Rectangle-of-influence triangulations

Transcription

1 CCCG 2016, Vancoer, British Colmbia, Ag 3 5, 2016 Rectangle-of-inflence trianglations Therese Biedl Anna Lbi Saeed Mehrabi Sander Verdonschot 1 Backgrond The concept of rectangle-of-inflence (RI) draings is old and ell-stdied in the area of graph draing. A graph has sch a draing if e can assign points to its ertices sch that for eery edge (, ) the spporting rectangle (i.e., the minimal closed axis-aligned rectangle R(, ) containing and ) contains no other points. In the original setp, the graph had to hae an edge for eery pair of points ith an empty spporting rectangle (the strong model, see e.g. [8]). Later papers focs on eak RI-draings, here for eery edge the spporting rectangle mst be empty, bt not all sch edges mst exist. Of particlar interest are planar eak RI-draings of planar graphs, since these can alays be deformed to reside on an n n-integer grid. See e.g. [9, 1]. Or Reslts: In this paper, e take to comptational geometry problems ho to trianglate a point set and ho to flip beteen to geometric trianglations and apply them in the setting of rectangle-of-inflence draings. In particlar, e sho that any point set can be trianglated (ith some exceptions near the conex hll edges, hich e sho to be necessary) sch that the reslting planar straightline graph (PSLG) is an RI-draing. Next, e trn to the problem of flipping among geometric trianglations, i.e., conerting one trianglation into another throgh the operation of flipping the diagonal of one qadrangle. We sho that any RI-trianglation can be conerted into any other RI-trianglation by O(n 2 ) sch flipping-operations (and Ω(n 2 ) flips are reqired for some RI-trianglations). Moreoer, all intermediate trianglations are also RI-trianglations. The main idea is that the L -Delanay-trianglation (defined belo) is an RI-trianglation; it hence sffices to find one flip-operation that gets the RI-trianglation closer to the L -Delanay-trianglation in some sense. We also stdy ho to flip from any trianglation to an RItrianglation hile getting closer. Existing literatre: Trianglating point sets and polygons is one of the standard problems in comptational geometry. Any set of n points can be tri- Daid R. Cheriton School of Compter Science, Uniersity of Waterloo, Waterloo, ON N2L 3G1, Canada, {albi,biedl, s2mehrabi}@aterloo.ca. TB and AL spported by NSERC. Comptational Geometry Lab, School of Compter Science, Carleton Uniersity, Ottaa, Ontario, Canada. sander@cg.scs.carleton.ca anglated in O(n log n) time (e.g. by compting the Delanay trianglation), and the interior of a polygon can be trianglated in O(n) time [4]. The Delanay trianglation has been generalied to other nit discs. In particlar, for any conex compact shape C, the C-Delanay-trianglation is a trianglation sch that for eery edge (, ) there exists a homothet of C that contains and and no other points [5]. Arenhammer and Palini [2] stdied a nmber of their properties. If C is a nit sqare (i.e., the nit-circle in the L -metric), then this trianglation is called the L -Delanay-trianglation. The L -Delanaytrianglation can be compted in O(n log n) time [5]. Obsere that the L -Delanay-trianglation is an RItrianglation, bt an RI-trianglation need not be a C-Delanay-trianglation for any C becase the spporting rectangles are not necessarily homothets of each other or expandable into sch. Flipping among trianglations is also a ell-stdied problem; see [3] for an oerie of many ariants and existing reslts. It is ery easy to see that any (geometric) trianglation T 1 can be flipped into any other trianglation T 2 ia the intermediary of the Delanay trianglation T D : We can alays find a flip that gets s closer to the Delanay trianglation (in the sense that some angle-sm increases), so keep flipping from T 1 ntil e reach T D. Also compte the flips from T 2 to T D, and reersing these flips and combining the to flip-seqences then gies the reslt. For C-Delanaytrianglations, a similar reslt holds: e can alays flip to get closer to the C-Delanay-trianglation [2]. Notation: Let P be a set of n points that e