On Flat Polyhedra deriving from Alexandrov's Theorem
OOn Flat Polyhedraderiving from Alexandrov’s Theorem
Joseph O’Rourke ∗ October 26, 2018
Abstract
We show that there is a straightforward algorithm to determine ifthe polyhedron guaranteed to exist by Alexandrov’s gluing theorem is adegenerate flat polyhedron, and to reconstruct it from the gluing instruc-tions. The algorithm runs in O ( n ) time for polygons whose gluings arespecified by n labels. A theorem of Alexandrov says that any “gluing” of polygons that satisfies threeconditions corresponds to a unique convex polyhedron. The theorem includesflat doubly covered convex polygons as among the possible “convex polyhedra”whose existence is guaranteed by the theorem. In this note we provide analgorithm that detects if a gluing will produce such a flat polyhedron, and if so,constructs it.Alexandrov’s 1941 theorem is described in his 1950 book
Convex Polyhedra ,recently translated into English [Ale05]. Descriptions may be found in [DO07,Sec. 23.3] and [Pak10, Sec. 37]. Here we give a brief statement of the theorem.Define an
Alexandrov gluing of a collection of polygons one as satisfying theseconditions:1. The gluing matches all the perimeters of the polygons by identifying whichpoints glue to which. A case of special interest is when there is just onepolygon, whose perimeter is glued to itself. Isolated points may haveno match, where the boundary “zips” closed in a neighborhood of thosepoints.2. The gluing creates no more than 2 π surface angle surrounding any pointof the resulting manifold.3. The gluing results in manifold that is homeomorphic to a sphere. ∗ Department of Computer Science, Smith College, Northampton, MA 01063, USA. [email protected] . a r X i v : . [ c s . C G ] S e p hese three conditions are obviously necessary for the manifold to be aconvex polyhedron. Alexandrov’s Theorem says that these conditions are alsosufficient: Theorem 1 (Alexandrov)
Any Alexandrov gluing corresponds to a uniqueconvex polyhedron (where a doubly covered polygon is considered a polyhedron).
Alexandrov’s proof is a difficult existence proof and gives little hint of thestructure of the polyhedron guaranteed by the theorem. Recently, Bobenkoand Izmestiev found an intricate but constructive proof of the theorem, whichcan be used to reconstruct the 3D polyhedron as the solution of a particulardifferential equation [BI08]. They have implemented an approximate numericalsolution of this equation in publicly available software. See [O’R07] for a high-level description of their proof.The flat polyhedra permitted by Alexandrov’s Theorem are necessary. Forexample, folding a square across a diagonal constitutes an Alexandrov gluing,and results in a flat doubly covered isosceles right triangle. The purpose of thisnote is to isolate the degenerate flat-polyhedron case of Alexandrov’s Theorem,and show that detection and reconstruction are possible by a straightforwardalgorithm that need not confront the complexities of the full theorem. n ? Before describing the algorithm, we first address a confusing issue concerningthe appropriate value of n for this question, the primary combinatorial countfor expressing complexity. There are four possible n ’s:1. n p : the total number of vertices in the collection of polygons.2. n g : the number of gluing labels defining the gluing instructions.3. n s : the number of vertices on the surface of the polyhedron P .4. n c : the number of corners or cone points on the surface of P .A cone point is a point surrounded by strictly less than 2 π of surface. Althoughwe are only interested in order of magnitudes when claiming a time complexityof O ( n ), it is not the case that all these possible n ’s are necessarily linearlyrelated.For n p , it is natural to count only vertices whose internal angle differs from π . But n g could be arbitrarily larger than n p . An example is shown in Fig-ure 1 (other examples may be found in [ADD+11]). Here a rectangle ( n p =4)is wrapped in a spiral to form a doubly covered trapezoid ( n c =4). But n g =12segments around the boundary of the rectangle are labeled to specify the Alexan-drov gluing, and it is clear that more spiraling could raise n g arbitrarily. Thegluing reduces the points on P delimiting the “faces” by possibly half, in this First brought to my attention by Anna Lubiw. b cba fed edc fb ba d d ffa e ecc ecb d frontback Figure 1: Top: rectangle with fold creases: n p =4, n g =12. Bottom: Two viewsof doubly covered trapezoid: n s =7, n c =4.case to n s =7. Three of these points on the trapezoid sides are not cone