d p i p j = i< j v 2/d p i p j see Fig.. The overall volume of the cyliders withi C d is at most Br d i, j d =. 2 p i p j d To boud this sum, we fix p i ad sum over p j. We use a d-dimesioal spherical Recall that Br d =πd/2 r d /Ɣd/2 + = r d, where Ɣ is the cotiuous geeralizatio of the factorial fuctio.

4 54 G. Barequet ad A. Shaikhet `ij 2a2 2=d jp ip j j G ij p j p i 2a2 2=d jp ip j j Fig.. A cylider i R d. packig argumet that exploits the fact that S v satisfies []. Specifically, we have j i /d p i p j d t= M t d /d a d, 2 t d where M t is the umber of poits of S v that lie i the d-dimesioal spherical shell cetered at p i with ier radius a t/ /d ad outer radius a t + / /d ; see Fig. 2. There are /d such spherical shells withi C d. Because of [], the umber of such =d 3 2 a =d p i Fig. 2. A spherical packig of balls i R d.

5 The -Lie Heilbro s Triagle Problem i d Dimesios 55 poits is M t = t d. This follows by a argumet of packig spheres of volume / withi a shell whose volume is t d /. Hece, the sum i 2 is. Summig this over all p i, we obtai a fial boud of v. Substitutig this i, we see that the total volume of the forbidde cyliders is v/ = Forbidde Prisms The forbidde regios where oe of the iequalities [3 ] is violated are v 3 d-dimesioal prisms ϕ ijk, for i < j < k v. The base area a portio of a two-dimesioal flat of ϕ ijk is at most d, ad its height is a d 2-dimesioal sphere of radius r i, j,k = 4d2 5d /dd d 2 p i p j p k The overall volume of the prisms withi C d is at most B d 2 i< j<k v r i, j,k d = i< j<k v. 4d2 5d /dd p i p j p k d 2. 3 To boud this sum, we fix p i, p j ad sum over p k. We use a d-dimesioal cylidrical packig argumet that exploits the fact that S v satisfies [] ad [2]. The cyliders are cetered at l ij ; see Fig. 3, where the lie l ij emaates from p i toward p j through the `ij =d p j 3 2 a =d p i Fig. 3. i R d. A d-dimesioal cylidrical packig a extruded d -dimesioal spherical packig of balls

6 56 G. Barequet ad A. Shaikhet dth dimesio. Specifically, we have k i, j p i p j p k N 0 2d 2/d /d d 2 a d t= 2 d 2 N t d 2/d a d 2, 4 t d 2 p i p j d 2 where N 0 is the umber of poits of S v that lie i the iermost d-dimesioal cylider of the packig cetered at l ij ad of radius a / /d, ad N t is the umber of poits of S v that lie i the cylidrical shell cetered at l ij with ier radius a t/ /d ad outer radius a t + / /d. bviously, N 0 = /d, sice the volume of the d -dimesioal cross-sectioal sphere of the iermost cylider is / d /d ad because of []. Also, we have N t = t d 2 /d. This follows by a argumet of packig spheres of volume / withi a shell whose volume is t d 2 / d /d. Hece, the quatity i 4 is 2d2 3d /dd + / p i p j d 2. Substitutig this i 3, we obtai the upper boud o the total volume of the forbidde prisms: + 2 3d2 4d /dd p i p j d 2 i< j v = v d2 4d /dd i< j v. 5 p i p j d 2 We boud the sum i the secod summad similarly to our boudig of the term i 2 i Sectio 2.3. We fix p i ad use a d-dimesioal spherical packig argumet withi spherical shells cetered at p i. Arguig as above, we obtai j i /d p i p j d 2 t= M t d 2/d a d 2 = t d 2 /d t= t d d 2/d a d 2 t d 2 =. Summig this over all p i, we obtai a fial boud of v. Substitutig this i 5, we see that the total volume of the forbidde prisms is v 2 + v =. 2 2d2 3d /dd 2.5. Geeral Forbidde Zoes I Sectios we computed the total volume of the forbidde zoes i which the respective iequalities [ ] [3 ] are violated. These zoes correspod to q = 2, 3, 4, respectively. I this sectio we aalyze the geeral case 4 < q d +. The smallest value of q for which this aalysis holds is 5, sice for q = 5 we eed coditios [4] ad [3], which are precisely the bottom of the rage i which the recursive relatio [q ] holds. The forbidde regios where oe of the iequalities [q ] is violated are v q d-dimesioal zoes ψ i i 2 i q for i < i 2 < < i q v ad 4 < q