assme to be in general position in the sense that no to points are on a horiontal or ertical line, and no 4 points are on a sqare. For any to points p, q P, define the spporting rectangle R(p, q) to be the minimal axis-aligned rectangle containing p and q. Define a spporting sqare S(p, q) to be a minimal axis-aligned sqare containing p and q; S(p, q) is not niqe. For any to points p, q P, e call a spporting rectangle/sqare of (p, q) empty if it contains no points of P other than p and q. An edge (p, q) beteen points of P is called an RIedge (L -edge) if R(p, q) is empty (resp., some spporting sqare S(p, q) is empty). Note that an L - edge is an RI-edge. An RI-polygon is a polygon for hich eery edge is an RI-edge. A planar straight-line graph (PSLG) is a trianglation if eery interior face

2 28 th Canadian Conference on Comptational Geometry, 2016 a Figre 1: (Left) A trianglation. (, ) is not locally RI. (, ) is locally L (therefore locally RI), bt not globally L. (Right) A polygon (solid) ith its trapeoidation (thin dashed). Edge (b, f) old be added ith the first method, edge (c, e) ith the second. of the PSLG is a triangle. An RI-trianglation (L - trianglation) is a trianglation for hich eery edge is an RI-edge (L -edge). We sometimes se nrestricted trianglation for a trianglation that need not be an RI-trianglation. A trianglation is maximal if it contains as many edges as possible hile staying ithin the additional reqirements that e impose. Ths, a maximal RI-trianglation is an RI-trianglation ith as many edges as possible hile haing only RI-edges, and similarly for maximal L -trianglations. For an edge (, ) in a trianglation, ertex is facing (, ) if there exists an interior face {,, }. Eery interior edge has exactly to ertices facing it. We say that (, ) is locally RI (resp. locally L ) if R(, ) (resp.. some spporting sqare S(, )) contains none of the ertices facing (, ). Sometimes e say that an RI-edge (L -edge) is globally RI (globally L ). 2 RI-trianglating an RI-polygon We first sho that any RI-polygon P can be RItrianglated, i.e., made into a trianglation by adding only RI-edges in its interior (presming no extra points are inside P ). To do so, first find a trapeoidation of P, i.e., extend ertical sbdiision lines from all ertices. This can be done in linear time [4]. We add RI-edges in to ays. First, check hether there is any trapeoid that has ertical sbdiision lines on both its left and right sides, and for hich the to ertices, that cased these lines are on opposite sides (top/bottom) of the trapeoid. If so, add the diagonal (, ). This is an RI-edge since R(, ) is contained ithin the spporting rectangles of the top and bottom edge of T and the interior of P, all of hich are empty. See edge (b, f) in Fig. 1. Secondly, if no sch trapeoid exists, then any remaining face is x-monotone, i.e., it consists of to x- monotone chains from left to right. Since the first method does not apply, one of the to chains is a single edge. Consider one sch piece ith (say) the bottom chain a single edge (, ). All ertices in the top chain f b c d e C IV C I C III Q C II Figre 2: The maximal-hll of a set of points, and ho to add corner-points and edges to them. are otside R(, ) and hence hae larger x-coordinate than,. Let be a local maximm in the top chain, and connect the neighbors of. The ne edge is an RI-edge, becase its spporting rectangle is contained in the nion of the ones of the edges incident to as ell as R(, ) and the interior of P. See edge (c, e) in Fig. 1 (ith =f, =e, =d). So e add RI-edges ntil all interior faces are triangles. This takes linear time (once the trapeoidation is fond), since finding the local minimm/maximm for the second rle can be done in O(1) amortied time. Theorem 1 Eery RI-polygon can be trianglated sing only RI-edges in linear time. 3 Oter face considerations The remaining sections deal ith trianglations of point sets, rather than polygons. Here there arise some complications at the oter face. For any maximal (nrestricted) trianglation of P, the oter face consists of the conex hll CH (P ). If some edge of CH (P ) is