points,having 2 π of surface surrounding them.For the algorithm described in the next section, it is n s that largely deter-mines the complexity. Because n s and n g are linearly related, and n g > n s , itis cleanest to define n = n g , the complexity of the gluing instructions. In manycases, all four n ’s are linearly related, but in general it could be that n g (and so n s ) are arbitrarily larger than n p and n c . We must have n g ≥ n p and n s ≥ n c .The relationship between n p and n c is quite close. Not each of the n p verticesof a polygon necessarily ends up as a vertex of P , because vertices of the polygonwhose angles sum to 2 π can be glued together. And not each of the n c conepoints of P derives from a polygon vertex, because one can create a fold point of angle π at the interior of a polygon edge. Thus there is no exact relationshipbetween n p and n c . However, there can be at most four fold-points [DO07,Lem. 25.3.1], because the Gauss-Bonnet Theorem limits the total curvature of P to 4 π ,. So we have n c ≤ n p + 4.In summary, n g dominates all the others, so by the choice n = n g , we haveall four n ’s are O ( n g ) = O ( n ). The result of gluing the polygons together according to the gluing instruc-tions results in P , which could be called an abstract polyhedral surface [Pak10,Ex. 39.11]. P has zero curvature everywhere except at its cone points. Forminga data structure representing P can be accomplished in O ( n g ) = O ( n ) time.3he next step of the algorithm is to identify the n c cone points of P , whichare vertices of the convex polyhedron P guaranteed by Alexandrov’s Theorem.(We are using P for the abstract surface and P for the geometric polyhedron.)Call them v , . . . , v n c in arbitrary order. This step can be achieved in O ( n c ) = O ( n ) time.The second step of the algorithm is to find the (cid:0) n c (cid:1) shortest paths on P from each v i to each v j . Call this set of shortest paths Σ. Note that herewe cannot be assured we can use an algorithm for finding shortest paths ona convex polyhedron P if that algorithm uses the 3D structure of P , becausewe only have available the abstract surface P . This excludes the use of thefastest known algorithm, the O ( n log n ) algorithm in [SS08], whose first stepbuilds an oct-tree data structure around P in 3D. However, the Chen and Hanalgorithm [CH96] works entirely intrinsic to the surface, and so can be applied to P . That algorithm assumes the surface is triangulated, and unfolds the surfacetriangle-by-triangle. Since triangulating a planar graph with n s vertices resultsin O ( n s ) triangles, the appropriate count here is n s = O ( n ) by our choice of n inSection 2. The Chen and Han algorithm has time complexity O ( n ). Repeatingthis for each vertex v i as source results in O ( n ) time for this step. It is possiblethis brute-force approach to computing all vertex-vertex shortest paths couldbe improved, but we make no attempt here.A key fact we use at this juncture is that every edge of a convex polyhedron P is the shortest path on P between the two endpoint vertices it connects.This follows because any other path between those endpoints is not a straightsegment in 3D, and so is strictly longer. Thus we know the unknown edges of P are among the O ( n ) shortest paths Σ on P .If indeed P is a flat polyhedron, then it is a doubly covered convex polygon,whose rim ρ contains all the vertices v , . . . , v n c . Moreover, the path on P thatconstitutes ρ must bisect the angle at each v i , because the half-angle on oneside is mirrored on the other side. So we look for such a path. We now showthat: Claim 1. If ρ exists, it can be found in O ( n ) time. Claim 2. If ρ is found, then P is uniquely identified as a flat polyhedron.Start with v , and look at the shortest path σ ( v , v j ) to each v j , j > σ ( v j , v k ) so that ( v , v j , v k )bisects the total angle at v j . If so, this a potential start to ρ , and so thispath should be followed. The path is now entirely determined by the bisectionproperty: the path through each vertex must bisect the total angle there. Ifbisection holds at each step, and all vertices are included into one loop, thenwe have found a candidate for ρ . If at any stage, an outgoing bisecting shortestpath is not available from Σ, that search can be abandoned, as it could neverproduce ρ . If the bisecting path closes into a loop without passing through everyvertex, again we may discard it. In the case that we have found a candidatefor ρ , we make one more test, for simplicity, non-self-intersection, for ρ must4
