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Part III: Polyhedra d: Alexandrov to Sabitov Joseph ORourke Smith College.

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Presentation on theme: "Part III: Polyhedra d: Alexandrov to Sabitov Joseph ORourke Smith College."— Presentation transcript:

1 Part III: Polyhedra d: Alexandrov to Sabitov Joseph ORourke Smith College

2 Outline: Reconstruction of Convex Polyhedra zCauchy to Sabitov (to an Open Problem) yCauchys Rigidity Theorem yAleksandrovs Theorem ySabitovs Algorithm

3 Steffens flexible polyhedron 14 triangles, 9 vertices

4 The Bellows Conjecture zPolyhedra can bend but not breathe [Mackenzie 98] zSettled in 1997 by Robert Connelly, Idzhad Sabitov, and Anke Walz zHerons formula for area of a triangle: A 2 = s(s-a)(s-b)(s-c) zFrancescas formula for the volume of a tetrahedron zSabitov: volume of a polyhedron is a polynomial in the edge lengths.

5 Aleksandrovs Theorem (1941) zFor every convex polyhedral metric, there exists a unique polyhedron (up to a translation or a translation with a symmetry) realizing this metric."

6 Pogorelovs version (1973) zFor every convex polyhedral metric, there exists a unique polyhedron (up to a translation or a translation with a symmetry) realizing this metric. zAny convex polyhedral metric given... on a manifold homeomorphic to a sphere is realizable as a closed convex polyhedron (possibly degenerating into a doubly covered plane polygon)."

7 Alexandrov Gluing (of polygons) Alexandrov Gluing (of polygons) zUses up the perimeter of all the polygons with boundary matches: xNo gaps. xNo paper overlap. xSeveral points may glue together. zAt most 2 angle at any glued point. zHomeomorphic to a sphere. Aleksandrovs Theorem unique polyhedron

8 Folding the Latin Cross

9 Cauchy vs. Aleksandrov zCauchy: uniqueness zAleksandrov: existence and uniqueness zCauchy: faces and edges specified zAleksandrov: gluing unrelated to creases

10 Uniqueness zCauchy: combinatorial equivalence + congruent faces => congruent zAleksandrov: Two isometric polyhedra are equivalent zThe sphere is rigid [Minding] zThe sphere is unique given its metric [Liebmann, Minkowski] zClosed regular surfaces are rigid [Liebmann, Blaschke, Weyl] zUniqueness of these w/in certain class [Cohn-Vossen] z... zIsometric closed convex surfaces are congruent [Pogorelov 73]

11 Alexandrov Existence 1 zInduction on the number of vertices n of the metric: yfrom realization of n-1 vertex metric to n vertex metric yby continuous deformation of metrics ytracked by polyhedral realizations

12 Existence 2 zThere are two vertices a and b with curvature less than. zConnect by shortest path. zCut manifold along and insert double that leaves curvature unchanged. zAdjust shape of until a and b both disappear: n-1 vertices. zRealize by induction, introduce nearby pseudovertex, track sufficiently close metrics. curvature = 2 – angs curvature

13 (a) a flattened; (b) b flattened.

14 D-Forms Smooth closed convex curves of same perimeter. Glue perimeters together. D-form D-form c1c1 c2c2

15 Fig 6.49, p.401 Helmut Pottmann and Johannes Wallner. Computational Line Geometry. Springer-Verlag, 2001.

16 Pita Forms

17 Pottmann & Wallner zWhen is a D-form is the convex hull of a space curve? zWhen is it free of creases? Always [Modulo a believable lemma not yet formally proved.]

18 Sabitovs Algorithm zGiven edge lengths of triangulated convex polyhedron, zcomputes vertex coordinates zin time exponential in the number of vertices.

19 Sabitov Volume Polynomial zV 2N + a 1 (l)V 2(N-1) + a 2 (l)V 2(N-2) a N (l)V 0 = 0 yTetrahedron: V 2 + a 1 (l) = 0 yl = vector of six edge lengths ya 1 (l) = ijk (l i ) 2 (l j ) 2 (l k ) 2 /144 yFrancescas formula zVolume of polyhedron is root of polynomial

20 2 N possible roots

21 Generalized Polyhedra zPolynomial represents volume of generalized polyhedra yany simplicial 2-complex homeomorphic to an orientable manifold of genus 0 ymapped to R 3 by continuous function linear on each simplex. zNeed not be embeddable: surface can self-intersect.

22 Sabitov Proof 1

23 Sabitov Proof 2 zvol(P) = vol(P) - vol(T), = 1 z... [many steps]... zpolynomial = 0 zunknowns: yedge lengths vector l yV = vol(P) yunknown diagonal d

24 Sabitov Proof 3 zunknowns: yedge lengths l [given] yV = vol(P) [known from volume polynomial] yunknown diagonal d zTry all roots for V, all roots for d: candidates for length of d & dihedral at e. zRepeat for all e. zCheck the implied dihedral angles for the convex polyhedron.

25 Open: Practical Algorithm for Cauchy Rigidty Find either a polynomial-time algorithm, or even a numerical approximation procedure, that takes as input the combinatorial structure and edge lengths of a triangulated convex polyhedron, and outputs coordinates for its vertices.


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