not an RI-edge, then it obiosly cannot be in an RItrianglation. We begin by characteriing the oter face of any RI-trianglation. The maxima-hll: We need some definitions that are closely related to the rectilinear conex hll (see e.g. [10]) and the maxima of a set of ectors (see e.g. [7]). Define a first qadrant of a point p to be the set {(x, y) : x x(p), y y(p)}. Define the first-qadrantchain C I to be all those points p in P for hich the first qadrant relatie to p contains no other points of P ; e sort these points by increasing x-coordinate (hence by decreasing y-coordinate), and connect them ith straight-line segments in this order. Similarly define three other chains C II, C III, C IV sing the three other types of qadrants. Note that C I and C II share one endpoint (the one ith maximm x-coordinate), and similarly for the other chains, so e can combine the for polygonal chains into one closed polygonal chain that e call the maxima-hll MH (P ). Note that some chains may hae edges in common, bt no to edges cross, so MH (P ) may self-oerlap bt it has a elldefined interior region. See Fig. 2. II III I IV

3 CCCG 2016, Vancoer, British Colmbia, Ag 3 5, 2016 In the appendix, e proe the folloing. Lemma 2 For any point set P, the maxima-hll consists of L -edges (hence RI-edges). Lemma 3 For any point set P, any RI-edge (, ) is ithin the region bonded by MH (P ). Combining them gies: Corollary 1 An RI-trianglation is maximal if and only if its oter face consists of the maxima-hll. Adding corner-points: It ill be cmbersome to deal directly ith edges that are on the conex hll bt not RI-edges. To simplify or life, e add points as follos. For any point set P, let P + be the set obtained by adding for corner-points I, II, III, IV that form an axis-aligned rectangle (e rotate it slightly to be in general position) and are otside the bonding box of P, ith i in the ith qadrant relatie to all points of P. See Fig. 2. Notice that the conex hll of P + consists of L -edges. Lemma 4 Let T + be a maximal RI-trianglation of P +, and let T := T + { I, II, III, IV }. Then T is a maximal RI-trianglation of T. Proof. Clearly any edge of T is an RI-edge, so e only need to arge maximality. Assme (p, I ) is an edge in T + for some p II, III, IV. Then R(p, I ) contains no other point, and is to the right and/or aboe II, III, IV. So expanding R(p, I ) into the first qadrant of p does not add points of P, hence p is on the maxima-hll. So remoing { I, II, III, IV } from T + leaes an RI-trianglation here the oter face is the maxima-hll. By Corollary 1 this is maximal. 4 RI-trianglating a point set In this section, e stdy ho to find a maximal RItrianglation of a gien point set. It is obios that this exists (for example the L -Delanay trianglation ill do), bt or algorithm is especially simple. Theorem 5 A maximal RI-trianglation of a point set P can be compted in O(n log n) time, or O(n) time if P is sorted by x-coordinate. Proof. As before, add corner-points I, II, III, IV to obtain point set P +. Sort the points by x- coordinates, and add an edge beteen any to consectie points; these are clearly RI-edges. Also add the cycle I, II, III, IV ; these are also RI-edges. No e hae a PSLG hose faces consist of RI-polygons. Trianglate each polygon ith Theorem 1. We obtain an RI-trianglation T + of P +, and it is clearly maximal since all conex-hll edges are in it. By Lemma 4, deleting the for corner-points gies the reslt. 5 Flipping and RI-trianglations In this section, e inestigate flipping hile maintaining RI-trianglations or (if e start ith an nrestricted one) getting closer to an RI-trianglation. The natral intermediary here is the L -Delanay trianglation, hich is an RI-trianglation. So e sho that any trianglation can be flipped to an RI-trianglation hile getting closer, and then that any RI-trianglation can be flipped to the L -Delanay trianglation hile remaining an RI-trianglation throghot. 