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## Stair-convexity

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**Stair-convexity**Gabriel Nivasch ETH Zürich Joint work with: Boris Bukh(Cambridge U.),JiříMatoušek(Charles U.)**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**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)**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.**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.**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 ∞?**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.**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**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**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**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?**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'**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.**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**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.**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.**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:**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**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).**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.**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.)**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.**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.**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**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.**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).**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…**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).**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:**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?**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.**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?**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 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).**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.**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).**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.