Unfolding Face-Neighborhood Convex Patches - Smith College ...

cs.smith.edu

Unfolding Face-Neighborhood Convex Patches - Smith College ...

25 th Canadian Conference on Computational Geometry, 2013petal unfolding. 2 The three positive results mentionedabove are all via petal unfoldings: the dome unfolding,the prismoid unfolding, and Pincu’s edge-neighborhoodpatch unfolding. Thus Fig. 1 without its base, whichis a edge-neighborhood patch, can be petal-unfolded:simply cut each lateral edge a i b i . We henceforth concentrateon petal unfoldings (until the final Sec. 4).New Results. Given the collection of partial resultsand unsolved problems reviewed above, it is naturalto explore petal unfoldings of vertex-neighborhoodpatches. Our results are as follows:1. Define a topless prismatoid as one with A removed;so it is a special (non-jagged) vertex-neighborhoodN v (B). We prove that every topless prismatoidwhose lateral faces are triangles has a petal unfoldingwithout overlap (Thm. 7). This shows that, insome sense, placing the top A is an obstruction tounfolding prismatoids.2. Via a counterexample polyhedron P (Fig. 8), weshow that not every vertex-neighborhood patchN v (F ) has a non-overlapping petal unfolding.3. However, if P is non-obtusely triangulated, N v (F )does have a non-overlapping petal unfolding for everyface of P (Thm. 8).4. This leads to a non-overlapping unfolding of a restrictedclass of prismatoids (Cor. 9).I am hopeful that the main proof technique—obtaininga result for flat patches and then lifting into z > 0—willlead to further results.We conclude in Section 4 with a conjecture that notevery edge-neighborhood has a non-overlapping “zipperunfolding.”2 Topless Prismatoid Petal UnfoldingLet P be a prismatoid, and assume all lateral faces aretriangles, the generic and seemingly most difficult case.Let A = (a 1 , a 2 , . . .) and B = (b 1 , b 2 , . . .). Call a lateralface that shares an edge with B a base or B-triangle,and a lateral face that shares an edge with A a top orA-triangle. A petal unfolding cuts no edge of B, andunfolds every base triangle by rotating it around its B-edge into the base plane. The collection of A-trianglesincident to the same b i vertex—the A-fan AF i —mustbe partitioned into two groups, one of which rotatesclockwise (cw) to join with the unfolded base triangleto its left, and the other group rotating counterclockwise(ccw) to join with the unfolded base triangle to itsright. Either group could be empty. Finally, the top Ais attached to one A-triangle. So a petal unfolding has2 Called a “volcano unfolding” in [6, p. 321].choices for how to arrange the A-triangles, and whichA-triangle connects to the top.It remains possible that every prismatoid has a petalunfolding: so far I have not been able to find a counterexample.Now we turn to our main result: everytopless prismatoid has a petal unfolding. An exampleof a petal unfolding of a topless prismatoid is shown inFig. 2.Figure 2: Unfolding of a topless prismatoid. A-fans arelightly shaded.Even topless prismatoids present challenges. For example,consider the special case when there is only oneA-triangle between every two B-triangles. Then theonly choice for placement of the A-triangles is whetherto turn each ccw or cw. It is natural to hope that rotatingall A-triangles consistently ccw or cw suffices toavoid overlap, but this can fail. A more nuanced approachwould turn each A-triangle so that its (at mostone) obtuse angle is not joined to a B-triangle, but thiscan fail also.The proof that topless prismatoids