We show that four of the five Platonic solids' surfaces may be cut open with
a Hamiltonian path along edges and unfolded to a polygonal net each of which
can "zipper-refold" to a flat doubly covered parallelogram, forming a rather
compact representation of the surface. Thus these regular polyhedra have
particular flat "zipper pairs." No such zipper pair exists for a dodecahedron,
whose Hamiltonian unfoldings are "zip-rigid." This report is primarily an
inventory of the possibilities, and raises more questions than it answers.Comment: 15 pages, 14 figures, 8 references. v2: Added one new figure. v3:
  Replaced Fig. 13 to remove a duplicate unfolding, reducing from 21 to 20 the
  distinct unfoldings. v4: Replaced Fig. 13 again, 18 distinct unfolding

O'Rourke, Joseph

English

arXiv

We show that four of the five Platonic solids’ surfaces may be cut open with a Hamiltonian path along edges and unfolded to a polygonal net each of which can “zipper-refold” to a flat doubly covered parallelogram, forming a rather compact representation of the surface. Thus these regular polyhedra have particular flat “zipper pairs. ” No such zipper pair exists for a dodecahedron, whose Hamiltonian unfoldings are “zip-rigid.” This report is primarily an inventory of the possibilities, and raises more questions than it answers

Joseph O&apos

CiteSeerX

Flat Zipper-Unfolding Pairs for Platonic Solids

We show that four of the five Platonic solids\u27 surfaces may be cut open with a Hamiltonian path along edges and unfolded to a polygonal net each of which can  zipper-refold  to a flat doubly covered parallelogram, forming a rather compact representation of the surface. Thus these regular polyhedra have particular flat  zipper pairs.  No such zipper pair exists for a dodecahedron, whose Hamiltonian unfoldings are  zip-rigid.  This report is primarily an inventory of the possibilities, and raises more questions than it answers

