xix spanish meeting on computational geometry book of

XIX Spanish Meeting on Computational Geometry

Book of abstracts

Universidad Politecnica de Madrid

Madrid July 5-7 2021

EGC21 preface

Preface

This book contains the abstracts of the invited talks contributed papers and contributed talksaccepted for presentation at the XIX Spanish Meeting on Computational Geometry (formerlyEncuentros de Geometrıa Computacional EGC) The current edition was organized in MadridSpain but held online on July 5-7 2021 due to the COVID-19 pandemic

This series of meetings focuses on current research topics in discrete and computationalgeometry Since the seminal edition in 1990 the Encuentros have combined a strong scien-tific program with a friendly atmosphere The intended audience ranges from experiencedresearchers to students facing their debut in the area The strong collaboration links of theSpanish community with foreign colleagues made advisable in 2011 to change the language ofthe meeting to English and have the submissions peer-reviewed by an international programcommittee

In this edition following the experience started in the previous one two lengths were possiblefor the contributed submissions 4 pages (rdquopaperrdquo) or 1 page (rdquotalkrdquo) We received a total of26 submissions consisting of 16 talks and 10 papers Among them one was withdrawn beforereviewing and the other 25 were finally accepted composing the core of this book Threerevisions were collected for all the submissions except for one passing two reviews

As each of them the current edition of the meeting is the result of the work and dedication ofa lot of people First thanks go to the authors for choosing EGC to share and disseminate theirwork Second I would like to thank the members of the program committee and the externalreviewers for accepting to contribute their expertise to this meeting carefully constructivelyand on time Thirdly I am truly thankful to the three excellent invited speakers for acceptingour invitation Fernando Blasco Kevin Buchin and Evanthia Papadopoulou Finally it is veryappreciated the support of the Departamento de Matematica Aplicada a las Tecnologıas de laInformacion y las Comunicaciones and the Escuela Tecnica Superior de Ingenierıa de SistemasInformaticos from the Universidad Politecnica de Madrid although this edition would not havebeen possible without the tough work of the organizing committee Guillermo Esteban JesusGarcıa and Alejandra Martınez Having to deal with the uncertainty of these times they didtheir best for the success of this meeting

July 2021Alcala de Henares

David OrdenProgram Committee Chair

i

EGC21 Program Committee

Program Committee

Sergio Cabello Dept of Mathematics FMF University of LjubljanaRuy Fabila-Monroy Departamento de Matematicas CinvestavStefan Felsner TU BerlinSilvia Fernandez California State University NorthridgeDelia Garijo University of SevilleClemens Huemer Universitat Politecnica de CatalunyaJoseph Mitchell Stony Brook UniversityBengt J Nilsson Malmo UniversityDavid Orden (Chair) Universidad de AlcalaIrene Parada TU EindhovenDavid Rappaport School of Computing Queenrsquos UniversityFrancisco Santos Universidad De CantabriaAndre Schulz FernUniversitat in HagenRodrigo Silveira Universitat Politecnica de CatalunyaJavier Tejel University of ZaragozaJorge Urrutia Universidad Nacional Autonoma de MexicoInmaculada Ventura Molina University of SevilleBirgit Vogtenhuber Graz University of Technology

ii


Organizing Committee

Guillermo Esteban Universidad de AlcalaJesus Garcıa Lopez de Lacalle (Chair) Universidad Politecnica de MadridAlejandra Martınez Universidad de Alcala

iii

EGC21 Table of Contents

Table of Contents

Initial sections

Preface i

Program Committee ii

Organizing Committee iii

Table of Contents iv

Invited talks

Algorithms for trajectory clustering and segmentation 1

Kevin Buchin

Abstract tree-like Voronoi diagrams and site-deletion in expected linear time 2

Evanthia Papadopoulou

Algorithms and ideas in mathematical magic 3

Fernando Blasco

Contributions Monday July 5th 1400-1500

Plane paths in simple drawings of complete graphs 4

Oswin Aichholzer Alfredo Garcia Javier Tejel Birgit Vogtenhuber and AlexandraWeinberger

Crossing-optimal extension of simple drawings 5

Robert Ganian Thekla Hamm Fabian Klute Irene Parada and Birgit Vogtenhuber

Computing the continuous mean distance for certain graph classes 6

Delia Garijo Alberto Marquez and Rodrigo Silveira


Shortest paths in weighted hexagonal tessellations 7

Guillermo Esteban Prosenjit Bose David Orden and Rodrigo Silveira

On prescribing total orders and preorders to pairwise distances of points in Euclideanspace 11

Vıctor Hugo Almendra-Hernandez and Leonardo Martınez-Sandoval

Applications of geometric graphs to sensory analysis of cookies 15

David Orden Encarnacion Fernandez-Fernandez Marino Tejedor-Romero andAlejandra Martınez-Moraian


Bounds on the Diameter of Graph Associahedra 16

Jean Cardinal Lionel Pournin and Mario Valencia-Pabon

Showing non-realizability of spheres by distilling a tree 17

Julian Pfeifle

iv


Computing the type cone of nestohedra 21

Arnau Padrol Vincent Pilaud and Germain Poullot

Contributions Tuesday July 6th 1400-1500

The Voronoi diagram of rotating rays with applications to floodlight illumination 22

Carlos Alegrıa Ioannis Mantas Evanthia Papadopoulou Marko Savic HendrikSchrezenmaier Carlos Seara and Martin Suderland

The edge labeling of higher order Voronoi diagrams 23

Merce Claverol Andrea de Las Heras Parrilla Clemens Huemer and AlejandraMartınez-Moraian

On Guillotine Cuts of Boundary Rectangles 27

Pablo Perez-Lantero and Carlos Seara


Faster distance-based representative skyline in the plane 31

Sergio Cabello

On the number of connected rectilinear convex 4-gons 32

Alejandra Martınez-Moraian and David Orden

No selection lemma for empty triangles 36

Ruy Fabila-Monroy Carlos Hidalgo-Toscano Daniel Perz and Birgit Vogtenhuber

On (α k)-sets and (α k)-hulls in the plane 37

Merce Claverol Luis H Herrera Pablo Perez-Lantero and Carlos Seara

Contributions Wednesday July 7th 1400-1500

Developable surfaces bounded by spline curves 41

Alicia Canton Leonardo Fernandez-Jambrina Marıa Eugenia Rosado Marıa andMarıa Jesus Vazquez-Gallo

Planar aesthetic curves 42

Alicia Canton Leonardo Fernandez Jambrina and Marıa Jesus Vazquez Gallo

Parallel Simulated Annealing for Continuous Dispersion Problems 43

Narcis Coll and Marta Fort


New variants of perfect non-crossing matchings 44

Ioannis Mantas Marko Savic and Hendrik Schrezenmaier

On Maximum-Sum Matchings of Points 45

Sergey Bereg Oscar Chacon-Rivera David Flores-Penaloza Clemens Huemer PabloPerez-Lantero and Carlos Seara

v


Minimum Color Spanning Circle in Imprecise Domain 49

Ankush Acharyya Ramesh K Jallu Vahideh Keikha Maarten Loffler and MariaSaumell


A discrete isoperimetric inequality 53

David Iglesias Lopez Eduardo Lucas Marın and Jesus Yepes Nicolas

Algorithmic geometry with infinite time computation 54

Clemens Huemer Moritz Muller Carlos Seara and Adrian Tobar Nicolau

Pattern recognition of homogenized standard sets of image patterns arising from Latinsquares 55

Raul Falcon

Final section

Author Index 59

vi

XIX Spanish Meeting on Computational Geometry Madrid July 5-7 2021

Algorithms for trajectory clustering and segmentation

Kevin Buchin

Eindhoven University of Technology

Nowadays more and more trajectory data is being collected of people animals and vehicles Analyzing suchdata requires efficient algorithms In this talk I will focus on geometric algorithms for two fundamental analysistasks clustering and segmenting trajectoriesClustering asks to group similar trajectories or subtrajectories for instance to identify common routes Themain clustering problem that I will consider is center-based clustering with respect to the Frechet distancewhere centers are required to have constant complexity For this problem I will present a 3-approximation anddiscuss subsequent developmentsSegmentation asks to partition a trajectory into movement phases such that within each phase the variation ofmovement characteristics low I will present geometric and model-based approaches to segmenting one or severaltrajectories

1


Abstract tree-like Voronoi diagrams and site-deletion inexpected linear time

Evanthia Papadopoulou1

1Faculty of Informatics Universita della Svizzera italiana Lugano Switzerland

Differences between classical Voronoi diagrams of points in the plane versus segments circles polygons orclusters of points are sometimes forgotten or underestimated As a result basic open problems exist even to dateupdating a Voronoi diagram after deletion of one site is such an example Although linear-time techniques forsite deletion in point Voronoi diagrams had been known to exist since the late 80rsquos the corresponding problemsfor non-point sites remained open until recently

In this talk I will address this problem in expected linear time under the framework of abstract Voronoi dia-grams simultaneously covering several concrete cases of non-point or weighted-point sites To pursue this goalI will introduce abstract Voronoi-like diagrams a relaxed Voronoi structure of independent interest which leadsto a very simple randomized incremental technique generalizing the one for points The algorithm extends tocomputing various other tree-like Voronoi diagrams such as constructing the farthest abstract Voronoi diagramafter the order of its regions at infinity is known constructing the order-(k+1) subdivision within an order-kVoronoi region and others The time analysis introduces a simple alternative to backwards analysis applicableto order-dependent structures

Parts of this talk are joint work with Kolja Junginger

2


Algorithms and ideas in mathematical magic

Fernando Blascolowast1

1Dep Matematica Aplicada ETSI Montes Forestal y del Medio Natural Universidad Politecnica de Madrid

Abstract

There are a few mathematical principles that arise in card magic Most of them are related to concepts thatappear in Discrete Mathematics such as modular arithmetic or permutations (in fact shuffling a deck of cardsis applying a permutation to it) Moreover binary system is related to faro shuffles and De Bruijn sequencesare not only related to Penrose tilings and quasicrystals but also they are related to amazing magic tricks

In this talk we shall make a review of different papers published in mathematical journals such as Mathe-matical Intelligencer or The American Mathematical Monthly as well as some books on the subject with ideasfrom Martin Gardner Brent Morris Ron Graham Persi Diaconis and Colm Mulcahy The talk will be partic-ipative showing in a first stage te magic effect and after that we shall comment the underlying mathematicsbehind the trick that for mathematical effects are often more interesting than the trick itself

Some ideas concerning the use of mathematical card magic in an usual course on Mathematics will be shownWe will also present programs and apps that use mathematical ideas in order to create magical effects

3


Plane paths in simple drawings of complete graphs

Oswin Aichholzerlowast1 Alfredo Garcıadagger2 Javier TejelDagger2 Birgit Vogtenhubersect1 and Alexandra Weinbergerlowast1

1Institute of Software Technology Graz University of Technology Austria2Departamento de Metodos Estadısticos IUMA Universidad de Zaragoza Spain

Simple drawings are drawings of graphs in the planesuch that vertices are distinct points in the planeedges are Jordan arcs connecting their endpoints andedges intersect at most once either in a proper crossingor in a shared endpoint

It is conjectured that every simple drawing of thecomplete graph with n vertices Kn contains a planeHamiltonian cycle and consequently a plane Hamil-tonian path However to the best of our knowledgeΩ((log n)

16 ) [4] is currently the best known lower bound

for the length of a plane path contained in any simpledrawing of Kn We improve this bound to Ω( logn

log logn )To prove our new bound we will use a special kind

of simple drawings We say that a simple drawingD is c-monotone if there is a point O such that anyray emanating from O intersects any edge of D atmost once A c-monotone drawing D is generalizedtwisted [1] if there exists a ray r emanating from Othat intersects every edge of D Note that given a c-monotone drawing D if there exists a ray r emanatingfrom O such that no edge of D crosses r then D isstrongly isomorphic to an x-monotone drawing (anyvertical line intersects any edge of the drawing at mostonce) See Figure 1 for some examples

It is well-known that any x-monotone drawing of Kn

contains a plane Hamiltonian path Besides we canshow that generalized twisted drawings of Kn alsocontain plane Hamiltonian paths Using DilworthrsquosTheorem about chains and anti-chains [2] we canprove that any c-monotone drawing of Kn contains asubdrawing of Kradicn that is either generalized twisted orstrongly isomorphic to an x-monotone drawing of KradicnAs a consequence any c-monotone drawing of Kn

contains a plane path of length at leastradicn

Using this result we prove the following theorem

Theorem 1 Any simple drawing D of Kn contains aplane path of length Ω( logn

log logn )

lowastEmails oaichweinbergeristtugrazat Supported bythe Austrian Science Fund (FWF) grant W1230daggerEmail olaverriunizares Supported by Gobierno Aragon

E41-17RDaggerEmail jtejelunizares Supported by project PID2019-

104129GB-I00 AEI 1013039501100011033 of the SpanishMinistry of Science and InnovationsectEmail bvogtisttugrazat Supported by the FWF grant

I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r

Figure 1 A generalized twisted drawing of K5 (left)and a c-monotone drawing of K5 that is strongly iso-morphic to an x-monotone drawing of K5 (right)

Sketch of the proof We take the star of a vertex v(the set of edges of D incident to v) and we extendthis star to a maximal plane subdrawing H whichmust be biconnected [3]

If there is a vertex w in H v that has degree at least(log n)2 in H then the subdrawing H prime of H inducedby v w and their at least (log n)2 common neighborsis weakly isomorphic to a c-monotone drawing so H prime

contains a plane path of length at least lognOtherwise the maximum degree in H v is less than

(log n)2 As H is biconnected H v contains a planetree T of order n minus 1 whose maximum degree is atmost (log n)2 Since the diameter of a d-ary tree oforder n is at least Ω( logn

log d ) T contains a plane path

of length at least Ω( lognlog logn )

A manuscript with proofs is added as an appendix

References

[1] O Aichholzer A Garcıa J Tejel B Vogtenhuberand A Weinberger Plane matchings in simple draw-ings of complete graphs submitted to CGYRF 2021

[2] R Dilworth A Decomposition Theorem for PartiallyOrdered Sets Annals of Mathematics 51 (1) 161ndash1661950

[3] A Garcıa A Pilz and J Tejel On Plane Subgraphsof Complete Topological Drawings accepted in ArsMathematica Contemporanea

[4] J Pach J Solymosi and G Toth Unavoidable con-figurations in complete topological graphs DiscreteComput Geometry 30 311ndash320 2003

4


Crossing-optimal extension of simple drawingslowast

Robert Ganiandagger1 Thekla Hammdagger1 Fabian KluteDagger2 Irene Paradasect3 and Birgit Vogtenhuberpara4

1Algorithms and Complexity Group TU Wien Austria2Utrecht University The Netherlands

3TU Eindhoven The Netherlands4Graz University of Technology Austria

We study the extension problem for simple draw-ings1 in the context of crossing minimization Our aimis to extend a given simple drawing with k new edgeswhile maintaining simplicity and restricting newlycreated crossings We consider the following problem

Simple Crossing-Minimal Edge Insertion (SCEI)

Input A graph G = (VE) along with a connectedsimple drawing G an integer ` and a set F of kedges of the complement of G

Question Can G be extended to a simple drawingGprime of the graph Gprime = (VEcupF ) such that the num-ber of crossings in Gprime involving an edge of F is atmost `

SCEI was recently shown to be NP-complete al-ready when |F | = 1 and ` ge |E| (meaning that theaim is merely to obtain a simple drawing) [1] Ourmain contribution is an FPT algorithm2

Theorem 1 SCEI is FPT with respect to k + `

On a high level our approach follows the generalstrategy used in [2] for extending 1-planar drawings

1 We preprocess G and a planarization of G to re-move parts of G which are too far away to interactwith our solution This is then translated into agraph representation of bounded treewidth

2 We identify a combinatorial characterization thatcaptures how the solution curves will be embed-ded into G Crucially the characterization hassize bounded by our parameters

lowastFull paper arXiv201207457RG FK and BV supported by the Austrian Science Fund(FWF) via projects P31336 J-4510 and I 3340 respectively

daggerEmails [rganian|thamm]actuwienacatDaggerEmail fmkluteuunlsectEmail imdeparadamunoztuenlparaEmail bvogtisttugrazat1In a simple drawing any two edges share at most one point2A fixed-parameter tractable (FPT) algorithm with respect

to a parameter κ runs in time f(κ) middot nO(1) where f is anycomputable function and n is the size of the input

3 We perform brute-force branching over all char-acterizations to pre-determine the behavior of asolution in G and for each such characteriza-tion we employ Courcellersquos theorem to determinewhether there exists a solution with this charac-terization

The specific implementation of each step of this strat-egy differs substantially from the previous work [2]By far the greatest challenge occurs in Step 1 No-tably removing the parts of G required to obtain abounded-treewidth graph representation creates holesin the drawing and these could disconnect edges in-tersecting these holes The graph representation canthen lose track of ldquowhich edge parts belong to eachotherrdquo which means we can no longer use it to deter-mine whether the extended drawing is simple

To handle this problem we employ an in-depth ge-ometric analysis combined with a careful use of thesunflower lemma and subroutines which invoke Cour-cellersquos theorem to construct a representation which(a) still has bounded treewidth and (b) contains par-tial information about which edge parts belong to thesame edge in G

Finally we note that a core ingredient in our ap-proach is the use of Courcellersquos theorem and hencethe algorithms underlying our tractability results willhave an impractical dependency on k However forthe special case of |F | = 1 (ie when inserting a singleedge) we use so-called representative sets to providea single-exponential fixed-parameter algorithm whichis tight under the exponential time hypothesis

Theorem 2 SCEI with |F | = 1 can be solved in timeO(2O(`) middot |G| middot log |E(G)|)

References

[1] A Arroyo F Klute I Parada R Seidel B Vogtenhu-ber and T Wiedera Inserting one edge into a simpledrawing is hard In Proc WG LNCS 12301 pages325ndash338 2020

[2] E Eiben R Ganian T Hamm F Klute andM Nollenburg Extending partial 1-planar drawingsIn Proc ICALP LIPIcs 168 pages 431ndash4319 2020

5


Computing the continuous mean distance for certain graph classeslowast

Delia Garijodagger1 Alberto MarquezDagger1 and Rodrigo I Silveirasect2

1Universidad de Sevilla Spain2Universitat Politecnica de Catalunya Spain

The mean distance of a connected unweightedgraph was first introduced in the context of archi-tecture to compare floor plans although a lot of in-terest came from chemical graph theory where theclosely related Wiener indexmdashthe sum of all pairwisedistances in the graphmdashhas been extensively studiedThe most usual way to define the mean distance is asthe arithmetic mean of all nonzero distances betweenvertices where distances are taken in the graph overall unordered pairs of vertices In the context of graphtheory Doyle and Graver [1] were the first to proposethe mean distance as a graph parameter Since thenit has been intensively studied

In a different direction Doyle and Graver [2 3] alsointroduced the mean distance of a shape defined forany weighted graph embedded in the plane Eachedge of the graph is iteratively subdivided into shorteredges so that the edge lengths approach zero Themean distance of the shape is then defined as thelimit of the mean distance of such a sequence of re-finements Doyle and Graver managed to compute itsexact value for seven specific types of simple graphs(ie a path a Y-shape an H-shape a cross and threemore) and six rather specific families of graphs themost general ones being cycles and stars with k edgesof length 1k [3]

In this work we continue in this direction studyingthe mean distance of weighted graphs in a continuoussetting Our main motivation arises from geometricgraphs undirected graphs where each vertex is a two-dimensional point and each edge is a straight line seg-ment between the corresponding two points Unlikeabstract graphs in geometric graphs distances are notonly defined for pairs of vertices but they exist for anytwo points on the graph including points on the inte-rior of edges Therefore the concept of mean distancegeneralizes naturally to geometric graphs defined asthe average distance between all pairs of points onedges of the graph

In this talk we will present novel computational

lowastThis work was supported by PID2019-104129GB-I00AEI1013039501100011033 Gen Cat 2017SGR1640 andBFU2016-74975-PdaggerEmail dgarijousesDaggerEmail almarusessectEmail rodrigosilveiraupcedu

results about the continuous mean distance Whilewe can show that the continuous mean distance ofa weighted graph with m edges can be computed inO(m2) time the focus of the talk will be on particulargraph classes for which the mean distance can be com-puted faster We will present several structural re-sults that allow a faster computation of the continuousmean distance for several classes of weighted graphsincluding complete graphs and families of graphs thathave a cut vertex In particular we will show the fol-lowing results which apply to geometric graphs andalso to some other non-geometric graphs

Proposition 1 The continuous mean distance of aweighted tree T` with n vertices can be computed inO(n) time

Proposition 2 The continuous mean distance of aweighted cactus graph with n vertices can be com-puted in O(n) time

Proposition 3 The continuous mean distance of thecomplete graph on n vertices where all edges havelength α is given by the following formula

α(9n2 minus 22n+ 12)

6 (n2 minus n)

All results to be presented in the talk can be foundin the full version of this work [4]

References

[1] J K Doyle and J E Graver Mean distance in agraph Discrete Math 17147ndash154 1977

[2] J K Doyle and J E Graver Mean distance forshapes J Graph Theory 6(4)453ndash471 1982

[3] J K Doyle and J E Graver A summary of re-sults on mean distance in shapes Environmentand Planning B Planning and Design 9177ndash17901 1982

[4] D Garijo A Marquez and R I Silveira Con-tinuous mean distance of a weighted graph 2021arXiv210311676

6


Shortest paths in weighted hexagonal tessellationslowast

Prosenjit Bosedagger1 Guillermo EstebanDagger1 2 David Ordensect2 and Rodrigo I Silveirapara3

1School of Computer Science Carleton University Canada2Departamento de Fısica y Matematicas Universidad de Alcala Spain

3Departament de Matematiques Universitat Politecnica de Catalunya Spain

Abstract

A continuous 2-dimensional space is often discretizedby being tessellated into a grid of weighted hexago-nal cells Two types of shortest paths between pointss and t can be defined namely the weighted short-est path SPw(s t) and the shortest path constrainedto the grid SGPw(s t) We prove that the ratioSGPw(st)SPw(st) is at most 3

2 for any weight assignment

1 Introduction

Finding optimal obstacle-avoiding paths from a start-ing point s to an ending point t is an important prob-lem in areas such as robotics [10] video-games [4]and geographical information systems (GIS) applica-tions [3] among others Some methods use regulargrids as navigation regions in two dimensions Theseapproaches are easier to implement [11] and are verycommon in video-games In a 2-dimensional spaceonly three types of regular polygons can be used totessellate continuous 2D environments namely trian-gles squares and hexagons

Often the cost of traversing space is not uni-form and going through different regions incurs dif-ferent costs This leads to the Weighted Region Prob-lem (WRP) [6 8] ie determining a shortest paththrough a weighted planar polygonal subdivision Ex-isting algorithms for the WRP are quite complex indesign and implementation or have very high time andspace complexities [7] Recently it has been proven [2]that the WRP cannot be solved in the Algebraic Com-putation Model over the Rational Numbers Thismakes approximation algorithms suitable and neces-sary Our work focuses on two possibilities previouslyconsidered in the literature for graphs in a regulargrid called k-corner grid graph and space graph

lowastResearch supported by NSERC Project PID2019-104129GB-I00 AEI 1013039501100011033 of the SpanishMinistry of Science and Innovation Gen Cat 2017SGR164and H2020-MSCA-RISE project 734922 - CONNECTdaggerEmail jitscscarletoncaDaggerEmail gestebanuahessectEmail davidordenuahesparaEmail rodrigosilveiraupcedu

v

(a) Neighbors of a ver-tex v in G3corner

v

(b) Neighbors of a ver-tex v in G12corner

Figure 1 Vertex v is connected to its neighbors in ahexagonal tessellation

In a k-corner grid graph (Gkcorner) the vertex set isthe set of corners of the tessellation Each vertex isconnected by an edge to all of its k neighboring ver-tices Different definitions of neighbor are considereddepending on the tessellation and the design decisionSee Figure 1 for 3-corner and 12-corner grid graphsin a hexagonal tessellation (Analogous k-corner gridgraphs can be defined for triangular and square tes-sellations) In the unrestricted space graph the vertexset is the set of all points in the plane and the edgescorrespond to segments between any pair of verticesof the graph

When the continuous space is tessellated eachcell Hi has a weight ωi isin Rgt0 and the cost ofa segment πi traversing cell Hi is given by ωiπiwhere middot is the Euclidean norm In the case wherea segment π goes along the boundary of two cells Hj

and Hk the cost is minωj ωkπDifferent types of shortest paths between two ver-

tices s and t travel along different geometric graphsfor which two main possibilities have been con-sidered in the literature a weighted shortest gridpath SGPw(s t) which is the optimal path whoseedges are edges of a k-corner grid graph and aweighted shortest path SPw(s t) which is the opti-mal path between s and t See red path SGPw(s t)and blue path SPw(s t) in Figure 2 for a comparisonof these two types of paths in the 3-corner grid graph

Geographic and spatial models provide approxi-mations of continuous 2D-spaces and high-qualityapproximate paths are favored over optimal pathsthat are expensive to compute Thus SGPw(s t)

7


ω1 =radic13

ω2 = 1s

t

Figure 2 SPw(s t) (blue) and a SGPw(s t) (red) be-tween two corners s and t in G3corner The cost ofeach path is 14 and 1642 respectively

is considered alternative to SPw(s t) and the ratio

R = SGPw(st)SPw(st) indicates the approximation factor

In this paper we focus on upper bounding the ratio Rin G3corner for a weighted hexagonal tessellation

2 Previous results

Almost all previous bounds for the ratio SGPw(st)SPw(st)

consider a limited set of weights for the cells Nash [9]considered only weights in the set 1infin and provedthat the weight of SGPw(s t) in hexagonal G6cornerhexagonal G12corner square G4corner square G8cornertriangle G6corner and triangle G3corner can be up toasymp115 asymp104 asymp141 asymp108 asymp115 and 2 times theweight of SPw(s t) respectively When the weightsof the cells are allowed to be in Rgt0 the only resultthat we are aware of is for square tessellations andanother type of shortest path with vertices at thecenter of the cells for which Jaklin [5] showed thatSGPw(st)SPw(st) le 2

radic2

3 SGPw(st)SPw(st) ratio in G3corner for hexagonal cells

We are interested in obtaining for two vertices s

and t an upper bound for the ratio SGPw(st)SPw(st)

in G3corner for hexagonal tessellations H We firstshow in Theorem 1 that given a path GP (s t) be-tween s and t whose edges are edges of the 3-corner

grid graph the ratio GP (st)SPw(st) of the whole path can

be upper bounded by the maximum among all the ra-

tios GP (uiui+1)SPw(uiui+1) where ui and ui+1 are two consecu-

tive crossing points between a GP (s t) and SPw(s t)Hereforth grid graphs will be 3-corner grid graphs un-less otherwise specified

Theorem 1 Let GP (s t) and SPw(s t) be respec-tively a grid path and a weighted shortest pathfrom s to t Let ui and ui+1 be two consecutive cross-ing points between GP (s t) and SPw(s t) Then the

ratio GP (st)SPw(st) is at most the maximum of all ratios

GP (uiui+1)SPw(uiui+1)

The key idea to prove our main result is to ob-tain a relation between the weights of cells adjacent

s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4

Figure 3 Weighted shortest path SPw(s t) (blue) andthe crossing path X(s t) (red) from s to t in G3corner

to SGPw(s t) To do so we will define a class of gridpaths called crossing paths X(s t) see the red pathin Figure 3

Let (H1 Hn) be the ordered sequence of con-secutive cells traversed by SPw(s t) in the tessella-tion H Let vi1 v

i6 be the six consecutive cor-

ners of the boundary of Hi 1 le i le n Let (s =a1 a2 an+1 = t) be the sequence of consecutivepoints where SPw(s t) traverses the cells boundaryin H In particular let ai and ai+1 be respectivelythe points where SPw(s t) enters and leaves Hi Thecrossing path X(s t) from a vertex s to a vertex t ispiecewise defined next

Definition 2 The crossing path X(s t) between twovertices s and t is defined as the path X1 cup cupXnwhere Xi is determined for the pair (ai ai+1) 1 lei le n as follows

bull If ai = vi1 and ai+1 = vi2 Xi = (vi1 vi2) if ai+1 =

vi3 Xi = (vi1 vi2 v

i3) and if ai+1 = vi4 Xi =

(vi1 vi2 v

i3 v

i4) see Figure 4(a) The cases when

ai isin vi5 vi6 are symmetric

bull If ai = vi1 and ai+1 isin (vi1 vi2) Xi = (vi1 v

i2)

if ai+1 isin (vi2 vi3) Xi = (vi1 v

i2 v

i3) see Figure

4(b) and if ai+1 isin (vi3 vi4) Xi = (vi1 v

i2 v

i3 v

i4)

see Figure 4(c) The cases when ai+1 isin(vi4 vi5) (vi5 v

i6) (vi6 v

i1) are symmetric

If ai belongs to the interior of an edge ei1 isin Hi let p bethe point different from ai where the line through aiperpendicular to ei1 intersects the boundary of Hi

bull If ai+1 is a corner of Hi to the left (right)of minusrarraip Xi is the set of ai+1 and the corners of Hi

to the left (right) of minusminusminusminusrarraiai+1 see Figure 4(d)

bull If ai+1 is a point to the left (right) of minusrarraip in theinterior of an edge ei2 isin Hi not parallel to ei1 Xi

is the set of the endpoint of ei1 to the left (right)of minusrarraip and both endpoints of ei2 see Figure 4(e)

bull If ai+1 is a point in the interior of an edge ei2parallel to ei1 and Ximinus1 contains the endpointof ei1 to the left (right) of minusrarraip Xi is the set ofcorners of Hi to the left (right) of minusminusminusminusrarraiai+1

8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi

Figure 4 Some of the positions of the intersectionpoints between SPw(s t) (blue) and the edges of HiThe subpath of the crossing path X(s t) created is inred

Let (s = u1 u2 u` = t) be the se-quence of consecutive crossing points betweenX(s t) and SPw(s t) see Figure 3 The unionof SPw(s t) and X(s t) between two consecutivecrossing points uj and uj+1 for 1 le j lt ` inducesa weakly simple polygon (see [1] for a formal defini-tion) Depending on the number of cells that sep-arate uj and uj+1 we distinguish different types ofweakly simple polygons The crossing points uj uj+1

could belong to the same cell see Figure 5 or to differ-ent cells see Figure 6 Observe that by the definitionof X(s t) the only weakly simple polygons that canarise are those defined in Definitions 3 and 4

Definition 3 A weakly simple polygon inducedby two consecutive crossings uj uj+1 isin Hi be-tween X(s t) and SPw(s t) is of type P 1

k 1 le k le 6if

bull For 1 le k le 5 X(uj uj+1) travels along k edgesof Hi

Let p be the intersection point different from uj be-tween the edges of Hi and the line through uj per-pendicular to the edge ei1 3 uj

bull For k = 6 uj+1 is a corner of Hi to theleft (right) of minusrarrujp and uj+1 is connected to ujin X(uj uj+1) through the corners of Hi to theleft (right) of minusrarrujp In addition SPw(s t) tra-verses Hi from ei1 to its parallel edge

Definition 4 Let (Hi Hm) m ge i + 1 be anordered sequence of consecutive cells whose interior istraversed by SPw(s t) Let uj isin Hi and uj+1 isin Hm

be two consecutive crossing points between SPw(s t)and X(s t) Let p isin Hmminus1 be the intersection pointbetween the edges of Hm and the line through uj

P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ

Figure 5 Some weakly simple polygons P 1k and the

subpath of the crossing path X(s t) (red) from ujto uj+1 The angle of incidence θ of SPw(s t) (blue)in the boundary of Hi obeys Snellrsquos law and for sim-plicity is only depicted in P 1

3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q

Figure 6 Two weakly simple polygons Pmminusi+11

and Pmminusi+15 and the subpaths of SPw(s t) (blue)

and X(s t) (red) from uj to uj+1

perpendicular to the edge uj belongs to Let q bethe point where SPw(uj uj+1) leaves Hm for the firsttime and let q be to the left (right) of minusrarrujp Let b bethe left (right) endpoint of the edge q belongs to withrespect to the SPw(s t) when considering SPw(s t)oriented from s to t A weakly simple polygon in-duced by uj and uj+1 is of type Pmminusi+1

k 1 le k le 5if

bull X(uj b) travels through the cornersof Hi Hm to the left (right) of minusrarrujpbull For 1 le k le 4 in addition X(b uj+1) travels

along k edges of Hm

In order to prove our main theorem we will need aseries of intermediate results

Let (Hi Hm) m ge i + 1 be an ordered se-quence of consecutive cells whose interior is traversedby SPw(s t) Let uj isin Hi and uj+1 isin Hm betwo consecutive crossing points between SPw(s t)and X(s t) According to Lemmas 5 and 6 an up-

per bound for the ratioX(uj uj+1)SPw(uj uj+1) in the weakly

simple polygons of type Pmminusi+1k k isin 1 4 and

Pmminusi+15 is obtained when m = i + 1 and m = i re-

spectively

9


Lemma 5 Let uj and uj+1 be two consecutivecrossing points between SPw(s t) and X(s t) ina weakly simple polygon of type Pmminusi+1

k k isin1 4 m ge i + 1 Then an upper bound for the

ratioX(uj uj+1)SPw(uj uj+1) is obtained when m = i+ 1

Lemma 6 Let uj and uj+1 be two consecutive cross-ing points between SPw(s t) and X(s t) in a weaklysimple polygon of type Pmminusi+1

5 m ge i Then an

upper bound for the ratioX(uj uj+1)SPw(uj uj+1) is obtained

when m = i

We can see thatX(uj uj+1)SPw(uj uj+1) = 1 in a P 1

1

A P 13 may contain two appearances of a P 1

2 ie SPw(uj uj+1) goes along the edges of the cell ujand uj+1 belong to If that is the case we can boundits ratio by the ratio in a P 1

2 Analogously in a P 22

if SPw(uj uj+1) goes along the edges of the cell uj+1

belongs to we can bound its ratio by the ratio in a P 21

and a P 12 In addition note that a P 1

4 and a P 15 con-

tain multiple appearances of a P 12 and a P 1

3 Hencethe ratios for them can be bounded by the ratios ina P 1

2 and a P 13 Also a P 2

3 and a P 24 contain multiple

appearances of a P 12 and a P 1

3 in the cell uj+1 belongsto and hence the ratios for them are bounded bythat for a P 2

