1. 1 / 41 Convex Hulls in 3-space Jason C. Yang

2. 2 / 41 Problem Statement Given P: set of n points in 3-space Return: –Convex hull of P: CH (P) –Smallest polyhedron s.t. all elements of P on or in the interior of CH (P).

3. 3 / 41 Algorithm Randomized incremental algorithm Steps: –Initialize the algorithm –Loop over remaining points Add p r to the convex hull of P r-1 to transform CH (P r-1 ) to CH (P r ) [for integer r  1, let P r :={p 1,…,p r }] Main Idea: Incrementally insert new points into the running/intermediate Convex Hull.

4. 4 / 41 Initialization Need a CH to start with Build a tetrahedron using 4 points in P –Start with two distinct points in P: p 1 and p 2 –Walk through P to find p 3 that does not lie on the line through p 1 and p 2 –Find p 4 that does not lie on the plane through p 1, p 2, p 3 –Special case: No such points exist? Compute random permutation p 5,…,p n of the remaining points Need a CH to start with Build a tetrahedron using 4 points in P –Start with two distinct points in P: p 1 and p 2 –Walk through P to find p 3 that does not lie on the line through p 1 and p 2 –Find p 4 that does not lie on the plane through p 1, p 2, p 3 –Special case: No such points exist? All points lie on a plane. Use planar CH algorithm! Compute random permutation p 5,…,p n of the remaining points

5. 5 / 41 Inserting Points into CH Add p r to the convex hull of P r-1 to transform CH (P r-1 ) to CH (P r ) [for integer r  1, let P r :={p 1,…,p r }] Two Cases: 1) P r is inside or on the boundary of CH (P r-1 ) Trivial: CH (P r ) = CH (P r-1 ) 2) P r is outside of CH (P r-1 )

6. 6 / 41 Case 2: P r outside CH (P r-1 ) Determine horizon of p r on CH (P r-1 ) –Closed curve of edges enclosing the visible region of p r on CH (P r-1 )

7. 7 / 41 Visibility Consider plane h f containing a facet f of CH (P r-1 ) f is visible from a point if that point lies in the open half-space on the other side of h f

8. 8 / 41 Rethinking the Horizon –Boundary of polygon obtained from projecting CH (P r-1 ) onto a plane with p r as the center of projection

9. 9 / 41 CH (P r-1 )  CH (P r ) Remove visible facets from CH (P r-1 ) Found horizon: Closed curve of edges of CH (P r-1 ) Form CH (P r ) by connecting each horizon edge to p r to create a new triangular facet

10. 10 / 41 Algorithm So Far… Initialization –Form tetrahedron CH (P 4 ) from 4 points in P –Compute random permutation of remaining pts. For each remaining point in P –p r is point to be inserted –If p r is outside CH (P r-1 ) then Determine visible region Find horizon and remove visible facets Add new facets by connecting each horizon edge to p r How do we determine the visible region?

11. 11 / 41 How to Find Visible Region Naïve approach: –Test every facet with respect to p r –O(n 2 ) Trick is to work ahead: Maintain information to aid in determining visible facets.

12. 12 / 41 Conflict Lists For each facet f maintain P conflict ( f )  {p r+1, …, p n } containing points to be inserted that can see f For each p t, where t > r, maintain F conflict (p t ) containing facets of CH (P r ) visible from p t p and f are in conflict because they cannot coexist on the same convex hull

13. 13 / 41 Conflict Graph G Bipartite graph –pts not yet inserted –facets on CH (P r ) Arc for every point-facet conflict Conflict sets for a point or facet can be returned in linear time At any step of our algorithm, we know all conflicts between the remaining points and facets on the current CH

14. 14 / 41 Initializing G Initialize G with CH (P 4 ) in linear time Walk through P 5-n to determine which facet each point can see f1f1 f2f2 p6p6 p5p5 p7p7 p6p6 p7p7 p5p5 f2f2 f1f1 G

