Week 7 Convex Hulls in 3D

Transcription

1 1 Week 7 Convex Hulls in 3D

2 2 Polyhedra A polyhedron is the natural generalization of a 2D polygon to 3D

3 3 Closed Polyhedral Surface A closed polyhedral surface is a finite set of interior disjoint polygons embedded in E3, such that the union of the polygons is a connected 2-manifold.

4 4 Polyhedron A polyhedron is a closed subset of E3 whose boundary can be expressed as a polyhedral surface.

5 5 Confused? These definitions are formal but confusing It is really very difficult to define exactly what is a polyhedron Any simple definition will allow some polyhedra which you don't want to be counted as polyhedra

6 6 A Simpler Definition Let's look at some non-polyhedral objects Disconnected!

7 7 A Simpler Definition Top face removed! not disc Not homeomorphic to a disk, not a 2-manifold!

8 8 A Simpler Definition Is a cylinder a polyhedron? Encloses 3D space Connected 2-manifold But it is curved! Cannot be partitioned into a finite number of polygons

9 9 A Simpler Definition Two 2-manifolds touching each other becomes a non-manifold

10 10 A Simpler Definition Self Intersections are not good either non-manifold

11 11 A Simpler Definition There is no simpler definition The polyhedral surface must be Connected 2-manifold Closed Consisting of a finite number of polygons

12 12 Faces, Edges and Vertices Faces maximal, interior connected, plane regions of the boundary Vertices Points where three or more faces meet Edges Line segments where two faces meet

13 13 Faces, Edges and Vertices vertex face face edge face

14 14 Regular Polygons and Polyhedra A regular polygon is one with equal sides and angles

15 15 Regular Polygons and Polyhedra A regular polyhedron is one with congruent regular faces and the number of faces at each vertex is the same hexahedron tetrahedron icosahedron octahedron dodecahedron

16 16 Regular Polyhedra Also called Platonic Solids There are only 5 of them Prove!

17 17 Platonic Solids Each face is a regular p-gon has p vertices sum of the interior angles is (p-2)π each face angle is p 2 p Sum of the angles meeting at a vertex < 2π if we have k faces meeting at the vertex k p 2 2 p k p 2 2 p pk 2 k 2 p 4 4 p 2 k 2 4

18 18 Platonic Solids p k (p-2)(k-2) Name 3 1 Tetrahedron 3 2 Cube 4 2 Octahedron 3 3 Dodecahedron 5 3 Icosahedron Description 3 triangles at each vertex 3 squares at each vertex 4 triangles at each vertex 3 pentagons at each vertex 5 triangles at each vertex p: number of vertices in each face k: number of faces meeting at each vertex

19 19 Euler's Formula Is there a relation between the number of faces the number of edges the number of vertices

20 20 Polygon A simple relation exists for polygons V: number of vertices E: number of edges V=E

21 21 Polyhedra V: Number of vertices E: Number of edges F: Number of faces V-E+F = = 2

22 22 Linearity Assume all faces are triangulated F faces, each has 3 edges Each edge is shared by 2 faces Therefore, 3F = 2E V E F =2 2E V E =2 3 E V 2= 3 E=3V 6 3V =O n 2E F= =2V 4 2V =O n 3

23 23 Hull Algorithms Gift Wrapping Similar to the 2D version At any time Choose a hull face Choose one edge, where the other face is not yet found Bend the plane until you hit a vertex Create a new hull face (triangular)

24 24 Gift Wrapping A clever implementation requires O(n2) time O(n) time to find each new vertex F = O(n) faces O(n) x O(n) = O(n2) Output sensitive If F is much smaller than n, O(nF)

25 25 Divide and Conquer The only algorithm to achieve O(n log n) bound Important and beautiful But difficult to implement

26 26 Divide and Conquer Algorithm Outline Sort the points in x-coordinates [O(n log n)] Divide into two sets A and B [O(n)] Construct the hull for each set [T(n/2)] Merge the two hulls [must be O(n)] T(n) = 2T(n/2) + O(n) = O(n log n)

27 27 Divide and Conquer A B

28 28 Divide and Conquer a b

29 29 Divide and Conquer c a b

30 30 Divide and Conquer c a b

31 31 Divide and Conquer c a b

32 32 Divide and Conquer c a b

33 33 Divide and Conquer

34 34 Divide and Conquer

35 35 Divide and Conquer

36 36 Divide and Conquer

37 37 Divide and Conquer If we can find each new triangle in constant time, the merge works in O(n) time overall Then, we must find c in O(1) time Lemma: c is always a neighbor of a or b Proof: due to convexity (exercise)

38 38 Divide and Conquer Simple idea Find the minimum angle neighbors at A and B Call them A-winner and B-winner c is either A-winner or B-winner If a or b has many neighbors, the search may take O(n) time For O(n) triangles, this leads to O(n2)

39 39 Divide and Conquer Fortunately, the following Lemma saves us If ai is the winner on A, and bi is the winner on B and ai is the next ultimate winner A then, bi+1, the B-winner in the next iteration is CCW of bi around b bi+1 bi ai a b B

40 40 Divide and Conquer Discarding Hidden Faces

41 41 Divide and Conquer Simple Idea On each convex hull, the shadow boundary is known Cut and remove all faces inside this boundary

42 42 Divide and Conquer Doesn't always work!

43 43 Divide and Conquer Resulting hull

44 44 Extra: Proof of Euler's Formula Red-Blue coloring Edge-Vertex graph is red Edge-Face graph is blue Since both graphs are trees EV-graph: E1 = V 1 EF-graph: E2 = F 1 Overall: E1+E2 = V+F 2 V+F-E=2

45 45 Extra: Proof of Euler's Formula Alternative Proof via Planar Graphs A polyhedral graph can be flattened on a plane Base case: A tetrahedron has 4 vertices, 6 edges and 4 faces Inductive Hypothesis: All polyhedral graphs with m edges V E+F=4 6+4=2 V E(=m) + F = 2 Inductive Step: Remove an edge from a graph with m+1 edges This removes a face as well And is covered by the hypothesis Adding the edge back increases both E and F by 1, therefore V (m+1) + (F+1) = 2