1 and a P 22

All this implies that the upper bound for the ratioX(uj uj+1)SPw(uj uj+1) is obtained in a weakly simple polygon

of type either (i) P 12 (ii) P 1

3 or P 22 when SPw(uj uj+1)

does not travel along the edges of a cell (iii) P 16 or

(iv) P 21 The proofs of the bounds in Lemmas 7 8

and 9 require multiple case analyses

Lemma 7 Let uj and uj+1 be two consecutive cross-ing points between a shortest path SPw(s t) and thecrossing path X(s t) An upper bound for the ratioX(uj uj+1)SPw(uj uj+1) in P 1

2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13

According to Lemmas 7 8 and 9 an upper bound

for the ratioX(uj uj+1)SPw(uj uj+1) in the five types of weakly

simple polygons is 32 Hence using Theorem 1 and

the fact that SGPw(s t) le X(s t) we obtain ourmain result

Theorem 10 In G3corner an upper bound for the ra-

tio SGPw(st)SPw(st) is 3

2

4 Conclusions

We proved an upper bound for the ratio betweenthe lengths of a weighted shortest grid path in the3-corner grid graph and a weighted shortest pathFollowing an analogous procedure we can obtain an

upper bound for the ratio SGPw(st)SPw(st) in G12corner

as well as for other tessellations such as triangularand square Along similar lines we can obtain up-per bounds for another type of grid graph where thevertices are cell centers instead of corners

References

[1] H C Chang J Erickson and C Xu Detectingweakly simple polygons In Proceedings of the twenty-sixth annual ACM-SIAM Symposium on Discrete Al-gorithms pages 1655ndash1670 SIAM 2014

[2] J L de Carufel C Grimm A MaheshwariM Owen and M Smid A note on the unsolvabilityof the weighted region shortest path problem Com-putational Geometry 47(7)724ndash727 2014

[3] L de Floriani P Magillo and E Puppo Applica-tions of computational geometry to geographic infor-mation systems Handbook of computational geome-try 7333ndash388 2000

[4] S Forum NAM traffic simulator and data view sup-port thread httpscommunitysimtropoliscomforumstopic29437-nam-traffic-simulator-

and-data-view-support-thread

[5] N S Jaklin On Weighted Regions and SocialCrowds Autonomous-agent Navigation in VirtualWorlds PhD thesis Utrecht University 2016

[6] J Mitchell Shortest paths among obstacles zero-costregions and roads Technical report Cornell Univer-sity Operations Research and Industrial Engineering1987

[7] J Mitchell D M Mount and C Papadimitriou Thediscrete geodesic problem SIAM Journal on Com-puting 16(4)647ndash668 1987

[8] J Mitchell and C Papadimitrou The weightedregion problem Finding shortest paths through aweighted planar subdivision Journal of the ACM38(1)18ndash73 1991

[9] A Nash Any-Angle Path Planning PhD thesis Uni-versity of Southern California 2012

[10] M Sharir and S Sifrony Coordinated motion plan-ning for two independent robots Annals of Mathe-matics and Artificial Intelligence 3(1)107ndash130 1991

[11] N Sturtevant A sparse grid representation for dy-namic three-dimensional worlds In Proceedings ofthe AAAI Conference on Artificial Intelligence andInteractive Digital Entertainment volume 6 2011

10


On prescribing total orders and preorders to pairwise distances of pointsin Euclidean space

Vıctor Hugo Almendra-Hernandezlowast1 and Leonardo Martınez-Sandovaldagger1

1Faculty of Sciences National Autonomous University of Mexico

Abstract

We show that any total preorder on a set with(n2

)elements coincides with the length order on pairwisedistances of some point set of size n in Rnminus1 Fortotal orders a set of n points in Rnminus2 suffices Wealso prove that the required dimensions in both casesare optimal We use tools from convexity and positivesemidefinite quadratic forms

1 Introduction

For a positive integer n we define [n] = 1 2 nLet P = pi i isin [n] be a set of n points in d-dimensional Euclidean space We assume that P is ingeneral position

Under these assumptions the point set P inducesa total preorder le on the family of pairs

Dn =

([n]

2

)= (i j) 1 le i lt j le n

given by (i1 i2) le (j1 j2) if and only if pi1 minus pi2 lepj1 minus pj2 When P induces pairwise distinct dis-tances this preorder is also antisymmetric and thusit is a total order on Dn We say that le is inducedby P

P1

P2

P3

P4

(1 2) le (2 3) equiv (1 3) le (3 4) equiv (2 4) le (1 4)

Figure 1 Example of induced preorder

Is every total preorder on Dn induced by a set ofpoints in d-dimensional Euclidean space What abouttotal orders

lowastEmail vhalmendrahcienciasunammx This work wassupported by UNAM-PAPIIT IA104621daggerEmail leomtzcienciasunammx This work was sup-

ported by UNAM-PAPIIT IA104621

We prove that any given total order or preorder isachievable if and only if d is large enough in terms ofn Our main result is an exact bound on the minimaldimension required for this to happen

Theorem 1 Let n ge 3 be an integer

bull The minimal dimension into which any total or-der on Dn can be induced by the pairwise dis-tances of a point set in Rd is d = nminus 2

bull The minimal dimension into which any total pre-order on Dn can be induced by the pairwise dis-tances of a point set in Rd is d = nminus 1

As a helpful reminder a total preorder on a set Xis a reflexive and transitive relation le in which everytwo elements of X are comparable We say that x lt yif x le y is in the relation but y le x is not We definex equiv y if and only if x le y and y le x It is immediatethat equiv is an equivalence relation on X and that leinduces a total order on the equivalence classes Ifle is antisymmetric then each equivalence class hasexactly one element so le is itself a total order on X

We divide the proof of Theorem 1 in two sectionsIn Section 2 we provide our lower bounds We ex-hibit for each n ge 3 a total order on Dn that cannotbe induced from a family of n points in Rnminus3 Wealso exhibit a preorder that cannot be induced froma family of n points in Rnminus2

In Section 3 we use a powerful lemma on Euclideandistances and positive semidefinite matrices to induceany given total order on Dn from a point set in Rnminus2To introduce our technique first we prove that anypreorder on Dn can be attainable in Rnminus1 We thenshow how to adapt the technique to reduce the re-quired dimension in the case of total orders

2 Lower bound

If n = 3 it is clear that we need d ge 1 to attain everypossible total order on D3 Thus the first non-trivialcase is n = 4 We claim that there is no point set inR that induces a total order on D4 with the followingrelations

11


P1 P2 P3 P4

Figure 2 Case n = 4

1 (1 4) is the (unique) maximal element of the totalorder

2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

The proof is simple in this case since (1 4) is themaximal element this forces p1 and p4 to be the ex-tremal points in the geometric configuration We mayassume without loss of generality that p1 is the left-most point and p4 is the rightmost point The secondrelation forces p2 to be closer to p1 than p3 but thiscontradicts the last relation

The constructions for larger values of n require acareful selection of prescribed relations and convexityarguments We begin by stating an auxiliary geomet-ric result and a corollary

Proposition 2 Let P = p1 pd+1 be a set ofd+1 affinely independent points in Rd For i isin [d+1]let Πi be the hyperplane spanned by P pi and Hi

the closed halfspace defined by Πi in which pi liesLet Π be a hyperplane such that p1 lies in one of

its open halfspaces which we call H+ and such thatP p1 is contained in the (closed) complement HminusThen the closure ∆prime of

H+ capH2 cap capHd+1

is a simplex contained in the simplex

∆ = H1 capH2 cap capHd+1

A proof for this proposition can be found in theappendix

By repeatedly applying the lemma above we havethe following consequence

Corollary 3 Let P = p1 pd+1 be a set of d+1affinely independent points in Rd For i isin [d + 1]let Πi be a hyperplane such that pi lies on one of itsopen halfspaces Hi and P pi is contained in thecomplement of Hi

Then the closure ∆prime of H1 cap H2 cap cap Hd+1 isa (possibly empty) simplex contained in the simplexspanned by P

Proposition 4 Let n ge 4 be an integer Then thereis no set of n points P in general position in Rnminus3 thatinduces a total order on Dn including the following inthe relation

1 For any pair (i j) in Dnminus3 and a pair (k l) inDn Dnminus3 we have (i j) gt (k l)

2 For any pair (i j) isin [nminus3]timesnminus2 nminus1 n anda pair (k l) in Dn ([n minus 3] times n minus 2 n minus 1 n)we have (i j) lt (k l)

P1 Pnminus3 Pnminus2 Pnminus1 Pn

lt lt

Figure 3 Relations for order in Dn

Proof We proceed by contradiction Suppose thatthere is a point set P = 1 2 n that induces atotal order le on Dn with the given relations Since leis a total order all pairwise distances of P are distinct

Let Π be the affine hyperplane spanned byp1 pnminus3 Among the points pnminus2 pnminus1 pn twoof them lie on the same open halfspace defined by ΠWithout loss of generality we may assume they arepnminus2 and pnminus1 We now show that pnminus1 lies in thesimplex ∆ spanned by p1 pnminus2

To do so for i isin [n minus 3] let Πi be the perpendic-ular bisector hyperplane to the segment pipnminus2 andΠnminus2 = Π (See Figure 4) For i in [n minus 2] let Hi bethe open halfspace of Πi on which pi lies

P1 P2

P3

P4

P5

Π = Π3

Π1

Π2

Figure 4 Proof for n = 5 in the plane

By the relations in (1) the distances among pointspi pj with i j isin [n minus 3] are the largest This impliesthat for distinct i j isin [n minus 3] we have pi minus pj gtpj minus pnminus2 so pj is on the opposite open halfspacedefined by Πi as pi This means that H1 Hnminus2

satisfy the hypothesis of Corollary 3 and therefore theclosure ∆prime of H1 cap capHnminus2 is contained in ∆ (forexample in Figure 4 this corresponds to the orange-shaded triangle lying inside the triangle P1 P2 P3)

To finish the proof of our claim note that the re-lations in (2) imply that the next largest distances

12


are among points pi pj with i j isin n minus 2 n minus 1 nso the remaining distances are smaller than these Inparticular for every i in [nminus 3] we have that

pnminus1 minus pnminus2 gt pnminus1 minus pi

and thus pnminus1 lies in Hi Since pnminus1 was originallychosen to be in Hnminus2 we conclude that pnminus1 is in∆prime sube ∆ as claimed

An analogous proof shows that pnminus2 lies in the sim-plex spanned by P pnminus2 pn We conclude thatpnminus2 = pnminus1 a contradiction to P having n distinctpoints

Now we focus on the lower bound for total pre-orders

Proposition 5 Let n ge 3 be an integer Then thereis no set of n points P in general position in Rnminus2

that induces a preorder on Dn in which (n minus 1 n) isa unique minimal element and the rest of the pairsbelong to a single and maximal class

A proof for this can be found in the appendix Thelower bound in the case of preorders can be obtainedin other ways For example Yugai [4] proves that themaximal number of times a diameter can appear in apoint set on n points in Rnminus2 is

(nminus1

2

)+ nminus 3 lt

(n2

)

Therefore any preorder that imposes more than thisnumber of diameters will be impossible to attain Fora deeper study on the maximal number of times adiameter of a point set can appear see [1] and thereferences therein

3 Upper bound

Our upper bounds rely on a powerful lemma bySchoenberg [3] that characterizes families of real num-bers that can appear as Euclidean distances inducedby a point set We refer the reader to the text by Ma-tousek [2] for a nice and short proof of the followingresult using linear algebra

Theorem 6 Let mij for i j in [n+1] be nonnegativereal numbers with mij = mji for all i j and mii = 0for all i Then there exist points p1 pn+1 in Rn

with pi minus pj = mij for all i j if and only if the ntimesnmatrix G with entries

gij =1

2(m2

(n+1)i +m2(n+1)j minusm

2ij)

for i j isin [n] is positive semidefinite

Note that Theorem 6 does not guarantee that thepoints p1 pn+1 are distinct To illustrate our tech-nique we begin by proving the upper bound in thecase of preorders

Proposition 7 Any total preorder le on Dn can beinduced by a set of n points in Rnminus1

Proof Let equiv be the equivalence relation induced onDn by le Since le induces a total order on the equiv-alence classes we may name them as follows

Q1 lt Q2 lt lt Qm

Let ε gt 0 be a sufficiently small real number to bedetermined later

We define the following numbers

mij =

0 if i = j

1 + kε if i lt j and (i j) isin Qk

1 + kε if i gt j and (j i) isin Qk

Note the mij are well defined since we have a totalpreorder for each pair (i j) isin Dn there exists k suchthat (i j) isin Qk From this definition it is clear thatmij = mji Consider now the (nminus1)times (nminus1) matrixG with entries

gij =1

2(m2

ni +m2nj minusm2

ij)

for i j isin [n] We claim that if ε is small enoughthen G is positive definite

Indeed the values gij depend continuously on thevalues mij and these in turn depend continuously onε As εrarr 0 we get that

2Grarr

2 1 1 middot middot middot 11 2 1 middot middot middot 11 1 2 middot middot middot 1

1 1 1 middot middot middot 2

The matrix on the right hand side corresponds to

the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i

which is positive definiteThe subset of Mn(R) consisting of positive definite

matrices is open So we may set ε gt 0 as a num-ber such that 2G is positive definite therefore G willbe positive definite too By Theorem 6 there arep1 pn in Rnminus1 such that pi minus pj = mij for alli j Since distances between distinct points are nonzero this set P has exactly n points We claim thatP induces the given preorder le on Dn

13


Indeed if (i j) le (k l) then there are indices m1 lem2 such that (i j) isin Qm1 and (k l) isin Qm2 and then

pi minus pj = 1 +m1ε le 1 +m2ε = pk minus pl

If it is not the case that (i j) le (k l) then (k l) lt(i j) so there are indices m2 lt m1 such that (i j) isinQm1

and (k l) isin Qm2 Thus we have

pi minus pj = 1 +m1ε gt 1 +m2ε = pk minus pl

which shows that the we do not have (i j) le (k l)in the induced relation

Therefore we recover exactly the relations given byle with the order of pairwise distances of P

A careful adaptation of the proof above yields thedesired result for total orders

Proposition 8 Any total order le on Dn can be in-duced by a set of points in Rnminus2

Proof (Sketch)Let le be a total order on Dn and let N =

(n2

)

Then all the elements in Dn can be listed as follows

(i1 j1) lt (i2 j2) lt lt (iN jN )

Without loss of generality we may assume that(i1 j1) = (nminus 1 n)

Let ε gt 0 be a sufficiently small real number to bedetermined later For (i j) 6= (nminus 1 n) we define thenumbers

mij =

0 if i = j

1 + kε if i lt j and (i j) = (ik jk)

1 + kε if i gt j and (j i) = (ik jk)

The mij are well defined since we have a total or-der for each pair (i j) isin Dn there exists k such that(i j) = (ik jk) We can find point sets p1 pnminus1

and q1 qnminus2 qn such that

pi minus pj = mij for i j isin [nminus 1]

qi minus qj = mij for i j isin [nminus 2] cup n

The point sets P prime = p1 pnminus2 and Qprime =q1 qnminus2 have the same pairwise distances sothere is an isometry that takes one to the other Thuswe may assume that pi = qi for i isin [nminus 2] Let π bethe hyperplane of Rnminus2 spanned by p1 pnminus2We may assume that pnminus1 and qn lie on the samehalfspace defined by π

As ε rarr 0 the point sets P prime and Qprime converge tobe the vertices of a unit regular (d minus 2)-dimensionalsimplex Therefore as ε rarr 0 we have thatpnminus1 minus qn rarr 0

We set pn = qn Note that pn 6= pnminus1 as otherwisewe would get the contradiction

m1(nminus1) = p1 minus pnminus1 = p1 minus pn = m1n

Therefore P = p1 pnminus1 pn is point set ofsize n It can be shown that this set induces the totalorder le on Dn

4 Discussion

We have shown that n minus 1 is the minimal dimensioninto which every total preorder on Dn can be inducedby the order of the pairwise distances of n distinctpoints In the case of total orders on Dn this minimaldimension can be reduced to nminus 2

The proof of Proposition 8 may give the impressionthat the hypothesis can be weakened to only requirea total preorder with a unique minimal element towhich we will associate the distance pnminus1 minus pn Theproof will fail as witnessed by the counterexample inthe remark on diameters after the proof of Proposition5

The main problem when replicating the argumentis that there will be no guarantee that the constructedpoints are distinct

Our study leads to the following open problem

Problem 9 For n ge d + 3 (resp n ge d + 2) char-acterize the linear orders (resp total preorders) onDn that can be induced from the ordering of pairwisedistances of a point set of size n in Rd

We expect a full characterization to be out of reachof current tools in the area as eg a full solution tothe total preorders problem would imply a solutionto the maximum number of diameter pairs problem[1] Nevertheless any partial progress would shed ad-ditional light on the complex behaviour of pairwisedistances of points in Euclidean space

References

[1] H Martini and V Soltan Antipodality properties offinite sets in Euclidean space Discrete Mathematics290(2)221 ndash 228 2005

[2] Jirı Matousek Thirty-three miniatures Mathematicaland Algorithmic applications of Linear Algebra Amer-ican Mathematical Society Providence RI 2010

[3] IJ Schoenberg Remarks to Maurice Frechetrsquos arti-cle rsquoSur la definition axiomatique drsquoune classe drsquoespacedistances vectoriellement applicable sur lrsquoespace deHilbertrsquo Annals of Mathematics 36(3)724 ndash 7321935

[4] SA Yugai On the largest number of diameters of apoint set in Euclidean space Investigations in Topo-logical and Generalized Spaces (Russian) pages 84ndash871988

14


Applications of geometric graphs to sensory analysis of cookieslowast

David Ordendagger1 Encarnacion Fernandez-FernandezDagger2 Marino Tejedor-Romerosect1 and AlejandraMartınez-Moraianpara1

1Departamento de Fısica y Matematicas Universidad de Alcala Madrid Spain2Area de Tecnologıa de Alimentos ETS de Ingenierıas Agrarias Universidad de Valladolid Palencia Spain

Sensory analysis is a very important branch ofscience for food industry where sensory profiling iscrucial in order to adapt the products to the con-sumersrsquo preferences While traditionally such a taskwas mainly performed by trained panels and thereforetime-consuming a number of rapid sensory methodshave arisen in the last two decades [4]

Among such methods Projective Mapping asks thetasters to position samples on the plane following asimple rule The more similar the samples the closerthey should be positioned (and vice versa) Despitethe geometric and two-dimensional nature of this pro-cess the data collected had only been treated by sta-tistical techniques until some of us recently proposedthe geometric method SensoGraph [1] using Gabrielgraphs for clustering the answers and the Kamada-Kawai graph drawing algorithm to produce an aver-age positioning

This talk will review the methods and results in [2]where distances between samples are considered asan alternative to the Gabriel graph and the resultingtwo geometric methods are compared to the statisti-cal standard [3] Multiple Factor Analysis (MFA) interms of quality and stability of the results A largenumber of 349 tasters participated in the study evalu-ating eight different commercial chocolate chip cook-ies (one best-selling brand and seven private labelsfrom different supermarkets) plus a blind duplicatesample (the best-selling brand)

Both the geometric and the statistical outputs pro-vided the same groups of samples with the two blindduplicates appearing as the most similar ones SeeFigure 1 Actually the identification of the duplicateswas clearer with the geometric methods On the con-trary the stability of the results (studied using boot-strapping resampling) was better for MFA As for thegeometric methods using distances provided greaterstability than the use of the Gabriel graph

Our results show the interest for sensory analysts of

lowastSupported by Project PID2019-104129GB-I00 AEI 1013039501100011033 of the Spanish Ministry of Science

daggerEmail davidordenuahesDaggerEmail encarnacionfernandezuvaessectEmail marinotejedoruahesparaEmail alejandramartinezmuahes

adding techniques based on geometric graphs to theirtoolbox

Figure 1 Output for SensoGraph using distancesSamples considered more similar appear closer andvice versa The similarity between samples is also rep-resented by the width and color of the connections

References

[1] D Orden E Fernandez-Fernandez JM Rodrıguez-Nogales J Vila-Crespo (2019) Testing SensoGrapha geometric approach for fast sensory evaluation FoodQuality and Preference 72 1ndash9

[2] D Orden E Fernandez-Fernandez M Tejedor-Romero A Martınez-Moraian (2021) Geometric andstatistical techniques for projective mapping of choco-late chip cookies with a large number of consumersFood Quality and Preference 87 104068

[3] Pages J (2005) Collection and analysis of perceivedproduct inter-distances using multiple factor analy-sis Application to the study of 10 white wines fromthe Loire Valley Food Quality and Preference 16(7)642ndash649

[4] P Varela and G Ares Novel techniques in sensorycharacterization and consumer profiling CRC Press2014

15


Bounds on the Diameter of Graph Associahedralowast

Jean Cardinaldagger1 Lionel PourninDagger2 and Mario Valencia-Pabonsect2

1Universite libre de Bruxelles (ULB)2LIPN Universite Sorbonne Paris Nord

Graph associahedra are generalized permutohedraarising as special cases of nestohedra and hyper-graphic polytopes [3 7] The graph associahedronof a connected graph G encodes the combinatoricsof search trees on G defined recursively by a root rtogether with search trees on each of the connectedcomponents of G minus r In particular the skeleton ofthe graph associahedron is the rotation graph of thosesearch trees two vertices form an edge if the corre-sponding trees differ by a rotation a local operationsimilar to rotations in binary search trees

The question of how large the diameter of a poly-hedron arises naturally from the study of linear pro-gramming and the simplex algorithm (see for in-stance [10] and references therein) The case of associ-ahedra [4 12]mdashwhose diameter is known exactly [8]mdashis particularly interesting Indeed the diameter ofthese polytopes is related to the worst-case complex-ity of rebalancing binary search trees [11] Classicalassociahedra are associahedra of paths of length nand their diameter is 2nminus 6 for a large enough n

We investigate the diameter of graph associahe-dra as a function of some parameters of the under-lying graph The results are relevant in the contextof data structures for searching in graphs [1] Thequestion has already been studied by Manneville andPilaud [5] by Pournin [9] in the special case of cy-clohedra and by Cardinal Langerman and Perez-Lantero [2] in the special case of tree associahedra

We aim at giving tighter bounds on the diameter ofgraph associahedra in terms of some structural invari-ants of the underlying graph in particular the path-width the treewidth and the tree-depth These in-variants play a key role in the theory of graph minorsand in graph algorithms The tree-depth of a graphG is the smallest height of a search tree on G wherea tree composed of a single vertex has height one [6]The tree-depth is definitely a natural invariant to con-sider as it is a function of the exact same objects thatform the vertices of the graph associahedron

lowastThis is a summary of a paper to appear in the Proceed-ings of the XI Latin and American Algorithms Graphs andOptimization Symposium (LAGOSrsquo21)daggerEmail jcardinulbacbeDaggerEmail lionelpourninuniv-paris13frsectEmail valencialipnuniv-paris13fr

We first prove that the lower bound of m from Man-neville and Pilaud [5] on the diameter of the associa-hedra of a graph on m edges is essentially tight for alltrivially perfect graphs Those graphs appear natu-rally here as they are maximal for a fixed tree-depthUsing this we give lower and upper bounds on themaximum diameter of associahedra of graphs on nvertices as a function of the pathwidth treewidthand tree-depth of the graph These bounds are near-linear in n middot p(G) where p(G) is the parameter Forthe tree-depth we get a tight linear bound Finallywe prove that the maximum diameter of associahedraof graphs of pathwidth two is already superlinear

References

[1] Prosenjit Bose Jean Cardinal John Iacono GrigoriosKoumoutsos and Stefan Langerman Competitive onlinesearch trees on trees In Proceedings of the Annual ACM-SIAM Symposium on Discrete Algorithms (SODA20)pages 1878ndash1891 2020

[2] Jean Cardinal Stefan Langerman and Pablo Perez-Lantero On the diameter of tree associahedra ElectrJ Comb 25(4)P418 2018

[3] Michael Carr and Satyan L Devadoss Coxeter complexesand graph-associahedra Topology and its Applications153(12)2155ndash2168 2006

[4] Jean-Louis Loday Realization of the stasheff polytopeArchiv der Mathematik 83(3)267ndash278 Sep 2004

[5] Thibault Manneville and Vincent Pilaud Graph prop-erties of graph associahedra Seminaire Lotharingien deCombinatoire B73d 2015

[6] Jaroslav Nesetril and Patrice Ossona de Mendez SparsityGraphs Structures and Algorithms chapter 6 pages 115ndash144 Springer 2012

[7] Alexander Postnikov Permutohedra associahedra andbeyond International Mathematics Research Notices2009(6)1026ndash1106 2009

[8] Lionel Pournin The diameter of associahedra Advancesin Mathematics 25913ndash42 2014

[9] Lionel Pournin The asymptotic diameter of cyclohedraIsrael Journal of Mathematics 219(2)609ndash635 Apr 2017

[10] Francisco Santos A counterexample to the Hirsch conjec-ture Annals of Mathematics 176383ndash412 2012

[11] Daniel Sleator Robert Tarjan and William Thurston Ro-tation distance triangulations and hyperbolic geometryJournal of the American Mathematical Society 1647ndash6811988

[12] James Dillon Stasheff Homotopy associativity of H-spaces I Transactions of the American MathematicalSociety 108(2)275ndash292 1963

16


Showing non-realizability of spheresby distilling a tree

Julian Pfeifle

Universitat Politecnica de Catalunya

Abstract

In [Zhe20a] Hailun Zheng constructs a combinato-rial 3-sphere on 16 vertices whose graph is the com-plete 4-partite graph K4444 Such a sphere seemsunlikely to be realizable as the boundary complex ofa 4-dimensional polytope but all known techniquesfor proving this fail because there are just too manypossibilities for the 16times 4 = 64 coordinates of its ver-tices Known results [PPS12] on polytopal realizabil-ity of graphs also do not cover multipartite graphs

In this paper we level up the old idea ofGrassmannndashPlucker relations and assemble them us-ing integer programming into a new and more pow-erful structure called positive GrassmannndashPluckertrees that proves the non-realizability of this exam-ple and many other previously inaccessible families ofsimplicial spheres See [Pfe20] for the full version

1 Introduction

A simplicial (d minus 1)-sphere Σ is a simplicial complexhomeomorphic to a (d minus 1)-dimensional sphere Wesay that Σ is non-realizable if there does not exista (necessarily simplicial) d-polytope whose boundarycomplex is isomorphic to Σ

Example 1 The following list of 19 facets defines a 3-sphere Σ on 8 vertices with 27 edges and 38 triangles

+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]

The signs define an orientation of Σ

How to prove that this 3-sphere is non-realizableIn this case the venerable GrassmannndashPlucker rela-tions suffice These are polynomial relations that aresatisfied by the determinants of any d+ 1 points of a

Email julianpfeifleupcedu The author was supportedby the grant PID2019-106188GB-I00 from the Spanish Ministryof Education (MEC)

realization of Σ in d-space The most basic ones arethe three-term GP relations Γ(S|ijkl) = 0 with

Γ(S|ijkl) = [Sij][Skl]minus [Sik][Sjl] + [Sil][Sjk] (1)

which are valid for any subset S sub 1 2 n of size(dminus1) and any four indices i j k l isin 1 2 nSA typical 3-term GP relation in our example is

0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]

By permuting the entries inside these determinantswe can change their signs mdash even permutations willleave the sign unchanged while odd permutations willflip it A particularly advantageous way of changingthe signs is

0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]

The advantage of rewriting Γ(045|1267) in this way isthat now all determinants are positive For example[01425] gt 0 because [01425] = minus[01245] and [01245]is the ldquosigned slackrdquo of the point x5 with respect tothe facet [0124] in the supposed convex realization ofΣ but the orientation of 0124 in Σ is negative by theabove list The other determinants can be similarlychecked to be positive

But this expresses zero as a positive combinationof positive numbers which is impossible thereforethere is no convex realization of this 3-sphere

2 The non-realizability of Zhengrsquos 3-sphere

To explain why Zhengrsquos combinatorial 3-sphere Z isimportant letrsquos fix definitions A (dminus 1)-dimensionalsimplicial complex is balanced if its 1-skeleton is d-colorable in the graph-theoretic sense ie its verticescan be colored with d colors in such a way that theendpoints of all edges receive different colors More-over a (d minus 1)-dimensional balanced simplicial com-plex Σ is balanced k-neighborly if each k-subset of thevertex set that contains at most one vertex of eachcolor class is actually a face of Σ

17


Now we can say why Zhengrsquos example Z is impor-tant mdash in fact it is important in at least two ways

First there has been a lot of work on analogiesbetween combinatorial data in the balanced and thenon-balanced settings [JM18 JMNS18 Ven19] Forexample one would like to have a balanced analogueof the celebrated Upper Bound Theorem by McMullenand Stanley For this in particular one would like bal-anced analogues of the extremal examples to even ex-ist ie one would like to construct infinite familiesof balanced k-neighborly simplicial spheres WhatZheng shows in [Zhe20a] however is that (i) thereis no balanced 2-neighborly homology 3-sphere on 12vertices (ii) there is no balanced 2-neighborly homol-ogy 4-sphere on 15 vertices (iii) but taking suspen-sions over Z yields a balanced 2-neighborly homology(3 +m)-sphere on 16 + 2m vertices for every m ge 0

The second reason why her example is importantlies in the fact that in [PPS12] the authors studywhich graphs are realizable as the 1-skeleton of poly-topes The case of multipartite graphs was not treatedthere and to date the only polytope whose graph isknown to be the multipartite graph K4444 is the 4-dimensional cross polytope

We can now show for the first time that Z is notrealizable as the boundary complex of a convex poly-tope and therefore that Z does not yield a new poly-tope whose graph is K4444

Theorem 2 Zhengrsquos balanced sphere Z is not poly-topal

Proof An orientation of the facets of Z is given bythe following list

-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]

We prove the non-realizability of Z in a similar way asin Example 1 but we allow GP relations in which wedo not have full control over the signs For examplewe can express

0 = Γ(18f|56bd)

= [18f56][18fbd] minus [18f5b][18f6d] + [18f5d][18f6b]

= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]

where all (black) signs are known to be positive in anyrealization of Z but the (red) sign with a questionmark can be either positive or negative For instance[16f8] is a positively oriented facet because the ori-entation of [168f] in the given list is negative andthis implies that all determinants of the form [16f8x]

must be positive because all points x lie on the sameside of the facet in any convex realization On theother hand no four-element subset of [18bdf] ap-pears in the list of facets of Z so the sign of thatdeterminant could be positive or negative

To balance this uncertainty we look for anotherGP relation that involves [18bdf] A candidate is

0 = Γ(1bf|48de)

= [1bf48][1bfde] minus [1bf4d][1bf8e] + [1bf4e][1bf8d]

= [14bf8][1bdef] + [14bfd][18bef] minus [14bfe][18bdf]

On the one hand we can eliminate the unknown sign[18bdf] by forming the polynomial combination

[14bfe] middot Γ(18f|56bd) + [16f85] middot Γ(1bf|48de)

but on the other hand we now have two additionalunknown signs to worry about

Somewhat surprisingly we are able to bring thisprocess to a closure by forming the following polyno-mial combination of GP relations

[36fb5]

([36fb4]

([14bf3]

([16f85]

([14bfd](minusΓ(18f|46be))

+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)

)+[16f84] [14bf8] [14bfe] [16f85](minusΓ(3bf|146d))

)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)

It is encoded in the positive GrassmannndashPlucker treein Figure 1 and multiplying it out as in Figure 2proves the non-realizability of Z

One can check that arranging the GP polynomialsinto a tree ie a graph without cycles guaranteesthat the final certificate does not depend on the orderin which the certificate is multiplied out

3 Finding positive GrassmannndashPlucker trees

How do we go about finding such certificates Firstwe restrict to a useful subclass of GP relations thatpermit algebraic elimination

Definition 3 A three-term GP relation as in (1) hasno adjacent unknown solids if no two determinantsthat are multiplied together have unknown sign

18


minusΓ(18f|46be)

Γ(18f|56bd)

Γ(1bf|34de)Γ(1bf|48de)

minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]

Figure 1 GrassmannndashPlucker tree proving the non-realizability of Zhengrsquos 3-sphere

0 = [36fb5]

([36fb4]

([14bf3]

([16f85]

([14bfd](minus[16f84][18bef] + [14bf8][16f8e] + [14e8f][16f8b])

+ [16f84]([14bf8][1bdef] + [14bfd][18bef] minus [14bfe][18bdf]))

+ [16f84] [14bfe]([16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]

))+ [16f84] [14bf8] [16f85]

(minus [14bf3][1bdef] + [13bdf][14bfe]minus [13bef][14bfd]

))+ [16f84] [14bf8] [14bfe] [16f85]

([14bf3][36fbd] + [36fb1][34dbf]minus [13bdf][36fb4]

))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]

([15fb3][36fbe] + [36fb1][35bef] + [13bef][36fb5]

)= [14bf3] [16f84] [14bf8] [14bfe] [16f85] [36fb5] [36fbd] + [14bf3] [16f84] [14bfe] [15fb8] [16f8d] [36fb4] [36fb5]

+ [14bf3] [16f84] [14bfe] [158df] [16f8b] [36fb4] [36fb5] + [14bf3] [14bf8] [14bfd] [16f85] [16f8e] [36fb4] [36fb5]

+ [14bf3] [14e8f] [14bfd] [16f85] [16f8b] [36fb4] [36fb5] + [15fb3] [16f84] [14bf8] [14bfd] [16f85] [36fb4] [36fbe]

+ [36fb1] [16f84] [14bf8] [14bfd] [16f85] [36fb4] [35bef] + [36fb1] [16f84] [14bf8] [14bfe] [16f85] [34dbf] [36fb5]

Figure 2 The certificate derived from Figure 1 The sign of each determinant with a ldquordquo can be different indifferent realizations but the certificate is arranged in such a way that they all cancel After multiplying out allsurviving determinants are known to be positive in any realization but the whole certificate must sum to zeroSince this canrsquot happen the realization cannot exist

Next we set up the GP graph in which we willsearch for our certificate Its nodes are

VΣ =

plusmnΓ(S|ijkl) Γ(S|ijk`) has no two adjacent

unknown solids

and the edges are labelled with the set S(Σ) ofnormal forms [Pfe20 Definition 32] of solids of ΣThere can be multiple edges between two nodes butthey have different labels two nodes ΓΓprime isin VΣ arejoined with an edge (π ΓΓprime) labelled π isin S(Σ)in EΣ sube S(Σ)times

(VΣ

2

)iff they share a solid π such that

σi = minusσprimei where σi σprimei are the canonical signs [Pfe20

Definition 44] of the terms containing π in ΓΓprimeSaid differently each GP polynomial is connected

to many other GP polynomials and each connect-ing edge is labelled with an unknown solid commonto both polynomials At each node there can be atmost 3 different labels but potentially thousands ofincident edges with those labels