have petal unfoldingsfollows this outline:1. An “altitudes partition” of the plane exterior to thebase unfolding (petal unfolding of N e (B)) is definedand proved to be a partition.2. It is shown that both P and this partition vary ina consistent manner with respect to the separationz between the A- and B-planes.3. An algorithm is detailed for petal unfolding the A-triangles for the “flat prismatoid” P(0), the limitof P(z) as z → 0, such that these A-triangles fitinside the regions of the altitude partition.4. It is proved that nesting within the partition regionsremains true for all z.2.1 Altitude PartitionWe use a i and b j to represent the vertices of P, andprimes to indicate unfolded images on the base plane.Let B i = △b i b i+1 a ′ j be the i-th base triangle. Saythat B ∪ = B ∪ ( ⋃ i B i) is the base unfolding, the petal


CCCG 2013, Waterloo, Ontario, August 8–10, 2013unfolding of N e (B) without any A-triangles. The altitudepartition partitions the plane exterior to B ∪ .Let r i be the altitude ray from a ′ j along the altitudeof B i . Finally, define R i to be the region of the planeincident to b i , including the edges of the B i−1 and B itriangles incident to b i , and bounded by r i−1 and r i .See Fig. 3.Thus the vertices a ′ j (z) of the base unfolding “rideout” along the altitude rays r i as z increases (see aheadto Fig. 6 for an illustration). Therefore the combinatorialstructure of the altitude partition is fixed, andR i only changes geometrically by the lengthening of theedges b i a ′ j and b i+1a ′ j and the change in the angle gapκ bi (z) at b i .r 3r 2R 4 a' 3 a' 2B 3bBb 2 4r 34B 4a' 4b B b5 2bB 1 B51a' 5a'r 15R r 1 1Figure 3: Partition exterior to B ∪ by altitude rays r i .Here both A and B are pentagons; in general therewould not be synchronization between the b i and a i indices.The A-triangles are not shown.Lemma 1 No pair of altitude rays cross in the baseplane, and so they define a partition of that plane exteriorto the base unfolding B ∪ .Our goal is to show that the A-fan AF i incident to b ican be partitioned into two groups, one rotated cw, oneccw, so that both fit inside R i .2.2 Behavior of P(z)We will use “(z)” to indicate that a quantity varies withrespect to the height z separating the A- and B-planes.Lemma 2 Let P(z) be a prismatoid with height z.Then the combinatorial structure of P(z) is independentof z, i.e., raising or lowering A above B retainsthe convex hull structure.We will call P(0) = lim z→0 P(z) a flat prismatoid.Each lateral face of P(0) is either an up-face or a downface,and the faces of P(z) retain this classification inthat their outward normals either have a positive or anegative vertical component.Lemma 3 Let P(z) be a prismatoid with height z, andB ∪ (z) its base unfolding. Then the apex a ′ j (z) of eachB i ′(z) triangle △b ib i+1 a ′ j (z) in B∪ (z) lies on the fixedline containing the altitude of B i ′(z).2.3 Structure of A-fansHenceforth we concentrate on one A-fan, which we alwaystake to be incident to b 2 , and so between B 1 =△b 1 b 2 a 1 and B 2 = △b 2 b 3 a k . The a-chain is the chainof vertices a 1 , . . . , a k . Note that the plane in R 3 containingface B 1 of P supports A at a 1 , and the planecontaining B 2 supports A at a k . Let β = β 2 be the baseangle at b 2 : β = ∠b 1 b 2 b 3 . We state here a few facts trueof every A-fan.1. An a-chain spans at most “half” of A, i.e., a portionbetween parallel supporting lines to A (because β >0).2. If an A-fan is unfolded as a unit to the base plane,the a-chain consists of