Edge unfolding Cut sets Source foldouts Proof and algorithm Complexity issues Aleksandrov unfolding? Unfolding polyhedra.

Transcription

1 Unfolding polyhedra Ezra Miller University of Minnesota University of Nebraska 27 April 2007

2 Outline 1. Edge unfolding 2. Cut sets 3. Source foldouts 4. Proof and algorithm 5. Complexity issues 6. Aleksandrov unfolding? 0

3 Edge unfolding spanning tree: a collection of edges that is spanning: hits every vertex a tree: connected and has no cycles complement of spanning tree is metrically flat locally homeomorphic to open disk Finding a spanning tree is easy; but: Question (Fukuda 1987): Does the complement of a spanning tree always unfold without overlap? 1

4 Edge unfolding spanning tree: a collection of edges that is spanning: hits every vertex a tree: connected and has no cycles complement of spanning tree is metrically flat locally homeomorphic to open disk Finding a spanning tree is easy; but: Question (Fukuda 1987): Does the complement of a spanning tree always unfold without overlap? Answer (Namiki): No. Question (Dürer 1525, Shephard 1975): Given a polyhedron, is there a spanning tree whose complement unfolds without overlap? 1

5 Edge unfolding spanning tree: a collection of edges that is spanning: hits every vertex a tree: connected and has no cycles complement of spanning tree is metrically flat locally homeomorphic to open disk Finding a spanning tree is easy; but: Question (Fukuda 1987): Does the complement of a spanning tree always unfold without overlap? Answer (Namiki): No. Question (Dürer 1525, Shephard 1975): Given a polyhedron, is there a spanning tree whose complement unfolds without overlap? Answer: Unknown. [Schevon 1987] experiments: probability that a random spanning tree yields unfolding without overlap 0 as number of vertices. Higher dimension: less well understood. 1

6 Cut sets P R d+1 compact convex polyhedron of dimension d + 1 = bounded intersection of finitely many closed halfspaces S = P polyhedral sphere of dimension d Goal: understand the metric on S. Def: A cut set is a union K S of polyhedra with dim K = d 1, and S\K isometric to open polyhedral ball in R d Why define this? Classical, but if d = 2 (Sharir Schorr 1986): Robot motion planning unfolding surfaces shortest path straight line in foldout Discrete geodesic problem Example (Aleksandrov unfolding, d = 2): Pick a source point in S. Let K = union of shortest paths from source point to vertices. Theorem (Aronov O Rourke 1992) Aleksandrov unfoldings for d = 2 are nonoverlapping. 2

7 Source foldouts Observation. Aleksandrov procedure fails to yield a cut set in dimensions d 3: union of shortest paths to faces of dimension d 2 is polyhedral, but the complement is not a ball. Def: Fix source point v S \ S d 2. The cut locus K v is the closure of {x S x has 2 shortest paths in S to v}. Riemannian geometry analogue: S = manifold v = source point exp : T v S locally: exp(ray) = shortest path globally:?? Theorem: For any polyhedron P and any source point v S \ S d 2, the cut locus is a cut set. 3

8 Proof and algorithm Proof: d = 2 abstractly: (Volkov Podgornova 1971) d = 2 algorithm: (Sharir Schorr 1986) d 3 both: (M Pak, to appear in Discrete & Comput. Geom.) Two key points: 1. Shortest paths to v don t pass through S d 2 exp : T v S surjective (false for general polyhedral manifolds!) S \ K v is an open star-shaped ball 2. Subdivision of every facet into regions consisting of points with same combinatorial type of shortest path to v is polyhedral (false for general polyhedral manifolds!) Algorithm: 1. Keep track of combinatorial types (facet sequences) 2. Extend (metrically) shortest available type onto adjacent facet. 3. Output subdivision of each facet into combinatorial types. 4

9 Complexity issues Theorem (M Pak) For fixed d, algorithm to source unfold S = P into R d is polynomial in #{facets} and #{combinatorial types of shortest paths from v}. Conjecture (M Pak) If P R d+1 has n facets, then #{combinatorial types of shortest paths from v} n O(d 2). Conjecture (M Pak) If v, w P and the polytope P has n facets, then #{combinatorial types of shortest paths from v to w} is polynomial in n. Note: False for general polyhedral manifolds, even if d = 2. Conjecture (M Pak) There is a fixed polynomial function f (d) such that if P R d+1 has n facets, then #{combinatorial types of shortest paths in S} n f (d). d = 2: O(n 7 ) (Sharir 1985) O(n 4 ) (Mount 1985) 5

10 Aleksandrov unfolding? Key to polynomiality: Aleksandrov unfolding in higher dimensions. Observation. Aleksandrov cut set in dimension d = 2 separates combinatorial types of shortest paths into convex regions: peels (Volkov Podgornova, 1971) How to construct in dimensions d 3: polyhedral vector fields. Aleksandrov cut set a positively curved generalization of real hyperplane arrangement, or oriented matroid; should retain enough structure to imply polynomially many regions. Potential further connections: combinatorial differentiable (CD) manifolds (via polyhedral vector fields and generalizations of matroids) metric combinatorics of polyhedral spheres construction of characteristic classes on polyhedral manifolds 6