Our goal is to find a tree in this graph in such a waythat each node has exactly one incident edge labelledwith each label occurring at that node The task istherefore to distill the lucky nodes out of this graph

and for each lucky node the up to three lucky edgesout of the thousands of candidates oh and wersquod likethe resulting tree to be as small as possible

For this we solve the integer program on the in-teger indicator variables xΓ Γ isin VΣ and xe e = (π ΓΓprime) isin EΣ defined in Figure 3 The in-equalities for these variables xΓ xe isin 0 1 have thefollowing interpretation

(2) ensures that both endpoints of an edgepresent in the solution are present

(3) ensures that at most one edge is selected be-tween two selected nodes

(4) forces the solution to be a tree with at leastone node

(5) ensures that if a node Γ with an unknownsign π is present in the solution then there isexactly one edge of that label incident to Γ

4 More results

We have implemented a search for positive GP trees inthe software framework polymake [GJ00] With this

19


minsum

ΓisinVΣ

xΓ st 2sum

π e=(πΓΓprime)isinEΣ

xe le xΓ + xΓprime for each ΓΓprime isin(VΣ

2

)(2)

sumπ e=(πΓΓprime)isinEΣ

xe le 1 for each ΓΓprime isin(VΣ

2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)

sumΓprime e=(πΓΓprime)isinEΣ

xe = xΓ for all Γ isin VΣ for all unknown π isin Γ (5)

Figure 3 The integer program for finding positive GrassmannndashPlucker trees

implementation we can prove the non-realizability ofseveral previously inaccessible families

41 Topological Prismatoids

In [CS19] Francisco Criado and Francisco Santosintroduced topological prismatoids a combinatorialabstraction of the geometric prismatoids used bySantos [San12] to construct counterexamples to theHirsch conjecture Criado and Santos constructfour combinatorially distinct non-d-step topological4-dimensional prismatoids on 14 vertices referred toas 1039 1963 2669 and 3513 which implythe existence of 8-dimensional spheres on 18 ver-tices whose combinatorial diameter exceeds the Hirschbound In [CS19] the question of polytopality of thesecombinatorial prismatoids was left open

Theorem 4 The topological prismatoids 10391963 2669 and 3513 are not polytopal

42 Novik and Zhengrsquos centrally symmetric neigh-borly d-spheres

In [NZ19] Novik and Zheng give several constructionsof centrally symmetric highly neighborly d-spheresThey are based on a family ∆d

n of cs-dd2e-neighborlycombinatorial d-spheres on 2n ge 2d + 2 verticeswhich arise as the case i = dd2e of an inductively con-structed family ∆di

n of cs-i-neighborly combinatoriald-spheres Each of those contains a certain combina-torial d-ball Bdiminus1

n which is the only part that getsdeleted in a step of the inductive construction Ford = 3 Novik and Zhengrsquos family ∆3

n n ge 4 isprecisely Jockuschrsquos family from [Joc95]

Theorem 5 For n ge 6 no member ∆4n of Novik and

Zhengrsquos family is realizable

Theorem 6 (with [Zhe20b]) For n minus 2 ge d ge 3 nomember ∆d

n of Novik and Zhengrsquos family is realizable

References

[CS19] Francisco Criado and Francisco Santos Topo-logical Prismatoids and Small Simplicial Spheresof Large Diameter Experimental Mathemat-ics 2019 httpsdoiorg10108010586458

20191641766

[GJ00] Ewgenij Gawrilow and Michael Joswigpolymake a framework for analyzing con-vex polytopes In Polytopesmdashcombinatorics andcomputation (Oberwolfach 1997) volume 29of DMV Sem pages 43ndash73 Birkhauser Basel2000

[JM18] Martina Juhnke-Kubitzke and Satoshi MuraiBalanced generalized lower bound inequality forsimplicial polytopes Sel Math New Ser24(2)1677ndash1689 2018

[JMNS18] Martina Juhnke-Kubitzke Satoshi Murai Is-abella Novik and Connor Sawaske A gener-alized lower bound theorem for balanced mani-folds Math Z 289(3-4)921ndash942 2018

[Joc95] William Jockusch An infinite family of nearlyneighborly centrally symmetric 3-spheres JComb Theory Ser A 72(2)318ndash321 1995

[NZ19] Isabella Novik and Hailun Zheng Highly neigh-borly centrally symmetric spheres 2019

[Pfe20] Julian Pfeifle Positive Plucker tree certificatesfor non-realizability httpsarxivorgabs

201211500 2020

[PPS12] Julian Pfeifle Vincent Pilaud and FranciscoSantos Polytopality and Cartesian products ofgraphs Isr J Math 192121ndash141 2012

[San12] Francisco Santos A counterexample to theHirsch conjecture Ann Math (2) 176(1)383ndash412 2012

[Ven19] Lorenzo Venturello Balanced triangulations onfew vertices and an implementation of cross-flips Electron J Comb 26(3)research paperp361 25 2019

[Zhe20a] Hailun Zheng Ear decomposition and balancedneighborly simplicial manifolds Electron JComb 27(1)research paper p110 17 2020

[Zhe20b] Hailun Zheng personal communication 2020

20


Computing the type cone of nestohedra

Arnau Padrollowast1 Vincent Pilauddagger2 and Germain PoullotDagger1

1IMJ-PRG Sorbonne Universite and Universite de Paris2CNRS amp LIX Ecole Polytechnique

Nestohedra are an important family of polytopesintroduced independently by Feichtner and Sturm-fels [1] and Postnikov [2] They can be seen asthe hypergraph generalization of Carr and Deva-dossrsquo graph associahedra [3] Their face lattice en-codes the combinatorial structure of the connectedinduced sub(hyper)graphs of a given (hyper)graphand was first introduced by De Concini and Procesiin their study of wonderful compactifications of sub-space arrangements [4] As particular examples theycontain several well known polytopes such as per-mutohedra associahedra cyclohedra stellohedra andPitman-Stanley polytopes

Postnikovrsquos nestohedra are generalized permutohe-dra which means that they can be obtained fromthe classical permutahedron by gliding its facets alongtheir normal vectors It is natural to ask for the wholeset of deformations of the permutahedron that realizea given nestohedron This is equivalent to studyingits type cone a notion introduced by McMullen [5]Given a polytope P with facet normal vectors givenin a matrix G the type cone TC(P ) is the set of vec-tors h such that P and x isin Rn Gx le h havethe same normal fan In other words it records howto move the facets of P along their normal vectorswithout changing the combinatorics of P

Our work is motivated by a recent realization ofthe associahedron by Arkani-Hamed et al in [6] thatarose in the study of scattering particles in mathemat-ical physics This realization was extended to othergeneralizations of the associahedron by Bazier-Matteet al in [7] and by Padrol et al in [8] and it can bereinterpreted in terms of the type cone In particularit is shown in [8] that a key ingredient of the originalconstruction resides in the fact that the type cone ofthe associahedron is simplicial

The type cone is classically described as a poly-hedral cone by a family of inequalities called wall-crossing inequalities but the vast majority of theseinequalities are redundant Our result gives non-redundant facet descriptions of the type cones of allnestohedra In our description each induced subhy-

lowastEmail aranaupadrolimj-prgfrdaggerEmail vincentpilaudlixpolytechniquefrDaggerEmail germainpoullotpolytechniqueedu

pergraph gives rise to few natural inequalities Es-pecially when the type cone is simplicial we can ex-plicitly associate each facet of the type cone to an in-duced subhypergraph Our description has several ap-plications we can determine the hypergraphs whosenestohedron has a simplicial type cone easily com-pute the number of facets of the type cone even inthe non-simplicial case and provide new Minkowskisum decompositions of nestohedra

References

[1] E M Feichtner and B Sturmfels Matroid polytopesnested sets and Bergman fans Portugaliae Mathe-matica Nova Serie Volume 62 Number 4 (2005)437ndash468

[2] A Postnikov Permutohedra Associahedra and Be-yond International Mathematics Research NoticesVolume 2009 Issue 6 (2009) 1026ndash1106

[3] M Carr and S Devadoss Coxeter complexes andgraph-associahedra Topology and its ApplicationsVolume 153 number 12 (2006) 2155ndash2168

[4] C De Concini and C Procesi Wonderful modelsof subspace arrangements Selecta Mathematica NewSeries Volume 1 number 3 (1995) 459ndash494

[5] P McMullen Representations of polytopes and poly-hedral sets Geometriae Dedicata Volume 2 (1973)83ndash99

[6] N Arkani-Hamed Y Bai S He and G Yan Scat-tering forms and the positive geometry of kinemat-ics color and the worldsheet Journal of High EnergyPhysics Number 5 (2018) 096

[7] V Bazier-Matte G Douville K Mousavand HThomas and E Yıldırım ABHY Associahedra andNewton polytopes of F -polynomials for finite typecluster algebras preprint 2018 arXiv180809986

[8] A Padrol Y Palu V Pilaud and P-G Plamon-don Associahedra for finite type cluster algebras andminimal relations between g-vectors preprint 2019arXiv190606861

21


The Voronoi diagram of rotating rays with applications to floodlightilluminationlowast

Carlos Alegrıa1 Ioannis Mantas2 Evanthia Papadopoulou2 Marko Savic3 Hendrik Schrezenmaier4 CarlosSeara5 and Martin Suderland2

1Dipartimento di Ingegneria Universita Roma Tre Italy2Faculty of Informatics Universita della Svizzera italiana Switzerland

3Department of Mathematics and Informatics Faculty of Sciences University of Novi Sad Serbia4Institut fur Mathematik Technische Universitat Berlin Germany


1 Introduction

We introduce the Rotating Rays Voronoi Diagram aVoronoi structure where the input is a set S of n raysand the distance from a ray r isin S to a point x isin R2

is the counterclockwise angular distance see Fig 1We denote the diagram by RVD(S)

Figure 1 Left Point x has distance α from r Thewedge is an α-floodlight Right The RVD of 4 rays

The diagram can be used to solve illuminationproblems where a domain has to be covered by α-floodlights ie wedges of aperture α see eg [2]More specifically given a set of n α-floodlights alignedto the rays of S we show how to compute the mini-mum angle αlowast required to illuminate a given domain

2 RVD in the plane

When the considered domain is the whole planewe show that RVD(S) has Ω(n2) worst case com-plexity even if the set of input rays are pairwisenon-intersecting Further a single Voronoi region ofRVD(S) can have Θ(n2) worst case complexity

Using the envelopes of distance functions in 3-spacewe derive that RVD(S) has O(n2+ε) complexity andit can be constructed in O(n2+ε) time forallε gt 0 The an-

lowastAn extended abstract was presented in the 37th EuropeanWorkshop on Computational Geometry (EuroCG 2021)

1Email carlosalegriauniroma3it2Email ioannismantas evanthiapapadopoulou mar-

tinsuderlandusich3Email markosavicdmiunsacrs4Email schrezenmathtu-berlinde5Email carlossearaupcedu

gle αlowast is realized at a vertex of RVD(S) Hence afterconstructing RVD(S) a simple traversal reveals αlowast

3 RVD of a convex polygon - Brocard illumination

If the target domain is an n-sided simple polygon withan α-floodlight aligned to each edge then the prob-lem of finding the angle αlowast is known as the Brocardillumination problem see Fig 2 For a convex poly-gon we improve upon previous results [1] computingαlowast in optimal Θ(n) time by means of the RVD

Given a convex polygon P we obtain a set SP ofedge-aligned rays The diagram RVD(SP) confinedto P is a tree of Θ(n) complexity see Fig 2 Wedescribe an algorithm to construct RVD(SP) cap P indeterministic Θ(n) time We first partition SP into4 subsets then construct the RVD of each subset inO(n) time and we finally merge the 4 diagrams Theangle αlowast is again realized at a vertex of RVD(SP)capPHence we can find αlowast in Θ(n) time

Figure 2 Left Illumination with α-floodlightsaligned to the edges of a polygon P Right RVD(SP)confined to P Highlighted are the rays realizing αlowast

References

[1] C Alegrıa D Orden C Seara and J Urrutia Illu-minating polygons by edge-aligned floodlights of uni-form angle (Brocard illumination) European Work-shop on Computational Geometry 2017

[2] J Urrutia Art gallery and illumination problemsHandbook of Computational Geometry 2000

The algorithm described in [1] is in fact incorrect but iteasily implies a O(n logn) time and O(n) space algorithm

22


The edge labeling of higher order Voronoi diagrams

Merce Claverollowast1 Andrea de las Heras ParrillaDagger1 Clemens Huemersect1 and Alejandra Martınez-Moraianpara2

1Departament de Matematiques Universitat Politecnica de Catalunya2Departamento de Fısica y Matematicas Universidad de Alcala

Abstract

We present an edge labeling of order-k Voronoi dia-grams Vk(S) of point sets S in the plane and studyproperties of the regions defined by them Amongthem we show that Vk(S) has a small orientable cycleand path double cover and we identify configurationsthat cannot appear in V3(S)

1 Introduction

Let S be a set of n points in general position in theplane (no three collinear and no four cocircular) andlet 1 le k le nminus 1 be an integer The order-k Voronoidiagram of S Vk(S) is a subdivision of the plane intocells also called faces such that the points in the samecell have the same k nearest points of S also calledk nearest neighbors Voronoi diagrams have appli-cations in a broad range of disciplines see eg [1]The most studied Voronoi diagrams of point sets Sare V1(S) the classic Voronoi diagram and Vnminus1(S)the furthest point Voronoi diagram which only hasunbounded faces Many properties of Vk(S) were ob-tained by Lee [8] we also mention [4 9 10 13] amongthe sources on the structure of Vk(S)

An edge that delimits a cell of Vk(S) is a (possiblyunbounded) segment of the perpendicular bisector oftwo points of S This well-known observation inducesa natural labeling of the edges of Vk(S) with the fol-lowing rule

bull Edge rule An edge of Vk(S) in the perpendic-ular bisector of points i j isin S has labels i and j Weput the label i on the side (half-plane) of the edgethat contains point i and label j on the other side

lowastEmail merceclaverolupcedu Research supported byPID2019-104129GB-I00 AEI 1013039501100011033 and byGen Cat DGR 2017SGR1640DaggerEmail andreadelasherasestudiantatupcedusectEmail clemenshuemerupcedu Research supported by

PID2019-104129GB-I00 AEI 1013039501100011033 and byGen Cat DGR 2017SGR1336paraEmail alejandramartinezmuahes Research supported

by PID2019-104129GB-I00 AEI 1013039501100011033

This work has received funding from the Euro-pean Unionrsquos Horizon 2020 research and innova-tion programme under the Marie Sk lodowska-Curie grant agreement No 734922

See Figure 1 Based on the edge rule we prove avertex rule and a face rule Vk(S) contains two typesof vertices denoted as type I and type II (also callednew and old vertices [8]) that are defined belowbull Vertex rule Let v be a vertex of Vk(S) and leti j ` isin S be the set of labels of the edges incident tov The cyclic order of the labels of the edges aroundv is i i j j ` ` if v is of type I and it is i j ` i j ` ifv is of type IIbull Face rule In each face of Vk(S) the edges that

have the same label i are consecutive and these labelsi are either all in the interior of the face or are all inthe exterior of the face

Note that when walking along the boundary of aface in its interior (exterior) a change in the labelsof its edges appears whenever we reach a vertex oftype II (type I) or possibly at consecutive unboundededges of an unbounded face see Figure 1

Edges with same label i enclose a region Rk(i) con-sisting of all the points of the plane that have pointi isin S as one of their k nearest neighbors from SThe union of all these regions Rk(i) is a k-fold cover-ing of the plane Rk(i) is related to the k-th nearestpoint Voronoi diagram (k minus NP VD) of S that as-signs to each point of the plane its k-nearest neighborfrom S [10] This diagram is also called k-th degreeVoronoi diagram in [4] The region of a point i ink minus NP V D is Rk(i)Rkminus1(i) In [5] it was provedthat Rk(i) is star-shaped We further observe thatR1(i) is contained in its kernel and we identify thereflex (convex) vertices on its boundary Bk(i) as ver-tices of type II (type I)

We also show that every Vk(S) admits an orientabledouble cover [7] of its edges using precisely the cyclesand paths

⋃iisinS Bk(i) A cycle and path double cover

of a graph G is a collection of cycles and paths Csuch that every edge of G belongs to precisely twoelements of C Paths are needed in the double coverC of Vk(S) due to the unbounded edges A doublecover C is orientable if an orientation can be assignedto each element of C such that for every edge e of Gthe two cycles resp paths that cover e are orientedin opposite directions through e [7]

The small cycle double cover conjecture states thatevery simple bridgeless graph on n vertices has a cy-cle double cover with at most n minus 1 cycles [2] Seyf-

23


farth [12] proved this conjecture for 4-connected pla-nar graphs and also proved that any simple bridgelessplanar graph of size n has a cycle double cover with atmost 3b(nminus 1)2c cycles [11] We show in Property 7that a higher order Voronoi diagram admits a muchsmaller cycle and path double cover compared to itsnumber of vertices

We present several more new properties of Vk(S)All of them rely on the edge labeling and on elemen-tary geometric arguments in the plane This tech-nique also allows us to obtain new proofs of known re-sults about Vk(S) Other techniques used previouslyalso apply projections of points to Rd and hyperplanearrangements among others

We finally focus on the edge labeling of V3(S) andshow that certain configurations cannot appear inV3(S) We omit all proofs in this abstract

We define some notation The points of S are1 n A set of k neighbors defining a cell of Vk(S)is denoted by Pk sub S and the corresponding face isdenoted by f(Pk) If the face is bounded we denoteit by f b(Pk) Note that not every possible Pk deter-mines a face in Vk(S) The (perpendicular) bisectorof two points i j isin S is denoted bij And an edge ofVk(S) on bij is denoted bij A vertex of Vk(S) is theintersection of three bisectors bab bbc and bca Equiv-alently it is the center of the circle through a b c isin SThere are two types of vertices in f(Pk) and hence inVk(S) If a isin Pk and b c isin S Pk we say that it is oftype I and if a b isin Pk and c isin S Pk we say that itis of type II It is well known that a vertex of type I inVk(S) is also a vertex of Vk+1(S) and a vertex of typeII in Vk(S) is also a vertex of Vkminus1(S) [8] Furtherthe three edges of Vk(S) around the vertex alternatein cyclic order with the three edges of Vk+1(S) (respVkminus1(S))

The edges and vertices of Vk(S) form (a drawingof) a graph which also contains unbounded edgesWhen considering the union of Vk(S) and Vk+1(S)(resp Vkminus1) the graph induced by Vk+1(S) (Vkminus1) ina cell f(Pk) of Vk(S) is the subgraph of Vk+1 (Vkminus1)whose vertices and edges are contained in f(Pk)

2 Properties of the labeling of Vk(S)

Property 1 For k gt 1 every bounded face f b(Pk) ofVk(S) contains at least two and at most nminusk verticesof type I and at least two and at most k vertices oftype II

The lower bounds in Property 1 were already givenin [8] In particular they imply that for k gt 1 Vk(S)does not contain triangles as also proved in [9] with adifferent method As for the upper bounds there existpoint sets S such that some face of Vk(S) has exactlynminus k vertices of type I and k vertices of type II

1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37

Figure 1 The edge labeling of V3(S) for a set S ofeight points in convex position Vertices of type I aredrawn in blue and vertices of type II in red

Vk(S)

Vk+1(S)

Vkminus1(S)

Figure 2 The graphs induced by Vkminus1(S) andVk+1(S) in a face f b(Pk) are trees Each of themcreates a subdivision of f b(Pk) into regions

The next property concerns the graphs induced byVk+1(S) and Vkminus1(S) inside f b(Pk) It is known thatthese graphs are trees [4 8 13] see Figure 2

Property 2 Let f b(Pk) be a bounded cell of Vk(S)for k gt 1 Then on its boundary not all vertices ofthe same type are consecutive Equivalently insidef b(Pk) there is an edge of Vk+1(S) that crosses anedge of Vkminus1(S)

The graphs induced by Vk+1(S) and Vkminus1(S) insidea bounded face f b(Pk) of Vk(S) determine a subdi-vision of f b(Pk) into regions see Figure 2 All theregions induced by Vk+1(S) in f b(Pk) have the samek nearest neighbors from S and a different (k + 1)-nearest neighbor Similarly all the regions inducedby Vkminus1(S) in f b(Pk) have the same k nearest neigh-bors and differ in the k-nearest neighbor Property 3describes the labeling of the edges in the boundaryof such regions from which the vertex rule in the in-troduction can be deduced There are two types ofboundary edges in the regions edges of f b(Pk) andedges of Vk+1(S) (resp Vkminus1(S))

Property 3 Let f b(Pk) be a bounded cell of Vk(S)k gt 1 and let R be a region in the subdivision off b(Pk) induced by Vk+1(S) (resp Vkminus1(S)) Theedges of Vk+1(S) (Vkminus1(S)) in the boundary of R have

24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)

Figure 3 V3(S) for S the point set in Figure 1 ineach face its three nearest neighbors are indicatedIn green the region R3(2) formed by all the cells ofV3(S) that have point 2 as one of their three nearestneighbors All the cells in R3(2) contain the label2 The boundary B3(2) of R3(2) is formed by all theedges that have the label 2 and this label is alwaysinside R3(2) Vertices of B3(2) with an incident edgelying in the interior of R3(2) are of type II and theother vertices of B3(2) are of type I

the unique (k + 1)-nearest neighbor (resp k-nearestneighbor) of the points of R as label inside (outside)R The edges of f b(Pk) in the boundary of R havethis label outside (inside) R

Figure 4 illustrates Property 3 and the vertex ruleWe now describe properties of the region Rk(i)

formed by all the faces of Vk(S) having point i isin Sas one of their defining points and of its boundaryBk(i) First we observe that for every k gt 1 and forevery point i isin S Rkminus1(i) sub Rk(i)

Property 4 Rk(i) is a connected region Further-more for any bounded face f b(Pk) contained in Rk(i)Rk(i) f b(Pk) is a connected region

Property 5 The boundary Bk(i) of a region Rk(i) isformed by all the edges of Vk(S) that have the label iand this label is always inside Rk(i) Bk(i) is either acycle or one or more paths whose first and last edgeare unbounded edges of Vk(S)

Property 5 is illustrated in Figure 3

Property 6 The vertices of Bk(i) that are incidentto an edge of Vk(S) lying in the interior of Rk(i) areof type II and the remaining vertices of Bk(i) are oftype I Moreover for k gt 1 if Bk(i) is a cycle thenit encloses at least three faces of Vk(S) If Bk(i) hasr reflex vertices then it encloses at least r faces ofVk(S)

Let C be the collection of paths and cycles in⋃iisinS Bk(i) of Vk(S) We prove that C forms a double

cover of Vk(S) Let finfink be the number of unboundedfaces (or unbounded edges) in Vk(S) also equal to thenumber of (kminus 1)-edges of S [4 13] It is known thatfinfink is at least 2k + 1 [6] and at most O(n 3

radick) [3] We

use precisely finfink paths for C

Property 7 Vk(S) which has finfink unbounded faceshas an orientable cycle and path double cover consist-ing of finfink paths and of at most maxnminus 2kminus 1 2kminusnminus 1 cycles

This double cover of a higher order Voronoi diagramis small compared to its number of vertices from [48] we deduce that Vk(S) has at least (2kminus 1)nminus 2k2

vertices for k lt (n minus 2)2 When S is in convexposition then Vk(S) has this number of vertices

The bound on the number of cycles in Property 7is attained for the point sets S from [6] which have2k + 1 (k minus 1)-edges Property 7 also shows that fork = bn2c and k = dn2e Vk(S) has an orientablepath double cover with finfink paths This also holds forpoint sets in convex position and any value of k

Property 8 Let S be a set of n points in convexposition Then Vk(S) has an orientable path doublecover consisting of n paths

Property 9 For every i isin S the region Rk(i) ofVk(S) is star-shaped Furthermore the face R1(i) ofV1(S) is contained in its kernel

Property 10 Let v be a vertex of Bk(i) If v is oftype I then it is a convex vertex of Bk(i) and if v isof type II it is a reflex vertex of Bk(i)

3 Not too many alternating hexagons in V3(S)

A hexagon in Vk(S) is called alternating if its verticesalternate between type I and type II see Figure 5 ByProperty 1 V2(S) contains no alternating hexagonsWe consider then V3(S) Figure 5 shows that it is notpossible to label the edges of the hexagons of V3(S)while complying with the properties of the labeling

Property 11 Let v be a vertex of type I in V3(S)Then at most two of the three incident faces to v arealternating hexagons

Property 12 Any alternating hexagon in V3(S) isadjacent to at most three other alternating hexagons

25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)

Figure 4 (a) The graph induced by Vk+1(S) in f b(Pk) divides f b(Pk) into regions Rmi delimited by bisectorsbetween the (k + 1)-nearest neighbor mi isin S of the points of the region and another point of S The cyclicorder of the labels of the edges around a vertex v incident with Rms

Rmrand Rmt

is mr ms ms mt mt mr(b) The graph induced by Vkminus1(S) in f b(Pk) divides f b(Pk) into regions Rij delimited by bisectors between thek-nearest neighbor ij isin S of the points of Rij and another point of S The cyclic order of the labels of the edgesaround v is ir is it ir is it

`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w

Figure 5 Three alternating hexagons incident to ver-tex v of type I The labels around w shown in red donot comply with the properties of the labeling

References

[1] F Aurenhammer Voronoi diagrams - a survey of afundamental geometric data structure ACM Com-puting Surveys 23(3) (1991) 345ndash405

[2] J A Bondy Small cycle double covers of graphsCycles and Rays NATO ASI Ser C Kluwer 199021ndash40

[3] T Dey Improved bounds for planar k-sets and re-lated problems Disc Comput Geom 19 (1989)373ndash382

[4] H Edelsbrunner Algorithms in Combinatorial Ge-ometry Springer 1987

[5] H Edelsbrunner M Iglesias-Ham Multiple coverswith balls I Inclusion-exclusion Computational Ge-ometry Theory and Applications 68 (2018) 119ndash133

[6] P Erdos L Lovasz A Simmons EG Straus Dis-section graphs of planar point sets A survey of com-binatorial theory North Holland 1973 139ndash149

[7] F Jaeger A survey on the cycle double cover con-jecture Ann Disc Math 27 North-Holland 19851ndash12

[8] DT Lee On k-nearest neighbor Voronoi diagrams inthe plane IEEE Trans Comput 31 (1982) 478ndash487

[9] JE Martınez-Legaz V Roshchina M Todorov Onthe structure of higher order Voronoi cells J OptimTheory Appl 183 (2019) 24ndash49

[10] A Okabe B Boots K Sugihara SN Chiu SpatialTessellations Concepts and Applications of Voronoidiagrams second edition Wiley 2000

[11] K Seyffarth Hajos conjecture and small cycle doublecovers of planar graphs Disc Math 101 (1992) 291ndash306

[12] K Seyffarth Small cycle double covers of 4-connectedplanar graphs Combinatorica 13 (1993) 477ndash482

[13] D Schmitt JC Spehner Order-k Voronoi diagramsk-sections and k-sets Proc of Japanese Conferenceon Discrete and Computational Geometry LectureNotes in Computer Science 1763 (1998) 290ndash304

26


On Guillotine Cuts of Boundary Rectangles

Pablo Perez-Lanterolowast1 and Carlos Searadagger2

1Universitat Politecnica de Catalunya Spain2Universidad de Santiago de Chile (USACH) Chile

Abstract

Let R be a set of n pairwise disjoint axis-aligned rect-angles each having at least one side contained in theboundary of a given rectangular domain In this notewe prove that any set R as above has a subset of atleast 3

4n minus O(1) rectangles that are separable by astraight guillotine partition of the plane and that thisbound is tight

1 Introduction

A guillotine partition of the plane is a subdivision ofthe plane into (possibly unbounded) cells using thefollowing recursive strategy We start with a sin-gle subdivision consisting of the whole plane as theunique cell At each step we take a cell C of thecurrent subdivision and a straight line ` through theinterior of C and subdivide C into the two new sub-cells C1 = Ccap`+ and C2 = Ccap`minus losing cell C where`+ and `minus are the two closed half-planes bounded by` respectively A set Rlowast of k pairwise disjoint rectan-gles is guillotine-separable if there exists a guillotinepartition P of k cells that separates Rlowast That is eachcell of P contains one rectangle of Rlowast (see Figure 1aand Figure 1b) In this paper all rectangles are con-sidered axis-aligned and close

Pach and Tardos [8] wrote that ldquoit seems plausi-blerdquo that there exists a constant c such that any setR of n pairwise disjoint rectangles in the plane has aguillotine-separable subset of at least c middot n rectanglesThey proved this statement for c = Ω(1 log n) nota constant and considered in the proof axis-alignedguillotine partitions We say that a guillotine par-tition is axis-aligned if it is built with axis-parallellines which subdivide the plane into (possibly un-bounded) rectangular cells Abed et al [1] provedthat the statement is true for axis-aligned guillotine

lowastEmail pabloperezlusachcl Research supported by DI-CYT 041933PL Vicerrectorıa de Investigacion Desarrollo e In-novacion USACH (Chile) and Programa Regional STICAM-SUD 19-STIC-02

daggerEmail carlossearaupcedu Research supported byprojects MTM2015-63791-R MINECO FEDER and Gen CatDGR 2017SGR1640


(a) (b)

Figure 1 (a) 6 guillotine-separable rectangles (b) 4rectangles that are not guillotine-separable

partitions when R is a set of squares (with c = 181recently improved to 140 [5]) More importantlythey noted that the existence of such a constant cfor any set of pairwise disjoint rectangles (when re-stricted to axis-aligned guillotine partitions) implies aO(n5)-time c-approximation algorithm for the Maxi-mum Independent Set of Rectangles (MISR) problemThe MIRS problem is a fundamental NP-hard combi-natorial optimization problem defined as Given a setof n rectangles finding a subset of them of maximumcardinality such that the rectangles in the subset arepairwise disjoint

Recently Mitchell [7] proved that any set R of npairwise disjoint rectangles in a rectangular domainhas a subset Rprime of size at least n10 for which thereexists a hierarchical rectilinear cut of the domain suchthat any leaf cell contains exactly one rectangle of Rprimeand every segment of every cut penetrates at mosttwo rectangles of Rprime Every cut is a polyline madeof O(1) axis-aligned segments so that every cell is anorthogonal polygon obtained by removing a set of upto four disjoint subrectangles from a given rectangleR with each subrectangle containing exactly one ofthe four corners of R Base on this Mitchell [7] gavethe first polynomial-time constant approximation al-gorithm to the MIRS problem with O(n34) runningtime and 110 approximation factor

We consider that all guillotine partitions are axis-aligned and study the next open question

Question 1 Do there exist constants c isin (0 1] andn0 gt 0 such that any set of n ge n0 pairwise disjointrectangles in the plane has a guillotine-separable sub-set of at least c middot n rectangles

27


We have several motivations to study Question 1First if it is true then we will directly obtain a O(n5)-time c-approximation algorithm for the MISR prob-lem Second we consider this question interestingby itself because it explores the combinatorial dis-tribution of finite sets of pairwise disjoint rectanglesFinally as the topic of this paper we can impose re-strictions to the set of rectangles such as The rectan-gles are contained in a rectangular region with a sidein the boundary of the region [2 6] or the rectanglesintersect the same diagonal line [3 4]

Ahmadinejad and Zarrabi-Zadeh [2] and Kong etal [6] studied the MISR problem when given a rect-angular domain D each of the n rectangles of theinput has at least one side contained in the boundaryof D solving the problem exactly in O(n4) time Weconsider Question 1 in this setting That is we have npairwise disjoint rectangles each having at least oneside contained in the boundary of the domain D Wecall such a set a boundary set of rectangles (see Fig-ure 2a) First we observe that Question 1 is true forc = 14 Namely there always exists one of the foursides of D such that at least n4 of the rectangleshave a side contained in that side of D This sub-set of k ge n4 rectangles is guillotine-separable bythe guillotine partition induced by kminus 1 parallel linesseparating the k rectangles Unless otherwise speci-fied all lines in this paper are considered axis-parallelSecond we prove in Section 2 that there always ex-ists a subset of at least 3

4nminusO(1) rectangles that areguillotine-separable and that this bound is tight

2 Boundary rectangles

Let D be a rectangular domain and letR be a bound-ary set of n rectangles in D Let Rt (resp Rb R`and Rr) be the subset of R with the rectangles whosetop (resp bottom left and right) side is contained inthe top (resp bottom left and right) side of D Notethat Rt Rb R` and Rr are all guillotine-separableby taking the guillotine partition induced by parallellines separating the rectangles For every rectangle Rlet top(R) bottom(R) left(R) and right(R) denotethe lines through the top bottom left and right sidesof R respectively

We say that R is a corner set of rectangles if eachrectangle of R has a side contained in the union oftwo fixed adjacent sides of D We say that R is aparallel set of rectangles if each rectangle of R has aside contained in the union of two parallel sides of D

Lemma 2 If R is a corner set of rectangles then Ris guillotine-separable

Proof Assume wlog that R = R` cup Rt We pro-ceed by induction If |R`| = 0 or |Rt| = 0 then allrectangles of R are trivially guillotine-separable So

D

(a)

A

(b)

A

B

(c)

Figure 2 (a) 8 guillotine-separable boundary rectangles(b-c) Proof of Lemma 2

consider that |R`| ge 1 and |Rt| ge 1 Let A be thebottommost rectangle of R` If top(A) does not cut(the interior of) any rectangle of Rt (see Figure 2b)then by the induction hypothesis we can assume thatRA is guillotine-separable by some guillotine par-tition G Since top(A) separates A from all rectanglesin RA by combining top(A) with G we can createa guillotine partition that separatesR Otherwise as-sume that top(A) does cut some rectangle of Rt andlet B be the leftmost rectangle of Rt cut by top(A)(see Figure 2c) Let Rprime sube Rt be the set of the rect-angles (including B) to the right of left(B) By theinduction hypothesis we have that R (Rprime cup A) isguillotine-separable by some guillotine partition GprimeSince left(B) separates all rectangles in Rprime from allin R Rprime and also top(A) separates A from all rect-angles in R (Rprime cup A) by combining left(B) thepart of top(A) to the left of left(B) and Gprime we cancreate a guillotine partition that separates R

Lemma 3 If R is a parallel set of rectangles thenR has a guillotine-separable subset of cardinality atleast 3