5.1 Flipping to an RI-trianglation Let P + be a point set for hich any conex hll edge is an RI-edge (e ill arge later ho to remoe this assmption). Let T be an arbitrary trianglation of P +. A bad triangle {,, } in T is a face {,, } sch that R(, ). After possible rotation, assme that is in the 2nd qadrant and is in the 4th qadrant relatie to, ith edge (, ) roted aboe. Define the special region of bad triangle {,, } to be all points p aboe (, ) ith x() x(p) x() (inclding ) and all points p to the right of (, ) ith y() x(p) y() (inclding ). See Fig. 3(a). The definition is symmetric for the other three possible rotations of a bad triangle. No define for any trianglation T the potential fnction Φ(T ) to be the sm, oer all bad triangles {,, }, of the nmber of points in the special region of {,, }. Lemma 6 Let T be any trianglation of P + ith Φ(T ) > 0. Then there exists an edge in T that e can flip so that the reslting trianglation T satisfies Φ(T ) < Φ(T ). Proof. Since Φ(T ) > 0, it mst hae at least one bad triangle, and hence edges that are not locally RI. Of all those edges, let (, ) be the one that maximies the L 1 -distance beteen its endpoints, ths x() x() + y() y() is maximm among all edges that are not locally RI. Since conex hll edges are RI-edges, e kno that (, ) is an interior edge and has to facing ertices. Let be a ertex facing (, ) that is in R(, ), and assme again that (after possible rotation) the bad triangle {,, } has (or ) in the 2nd (4th) qadrant of ith (, ) aboe. Let be the other ertex facing (, ). We claim that x() > x(). We kno that is aboe the line throgh (, ) since {,, } is a face. If e had x() < x(), then R(, ), and so (, ) is not locally RI bt its L 1 -distance is longer than the one of (, ), contradicting or choice of edge (, ). Therefore e kno x() > x(), and symmetrically one arges y() > y(). Notice that therefore qadrilateral {,,, } is conex and so edge (, ) is flippable. We claim that regardless of the position of, doing this

4 28 th Canadian Conference on Comptational Geometry, 2016 Lemma 7 Let T be a trianglation sch that all edges are locally RI and all oter face edges are globally RI. Then all edges are globally RI. (a) (b) Proof. Sppose that some edges of T are not globally RI, hence contain points inside their spporting rectangles. Let e = (, ) be the edge ith the closest sch point, say. Since e is an interior edge by assmption, it has to facing ertices; let be the ertex facing e that is on the same side of the spporting line of e as is. Sppose that and lie right of e, and is the left endpoint of e; see Fig. 1. Obsere that mst lie either aboe R(, ) (ithin the same x-range) or to the right of R(, ) (ithin the same y-rage), else some edge of {,, } old not be locally RI or {,, } old not be a face. Assme lies strictly aboe R(, ), ithin the same x-range. Then (, ) is also not globally RI, since lies in R(, ). Bt is closer to (, ) than to e, contradicting or choice of e. (c) Figre 3: (a) The special region (shaded) of bad triangle {,, }. (b-d) Possible positions of the other facing ertex. Light gray regions ere in the special region of the bad triangle {,, }. (d) flip improes the potential fnction. To sho this, e distingish here is located x() > x(), y() > y() (see Fig. 3(b)): In this case, neither of the to ne triangles {,, } and {,, } is bad. Since triangle {,, } sed to be bad, Φ decreases by at least 2. x() > x(), y() < y() (see Fig. 3(c)): In this case, the ne triangle {,, } is bad, bt {,, } is not. The special region of triangle {,, } is a strict sbset of the one for triangle {,, }, and in particlar, excldes. Hence Φ decreases. R(, ) (see Fig. 3(d)): In this case, both ne triangles {,, } and {,, } are bad. Bt both triangles {,, } and {,, } that ere remoed ere bad, and the special regions of the ne triangles are strict sbsets of the special regions of the old triangles that, in particlar, contain and only once, instead of tice. So again Φ decreases. The cases for x() < x(), y() > y() and for R(, ) are symmetric. Note in particlar that if Φ(T ) = 0, then it has no bad triangles (becase any bad triangle has at least to points in its special region); therefore it has no edge that is not locally RI. Ptting the to lemmas together, e can hence flip from any trianglation to an RI-trianglation