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

The stretched grid – in the plane The stretched grid is an n x n grid suitably streched in the y-direction: 1 δ1δ1 δ2δ2 δ3δ3 δ4δ4 … δ5δ5 δ 1 << δ 2 << δ 3 << δ 4 << δ 5 << … n = suitable parameter (large enough) n columns n rows

The stretched grid – in the plane δiδi Specifically: We choose δ i large enough that the segment from point (1,1) to point (n, i+1) passes above point (2, i): (1, 1) (n, i+1) (2, i)

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

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

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 ∞?

Stair-convexity – in the plane 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 b a b We want to understand convexity in the stretched grid, in the limit n  ∞. A set S in R 2 is stair-convex if for every pair of points of S, the stair-path between them is entirely contained in S.

Stair-convexity – in the plane As the grid size n 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

The stretched grid – in R d Growth much faster than in dimension d–1 (d–1)-dimensional stretched grid δ1δ1 δ2δ2 δ3δ3 δ4δ4 … δ 1 << δ 2 << δ 3 << δ 4 << …

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

The stretched grid – in R d 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?

Stair-convexity – in R d Stair-path between two points: Let a and b be points in R d, with a lower 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' x y z a b b

Stair-convexity – in R d A set S in R d 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 R d : Every horizontal slice (by a hyperplane perpendicular to the last axis) is stair- convex in R d–1. The slice grows as we slide the hyperplane up.

Stair-convexity – in R d 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

Properties of stair-convexity Let P be a set of points in the plane. sconv(P) = stair-convex hull of P. 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. sconv(P) x P

Properties of stair-convexity Generalization to R d (“stair-convexity lemma”): A: Divide space into d+1 regions as follows: When is point x contained in sconv(P)? Then, x is in sconv(P) iff P contains one point in each region. 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.

First Selection Lemma Let S be a set of n points in R d. S defines d-dimensional simplices. Lemma: There exists a point in R d that is contained in many simplices: At least c d n d+1 – O(n d ) simplices. Lower bounds:Upper bounds: c 2 ≥ 1/27 [BF ’84]c 2 ≤ 1/27 + 1/729 [BF ’84] [Wagner ‘03] [Bárány ‘82] (“Trivial”) We show: c 2 ≤ 1/27

First Selection Lemma Claim: c d ≤ (d+1) –(d+1). Proof: Let S be the stretched grid in R d of side n 1/d. (Then |S| = n.) Map into the unit cube: 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. At most (n / (d+1)) d+1 d-simplices.QED Let x be an arbitrary point in R d. In the worst case, all parts have equal size n/(d+1). 1 1 x n/3 S

Diagonal of the stretched grid The diagonal of the stretched grid… D Map into the unit cube …is another useful point set. D can be alternatively defined as follows: Take a single fast-growing sequence of length dn: x 11 << x 12 << … << x 1n << x 21 << x 22 << … << x 2n << … << x d1 <<… << x dn Then let p i = (x 1i, x 2i, …, x di ) for i = 1, 2, …, n, and D = (p 1, p 2, …, p n ).

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

Variants of the First Selection Lemma We recently proved [BMN’08]: Let S be a set of n points in R 3. Then there exists a line that stabs at least n 3 / 25 – o(n 3 ) triangles spanned by S. [Bukh]: The stretched grid in R 3 gives a matching upper bound for this. No line stabs more than n 3 / 25 + o(n 3 ) triangles. (Complicated calculation, which seems hard to generalize.) (We would like to find a k-flat that stabs many j-dimensional simplices in R d, in general. We think the stretched grid gives tight upper bounds.)

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. [NS ‘09, fixing Eppstein ‘93]: Ω(m 3 / (n 6 log 2 n)) triangles. [ACEGSW ’91]: Ω(m 3 / (n 6 log 5 n)) triangles.

Second Selection Lemma Upper bound for the Second Selection Lemma in the plane: [Eppstein ‘93]: Sometimes you cannot intersect more than O(m 2 / n 3 ) 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(m 2 /(n 3 log n)) triangles.

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(m 2 /(n 3 log n)) triangles. Our set S: The stretched grid. Our set T (roughly speaking): All “increasing” triangles whose area is ≤ δ. 1 1 ≤ δ (δ is chosen so that |T| = m.)

Weak epsilon-nets Let S be a finite point set in R d. We want to stab all “large” convex hulls in S. 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. Problem: Construct N of minimal size. Let ε < 1 be a parameter. N is called a weak ε-net for S. Namely: SN (“Weak”: we don’t require.)

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(r 2 ) [ABFK’92]. Every point set S in R d has a weak 1/r-net of size O(r d 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 R d (of side sufficiently large in terms of r) then every weak 1/r-net for S has size Ω(r log d–1 r).

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

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. Or in other words: Let N be any set of n points in the unit square. Then there’s an unstabbed stair-convex set of area Ω((log n) / n). Claim: Such a set N must have Ω(r log r) points. N

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: 1 st level rectangle: 2 nd level rectangle: 3 rd level rectangle: (log 2 n)-th level rectangle: … x = 1/2 y = 1/(4n) Each rectangle has area 1/(8n) x/2 2y2y x/4 4y4y still ≤ 1/2

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

Weak epsilon-nets Each point of N invalidates a region of area at most 1/(8n). Q Q has area 1/4. N  At least half of Q is k-safe. N has n points. For every k, a random point of Q has probability 1/2 of being k-safe.

Weak epsilon-nets For a point p in Q, the fan of p is the set of rectangles of level 1, 2, 3, …, log 2 n 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?

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

Weak epsilon-nets Tightness: Claim: There exists a set of O(r log d–1 r) 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 log d–1 r).

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. D (α(n) grows slower than log* n, log** n, …) We show: A weak 1/r-net for D must have size Ω(r 2 poly(α(r)) ). very slow-growing degree of poly ≈ d/2.

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

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

THANKS!