4 |R| Furthermore this lower bound is tight

Proof Assume wlog that R = Rt cup Rb We pro-ceed by induction based on the idea of sweepingthe elements of R from left to right If |Rt| = 0or |Rb| = 0 then all rectangles of R are triviallyguillotine-separable So consider that |Rt| ge 1 and|Rb| ge 1 Let A isin R be the rectangle such thatright(A) is leftmost among all rectangles of R andassume wlog that A isin Rb Let n = |R| Theproof follows into the following seven disjoint casesenumerated from (a) to (g) (see Figure 3)

(a) right(A) does not cut any rectangle in Rt (seeFigure 3a) All rectangles in R A are to the rightof right(A) Then by the inductive hypothesis wehave that RA has a guillotine-separable subset Rprimesuch that |Rprime| ge 3

4 |R A| = 34 (n minus 1) Combining

right(A) with the guillotine partition that separates

28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)

Figure 3 Proof of Lemma 3

Rprime we have that AcupRprime is guillotine-separable and|A cup Rprime| = 1 + |Rprime| ge 1 + 3

4 (nminus 1) ge 34n

(b) right(A) cuts a rectangle B isin Rt and right(B)does not cut any rectangle in Rb (see Figure 3b) LetRprime sube R denote the rectangles to the left of right(B)which includes A and B Let t = |Rprime| ge 2 Observethat B must be the leftmost rectangle in Rt and alsothe unique rectangle ofRt that is inRprime Furthermorebottom(B) separates B from all rectangles inRprimeBHence the t rectangles of Rprime are guillotine-separableBy the inductive hypothesis the n minus t rectangles ofR Rprime have a guillotine-separable subset R2 of atleast 3

4 (n minus t) rectangles Combining right(B) withthe guillotine partitions of Rprime and R2 respectivelywe obtain that Rprime cup R2 is guillotine-separable andsatisfies |Rprime cupR2| = |Rprime|+ |R2| ge t + 3

4 (nminus t) ge 34n

(c) right(A) cuts a rectangle B isin Rt right(B)cuts a rectangle C isin Rb and there is a rectangleR isin Rb located between right(A) and left(C) (seeFigure 3c) Let Rprime sub R be the set of the rectanglesto the left of right(B) which satisfies ABR isin Rprimeand are guillotine-separable (using arguments similarto those of previous cases) and let t = |Rprime| ge 3 Bythe inductive hypothesis the n minus t minus 1 rectangles tothe right of right(B) has a guillotine-separable subsetRc such that |Rc| ge 3

4 (n minus t minus 1) Note that Rprime cupRc is guillotine-separable since right(B) separatesall rectangles in Rprime from all rectangles in Rc andsatisfies |Rprime cupRc| = |Rprime|+ |Rc| ge t + 3

4 (nminus tminus 1) =34n + t4minus 34 ge 3

4n

(d) right(A) cuts a rectangle B isin Rt right(B) cutsa rectangle C isin Rb and ABC are the only rect-angles of R to the left of right(C) (see Figure 3d)

ABC is guillotine-separable and by the induc-tive hypothesis the n minus 3 rectangles to the right ofright(C) have a guillotine-separable subset Rd suchthat |Rd| ge 3

4 (nminus3) We have that ABCcupRd isguillotine-separable and satisfies |ABC cup Rd| ge3 + 3

4 (nminus 3) ge 34n

(e) right(A) cuts a rectangle B isin Rt right(B) cutsa rectangle C isin Rb there is no rectangle inRb locatedbetween A and C from left to right right(C) doesnot cut any rectangle in Rt and there exists at leasta rectangle R isin Rt located between right(B) andright(C) (see Figure 3e) LetRprime sube R be the set of therectangles to the left of right(C) without includingC and let t = |Rprime| We have that ABR isin Rprime thust ge 3 Furthermore Rprime is guillotine-separable Bythe inductive hypothesis the n minus t minus 1 rectangles ofR to the right of right(C) have a guillotine-separablesubset Re such that |Re| ge 3

4 (n minus t minus 1) We finallyhave that Rprime cupRe is guillotine-separable and satisfies|Rprime cupRe| ge t + 3

4 (nminus tminus 1) = 34n + t4minus 34 ge 3

4n(f) right(A) cuts a rectangle B isin Rt right(B) cuts

a rectangle C isin Rb right(C) cuts a rectangle D isin Rtand ABC are the only rectangles of R to the leftof right(C) (see Figure 3f) ABC is guillotine-separable and by the inductive hypothesis the nminus 4rectangles to the right of right(C) have a guillotine-separable subset Rf such that |Rf | ge 3

4 (n minus 4) Wehave that ABC cup Rf is guillotine-separable andsatisfies |ABC cup Rf | ge 3 + 3

4 (nminus 4) = 34n

(g) right(A) cuts a rectangle B isin Rt right(B)cuts a rectangle C isin Rb right(C) cuts a rectangleD isin Rt there is no rectangle of Rb located betweenright(A) and left(C) and there is at least a rectangleR isin Rt located between right(B) and left(D) (seeFigure 3g) Let Rprime sub R be the set of rectangles to theleft of left(D) and let t = |Rprime| Since ABR isin Rprimewe have t ge 3 Using right(B) as the first cut wehave that Rprime is guillotine-separable Furthermore bythe inductive hypothesis the nminustminus1 rectangles to theright of left(D) have a guillotine-separable subset Rg

such that |Rg| ge 34 (nminus tminus 1) We have that Rprime cupRg

is guillotine-separable and satisfies |Rprime cup Rg| ge t +34 (nminus tminus 1) ge 3

4nHence the first part of the lemma follows To see

that the bound 34n is tight refer to Figure 4 We

have n4 groups of 4 rectangles each In each groupat most 3 rectangles are separated by any guillotinepartition Thus any guillotine partition will separateat most 3

4n rectangles

Theorem 4 If R is a boundary set of rectanglesthen R has a guillotine-separable subset of size atleast 3

4 |R| minus O(1) Furthermore this lower bound inthe cardinality is tight up to the additive term O(1)

Proof Assume wlog that none of the sets Rt RbR` and Rr is empty So let L and R be the rectan-

29


1 2 n4


gles of maximum width in R` and Rr respectivelySimilarly let T and B be the rectangles of maximumheight in Rt and Rb respectively Further assumewlog that L 6= R and T 6= B The proof is dividedinto the following two disjoint cases

(a) right(L) is to the left of left(R) or bottom(T )is above top(B) By symmetry assume wlog thatright(L) is to the left of left(R) (see Figure 5a)Note that this case generalizes the cases in which RtRb R` or Rr is empty L = R or T = B LetR1 sub R be the rectangles to the left of right(L)where L isin R1 The rectangles of R1 above L area corner set of rectangles then guillotine-separableby Lemma 2 Similarly the rest of the rectanglesof R1 those below L are a corner set of rectanglesthen guillotine-separable Hence R1 is guillotine-separable Similarly R3 is guillotine-separable whereR3 sub R is the set of the rectangles to the right ofleft(R) Let R2 be the set of rectangles betweenright(L) and left(R) which is a parallel set of rectan-gles and then has a guillotine-separable subset Rprime2 ofcardinality at least 3

4 |R2| by Lemma 3 By the linesright(L) and left(R) we have that R1 cup Rprime2 cup R3 isguillotine-separable Let t = |R1cupR3| ge 2 Note that|R2| ge nminus tminus 4 since lines right(L) and left(R) cutat most four rectangles of R We then obtain that|R1 cup Rprime2 cup R3| = |R1 cup R3| + |Rprime2| ge t + 3

4 |R2| get + 3


4nminus 52

(b) right(L) is to the right of left(R) andbottom(T ) is below top(B) (see Figure 5b) The mainobservation is that there are at most 4 rectangles ofR which can be cut by more than one line amongright(L) left(R) bottom(T ) and top(B) (the rect-angles denoted Lprime Rprime T prime and Bprime in the figure) Fur-thermore each of these lines cuts at most two rect-angles among Lprime Rprime T prime and Bprime Hence there mustbe a line among these four ones that cuts at most2 + (n minus 4)4 = n4 + 1 rectangles of R Assumewlog that this line is right(L) and let Rprime sub Rbe the rectangles not cut by right(L) which satisfies|Rprime| ge 3

4nminus 1 Using arguments similar to those usedin case (a) we have that the rectangles of Rprime to theleft of right(L) are guillotine-separable (since L isin Rprimeseparates two corner sets of rectangles) and the rect-angles of Rprime to the right of left(R) are also guillotine-separable (since R isin Rprime separates two corner sets ofrectangles) Hence Rprime is guillotine-separable

L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)

Figure 5 Proof of Theorem 4

The tightness of the bound follows from the secondpart of Lemma 3

References

[1] F Abed P Chalermsook J R Correa A Karren-bauer P Perez-Lantero J A Soto and A Wiese Onguillotine cutting sequences In APPROXRANDOMpages 1ndash19 2015

[2] A Ahmadinejad and H Zarrabi-Zadeh Finding max-imum disjoint set of boundary rectangles with applica-tion to PCB routing IEEE Transanction on CAD ofIntegrated Circuits and Systems 36(3)412ndash420 2017

[3] S Bandyapadhyay A Maheshwari S Mehrabi andS Suri Approximating dominating set on intersec-tion graphs of rectangles and L-frames Comp Geom8232ndash44 2019

[4] J R Correa L Feuilloley P Perez-Lantero and J ASoto Independent and hitting sets of rectangles inter-secting a diagonal line Algorithms and complexityDiscrete Comput Geom 53(2)344ndash365 2015

[5] A Khan and M R Pittu On guillotine separabilityof squares and rectangles In APPROXRANDOM2020

[6] H Kong Q Ma T Yan and M D Wong An op-timal algorithm for finding disjoint rectangles and itsapplication to PCB routing In DAC pages 212ndash2172010

[7] J S Mitchell Approximating maximum indepen-dent set for rectangles in the plane arXiv preprintarXiv210100326 2021

[8] J Pach and G Tardos Cutting glass Discrete Com-put Geom 24(2-3)481ndash496 2000

30


Faster Distance-Based Representative Skyline in the Plane

Sergio Cabellolowast12

1University of Ljubljana Slovenia2Institute of Mathematics Physics and Mechanics Slovenia

A point p isin R2 dominates a point q isin R2 ifx(p) ge x(q) and y(p) ge y(q) For a set of pointsP sub R2 its skyline is the subset of points of P thatare not dominated by any other point of P We denoteby sky(P ) the set of skyline points of P It is well-known [4] that sky(P ) can be computed in O(n log h)time where n = |P | and h = | sky(P )|

For each subset of points Q sube sky(P ) we define

ψ(QP ) = maxpisinsky(P )

minqisinQ

d(p q)

where d(p q) denotes the Euclidean distance be-tween p and q An alternative point of view forψ(QP ) is the following it is the smallest value λsuch that disks centered at the points of Q with ra-dius λ cover the whole sky(P )

Our aim is to obtain efficient algorithms for thefollowing optimization problem for a given point setP in the plane and a positive integer k compute

opt(P k) = minψ(QP ) | Q sube sky(P ) and |Q| le k

The problem of computing opt(P k) and an op-timal solution was introduced in the context ofdatabases by Tao et al [5] under the name of distance-based representative of sky(P ) They showed that theproblem can be solved in O(kh2) time assuming thatthe skyline is available and sorted by x-coordinateThus it takes O(n log h + kh2) time In the unpub-lished full version of their work they improved thetime bound of the second phase to O(kh) which im-plies a time bound of O(n log h+ kh)

The very same problem was considered in the con-text of optimization by Dupin Nielsen and Talbi [3]who noted the connection to clustering multiobjec-tive optimization and decision analysis They no-ticed that opt(P k) is the (discrete) k-center prob-lem for sky(P ) and solved the opt(P k) problem inO(kh log2 h) time again assuming that the skyline isavailable and sorted Thus starting from P the timebound is O(n log h + kh log2 h) They also provide alinear-time algorithm for opt(P 1) and an algorithm

lowastEmail sergiocabellofmfuni-ljsi Research supported bythe Slovenian Research Agency (P1-0297 J1-9109 J1-1693 J1-2452)

with running time O(h log h) for opt(P 2) again as-suming that the skyline is available and sorted by x-coordinate

Our contribution We provide a quite simple al-gorithm for computing opt(P k) in O(n log h) timeThis improves all previous results

At first this may seem optimal because comput-ing the skyline of P takes Ω(n log h) time [4] How-ever do we really need to compute the skyline Afterall we only need to select a (particular) subset of kpoints from the skyline and this has a lower boundof Ω(n log k) time We show that the decision prob-lem for opt(P k) can be solved in O(n log k) timewhich is asymptotically optimal if we want to find asolution The algorithm uses the technique employedby Chan [2] for the convex hull in the plane split Pinto t sets compute the skyline of each of them sep-arately and use this to compute certain points alongthe skyline as needed merging information from the tskylines

We further combine the decision problem withparametric search and selection in sorted arrays toshow that opt(P k) can be computed in O(n log k +n loglog n) time This is asymptotically optimal whenlog k = Ω(loglog n) if we also want to have an optimalsolution We also show that opt(P 1) can be solvedin linear time without computing sky(P )

The full version of this work is available at [1]

References

[1] S Cabello Faster distance-based representative sky-line and k-center along Pareto front in the planePreprint at httpsarxivorgabs201215381

[2] TM Chan Optimal output-sensitive convex hull al-gorithms in two and three dimensions Discret Com-put Geom 16(4)361ndash368 1996

[3] N Dupin F Nielsen EG Talbi Unified polynomialdynamic programming algorithms for p-center vari-ants in a 2d Pareto front Mathematics 9 2021

[4] DG Kirkpatrick R Seidel Output-size sensitive al-gorithms for finding maximal vectors In SoCG 198589ndash96

[5] Y Tao L Ding X Lin J Pei Distance-based rep-resentative skyline In ICDE 2009 892ndash903

31


On the number of connected rectilinear convex 4-gons

Alejandra Martınez-Moraianlowast1 and David Ordendagger1

1Departamento de Fısica y Matematicas Universidad de Alcala Spain

Abstract

We formulate an Erdos-Szekeres problem for sets ofpoints in the plane replacing usual convexity by recti-linear convexity We observe that the minimum num-ber of connected rectilinear convex k-gons is zero Wethen prove that the maximum number of connectedrectilinear convex 4-gons in a set of n points in theplane is at least A(nminus 3) where A(n) is the sequenceA096338 in the OEIS We conjecture that this is theexact maximum number of such 4-gons in a set of npoints

1 Introduction

Let P = p1 pn be a set of n points in general(orthogonal) position in the plane that is no twopoints are in the same verticalhorizontal line Theconvex hull of P denoted as CH(P ) is the intersec-tion of all convex subsets of the plane which contain P

It can be computed as R2 ⋃

HisinHH where H is the set

of all half-planes of R2 that pass through two pointsof P and do not contain any point of P in their inte-rior [2] The convex hull of a subset S sube P is called aconvex k-gon if CH(S) has exactly k le n points of Sas vertices and hence those k points are the verticesof a convex polygon

Classical Erdos-Szekeres problems deal with the ex-istence and number of convex k-gons in a set of npoints [7] One of the main questions in this area isto find the minimum number of convex k-gons thatcan be determined by a set of n points [3]

Here we study a variation of this problem in whichconvexity is replaced by rectilinear convexity or ortho-convexity where the rectilinear convex sets are thosesets whose intersection with any vertical or horizon-tal line is connected [1 4] A related algorithmicalproblem has been recently studied in [5]

lowastEmail alejandramartinezmuahes Research sup-ported by Project PID2019-104129GB-I00 AEI 1013039501100011033 of the Spanish Ministry of Science andInnovation

daggerEmail davidordenuahes Research sup-ported by Project PID2019-104129GB-I00 AEI 1013039501100011033 of the Spanish Ministry of Science andInnovation

An open quadrant of the plane is the intersectionof two open half-planes whose supporting lines areparallel to the x- and y- axes A quadrant is P -freeif it does not contain any point of P Let Q be theset of all P -free open quadrants of the plane Therectilinear convex hull of P is the set

RH(P ) = R2 ⋃

QisinQQ

Note that a rectilinear convex hull is a closed setthat might be disconnected or have no area see Fig-ure 1 Also note that the rectilinear convex hull is asubset of the usual convex hull Thus

RH(P ) = CH(P ) ⋃

QisinQQ

With this definition it is easy to see that it is enoughto remove the P -free open quadrants that are definedby two points of P that is the four quadrants Qi

pq

of the form u isin R2 ux ≶ px uy ≶ qy for p q isin P numbered according to the usual numbering of quad-rants in the cartesian plane Then

RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)

Figure 1 Three examples of rectilinear hulls

The rectilinear convex hull of a subset S sube P isa rectilinear convex k-gon if the boundary of RH(S)contains exactly k le n points of S Although a rec-tilinear convex k-gon might be disconnected or haveno area it is natural to be interested in those whichare connected what also implies that they have pos-itive area See Figure 2 In this framework one canconsider the next Erdos-Szekeres question

32


(a) (b) (c)

Figure 2 Examples of a connected rectilinear convex4-gon (a) 5-gon (b) and 8-gon (c)

Problem 1 What is the minimum number of con-nected rectilinear convex k-gons that can be deter-mined by a set of n points in general position in theplane

However for every natural number n and k ge 4there exist sets of n points in general position in thespirit of Figure 1c that contain no connected rectilin-ear convex k-gon Note that for k isin 2 3 connectedrectilinear convex k-gons do not exist Therefore weaddress the problem of finding the maximum numberof connected rectilinear convex k-gons instead

Problem 2 What is the maximum number of con-nected rectilinear convex k-gons k ge 4 that can bedetermined by a set of n points in general position inthe plane

This was a trivial problem in the classic frameworkas it is now the minimum problem in the rectilinearconvex setting But this maximum problem in therectilinear convex setting is non trivial even for k = 4

2 The maximum number of connected rectilinearconvex 4-gons

Hereforth the term ldquok-gonrdquo must be interpreted asldquoconnected rectilinear convex k-gonrdquo As an initialstep we study the case k = 4

Notation 3 For a set of points in the plane we callN (north) the point with the largest y-coordinate inthe set S (south) E (east) and W (west) are definedsimilarly Note that a point may have two such labelsAs usual qx and qy denote respectively the x- andthe y-coordinate of a point q

Lemma 4 Let P = p1 p2 p3 p4 be a set of 4 pointsin general orthogonal position Then area(RH(P )) gt0 if and only if there exists a point q such that thevertical and horizontal lines through q separate thepoints in P in 4 different regions

Proof See Figure 3 If area(RH(P )) gt 0take as q any point in the interior of RH(P )For the other implication take 0 lt ε lt

miniisin14

mind(qx pix) d(qy piy ) Then the ball

B(q ε) with center q and radius ε does not containthe apex of any P -free quadrant Hence B(q ε) subRH(P )

q

Figure 3 The vertical and horizontal lines through qseparate the four red points into different regions

Proposition 5 Let P be a set of four points in theplane RH(P ) is a 4-gon if and only if the cardinalityof the set NSEW is exactly 4 and either Nx gtSx and Wy gt Ey or Nx lt Sx and Wy lt Ey

Proof Let P = p q r s be a set of 4 points in gen-eral orthogonal position in the plane Suppose with-out loss of generality that there is one point say psuch that N = E = p (the other cases are symmet-ric) Consider the points Eprime and N prime that maximizerespectively the x- and y-coordinates in the set P pIf again N prime = Eprime = q then the quadrants u isin R2 ux lt px uy gt qy and u isin R2 ux gt qx uy lt pyare P -free and overlap disconnecting point p from therest of points in RH(P ) Hence RH(P ) cannot be aconnected 4-gon If N prime = q and Eprime = r then theargument is the same see Figure 4

p

q

r

s

Figure 4 The overlapping of the purple and the greenquadrants disconnects point p = N = E from the restof the rectilinear convex hull of the set p q r s

The definitions of N S E and W completely de-termine the relative positions of these four points ex-cept for two aspects W might be above or belowE and N might be to the left or to the right of SSuppose Wy gt Ey Then among the two possibilitiesNx ≶ Sx only the option Nx gt Sx holds by Lemma 4It is easy to check that this option produces a 4-gonThe case Wy lt Ey is analogous See Figure 5

From Proposition 5 there are 2 types of 4-gonswhich can be seen as clockwise and counterclockwisewindmills see Figure 5 Let Gk(n) denote the max-imum number of connected rectilinear convex k-gons

33


(a) (b)

Figure 5 The two types of 4-gons considered

over all sets of n points in the plane n ge 4 Wechecked that G4(4) = 1 and G4(5) = 2

Let A(i) be defined for i ge 1 as

A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd

These A(i) form the sequence A096338 in theOEIS [6] with initial terms 1 2 6 10 20 30

Theorem 6 G4(n) ge A(nminus 3) for every n ge 4

Proof Let P be a set of n points placed on the ver-tices of an n-sided regular polygon after a tiny clock-wise rotation of angle ε so that no two points are ona common horizontalvertical line see Figure 6 Weprove that such a P determines A(n minus 3) different4-gons

N

E

W

S

ε

N or W

N or E

S or ES or W

N

S

Figure 6 A configuration of n = 16 points containingA(13) = 336 connected rectilinear convex 4-gons

Suppose that n equiv 0 (mod 4) the cases n equiv 1 2 3(mod 4) are analogous Observe that the points N SE and W divide in cyclic order the remaining pointsof P into four sets of nminus4

4 points Also note that thepoints in the north-west set can only act either as thenorth or the west of a 4-gon Equivalent propertieshold for the three remaining sets of nminus4

4 points When

a point of P not necessarily the point N acts as thenorth of a subset of P for example a 4-gon we denoteit as N We proceed analogously for S W and E

The process of enumeration of 4-gons in P willwork as follows We consider all the possiblepairs N -S where N must be one of the upperhalf nminus2

2 points different from W and S must beone of the remaining lower half points differentfrom E see Figure 6 For every such pair theonly feasible choices for W and E are the pointsu isin P Sy lt uy lt Ny ux lt minNx Sx

and

u isin P Sy lt uy lt Ny ux gt maxNx Sx

re-

spectively In Figure 6 for the two points N and Sthese two sets are determined by the blue dashedlines The possible choices for W are the points in theleft-middle region and the possible choices for E arein the right-middle region But not all the possiblepairs W -E that we obtain this way determine a 4-gonwith the pair N -S because we still need the 4 pointsto fulfill Proposition 5

Consider first N = N and S = S Since Nx gt Sxthen we need to look for the pairs W -E such thatWy gt Ey There are nminus2

2 possibilities for W Westart considering the closest one to S and continueclockwise until the closest to N For the first choiceof W there is only one E below In the next stepsbecause of the nature of the point configuration P every time we switch to the next W we gain onechoice of E (with Ey between the previous and the

current Wy) and we do not lose the choices from the

previous steps For the last step the W closest to N we have nminus2

2 choices of E Then the total number of

pairs W -E that form a 4-gon with N and S is

1 + 2 + 3 + middot middot middot+ nminus 2

2 (1)

Next fix N = N and make S vary among the pointsin the lower half of P For the choices of S in thesouth-west set Nx gt Sx There are nminus4

4 such choicesStarting from the closest one to S and moving clock-wise in every step we lose 2 choices of W one becauseit becomes the new S and the other because a pointin the north-west set that was a suitable choice for W becomes now to the right of the new S and hence itis not a suitable choice anymore We obtain

(1)︸︷︷︸S closest to W

+ (1 + 2 + 3) + middot middot middot

+

(1 + 2 + middot middot middot+ nminus 2

2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34


4-gons where each parenthesized summand in Equa-tion (2) corresponds to a choice of S and is obtainedas explained above for Equation (1) With an analo-gous procedure for the possibilities of S to the rightof S we obtain

(1 + 2)︸︷︷︸S closest to E

+ (1 + 2 + 3 + 4) + middot middot middot

+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)

additional 4-gonsSimilar arguments work in the cases where S = S

and we move N k positions clockwise from N Ob-serve that when doing so we lose 2k choices of E ByProposition 5 we obtain

nminus44sum

k=1

nminus22 minus2ksumi=1

i (4)

4-gons On the other hand when we move N k posi-tions counterclockwise from N we lose 2k minus 1 choicesof W The number of 4-gons obtained is

nminus44sum

k=1

nminus22 minus2k+1sum

i=1

i (5)

If we move both N k positions clockwise and S jpositions counterclockwise there are two possibilitiesj le k and j gt k The number of 4-gons obtained inthe first case is

nminus44sum

k=1

ksumj=1


i (6)

and in the second case it is

nminus44sum

k=1

nminus44sum

j=k+1

nminus22 minus2j+1sum

i=1

i (7)

Similarly if we move N and S clockwise k positionsfor N and j positions for S there are two possibilitiesIf j le k the number of 4-gons obtained is

nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1

nminus22 minus2jsumi=1

i (9)

when we move N and S counterclockwise k positionsfor N and j positions for S if j lt k we have

nminus44sum

k=1

kminus1sumj=1


i=1

i (10)

4-gons and if j ge k we havenminus44sum

k=1

nminus44sum

j=k


i=1

i (11)

Finally if N is moved k counterclockwise and S ismoved j clockwise with j lt k we get

nminus44sum

k=1

kminus1sumj=1


i=1

i (12)

4-gons and if j ge k we getnminus44sum

k=1

nminus44sum

j=k


i (13)

Summing up equations (1)-(13) gives A(n minus 3) astotal number of 4-gons in P

3 Concluding comments

Based on computer simulations we conjecture thatG4(n) = A(nminus3) for every n ge 4 It would be nice tofind a correspondence between the maximum numberof connected rectilinear convex k-gons and other ob-jects counted by the sequence A(n) In addition wewant to study Gk(n) for k gt 4

References

[1] C Alegrıa D Orden C Seara and J Urrutia Effi-cient computation of minimum-area rectilinear con-vex hull under rotation and generalizations Journalof Global Optimization 79 (2020) 1ndash28

[2] M de Berg O Cheong M van Kreveld and MOvermars Computational geometry Introductionin Computational geometry Algorithms and appli-cations 1ndash17 Springer 2008

[3] P Erdos and R Guy Crossing number problems TheAmerican Mathematical Monthly 8 (1973) 52ndash58

[4] E Fink and D Wood Restricted-orientation convex-ity Springer Science amp Business Media 2012

[5] H Gonzalez-Aguilar D Orden P Perez-LanteroD Rappaport C Seara J Tejel and J UrrutiaMaximum rectilinear convex subsets SIAM Journalon Computing 501 (2021) 141ndash170

[6] OEIS Foundation Inc (2021) The On-Line En-cyclopedia of Integer Sequences httpoeisorg

A096338 Accessed 2021-01-01

[7] B Vogtenhuber Combinatorial aspects of [colored]point sets in the plane PhD thesis Institute forSoftware Technology Graz University of TechnologyGraz Austria 2011

35


No selection lemma for empty triangleslowast

Ruy Fabila-Monroydagger1 Carlos Hidalgo-ToscanoDagger1 Daniel Perzsect2 and Birgit VogtenhuberDagger2

1Departamento de Matematicas Cinvestav Ciudad de Mexico Mexico2Institute of Software Technology Graz University of Technology Graz Austria

Let S be a set of n points in general position inthe plane A triangle of S is a triangle whose verticesare points of S We consider the following questionGiven an n-point set S and a family F of triangles ofS how many triangles of F contain a common pointof the plane in their interiors Boros and Furedi [3]showed that if F is the family of all

(n3

)triangles of

S then Θ(n3) of them have a common point in theirinteriors Barany Furedi and Lovasz [1] extendedthis to any family F of size Θ(n3) a result known asSecond Selection Lemma

Among several results for families of smaller sizeEppstein [4] constructed point sets S and families ofn3minusα triangles of S such that every point of the planeis in at most n3minusα(2nminus 5) triangles for α ge 1 and inat most n3minus2α triangles for 0 le α le 1 He also showedlower bounds for any family of n3minusα triangles

In this work we study the above question for thefamily of all empty triangles of S A triangle of S isempty if it does not contain any points of S in its inte-rior While Eppsteinrsquos lower bounds carry over to oursetting his upper bound constructions do not apply(as they are about different families) Surprisingly wecan still show the same upper bounds

Theorem 1 For every integer n and every 0 le α le 1there exist sets S of n points with Θ(n3minusα) emptytriangles where every point of the plane lies in theinterior of at most Θ(n3minus2α) empty triangles of S

We prove Theorem 1 by constructing point sets withthese properties Our construction is based on Hortonsets and squared Horton setsHorton sets were first studied by Horton [5] and

later generalized by Valtr [6] They are recursivelydefined as follows A single point is always a Hortonset Further if H0 and H1 are Horton sets of samecardinality such that any line spanned by two pointsof H0 is below H1 and any line spanned by two pointsof H1 is above H0 then H0 cupH1 is also a Horton set

An ε-perturbation of G is a perturbation of G whereevery point p of G is replaced by a point at distance at

lowastSupports FWF grant I 3340-N35 EU H2020 grant 734922daggerEmail ruyfabilamathcinvestavedumxDaggerEmail cmhidalgomathcinvestavmxsectEmails daperz bvogtisttugrazat

most ε to p A squared Horton set H first defined byValtr [6] is a specific ε-perturbation of an integer gridG such that triples of non-collinear points in G keeptheir orientations in H and such that points along eachline in G are perturbed to points forming a Hortonset in H Barany and Valtr [2] showed that squaredHorton sets of size n span only Θ(n2) empty trianglesFor squared Horton sets we show the following results

Theorem 2 Let H be a squared Horton set of npoints i) Every point of the plane is in the inte-rior of O(n) empty triangles of H ii) Every point ofH is incident to O(n) empty triangles of H

Construction for Theorem 1 We denote by 5 apoint set which is obtained by placing four points onthe corners of a square and adding further points alongfour slightly concave arcs between adjacent cornerssuch that on each arc there is almost the same numberof points A 5-squared Horton set H5 is the set weobtain by replacing every point of a squared Hortonset H with m points by a small 5 with k points ThusH5 has n = km points We show that the numberof empty triangles in H5 is Θ(m2k3) On the otherhand using Theorem 2 we prove that every point ofthe plane is in the interior of O(mk3) empty trianglesof H5 Theorem 1 then follows from those two resultsand by setting m = nα and k = n1minusα

References

[1] I Barany Z Furedi and L Lovasz On the numberof halving planes Combinatorica 10 (1990) 175ndash183

[2] I Barany and P Valtr Planar point sets with a smallnumber of empty convex polygons Studia Sci MathHungar 41 (2004) 243ndash266

[3] E Boros and Z Furedi The number of trianglescovering the center of an n-set Geometriae Dedicata17 (1984) 69ndash77

[4] D Eppstein Improved bounds for intersecting trian-gles and halving planes J Combin Theory Ser A 62(1993) 176ndash182

[5] J D Horton Sets with No Empty Convex 7-GonsCanadian Mathematical Bulletin 4 (1983) 482-ndash484

[6] P Valtr Convex independent sets and 7-holes inrestricted planar point sets Disc Comp Geom 7(1992) 135ndash152

36


On (α k)-sets and (α k)-hulls in the plane

Merce Claverollowast1 Luis H Herreradagger2 Pablo Perez-LanteroDagger2 and Carlos Searasect1

1Universitat Politecnica de Catalunya (Spain)2Universidad de Santiago de Chile (Chile)

Abstract

This abstract reports first the study of upper andlower bounds for the maximum number of all the com-binatorially different (α k)-sets of an n-point set P inthe plane 0 lt α le π and 0 lt 2k lt n depending onthe (fixedvariable) values of α and k relating themwith the known bounds for the maximum number ofk-sets the O(n 3

radick) upper bound from Dey [5] and the

neΩ(radic

log k) lower bound from Toth [7] and showingalso efficient algorithms for generating all of them

Second we study the depth of a point p isin P ac-cording to the (α k)-set criterion (instead of the k-setcriterion) We compute the depths of all the points ofP for a given angle α and also design a data structurefor reporting the angle-interval(s) of a given depth fora point of P in O(log n) time (if it exists)

Finally we define the (α k)-hull of P for fixed val-ues of α and k and design an algorithm for computingthe (α k)-hull of P for given values of α and k To dothat we follow the relevant ideas and techniques fromCole et al [4] Unfortunately the algorithm is still noso efficient as we wish and we believe that their com-plexities strong depends on the fixed values for theparameters α and k more concretely as α is closer toπ the time complexity is close to the optimal

1 Preliminaries

Let P be a set of n points in the plane in general po-sition ie no three points are colinear A wedge isthe convex region bounded by two rays with commonorigin with aperture angle α 0 lt α le π and denotedby an α-wedge An (α k)-set of P is a k-subset K of

lowastEmail merceclaverolupcedu Supported by MICINNPID2019-104128GB-I00 AEI 1013039501100011033 GenCat DGR2017SGR1640daggerEmail luisherrerabusachclDaggerEmail pabloperezlusachcl Supported by DICYT

041933PL Vicerrectorıa de Investigacion Desarrollo e Inno-vacion USACH (Chile) and Programa Regional STICAMSUD19-STIC-02sectEmail carlossearaupcedu Supported by MICINN

PID2019-104129GB-I00 AEI 1013039501100011033 GenCat DGR2017SGR1640


P inside an α-wedge which contain no other point ofP and this α-wedge is denoted by (α k)-wedge Each(α k)-set K has (a not-unique and directed) associ-ated line defined by a point of CH(K) and a pointof P and which contains a ray of the α-wedge Theassociated line facilitates the counting of the numberof combinatorially different (α k)-sets of P

To study all the different cases for the parameters αand k we use the notation α0 when α is fixed and k0

when k is fixed When these parameters are variablewe simply use α and k By fα0

k0(n) we denote the

maximum number of (α0 k0)-set of P overall n-pointsets and analogously for fα0

k (n) fαk0(n) and fαk (n)Related works The first study of the (α k)-setswas done by Claverol [1 2] by determining upper andlower bounds on the number of (α k)-sets for P Herewe reproduce part of the results Later Erickson etal [6] considered generalizations of the CenterpointTheorem in which the half-spaces are replaced withwedges (or cones) of angle α There are other papersin the literature focusing in this topic [5 7]

2 Upper and lower bounds

The results presented here about the upper and lowerbounds for fαk (n) are summarized in Figure 1 andclassified into the four cases for the values (fixed orvariable) of α and k Almost all of them were obtainedin [1 2] Due to the lack of space we only illustratethe lower bound for α0 and k0

α k Upper bound Lower boundα0 k0 O(n2) fα0

k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)

Figure 1 Upper and lower bounds for fαk (n) withfα0

k0(n) being a function of α0 and k0

Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)

(n(k0 minus 1)minus π

2α0(k0 minus 1)2

))

Proof We construct a set P of n points in convex po-sition as follows First put m+1 points s0 s1 sm