hile steadily improing the potential-fnction. Initially there are O(n) bad triangles, each of hich defines a special region containing at most n points, so Φ(T ) O(n 2 ). Each flip improes the fnction, and e are done hen it is 0, hich means that the nmber of flips is O(n 2 ). We smmarie: Lemma 8 Any maximal trianglation of P + can be conerted into an RI-trianglation of P + sing O(n 2 ) flips. We arge in the appendix that this bond is sometimes tight. Lemma 9 There are trianglations that cannot be conerted into an RI-trianglation ith o(n 2 ) flips. Proof. Consider a set of points spread eenly oer to opposing conex chains, sch that the only possible RIedges beteen the to chains connect point i on the first chain to point i and i + 1 on the opposing chain (see Fig. 4). The edges connecting consectie points on each chain are not intersected by any other possible edge, hich implies that these edges are present in eery trianglation and can neer be flipped. Ths, the region beteen the to chains is independent of the otside regions. Frther, by constrction, this center region has a niqe RI-trianglation (Fig. 4, right). No consider a trianglation that incldes the long diagonal connecting one end of the first conex chain to the opposite end of the other chain (Fig. 4, left). Transforming this trianglation into the niqe RI-trianglation reqires Ω(n 2 ) flips, by an argment analogos to the one gien by Hrtado et al. [6, Theorem 3.8]

5 CCCG 2016, Vancoer, British Colmbia, Ag 3 5, 2016 Figre 4: Trning the trianglation on the left into an RI-trianglation reqires Ω(n 2 ) flips for the region beteen the chains to become the only possible RItrianglation shon on the right. Arbitrary point sets: It remains to arge ho to handle point sets P here not all conex-hll edges are RI-edges. Assme e are gien a maximal trianglation T of P, and e old like to flip that to a maximal RItrianglation T R. Add corner-points I, II, III, IV as before to obtain point set P + and connect them in a cycle. Connect i (for i {I, II, III, IV }) to all points p on the conex hll of P for hich qadrant i contains only p and i. This gies a trianglation T + of P + becase the conex hll is the oter face T. See Fig. 2. The conex hll of P + consists of RI-edges, so there exists a seqence σ of flips that trns T + into a maximal RI-trianglation T + R. Lemma 10 Throghot flip-seqence σ, all edges incident to corner-points are RI-edges. Therefore any edge being flipped is not incident to a corner-point. Proof. The first claim implies the second becase flipped edges ere not locally RI. We proe the first claim for I only. Apart from the edges to II and IV, ertex I has an edge only to a point p for hich the first qadrant is empty, so R( I, p) is empty and the claim holds. We no sho that any time e flip an edge (, ) sch that the ne edge is (, I ) for some,,, this ne edge is an RI edge. Since I is otside the bonding box of P, it is otside R(, ). Since (, ) as not locally RI (by or choice of edges to flip), therefore R(, ). In the naming of Fig. 3, e hae I = and the case of Fig. 3(b) applies since I is in the first qadrant of. We already kne that in this case (, ) is locally RI. Bt since = I, edge (, ) is globally RI: R(, I ) lies ithin the nion of R(, I ) and R(, I ) (both empty, becase these edges are incident to I and hence RI-edges), and the interior of triangles {,, } and {, I, } (hich are faces). We no create a seqence of flips and edge deletions for T that leads to a maximal RI-trianglation by mirroring the flip-seqence of T +. We maintain the claim that at any time trianglation T eqals T + ith the corner-points remoed. Clearly this holds initially. Say the next flip for T + as to flip (, ) to (, ). We kno, P. If, P, then (by indction) the 4-cycle,,, that exists in T + also exists in T, and so e can do the exact same flip in T and the claim holds. Else, one of, is a corner-point. Do an edge-deletion in T, i.e., remoe edge (, ) ithot adding a ne one. The claim still holds since one end of (, ) is not in P. We end ith a trianglation T R of P that eqals T + R ith the corner-points remoed. By Lemma 4, this is a maximal RI-trianglation. Theorem 11 We can conert any maximal trianglation into an RI-trianglation by doing O(n 2 ) flips and O(n) edge-deletions. 