132 13 2 1 3 a cb a cb (a) (b)
Figure 2: A regular hexagon (a) that folds (toward the viewer) to a doublycovered equilateral triangle (b). The edge labels indicate gluing instructions.be a simple closed path. Although it is unclear if there can be a bisecting self-crossing path through every vertex, in the absence of a proof of non-existence,we simply check for this condition.Let us illustrate before proceeding with the description. Figure 2(a) shows aregular hexagon, whose gluing instructions fold it to an equilateral triangle (b). In this simple example, the bisecting path ρ = ( a, b, c ) would be found immedi-ately.Figure 3 shows a slightly more complex example, based on [DO07, Fig. 25.24].Note the identification of the four vertices of P in (a) of the figure. Suppose v = a and v j = c . The path ( a, c ) can indeed be extended through c to bisectthe angle of π there, but only by prematurely returning to a (and then it doesnot bisect at a ). So this path would be abandoned by the algorithm. The pathbeginning ( a, b ) continues to ρ = ( a, b, c, d ).Returning to the argument, let us count up the worst-case complexity offollowing one ( v , v j ) path, without attempting sophisticated algorithms. Letus assume the shortest paths in Σ are maintained in sorted order around eachsource vertex, which is easily returned by the Chen and Han algorithm. We startwith σ ( v , v j ), and search in Σ for a bisecting extension σ ( v j , v k ) in O (log n )time. This same cost is incurred at each successive step. We also must checkat each step if we have prematurely closed a loop, which can be accomplishedwith a constant-time array lookup of the previously visited vertices. So a fullpath ρ = ( v , v j , . . . , v ) can be found in O ( n log n ) time. Finally, we need tocheck ρ for simplicity. Although it seems possible this could be accomplishedin O ( n log n ) time, or even in O ( n ) (because determining whether a polygonis simple can be accomplished in O ( n ) time [Cha91]), let us just count this as O ( n ) by a brute-force comparison of every pair of edges.We repeat this procedure for the n − v , v j ), and so spend O ( n ) time overall either finding a ρ , or determining that no such ρ exists. Ifthere is such a ρ , we must find it by this procedure, because it must pass through This example was used by Daniel Mehkeri in a
Math Overflow question 4 July 10.
176 54 3 21
65 4 3 a dcb ba dcb ad c b (a) (c)(b) Figure 3: Folding of the Latin cross (a) (away from the viewer) to a doubly cov-ered quadrilateral, (b) and (c). The number labels indicate gluing instruction.6 . So we have established Claim 1 above.Claim 2 is that if we find ρ , then indeed P must be the doubly coveredflat convex polygon whose boundary is ρ . Here we can employ this resultfrom [IOV10, Cor. 4]: Lemma 1
A convex polyhedral manifold with convex boundary and with nointerior curvature is isometric to a planar convex polygon.
The proof of this lemma uses both Alexandrov’s Theorem and a separate lemmaof Alexandrov. Because ρ includes all vertices, the interior is indeed curvature-free, so each “half” of P bounded by ρ is isometric to a planar convex polygon.So we know we have a doubly covered convex polygon P . Finally, Alexandrov’sTheorem establishes that P is unique, so there is no need to seek another ρ . Although reconstructing the 3D structure of the polyhedron guaranteed to ex-ist by Alexandrov’s Theorem is a challenging problem, it is relatively easy todetect the degenerate flat-polyhedron case of the theorem, and to reconstructthe doubly covered convex polygon: Just follow vertex-to-vertex shortest pathsseeking the rim. Although the algorithm described has cubic time complexity,it seems possible that it could be reduced to near-quadratic complexity.Returning to the different n ’s discussed in Section 2, the time complexity is O ( n g ) to form the abstract surface P , O ( n s n c ) to find the shortest paths, and O ( n c ) to find ρ . The factor that dominates is n s , so it could be worthwhileto reduce n s by merging coplanar faces prior to running the shortest-pathsalgorithm.Perhaps a more interesting direction for future research is to explore whetherother special classes of polyhedra P might be reconstructable from an Alexan-drov gluing without all the machinery of [BI08]. Acknowledgments.
The idea for this note resulted from a stimulating work-shop conversation with Alexander Bobenko, Ivan Izmestiev, and Konrad Polth-ier in 2007. I thank Anna Lubiw for raising the wrapping issue discussed inSection 2.
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