convex, reflex, and convexportions, any of which may be empty. So, excludingthe first and last vertices, the interior vertices of thechain have convex angles, then reflex, then convex.3. Correspondingly, an A-fan consists of down-facesfollowed by up-faces followed by down-faces, whereagain any (or all) of these three portions could beempty.4. All four possible combinations of up/down are possiblefor the B 1 and B 2 triangles.The second fact above is not so easy to see. The intuitionis that there is a limited amount of variationpossible in an a-chain. It is the third fact that we willuse essentially; it will become clear shortly.2.4 Flat Prismatoid Case AnalysisHow the A-fan is proved to sit inside its altitude regionR for P(0) depends primarily on where b 2 sits withrespect to A, and secondarily on the three B-vertices(b 1 , b 2 , b 3 ). Fig. 4 illustrates one of the easiest cases,when b 2 is in C, the convex region bounded by the a-chain and extensions of its extreme edges. Then all theA-faces are down-faces, the a-chain is convex, one of thetwo B-faces is a down-face (B 2 in the illustration), andwe simply leave the A-fan attached to that B down-face.A second case occurs when b 2 is on the reflex side ofA. An instance when both B-triangles are down-facesis illustrated in Fig. 5. Now the A-fan consists of downfacesand up-faces, the up-faces incident to the reflexside of the a-chain. These up-faces must be flipped in


25 th Canadian Conference on Computational Geometry, 2013R a 4a 5a 6=a ka 3a 2a 1b 2r 1A' sta' 1r 2CB 2b 3=b 1Figure 4: Case 1b. Here we have illustrated b 1 = b 3 toallow for the maximum a-chain extent.the unfolding, reflected across one of the two tangentsfrom b 2 to A. A key point is that not always will bothflips be “safe” in the sense that they stay inside the altituderegion. Fortunately, one of the two flips is alwayssafe:a' 3 a' 6a' 5 a'4a' b42ba 21 aa' 45Ba 5 1B 1a 6=a ka 6=aB k2B 2(a)a 3aa 32a 2a 1Figure 5: Case 2a. The A-triangles between the tangentsb 2 to a 3 and b 2 to a 6 are up-faces. (a) shows theup-faces flipped over the left tangent b 2 a 6 , and (b) whenflipped over the right tangent b 2 a 3 .Lemma 4 Let b 2 have tangents touching a s and a t ofA. Then either reflecting the enclosed up-faces acrossthe left tangent, or across the right tangent, is “safe” inthe sense that no points of a flipped triangle falls outsidethe rays r 1 or r k .The remaining cases are minor variations on those illustrated,and will not be further detailed.2.5 Nesting in P(z) regionsThe most difficult part of the proof is showing that thenesting established above for P(0) holds for P(z). Akey technical lemma is this:Lemma 5 Let △b, a 1 (z), a 2 (z) be an A-triangle, withangles α 1 (z) and α 2 (z) at a 1 (z) and a 2 (z) respectively.Then α 1 (z) and α 2 (z) are monotonic from their z = 0values toward π/2 as z → ∞.(b)L xa k(z)a x(z)r2a t(z)α ta kBB 1a'' 12a(a)1b 3a xa ta' 1(z)a' 2(z)A' st(z)a' 3(z)a ta' 1b 2a 4a 3a 2 a''b 11(b)b 2b 1a xα tr 1B' 1a kA x(z)a 1a'' 1(z)Figure 6: (a) z = 0. △a t a x a k encloses the convex section,and △a 1 b 2 a t encloses the reflex section. (b) z > 0.Reflex angle α t (z) decreases as z increases.I should note that it is not true, as one mighthope, that the apex angle at b of that A-triangle,∠a 1 (z), b, a 2 (z), shrinks monotonically with increasingz, even though its limit as z → ∞ is zero. Nor is theangle gap κ b (z) necessarily monotonic. These nonmonotonicangle variations complicate the proof.Another important observation is that the sorting ofba i edges by length