Half-Space Intersections

1. Half-Space Intersections Randomized algorithm and expected run time analysis

2. Problem Definition – 2D • Given n half-planes as linear inequalities, we wish to find a boundary where they all intersect

3. Problem Definition – 3D • We extend our problem to 3D • The intersection is a convex polyhedron • This polyhedron is represented as a graph with degree 3 • inter(S): the intersection of a set S of half-spaces http://wpcontent.answers.com/wikipedia/commons/thumb/6/65/Half_Space.svg/300px-Half_Space.svg.png

4. Further Assumptions • inter(S) is bounded. If not – there’s a point at infinity which bounds it • No 3 vertices are collinear • Intersection of planes and spaces can be computed in constant time • Deciding if a point is inside a half-space can be computed in constant time

5. Algorithm Outline • Randomize a permutation of the set S of half spaces • For : • Add half space hito inter(Si-1) by • removing all the vertices in • adding new vertices on the boundary of

6. Addition of a Half Space • At step i we have a bidirectional pointer from every to some • When we add hi we start at the conflicting vertex and search the graph until we enter hi • We destroy all the vertices we encounter and create new ones on the face of

7. Addition of a Half Space

8. Pointer Update • Recall: the pointer is bidirectional • When we delete a vertex v pointing to a half-space h ∈ S \ Siwe have to find a new vertex w, such that • We perform BFS along until we encounter vertices inside inter(Si). This is our w.

9. Number of Vertices Created • Backwards analysis: suppose we delete vertices from inter(Si) and get inter(Si-1) • The convex polyhedron inter(Si) has i faces and O(i) vertices (because the degree is bounded by 3) • If we choose one of i half-spaces to delete, the expected number of vertices we delete is • Hence: the expected number of vertices created at the addition of hi is constant

10. Search Cost Analysis • When adding a half-space, we perform a BFS starting from the pointed vertex • Time – linear in # of deleted vertices • Each deleted vertex was created once: we only count creations • How many vertices are created in total?

11. Total # of Vertices Created • Total vertices created when adding hi: • c(H,h):# of conflict vertices between inter(H) and h • Backwards analysis – the expectation over all hi is bounded by: since the degree of each vertex is bounded by 3

12. Total # of Vertices Created • hi+1 is chosen randomly, so: • The expected # of vertices is bounded by: • We saw the algorithm runs in time linear to the number of vertices created, so this expression also bounds the run time of the algorithm

13. Bounding the Total Run Time • c(Si,hi+1) is also the expected number of vertices deleted by the addition of hi+1

14. Bounding the Total Run Time • Let tc(v) be the time v is created and td(v) be the time v is destroyed • Then: for all the vertices v created in the algorithm

15. Bounding the Total Run Time • Obviously, tc(v) ≤ td(v) – 1. Therefore: • Recall that the expected number of vertices created at each iteration is constant! Constant

16. Bounding the Total Run Time • The total run time of the algorithm is:

17. Convex Hull in 3D • Using duality we can reduce the problem of convex hull in 3D to finding half-space intersection (in 3D) • We just saw how to compute the half-space intersection in O(nlogn) time • 3D Convex Hull problem takes also O(nlogn) time to compute

18. Thank You!