Stair-convexity

1. Stair-convexity Gabriel Nivasch ETH Zürich Joint work with: Boris Bukh(Cambridge U.),JiříMatoušek(Charles U.)

2. The stretched grid – in the plane The stretched grid is an n x n grid suitably streched in the y-direction: n = suitable parameter (large enough) … δ5 δ4 nrows δ1 << δ2 << δ3 << δ4 << δ5 << … δ3 δ2 δ1 1 ncolumns

3. The stretched grid – in the plane 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 (2, i) (1, 1)

4. The stretched grid – in the plane 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.

5. The stretched grid – in the plane What happens to straight segments under the bijection π? 1 1 If the grid is very dense, then they look almost like upside-down L’s.

6. The stretched grid – in the plane We want to understand convexity in the stretched grid. (We always look at its image under the bijection π.) Let’s take a few grid points and take their convex hull: What happens when n (grid side) tends to ∞?

7. b b a a Stair-convexity – in the plane We want to understand convexity in the stretched grid, in the limit n ∞. We introduce a 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 is an upside-down L, going first up, and then left or right. A set S in R2 is stair-convex if for every pair of points of S, the stair-path between them is entirely contained in S.

8. Stair-convexity – in the plane As the grid sizen tends to infinity, convexity in the stretched grid reduces to stair-convexity in the unit square (under the bijection π): Further: area of the stair-convex set ≈ fraction of grid points contained

9. The stretched grid – in Rd … δ4 δ1 << δ2 << δ3 << δ4 << … Growth much faster than in dimension d–1 δ3 δ2 (d–1)-dimensional stretched grid δ1

10. The stretched grid – in Rd Specifically: We place the (i+1)-st layer high enough such that: point in (i+1)-st layer i-th layer 1st layer

11. The stretched grid – in Rd Define a bijection π that maps the stretched grid into a uniform grid in the unit d-cube: Again let the grid side n ∞. What happens to convexity?

12. b b a z y x Stair-convexity – in Rd Stair-path between two points: • 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'

13. Stair-convexity – in Rd 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 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.

14. Stair-convexity – in Rd Let the grid side n ∞. Then, under the bijection π, line segments look like stair-paths… …and convex sets look like stair-convex sets. Further: volume of stair-convex set ≈ fraction of grid points contained

15. Properties of stair-convexity Let P be a set of points in the plane. sconv(P) = stair-convex hull of P. P sconv(P) x Let x be another point. When is x contained in sconv(P)? A: Divide the plane into 3 regions as follows… x is in sconv(P) iff P contains one point in each region.

16. Properties of stair-convexity Generalization to Rd (“stair-convexity lemma”): When is point x contained in sconv(P)? A: Divide space into d+1 regions as follows: region 0: (–, –, –, –, …, –, –) region 1: (+, –, –, –, …, –, –) region 2: (?, +, –, –, …, –, –) region 3: (?, ?, +, –, …, –, –) … region d: (?, ?, ?, ?, …, ?, +) – smaller than x in this coord. + greater than x in this coord. ? doesn‘t matter. Then, x is in sconv(P) iff P contains one point in each region.

17. First Selection Lemma S defines d-dimensional simplices. Let S be a set of n points in Rd. Lemma: There exists a point in Rd that is contained in many simplices: At least cd nd+1 – O(nd) simplices. Lower bounds: Upper bounds: c2 ≥ 1/27 [BF ’84] c2 ≤ 1/27 + 1/729 [BF ’84] [Bárány ‘82] [Wagner ‘03] (“Trivial”) c2 ≤ 1/27 We show:

18. First Selection Lemma Claim:cd ≤ (d+1)–(d+1). Proof: Let S be the stretched grid in Rd of side n1/d. (Then |S| = n.) S n/3 Map into the unit cube: x n/3 n/3 Let x be an arbitrary point in Rd. 1 Then x defines a partition of S into d+1 disjoint subsets (stair-conv lemma). A d-simplex defined by S contains x iff each vertex lies in a different subset. In the worst case, all parts have equal size n/(d+1). 1 At most (n / (d+1))d+1d-simplices. QED

19. D Diagonal of the stretched grid Map into the unit cube The diagonal of the stretched grid… …is another useful point set. D can be alternatively defined as follows: Take a single fast-growing sequence of length dn: x11 << x12 << … << x1n << x21 << x22 << … << x2n << … << xd1 <<… << xdn Then let pi = (x1i, x2i, …, xdi) for i = 1, 2, …, n, and D = (p1, p2, …, pn).