7 The -Lie Heilbro s Triagle Problem i d Dimesios 57 d +, whose bases are portios of q 2-dimesioal flats with volume at most d q 2/2. The height of the zoe ψ i i 2 i q is a d q +2-dimesioal sphere of radius r i,...,i q = V q / p i p i2 p iq. The total volume of the zoes withi C d isat most V d q+2 Br d q+2 i d q 2/2 q =. 6,...,i q p i p i2 p iq d q+2 i < <i q v i < <i q v To boud this sum, we fix p i, p i2,...,p iq 2 ad sum over p iq. We use a packig argumet that exploits the fact that S v satisfies [] [q 2]. The packig cosists of the Cartesia product of the q 3-dimesioal flat π = π i i 2 i q 2 that passes through p i, p i2,...,p iq 2, ad spheres whose ceters belog to π ad exted to the d q + 3- dimesioal space orthogoal to π. Specifically, we have i q i,...,i q 2 p i p i2 p iq d q+2 Z 0 /d /d d q+2 V d q+2 + Z t, 7 a q 2 t= t p i p i2 p iq 2 where Z 0 is the umber of poits of S v that lie i the iermost shape of the packig cetered at the flat π ad of radius a / /d, ad Z t is the umber of poits of S v that lie i the shell cetered at π with ier radius a t/ /d ad outer radius a t + / /d. bviously, Z 0 = q 3/d, sice the volume of the iermost shape is / d q+3/d ad because of []. Also, we have Z t = t d q+2 q 3/d. This follows by a argumet of packig spheres of volume / withi a shell whose volume is t d q+2 / d q+3/d. Hece, the sum i 7 is i <...<i q 2 v q 3/d V d q+2 q 2 + V d q+2 q 2. 8 p i p i2 p iq 2 d q+2 Substitutig this i 6, we obtai the upper boud o the total volume of the forbidde zoes: V d q+2 q 3/d q + p i p i2 p iq 2 d q+2 = Combiig this with the equality d q+2 q 3/d v q 2 Vq + V q 2 V q = i < <i q 2 v a q V q 2 a q 2 dq 2+q 3/dd q+2, Vq V q 3 d q+2.

8 58 G. Barequet ad A. Shaikhet we see that the total forbidde volume is + i < <i q 2 v Vq V q 3 d q+2. 9 I order to show that the boud i 9 is, it remais to prove that the secod summad i it is. From [q ] ad [q 2] we kow that ad V q = V q 2 = a q V q 2 a q 2 dq 2+q 3/dd q+2 a q 2 V q 3 a q 3 dq 3+q 4/dd q+3, respectively. By substitutig this i the secod summad of 9, we obtai d q+2 Vq i < <i q 2 V v q 3 aq = < a q 3 aq a q 3 d q+2 Thus, it suffices to prove that i.e. kowig that v, that i < <i q 2 v 2q 3d2 + 2q 2 +3q 22d 2q 2 +2q 7/dd q+3 d q+2 v q 2 2q 3d2 + 2q 2 +3q 22d 2q 2 +2q 7/dd q q 3d2 + 2q 2 +3q 22d 2q 2 +2q 7/dd q+3 >v q 2, 2q 3d 2 + 2q 2 + 3q 22d 2q 2 + 2q 7 dd q + 3 which, after simple maipulatios, is However, it is easily verified that q 4d 2 q 4 2 d 2q 2 + 2q 7 > 0. > q 2, q 4d 2 q 4 2 d 2q 2 + 2q 7 = q 4d q + 2d > 0, usig the iequalities q 5, d q, ad d 4.