37


in counterclockwise order on the unit circle C (redpoints in Figure 2) where m + 1 = b π

2α0c These

points are equally spaced on C such that the chordsisi+1 subtends the angle α0 for i = 0 1 mminus 1

We add to P the sets Si i = 0 1 m eachformed by si together with k0 minus 2 points of P allof them equally spaced and close enough to the pointsi ie the distance between consecutive points in-cluding si is ε for a small enough ε gt 0 Noticethat the length of the arc of the unit circle subtendedby a central angle 2α0 is exactly 2α0 so we can takeε small enough such that all the points in Si are inan arc of C of length less than α0 In total we areadding (m+ 1)(k0minus 1) points to P Finally add to Pa set T of nminus (m+ 1)(k0 minus 1) equally spaced pointsclose enough between them to complete the total ofn points (see Figure 2) Notice that this constructionworks for small angles α0 such that m + 1 ge 2 andthus π

2α0ge 2 ie α0 le π

4 Thus we have α0-wedges containing one point from

T and k0minus 1 points from P T ie k0minus 1minusu pointsfrom Si and u points from Si+1 for some 0 le i le mminus1and 0 le u le k0 minus 2 (see Figure 2) Notice that weare not selecting the (k0 minus 1)-subset Sm when u = 0Then for the set P we know that the total numberof combinatorially different (α0 k0)-sets is

Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0

k0(n) grows up with k0 Thus

for k0 = n4 + 1 fα0

n4+1(n) isin Ω(n2) which is almost

the obtained upper bound We assume that k0 ge 3since for k0 = 2 we have

(n2

)combinatorially different

(α0 2)-sets for some small value of α0

3 The α-wedge depth with respect to P

The α0-wedge depth of x isin R2 with respect to P withthe (α0 k)-set criterion α0-depth for short denotedby DepthPα0

(x) is defined as follows

DepthPα0(x) = min

1leklen2k = |P capW |

for any possible closed α0-wedge W with apex at xThis definition is an extension of the depth with

the k-set criterion where the k-line passes throughpoint x In fact if α0 = π we obtain the depth withthe k-set criterion Thus for any pi isin P and α00 lt α0 le π DepthPα0

(pi) le n2 The points pi isin Pin the boundary of the convex hull of P CH(P ) haveDepthPα0

(pi) = 1 By the definition if x is either inthe exterior of CH(P ) or it belongs to the boundary ofCH(P ) but x isin P and α0 lt π then DepthPα0

(x) = 0Moreover for α0 small enough all the points pi isin Phave DepthPα0

(pi) = 1 and there are no points x withDepthPα0

(x) ge 2

k0 minus 1

k0 minus 1 k0 minus 1

k0 minus 1

α0

α0

sm

|T | = nminus (m + 1)(k0 minus 1)

α0

α0

s0

s1

C

k0 minus 1

T

Figure 2 Equally spaced groups of points Si on C

Theorem 2 Given P and α0 the sorted list of thevalues DepthPα0

(pi) i = 1 n can be computed inO(n2 log n) time and O(n) space Then DepthPα0

(pi)can be computed in O(log n) time

31 Report α such that DepthPα (pi) = k

We consider the problem Pre-compute a data struc-ture such that given pi isin P and k 1 le k le n2 thenthe angular interval (αi1 α

i2) sube [0 2π) such that for

any α isin (αi1 αi2) the DepthPα (pi) = k can be reported

in O(log n) time

Theorem 3 In O(n2 log n) time and O(n2) space wecan compute a data structure such that for a givenpi isin P and k 1 le k le n2 in O(log n) time we

can reported the angular interval (δki δ(k+1)i ) sube [0 2π)

such that for any α isin (δki δ(k+1)i ) the DepthPα (pi) = k

4 The (α0 k0)-hulls

A point x inside CH(P ) can be characterized by theproperty that any line through x has at least a pointof P in each of the closed half-planes determined bythe line Generalizing this definition Cole et al [4]defined the k-hull of P for positive k as the set ofpoints x such that for any line through x there areat least k points of P in each closed half-plane Itis clear that the k-hull contains the (k + 1)-hull andif k is greater than dn2e then the k-hull is emptyWe extend this definition to the (α0 k0)-hull(P ) for0 lt α0 le π and 1 le k0 le n2 as follows

38


Definition 4 The (α0 k0)-hull(P ) is the set of pointsx isin R2 such that any closed α0-wedge with apex at xcontains at least k0 points of P

Notice that x has to be inside CH(P ) and then the(α0 k0)-hull(P ) is contained in CH(P ) For α0 = πDefinition 4 is equivalent to the k0-hull of P Fromthe same definition we can easily conclude that the(α0 k0)-hull(P ) contains the (α0 (k0 + 1))-hull(P )

41 The (α0 k0)-hulls for points in convex position

For points in convex position we select n consecu-tive sets of k0 consecutive points in CH(P ) say Kand compute the corresponding arcs aij of adjointcircles defined by apices of the α0-wedges containingK supported in two points pi and pj of CH(K)The endpoints of the arcs occurs when a ray of theα0-wedge bumps a point of CH(K) There are O(n)arcs and the number of intersections between thosearcs is at most O(n2) Doing a sweep-line we computethe (α0 k0)-hull(P ) If α0 is close to π the number ofintersections between the arcs is O(n) and the com-plexities decrease accordingly See Figure 3

Theorem 5 The (α0 k0)-hull(P ) can be computedin O(n2 log n) time and O(n2) space

Figure 3 Left (90 i)-hull(P ) contains (90 i + 1)-hull(P ) i = 1 2 3 4 and (90 5)-hull(P ) is an emptyRight the (90 3)-hull(P ) as the intersection of re-gions defined by the three cycles

42 Computing the (α0 1)-hull(P )

Let ei = pipi+1 be the edges of the boundary ofCH(P ) By Definition 4 the vertices CH(P ) belongto the (α0 1)-hull(P ) The two rays of any α0-wedgewith apex at x inside CH(P ) which containing ex-actly one point of P not in CH(P ) has to intersectthe same edge ei of CH(P ) Thus we have the nextfact about the (α0 1)-hull(P )

Fact 1 The apices of the α0-wedges containing ex-actly one point which rays cross the edge ei define apolygonal-curve fi from pi to pi+1 By definition fi

is not self-intersecting We call Ri the (closed) re-gion defined by ei and fi where Ri = CH(P ) Ri(see Figure 4 Up) The (α0 1)-hull(P ) is the region⋂i=1mRi (see Figure 4 Down)

pipi+1

ei

ei

fi

pi

pi+1

Figure 4 Up Region Ri in red α0 = 90 Down(90 1)-hull(P )=

⋂i=1mRi in red In blue the

polygonal-curve fi

The (α0 1)-hull(P ) can have disconnected regionsas the points in CH(P ) cap P in Figure 4 In fact ifα0 lt π and p isin CH(P ) cap P p is an isolated point of(α0 1)-hull(P ) Any p isin P belongs to (α0 1)-hull(P )because a α0-wedge with apex at p always contains atleast p Based on Fact 1 we describe the steps of analgorithm for computing (α0 1)-hull(P ) for a set P ofn points in general position

1 If π2 le α0 lt π in O(n log n) time and O(n)

space we compute the (α0 1)-hull(P ) as followscompute the α0-maximal points of P and thepolygonal-curves fi for the edges ei (see Fig-ure 4 Right) Alegrıa-Galicia et al [3] showedthe algorithm for computing the sequence ofO(n)arcs forming all the fi i = 1 m where mis the number of edges of CH(P ) Then inO(n log n) time and O(n) space we can do a line-sweep of the arrangement of all fi and com-pute (α0 1)-hull(P ) formed by a (possible dis-connected) region defined by the intersection ofall Ri i = 1 m and CH(P ) See Figure 4

2 If 0 lt α0 lt π2 the computation of all the fi

i = 1 m can be done in O( nα0log n) time and

O( nα0) space But the the computation of (α0 1)-

hull(P ) can be done in O(n2) time and space be-cause we compute the (at most) O(n2) intersec-tion points inside CH(P ) between all the fi

Theorem 6 If π2 le α0 le π the (α0 1)-hull(P )can be computed in O(n log n) time and O(n) spaceIf 0 lt α0 lt π2 the (α0 1)-hull(P ) can be computedin O(n2) time and space

39


43 The (α0 k0)-hulls for points in general position

We adapt ideas in [4] to the (α0 k0)-hull(P ) conceptas follows A (directed) line ` is a k0-divider for P if `has at most k0minus1 points of P strictly to its right andat most nminus k0 points of P strictly to its left For anyorientation θ isin [0 2π) of ` there is a unique k0-dividerdenoted by `θ A special k0-divider is a k0-divider thatcontains at least two points The half-space to the leftof a k0-divider is a special half-space The k0-hull isthe intersection of the special half-spaces

The direction of an α0-wedge with apex at x isin R2

is defined by the direction of its right ray (in the clock-wise rotation from its apex x) and it is given by theangle θ formed by X-axis with the line containing theright ray Let W θ

α0denote a directed α0-wedge

Definition 7 A directed α0-wedge W θα0

is a directed(α0 k0)-divider for P if W θ

α0contains at most k0 minus 2

points of P strictly in its interior and at most nminus k0

points of P strictly in its exterior The boundary ofW θα0

must contain at least two points of P A spe-cial (α0 k0)-divider is an (α0 k0)-divider for P thatcontains at least three points on the boundary

Given θ isin [0 2π) there are at most O(n) differentdirected (α0 k0)-dividers W θ

α0 eg a set P with O(n)

points on the (almost vertical) right chain of CH(P )k0 = 4 α0 = π4 and θ = 0 see Figure 5 For α0 = πand fixed θ there are only two (π k0)-dividers

1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y

Figure 5 Different directed (α0 k0)-dividers W θα0

Rotation process An (α0 k0)-divider passing throughpi and pj can be rotated anchored at pi and pj whileits apex traces an arc aij on the adjoint circle definedby α0 and segment pipj until the wedge bumps apoint pk see Figure 6 The poly-curve defined by arcsrays and segments from the wedges at the endpointsdefine an unbounded region denoted by Aij Theregion Aij has the property that for any point x in itsinterior there always exists an α0-wedge with apex atx and direction in the rank between the directions ofthe extreme wedges which contains at most k0 pointssee Figure 6 Starting with an orientation say θ = 0

the rotation process end at the initial (α0 k0)-dividerand the apices of the α0-wedges trace a cycle whichinterior is the intersection of the complementary ofthe union of all the regions Aij

pi

pj

Aij cupAjk cupAkl

Figure 6 Three consecutive (colored) arcs

Sketch of the algorithm First take the orientationθ = 0 and compute the list L of the (α0 k0)-dividerswith this orientation For each (α0 k0)-divider applythe rotation process above and compute the corre-sponding cycle checking the used (α0 k0)-dividers inL There are at most O(n) different cycles formedby sets of arcs aij The total number of arcs aij isbounded by the number of (α0 k0)-sets together withtheir rotations The still no fixed question is the up-per bound of the number of intersection between thosearcs depending on α0 In any case there are at mostO(n4k0

2) We do a line-sweep of the arrangement ofcycles and compute the region(s) of (α0 k0)-hull(P )

Theorem 8 The (α0 k0)-hull(P ) can be computedin O(n4k0

2 log n) time and O(n2k0) space

References

[1] M Claverol Problemas geometricos en morfologıacomputacional PhD Thesis UPC

[2] M Abellanas M Claverol F Hurtado C Seara(α k)-sets in the plane Proc Segundas Jornadas deMatematica Discreta 2002

[3] C Alegrıa-Galicia D Orden C Seara J Urru-tia Efficient computation of minimum-area rectilin-ear convex hull under rotation and generalizationsJournal of Global Optimization Vol 79(3) 2021

[4] R Cole M Sharir C K Yap On k-hulls and relatedproblems SIAM J Comput 16 1987

[5] T K Dey Improved bounds for k-sets related prob-lems Discrete Computational Geometry 19 1998

[6] J Erickson F Hurtado P Morin Centerpoint theo-rems for wedges Discrete Mathematics and Theoret-ical Computer Science 111 2009

[7] G Toth Points sets with many k-sets Discrete andComputational Geometry 26 2001

40


Developable surfaces bounded by spline curves

A Cantonlowast1 L Fernandez-Jambrinadagger1 ME Rosado MarıaDagger2 and MJ Vazquez-Gallosect1

1Matematica e Informatica Aplicadas Universidad Politecnica de Madrid Spain2Matematica Aplicada Universidad Politecnica de Madrid Spain

Abstract

In this talk we revisit the problem of constructing adevelopable surface patch bounded by two rational orNURBS (Non-Uniform Rational B-spline) curves [1]NURBS curves are curves which are piecewise ra-

tional That is they are a generalisation of splinecurves which are piecewise polynomial curves Simi-larly we define NURBS surfaces and solids NURBScurves are the standard in Computer Aided DesignDevelopable surfaces are ruled surfaces with null

Gaussian curvature This implies that they can beconstructed from planar surfaces by just cuttingrolling and folding so that metric properties such aslengths and angles between curves and areas are pre-served These geometric properties are of great inter-est for steel and textile industry since there are piecesdesigned in the plane and then curved into spaceThis problem has been addressed in several ways

but the key drawback is that when we require thedevelopable surface to be NURBS and bounded byNURBS curves the possibilities are restricted [2]For this reason our proposal is to consider devel-

opable surface patches which are not NURBS thoughbounded by NURBS curves [1] In fact we are able toobtain every possible solution to this problem in ourframeworkWe start with a ruled surface parametrised by

b(t v) and bounded by two parametrised curves c(t)d(t)

b(t v) = (1minus v)c(t) + vd(t) t v isin [0 1]

For given c(t) and d(t) this ruled surface will not bedevelopable in general Our contribution to deal withthis problem is based on reparametrisation of one ofthe bounding curves by a function T (t)

b(t v) = (1minus v)c(t) + vd(T (t))

and require b(t v) to satisfy the null Gaussian cur-vature condition This condition can be seen to be

lowastEmail aliciacantonupmesdaggerEmail leonardofernandezupmesDaggerEmail eugeniarosadoupmessectEmail mariajesusvazquezupmes

The authors have been partially supported by the Spanish Mi-

nisterio de Economıa y Competitividad through research grant

TRA2015-67788-P

algebraic in T (t) since the dependence on the deriva-tive T prime(t) is factored out

det(

cprime(t) d(T ) d(T )minus c(t))

= 0

and is of degree 2n minus 2 if the bounding curves c(t)d(t) are of degree n The dot and the comma standfor derivation with respect to T and tThe price to pay is that solutions of this algebraic

equation will not be rational or polynomial in generaland b(t v) will no longer be NURBSThere is an important case which is even simpler to

handle If the bounding curves c(t) d(t) are not ratio-nal or piecewise rational (just polynomial or piecewisepolynomial) and lie on parallel planes the degree maybe seen to decrease to nminus 1Since the condition on the reparametrisation is al-

gebraic the number of possible solutions is finite butnot all of them are geometrically acceptable For be-ing a reparametrisation T (t) must be a monotonicallyincreasing function This can be checked with the helpof

T prime(t) =det

(

cprimeprime(t) d(T ) d(T )minus c(t))

det(

d(T ) cprime(t) d(T )minus c(t))

∣

∣

∣

∣

∣

∣

T=T (t)

which we derive from the null Gaussian curvature con-ditionThis implies that monotonicity is granted if

sgn (cprimeprime(t) middot ν(t)) = sgn(

d(T ) middot ν(t))∣

∣

∣

T=T (t)

where ν(t) is the unitary normal to the ruled surfacealong the segment at t This means that the normalcurvatures of both curves must have the same sign foreach value of tHence acceptable solutions just appear if both

curves are qualitatively similar regarding their cur-vature

References

[1] L Fernandez-Jambrina F Perez-Arribas Devel-opable surfaces bounded by NURBS curves Journalof Computational Mathematics 38 (2020) 693-709

[2] L Fernandez-Jambrina Bezier developable surfacesComputer Aided Geometric Design 55 (2017) 15ndash28

41


Planar aesthetic curves

A Canton1 L Fernandez-Jambrina1 and MJ Vazquez-Gallo1

1Matematica e Informatica Aplicadas Universidad Politecnica de Madrid Spain

Figure 1 An aesthetic typical curve

Abstract

In Geometric Design it is of interest the represen-tation of curves and surfaces that are aestheticallypleasing In order to have a notion amenable to im-plementation in CAGD there have been defined aes-thetic curves as those with monotonic curvature andfor spatial curves monotonic torsion There are sev-eral approaches to obtain aesthetic Bezier curves butwe will follow the lead of [4] and [3]

In [4] Mineur Lichah Castelain and Giaume ob-tain the edges of the control polygon of a planar Bezierspiral by a rotation and a dilation of the previousedge in the control polygon of the curve what the au-thors name typical curve Certain relations betweenthe scaling factor and the rotation angle give rise toaesthetic Bezier spirals starting with any initial edgeof the control polygon and for any degree of the Beziercurve (see Figure 1)

Inspired by this work in [3] Farin extends themethod by considering Bezier curves whose controlpolygon is obtained by the action of a given matrixon the previous edge of the control polygon His in-sight is to exploit the invariance of the curvature andtorsion under subdivision to give some conditions onthe matrix and its singular values that give rise toaesthetic Bezier curves for any initial edge what hecalls Class A matrices and Class A Bezier curves

Email aliciacantonupmesEmail leonardofernandezupmesEmail mariajesusvazquezupmes

The three authors have been partially supported by the Span-ish Ministerio de Economıa y Competitividad through researchgrant TRA2015-67788-P

Figure 2 An aesthetic curve generated by a symmet-ric matrix

However counterexamples to Farinrsquos conditionshave been produced (see [2] and [5]) that is matricesfor which these conditions hold but they do not gener-ate curves with monotonic curvature Moreover in [2]Cao and Wang give conditions on the eigenvalues ofa (2times 2 or 3times 3) symmetric matrix that generates anaesthetic (planar or spatial) Bezier curve (see Figure2)

In this talk we present a simple explicit formulafor the curvature of planar Bezier curves generatedby Farinrsquos method This formula is easily obtained bythe invariance under subdivision property and fromit there can be derived conditions on the eigenvaluesof a general matrix and the initial edge of the con-trol polygon that give rise to aesthetic Bezier curvesThis approach gives a common framework to the pre-vious works and recovers the results in [4] and [2] asparticular cases For more details we refer to [1]

References

[1] A Canton L Fernandez-Jambrina MJ Vazquez-Gallo Curvature of planar aesthetic curves Jour-nal of Computational and Applied Mathematics 381(2021) 113042 19 pp

[2] J Cao G Wang A note on Class A Bezier curvesComputer Aided Geometric Design 25 (7) (2008)523ndash528

[3] G Farin Class A Bezier curves Computer Aided Ge-ometric Design 23 (7) (2006) 573ndash581

[4] Y Mineur T Lichah H Castelain JM GiaumeA shape controlled fitting method for Bezier curvesComputer Aided Geometric Design 15 (1998) 879ndash891

[5] A Wang G Zhao Counter examples of ldquoClass ABezier curvesrdquo Computer Aided Geometric Design61 (2018) 6ndash8

42


Parallel Simulated Annealing for Continuous Dispersion Problems

Narcıs Colllowast1 and Marta Fortdagger1

1Graphics and Imaging Laboratory Universitat de Girona

Figure 1 a) Maximizing the covering b) Maximizing the minimum distance c) Minimizing interactions

Abstract

Many problems in location theory deal with theplacement of desirable facilities (hospitals fire sta-tions cellular antennas etc) in a region domaintrying to minimize some objective function usually re-lated to the distances between the facilities and someelements also placed in the region However thereare situations in which proximity between facilitiesis undesirable and the facilities must be located bytrying to maximize an objective function related tothe distances between them These location problemsare named dispersion problems and the facilities arecalled obnoxious [3] Examples of this type of facilityare nuclear power plants or garbage dumps More-over the pandemic turned the location of several peo-ple in a closed room into a location problem Peoplecan be seen as rdquoobnoxious facilitiesrdquo as they have tokeep far enough from the other In general the ele-ments to be located both people or facilities occupya specific space in the region and a simple discreteset of points cannot model them

In [1] we presented an approach based on simu-lating annealing on GPUs to determine where to lo-cate k disks of fixed-radius so that they globally coveras much area of a polygonal domain as possible (seeFigure 1(a)) In this work we extend that approachto the two following continuous dispersion problemsdealing with obnoxious elements represented by disksof radius r that in these cases have to be fully con-tained in the polygonal domain

lowastEmail collimaeudgedu Research supported byPID2019-106426RB-C31 of the Spanish Government

daggerEmail mfortimaeudgedu Research supported byPID2019-106426RB-C31 of the Spanish Government

1 The classic continuous k-dispersion problem [2]considering that the elements to be located aredisk-shaped That is place k disks within apolygonal domain maximizing the minimum dis-tance between any pair (see Figure 1(b))

2 Obtain the best location for k elements occupy-ing a specific space (disk of radius r) in a closedenvironment trying to minimize possible interac-tions within a radius R gt r That is to placek disks inside a polygonal domain such that thetotal overlap between the disks expanded to theradius R is minimized (see Figure 1(c))

To solve these problems we pre-compute a uniformgrid covering the polygon bounding box Its cells storewhether an r-radius disk centered in the cell centeris contained or not in the polygon The simulatedannealing perturbation tries to randomly move eachdisk away from its nearest disk using the informationpre-stored in the grid The strategy whose two stagesrun in parallel provides a good enough location forthe disks and can be easily extended to maximize thesum of the distances between them

References

[1] N Coll M Fort M Saus Parallel Simulated An-nealing for coverage area maximization in Book ofAbstracts of the XVIII Spanish Meeting on Compu-tational Geometry Girona 2019 13ndash16

[2] Z Drezner E Erkut Solving the Continuous p-Dispersion Problem Using Non-Linear ProgrammingThe Journal of the Operational Research Society46(4) (1995) 516ndash520

[3] E Erkut S Neuman Analytical models for locat-ing undesirable facilities Journal of Operational Re-search 40(3) (1989) 275ndash291

43


New variants of perfect non-crossing matchingslowast

Ioannis Mantasdagger1 Marko SavicDagger2 and Hendrik Schrezenmaiersect3

1Faculty of Informatics Universita della Svizzera italiana Lugano Switzerland2Department of Mathematics and Informatics Faculty of Sciences University of Novi Sad Novi Sad Serbia

3Institut fur Mathematik Technische Universitat Berlin Berlin Germany

Abstract

Given a set P of 2n points in R2 we are interestedin matching them with line segments We considerperfect (all points are matched) non-crossing (all linesegments are pairwise disjoint) matchings We studyboth the monochromatic setting ie all points areallowed to match and the bichromatic ie P is par-titioned into sets B and R with |B| = |R| = n andonly points of different sets are allowed to be matched

In the above settings a perfect non-crossing match-ing can always be found in O(n log n) time using ham-sandwich cuts Often though not any matching is suf-ficient and the interest lies in finding a matching withrespect to some optimization criterion Many criteriahave been considered a popular one is the MinMaxor bottleneck matching where the goal is to minimizethe length of the longest edge see eg [1 2 4 5]

We extend work by looking into three new optimiza-tion criteria namely MaxMin MinMin and Max-Max defined analogously to MinMax We considerthe input P in different configurations Our resultstogether with prior work are summarized in Table 1

a) b)

c) d)

Figure Optimal monochromatic a) MinMin b)MaxMax c) MinMax and d) MaxMin matchings

lowastThis work was presented in the 7th Annual InternationalConference on Algorithms and Discrete Applied Mathematics(CALDAM 2021) A full version of this work can be found inarXiv [3]daggerEmail ioannismantasusichDaggerEmail markosavicdmiunsacrssectEmail schrezenmathtu-berlinde

General Position Convex Circle

MinMin1 O(nh + n log n) O(n) O(n)

MaxMax1 O(n1+ε + n23h43 log3 n) O(n) O(n)MinMax1 NP-hard [1] O(n2) [4] O(n)MaxMin1 O(n3) O(n)

MinMin2 O(n) O(n)MaxMax2 O(n) O(n)MinMax2 NP-hard [2] O(n2) [5] O(n) [5]MaxMin2 O(n3) O(n3)

Table 1 Time needed to find a matching in different vari-ants Index 1 eg MaxMax1 refers to the monochro-matic variant and index 2 to the bichromatic h is thesize of the convex hull of P ε gt 0 is an arbitrarily smallconstant Results without reference are given in this work

Results When P is monochromatic we identifyan edge feasibility criterion verifiable in O(1) timeUsing this criterion we give two algorithms forMinMin1 and MaxMax1 one based on weak radialorderings and one on halfplane range queries

If P is convex all variants can be solved in O(n3)time using dynamic programming see eg [2] ForMinMin1 resp MaxMax1 we reduce the prob-lem to finding the shortest resp longest edge be-tween two convex polygons leading to O(n)-time al-gorithms We extend this algorithm to MinMin2resp MaxMax2 using the notion of orbits [5]

Finally in the special case that P lies on a circlewe provide an O(n)-time algorithm for MaxMin1

References

[1] A K Abu-Affash P Carmi M J Katz and Y Tra-belsi Bottleneck non-crossing matchings in the planeComputational Geometry 47(3A) 447ndash457 2014

[2] J G Carlsson B Armbruster S Rahul and HBellam A bottleneck matching problem with edge-crossing constraints Int Journal of ComputationalGeometry amp Applications 25(4) 245ndash261 2015

[3] I Mantas M Savic and H Schrezenmaier Newvariants of perfect non-crossing matchings arXivpreprint arXiv200103252 2020

[4] M Savic and M Stojakovic Faster bottleneck non-crossing matchings of points in convex position Com-putational Geometry 65 27ndash34 2017

[5] M Savic and M Stojakovic Structural properties ofbichromatic non-crossing matchings arXiv preprintarXiv180206301 2018

44


On Maximum-Sum Matchings of Points

S Bereg1 O Chacon-Rivera2 D Flores-Penaloza3 C Huemersect4 P Perez-Lanterodagger2 and C Searasect4

1University of Texas at Dallas USA2Universidad de Santiago de Chile (USACH) Chile3Universidad Nacional Autonoma de Mexico Mexico

4Universitat Politecnica de Catalunya Spain

Abstract

Huemer et al (Discrete Math 2019) proved that forany two point sets R and B with |R| = |B| the per-fect matching that matches points of R with pointsof B and maximizes the total squared Euclidean dis-tance of the matched pairs has the property that allthe disks induced by the matching have a commonpoint In this work we study the perfect matchingthat maximizes the total Euclidean distance Firstwe prove that this setting does not always ensure thecommon intersection property of the disks Secondwe extend the study for sets of 2n uncolored pointsAs the main result we prove that in this case all disksof the matching do have a common point

1 Introduction

Let R and B be two disjoint point sets in the planewith |R| = |B| = n n ge 2 The points in R are redand those in B are blue A matching of R cup B is apartition of RcupB into n pairs such that each pair con-sists of a red and a blue point A point p isin R and apoint q isin B are matched if and only if the (unordered)pair (p q) is in the matching For every p q isin R2 weuse pq to denote the segment connecting p and q andp minus q to denote its length which is the Euclideannorm of the vector pminusq Let Dpq denote the disk withdiameter equal to pminusq that is centered at the mid-point p+q

2 of the segment pq For any matchingM weuse DM to denote the set of the disks associated withthe matching that is DM = Dpq (p q) isinM

Huemer et al [4] proved that if M is any match-ing that maximizes the total squared Euclidean dis-tance of the matched points ie it maximizessum

(pq)isinM pminus q2 then all disks of DM have a point

Email besputdallaseduEmail oscarchacon pabloperezlusachclEmail dflorespenalozagmailcomsectEmail clemenshuemer carlossearaupcedu


in commonIn this article we consider the max-sum matching

E as the matching that maximizes the total Euclideandistance of the matched points For any matchingMlet cost(M) denote

sum(pq)isinM pminusq Thus E is such

that cost(E) is maximum among all matchings Asour first result we prove in Section 3 that every pairof disks in E have a common point but it cannot beguaranteed that all disks have a common point

We also consider max-sum matchings of sets of 2nuncolored points in the plane where a matching isjust a partition of the points into n pairs As the mainresult we prove in Section 4 that for any point set Pof 2n uncolored points in the plane and a max-summatching M of P all disks in DM have a commonintersection We use Hellyrsquos theorem that is we provethe claim for n = 3

Theorem 1 (Helly 1923) Let F be a finite familyof closed convex sets in Rn such that every subfamilyof n + 1 sets of F has nonempty intersection Thenall sets in F have nonempty intersection

The extended version of this article can be foundin [1] Any omitted proof and further details can befound there

2 Related works

Fingerhut [2] motivated by a problem in designingcommunication networks [3] conjectured that given aset P of 2n uncolored points in the plane and a max-sum matching (ai bi) i = 1 n of P there existsa point o of the plane not necessarily a point of P such that

ai minus o+ bi minus o le (2radic

3) middot ai minus bi (1)

for all i isin 1 n The statement of equa-tion (1) is equivalent to stating that the intersectionEa1b1 cap Ea2b2 cap middot middot middot cap Eanbn is not empty where Epq

is the region bounded by the ellipse with foci p andq and semimajor axis length (1

radic3) middot p minus q [2]

Then by Hellyrsquos Theorem it is sufficient to prove

45


equation (1) for n le 3 The factor 2radic

3 is the min-imum possible [2] it is enough to consider an equi-lateral triangle where at each vertex two points arelocated The max-sum matching of the six points ismade of pairs of vertex-opposed points and the re-gions defined by the three ellipses have exactly onepoint in common which is the center of the triangle

Eppstein [2] proved that the result holds with 25instead of 2

radic3 Let o be the midpoint of the shortest

edge in the matching Namely for all i isin 1 n

ai minus o+ bi minus o le 25 middot ai minus bi (2)

Our main result stating that for any point set Pof 2n uncolored points in the plane and a max-summatching E = (ai bi) i = 1 n of P all disksin DE have a common intersection implies that anypoint o in the common intersection satisfies

ai minus o+ bi minus o leradic

2 middot ai minus bi

for all i isin 1 n

3 Common intersection might fail for red-bluematchings

Lemma 2 Every pair of disks in DE have a non-empty intersection

Theorem 3 There exist disjoint point sets R cup Bwith |R| = |B| = 3 such that for any max-summatching E of R and B the intersection of the disksof DE is the empty set

Proof Let R = a b c and B = aprime bprime cprime witha = (minus1 0) b = (1 0) c = (0

radic3) cprime = (0 3)

and aprime isin bc and bprime isin ac such that c minus aprime =c minus bprime = ε for a parameter ε gt 0 that ensuresthat E = (a aprime) (b bprime) (c cprime) is the only maximummatching of R cupB (see Figure 1)

Note that

aminus bprime+ bminus cprime+ cminus aprime= aminus cprime+ bminus aprime+ cminus bprime=radic

10 + (2minus ε) + ε

= 2 +radic

10

and we need to ensure that

2 +radic

10 lt aminus aprime+ bminus bprime+ cminus cprime = cost(E)

That is the matching (a aprime) (b bprime) (c cprime) has largertotal Euclidean distance than (a bprime) (b cprime) (c aprime)and (a cprime) (b aprime) (c bprime) Since

aminus aprime+ bminus bprime+ cminus cprime= 2aminus aprime+ cminus cprime= 2aminus aprime+ (3minus

radic3)

gt 2(aminus c minus ε) + (3minusradic

3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime


it suffices to ensure

2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since

aminus cprime+ bminus bprime+ cminus aprime= aminus aprime+ bminus cprime+ cminus bprimelt (2 + ε) +

radic10 + ε

= 2 +radic

10 + 2ε

to ensure that (a aprime) (b bprime) (c cprime) has larger to-tal Euclidean distance than (a cprime) (b bprime) (c aprime) and(a aprime) (b cprime) (c bprime) it suffices to guarantee that

2 +radic

10 + 2ε lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)

Hence any ε gt 0 satisfying (4) (and then also (3)) issuch that E = (a aprime) (b bprime) (c cprime) is the only max-imum matching of R cup B It remains to show thatDaaprime capDbbprime capDccprime = empty To see that it is straightfor-ward to show (based on the fact that c isin Daaprime andc isin Dbbprime) that all points of Daaprime cap Dccprime have nega-tive x-coordinates and all points of Dbbprime capDccprime havepositive x-coordinates

Theorem 4 For any n ge 4 there exist disjoint pointsets R cup B with |R| = |B| = n such that for anymax-sum matching E of R and B the intersection ofthe disks of DE is the empty set

46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)

Figure 2 The ten different relative positions

4 Common intersection on monochromaticmatchings

It is not difficult to show that from Lemma 2 it fol-lows that the constant 25 can be improved to

radic5

for bichromatic maximum-sum matchings For themonochromatic case this bound can be further im-proved to

radic2 as we now show

Lemma 5 Let a b c d be a set of four pointssuch that (a b) (c d) is a max-sum matching ofa b c d and d is in the interior of ∆abc Then

d is in the interior of disk Dab ie ~cd points to ab

Since every pair of segments of a max-sum match-ing cross or one of them points to the other one weidentify ten cases of relative position of the three seg-ments as shown in Figure 2 listed from (A) to (J)

Lemma 6 If the segments of a max-sum matching ofsix points fall in one of the cases from (A) to (G)then the three disks of the matching have a commonintersection

Proof Let a b c aprime bprime cprime be a 6-point set and letM = (a aprime) (b bprime) (c cprime) be a max-sum matching

Case (A) At least one altitude of the triangle Tbounded by the three segments goes through the in-terior of T Let u be the vertex of such an altitudein a side of T By Thalesrsquo theorem each of the threedisks Daaprime Dbbprime and Dccprime contains u

Case (B) Let u be the intersection point betweenbbprime and ccprime If Dbbprime contains a then we are donesince a isin Dccprime because ~aprimea points to ccprime (Lemma 5)Similarly if Dccprime contains aprime then we are done sinceaprime isin Dbbprime because ~aaprime points to bbprime Otherwise ifa isin Dbbprime and aprime isin Dccprime then the triangle ∆aaprimeu issuch that the interior angles at a and aprime respectivelyare both acute Hence the altitude h from vertex ugoes through the interior of ∆aaprimeu and let v isin aaprime

be the other vertex of h Since ~aprimea points to ccprime and~aaprime points to bbprime each of the disks Daaprime Dbbprime andDccprime contains v by Thalesrsquo theorem The proof forCase (C) is both similar and simpler

a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8

Case (D) If Dbbprime contains cprime then we are done

since cprime isin Daaprime because ~ccprime points to aaprime (Lemma 5)Similarly if Dccprime contains b then we are done since b isinDaaprime because ~bprimeb points to aaprime Otherwise if cprime isin Dbbprime

and b isin Dccprime then the triangle ∆cprimebu is such that theinterior angles at cprime and b respectively are both acuteHence the altitude h from vertex u goes through theinterior of ∆cprimebu and let v isin cprimeb be the other vertexof h We have v isin Dbbprime cap Dccprime by Thalesrsquo theoremFurthermore since cprime b isin Daaprime we have that segmentcprimeb is contained in Daaprime Hence v isin DaaprimecapDbbprimecapDccprime

