Abstract Improved Upper Bounds on the Reflexivity of Point Sets

Improved Upper Bounds on the Reflexivity of Point Sets Eyal Ackerman∗Oswin Aichholzer†Bal´a zs Keszegh‡AbstractGiven a set S of n points in the plane,the reflexivity ofS,ρ(S),is the minimum number of reflex vertices in asimple polygonalization of S.Arkin et al.[4]proved thatρ(S)≤⌈n/2⌉for any set S,and conjectured that thetight upper bound is⌊n/4⌋.We show that the reflexivityof any set of n points is at most312n+O(1)≈0.4167n.1IntroductionGiven a set S of n≥3points in the plane,a polyg-onalization of S is a simple polygon P whose vertices are the points of S.Throughout this paper we assume that the points are in general position,that is,no three of them are collinear.A vertex of a simple polygon is reflex if the(interior)angle of the polygon at that ver-tex is greater thanπ.We denote byρ(P)the number of reflex vertices of a polygon P.The reflexivity of a set of points S,ρ(S),is the smallest number of reflex vertices any polygonalization of S must have.Further, we denote byρ(n)the maximum valueρ(S),such that S is a set of n points.Table1listsρ(n)for n≤10. These values were verified using a computer[2,4].n468100123∗School of Computing Science,Simon Fraser University,Burn-aby,BC,Canada,eyal@cs.sfu.ca†Institute for Software Technology,Graz University of Tech-nology,Austria,oaich@ist.tugraz.at Supported by the Austrian FWF Joint Research Project’Industrial Geometry’S9205-N12.‡Department of Mathematics and its Applications,Central European University,Budapest,Hungary,tphkeb01@phd.ceu.hu This work was done while visiting the School of Computing Sci-ence at Simon Fraser University.notion of reflexivity,its applications,and related prob-lems.Our main result is the following improvement for the upper bound ofρ(n).Theorem1ρ(n)≤3⌊n−212⌋+4.2Modified Reflexivity and Iterative Subdivision Recall that the convex hull of afinite set S of points, CH(S),is composed of the boundary and the interior of a convex polygon.By abuse of notation we will make no distinction between that polygon and CH(S).To prove a stronger variant of Theorem1wefirst introduce some notation.Let S be a set of points and let e be an edge of of CH(S).We denote byρe(S)the minimum possible number of reflex vertices in a polygonalization P of S, such that e is an edge of P.Similarly,¯ρ(S)=maxe is an edge of CH(S)ρe(S)¯ρ(n)=max|S|=n¯ρ(S)Obviouslyρ(n)≤¯ρ(n),so our goal is to derive good upper bounds for¯ρ(n).To this end wefirst provide a central lemma,which allows us to subdivide a point set in a way that we can consider the polygonalizations of the subsets rather independently.Lemma3Given an integer k>2,a set S of n>k points,and two points p,q∈S,such that pq is an edge of CH(S).Then,there exists a point t∈S\{p,q}and two sets L,R⊂S such that:1.L∪R=S,L∩R={t},q∈R,and p∈L;2.The triangle△pqt contains no other points from S;3.CH(R)∩CH(L)={t};1and4.|R|=k.Proof.Assume,w.l.o.g.,that p and q lie on the x-axis, such that p is to the left of q and all the remaining points are above the x-axis.Let t1be the point of S such that the angle∠t1pq is the smallest.Let S1be the set of points to the right of the line e1determined by q and t1.If|S1|>k−2,then define r1∈S1to be the point creating the(k−1)st smallest angle∠r1t1q, and denote by f1the line through t1and r1.Otherwise, if|S1|≤k−2let f1=e1.Set R1={q,t1}∪{p′∈S1|p′is to the right of f1}and L1=(S\R1)∪{t1}. We claim that t1,R1,and L1satisfy properties(1)–(3) of the lemma:(1)This property holds by the definition of R1and L1;(2)By the choice of t1the triangle△pqt1 is empty;(3)All the points in R1are to the right of f1, except for t1and possibly q.All the points in L1are to the left of f1,except for t1and possibly r1.However q and r1cannot both lie on f1.If|S1|≥k−2,then we also have that|R1|=k(either by the choice of