9 The -Lie Heilbro s Triagle Problem i d Dimesios Summatio We are ow ready to boud Hd o-lie, the maximum possible volume of the smallest simplex defied by some d + ofthe poits put i the d-dimesioal uit cube. I other words, we wat to boud V d from below. For this purpose we use its recursive defiitio ad write d+ a q a d V d = V a q=4 q 2 dq 2+q 3/dd q+2 2 = d+. dq 2+q 3/dd q+2+2/d q=4 Let us boud from above the power of i : d+ q=4 dq 2 + q 3 dd q + 2 q=4 + 2 d d+ q /d = d q + 2 d = d 2 t= d + /d t t d+ q=4 d q d q + 2 d 2/d + 2 d < d + /dld d 2 2/d+2/d = d + ld d ld 2.265/d +2/d < 2 d + ld d , where we use the facts which ca be verified by elemetary calculus that. k t= /t < lk for k 3 that is, d 5, ad 2. 2/d ld 2.265/d < for d 5. We see that V d > a d / d+ l d d Note that whe d teds to ifiity, some terms vaish ad d 2 t= /t l d 2 approaches the Euler Mascheroi costat, γ = Thus, for large-eough values of d,wehavev d >a d / d+ l d d It remais to show that the costats a, a 2,...,a d ca be fixed so that the total volume of the forbidde zoes is strictly less tha. To this aim ote that amog these costats, the total volume of the forbidde balls depeds solely o a, the total volume of the forbidde cyliders depeds oly o a, a 2, ad so o. This allows us to fix the values of the costats sequetially so that the total volume of ay type of forbidde zoe is strictly less tha /d. Specifically, we fix a to make the total volume of the forbidde balls less tha /d. The, havig a already fixed, we fix a 2 to make the total volume of the forbidde cyliders less tha /d, ad so o. The grad total of the volume of the forbidde zoes is, thus, less tha d/d =. See [B2] for the implemetatio of this techique for d = 3, 4. Note that the values of a, a 2,...,a d deped oly o d ad ot o. This allows the iterative procedure to cotiue ad ifiitum, while the forbidde zoes of each iteratio properly cotai the respective forbidde zoes of the precedig iteratio.

10 60 G. Barequet ad A. Shaikhet This completes the proof of the mai theorem: Theorem. H o-lie d = / d+ l d d for d Coclusio I this paper we show, by usig ested packig argumets, that Hd o-lie = / d+ l d d for all d 5. For large-eough values of d we have Hd o-lie = / d+ l d d This compares favorably with the bestkow lower boud [L] i the off-lie case Hd off-lie = log / d. Ackowledgmet The authors thak Micha Sharir for helpful discussios o the triagle problem ad o ested packig argumets. Refereces [B] G. Barequet, A lower boud for Heilbro s triagle problem i d dimesios, SIAM J. Discrete Math., 4 200, [B2] G. Barequet, The o-lie Heilbro s triagle problem, Discrete Math., , 7 4. [BN] G. Barequet ad J. Naor, Large k-d simplices i the d-dimesioal uit cube, Far East J. Appl. Math., , [JLV] T. Jiag, M. Li, ad P. Vitáyi, The average-case area of Heilbro-type triagles, Radom Structures Algorithms, , [KPS] J. Komlós, J. Pitz, ad E. Szemerédi, Heilbro s triagle problem, J. Lodo Math. Soc. 2, 24 98, [KPS2] J. Komlós, J. Pitz, ad E. Szemerédi, A lower boud for Heilbro s problem, J. Lodo Math. Soc. 2, , [L] H. Lefma, Heilbro s problem i higher dimesio, Combiatorica, , [L2] H. Lefma, Large triagles i the d-dimesioal uit-cube, Proc. 0th A. It. Computig ad Combiatorics Cof., Jeju Islad, South Korea, pp , Lecture Notes i Computer Sciece, 306, Spriger-Verlag, Berli, August [L3] H. Lefma, Distributios of poits i d dimesios ad large k-poit simplices, Proc. th A. It. Computig ad Combiatorics Cof., Kumig, Chia, pp , Lecture Notes i Computer Sciece, 3595, Spriger-Verlag, Berli, August [R] K.F. Roth, a problem of Heilbro, Proc. Lodo Math. Soc., 26 95, [S] W.M. Schmidt, a problem of Heilbro, J. Lodo Math. Soc. 2, 4 97, Received September 6, 2006, ad i revised form Jauary 9, lie publicatio May 8, 2007.

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