Cases (E) (F) and (G) In each of these casesthe same oriented segment points to each of the othertwo ones Say segment ~aaprime points to both bbprime and ccprimeThen we have that aprime isin Dbbprime cap Dccprime by Lemma 5Hence aprime isin Daaprime capDbbprime capDccprime

The following lemma guarantees that if we extendone segment by moving one of the points then the re-sulting segments correspond to a max-sum matchingof the resulting point set

Lemma 7 LetM = (ai bi) i = 1 n be a max-sum matching of the set P of 2n uncolored points andlet c isin P be a point such that b1 belongs to the interiorof the segment a1c Then Mlowast = (M (a1 b1)) cup(a1 c) is a max-sum matching of (P b1) cup c

Lemma 8 Let a b c aprime bprime cprime and z be seven pointssuch that c is to the left of line `(a b) segments ~aaprime~bbprime and ~ccprime point to bbprime ccprime and aaprime respectivelyand for each u isin a b c point z is to the left of line`(u uprime) and vectors uminus z and uprime minus z are orthogonalRefer to Figure 3 Then (a aprime) (b bprime) (c cprime) is nota max-sum matching of point set a b c aprime bprime cprime

Lemma 9 Let a b c aprime bprime cprime and z be seven pointssuch that c is to the left of line `(a b) segments ~aaprime

and ~bbprime point to bbprime and ccprime respectively segmentsaaprime and ccprime have a common point with a and aprime tothe right and left of line `(c cprime) respectively and foreach u isin a b c point z is to the left of line `(u uprime)and vectors uminus z and uprime minus z are orthogonal Refer toFigure 4 Then (a aprime) (b bprime) (c cprime) is not a max-sum matching of point set a b c aprime bprime cprime

47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9

Lemma 10 Let a b c aprime bprime cprime and z be sevenpoints such that none of them is to the right ofline `(a aprime) segments ~bprimeb ~bbprime and ~ccprime point to aaprimeccprime and aaprime respectively and for each u isin a b cpoint z is to the left of line `(u uprime) and vectors uminus zand uprime minus z are orthogonal Refer to Figure 5 Then(a aprime) (b bprime) (c cprime) is not a max-sum matching ofa b c aprime bprime cprime

Lemma 11 If the segments of a max-sum matchingof six points fall in one of the cases from (H) to (J)then the three disks of the matching have a commonintersection

Proof Suppose by contradiction that the threedisks denoted D1 D2 and D3 intersect pairwisebut without a common intersection Let u12 u23and u31 be the vertices of the pairwise disjoint lensesD1 capD2 D2 capD3 and D3 capD1 respectively locatedin the triangle with vertices the centers of D1 D2and D3 respectively

The idea is to use Lemma 7 in combination withLemmas 8 9 and 10 such that the point z of theselemmas is among u12 u23 and u31 To this endwe need to guarantee that point z is not an extremepoint of some segment of the matching

Note that two vertices among u12 u23 and u31

cannot be the extreme points of a same segment of thematching Furthermore if each of the three verticesis an extreme point of some segment of the matchingthen at least one pair of disjoint segments violatesLemma 5 That is the extreme point of one segmentin the interior of the convex hull of the four involvedpoints is not in the interior of the disk correspondingto the other segment Hence we can assume that atleast one vertex among u12 u23 and u31 is not anextreme point of a segment of the matching say ver-tex u12 This implies that we can extend the segmentof disk D3 by moving one of its extreme points suchthat the new three matching disks have a singletoncommon intersection at u12 Let z = u12 where z isdistinct from all the new six points

Let the new six points be denoted as a b caprime bprime and cprime in such a way that the new seg-

b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10

ments are precisely aaprime bbprime and ccprime and for eachu isin a b c point z is to the left of line `(u uprime) ByLemma 7 (a aprime) (b bprime) (c cprime) is a max-sum match-ing of a b c aprime bprime cprime

If aaprime bbprime and ccprime fall in Case (H) then byLemma 8 (a aprime) (b bprime) (c cprime) is not max-sumIf they are in in Case (I) then by Lemma 9(a aprime) (b bprime) (c cprime) is not max-sum Otherwise ifthey are in in Case (J) then by Lemma 10 we havethat (a aprime) (b bprime) (c cprime) is not max-sum There ex-ists a contradiction in each of the cases thus the orig-inal three disks must have a common intersection

Theorem 12 Let P be a set of 2n (uncolored) pointsin the plane Any max-sum matchingM of P is suchthat all disks of DM have a common intersection

Theorem 13 Let P be a set of 2n (uncolored) pointsin the plane and let (ai bi) i = 1 n be a max-sum matching of P Then there exists a point o ofthe plane such that for all i isin 1 n we have



References

[1] S Bereg O Chacon-Rivera D Flores-PenalozaC Huemer P Perez-Lantero and C Seara Onmaximum-sum matchings of points arXiv1911106102019

[2] D Eppstein Geometry Junkyard httpswwwicsuciedu~eppsteinjunkyardmaxmatchhtml

[3] J A Fingerhut S Suri and J S Turner Designingleast-cost nonblocking broadband networks J Algo-rithms 24(2)287ndash309 1997

[4] C Huemer P Perez-Lantero C Seara and R I Sil-veira Matching points with disks with a common in-tersection Discrete Mathematics 342(7)1885ndash18932019

48


Minimum Color Spanning Circle in Imprecise Domain

Ankush Acharyya1 Ramesh K Jallu1 Vahideh Keikha1 Maarten Loffler2 and Maria Saumell13

1The Czech Academy of Sciences Institute of Computer Science Czech Republicacharyyajallukeikhasaumellcscascz

2Department of Information and Computing Sciences Utrecht University Netherlands mloffleruunl3Department of Theoretical Computer Science Faculty of Information Technology Czech Technical University

in Prague Czech Republic

Abstract

Let R be a set of n colored imprecise points whereeach point is colored by one of k colors Each im-precise point is specified by a unit disk in which thepoint lies We study the problem of computing thesmallest and the largest possible minimum color span-ning circle among all possible choices of points insidetheir corresponding disks We present an O(nk log n)time algorithm to compute a smallest minimum colorspanning circle Regarding the largest minimum colorspanning circle we show that the problem is NP-Hardand present a 1

3 -factor approximation algorithm Weimprove the approximation factor to 1

2 for the casewhere no two disks of distinct color intersect

1 Introduction

Recognition of color spanning objects of optimumsize in the classical (precise) setting is a well-studiedproblem in the literature [1 2 3 5] The simplesttype of two-dimensional problem considered in thissetup is the minimum color spanning circle (MCSC)problem defined as follows Given a colored point setP in the plane such that each point in P is coloredwith one of k possible colors compute a circle of min-imum radius that contains at least one point of eachcolor (see Figure 1a) The minimum color spanningcircle can be computed in O(nk log n) time using theupper envelope of Voronoi surfaces [5]

In this work the exact coordinates of the inputpoints in P are unknown Instead we are given aset R = R1 R2 Rn of n unit disks of diame-ter 1 in the plane where each disk is colored withone of k possible colors A colored point set P is arealization of R if there exists a color-preserving bi-jection between P and R such that each point in P iscontained in the corresponding disk in R Each real-ization of R gives a MCSC of certain radius We areinterested in finding realizations of R such that thecorresponding MCSC has the smallest (S-MCSC) andlargest (L-MCSC) possible radius (see Figure 1bc)

Related work The motivation of the problem stemsfrom many real-life situations where the locations ofthe points are subject to errors and their exact coor-dinates are unknown Such a set of points is known asan imprecise or uncertain point set and the set of allpossible locations of a point is called its region [8 10]In the literature different variations have been con-sidered where the regions are modelled as simple ge-ometric objects such as disks or squares In particu-lar when the regions are modelled as disks Jadav etal [6] proposed an algorithm to compute the small-est enclosing circle that contains at least one pointfrom each region Robert and Toussaint [9] studiedthe problem of computing the smallest width corri-dor intersecting a set of convex regions Loffler andvan Kreveld [8] considered the problem of computingthe smallest and largest possible axis-parallel bound-ing box and circle of a set of regions modelled as cir-cles or squares Colored variations of other geometricproblems have also been studied in the context of im-precise points [4] To the best of our knowledge thereis no prior result on the minimum color spanning cir-cle problem for imprecise point sets

2 The smallest MCSC (S-MCSC) problem

Given a set R of n imprecise points modeled as unitdisks we present an algorithm that finds a S-MCSCdenoted by Copt and the realization of R achievingit Let ropt be the radius of Copt

Let C = c1 cn be the set of center points ofthe disks in R Let Cc be a MCSC of the colored set Cand let rc be its radius The following relation holds

Lemma 1 If rc gt12 then rc = ropt + 1

2

We compute rc If rc gt12 ropt = rcminus 1

2 Otherwisethe center of Cc lies in the intersection of k distinctcolored disks resulting in a MCSC of ldquozerordquo radius

Theorem 2 A smallest minimum color spanning cir-cle of R can be computed in O(nk log n) time

49


(a) (b) (c)

Figure 1 (a) MCSC for precise colored point set (b-c) S-MCSC and L-MCSC for imprecise colored point set

3 The largest MCSC (L-MCSC) problem

In this section we consider the L-MCSC problemwhere for the given set R the goal is to find a realiza-tion such that any MCSC is as large as possible Weshow that the problem is NP-Hard using a reductionfrom planar 3-SAT [7]

Given a planar 3-SAT instance we construct a set ofcolored unit disks with the following property Thereexists a realization such that any MCSC has diameterc if and only if the 3-SAT instance is satisfiable wherec = 9

8 We use disks of only two colors red and blueThus a point set having any MCSC of diameter atleast c is equivalent to saying that there is no red-blue pair of points at distance less than c We denotethe family of realizations with this property by Pc

cc

c c(a) (b) (c)

38

38

Figure 2 (a) A stack of disks (b-c) the placementsof points with red-blue distances equal to c

A stack of disks is a set of three vertically aligneddisks of alternating colors As shown in Figure 2a fora blue-red-blue stack of disks the distance betweenthe centers of the blue disks and the center of the reddisk is 3

8 For a realization of the stack of disks inPc it is easy to see that the following holds The redpoint can only be placed at one of the left or rightextreme positions of the red disk (see Figure 2bc)and such a placement forces the placement of pointsin the blue disks at distance c Thus there exists onlytwo possible valid placements shown in Figure 2bc

Variable Gadget Our variable gadget (see Figure 3)is an alternating chain of red and blue disks placed ona hexagonal tiling of the plane The distance betweenthe centers of two consecutive red and blue disks alongthe same edge of the hexagon is c Each edge ofthe hexagon contains two stacks of disks placed nearthe endpoints and every pair of consecutive edges isjoined by a blue disk In the following descriptionwe say that pi and pprimei are the leftmost and rightmostpoints of disk Ri if they are its leftmost and right-most points after the hexagon has rotated so that the

edge containing the center of Ri is horizontal and thecenter of the hexagon is below the edge

At the top-left corner of the variable gadgets thedisks are placed as follows (the other corners are con-structed similarly) Let Ri be the last disk in clock-wise order along the top-left edge of the hexagon andlet Rj and Rk be the first and second disks along thetop edge (see Figure 3) The placements of Rj andRk are fixed because their centers are at distance cRegarding Ri it is placed in such a way that the lowerblue disk of its stack contains a point z which is atdistance c from both pk and pi (see Figure 3) Noticethat if a realization in Pc chooses pprimei the choice forRj is not unique however none of the points in Rjat distance at least c from pprimei is compatible with thechoice of pk for Rk Therefore the choice of pprimei forcesthe choice of pprimek and clearly the choice of pk forcesthe choice of pi

c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej

Figure 3 A variable gadget with zoomed in view forthe top-left corner

For a realization in Pc of a variable gadget thefollowing holds The stack containing Rk is con-strained to choose either pk or pprimek Let us assumethat it chooses pprimek This choice propagates to the rightthrough the chain of disks in the top edge The reddisk of the stack on the right of the edge also choosesits rightmost point and this forces the red disk of the

50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi

Figure 4 A clause gadget (left) Connection gadgetfor a positive variable in a clause with truth value T(right)

first stack of the top right edge to choose its right-most point too Therefore the choice of pprimek propa-gates through the whole hexagon If Rk chooses pkthe same phenomenon occurs We conclude

Lemma 3 For any realization in Pc of a variablegadget either all disks centered at the edges of thehexagon except for the corners of the hexagon choosetheir rightmost point or they all choose their leftmostpoint

Clause gadget Clause gadgets are illustrated inFigure 4 (left) We consider an equilateral triangleof side 35 and center c0 and we place one red diskat every corner of the triangle in such a way that thecenter of the disk is aligned with c0 and its nearestcorner of the triangle Then we place a blue diskcentered at c0 Each red disk of a clause gadget is as-sociated to one of the literals occurring in the clauseand is connected to the corresponding variable gadgetvia a connection gadget Intuitively to decide if thereexists any realization in Pc each red disk Rτ of theclause gadget has essentially two relevant placementscalled tRτ

and fRτ(see Figure 4 (left)) As we will

see when the associated literal is set to true we canchoose the placement tRτ

and when it is set to falsewe are forced to choose fRτ It is easy to see thatthere exists a realization in Pc of the clause gadget ifand only if tRτ

is chosen for at least one of the disksRτ

Connection gadget A variable gadget is connectedto each of its corresponding clause gadgets with thehelp of a connection gadget A connection gadget con-sists of an alternating chain of red and blue disks to-gether with some stacks of disks (see Figure 4 (right))

The location of a connection (between a variablegadget and its associated clause gadget) depends onwhether the variable in the clause is positive or neg-ative For a positive variable the connection is es-tablished through a pair of a red disk Ri and a blue

RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj

Figure 5 Point-placements corresponding to (a) posi-tive and (b) negative variable in a clause at the inter-section of a connection gadget and a variable gadget

disk Rj which appear consecutive along an edge ofthe hexagon and such that none of them is a cornerof the hexagon and Ri comes before Rj in clockwiseorder (see Figure 5a) Let Rk be the first red disk ofthe connection gadget The top-most point of Rk isat distance 9

8 from pj and its bottom-most point is atdistance 9

8 from pprimej For a negative variable the con-nection is established through a pair of blue-red disksin the variable gadget The placement of the first reddisk of the connection gadget is analogous to the onein the red-blue configuration (see Figure 5b)

The truth value T of a variable is associated withthe choice of the rightmost points of the disks in thevariable gadget If the variable appears positive ata clause this allows the choice of the bottom-mostpoint of Rk and this propagates through the connec-tion gadget and eventually allows the choice of theassociated point tRτ

in the clause gadget If the truthvalue is F pj is selected which forces the choice of apoint in a close vicinity of the top-most point of Rkand eventually of fRτ

The analysis of the other casesare similar

The chain in a connection gadget might be bentby 120 (in standard coordinate system) maintain-ing the planarity and the distance constraint c Theplacement of disks at each bend of a connection gad-get is similar to the placements at the corners of avariable gadget Stacks of disks are used around thebends and next to the first disk Rk

In summary for a given planar 3-SAT instance weembed its dependency graph into a hexagonal gridand then replace the vertices by variable and clausegadgets which are connected using the connectiongadgets as described above Thus we have the fol-lowing

Lemma 4 The planar 3-SAT formula has a satisfyingassignment if and only if there exists a realization ofthe disks in Pc

Theorem 5 The problem of finding the largest min-imum color spanning circle of R is NP-Hard

51


Ri

pici

Q

Figure 6 The tilted grid We choose pi as the redcorner of Q contained in Ri

4 Approximation algorithms

In this section we provide approximation algorithmsfor the L-MCSC problem Let ropt denote the radiusof a largest possible minimum color spanning circle ofR We first prove bounds on ropt

Lemma 6 ropt ge 14

Proof It is easy to see that it is enough to prove theresult for the case where k = 2 We show that thebound holds when k = 2 by providing a realization Pwhose MCSC achieves the bound Consider a regularsquare grid rotated by π4 such that the side of everycell of the grid has length 12 We color the corners ofthe cells in red or blue in such a way that all cornerslying in some vertical line are colored red all cornerslying in the next vertical line are colored blue and soon (see Figure 6) Now let Ri isin R have red color andlet Q be the cell of the grid containing the center ofRi (if the center of Ri lies on an edge or vertex of thegrid we assign it to any of the adjacent cells) Noticethat at least one of the two red corners of Q lies insideRi We choose such a corner as pi isin P Similarly forevery Rj isin R of blue color P contains one of theblue corners of a cell containing the center of Rj Weobtain that P is a subset of the grid corners Sincethe distance between any pair of red and blue cornersis at least 12 the radius of any MCSC is at least14

Lemma 7 ropt le rc + 12

Input A set R of n unit disksOutput A MCSC of a realization of R withradius at least ropt3compute Ccif rc ge 14 then

return Ccelse

return a MCSC of P g

Algorithm 1 13 -factor approximation algo-

rithm for the L-MCSC problem

Let P g denote the realization of R described in theproof of Lemma 6 Our algorithm to compute theapproximate L-MCSC is presented in Algorithm 1

Theorem 8 A 13 -factor approximation for the

L-MCSC problem can be computed in O(nk log n)time If no two distinct colored disks of R intersectthe approximation factor becomes 1

2

Acknowledgements

The authors would like to thank Irina Kostitsyna forkey discussions on the hardness reduction and HansRaj Tiwary for the proof of Lemma 6 AA RJ VKand MS are supported by the Czech Science Foun-dation grant number GJ19-06792Y and with insti-tutional support RVO67985807

References

[1] A Acharyya S C Nandy and S Roy Minimumwidth color spanning annulus Theor Comput Sci72516ndash30 2018

[2] S Das P P Goswami and S C Nandy Small-est color-spanning object revisited Intl J ComputGeom Appl 19(05)457ndash478 2009

[3] M de Berg J Gudmundsson M J Katz C Lev-copoulos M H Overmars and A F van der Stap-pen TSP with neighborhoods of varying size JAlgorithms 57(1)22ndash36 2005

[4] M Dror and J B Orlin Combinatorial optimizationwith explicit delineation of the ground set by a collec-tion of subsets SIAM J Discret Math 21(4)1019ndash1034 2008

[5] D P Huttenlocher K Kedem and M Sharir Theupper envelope of Voronoi surfaces and its applica-tions Discret Comput Geom 9(3)267ndash291 1993

[6] S Jadhav A Mukhopadhyay and B BhattacharyaAn optimal algorithm for the intersection radius of aset of convex polygons J Algorithms 20(2)244ndash2671996

[7] D Lichtenstein Planar formulae and their usesSIAM J Comput 11(2)329ndash343 1982

[8] M Loffler and M van Kreveld Largest boundingbox smallest diameter and related problems on im-precise points ComputGeom 43(4)419ndash433 2010

[9] J-M Robert and G Toussaint Computational ge-ometry and facility location In Proc InternationalConference on Operations Research and ManagementScience pages 11ndash15 1990

[10] D Salesin J Stolfi and L Guibas Epsilon geometrybuilding robust algorithms from imprecise computa-tions In SoCG pages 208ndash217 ACM 1989

52


A discrete isoperimetric inequality

David Iglesiaslowast Eduardo Lucasdagger and Jesus Yepes NicolasDagger

Departamento de Matematicas Universidad de Murcia Campus de Espinardo 30100 Murcia Spainsect

1 Introduction

The isoperimetric inequality in its classical form datesback to antiquity and states that circles are the onlyclosed plane curves minimizing the length for a pre-scribed enclosed area This can be succinctly ex-pressed as L2 ge 4πA where L is the length of thecurve and A is the enclosed area

This result was eventually generalized to arbitrarydimension in the 19th century Its form for convexbodies in Rn can be stated by saying that the volumevol(middot) and surface area S(middot) (Minkowski content) ofany n-dimensional convex body K satisfy(

S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)

where Bn denotes the Euclidean (closed) unit ball

2 Discretizing the isoperimetric inequality

In order to discretize the isoperimetric inequality wemay consider the following ldquoneighbourhood formrdquo forany n-dimensional convex bodies KE sub Rn and allt ge 0 we have

vol(K + tE) ge vol(rE + tE) (2)

where r gt 0 is such that vol(rE) = vol(K) Theisoperimetric inequality (1) is equivalent to (2) forE = Bn The advantage of using the volume of aneighbourhood of K instead of its surface area isthat it can be extended to other spaces in which thelatter notion makes no sense

Recently in [2] a discrete isoperimetric inequalityhas been derived for the integer lattice Zn endowedwith the Linfin norm and the cardinality measure | middot |To this aim a suitable extension of lattice cubes (iethe intersection of cubes [a b]n with Zn) is consid-ered they define a well-order in Zn which essentiallyconcentrates the points around the origin and con-sider the initial segments in that order ie the first

lowastEmail davidiglesiasumesdaggerEmail eduardolucasumesDaggerEmail jesusyepesumessectThe work is partially supported by MICINNFEDER

project PGC2018-097046-B-I00 and by ldquoPrograma de Ayudasa Grupos de Excelencia de la Region de Murciardquo FundacionSeneca 19901GERM15

1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23

Figure 1 The extended lattice cube I23 in Z2

(left) and the corresponding extended cube C23 in R2

(right)

r points in the order We will call these sets extendedlattice cubes Ir A certain modificaction of these setsthe so-called extended cubes Cr (cf Figure 1) will playa crucial role

3 Main results

In this talk we study an analogue of the discreteisoperimetric inequality obtained in [2 Theorem 1]in the setting of arbitrary non-empty bounded sets inRn endowed with (the Linfin norm and) the lattice pointenumerator Gn(K) = |K cap Zn| In this way one mayconsider neighbourhoods of a given set at any distancet ge 0 not necessarily integer

Theorem 1 ([1 Theorem 12]) Let K sub Rn be abounded set with Gn(K) gt 0 and let r isin N be suchthat Gn(Cr) = Gn(K) Then for all t ge 0

Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)

Theorem 2 ([1 Theorem 14]) The discrete iso-perimetric inequality (3) implies the isoperimetric in-equality (2) with E = [minus1 1]n for non-empty com-pact sets

References

[1] D Iglesias E Lucas and J Yepes Nicolas On dis-crete Brunn-Minkowski and isoperimetric type in-equalities Submitted

[2] A J Radcliffe and E Veomett Vertex IsoperimetricInequalities for a Family of Graphs on Zk Electr JComb 19 (2) (2012) P45

53


Algorithmic geometry with infinite time computation

Clemens Huemer1 Moritz Muller1 Carlos Seara1 and Adrian Tobar Nicolausect1

1Universitat Politecnica de Catalunya

This abstract reports results of the fourth authorrsquosMaster Thesis [4] We present an algorithmic study ofproblems from computational geometry with count-ably infinite input especially countable sets in RnTo do so we use the infinite time Blum-Shub-Smale(ITBSS) machine due to Koepke and Seyfferth [2]which is an extension of the classical Blum-Shub-Smale (BSS) machine [1] to transfinite ordinal timeA BSS-machine works with a finite number of regis-ters each holding a real number computation stepsupdate these reals by applying a rational function ortest the positivity of some register see [1] for detailsEquivalently BSS-machines can be seen as unit-costTuring machines over R as an ordered field in the senseof [3] An ITBSS machine extends the computationsto transfinite time at a limit ordinal time each reg-ister content is updated to the limit of the registercontents The computation breaks if the contents ofsome register do not converge We refer to [2] fordetails We remark that ITBSS machines model lsquofea-siblersquo or lsquoefficientrsquo computations and refer to [5] for adiscussionIn order to treat geometric problems with the ITBSSmachine one needs to verify that basic operationssuch as storage access and modifications of count-able point sets can be done One way of handlingthis is explained in the thesis [4] We also refer thereader to [4] for the detailed exposition of the follow-ing problemsbull The accumulation points problem Accumula-tion points as limit points of a set in Rn are a properproblem of infinite input computation The accumu-lation points problem that we study is how to find theaccumulation points of a set in Rn The main diffi-culty that emerges from this problem is that the ac-cumulation points of a set do not need to be elements

Email clemenshuemerupcedu Research supported byPID2019-104129GB-I00 AEI 1013039501100011033 and byGen Cat DGR 2017SGR1336

Email moritzcsupceduEmail carlossearaupcedu Research supported by

PID2019-104129GB-I00 AEI 1013039501100011033 and byGen Cat DGR 2017SGR1640

sectEmail adriantobarupces


in the set otherwise an easy computation could checkfor each point in the set if it is an accumulation pointand give the solution We propose a solution for thisproblem when the input set contains a finite numberof accumulation points

Theorem 1 There is an ITBSS machine that givenas input an at most countable set of points in R2 withfinitely many accumulation points computes these ac-cumulation points

bull The convex hull problem The convex hullproblem studies how to find the intersection of allconvex sets containing a given input point set It isnot always possible to report all the points on theboundary of the convex hull with the ITBSS machinebecause this boundary can be uncountable To over-pass this limitation we use a countable family of half-spaces with common intersection the solution set asa method to report the closure of the convex hull Arational halfspace in Rn can be represented by a pointin Qn+1 By computing a set of halfspaces of Rn wemean computing a set of their representations

Theorem 2 There is an ITBSS-machine that givenas input an at most countable set X sube Rn computesan at most countable set of rational halfspaces whoseintersection is the closure of the convex hull of X

References

[1] L Blum On a theory of computation and complexityover the real numbers NP-completeness recursivefunctions and universal machines Bull Amer MathSoc 21 (1989) 1ndash46

[2] P Koepke Towards a theory of infinite time Blum-Shub-Smale machines in How the world com-putes Lecture Notes in Computer Science vol 7318Springer Berlin Heidelberg 2012 405ndash415

[3] B Poizat Les petits cailloux Une approche modele-theorique de lrsquoalgorithmie vol 3 Nur al Mantiq wal-Marsquorifah Aleas Lyon 1995

[4] A Tobar Nicolau Algorithmic Geometry with Infi-nite Time Computation Master Thesis UniversitatPolitecnica de Catalunya 2020 httpsupcommonsupceduhandle2117328100

[5] P Welch Discrete transfinite computation in Tur-ingrsquos Revolution Birkhauser Cham 2015 161-185

54


Pattern recognition of homogenized standard sets of image patternsarising from Latin squares

Raul M Falconlowast1

1Departamento de Matematica Aplicada I Universidad de Sevilla Spain

Abstract

In 2007 Dimitrova and Markovski described a graph-ical representation of quasigroups by means of fractalimage patterns It is based on the construction ofpseudo-random sequences arising from the multipli-cation table of a quasigroup that is from a Latinsquare In particular isomorphic quasigroups giverise to the same fractal image pattern up to permu-tation of underlying colors This possible differencemay be avoided by homogenizing the standard setsrelated to these patterns Based on the differentialbox-counting method the mean fractal dimension ofhomogenized standard sets constitutes a Latin squareisomorphism invariant which is analyzed in this paperin order to distribute Latin squares of the same orderinto isomorphism classes

1 Introduction

A Latin square of order n is an ntimesn array with entrieschosen from a set of n distinct symbols so that norepetition of symbol exists in the same row or in thesame column From here on let Ln denote the set ofLatin squares of order n based on the set of symbols[n] = 1 n Every Latin square L = (lij) isin Lnis uniquely identified with its set of entries

Ent(L) = (i j lij) 1 le i j le n

Let Sn be the symmetric group on the set [n] Everypermutation π isin Sn acts on the Latin square L bygiving rise to its isomorphic Latin square Lπ isin Lnwhere Ent(Lπ) = (π(i) π(j) π(lij)) 1 le i j le nAs such the permutation π is a Latin square isomor-phism To be isomorphic is an equivalence relationamong Latin squares Currently the distribution ofLatin squares into isomorphism classes is only known[7] for order n le 11 In order to deal with higherorders new Latin square isomorphism invariants arebeing introduced in the recent literature [2 6 14]This paper delves into this topic by focusing on themean fractal dimension of the homogenized standardset of image patterns associated to any given Latinsquare

lowastEmail rafalganuses

Every Latin square in Ln constitutes the multiplica-tion table of a quasigroup ([n] middot) where middot is a binaryoperation on the set [n] so that both left and rightdivisions are feasible Two quasigroups are isomor-phic if and only if their associated Latin squares areIn 1997 Markovski et al [9] (see also [10 11] pro-posed the construction of pseudo-random sequencesarising from a quasigroup ([n] middot) and a plaintext T =t1 tm with m isin N and ti isin [n] for all i le mMore specifically for each s isin [n] it is defined as theencrypted string Es(T ) = e1 emminus1 where

ei =

s middot t1 if i = 1

eiminus1 middot ti otherwise

In 2007 Dimitrova and Markovski [3] realized that aniterative implementation of this encryption describesa graphical representation of quasigroups by means ofimage patterns with a certain fractal character Morespecifically if r ge 2 is a positive integer and S =(s1 srminus1) is an (r minus 1)-tuple of positive integersin the set [n] then the rtimesm image pattern based onthe multiplication table L of a quasigroup ([n] middot) isthe r timesm array (pij) such that for each j le m

pij =

tj if i = 1

siminus1 middot piminus11 if i gt 0 and j = 1

pijminus1 middot piminus1j otherwise

Each one of its cells constitutes a pixel of the im-age pattern under consideration The symbol withineach pixel may uniquely be identified with a color of agiven palette of n colors Further the image patternjust described is called s-standard [4] with s isin [n]if S is the constant (r minus 1)-tuple (s s) and T isthe constant plaintext s s of length m From hereon let Prms(L) denote this array (It is denotedPrs(L) when r = m) The standard set of r times mimage patterns associated to the Latin square L isthe set Prms(L) s isin [n] It is so that isomorphicquasigroups give rise to the same standard set of rtimesmimage patterns up to permutation of underlying col-ors Due to it the recognition and analysis of imagepatterns based on quasigroups have recently arising asan efficient new approach for classifying quasigroupsand related structures into isomorphism classes [4 5]

55


Furthermore the computational analysis of thesepatterns enables the distribution of quasigroups intofractal and non-fractal classes Particularly fractalquasigroups have turned out to play a relevant rolefor designing error detecting codes [8] whereas non-fractal quasigroups are recommended for designingcryptographic primitives [1 12]

The paper is organized as follows Section 2 dealswith the concept of homogenized standard sets ofr times m image patterns which enables one to avoidthe possible difference of colors in standard sets basedon isomorphic quasigroups Based on the differen-tial box-counting method and in order to distinguishhomogenized standard sets of image patterns arisingfrom non-isomorphic quasigroups it is introduced thenotion of mean fractal dimension of any given homog-enized standard set

2 Homogenized standard sets of image patterns

Let L isin Ln be the Cayley table of a quasigroup Inaddition let Pn = c1 cn be a palette of n dis-tinct colors so that the gray-level intensity of the colorci is in (In this way the color cn is always white)Then an s-standard rtimesm image pattern Prms(L) issaid to be homogenized if the colors of the palette Pn

appear in natural order (according to their intensity)when the image pixels are read row by row then col-umn by column In addition a standard set of r timesmimage patterns based on L is said to be homogenizedif all its r times m image patterns are homogenized Inthis way isomorphic Latin squares give rise to exactlythe same homogenized standard set of r times m imagepatterns with respect to a given palette

Example 1 Let us consider the quasigroups havingas respective multiplication tables the following threeLatin squares in L4

1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3

The 3times4 collage in Figure 1 shows their homogenizedstandard sets of 90times90 image patterns The cell (i j)represents the j-standard 90times90 image pattern of theLatin square Li for all i le 3 and j le 4 C

Figure 1 enables us to ensure visually that L3 is iso-morphic neither to L1 nor to L2 However it is not soobvious that the homogenized standard sets of L1 andL2 are distinct In order to distinguish homogenizedstandard sets we study the fractal dimension of theirimage patterns

Figure 1 Homogenized standard sets of 90 times 90 im-age patterns of the Latin squares L1 (upper row) L2

(second row) and L3 (lower row)

Let Hrm(L) denote the homogenized standard setof rtimesm image patterns of a Latin square L isin Ln (Itis denotedHr(L) when r = m) Further let Div(rm)be the set of common divisors of both parameters rand m Since each positive integer k isin Div(rm) isin compliance with the image dimensions of every im-age pattern Prms(L) isin Hrm(L) with s isin [n] it isalways possible to cover the latter by an r

k timesmk grid

formed by two-dimensional boxes of side length k LetIijk(Prms(L)) denote the range of gray-level inten-sities within the region of Prms(L) bounded by thecell (i j) of that grid Then we consider the value

Ik(Prms(L)) =sum

(ij)isin[ rk ]times[m

k ]

(1 + Iijk(Prms(L)))

Based on the differential box-counting method [13] fordetermining the fractal dimension of a given grayscaleimage let us define the differential box-counting frac-tal dimension DB(Prms(L)) of the image patternPrms(L) as the slope of the linear regression line ofthe set of points

(ln(Ik(Prms(L))) ln(1k)) k isin Div(rm)

In addition let DB(Hrm(L)) denote the mean valueof this fractal dimension averaged over all the positiveintegers k isin Div(rm) It constitutes the mean fractaldimension of the homogenized standard set Hrm(L)The following result follows straightforwardly

Proposition 2 Let L1 and L2 be two Latin squaresin Ln If DB(Hrm(L1)) 6= DB(Hrm(L2)) for somepositive integers r and m then L1 and L2 are notisomorphic

Table 1 enumerates both the differential box-counting dimension and the mean fractal dimension ofeach one of the three homogenized standard sets de-scribed in Example 1 Notice in particular that theirmean fractal dimensions are pairwise distinct whichenables one to ensure that the Latin squares L1 L2

and L3 correspond to different isomorphism classes

56


LL1 L2 L3

DB(P901(L)) 200000 200000 200000DB(P902(L)) 195165 195165 192136DB(P903(L)) 18877 188873 192331DB(P904(L)) 18877 188873 190088DB(H90(L)) 19317625 19322775 19363875

Table 1 Differential box-counting and mean fractaldimensions of the homogenized standard sets of 90times90 image patterns described in Example 1