15. 15 / 41 Updating G Discard visible facets from p r by removing neighbors of p r in G Remove p r from G Determine new conflicts p6p6 p7p7 p5p5 f2f2 f1f1 G f1f1 f2f2 f3f3 f3f3 p6p6 p7p7 p5p5 f2f2 f1f1 G p6p6 p7p7 p5p5 f2f2 f1f1 G p5p5 p7p7 p6p6

16. 16 / 41 Determining New Conflicts If p t can see new f, it can see edge e of f. e on horizon of p r, so e was already in and visible from p t in CH (P r-1 ) If p t sees e, it saw either f 1 or f 2 in CH (P r-1 ) P t was in P conflict (f 1 ) or P conflict (f 2 ) in CH (P r-1 ) ptpt

17. 17 / 41 Determining New Conflicts Conflict list of f can be found by testing the points in the conflict lists of f 1 and f 2 incident to the horizon edge e in CH (P r-1 ) ptpt

18. 18 / 41 What About the Other Facets? P conflict ( f ) for any f unaffected by p r remains unchanged ptpt Deleted facets not on horizon already accounted for

19. 19 / 41 Final Algorithm Initialize CH (P 4 ) and G For each remaining point –Determine visible facets for p r by checking G –Remove F conflict (p r ) from CH –Find horizon and add new facets to CH and G –Update G for new facets by testing the points in existing conflict lists for facets in CH (P r-1 ) incident to e on the new facets –Delete p r and F conflict (p r ) from G

20. 20 / 41 Fine Point Coplanar facets –p r lies in the plane of a face of CH (P r-1 ) f is not visible from p r so we merge created triangles coplanar to f New facet has same conflict list as existing facet

21. 21 / 41 Analysis

22. 22 / 41 Complexity of CH for n points in 3-space is O(n) Number of edges of a convex polytope with n vertices is at most 3n-6 and the number of facets is at most 2n-4 From Euler’s formula: n – n e + n f = 2 Complexity

23. 23 / 41 Complexity Each face has at least 3 arcs Each arc incident to two faces 2n e  3n f Using Euler n f  2n-4n e  3n-6

24. 24 / 41 Expected Number of Facets Created Will show that expected number of facets created by our algorithm is at most 6n-20 Initialized with a tetrahedron = 4 facets

25. 25 / 41 Expected Number of New Facets Backward analysis: –Remove p r from CH (P r ) –Number of facets removed same as those created by p r –Number of edges incident to p r in CH (P r ) is degree of p r : deg(p r, CH (P r ))

26. 26 / 41 Expected Degree of p r Convex polytope of r vertices has at most 3r-6 edges Sum of degrees of vertices of CH (P r ) is 6r-12 Expected degree of p r bounded by (6r-12)/r

27. 27 / 41 Expected Number of Created Facets 4 from initial tetrahedron Expected total number of facets created by adding p 5,…,p n

28. 28 / 41 Running Time Initialization  O(nlogn) Creating and deleting facets  O(n) –Expected number of facets created is O(n) Deleting p r and facets in F conflict (p r ) from G along with incident arcs  O(n) Finding new conflicts  O(?)

29. 29 / 41 Total Time to Find New Conflicts For each edge e on horizon we spend O(card(P(e)) time where P(e)  P confict (f 1 )  P conflict (f 2 ) Total time is O(  e  L card(P(e))) bounded by expected value of  card(P(e)) Lemma 11.6 The expected value of  e card(P(e)), where the summation is over all horizon edges that appear at some stage of the algorithm is O(nlogn)

30. 30 / 41 Randomized Insertion Order

31. 31 / 41 Running Time Initialization  O(nlogn) Creating and deleting facets  O(n) Updating G  O(n) Finding new conflicts  O(nlogn) Total Running Time is O(nlogn)

32. 32 / 41 Higher Dimensional Convex Hulls Upper Bound Theorem: The worst-case combinatorial complexity of the convex hull of n points in d-dimensional space is  (n  d/2  ). Our algorithm generalizes to higher dimensions with expected running time of  (n  d/2  )

33. 33 / 41 Higher Dimensional Convex Hulls Best known output-sensitive algorithm for computing convex hulls in R d is: O(nlogk +(nk) 1-1/(  d/2  +1) log O(n) ) where k is complexity