in P(0) remains the same for allP(z), z > 0. More precisely, let |ba i | > |ba j | for twolateral edges connecting vertex b ∈ B to vertices a i , a j ∈A in P(0). Then |ba i (z)| > |ba j (z)| remains true for allP(z), z > 0.For the nesting proof, I will rely on a high-level description,and one difficult instance. At a high level,each of the convex or reflex sections of the a-chain areenclosed in a triangle, which continues to enclose thatportion of the a-chain for any z > 0 (by Fact 1, Sec. ??).The reflex enclosure is determined by the tangents fromb 2 to A: △a s b 2 a t . So then the task is to prove these (atmost three) triangles remain within R(z). Fig. 6 shows


CCCG 2013, Waterloo, Ontario, August 8–10, 2013a case where there is both a convex and a reflex section.Were there an additional convex section, it wouldremain attached to B 1 (z) and would not increase thechallenge.Lemma 6 If the a-chain consists of a convex and areflex section, and the safe flip (by Lemma 4) is to aside with a down-face (B 2 in the figure), then AF ′ (z) ⊂R(z): the A-fan unfolds within the altitude region forall z.W 3c 2a 2c 1A 2W 2b 3 ba 1 2A 1 W 1Bzb 1I have been unsuccessful in unifying the cases in theanalysis, despite their similarity. Nevertheless, the conclusionis this theorem:Theorem 7 Every triangulated topless prismatoid hasa petal unfolding.It is natural to hope that further analysis will lead toa safe placement of the top A (which, alas, might notfit into any altitude-ray region).3 Unfolding Vertex-NeighborhoodsWe now return to arbitrary face-neighborhoods. Asmentioned previously, Pincu proved that the petal unfoldingof N e (B) avoids overlap for any face B of aconvex polyhedron. Here we show that the vertexneighborhoodN v (B) does not always have a nonoverlappingpetal unfolding, even when all faces in theneighborhood are triangles.A portion of the a 9-vertex example P that establishesthis negative result is shown in Fig. 7. The b 1 b 3edge of B lies on the horizontal xy-plane. The vertices{b 2 , a 1 , a 2 , c 1 , c 2 } all lie on a parallel plane at height z,with b 2 directly above the origin: b 2 = (0, 0, z).All of N v (B) is shown in Fig. 8. The structure inFig. 7 is surrounded by more faces designed to minimizecurvatures at the vertices b i of B. Finally, P is theconvex hull of the illustrated vertices, which just addsa quadrilateral “back” face (p 1 , c 1 , c 2 , p 3 ) (not shown).The design is such that there is so little rotation possiblein the cw and ccw options for the triangles incidentto vertex b 2 of B, that overlap is forced: see Figs. 9,10, and 11. The thin △b 2 a 1 a 2 overlaps in the vicinityof a 1 if rotated ccw, and in the vicinity of a 2 is cw(illustrated).One can identify two features of the polyhedron justdescribed that lead to overlap: low curvature vertices(to restrict freedom) and obtuse face angles (at a 1 anda 2 ) (to create “overhang”). Both seem necessary ingredients.Here I pursue excluding obtuse angles:Figure 7: Faces of P in the immediate vicinity of B.Bb 1p 3Figure 8: All faces incident to N v (B), and one more,the purple quadrilateral (a 1 , c 1 , c 2 , a 2 ). The red vectorsare normal to B and to △b 1 p 1 c 1 .c 2c 1a 2 a 1b 2b 3 b 1p 3 p 1Figure 9: Complete unfolding of all faces incident to B.c 2c 1a 2a 1b 2b 30.8ºFigure 10: Zoom of Fig. 9.a 2A 2B2.8ºFigure 11: Zoom of Fig. 10 in vicinity of a 2 overlap.The angle gap at b 3 is 0.8 ◦ , and the gap at b 2 is 2.8 ◦ .b 1p 1Theorem 8 If P is nonobtusely triangulated, then forevery face B, N v (B) has a petal unfolding.