5.2 Flipping beteen RI-trianglations As explained earlier, to flip beteen maximal RItrianglations hile maintaining an RI-trianglation, it sffices to sho that eery maximal RI-trianglation can be flipped into the L -Delanay-trianglation. To proe this, e se again a potential-fnction argment, bt ith a different fnction. We need the folloing: Lemma 12 [2] Let T be a maximal trianglation here all edges are locally L. Then all edges are globally L. Or potential fnction depends on haing fixed, for eery edge (, ) of the crrent trianglation, a particlar spporting sqare S(, ), and conting the nmber of points of P in it. We define Ψ(T ) to be the sm, oer all edges (, ) of the nmber of points in S(, ). When e flip, e are free to choose a spporting sqare for the ne edge. Lemma 13 Let T be a maximal RI-trianglation that is not the L -Delanay trianglation. There exists an edge e sch that flipping e reslts in an RI-trianglation T and e can assign a spporting sqare to e sch that Ψ(T ) < Ψ(T ). Proof. By Lemma 12 T has some edge (, ) that is not locally L. Up to symmetry, e may assme that the height Y of R(, ) is no smaller than its idth. All spporting sqares of (, ) are then in the horiontal strip H of height Y beteen and. Since T is a maximal RI-trianglation, its oter face is the maxima-hll and consists of L -edges. So (, ) is an interior edge. Let and the to ertices facing (, ). We claim that both and mst be in strip H. To see this, recall that e hae an RI-trianglation, and hence rectangle R(, ) is empty. This rectangle bisects strip H. So if, say, is not in H, then at most one facing ertex is in H, hich means that to one side of R(, ) in H there is no facing ertex of (, ). We cold hence pick a spporting sqare S of (, ) that consists of

6 28 th Canadian Conference on Comptational Geometry, 2016 to the L -Delanay-trianglation hile maintaining an RI-trianglation. H S(, ) This bond is tight. Fig. 6 shos to RItrianglations of a point set that forms to conex chains. Hrtado et al. [6, Theorem 3.8] shoed that their flip distance is Ω(n 2 ) een ithot the restriction of sing only RI-trianglations. R Figre 5: Finding a spporting sqare for (, x). R(, ) extended to that side. Then S contains neither nor, and (, ) old be locally L, a contradiction. So both and are in strip H. By planarity they mst be on opposite sides of R(, ), say is to the left and is to the right. Qadrangle {,,, } hence is dran conex and edge (, ) is flippable. Define R be the minimm rectangle containing,,,, and notice that it is contained in the nion of the spporting rectangles of the edges (, ), (, ), (, ), (, ), (, ). Since e had an RI-trianglation, therefore R contains no points other than these 4. Also,,, are all on the bondary of R. Therefore R(, ) is empty and the ne edge (, ) is an RI-edge. No e explain ho to find a spporting sqare for (, ). Let X be the idth of R, and notice that X < Y, since X = Y old imply for points on a sqare and X > Y old mean that some sqare ithin R spports (, ) and does not contain,, contradicting that (, ) is not locally L. No consider the nion of R and the sqare S(, ) that as sed as spporting sqare for (, ). Since (, ) as not locally L, at least one of {, } is in S(, ), hence S(, ) R contains at most one more point than S(, ). Let R be the rectangle obtained by shrinking S(, ) R to height Y ε in sch a ay that, R. We choose ε so small that and remain in R and sch that Y ε > X. So R contains at least one feer points than S(, ). Finally shrink R in idth ntil it is a sqare; e can do this and retain and in it, since R is taller than R is ide. Using the reslting sqare for S(, ) decreases Ψ as desired. Theorem 14 Any maximal RI-trianglation T can be conerted into any other maximal RI-trianglation T by doing O(n 2 ) flips, and all intermediate trianglations are maximal RI-trianglations. Proof. As before it sffices to arge this if T is the L -Delanay trianglation. Compte an arbitrary set of spporting sqares for