…and stair-convexity

1. Lower bounds for weak epsilon-nets… …and stair-convexity Boris Bukh Princeton U. JiříMatoušek Charles U. Gabriel Nivasch Tel Aviv U.

2. Weak epsilon-nets S N Let S be a finite point set in Rd. We want to stab all “large”convex hulls in S. Namely: Let ε < 1 be a parameter. We want to construct another point set N such that, for every subset S' of at least an ε fraction of the points of S, the convex hull of S' contains at least one point of N. N is called a weak ε-net for S. Problem: Construct N of minimal size.

3. Weak epsilon-nets Known upper bounds for weak epsilon-nets: • Every point set S in the plane has a weak 1/r-net of size O(r2) [ABFK ’92]. • Every point set S in Rd has a weak 1/r-net of size O(rd polylog r) [CEGGSW ’95, MW ’04]. S Known upper bounds for specific cases: • If S is in convex position in the plane, then S has a weak 1/r-net of size O(rα(r)) [AKNSS ’08]. Inverse Ackermann function – extremely slow-growing!

4. Weak epsilon-nets Generalization to Rd: S A convex curve in Rd is a curve that is intersected at most d times by every hyperplane. (E.g.: the moment curve.) If Slies on a convex curvein Rd, then S has a weak 1/r-net of size at most r * 2poly(α(r)), where the poly depends on d [AKNSS ’08].

5. Weak epsilon-nets Known lower bounds for weak epsilon-nets: • For fixed d, only the trivial bound is known! Ω(r) • (For fixed r as a function of d, Matoušek [’02] showed an exponential lower bound of Ω(e√(d/2)) for r = 50.) Our results: • Lower bound of Ω(r log r) for the plane. • Generalizes to Ω(r logd–1r) for Rd. • Lower bound of r * 2poly(α(r)) for a point set on a convex curve in Rd, for every d≥ 3 (with a smaller poly than in the upper bound).

6. b b a a b b a z y x Stair-convexity We first introduce a cute notion, called stair-convexity. Stair-path between two points: • If a and b are points in the plane, the stair-path between a and b goes first up, and then left or right. • Let a and b be points in Rd, with alower than b in last coordinate. • Let a' be the point directly above a at the same height as b. • The stair-path between a and b is the segment aa', followed by the stair-path from a' to b (by induction one dimension lower). a'

7. Stair-convexity A set S in Rd is stair-convex if for every pair of points of S, the stair-path between them is entirely contained in S. Stair-convex set in the plane: • Stair-convex set in Rd: • Every horizontal slice (by a hyperplane perpendicular to the last axis) is stair-convex in Rd–1. • The slice grows as we slide the hyperplane up.

8. The lower-bound construction Our lower bound in the plane is achieved by an n x n grid suitably streched in the y-direction: … δ5 δ4 δ1 << δ2 << δ3 << δ4 << δ5 << … δ3 δ2 δ1 1

9. The lower-bound construction Specifically: We choose δilarge enough that the segment from point (1,1) to point (n, i+1) passes above point (2, i): (n, i+1) δi Claim: A weak 1/r-net for the stretched grid must have size Ω(r log r). (2, i) (1, 1)

10. The lower-bound construction Let B be the bounding box of the stretched grid. Let B' be the unit square. B π B' 1 1 Let π be a bijection between B and B' that preserves order in each coordinate, and maps the stretched grid into a uniform grid.

11. The lower-bound construction What happens to straight segments under the bijection π? 1 1 If the grid is very dense, then they look almost like stair-paths.

12. The lower-bound construction A convex set under π… …looks almost like a stair-convex set.

13. N The lower-bound construction As the grid sizen tends to infinity, constructing a weak ε-net for the stretched grid becomes equivalent to… …constructing a set of points that stabs all stair-convex sets of area εin the unit square. (This equivalence has a (boring) formal proof…)

14. N The lower bound Problem: Given ε = 1/r, construct a set of points N that stabs all stair-convex sets of area 1/r in the unit square. Claim: Such a set N must have Ω(r log r) points. Equivalent claim: Let N be any set of n points in the unit square. Then there’s an unstabbed stair-convexset of area Ω((log n) / n).

15. The lower bound Claim: Let N be a set of n points in the unit square. Then there’s an unstabbed stair-convex set of area Ω((log n) / n). Proof: Define rectangles: x = 1/2 y = 1/(4n) 1st level rectangle: x/2 2y 2nd level rectangle: Each rectangle has area 1/(8n) x/4 still ≤ 1/2 3rd level rectangle: 4y … (log2n)-th level rectangle:

16. The lower bound Let Q be the upper-left quarter of the unit square. Q p N Call a point p in Qk-safe if the k-th level rectangle with p as upper-left corner is not stabbed by any point of N. How much of Q is k-safe?