It is readily verified that all the isomorphism classesof Latin squares of order n le 4 are indeed character-ized by their corresponding mean fractal dimensionof homogenized standard sets of 90 times 90 image pat-terns Figure 2 illustrates their values in increasingorder for the five isomorphism classes of the set L3

and the 35 isomorphism classes of the set L4 (No-tice the existence of only one isomorphism class forall n isin 1 2)

Figure 2 Mean fractal dimensions of the homogenizedstandard sets of 90 times 90 image patterns of each iso-morphism class of Latin squares of order n isin 3 4

Notice in particular the existence of exactly oneisomorphism class associated to the maximum meanfractal dimension 2 in each one of the sets L3 and L4Their representatives are the Latin squares

1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4

Both of them are multiplication tables of idempotentquasigroups That is the cell (i i) contains the sym-bol i for all i isin [n] In fact the following result isreadily verified

Proposition 3 The mean fractal dimension of thehomogenized standard set of r times m image patternsbased on the multiplication table of an idempotentquasigroup is 2 for every pair of positive integers rand m

The existence of non-isomorphic idempotent quasi-groups of order five implies therefore that the meanfractal dimension is not definitive for characterizingisomorphism classes of Latin squares of higher or-ders It is the case of the following two non-isomorphicLatin squares in L5

1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5

Concerning the computational efficiency of usingthe mean fractal dimension as Latin square isomor-phism invariant notice that the maximum runningtime that is required to compute any of the mean frac-tal dimensions associated to Figure 2 is less than onesecond in an Intel Core i7-8750H CPU (6 cores) witha 22 GHz processor and 8 GB of RAM In the samecomputer system the mean fractal dimension of thehomogenized standard sets of 90times 90 image patternsassociated to the Latin square of order 256 describedin [4] is obtained in 81 63 seconds Its mean fractal di-mension is 1 88926 It is therefore computationallyfeasible to make use of this new invariant to deal withthe possible characterization of isomorphism classesof Latin squares of order n = 256 which are the mostcommonly used in the literature for designing codesand cryptographic primitives

3 Image patterns arising from random Latinsquares

Let us focus now on the problem of distributing ran-dom Latin squares into isomorphism classes by mak-ing use of the mean fractal dimension described inthe previous section To this end we choose the ran-domization method described in [2] which consistsof sequentially adding a set of feasible random en-tries to an empty n times n array until a Latin square isreached The computation of the mean fractal dimen-sion of the homogenized set of 90times90 image patternsof each one of these random Latin squares allows todistinguish non-isomorphic classes among them Bymeans of this procedure the five isomorphism classesof L3 have been obtained after six attempts Fur-thermore the 35 isomorphism classes of L4 have beenobtained after 326 attempts Figure 3 illustrates thecomputational progression for obtaining such classesin this last case In a similar way Figure 4 illustratesthe case n = 5 After 20 000 attempts 1 404 of the1 411 isomorphism classes have been distinguishedAll the mean fractal dimensions under considerationhave been obtained in less than one second by ourcomputer system

57


Figure 3 Computational progression for obtainingthe 35 isomorphism classes of L4

Figure 4 Computational progression concerning theobtention of isomorphism classes of L5

4 Conclusion and further work

The recognition and analysis of standard sets of im-age patterns associated to Latin squares has recentlyarising as an efficient way for distinguishing even vi-sually distinct isomorphism classes of Latin squaresThis paper has dealt with the concept of homoge-nized standard sets which avoids possible discrepan-cies concerning the underlying colors of these imagesBased on the differential box-counting method themean fractal dimension of these homogenized stan-dard sets turns out to be an efficient invariant fordistributing Latin squares into isomorphism classesIt has been shown to be computationally feasible fordealing with Latin squares of order n = 256 whichare of particular interest in Cryptography The studyof these Latin squares is a subject of future work As apreliminary stage the computational progression forobtaining the isomorphism classes of Latin squares oforder n le 5 has been shown by computing the meanfractal dimension of random Latin squares Computa-tional experiments concerning higher orders are cur-rently in progress Finally the study of algebraic andcombinatorial properties of those isomorphism classesof Latin square whose standard sets of image patternsare associated to the same mean fractal dimension issubject of future work

Acknowledgements

The author wants to express his gratitude to theanonymous referees for their pertinent comments andsuggestions which helped improve the manuscript

References

[1] V Bakeva A Popovska-Mitrovikj D MechkaroskaV Dimitrova B Jakimovski V Ilievski Gaussianchannel transmission of images and audio files usingcryptcoding IET Commun 13 (2019) 1625ndash1632

[2] E Danan R M Falcon D Kotlar T G MarbachR J Stones Refining invariants for computing au-totopism groups of partial Latin rectangles DiscreteMath 343 (2020) article 111812 21 pp

[3] V Dimitrova S Markovski Classification of quasi-groups by image patterns in Proceedings of the FifthInternational Conference for Informatics and Infor-mation Technology Bitola Macedonia 2007 152ndash160

[4] R M Falcon Recognition and analysis of image pat-terns based on Latin squares by means of Compu-tational Algebraic Geometry Mathematics 9 (2021)paper 666 26 pp

[5] R M Falcon V Alvarez F Gudiel A Compu-tational Algebraic Geometry approach to analyzepseudo-random sequences based on Latin squaresAdv Comput Math 45 (2019) 1769ndash1792

[6] R M Falcon R J Stones Partial Latin rectanglegraphs and autoparatopism groups of partial Latinrectangles with trivial autotopism groups DiscreteMath 340 (2017) 1242ndash1260

[7] A Hulpke P Kaski P R J Ostergard The numberof Latin squares of order 11 Math Comp 80 (2011)1197ndash1219

[8] N Ilievska V Bakeva A model of error-detectingcodes based on quasigroups of order 4 in Proceed-ings of the Sixth International Conference for Infor-matics and Information Technology Bitola Macedo-nia 2008 7ndash11

[9] S Markovski D Gligoroski S Andova Using quasi-groups for one-one secure encoding in Proceedingsof the Eight Conference Logic and Computer Science(LIRA) Novi Sad Serbia 1997 157ndash162

[10] S Markovski D Gligoroski V Bakeva Quasigroupstring processing Part 1 Contributions Sec MathTech Sci MANU XX 1-2 (1999) 13ndash28

[11] S Markovski V Kusakatov Quasigroup string pro-cessing Part 2 Contributions Sec Math Tech SciMANU XXI 1-2 (2000) 15ndash32

[12] A Popovska-Mitrovikj S Markovski V BakevaSome new results for random codes based on quasi-groups in Proceedings of the Tenth Conferencefor Informatics and Information Technology (CIIT2013) Bitola Macedonia 2013 178ndash181

[13] N Sarkar B B Chaudhuri An efficient differentialbox-counting approach to compute fractal dimensionof image IEEE Trans Syst Man Cybern 24 (1994)115ndash120

[14] R J Stones R M Falcon D Kotlar T G MarbachComputing autotopism groups of partial Latin rect-angles J Exp Algorithmics 25 (2020) article 11239 pp

58

EGC21 Author Index

Author Index

Acharyya Ankush 49Aichholzer Oswin 4Alegrıa Carlos 22Almendra-Hernandez Vıctor Hugo 11

Bereg Sergey 45Blasco Fernando 3Bose Prosenjit 7Buchin Kevin 1

Cabello Sergio 31Canton Alicia 41 42Cardinal Jean 16Chacon-Rivera Oscar 45Claverol Merce 23 37Coll Narcis 43

de Las Heras Parrilla Andrea 23

Esteban Guillermo 7

Fabila-Monroy Ruy 36Falcon Raul 55Fernandez-Fernandez Encarnacion 15Fernandez-Jambrina Leonardo 41 42Flores-Penaloza David 45Fort Marta 43

Ganian Robert 5Garcia Alfredo 4Garijo Delia 6

Hamm Thekla 5Herrera Luis H 37Hidalgo-Toscano Carlos 36Huemer Clemens 23 45 54

Iglesias Lopez David 53

Jallu Ramesh K 49

Keikha Vahideh 49Klute Fabian 5

59

EGC21 Author Index

Loffler Maarten 49Lucas Marın Eduardo 53

Mantas Ioannis 22 44Marquez Alberto 6Martınez-Moraian Alejandra 15 23 32Martınez-Sandoval Leonardo 11Muller Moritz 54

Orden David 7 15 32

Padrol Arnau 21Papadopoulou Evanthia 2 22Parada Irene 5Perz Daniel 36Pfeifle Julian 17Pilaud Vincent 21Poullot Germain 21Pournin Lionel 16Perez-Lantero Pablo 27 37 45

Rosado Marıa Marıa Eugenia 41

Saumell Maria 49Savic Marko 22 44Schrezenmaier Hendrik 22 44Seara Carlos 22 27 37 45 54Silveira Rodrigo 6 7Suderland Martin 22

Tejedor-Romero Marino 15Tejel Javier 4Tobar Nicolau Adrian 54

Valencia-Pabon Mario 16Vogtenhuber Birgit 4 5 36Vazquez-Gallo Marıa Jesus 41 42

Weinberger Alexandra 4

Yepes Nicolas Jesus 53

60

Portada

Actas_sin_portadapdf

Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results

ratio in for hexagonal cells

Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction

The non-realizability of Zhengs 3-sphere

Finding positive GrassmannndashPluumlcker trees

More results

Topological Prismatoids

Novik and Zhengs centrally symmetric neighborly d-spheres

paper_09

paper_10

Introduction

RVD in the plane

RVD of a convex polygon - Brocard illumination

paper_11

Introduction

Properties of the labeling of Vk(S)

Not too many alternating hexagons in V3(S)

paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction

The maximum number of connected rectilinear convex 4-gons

Concluding comments

paper_15

paper_16

Preliminaries

Upper and lower bounds

The -wedge depth with respect to P

Report such that DepthP(pi)=k

The (0k0)-hulls

The (0k0)-hulls for points in convex position

Computing the (01)-hull(P)

The (0k0)-hulls for points in general position

paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works

Common intersection might fail for red-blue matchings

Common intersection on monochromatic matchings

paper_22

Introduction

The smallest MCSC (S-MCSC) problem

The largest MCSC (L-MCSC) problem

Approximation algorithms

paper_23

Introduction

Discretizing the isoperimetric inequality

Main results

paper_24

paper_25

Introduction

Homogenized standard sets of image patterns

Image patterns arising from random Latin squares

Conclusion and further work

2_author_index

EGC21 preface

Preface

This book contains the abstracts of the invited talks contributed papers and contributed talksaccepted for presentation at the XIX Spanish Meeting on Computational Geometry (formerlyEncuentros de Geometrıa Computacional EGC) The current edition was organized in MadridSpain but held online on July 5-7 2021 due to the COVID-19 pandemic

This series of meetings focuses on current research topics in discrete and computationalgeometry Since the seminal edition in 1990 the Encuentros have combined a strong scien-tific program with a friendly atmosphere The intended audience ranges from experiencedresearchers to students facing their debut in the area The strong collaboration links of theSpanish community with foreign colleagues made advisable in 2011 to change the language ofthe meeting to English and have the submissions peer-reviewed by an international programcommittee

In this edition following the experience started in the previous one two lengths were possiblefor the contributed submissions 4 pages (rdquopaperrdquo) or 1 page (rdquotalkrdquo) We received a total of26 submissions consisting of 16 talks and 10 papers Among them one was withdrawn beforereviewing and the other 25 were finally accepted composing the core of this book Threerevisions were collected for all the submissions except for one passing two reviews

As each of them the current edition of the meeting is the result of the work and dedication ofa lot of people First thanks go to the authors for choosing EGC to share and disseminate theirwork Second I would like to thank the members of the program committee and the externalreviewers for accepting to contribute their expertise to this meeting carefully constructivelyand on time Thirdly I am truly thankful to the three excellent invited speakers for acceptingour invitation Fernando Blasco Kevin Buchin and Evanthia Papadopoulou Finally it is veryappreciated the support of the Departamento de Matematica Aplicada a las Tecnologıas de laInformacion y las Comunicaciones and the Escuela Tecnica Superior de Ingenierıa de SistemasInformaticos from the Universidad Politecnica de Madrid although this edition would not havebeen possible without the tough work of the organizing committee Guillermo Esteban JesusGarcıa and Alejandra Martınez Having to deal with the uncertainty of these times they didtheir best for the success of this meeting

July 2021Alcala de Henares

David OrdenProgram Committee Chair

i


Program Committee


ii




iii


Table of Contents

Initial sections

Preface i




Invited talks


Kevin Buchin




Fernando Blasco



















Julian Pfeifle

iv













Sergio Cabello



















v










Raul Falcon

Final section

Author Index 59

vi



Kevin Buchin



1








2





Abstract




3
















log logn )




I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r









References





4























References



5















6 (n2 minus n)


References





6






Abstract



1 Introduction




v


v








7


ω1 =radic13

ω2 = 1s

t





2 Previous results



radic2













s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4








vi3 Xi = (vi1 vi2 v


(vi1 vi2 v

i3 v




i2)


i2 v

i3) see Figure


i2 v

i3 v

i4)


i6) (vi6 v

i1) are symmetric







8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi





k 1 le k le 6if






P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ



3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q





k 1 le k le 5if


along k edges of Hm






spectively

9






5 m ge i Then an


when m = i


1







4 and a P 15 con-







1 and a P 22








2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13







2

4 Conclusions




References













10





Abstract



1 Introduction



Dn =

([n]

2



P1

P2

P3

P4













2 Lower bound


11


P1 P2 P3 P4

Figure 2 Case n = 4


2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

















lt lt





P1 P2

P3

P4

P5

Π = Π3

Π1

Π2





12










(nminus1

2

)+ nminus 3 lt

(n2

)


3 Upper bound




gij =1

2(m2


2ij)





Q1 lt Q2 lt lt Qm



mij =

0 if i = j




gij =1

2(m2

ni +m2nj minusm2

ij)



2Grarr




the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i



13












(n2

)





mij =

0 if i = j












4 Discussion







References





14














References





15











References













16



Julian Pfeifle


Abstract



1 Introduction



+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]







0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]


0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]





17








-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]


0 = Γ(18f|56bd)


= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]




0 = Γ(1bf|48de)







[36fb5]

([36fb4]

([14bf3]

([16f85]


+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)


)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)






18


minusΓ(18f|46be)

Γ(18f|56bd)


minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]


0 = [36fb5]

([36fb4]

([14bf3]

([16f85]




))+ [16f84] [14bf8] [16f85]


))+ [16f84] [14bf8] [14bfe] [16f85]


))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]








VΣ =


unknown solids


(VΣ

2












4 More results


19


minsum

ΓisinVΣ

xΓ st 2sum



2

)(2)



2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)


















References


20191641766







201211500 2020






20











References









21







1 Introduction





2 RVD in the plane











References




22





Abstract


1 Introduction






by PID2019-104129GB-I00 AEI 1013039501100011033










23










1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37


Vk(S)

Vk+1(S)

Vkminus1(S)






24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)











radick) [3] We













25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)


Rmrand Rmt


`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w


References














26





Abstract



1 Introduction






(a) (b)






27










D

(a)

A

(b)

A

B

(c)









28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)



4 (nminus 1) ge 34n



4 (nminus t) ge 34n




4n




4 (nminus 3) ge 34n







4 (nminus 4) = 34n







4n rectangles




29


1 2 n4





4 |R2| get + 3


4nminus 52



L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)



References









30








minqisinQ

d(p q)












References






31





Abstract


1 Introduction










RH(P ) = R2 ⋃

QisinQQ


RH(P ) = CH(P ) ⋃

QisinQQ


pq


RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)



32


(a) (b) (c)











miniisin14



q




p

q

r

s




33


(a) (b)




A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd




N

E

W

S

ε

N or W

N or E

S or ES or W

N

S




4 points When




and


re-









2 (1)





+


2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34





+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)



nminus44sum

k=1


i (4)


nminus44sum

k=1


i=1

i (5)


nminus44sum

k=1

ksumj=1


i (6)


nminus44sum

k=1

nminus44sum

j=k+1


i=1

i (7)


nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1


i (9)


nminus44sum

k=1

kminus1sumj=1


i=1

i (10)


k=1

nminus44sum

j=k


i=1

i (11)


nminus44sum

k=1

kminus1sumj=1


i=1

i (12)


k=1

nminus44sum

j=k


i (13)




References







A096338 Accessed 2021-01-01


35






(n3

)triangles of












References







36





Abstract



neΩ(radic




1 Preliminaries









k0(n) we denote the






k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)



Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))


37



2α0c These






Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0


for k0 = n4 + 1 fα0



(n2






DepthPα0(x) = min








(x) ge 2

k0 minus 1


k0 minus 1

α0

α0

sm


α0

α0

s0

s1

C

k0 minus 1

T









in O(log n) time






38












pipi+1

ei

ei

fi

pi

pi+1



polygonal-curve fi









39
















1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y




pi

pj

Aij cupAjk cupAkl






References








40





Abstract














TRA2015-67788-P


det(


= 0





T prime(t) =det

(


det(


∣

∣

∣

∣

∣

∣

T=T (t)




∣

∣

T=T (t)



References



41






Abstract









References






42






Abstract








References




43






Abstract




a) b)

c) d)











References






44






Abstract


1 Introduction














2 Related works








45











for all i isin 1 n







Note that


10 + (2minus ε) + ε

= 2 +radic

10


2 +radic




radic3)


3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime



2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since


radic10 + ε

= 2 +radic

10 + 2ε


2 +radic


3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)



46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)




radic5











a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8










47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9








b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10







References





48







Abstract




1 Introduction








2




49


(a) (b) (c)






cc

c c(a) (b) (c)

38

38







c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej



50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi











RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj












51


Ri

pici

Q




Lemma 6 ropt ge 14




return Ccelse







2

Acknowledgements


References











52





1 Introduction



S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)









1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23



(right)


3 Main results



Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)


References



53















References






54





Abstract


1 Introduction






ei =




pij =

tj if i = 1




55








1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3






k timesmk grid


Ik(Prms(L)) =sum


k ]

(1 + Iijk(Prms(L)))







56


LL1 L2 L3







1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4




1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5




57






Acknowledgements


References















58

EGC21 Author Index

Author Index





Esteban Guillermo 7





Jallu Ramesh K 49


59

EGC21 Author Index



Orden David 7 15 32








60

Portada


Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results


Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction



More results



paper_09

paper_10

Introduction

RVD in the plane


paper_11

Introduction



paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction


Concluding comments

paper_15

paper_16

Preliminaries




The (0k0)-hulls




paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works



paper_22

Introduction




paper_23

Introduction


Main results

paper_24

paper_25

Introduction




2_author_index


Program Committee


ii




iii


Table of Contents

Initial sections

Preface i




Invited talks


Kevin Buchin




Fernando Blasco



















Julian Pfeifle

iv













Sergio Cabello



















v










Raul Falcon

Final section

Author Index 59

vi



Kevin Buchin



1








2





Abstract




3
















log logn )




I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r









References





4























References



5















6 (n2 minus n)


References





6






Abstract



1 Introduction




v


v








7


ω1 =radic13

ω2 = 1s

t





2 Previous results



radic2













s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4








vi3 Xi = (vi1 vi2 v


(vi1 vi2 v

i3 v




i2)


i2 v

i3) see Figure


i2 v

i3 v

i4)


i6) (vi6 v

i1) are symmetric







8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi





k 1 le k le 6if






P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ



3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q





k 1 le k le 5if


along k edges of Hm






spectively

9






5 m ge i Then an


when m = i


1







4 and a P 15 con-







1 and a P 22








2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13







2

4 Conclusions




References













10





Abstract



1 Introduction



Dn =

([n]

2



P1

P2

P3

P4













2 Lower bound


11


P1 P2 P3 P4

Figure 2 Case n = 4


2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

















lt lt





P1 P2

P3

P4

P5

Π = Π3

Π1

Π2





12










(nminus1

2

)+ nminus 3 lt

(n2

)


3 Upper bound




gij =1

2(m2


2ij)





Q1 lt Q2 lt lt Qm



mij =

0 if i = j




gij =1

2(m2

ni +m2nj minusm2

ij)



2Grarr




the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i



13












(n2

)





mij =

0 if i = j












4 Discussion







References





14














References





15











References













16



Julian Pfeifle


Abstract



1 Introduction



+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]







0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]


0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]





17








-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]


0 = Γ(18f|56bd)


= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]




0 = Γ(1bf|48de)







[36fb5]

([36fb4]

([14bf3]

([16f85]


+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)


)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)






18


minusΓ(18f|46be)

Γ(18f|56bd)


minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]


0 = [36fb5]

([36fb4]

([14bf3]

([16f85]




))+ [16f84] [14bf8] [16f85]


))+ [16f84] [14bf8] [14bfe] [16f85]


))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]








VΣ =


unknown solids


(VΣ

2












4 More results


19


minsum

ΓisinVΣ

xΓ st 2sum



2

)(2)



2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)


















References


20191641766







201211500 2020






20











References









21







1 Introduction





2 RVD in the plane











References




22





Abstract


1 Introduction






by PID2019-104129GB-I00 AEI 1013039501100011033










23










1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37


Vk(S)

Vk+1(S)

Vkminus1(S)






24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)











radick) [3] We













25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)


Rmrand Rmt


`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w


References














26





Abstract



1 Introduction






(a) (b)






27










D

(a)

A

(b)

A

B

(c)









28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)



4 (nminus 1) ge 34n



4 (nminus t) ge 34n




4n




4 (nminus 3) ge 34n







4 (nminus 4) = 34n







4n rectangles




29


1 2 n4





4 |R2| get + 3


4nminus 52



L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)



References









30








minqisinQ

d(p q)












References






31





Abstract


1 Introduction










RH(P ) = R2 ⋃

QisinQQ


RH(P ) = CH(P ) ⋃

QisinQQ


pq


RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)



32


(a) (b) (c)











miniisin14



q




p

q

r

s




33


(a) (b)




A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd




N

E

W

S

ε

N or W

N or E

S or ES or W

N

S




4 points When




and


re-









2 (1)





+


2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34





+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)



nminus44sum

k=1


i (4)


nminus44sum

k=1


i=1

i (5)


nminus44sum

k=1

ksumj=1


i (6)


nminus44sum

k=1

nminus44sum

j=k+1


i=1

i (7)


nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1


i (9)


nminus44sum

k=1

kminus1sumj=1


i=1

i (10)


k=1

nminus44sum

j=k


i=1

i (11)


nminus44sum

k=1

kminus1sumj=1


i=1

i (12)


k=1

nminus44sum

j=k


i (13)




References







A096338 Accessed 2021-01-01


35






(n3

)triangles of












References







36





Abstract



neΩ(radic




1 Preliminaries









k0(n) we denote the






k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)



Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))


37



2α0c These






Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0


for k0 = n4 + 1 fα0



(n2






DepthPα0(x) = min








(x) ge 2

k0 minus 1


k0 minus 1

α0

α0

sm


α0

α0

s0

s1

C

k0 minus 1

T









in O(log n) time






38












pipi+1

ei

ei

fi

pi

pi+1



polygonal-curve fi









39
















1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y




pi

pj

Aij cupAjk cupAkl






References








40





Abstract














TRA2015-67788-P


det(


= 0





T prime(t) =det

(


det(


∣

∣

∣

∣

∣

∣

T=T (t)




∣

∣

T=T (t)



References



41






Abstract









References






42






Abstract








References




43






Abstract




a) b)

c) d)











References






44






Abstract


1 Introduction














2 Related works








45











for all i isin 1 n







Note that


10 + (2minus ε) + ε

= 2 +radic

10


2 +radic




radic3)


3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime



2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since


radic10 + ε

= 2 +radic

10 + 2ε


2 +radic


3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)



46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)




radic5











a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8










47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9








b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10







References





48







Abstract




1 Introduction








2




49


(a) (b) (c)






cc

c c(a) (b) (c)

38

38







c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej



50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi











RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj












51


Ri

pici

Q




Lemma 6 ropt ge 14




return Ccelse







2

Acknowledgements


References











52





1 Introduction



S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)









1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23



(right)


3 Main results



Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)


References



53















References






54





Abstract


1 Introduction






ei =




pij =

tj if i = 1




55








1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3






k timesmk grid


Ik(Prms(L)) =sum


k ]

(1 + Iijk(Prms(L)))







56


LL1 L2 L3







1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4




1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5




57






Acknowledgements


References















58

EGC21 Author Index

Author Index





Esteban Guillermo 7





Jallu Ramesh K 49


59

EGC21 Author Index



Orden David 7 15 32








60

Portada


Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results


Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction



More results



paper_09

paper_10

Introduction

RVD in the plane


paper_11

Introduction



paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction


Concluding comments

paper_15

paper_16

Preliminaries




The (0k0)-hulls




paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works



paper_22

Introduction




paper_23

Introduction


Main results

paper_24

paper_25

Introduction




2_author_index




iii


Table of Contents

Initial sections

Preface i




Invited talks


Kevin Buchin




Fernando Blasco



















Julian Pfeifle

iv













Sergio Cabello



















v










Raul Falcon

Final section

Author Index 59

vi



Kevin Buchin



1








2





Abstract




3
















log logn )




I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r









References





4























References



5















6 (n2 minus n)


References





6






Abstract



1 Introduction




v


v








7


ω1 =radic13

ω2 = 1s

t





2 Previous results



radic2













s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4








vi3 Xi = (vi1 vi2 v


(vi1 vi2 v

i3 v




i2)


i2 v

i3) see Figure


i2 v

i3 v

i4)


i6) (vi6 v

i1) are symmetric







8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi





k 1 le k le 6if






P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ



3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q





k 1 le k le 5if


along k edges of Hm






spectively

9






5 m ge i Then an


when m = i


1







4 and a P 15 con-







1 and a P 22








2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13







2

4 Conclusions




References













10





Abstract



1 Introduction



Dn =

([n]

2



P1

P2

P3

P4













2 Lower bound


11


P1 P2 P3 P4

Figure 2 Case n = 4


2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

















lt lt





P1 P2

P3

P4

P5

Π = Π3

Π1

Π2





12










(nminus1

2

)+ nminus 3 lt

(n2

)


3 Upper bound




gij =1

2(m2


2ij)





Q1 lt Q2 lt lt Qm



mij =

0 if i = j




gij =1

2(m2

ni +m2nj minusm2

ij)



2Grarr




the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i



13












(n2

)





mij =

0 if i = j












4 Discussion







References





14














References





15











References













16



Julian Pfeifle


Abstract



1 Introduction



+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]







0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]


0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]





17








-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]


0 = Γ(18f|56bd)


= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]




0 = Γ(1bf|48de)







[36fb5]

([36fb4]

([14bf3]

([16f85]


+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)


)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)






18


minusΓ(18f|46be)

Γ(18f|56bd)


minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]


0 = [36fb5]

([36fb4]

([14bf3]

([16f85]




))+ [16f84] [14bf8] [16f85]


))+ [16f84] [14bf8] [14bfe] [16f85]


))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]








VΣ =


unknown solids


(VΣ

2












4 More results


19


minsum

ΓisinVΣ

xΓ st 2sum



2

)(2)



2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)


















References


20191641766







201211500 2020






20











References









21







1 Introduction





2 RVD in the plane











References




22





Abstract


1 Introduction






by PID2019-104129GB-I00 AEI 1013039501100011033










23










1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37


Vk(S)

Vk+1(S)

Vkminus1(S)






24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)











radick) [3] We













25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)


Rmrand Rmt


`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w


References














26





Abstract



1 Introduction






(a) (b)






27










D

(a)

A

(b)

A

B

(c)









28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)



4 (nminus 1) ge 34n



4 (nminus t) ge 34n




4n




4 (nminus 3) ge 34n







4 (nminus 4) = 34n







4n rectangles




29


1 2 n4





4 |R2| get + 3


4nminus 52



L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)



References









30








minqisinQ

d(p q)












References






31





Abstract


1 Introduction










RH(P ) = R2 ⋃

QisinQQ


RH(P ) = CH(P ) ⋃

QisinQQ


pq


RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)



32


(a) (b) (c)











miniisin14



q




p

q

r

s




33


(a) (b)




A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd




N

E

W

S

ε

N or W

N or E

S or ES or W

N

S




4 points When




and


re-









2 (1)





+


2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34





+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)



nminus44sum

k=1


i (4)


nminus44sum

k=1


i=1

i (5)


nminus44sum

k=1

ksumj=1


i (6)


nminus44sum

k=1

nminus44sum

j=k+1


i=1

i (7)


nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1


i (9)


nminus44sum

k=1

kminus1sumj=1


i=1

i (10)


k=1

nminus44sum

j=k


i=1

i (11)


nminus44sum

k=1

kminus1sumj=1


i=1

i (12)


k=1

nminus44sum

j=k


i (13)




References







A096338 Accessed 2021-01-01


35






(n3

)triangles of












References







36





Abstract



neΩ(radic




1 Preliminaries









k0(n) we denote the






k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)



Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))


37



2α0c These






Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0


for k0 = n4 + 1 fα0



(n2






DepthPα0(x) = min








(x) ge 2

k0 minus 1


k0 minus 1

α0

α0

sm


α0

α0

s0

s1

C

k0 minus 1

T









in O(log n) time






38












pipi+1

ei

ei

fi

pi

pi+1



polygonal-curve fi









39
















1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y




pi

pj

Aij cupAjk cupAkl






References








40





Abstract














TRA2015-67788-P


det(


= 0





T prime(t) =det

(


det(


∣

∣

∣

∣

∣

∣

T=T (t)




∣

∣

T=T (t)



References



41






Abstract









References






42






Abstract








References




43






Abstract




a) b)

c) d)











References






44






Abstract


1 Introduction














2 Related works








45











for all i isin 1 n







Note that


10 + (2minus ε) + ε

= 2 +radic

10


2 +radic




radic3)


3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime



2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since


radic10 + ε

= 2 +radic

10 + 2ε


2 +radic


3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)



46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)




radic5











a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8










47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9








b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10







References





48







Abstract




1 Introduction








2




49


(a) (b) (c)






cc

c c(a) (b) (c)

38

38







c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej



50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi











RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj












51


Ri

pici

Q




Lemma 6 ropt ge 14




return Ccelse







2

Acknowledgements


References











52





1 Introduction



S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)









1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23



(right)


3 Main results



Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)


References



53















References






54





Abstract


1 Introduction






ei =




pij =

tj if i = 1




55








1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3






k timesmk grid


Ik(Prms(L)) =sum


k ]

(1 + Iijk(Prms(L)))







56


LL1 L2 L3







1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4




1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5




57






Acknowledgements


References















58

EGC21 Author Index

Author Index





Esteban Guillermo 7





Jallu Ramesh K 49


59

EGC21 Author Index



Orden David 7 15 32








60

Portada


Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results


Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction



More results



paper_09

paper_10

Introduction

RVD in the plane


paper_11

Introduction



paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction


Concluding comments

paper_15

paper_16

Preliminaries




The (0k0)-hulls




paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works



paper_22

Introduction




paper_23

Introduction


Main results

paper_24

paper_25

Introduction




2_author_index


Table of Contents

Initial sections

Preface i




Invited talks


Kevin Buchin




Fernando Blasco



















Julian Pfeifle

iv













Sergio Cabello



















v










Raul Falcon

Final section

Author Index 59

vi



Kevin Buchin



1








2





Abstract




3
















log logn )




I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r









References





4























References



5















6 (n2 minus n)


References





6






Abstract



1 Introduction




v


v








7


ω1 =radic13

ω2 = 1s

t





2 Previous results



radic2













s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4








vi3 Xi = (vi1 vi2 v


(vi1 vi2 v

i3 v




i2)


i2 v

i3) see Figure


i2 v

i3 v

i4)


i6) (vi6 v

i1) are symmetric







8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi





k 1 le k le 6if






P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ



3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q





k 1 le k le 5if


along k edges of Hm






spectively

9






5 m ge i Then an


when m = i


1







4 and a P 15 con-







1 and a P 22








2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13







2

4 Conclusions




References













10





Abstract



1 Introduction



Dn =

([n]

2



P1

P2

P3

P4













2 Lower bound


11


P1 P2 P3 P4

Figure 2 Case n = 4


2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

















lt lt





P1 P2

P3

P4

P5

Π = Π3

Π1

Π2





12










(nminus1

2

)+ nminus 3 lt

(n2

)


3 Upper bound




gij =1

2(m2


2ij)





Q1 lt Q2 lt lt Qm



mij =

0 if i = j




gij =1

2(m2

ni +m2nj minusm2

ij)



2Grarr




the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i



13












(n2

)





mij =

0 if i = j












4 Discussion







References





14














References





15











References













16



Julian Pfeifle


Abstract



1 Introduction



+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]







0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]


0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]





17








-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]


0 = Γ(18f|56bd)


= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]




0 = Γ(1bf|48de)







[36fb5]

([36fb4]

([14bf3]

([16f85]


+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)


)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)






18


minusΓ(18f|46be)

Γ(18f|56bd)


minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]


0 = [36fb5]

([36fb4]

([14bf3]

([16f85]




))+ [16f84] [14bf8] [16f85]


))+ [16f84] [14bf8] [14bfe] [16f85]


))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]








VΣ =


unknown solids


(VΣ

2












4 More results


19


minsum

ΓisinVΣ

xΓ st 2sum



2

)(2)



2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)


















References


20191641766







201211500 2020






20











References









21







1 Introduction





2 RVD in the plane











References




22





Abstract


1 Introduction






by PID2019-104129GB-I00 AEI 1013039501100011033










23










1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37


Vk(S)

Vk+1(S)

Vkminus1(S)






24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)











radick) [3] We













25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)


Rmrand Rmt


`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w


References














26





Abstract



1 Introduction






(a) (b)






27










D

(a)

A

(b)

A

B

(c)









28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)



4 (nminus 1) ge 34n



4 (nminus t) ge 34n




4n




4 (nminus 3) ge 34n







4 (nminus 4) = 34n







4n rectangles




29


1 2 n4





4 |R2| get + 3


4nminus 52



L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)



References









30








minqisinQ

d(p q)












References






31





Abstract


1 Introduction










RH(P ) = R2 ⋃

QisinQQ


RH(P ) = CH(P ) ⋃

QisinQQ


pq


RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)



32


(a) (b) (c)











miniisin14



q




p

q

r

s




33


(a) (b)




A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd




N

E

W

S

ε

N or W

N or E

S or ES or W

N

S




4 points When




and


re-









2 (1)





+


2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34





+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)



nminus44sum

k=1


i (4)


nminus44sum

k=1


i=1

i (5)


nminus44sum

k=1

ksumj=1


i (6)


nminus44sum

k=1

nminus44sum

j=k+1


i=1

i (7)


nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1


i (9)


nminus44sum

k=1

kminus1sumj=1


i=1

i (10)


k=1

nminus44sum

j=k


i=1

i (11)


nminus44sum

k=1

kminus1sumj=1


i=1

i (12)


k=1

nminus44sum

j=k


i (13)