O\u27Rourke, Joseph

Smith College: Smith ScholarWorks

Masthead Logo Smith ScholarWorksComputer Science: Faculty Publications Computer Science10-20-2010Flat Zipper-Unfolding Pairs for Platonic SolidsJoseph O'RourkeSmith College, jorourke@smith.eduFollow this and additional works at: https://scholarworks.smith.edu/csc_facpubsPart of the Computer Sciences Commons, and the Geometry and Topology CommonsThis Article has been accepted for inclusion in Computer Science: Faculty Publications by an authorized administrator of Smith ScholarWorks. Formore information, please contact scholarworks@smith.eduRecommended CitationO'Rourke, Joseph, "Flat Zipper-Unfolding Pairs for Platonic Solids" (2010). Computer Science: Faculty Publications, Smith College,Northampton, MA.https://scholarworks.smith.edu/csc_facpubs/36Flat Zipper-Unfolding Pairs for Platonic SolidsJoseph O’Rourke∗October 21, 2010AbstractWe show that four of the five Platonic solids’ surfaces may be cutopen with a Hamiltonian path along edges and unfolded to a polygonalnet each of which can “zipper-refold” to a flat doubly covered parallelo-gram, forming a rather compact representation of the surface. Thus theseregular polyhedra have particular flat “zipper pairs.” No such zipper pairexists for a dodecahedron, whose Hamiltonian unfoldings are “zip-rigid.”This report is primarily an inventory of the possibilities, and raises morequestions than it answers.1 IntroductionIt has been known since the time of Alexandrov—and it was certainly knownto him—that the surface of a polyhedron could sometimes be cut open to a netand refolded to a doubly covered polygon, which we will henceforth call a flatpolyhedron. Such flat polyhedra are explicitly countenanced in Alexandrov’s1941 gluing theorem.1 Perhaps the first specific example of this possibilityoccurred in [LO96], which included the example illustrated in Figure 1: thefamiliar Latin-cross unfolding of the cube may be refolded to a flat convexquadrilateral polyhedron. This is one of the two flat convex polyhedron thatmay be folded from the Latin cross [DO07, Fig. 25.32].Let us say that two polyhedra Q1 and Q2 form a net pair if they may beunfolded to a common polygonal net. In Figure 1, the cube is cut along edgesto unfold to the Latin cross polygon, but the flat quadrilateral must have facecuts through the interior of its two faces to unfold to the same Latin cross.In general there is little understanding of which polyhedra form net pairs.See, for example, Open Problem 25.6 in [DO07]. Here we explore a narrowquestion on net pairs, narrow enough to obtain a complete answer.The cuts to unfold a convex polyhedron to a single polygon form a spanningtree of the polyhedron’s vertices [DO07, Sec. 22.1.3]. Shephard explored thespecial case where the spanning tree is a Hamiltonian path of the 1-skeleton∗Department of Computer Science, Smith College, Northampton, MA 01063, USA.orourke@cs.smith.edu.1 See [DO07, Sec. 23.3] and [Pak10, Sec. 37] for descriptions of this theorem.1arXiv:1010.2450v4  [cs.CG]  20 Oct 2010(a) (b)33 3334415 521 5242Figure 1: Folding the Latin cross cube net to a flat quadrilateral polyhedron.Points with the same label in (a) are identified in the refolding (b).of the polyhedron, i.e., all cuts are along polyhedron edges [She75]. The resultis a Hamiltonian unfolding of the polyhedron. (Note the cube unfolding thatproduces Figure 1(a) is not a Hamiltonian unfolding: the cut tree has fourleaves.) Some combinatorial questions on Hamiltonian unfoldings were exploredin [DDLO02]; see [DO07, Fig. 25.59 ]. In particular, there are polyhedra thathave an exponential number of combinatorially distinct Hamiltonian unfoldings:2Ω(n) for a polyhedron with n vertices.Another variant is provided by the class of perimeter-halving foldings [DO07,Sec. 25.1.2 ], which correspond to spanning cut paths that may employ face cutsrather than solely following polyhedron edges. In [LDD+10] these paths werememorably rechristened as zipper paths, producing zipper unfoldings. We willadopt that nomenclature, including the verbs zip and unzip to mean folding andunfolding (respectively) along zipper paths. We reserve Hamiltonian path to bea zipper path along polyhedron edges. Finally, if two polyhedra each unzip toa common polygonal net, we say they form a zipper pair.The narrow question we explore is this:Question: Does each of the Platonic solids form a zipper pair witha flat convex polyhedron, with the zipper path on the regular poly-hedron forming a Hamiltonian path of its edges?We show that the tetrahedron,2 the cube, the octahedron, and the icosahe-dron all form such zipper pairs with flat parallelogram polyhedra. The dodeca-hedron has no such zipper mate. Note that it would be too restrictive to insist2 We drop the modifier “regular” to shorten the names of the five regular polyhedra.2that both zippers are Hamiltonian paths of the 1-skeletons, because for a flatpolyhedron, the 1-skeleton is the single cycle bounding the polygon, and so aHamiltonian unfolding is just two copies of the convex polygon joined along oneedge.2 Flat Zipper Pairs2.1 TetrahedronThe regular tetrahedron has only one Hamiltonian path (up to symmetries),which unfolds to the 2 × 1 parallelogram shown in Figure 2(b) (in Fig. 2in [LDD+10]). Because this net is a convex polygon, Thm. 25.1.4 in [DO07]establishes that it has an infinite number of zippings to various convex polyhe-dra. The zipping shown in Figure 2(c) folds it to a doubly covered rhombus.1 22xy112xy(b)(a)(c)Figure 2: (a) The