r1or because|S1|=k−2,see Figure1for an illustration of the former case).Figure1:More than k−2points to the right of e1Figure2:Exactly k−2points to the right of e j Suppose now that|S1|<k−2.We define t i,e i, and S i for i>1and|S i−1|<k−2recursively.Let t i be the point that minimizes the angle∠t i pq among the points in L i−1\{t i−1}(note that this set of points is not empty if|S i−1|<k−2).Let e i be the line through q and t i,and let S i be the set of points to the right of e i.Similarly,we define r i,f i,R i,and L i.If |S i|>k−2define r i∈S i to be the point creating the (k−1)st smallest angle∠r i t i q,and denote by f i the line through t i and r i.Otherwise,if|S i|≤k−2set f i=e i. Set R i={q,t i}∪{p′∈S i|p′is to the right of f i}and L i=(S\R i)∪{t i}.Let j be the smallest integer such that|S j−1|<k−2and|S j|≥k−2(observe that j≤k−1since S i−1∪{t i−1}⊆S i if|S i|<k−2). We claim that t j,R j,and L j satisfy properties(1)–(4). Properties(1)–(3)are satisfied for the same arguments used for the case i=1.Notice that p must be to the left of f j by the choice of t j and since|S j−1|<k−2.By the choice of j we have|S j|≥k−2and thus,either there are exactly k−2points to the right of e j(see Figure2), or the choice of r j guarantees that|R j|=k. Note that Lemma3implies that pt is an edge of CH(L)and tq is an edge of CH(R),ing this fact we will apply the suggested subdivision in the next section in order to obtain ourfirst main result.3A New Upper BoundFigure3illustrates the subdivision obtained in the pre-vious section.The idea to prove an upper bound on ¯ρ(n)is to iteratively split a set into subsets of constant size,to obtain good polygonalizations for these sets,and then to combine them according to Lemma3.The basecase is covered by the following result.Figure3:The subdivision of S guaranteed by Lemma3 Lemma4Let S be a set of at most8points in the plane.Then¯ρ(S)≤2.Proof.The claim is clearly true for n≤5since any point on CH(S)is a convex vertex of any polygonaliza-tion of S.For6≤n≤8we prove the statement by a case-analysis over the size of the onion layers of S;see the full version of this paper for details.The correctness of the statement was also verified using a computer by checking all possible configurations of at most8points in general position. We are now ready for afirst upper bound on¯ρ(n).Theorem5¯ρ(n)≤3⌊n−27⌋+2=3⌊n−27⌋−1+1=3⌊n−2k−1⌋+l≤l+1111315ρ(n) 3..4 4..5 4..644 4..512⌋+4which implies Theorem2.Obviously determining¯ρ(n) for n≥14could further improve the constant of Corol-lary7,and we leave this for future research.5Further ResultsConjecture3.4in[4]states thatρ(n)=⌊n4⌋≤ρ(n)≤⌈n12⌋+2can be found in O(n log n)time.See the full version for a proof of this claim and an algorithm to generate the required subdivisions. Corollary6makes the following conjecture interest-ing.Conjecture2There is a constant c0such that¯ρ(n)≤ρ(n)+c0.Note that the stronger statement that¯ρ(S)≤ρ(S)+ O(1)for any set S might also hold.Conjecture2,if true,would mean that it is possible to get arbitrarily close to the best possible linear upper bound by checking onlyfinitely many small cases.In other words,suppose the conjecture holds and c is a constant such thatρ(n)≤cn.Then,for anyǫ>0 there is k=k(ǫ)such that if we verify thatρ(k)≤ck, then for n>k we haveρ(n)≤(c+ǫ)n+O(1).Indeed, k large enough such that ck+c0+112⌋+4reflex vertices,and such a polygo-nalization can be found in O(n log n)time.The basic idea of the proof is that by Lemma3we can subdivide S into somefixed-size parts and use a stronger result on each of these parts.It would be interesting tofind other applications of the subdivision suggested in Lemma3. It is challenging to determine the structure of sets maximizing the reflexivity forfixed cardinality.On the one hand we have the sets used in[4]to provide the bound ofρ(n)≥⌊n。