17. The lower bound Each point of Ninvalidates a region of area at most 1/(8n). Q N has n points. N Q has area 1/4.  At least half of Q is k-safe. For every k, a random point of Q has probability 1/2 of being k-safe.

18. The lower bound For a point p in Q, the fan of p is the set of rectangles of level 1, 2, 3, …, log2n with p as left corner. P If p is randomly chosen, the expected fraction of rectangles in the fan of p that are not stabbed by any point of N is at least 1/2. There is a p that achieves this expectation.  Its fan has Ω(log n) non-stabbed rectangles. Their union is a stair-convex set. What is the area of this set?

19. The lower bound The lower-right quarters of the rectangles in the fan of p are pariwise disjoint: P  Each rectangle contributes area Ω(1/n). We have found an unstabbed stair-convex set of area Ω((log n) / n). QED

20. Generalization to arbitrary d Our construction generalizes to arbitrary dimension d: … δ4 δ1 << δ2 << δ3 << δ4 << … δ3 δ2 (d–1)-dimensional stretched grid δ1

21. Generalization to arbitrary d We place the (i+1)-st layer high enough such that: point in (i+1)-st layer i-th layer 1st layer

22. Generalization to arbitrary d Let π be a bijection that maps the stretched grid into a uniform grid in the unit d-cube. Then, under π, line segments look like stair-paths… …and convex sets look like stair-convex sets. Claim: In order to stab all stair-convex sets of volume 1/r in the unit d-cube, we need Ω(r logd–1r) points.

23. Tightness We cannot get more than Ω(r logd–1r) from our construction. Claim: There exists a set of O(r logd–1r) points that stabs all stair-convex sets of volume 1/rin the unit d-cube.

24. π(D) Point sets on convex curves • Recall: • A convex curve in Rd is a curve that intersects every hyperplane at most d times. • A point setthat lies on a convex curve in Rdhas a weak 1/r-net of size at most r * 2poly(α(r)) [AKNSS’08]. We show: This bound is not too far from truth in the worst case. Specifically: Let D be the diagonal of the stretched grid, for d ≥ 3. • Then: • D lies on a convex curve. • Every weak 1/r-net for D must have size at leastr * 2poly(α(r)) (with a smaller poly).

25. Point sets on convex curves • Dlies on a convex curve. In fact, D lies on the curve γ = { (c1t, c2t, …, cdt) : t in R }. for some constants c1, …, cd. • Every weak 1/r-net for D must have size at leastr * 2poly(α(r)). This follows by the same technique of [AKNSS]: Reduction to stabbing interval chains.

26. Point sets on convex curves Claim: The curve γ = { (c1t, c2t, …, cdt) : t in R } intersects every hyperplane at most d times. Proof: A hyperplane has the form { (x1, …, xd) : α1x1 + … + αdxd + αd+1 = 0 }, for some parameters α1, …, αd+1. So we need to prove that f(t) = α1c1t + ... + αdcdt + αd+1 has at most dzeros. Enough to prove that f'(t) = β1c1t + ... + βdcdthas at most d–1 zeros. = c1t ( β1 + β2(c2/c1)t + ... + βd(cd/c1)t ) By induction. QED

27. Other results The stretched grid in the plane yields an improved upper bound for the Second Selection Lemma. Second Selection Lemma: Let S be a set of n points in the plane, and let T be a set of m triangles spanned by S. Then there exists a point in the plane that stabs “many” triangles of T. • Current lower bound: Ω(m3 / (n6 log2n)) [Eppstein ’93, corrected by NS’08] • Current upper bound: O(m2 / n3) [Eppstein ’93] • (Actually for everyS there is a T that achieves this.) We show: Upper bound of O(m2 / (n3 log n)), taking S = stretched grid.

28. Other results We recently proved [BMN’08]: Let S be a set of n points in R3. Then there exists a line that stabs at least n3 / 25 – o(n3) triangles spanned by S. We showed: The stretched grid in R3 gives a matching upper bound for this. No line stabs more than n3 / 25 + o(n3) triangles. (Complicated calculation, which seems hard to generalize.)

29. Open problems • For weak epsilon-nets: • Close the gap between Ω(r log r) and O(r2) in the plane. • Close the gap between Ω(r) and O(rα(r)) in convex position in the plane. • Close the gaps for general d. • For the stretched grid: • Calculate, in general, the maximum number of k-simplices of the d-dimensional stretched grid that can be stabbed by a j-flat. (We did the case d = 3, k = 2, j = 1 by a complicated calculation.) THANK YOU!