Convex Hulls (3D)

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Convex Hulls (3D) Convex Hulls (3D) O’Rourke, Chapter 4 Outline • Polyhedra Polytopes Euler Characteristic • (Oriented) Mesh Representation Polyhedra Definition: A polyhedron is a solid region in 3D space whose boundary is made up of planar polygonal faces comprising a connected 2D manifold. Polyhedra The boundary of a polyhedron can be expressed as a combination of vertices, edges, and faces: Intersections are proper: » Elements don’t overlap, or » They share a single vertex, or » They share an edge and the two vertices Polyhedra The boundary of a polyhedron can be expressed as a combination of vertices, edges, and faces: Intersections are proper Locally manifold: » Edges around a vertex can be sorted to match their incidence on faces. Polyhedra The boundary of a polyhedron can be expressed as a combination of vertices, edges, and faces: Intersections are proper Locally manifold: Alternatively,» Edges the around subgraph a vertex of the can dual be obtained sorted to by restricting tomatch the adjacent their incidence faces (the on faces. link) is connected. Polyhedra The boundary of a polyhedron can be expressed as a combination of vertices, edges, and faces: Intersections are proper Locally manifold: » Edges around a vertex can be sorted to match their incidence on faces. » Exactly two faces meet at each edge. Polyhedra The boundary of a polyhedron can be expressed as a combination of vertices, edges, and faces: Intersections are proper Locally manifold Globally connected Definition Definition: Given an edge on a polyhedron, the dihedral angle of the edge is the internal angle between the two adjacent faces. 휋/2 Aside: The dihedral angle is a discrete measure of mean curvature. Definition Definition: Given a vertex on a polyhedron, the deficit angle at the vertex is 2휋 minus the sum of angles around the vertex. 휋/2 ⇒ 휋/2 휋/2 휋/2 Aside: The deficit angle is a discrete measure of Gauss curvature. Polytopes A convex polyhedron is a polytope: Non-negative mean curvature: All dihedral angles are less than or equal to 휋. (Necessary and sufficient.) Non-negative Gaussian curvature: Sum of angles around a vertex is at most 2휋. (Necessary but not sufficient). Platonic Solids Definition: A regular polygon is a polygon with equal sides and equal angles. … Platonic Solids Definition: A regular polygon is a polygon with equal sides and equal angles. A regular polyhedron is a convex polyhedron, with all faces congruent regular polygons and vertices having the same valence. Platonic Solids Claim: The five platonic solids are the only regular polyhedra. [Images courtesy of Wikipedia] Platonic Solids Proof: Assume each face is 푝-sided: ⇒ The sum of angles in a face is 휋 푝 − 2 ⇒ The angle at each vertex is 휋(1 − 2/푝) Assume each vertex has valence 푣: ⇒ The angle-sum at a vertex is 푣휋 (1 − 2/푝) Platonic Solids Proof: Since the polyhedron is convex: 푣휋 1 − 2 푝 < 2휋 ⇔ 푣 1 − 2 푝 < 2 ⇔ 푣 푝 − 2 < 2푝 ⇔ 푣푝 − 2푣 − 2푝 < 0 ⇔ 푝 − 2 푣 − 2 − 4 < 0 Platonic Solids Proof: Since the polyhedron is convex: 푝 − 2 푣 − 2 − 4 < 0 Since 푝, 푣 ≥ 3, valid options are 푝, 푣 : (3,3) (3,4) (4,3) (3,5) (5,3) Platonic Solids The platonic solids come in dual pairs, where one solid is obtained from the other by replacing faces with vertices: Cube ↔ Octahedron Icosahedron ↔ Dodecahedron Tetrahedron ↔ Tetrahedron (3,3) (3,4) (4,3) (3,5) (5,3) Topological Polyhedra The boundary of a polyhedron can be expressed as a combination of vertices, edges, and faces: Intersections are proper Geometric Locally manifold Topological Globally connected Topological Polyhedra If we ignore the vertex positions, we get a combinatorial structure composed of faces (cells), edges, and vertices.* [Nivoliers and Levy, 2013] *These are CW complexes. (And, if faces are triangles, these are simplicial complexes). Topological Polyhedra Properties (CW Complex): Faces intersect at edges and vertices. Edges are topologically line segments and intersect at vertices. Interiors of faces have disk-topology and the boundary is a polygon made up of edges. Topological Polyhedra Properties (Manifold): Each vertex is on the boundary of some edge. Each edge is on the boundary of some face. An edge is on the boundary of two faces. Edges around a vertex can be sorted. Topological Polyhedra Note: Given a topological polygon 푃, and given an edge 푒 ∈ 푃 that only occurs once on 푃: For any vertices 푣1, 푣2 ∈ 푃 there is a path from 푣1 to 푣2 that doesn’t pass through 푒. 