References







A096338 Accessed 2021-01-01


35






(n3

)triangles of












References







36





Abstract



neΩ(radic




1 Preliminaries









k0(n) we denote the






k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)



Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))


37



2α0c These






Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0


for k0 = n4 + 1 fα0



(n2






DepthPα0(x) = min








(x) ge 2

k0 minus 1


k0 minus 1

α0

α0

sm


α0

α0

s0

s1

C

k0 minus 1

T









in O(log n) time






38












pipi+1

ei

ei

fi

pi

pi+1



polygonal-curve fi









39
















1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y




pi

pj

Aij cupAjk cupAkl






References








40





Abstract














TRA2015-67788-P


det(


= 0





T prime(t) =det

(


det(


∣

∣

∣

∣

∣

∣

T=T (t)




∣

∣

T=T (t)



References



41






Abstract









References






42






Abstract








References




43






Abstract




a) b)

c) d)











References






44






Abstract


1 Introduction














2 Related works








45











for all i isin 1 n







Note that


10 + (2minus ε) + ε

= 2 +radic

10


2 +radic




radic3)


3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime



2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since


radic10 + ε

= 2 +radic

10 + 2ε


2 +radic


3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)



46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)




radic5











a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8










47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9








b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10







References





48







Abstract




1 Introduction








2




49


(a) (b) (c)






cc

c c(a) (b) (c)

38

38







c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej



50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi











RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj












51


Ri

pici

Q




Lemma 6 ropt ge 14




return Ccelse







2

Acknowledgements


References











52





1 Introduction



S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)









1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23



(right)


3 Main results



Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)


References



53















References






54





Abstract


1 Introduction






ei =




pij =

tj if i = 1




55








1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3






k timesmk grid


Ik(Prms(L)) =sum


k ]

(1 + Iijk(Prms(L)))







56


LL1 L2 L3







1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4




1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5




57






Acknowledgements


References















58

EGC21 Author Index

Author Index





Esteban Guillermo 7





Jallu Ramesh K 49


59

EGC21 Author Index



Orden David 7 15 32








60

Portada


Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results


Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction



More results



paper_09

paper_10

Introduction

RVD in the plane


paper_11

Introduction



paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction


Concluding comments

paper_15

paper_16

Preliminaries




The (0k0)-hulls




paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works



paper_22

Introduction




paper_23

Introduction


Main results

paper_24

paper_25

Introduction




2_author_index













Sergio Cabello



















v










Raul Falcon

Final section

Author Index 59

vi



Kevin Buchin



1








2





Abstract




3
















log logn )




I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r









References





4























References



5















6 (n2 minus n)


References





6






Abstract



1 Introduction




v


v








7


ω1 =radic13

ω2 = 1s

t





2 Previous results



radic2













s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4








vi3 Xi = (vi1 vi2 v


(vi1 vi2 v

i3 v




i2)


i2 v

i3) see Figure


i2 v

i3 v

i4)


i6) (vi6 v

i1) are symmetric







8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi





k 1 le k le 6if






P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ



3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q





k 1 le k le 5if


along k edges of Hm






spectively

9






5 m ge i Then an


when m = i


1







4 and a P 15 con-







1 and a P 22








2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13







2

4 Conclusions




References













10





Abstract



1 Introduction



Dn =

([n]

2



P1

P2

P3

P4













2 Lower bound


11


P1 P2 P3 P4

Figure 2 Case n = 4


2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

















lt lt





P1 P2

P3

P4

P5

Π = Π3

Π1

Π2





12










(nminus1

2

)+ nminus 3 lt

(n2

)


3 Upper bound




gij =1

2(m2


2ij)





Q1 lt Q2 lt lt Qm



mij =

0 if i = j




gij =1

2(m2

ni +m2nj minusm2

ij)



2Grarr




the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i



13












(n2

)





mij =

0 if i = j












4 Discussion







References





14














References





15











References













16



Julian Pfeifle


Abstract



1 Introduction



+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]







0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]


0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]





17








-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]


0 = Γ(18f|56bd)


= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]




0 = Γ(1bf|48de)







[36fb5]

([36fb4]

([14bf3]

([16f85]


+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)


)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)






18


minusΓ(18f|46be)

Γ(18f|56bd)


minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]


0 = [36fb5]

([36fb4]

([14bf3]

([16f85]




))+ [16f84] [14bf8] [16f85]


))+ [16f84] [14bf8] [14bfe] [16f85]


))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]








VΣ =


unknown solids


(VΣ

2












4 More results


19


minsum

ΓisinVΣ

xΓ st 2sum



2

)(2)



2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)


















References


20191641766







201211500 2020






20











References









21







1 Introduction





2 RVD in the plane











References




22





Abstract


1 Introduction






by PID2019-104129GB-I00 AEI 1013039501100011033










23










1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37


Vk(S)

Vk+1(S)

Vkminus1(S)






24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)











radick) [3] We













25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)


Rmrand Rmt


`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w


References














26





Abstract



1 Introduction






(a) (b)






27










D

(a)

A

(b)

A

B

(c)









28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)



4 (nminus 1) ge 34n



4 (nminus t) ge 34n




4n




4 (nminus 3) ge 34n







4 (nminus 4) = 34n







4n rectangles




29


1 2 n4





4 |R2| get + 3


4nminus 52



L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)



References









30








minqisinQ

d(p q)












References






31





Abstract


1 Introduction










RH(P ) = R2 ⋃

QisinQQ


RH(P ) = CH(P ) ⋃

QisinQQ


pq


RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)



32


(a) (b) (c)











miniisin14



q




p

q

r

s




33


(a) (b)




A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd




N

E

W

S

ε

N or W

N or E

S or ES or W

N

S




4 points When




and


re-









2 (1)





+


2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34





+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)



nminus44sum

k=1


i (4)


nminus44sum

k=1


i=1

i (5)


nminus44sum

k=1

ksumj=1


i (6)


nminus44sum

k=1

nminus44sum

j=k+1


i=1

i (7)


nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1


i (9)


nminus44sum

k=1

kminus1sumj=1


i=1

i (10)


k=1

nminus44sum

j=k


i=1

i (11)


nminus44sum

k=1

kminus1sumj=1


i=1

i (12)


k=1

nminus44sum

j=k


i (13)




References







A096338 Accessed 2021-01-01


35






(n3

)triangles of












References







36





Abstract



neΩ(radic




1 Preliminaries









k0(n) we denote the






k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)



Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))


37



2α0c These






Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0


for k0 = n4 + 1 fα0



(n2






DepthPα0(x) = min








(x) ge 2

k0 minus 1


k0 minus 1

α0

α0

sm


α0

α0

s0

s1

C

k0 minus 1

T









in O(log n) time






38












pipi+1

ei

ei

fi

pi

pi+1



polygonal-curve fi









39
















1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y




pi

pj

Aij cupAjk cupAkl






References








40





Abstract














TRA2015-67788-P


det(


= 0





T prime(t) =det

(


det(


∣

∣

∣

∣

∣

∣

T=T (t)




∣

∣

T=T (t)



References



41






Abstract









References






42






Abstract








References




43






Abstract




a) b)

c) d)











References






44






Abstract


1 Introduction














2 Related works








45











for all i isin 1 n







Note that


10 + (2minus ε) + ε

= 2 +radic

10


2 +radic




radic3)


3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime



2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since


radic10 + ε

= 2 +radic

10 + 2ε


2 +radic


3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)



46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)




radic5











a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8










47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9








b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10







References





48







Abstract




1 Introduction








2




49


(a) (b) (c)






cc

c c(a) (b) (c)

38

38







c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej



50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi











RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj












51


Ri

pici

Q




Lemma 6 ropt ge 14




return Ccelse







2

Acknowledgements


References











52





1 Introduction



S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)









1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23



(right)


3 Main results



Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)


References



53















References






54





Abstract


1 Introduction






ei =




pij =

tj if i = 1




55








1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3






k timesmk grid


Ik(Prms(L)) =sum


k ]

(1 + Iijk(Prms(L)))







56


LL1 L2 L3







1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4




1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5




57






Acknowledgements


References















58

EGC21 Author Index

Author Index





Esteban Guillermo 7





Jallu Ramesh K 49


59

EGC21 Author Index



Orden David 7 15 32








60

Portada


Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results


Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction



More results



paper_09

paper_10

Introduction

RVD in the plane


paper_11

Introduction



paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction


Concluding comments

paper_15

paper_16

Preliminaries




The (0k0)-hulls




paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works



paper_22

Introduction




paper_23

Introduction


Main results

paper_24

paper_25

Introduction




2_author_index










Raul Falcon

Final section

Author Index 59

vi



Kevin Buchin



1








2





Abstract




3
















log logn )




I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r









References





4























References



5















6 (n2 minus n)


References





6






Abstract



1 Introduction




v


v








7


ω1 =radic13

ω2 = 1s

t





2 Previous results



radic2













s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4








vi3 Xi = (vi1 vi2 v


(vi1 vi2 v

i3 v




i2)


i2 v

i3) see Figure


i2 v

i3 v

i4)


i6) (vi6 v

i1) are symmetric







8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi





k 1 le k le 6if






P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ



3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q





k 1 le k le 5if


along k edges of Hm






spectively

9






5 m ge i Then an


when m = i


1







4 and a P 15 con-







1 and a P 22








2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13







2

4 Conclusions




References













10





Abstract



1 Introduction



Dn =

([n]

2



P1

P2

P3

P4













2 Lower bound


11


P1 P2 P3 P4

Figure 2 Case n = 4


2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

















lt lt





P1 P2

P3

P4

P5

Π = Π3

Π1

Π2





12










(nminus1

2

)+ nminus 3 lt

(n2

)


3 Upper bound




gij =1

2(m2


2ij)





Q1 lt Q2 lt lt Qm



mij =

0 if i = j




gij =1

2(m2

ni +m2nj minusm2

ij)



2Grarr




the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i



13












(n2

)





mij =

0 if i = j












4 Discussion







References





14














References





15











References













16



Julian Pfeifle


Abstract



1 Introduction



+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]







0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]


0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]





17








-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]


0 = Γ(18f|56bd)


= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]




0 = Γ(1bf|48de)







[36fb5]

([36fb4]

([14bf3]

([16f85]


+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)


)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)






18


minusΓ(18f|46be)

Γ(18f|56bd)


minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]


0 = [36fb5]

([36fb4]

([14bf3]

([16f85]




))+ [16f84] [14bf8] [16f85]


))+ [16f84] [14bf8] [14bfe] [16f85]


))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]








VΣ =


unknown solids


(VΣ

2












4 More results


19


minsum

ΓisinVΣ

xΓ st 2sum



2

)(2)



2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)


















References


20191641766







201211500 2020






20











References









21







1 Introduction





2 RVD in the plane











References




22





Abstract


1 Introduction






by PID2019-104129GB-I00 AEI 1013039501100011033










23










1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37


Vk(S)

Vk+1(S)

Vkminus1(S)






24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)











radick) [3] We













25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)


Rmrand Rmt


`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w


References














26





Abstract



1 Introduction






(a) (b)






27










D

(a)

A

(b)

A

B

(c)









28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)



4 (nminus 1) ge 34n



4 (nminus t) ge 34n




4n




4 (nminus 3) ge 34n







4 (nminus 4) = 34n







4n rectangles




29


1 2 n4





4 |R2| get + 3


4nminus 52



L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)



References









30








minqisinQ

d(p q)












References






31





Abstract


1 Introduction










RH(P ) = R2 ⋃

QisinQQ


RH(P ) = CH(P ) ⋃

QisinQQ


pq


RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)



32


(a) (b) (c)











miniisin14



q




p

q

r

s




33


(a) (b)




A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd




N

E

W

S

ε

N or W

N or E

S or ES or W

N

S




4 points When




and


re-









2 (1)





+


2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34





+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)



nminus44sum

k=1


i (4)


nminus44sum

k=1


i=1

i (5)


nminus44sum

k=1

ksumj=1


i (6)


nminus44sum

k=1

nminus44sum

j=k+1


i=1

i (7)


nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1


i (9)


nminus44sum

k=1

kminus1sumj=1


i=1

i (10)


k=1

nminus44sum

j=k


i=1

i (11)


nminus44sum

k=1

kminus1sumj=1


i=1

i (12)


k=1

nminus44sum

j=k


i (13)




References







A096338 Accessed 2021-01-01


35






(n3

)triangles of












References







36





Abstract



neΩ(radic




1 Preliminaries









k0(n) we denote the






k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)



Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))


37



2α0c These






Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0


for k0 = n4 + 1 fα0



(n2






DepthPα0(x) = min








(x) ge 2

k0 minus 1


k0 minus 1

α0

α0

sm


α0

α0

s0

s1

C

k0 minus 1

T









in O(log n) time






38












pipi+1

ei

ei

fi

pi

pi+1



polygonal-curve fi









39
















1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y




pi

pj

Aij cupAjk cupAkl






References








40





Abstract














TRA2015-67788-P


det(


= 0





T prime(t) =det

(


det(


∣

∣

∣

∣

∣

∣

T=T (t)




∣

∣

T=T (t)



References



41






Abstract









References






42






Abstract








References




43






Abstract




a) b)

c) d)











References






44






Abstract


1 Introduction














2 Related works








45











for all i isin 1 n







Note that


10 + (2minus ε) + ε

= 2 +radic

10


2 +radic




radic3)


3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime



2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since


radic10 + ε

= 2 +radic

10 + 2ε


2 +radic


3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)



46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)




radic5











a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8










47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9








b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10







References





48







Abstract




1 Introduction








2




49


(a) (b) (c)






cc

c c(a) (b) (c)

38

38







c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej



50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi











RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj












51


Ri

pici

Q




Lemma 6 ropt ge 14




return Ccelse







2

Acknowledgements


References











52





1 Introduction



S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)









1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23



(right)


3 Main results



Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)


References



53















References






54





Abstract


1 Introduction






ei =




pij =

tj if i = 1




55








1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3






k timesmk grid


Ik(Prms(L)) =sum


k ]

(1 + Iijk(Prms(L)))







56


LL1 L2 L3







1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4




1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5




57






Acknowledgements


References















58

EGC21 Author Index

Author Index





Esteban Guillermo 7





Jallu Ramesh K 49


59

EGC21 Author Index



Orden David 7 15 32








60

Portada


Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results


Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction



More results



paper_09

paper_10

Introduction

RVD in the plane


paper_11

Introduction



paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction


Concluding comments

paper_15

paper_16

Preliminaries




The (0k0)-hulls




paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works



paper_22

Introduction




paper_23

Introduction


Main results

paper_24

paper_25

Introduction




2_author_index



Kevin Buchin



1








2





Abstract




3
















log logn )




I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r









References





4























References



5















6 (n2 minus n)


References





6






Abstract



1 Introduction




v


v








7


ω1 =radic13

ω2 = 1s

t





2 Previous results



radic2













s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4








vi3 Xi = (vi1 vi2 v


(vi1 vi2 v

i3 v




i2)


i2 v

i3) see Figure


i2 v

i3 v

i4)


i6) (vi6 v

i1) are symmetric







8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi





k 1 le k le 6if






P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ



3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q





k 1 le k le 5if


along k edges of Hm






spectively

9






5 m ge i Then an


when m = i


1







4 and a P 15 con-







1 and a P 22








2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13







2

4 Conclusions




References













10





Abstract



1 Introduction



Dn =

([n]

2



P1

P2

P3

P4













2 Lower bound


11


P1 P2 P3 P4

Figure 2 Case n = 4


2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

















lt lt





P1 P2

P3

P4

P5

Π = Π3

Π1

Π2





12










(nminus1

2

)+ nminus 3 lt

(n2

)


3 Upper bound




gij =1

2(m2


2ij)





Q1 lt Q2 lt lt Qm



mij =

0 if i = j




gij =1

2(m2

ni +m2nj minusm2

ij)



2Grarr




the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i



13












(n2

)





mij =

0 if i = j












4 Discussion







References





14














References





15











References













16



Julian Pfeifle


Abstract



1 Introduction



+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]







0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]


0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]





17








-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]


0 = Γ(18f|56bd)


= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]




0 = Γ(1bf|48de)







[36fb5]

([36fb4]

([14bf3]

([16f85]


+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)


)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)






18


minusΓ(18f|46be)

Γ(18f|56bd)


minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]


0 = [36fb5]

([36fb4]

([14bf3]

([16f85]




))+ [16f84] [14bf8] [16f85]


))+ [16f84] [14bf8] [14bfe] [16f85]


))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]








VΣ =


unknown solids


(VΣ

2












4 More results


19


minsum

ΓisinVΣ

xΓ st 2sum



2

)(2)



2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)


















References


20191641766







201211500 2020






20











References









21







1 Introduction





2 RVD in the plane











References




22





Abstract


1 Introduction






by PID2019-104129GB-I00 AEI 1013039501100011033










23










1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37


Vk(S)

Vk+1(S)

Vkminus1(S)






24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)











radick) [3] We













25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)


Rmrand Rmt


`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w


References














26





Abstract



1 Introduction






(a) (b)






27










D

(a)

A

(b)

A

B

(c)









28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)



4 (nminus 1) ge 34n



4 (nminus t) ge 34n




4n




4 (nminus 3) ge 34n







4 (nminus 4) = 34n







4n rectangles




29


1 2 n4





4 |R2| get + 3


4nminus 52



L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)



References









30








minqisinQ

d(p q)












References






31





Abstract


1 Introduction










RH(P ) = R2 ⋃

QisinQQ


RH(P ) = CH(P ) ⋃

QisinQQ


pq


RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)



32


(a) (b) (c)











miniisin14



q




p

q

r

s




33


(a) (b)




A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd




N

E

W

S

ε

N or W

N or E

S or ES or W

N

S




4 points When




and


re-









2 (1)





+


2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34





+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)



nminus44sum

k=1


i (4)


nminus44sum

k=1


i=1

i (5)


nminus44sum

k=1

ksumj=1


i (6)


nminus44sum

k=1

nminus44sum

j=k+1


i=1

i (7)


nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1


i (9)


nminus44sum

k=1

kminus1sumj=1


i=1

i (10)


k=1

nminus44sum

j=k


i=1

i (11)


nminus44sum

k=1

kminus1sumj=1


i=1

i (12)


k=1

nminus44sum

j=k


i (13)




References







A096338 Accessed 2021-01-01


35






(n3

)triangles of












References







36





Abstract



neΩ(radic




1 Preliminaries









k0(n) we denote the






k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)



Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))


37



2α0c These






Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0


for k0 = n4 + 1 fα0



(n2






DepthPα0(x) = min








(x) ge 2

k0 minus 1


k0 minus 1

α0

α0

sm


α0

α0

s0

s1

C

k0 minus 1

T









in O(log n) time






38












pipi+1

ei

ei

fi

pi

pi+1



polygonal-curve fi









39
















1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y




pi

pj

Aij cupAjk cupAkl






References








40





Abstract














TRA2015-67788-P


det(


= 0





T prime(t) =det

(


det(


∣

∣

∣

∣

∣

∣

T=T (t)




∣

∣

T=T (t)



References



41






Abstract









References






42






Abstract








References




43






Abstract




a) b)

c) d)











References






44






Abstract


1 Introduction














2 Related works








45











for all i isin 1 n







Note that


10 + (2minus ε) + ε

= 2 +radic

10


2 +radic




radic3)


3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime



2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since


radic10 + ε

= 2 +radic

10 + 2ε


2 +radic


3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)



46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)




radic5











a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8










47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9








b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10







References





48







Abstract




1 Introduction








2




49


(a) (b) (c)






cc

c c(a) (b) (c)

38

38







c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej



50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi











RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj












51


Ri

pici

Q




Lemma 6 ropt ge 14




return Ccelse







2

Acknowledgements


References











52





1 Introduction



S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)









1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23



(right)


3 Main results



Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)


References



53















References






54





Abstract


1 Introduction






ei =




pij =

tj if i = 1




55








1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3






k timesmk grid


Ik(Prms(L)) =sum


k ]

(1 + Iijk(Prms(L)))







56


LL1 L2 L3







1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4




1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5




57






Acknowledgements


References















58

EGC21 Author Index

Author Index





Esteban Guillermo 7





Jallu Ramesh K 49


59

EGC21 Author Index



Orden David 7 15 32








60

Portada


Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results


Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction



More results



paper_09

paper_10

Introduction

RVD in the plane


paper_11

Introduction



paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction


Concluding comments

paper_15

paper_16

Preliminaries




The (0k0)-hulls




paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works



paper_22

Introduction




paper_23

Introduction


Main results

paper_24

paper_25

Introduction




2_author_index








2





Abstract




3
















log logn )




I 3340-N35

v1

v3

v2v4

v5

O

r

v1

v3

v2v4v5

O

r









References





4























References



5















6 (n2 minus n)


References





6






Abstract



1 Introduction




v


v








7


ω1 =radic13

ω2 = 1s

t





2 Previous results



radic2













s=a1=u1

t=a5=u4

a2=u2

a3

a4=u3

H1

H2 H3

H4








vi3 Xi = (vi1 vi2 v


(vi1 vi2 v

i3 v




i2)


i2 v

i3) see Figure


i2 v

i3 v

i4)


i6) (vi6 v

i1) are symmetric







8


ai+1

ai = vi1

ai+1

aiai

ai+1

(d)

(c)

(e)

vi4

ei1

ai+1 = vi4

ai = vi1

(a)

vi3

vi5

vi6

vi2

ai+1ai = vi1

(b)

vi4

vi3

vi5

vi6

vi2

ei1

ei2

p p

vi2

vi3

Hi Hi Hi

Hi Hi





k 1 le k le 6if






P 11 P 1

3P 12

Hi

P 15P 1

4

ujuj

uj+1

uj+1

uj+1

P 16

ei1

p

uj+1

uj+1

uj+1

uj

uj uj

uj

HiHi

Hi Hi Hi

θ

θ



3

Pmminusi+11

uj

uj+1 = q

Hi

Hi+1

b

Hm

Pmminusi+15

uj

uj+1 = b

Hi

Hi+1

Hm

p p

q





k 1 le k le 5if


along k edges of Hm






spectively

9






5 m ge i Then an


when m = i


1







4 and a P 15 con-







1 and a P 22








2 is 2radic3


3 P 16 and P 2

1 is 32


2 is 5radic13







2

4 Conclusions




References













10





Abstract



1 Introduction



Dn =

([n]

2



P1

P2

P3

P4













2 Lower bound


11


P1 P2 P3 P4

Figure 2 Case n = 4


2 (1 2) lt (1 3)

3 (2 4) lt (3 4)

















lt lt





P1 P2

P3

P4

P5

Π = Π3

Π1

Π2





12










(nminus1

2

)+ nminus 3 lt

(n2

)


3 Upper bound




gij =1

2(m2


2ij)





Q1 lt Q2 lt lt Qm



mij =

0 if i = j




gij =1

2(m2

ni +m2nj minusm2

ij)



2Grarr




the quadratic form

(x1 xn) 7rarr 2

nsumi=1

x2i +

sum1leiltjlen

xixj

=

(sumi=1

xi

)2

+nsum

i=1

x2i



13












(n2

)





mij =

0 if i = j












4 Discussion







References





14














References





15











References













16



Julian Pfeifle


Abstract



1 Introduction



+[0123] -[0124] +[0135] -[0146]

+[0157] -[0167] -[0234] +[0345]

-[0456] +[0567] +[1237] -[1246]

-[1267] +[1357] -[2347] +[2456]

-[2457] -[2567] +[3457]







0 = Γ(045|1267)

= [04512][04567]minus [04516][04527] + [04517][04526]


0 = Γ(045|1267)

= (minus1)[01425](minus1)[05674]

minus [04651](minus1)[24750]

+ [01574] [24560]

= [01425][05674] + [04651][24750] + [01574][24560]





17








-[048c] +[048e] +[049c] -[049d] +[04ad]

-[04ae] +[059d] -[059f] -[05ad] +[05ae]

-[05be] +[05bf] +[068c] -[068e] -[069c]

+[069e] -[079e] +[079f] +[07be] -[07bf]

+[148c] -[148e] +[14ae] -[14af] -[14bc]

+[14bf] -[158c] +[158d] -[159d] +[159f]

+[15bc] -[15bf] +[168e] -[168f] -[16ae]

+[16af] -[178d] +[178f] +[179d] -[179f]

-[24ad] +[24af] +[24bd] -[24bf] +[258c]

-[258d] -[25ac] +[25ad] -[268c] +[268d]

+[269c] -[269e] +[26ae] -[26af] -[26bd]

+[26bf] -[279c] +[279e] +[27ac] -[27ae]

-[349c] +[349d] +[34bc] -[34bd] +[35ac]

-[35ae] -[35bc] +[35be] -[368d] +[368f]

+[36bd] -[36bf] +[378d] -[378f] +[379c]

-[379d] -[37ac] +[37ae] -[37be] +[37bf]


0 = Γ(18f|56bd)


= [16f85][18bdf] + [15fb8][16f8d] + [158df][16f8b]




0 = Γ(1bf|48de)







[36fb5]

([36fb4]

([14bf3]

([16f85]


+[16f84] Γ(1bf|48de))

+[16f84] [14bfe] Γ(18f|56bd))

+[16f84] [14bf8] [16f85] Γ(1bf|34de)


)+ [16f84] [14bf8] [14bfd] [16f85] [36fb4] Γ(3bf|156e)






18


minusΓ(18f|46be)

Γ(18f|56bd)


minusΓ(3bf|146d)

Γ(3bf|156e)

[13bdf]

[13bef][18bdf]

[18bef]

[1bdef]


0 = [36fb5]

([36fb4]

([14bf3]

([16f85]




))+ [16f84] [14bf8] [16f85]


))+ [16f84] [14bf8] [14bfe] [16f85]


))+ [16f84] [14bf8] [14bfd] [16f85] [36fb4]








VΣ =


unknown solids


(VΣ

2












4 More results


19


minsum

ΓisinVΣ

xΓ st 2sum



2

)(2)



2

) (3)

1 +sumeisinEΣ

xe =sum

ΓisinVΣ

xΓ (4)


















References


20191641766







201211500 2020






20











References









21







1 Introduction





2 RVD in the plane











References




22





Abstract


1 Introduction






by PID2019-104129GB-I00 AEI 1013039501100011033










23










1

6

7

84

5

2

38 1 43

25

36

478 5

16

27

6 5626

461

17

7 627

24

7 45

35

4

34

378

17 8

11 37

3886

62

2136

3228

26

38 35

7 5

37


Vk(S)

Vk+1(S)

Vkminus1(S)






24


1

6

7

128

178

278127

237

123234

236

346345

456

356367

267

678

567

84

5

2

3

2

2

22

2 2

2

25

64

7386

1

7

62

2

R3(2)











radick) [3] We













25


Rmr

mr

Rmr

mr

mr

mr

mr

mr

mr

msmsmt

mt

Rmq

fb(Pk)

Rms

Rmt

v

v

Rms

Rmr

Rmt

v

bmrmsbmsmt

bmtmr

(a)

Vkminus1(S)

Vk(S)

Vk+1(S)

ir

irir

it

irit is

is

f b(Pk)

Rir

Rit

Ris

ir

irir

Rir

v

u

vRit

Rir

Ris

v

bisit

bitir

biris

(b)


Rmrand Rmt


`

``

``

j

j

j

j

j

jj

j

mm

m

mm

i

i

i

ii

n

n

n

n

n

`

`

j

`

`

v

w


References














26





Abstract



1 Introduction






(a) (b)






27










D

(a)

A

(b)

A

B

(c)









28


A

(a)

A

B

(b)

CA

B

R

(c)

CA

B

emptyempty

(d)

A

B

C

R

empty

(e)

D

CA

B

emptyempty

(f)

A

B

C

empty

DR

(g)



4 (nminus 1) ge 34n



4 (nminus t) ge 34n




4n




4 (nminus 3) ge 34n







4 (nminus 4) = 34n







4n rectangles




29


1 2 n4





4 |R2| get + 3


4nminus 52



L R

(a)

L

RT

B

LprimeRprime

T prime

Bprime

(b)



References









30








minqisinQ

d(p q)












References






31





Abstract


1 Introduction










RH(P ) = R2 ⋃

QisinQQ


RH(P ) = CH(P ) ⋃

QisinQQ


pq


RH(P ) = CH(P ) ⋃

pqisinPiisin14

Qipq is P -free

Qipq

(a) (b) (c)



32


(a) (b) (c)











miniisin14



q




p

q

r

s




33


(a) (b)




A(i) =

i2sum

j=1

2

(j + 2

3

)if i is even

iminus12sum

j=1

2

(j + 2

3

)+

( i+12 + 2

3

)if i is odd




N

E

W

S

ε

N or W

N or E

S or ES or W

N

S




4 points When




and


re-









2 (1)





+


2minus 4

)+


2minus 2

)︸︷︷︸

S closest to S

(2)

34





+


2minus 3

)+


2minus 1

)︸︷︷︸

S closest to S

(3)



nminus44sum

k=1


i (4)


nminus44sum

k=1


i=1

i (5)


nminus44sum

k=1

ksumj=1


i (6)


nminus44sum

k=1

nminus44sum

j=k+1


i=1

i (7)


nminus44sum

k=1

ksumj=1


i (8)

and if j gt k it is

nminus44sum

k=1

nminus44sum

j=k+1


i (9)


nminus44sum

k=1

kminus1sumj=1


i=1

i (10)


k=1

nminus44sum

j=k


i=1

i (11)


nminus44sum

k=1

kminus1sumj=1


i=1

i (12)


k=1

nminus44sum

j=k


i (13)




References







A096338 Accessed 2021-01-01


35






(n3

)triangles of












References







36





Abstract



neΩ(radic




1 Preliminaries









k0(n) we denote the






k0(n)

α0 k O(n3) Ω(n3)α k0 O(n3 3

radick0) Ω(n2k0)

α k O(n4) Ω(n4)



Theorem 1

fα0

k0(n) isin Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))


37



2α0c These






Ω

((π

2α0minus 1)


2α0(k0 minus 1)2

))

As π2α0

is constant fα0


for k0 = n4 + 1 fα0



(n2






DepthPα0(x) = min








(x) ge 2

k0 minus 1


k0 minus 1

α0

α0

sm


α0

α0

s0

s1

C

k0 minus 1

T









in O(log n) time






38












pipi+1

ei

ei

fi

pi

pi+1



polygonal-curve fi









39
















1

2

3

4

empty

empty

empty

emptyempty

emptyempty

empty

empty

empty

empty

empty

x

y




pi

pj

Aij cupAjk cupAkl






References








40





Abstract














TRA2015-67788-P


det(


= 0





T prime(t) =det

(


det(


∣

∣

∣

∣

∣

∣

T=T (t)




∣

∣

T=T (t)



References



41






Abstract









References






42






Abstract








References




43






Abstract




a) b)

c) d)











References






44






Abstract


1 Introduction














2 Related works








45











for all i isin 1 n







Note that


10 + (2minus ε) + ε

= 2 +radic

10


2 +radic




radic3)


3)

= 7minusradic

3minus 2ε

a(minus1 0) b(1 0)

c(0radic3)

cprime(0 3)

bprime aprime

Dccprime

Daaprime Dbbprime



2 +radic

10 lt 7minusradic

3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)2 asymp 00528 (3)

Furthermore since


radic10 + ε

= 2 +radic

10 + 2ε


2 +radic


3minus 2ε

that is

ε lt (5minusradic

10minusradic

3)4 asymp 00264 (4)



46


(A) (B) (C) (D)

(E) (F) (G)

(H) (I) (J)




radic5











a b

c

z

aprime

bprime

cprime

Figure 3 Lemma 8










47


a

b

c

aprime

bprime

cprime

z

Figure 4 Lemma 9








b

a

cprime

bprime

aprime

z

c

o

Figure 5 Lemma 10







References





48







Abstract




1 Introduction








2




49


(a) (b) (c)






cc

c c(a) (b) (c)

38

38







c

Ri

Rj

pi

pj pprimekpk

z

120

Rk

pprimei

pprimej



50


Ri

Rj

Rk

tRi

tRj

fRj

fRifRk

tRk

c0

RjRi

Rk

xi











RjRi

Rk

(a) positive

(b) negative

Rk

Ri Rj












51


Ri

pici

Q




Lemma 6 ropt ge 14




return Ccelse







2

Acknowledgements


References











52





1 Introduction



S(K)

S(Bn)

)n

ge(

vol(K)

vol(Bn)

)nminus1

(1)









1

2

3

4

5 6

7

8

9

10 1112

13

14

15

16

17 1819 20

21

22

23



(right)


3 Main results



Gn

(K + t[minus1 1]n

)ge Gn

(Cr + t[minus1 1]n

) (3)


References



53















References






54





Abstract


1 Introduction






ei =




pij =

tj if i = 1




55








1 2 3 42 1 4 34 3 1 23 4 2 1

1 2 4 32 1 3 43 4 1 24 3 2 1

1 2 3 43 1 4 24 3 2 12 4 1 3

L1 L2 L3






k timesmk grid


Ik(Prms(L)) =sum


k ]

(1 + Iijk(Prms(L)))







56


LL1 L2 L3







1 3 23 2 12 1 3

and

1 3 4 24 2 1 32 4 3 13 1 2 4




1 3 2 5 44 2 5 1 35 4 3 2 13 5 1 4 22 1 4 3 5

1 3 4 5 25 2 1 3 44 5 3 2 12 1 5 4 33 4 2 1 5




57






Acknowledgements


References















58

EGC21 Author Index

Author Index





Esteban Guillermo 7





Jallu Ramesh K 49


59

EGC21 Author Index



Orden David 7 15 32








60

Portada


Actas_1pdf

i_Prefacepdf

ii_Comites

iii_Indice

Actas_2

1_Ponenciaspdf

invited_paper_1

_Ponenciaspdf

invited_paper_1pdf

invited_paper_3

paper_01

paper_02

paper_03

paper_04

Introduction

Previous results


Conclusions

paper_05

Introduction

Lower bound

Upper bound

Discussion

paper_06

paper_07

paper_08

Introduction



More results



paper_09

paper_10

Introduction

RVD in the plane


paper_11

Introduction



paper_12

Introduction

Boundary rectangles

paper_13

paper_14

Introduction


Concluding comments

paper_15

paper_16

Preliminaries




The (0k0)-hulls




paper_17

paper_18

paper_19

paper_20

paper_21

Introduction

Related works



paper_22

Introduction




paper_23

Introduction


Main results

paper_24

paper_25

Introduction




2_author_index

xix spanish meeting on computational geometry book of

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