20. Diagonal of the stretched grid Alternative proof of our upper bound for the FSL… (map into the unit cube) n/3 x n/3 D Worst case: n/3 …using the diagonal of the stretched grid.

21. Variants of the First Selection Lemma 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. [Bukh]: 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.) (We would like to find a k-flat that stabs many j-dimensional simplices in Rd, in general. We think the stretched grid gives tight upper bounds.)

22. Second Selection Lemma 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. • [ACEGSW ’91]: Ω(m3 / (n6 log5n)) triangles. • [NS ‘09, fixing Eppstein ‘93]: Ω(m3 / (n6 log2n)) triangles.

23. Second Selection Lemma Upper bound for the Second Selection Lemma in the plane: [Eppstein ‘93]: Sometimes you cannot intersect more than O(m2 / n3) triangles by any point. (In fact, he showed that for every set S of points, there is a set T of triangles that achieves this upper bound.) We show: There is a set S of n points, and a set T of m triangles, such that no point intersects more than O(m2 /(n3 log n)) triangles.

24. Second Selection Lemma Claim: There is a set S of n points, and a set T of m triangles, such that no point intersects more than O(m2 /(n3 log n)) triangles. Our set S: The stretched grid. Our set T (roughly speaking): All “increasing” triangles whose area is ≤ δ. ≤ δ 1 (δ is chosen so that |T| = m.) 1

25. 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. (“Weak”: we don’t require .) Problem: Construct N of minimal size.

26. 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]. Known lower bounds : • For fixed d, only the trivial bound was known until now! 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 result: If S is the stretched grid in Rd (of side sufficiently large in terms of r) then every weak 1/r-net for S has size Ω(r logd–1r).

27. Weak epsilon-nets Claim: Every weak 1/r-net for the stretched grid in Rd must have size Ω(r logd–1r). Proof in the plane…

28. N Weak epsilon-nets Equivalent 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. Or in other words: 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).

29. Weak epsilon-nets 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:

30. Weak epsilon-nets 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?

31. Weak epsilon-nets 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.

32. Weak epsilon-nets 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?

33. Weak epsilon-nets 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

34. Weak epsilon-nets Tightness: Claim: There exists a set of O(r logd–1r) points that stabs all stair-convex sets of volume 1/r in the unit d-cube. The stretched grid does have a weak 1/r-net of size O(r logd–1r).

35. D Weak epsilon-nets What can we say about weak epsilon-nets for the diagonal of the stretched grid? A: For d ≥ 3, a weak 1/r-net for D must have size superlinear in r. But just barely superlinear! Let α(n) denote the extremely slow-growing inverse Ackermann function. (α(n) grows slower than log* n, log** n, …) very slow-growing We show: A weak 1/r-net for D must have size Ω(r 2poly(α(r))). degree of poly ≈ d/2.

36. D Weak epsilon-nets A weak 1/r-net for D must have size Ω(r 2poly(α(r))). degree of poly ≈ d/2. Tightness: Ddoes have a weak 1/r-net of the same size O(r 2poly(α(r))) (up to lower-order terms of the poly).

37. D Weak epsilon-nets A weak 1/r-net for D must have size Ω(r 2poly(α(r))). degree of poly ≈ d/2. Why is this interesting? • One can show that D lies on a convex curve. • (A convex curve is a curve in Rd that intersects every hyperplane at most d times.) • [AKNSS ‘08] had shown: If a set S in Rd lies on a convex curve, then S has a weak 1/r-net of size O(r 2poly(α(r))). degree of poly ≈ d2/4.  The set D shows that [AKNSS ‘08] are not far from the truth in the worst case.

38. THANKS!