푣2 푣1 Topological Polyhedra Claim: If 푓1 and 푓2 are distinct faces of a topological polyhedron which share an edge 푒, then: replacing 푓1 and 푓2 with 푓1 ∪ 푓2, and removing 푒 from the edge list, we still have a valid topological polyhedron. 푓1 푓2 푒 Topological Polyhedra Proof (CW Complex): The edges/vertices of 푓1 ∪ 푓2 are in the complex (since 푒 is not on the boundary). * Since the intersection 푓1 ∩ 푓2 is connected and the interiors of 푓1 and 푓2 have disk- topology, the interior of 푓1 ∪ 푓2 also has disk- topology. *This is just a sketch of the proof. Topological Polyhedra Proof (CW Complex): The boundary of 푓1 ∪ 푓2 is connected. • Let 푣 ∈ 푒 be an end-point. • For 푣1, 푣2 ∈ 푓1 ∪ 푓2, there is a curve connecting 푣 to each 푣푖 that does not contain the edge 푒. • Concatenating the two curves we connect 푣1 to 푣2 along the boundary of 푓1 ∪ 푓2. Topological Polyhedra Proof (Manifold): The smaller polyhedron still passes through all the vertices. The edge 푒 is removed and all other edges remain adjacent to a face. Topological Polyhedra Proof (Manifold Edges): The old edges still have only two faces on them (or one face twice). Topological Polyhedra Proof (Manifold Vertices): If 푣 ∉ 푒, we can use the old edge ordering. 푓2 푒 푒 푒1 2 푓1 푣 푒4 푒3 Topological Polyhedra Proof (Manifold Vertices): If 푣 ∉ 푒, we can use the old edge ordering. 푓 ∪ 푓 1 2 푒 푒2 1 푣 푒4 푒3 Topological Polyhedra Proof (Manifold Vertices): If 푣 ∉ 푒, we can use the old edge ordering. If 푣 ∈ 푒 let {푒1, 푒2, … , 푒푘} be the old ordered edges around 푣, shifted so that 푒1 = 푒. Then 푒푘 and 푒2 are consecutive edges on 푓1 ∪ 푓2 so {푒2, … , 푒푘} is a valid ordering. 푓2 푒 푒2 푓1 푣 푒4 푒3 Topological Polyhedra Proof (Manifold Vertices): If 푣 ∉ 푒, we can use the old edge ordering. If 푣 ∈ 푒 let {푒1, 푒2, … , 푒푘} be the old ordered edges around 푣, shifted so that 푒1 = 푒. Then 푒푘 and 푒2 are consecutive edges on 푓1 ∪ 푓2 so {푒2, … , 푒푘} is a valid ordering. 푓1 ∪ 푓2 푒2 푣 푒4 푒3 Curves A (connected) curve on a topological polyhedron is a list of edges such that the ending vertex of one edge is the starting vertex of the next. Curves A (connected) curve on a topological polyhedron is a list of edges such that the ending vertex of one edge is the starting vertex of the next. A closed curve is a curve whose starting and ending points are the same. Genus-0 Polyhedra A polyhedron is genus-0 (or simply connected) if every non-trivial closed curve disconnects the faces of the polyhedron. Genus-0 Polyhedra Aside: The definition can be extended to surfaces with boundary if we curves that start and end the boundary are also considered closed. Genus-0 Polyhedra Equivalently, given a topological polyhedron 푃 we can define the dual graph 푃∗ = (푉∗, 퐸∗). ⇒ A curve 퐶 ⊂ 퐸 corresponds to a set of dual edges 퐶∗ ⊂ 퐸∗ of the dual. ⇒ 푃 is genus-0 if removing 퐶∗ disconnects 푃∗. Genus-0 Polyhedra 1. There is a continuous map from a polytope to a sphere. (e.g. Put the center of mass at the origin and normalize the positions.) 2. By the Jordan Curve Theorem the sphere is genus-zero. One Can Show: ⇒ The polytope must also be genus-0. Euler’s Formula For a genus-0 polyhedron 푃, the number of vertices, |푉|, the number of edges, |퐸|, and the number of faces, |퐹|, satisfy: |푉| − |퐸| + |퐹| = 2 Euler’s Formula (by Induction on |푭|) Base case: 퐹 = 1 We have: 푉 = 푣1, … , 푣푛 * The edges on the boundary of 푣6 푣5 the face form a connected 푣7 푣4 tree (otherwise there is a closed loop and the interior of 푣1 푣 the face is disconnected). 푣2 3 Then there are 푛 − 1 edges: 푉 − 퐸 + 퐹 = 푛 − 푛 − 1 + 1 = 2 *This is just a sketch of the proof. Euler’s Formula (by Induction on |푭|) Induction: Assume true for 퐹 = 푛 − 1 Find 푒 ∈ 퐸 shared by two distinct faces. If no such 푒 exists, then all faces are adjacent to themselves, which contradicts the assumption that the polyhedron is connected. Euler’s Formula (by Induction on |푭|) Induction: Assume true for 퐹 = 푛 − 1 Find 푒 ∈ 퐸 shared by two distinct faces. Remove 푒 and merge the two adjoining faces. Claim: The new polyhedron, 푃′, is still genus-0. Euler’s Formula (by Induction on |푭|) Proof (푃′ is genus-zero): Let 퐶 be a non-trivial curve on 푃′. ⇒ 퐶 is a non-trivial curve on 푃 with 푒 ∉ 퐶. ⇒ 푓1 and 푓2 are in the same component. ⇒ 퐶 disconnects 푓1 ∪ 푓2 from a face 푔 on 푃. ⇒ 퐶 disconnects 푓1 ∪ 푓2 from 푔 in 푃′. Euler’s Formula (by Induction on |푭|) Induction: Assume true for 퐸 = 푛 − 1 Find 푒 ∈ 퐸 shared by two distinct faces. Remove 푒 and merge the two adjoining faces. 푃′ is genus-0 with 퐸 − 1 edges, 퐹 − 1 faces, and |푉| vertices.
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