25 th Canadian Conference on Computational Geometry, 2013A nonobtuse triangle is one whose angles are each≤ π/2. It is known that any polygon of n vertices hasa nonobtuse triangulation by O(n) triangles, which canbe found in O(n log 2 n) time [3]. Open Problem 22.6 [6,p. 332] asked whether every nonobtusely triangulatedconvex polyhedron has an edge-unfolding. One can viewTheorem 8 as a (very small) advance on this problem. 3A little more analysis leads to a petal unfolding of a(very special) class of prismatoids (including their tops):Corollary 9 Let P be a triangular prismatoid all ofwhose faces, except possibly the base B, are nonobtusetriangles, and the base is a (possibly obtuse) triangle.Then every petal unfolding of P avoids overlap.It seems quite possible that this corollary still holds withB an arbitrary convex polygon, but the proof wouldneed significant extension.4 DiscussionI believe that unfolding convex patches may be a fruitfulline of investigation. For example, notice that the edgescut in a petal unfolding of a vertex-neighborhood of aface form a disconnected spanning forest rather than asingle spanning tree. One might ask: Does every convexpatch have an edge-unfolding via a single spanning cuttree? The answer is no, already provided by the bandedhexagon example in Fig. 1. For such a tree can onlytouch the boundary at one vertex (otherwise it wouldlead to more than one piece), and then it is easy to runthrough the few possible spanning trees and show theyall overlap.The term zipper unfolding was introduced in [5] fora non-overlapping unfolding of a convex polyhedronachieved via Hamiltonian cut path. They studied zipperedge-paths, following edges of the polyhedron, butraised the interesting question of whether or not everyconvex polyhedron has a zipper path, not constrainedto follow edges, that leads to a non-overlapping unfolding.This is a special case of Open Problem 22.3 in [6,p. 321] and still seems difficult to resolve.Given the focus of this work, it is natural to specializethis question further, to ask if every convex patch hasa zipper unfolding, using arbitrary cuts (not restrictedto edges). I believe the answer is negative: a versionof the banded hexagon shown in Fig. 12, a bottomlessprismoid, has no zipper unfolding. My argument forthis is long and seems difficult to formalize, so I leave theclaim as a conjecture. It would constitute an interestingcontrast to the recent result that all “nested” prismoidshave a zipper edge-unfolding [4].3 It can also be used to slightly improve Pincu’s “fewest nets”result for this class of polyhedra.Figure 12: The banded hexagon with a thin band.References[1] G. Aloupis. Reconfigurations of Polygonal Structures.PhD thesis, McGill Univ., Sch. Comput. Sci., 2005.[2] G. Aloupis, E. D. Demaine, S. Langermann, P. Morin,J. O’Rourke, I. Streinu, and G. Toussaint. Edgeunfoldingnested polyhedral bands. Comput. Geom.Theory Appl., 39(1):30–42, 2007.[3] M. W. Bern, S. Mitchell, and J. Ruppert. Linear-sizenonobtuse triangulation of polygons. Discrete Comput.Geom., 14:411–428, 1995.[4] E. Demaine, M. Demain, and R. Uehara. Zipper unfoldabilityof domes and prismoids. In Proc. 25th Canad.Conf. Comput. Geom., Aug. 2013.[5] E. Demaine, M. Demaine, A. Lubiw, A. Shallit, andJ. Shallit. Zipper unfoldings of polyhedral complexes. InProc. 22nd Canad. Conf. Comput. Geom., pages 219–222, Aug. 2010.[6] E. D. Demaine and J. O’Rourke. Geometric FoldingAlgorithms: Linkages, Origami, Polyhedra. CambridgeUniversity Press, July 2007. http://www.gfalop.org.[7] H. Guo, A. Maheshwari, D. Nussbaum, and J.-R. Sack.Shortest path queries between geometric objects on surfaces.In Comput. Sci. Appl.–ICCSA 2007, pages 82–95.Springer, 2007.[8] J. O’Rourke. Band unfoldings and prismatoids: A counterexample.Technical Report 087, Smith College, Oct.2007. arXiv:0710.0811v2 [cs.CG]; http://arxiv.org/abs/0710.0811.[9] J. O’Rourke. Tetrahedron angles sum to π: Bisectorplane. MathOverflow: http://mathoverflow.net/questions/94586/, Apr. 2012.[10] J. O’Rourke. Unfolding prismatoids as convex patches:Counterexamples and positive results. arXiv:1205.2048[cs.CG]. http://arxiv.org/abs/1205.2048, 2012.[11] J. O’Rourke. Dürer’s problem. In M. Senechal, editor,Shaping Space: Exploring Polyhedra in Nature,Art, and the Geometrical Imagination, pages 77–86.Springer, 2013.[12] V. Pincu. On the fewest nets problem for convex polyhedra.In Proc. 19th Canad. Conf. Comput. Geom.,pages 21–24, 2007.

More magazines by this user
Similar magazines