Hamiltonian cut path on a tetrahedron leads to the Hamilto-nian unfolding (b), which zips from x to y (identifying the labeled points) to aflat rhombus polyhedron of side length 1 (c).2.2 CubeThe cube has three distinct Hamiltonian unfoldings (Fig. 1 in [LDD+10]): onewith the path endpoints at opposite cube corners, and two with the path end-points at either end of a cube edge. One of the latter (shown in Figure 3)produces a ‘T’-shape that has no zippings except back to the cube. We callsuch a zipper unfolding zip-rigid. We defer an explanation of how it is knownthat this unfolding is zip-rigid to Section 2.3 below.3(b)(a)Figure 3: (a) The first Hamiltonian cut path leads to (b) a zip-rigid Hamiltonian‘T’-unfolding of the cube.The other two Hamiltonian unfoldings of the cube, which we call the ‘S’-and the ‘Z’-unfoldings, both zip to the same doubly covered parallelogram, asshown in Figures 4 and 6. An animation of the ‘S’-folding is shown in Figure 5.2.3 A Zipping AlgorithmLet P be a polygon, the polygonal net corresponding to a zip-pair of polyhedraQ1 and Q2. Each of the two zippings of P are perimeter-halving foldings,with the endpoints of the zip path bisecting the perimeter. If we normalize theperimeter of P to 1 and parametrize it from 0 to 1, we can view the two zippingsabstractly as in Figure 7. One of the zip-path endpoints are at 0 and 12 , andthe other zip-path endpoints are at x and y = x + 12 . We seek to find all thelocations x that determine a zipping to some convex polyhedron.As previously mentioned, if P is convex, then every x determines a convexpolyhedron (Thm. 25.1.4 in [DO07]), so we henceforth exclude that case. If Pis not convex, it has at least one reflex vertex v with internal angle β > π. Nowthere are only two options at v: (1) v can serve as x, so the zipping starts atv = x; or (2) Some strictly convex vertex ui whose internal angle αi satisfiesαi + β ≤ 2π is glued to v. If more than one vertex is glued to v, then thefolding would not be a zipping, as v would then constitute a junction of degree> 2 in the gluing tree ([DO07, Sec. 25.3]). Note that if ui glues to v, then xis determined: halfway between ui and v along the perimeter of P . Thus weonly need try each ui in turn, and check that Alexandrov’s conditions hold forthe uniquely determined zipping [DO07, Thm. 23.3.1]). This incidentally showsthat any P with a reflex vertex admits only O(n) zippings.For example, applying this algorithm to the cube ‘Z’-unfolding in Figure 6results in six zippings: two copies of the one shown in that figure, two copies ofa tetrahedron, one 5-vertex and one 6-vertex polyhedron.Although this provides a linear-time algorithm for determining all zippingsof P , it does not tell us which of these zippings lead to flat polyhedra. Although413215632544xy616324xy5(b)(a)(c)Figure 4: (a) The second Hamiltonian cut path on a cube. (b) The resultingHamiltonian ‘S’ unfolding. (c) Zipped according to the indicated point identifi-cations to a parallelogram polyhedron of side lengths 1 and 3√2.5Figure 5: Snapshots from an animation folding the parallelogram in Fig-ure 4(b,c).61124 5435 6xy6(b)(a)(c)2 31 254xy63Figure 6: (a) The third Hamiltonian cut path on a cube. (b) The resultingHamiltonian ‘Z’ unfolding. (c) Zipped according to the indicated point identi-fications to a parallelogram polyhedron of side lengths 1 and 3√2.xy½ 0Figure 7: A zip-pair, abstractly. The perimeter has been normalized to 1.7there is an O(n3) algorithm for deciding if an Alexandrov gluing is flat [O’R10],this remains unimplemented. We resorted to manual folding of the zippings.2.4 OctahedronThe octahedron also has three distinct Hamiltonian paths,3 one between the topand bottom vertices (separated by distance 2 in the 1-skeleton), and two pathsbetween adjacent (distance-1) vertices. The first Hamiltonian unfolding bothzips to a rectangle as shown in Figure 8, and zips to a parallelogram, Figure 9.I find the rectangle zipping especially surprising, as it derives from a shape allof whose angles are multiples of π/3 = 60◦.132154321542543xyxy(b)(c)(a)Figure 8: (a) Hamiltonian cut path on an octahedron. (b) Its correspondingHamiltonian unfolding. (c) Zipping folds it to a flat doubly covered rectangleof dimensions 12 × 2√3.One of the other Hamiltonian unfoldings of the octahedron, shown in Fig-ure 10, zips to a parallelogram. The other Hamiltonian unfolding does not have3 This is natural because the cube and octahedron are duals. However, it is shownin [LDD+10, Fig. 4] that the dual of a Hamiltonian unfolding is not necessarily a Hamil-tonian path through the faces of that unfolding.8122143314243xyxy(a)(b)Figure 9: (a) Another zipping of the same unfolding from Figure 8 leads to (b) a1×√3 parallelogram polyhedron.a flat zipping, although it does have zippings, e.g., to a tetrahedron all four ofwhose vertices have curvature π.I cannot resist mentioning that this last net folds to a flat rectangular poly-hedron, whose cut tree, however, is not a zipping: Figure 11.2.5 DodecahedronEvery Hamiltonian unfolding of the dodecahedron is zip-rigid, and therefore ithas no flat zip pair in the sense posed in our Question above. The reason is asfollows. Let x and y be the endpoints of the Hamiltonian path that unfolds thedodecahedron. Then the reflex angle of the net at x and y is 3· 35π = 324◦, leavingan external angle of 36◦ there. The smallest convex angle in any edge unfoldingof the dodecahedron is 35π = 108◦, so no vertex can glue into x or y. Therefore,a zipping must zip