T. Initially there are O(n) edges in the trianglation, each of hich has at most n points in its spporting sqare, so Ψ(T ) O(n 2 ). Applying the aboe flip means that ith O(n 2 ) flips e get Figre 6: A pair of RI-trianglations sch that Ω(n 2 ) flips are reqired to transform one into the other. References [1] S. Alamdari and T. Biedl. Open rectangleof-inflence draings of non-trianglated planar graphs. In Proc. GD 12, pages , [2] F. Arenhammer and G. Palini. On shape Delanay tessellations. Inf. Process. Lett., 114(10): , [3] P. Bose and F. Hrtado. Flips in planar graphs. Compt. Geom., 42(1):60 80, [4] B. Chaelle. Trianglating a simple polygon in linear time. Discrete Compt. Geom., 6: , [5] R. Drysdale. A practical algorithm for compting the Delanay trianglation for conex distance fnctions. In Proc. SODA 90, pages , [6] F. Hrtado, M. Noy, and J. Urrtia. Flipping edges in trianglations. Discrete Compt. Geom., 22(3): , [7] H. T. Kng, F. Lccio, and F. P. Preparata. On finding the maxima of a set of ectors. J. ACM, 22(4): , [8] G. Liotta, A. Lbi, H. Meijer, and S. Whitesides. The rectangle of inflence draability problem. Compt. Geom., 10(1):1 22, [9] K. Mira, T. Matsno, and T. Nishieki. Open rectangle-of-inflence draings of inner trianglated plane graphs. Discrete Compt. Geom., 41(4): , [10] T. Ottmann, E. Soisalon-Soininen, and D. Wood. On the definition and comptation of rectlinear conex hlls. Inf. Sci., 33: , 1984.

7 CCCG 2016, Vancoer, British Colmbia, Ag 3 5, 2016 A Proof of Lemma 2 We first need a helper-reslt. Lemma 15 Let (, ) be any edge on the first-qadrant chain C I, say x() < x(). Let Q be the first qadrant of point (x(), y()), i.e., it has and on its bondary. Then no point (other than, ) is in Q. R(, ) (recall that (, ) is an RI-edge), they in fact mst hae y-coordinate at least y(), hence be aboe R(, ). In conseqence the only edge of C that can intersect R(, ) is the edge incident to p, bt this edge is also aboe (, ) since p is aboe. So all of C is aboe (, ) as desired. Proof. See Figre 2. Assme to the contrary that Q contained points other than,, and let be the one that maximies the x-coordinate. Since the first qadrants relatie to and are empty, mst be in R(, ). By general position e hae x() < x() < x() and y() < y() < y(). The first qadrant of cannot contain any other point, for all sch points old be to the right of (contradicting the choice of ). So point shold hae been in C I, contradicting that, ere consectie in C I. p C p To complete the proof of Lemma 2, note that for any edge (, ) on C I, rectangle R(, ) is part of this firstqadrant, so it is empty, and e can expand it into a sqare spporting, that is empty. So any edge on C I is an L -edge. Similar argments hold for the other three qadrant-chains, hich proes Lemma 2. Figre 7: For the proof of Lemma 3. B Proof of Lemma 3 Proof. We aim to sho that any RI-edge (, ) is on or belo C I C II. Similar proofs in the other three directions sho that (, ) is either on or enclosed by the maxima-hll as desired. Up to renaming and symmetry, e may assme that is left of and higher than. Consider the maximal ertical strip S containing (, ). This strip mst intersect C I C II, since the rightmost point of P is in C I and the leftmost of P is in C II. Let C be the part of C I C II ithin S, say its ends are p (on a ertical line ith ) and p (on a ertical line ith ). Applying Lemma 15 (or the eqialent for second qadrants) to the edge containing p shos that the ertical ray pard from p contains no other points of P. So either p =, or p is aboe. Similarly p = or p is aboe. In hat follos, e se the term ertex of C for a point on C that is also a point of P, hile point of C refers to an arbitrary point that belongs to C. If C has no ertices, then it is a single line segment p p, and by the aboe (, ) is belo that. So assme that C has ertices. Since C (as part of C I C II ) consists of an increasing chain folloed a decreasing chain, its minima (ith respect to y-coordinate) appear at the ends. Since p is aboe, therefore all ertices of C hae y-coordinate at least y(). Since none of these ertices are inside