at x and y, leading directly back to the dodecahedron.We should mention that loosening the criteria posed in our Question leads toa flat refolding of a Hamiltonian net for the dodecahedron. Figure 12 illustratesone such, using the unfolding in Fig. 2 in [LDD+10]. Here the refolding inFigure 12(c) is neither convex nor a zipper folding.931432142xy(b)(c)(a)1324xyFigure 10: (a) Hamiltonian cut path on an octahedron. (b) Its correspondingHamiltonian unfolding. (c) Zipping folds it to a flat doubly covered parallelo-gram of dimensions 1× 2√3.102441123(a)(b)32143Figure 11: The same Hamiltonian unfolding from Figure 10 folds to a√32 × 2doubly covered rectangle, but this folding is not a zipping.11xyx ya b(b)(c)(a)xyFigure 12: (a) Hamiltonian cut path on a dodecahedron. (b) Its correspondingHamiltonian unfolding. (c) A non-zipper refolding to a doubly covered flatnonconvex polygon. The cut tree has degree 3 at vertices a and b.2.6 IcosahedronFor the tetrahedron, cube, and octahedron, it was easy to explore all the Hamil-tonian unfoldings, because there are so few (1, 3, and 3 respectively). The icosa-hedron, however, has hundreds of Hamiltonian unfoldings. At this writing, I donot know precisely how many geometrically distinct Hamiltonian unfoldings itpossesses.The diameter of the icosahedral graph is 3, so the end points of a Hamil-tonian path are a distance 1, 2, or 3 apart. Fixing two vertices separated bya distance d ∈ {1, 2, 3}, I found that there are, respectively, 512, 608, and 720labeled Hamiltonian paths between them.4 Of course not all these labeled pathsare distinct geometric paths because of symmetries. However, I have not carriedout the more difficult enumeration of the number of geometrically distinct (in-congruent as paths in R3) Hamiltonian paths on an icosahedron. But certainlythis number is less than 512 + 608 + 720 = 1840.For each of these 1840 Hamiltonian unfoldings, I ran the zipping algorithmin Section 2.3, which determined that 82 of the unfoldings had at least onezipping, while all the others are zip-rigid (82 = 12 + 20 + 50 in the three classes,respectively). By visual inspection, 18 of these Hamiltonian unfoldings are4 It is a curious fact that the number of labeled Hamiltonian cycles through any fixed edgeis 29 = 512. The simplicity of this expression suggests there might be a combinatorial expla-nation, a question I asked on MathOverflow, http://mathoverflow.net/questions/37788/.12distinct; they are displayed in Figure 13.Figure 13: The 18 distinct Hamiltonian unfolding of the icosahedron that eachhave at least one zipping to another convex polyhedron (Not all are displayedto the same scale.)Exactly one of these unfoldings (the leftmost in the first row) zips to aparallelogram, as shown in Figure 14. None of the other zips to a parallelogram(but it remains possible that there is some zipping to flat polyhedron differentfrom a doubly covered parallelogram).3 Future WorkAs is evident from the foregoing, there is little theory behind the unfoldingsdetailed here. The central open problem is to gain more insight into whichpolyhedra are net pairs, or more specifically, zipper pairs. Perhaps intuitioncan be strengthened by tackling specific subquestions that fall under this generalumbrella. It is easy to list such questions, all of are open because of the lack ofa general theory. For example, the Hamiltonian unfoldings of the Archimedeansolids detailed in [LDD+10] could be explored.An interesting specific but tangential question raised by this work is to deter-mine the exact number of geometrically distinct Hamiltonian paths on a regularicosahedron.Acknowledgments. I thank Stephanie Annessi and Katherine Lipow for helpin enumerating and folding the icosahedron Hamiltonian unfoldings.1363715432154298xy1067 98 10371542xy69810(b)(a)(c)Figure 14: (a) Hamiltonian cut path on an icosahedron. (b) Its correspondingHamiltonian unfolding. (c) Rezipping folds it to a flat doubly covered parallel-ogram of side lengths√3 and 5.14References[DDLO00] Erik D. Demaine, Martin L. Demaine, Anna Lubiw, and JosephO’Rourke. Examples, counterexamples, and enumeration re-sults for foldings and unfoldings between polygons and polytopes.Technical Report 069, Smith College, Northampton, July 2000.arXiv:cs.CG/0007019.[DDLO02] Erik D. Demaine, Martin L. Demaine, Anna Lubiw, and JosephO’Rourke. Enumerating foldings and unfoldings between polygonsand polytopes. Graphs and Combin., 18(1):93–104, 2002. Seealso [DDLO00].[DO07] Erik D. Demaine and Joseph O’Rourke. Geometric Folding Algo-rithms: Linkages, Origami, Polyhedra. Cambridge University Press,July 2007. http://www.gfalop.org.[LDD+10] Anna Lubiw, Erik Demaine, Martin Demaine, Arlo Shallit, andJonah Shallit. Zipper unfoldings of polyhedral complexes. In Proc.22nd Canad. Conf. Comput. Geom., pages 219–222, August 2010.[LO96] Anna Lubiw and Joseph O’Rourke. When can a polygon fold to apolytope? Technical Report 048, Dept. Comput. Sci., Smith College,June 1996. Presented at Amer. Math. Soc. Conf., 5 Oct. 1996.[O’R10] Joseph O’Rourke. On flat polyhedra deriving from alexandrov’s the-orem. http://arxiv.org/abs/1007.2016, July 2010.[Pak10] Igor Pak. Lectures on discrete and polyhedral geometry. http://www.math.ucla.edu/~pak/book.htm, 2010.[She75] Geoffrey C. Shephard. Convex polytopes with convex nets. Math.Proc. Camb. Phil. Soc., 78:389–403, 1975.15

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Flat Zipper-Unfolding Pairs for Platonic Solids

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