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

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 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 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(r α(r)) [AKNSS ’08]. S Inverse Ackermann function – extremely slow-growing!

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

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 log d–1 r) for R d. Lower bound of r * 2 poly(α(r)) for a point set on a convex curve in R d, for every d ≥ 3 (with a smaller poly than in the upper bound).

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 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' a b x y z a b a b b

Stair-convexity 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 the plane: 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.

The lower-bound construction Our lower bound in the plane is achieved by 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 << …

The lower-bound construction 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) Claim: A weak 1/r-net for the stretched grid must have size Ω(r log r).

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

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

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

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: 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

The lower bound 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?

The lower bound 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.

The lower bound 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?

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

Generalization to arbitrary d Our construction generalizes to arbitrary dimension d: (d–1)-dimensional stretched grid δ1δ1 δ2δ2 δ3δ3 δ4δ4 … δ 1 << δ 2 << δ 3 << δ 4 << …

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

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 log d–1 r) points.

Tightness We cannot get more than Ω(r log d–1 r) from our construction. 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.

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

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

Point sets on convex curves Claim: The curve γ = { (c 1 t, c 2 t, …, c d t ) : t in R } intersects every hyperplane at most d times. Proof: A hyperplane has the form { (x 1, …, x d ) : α 1 x 1 + … + α d x d + α d+1 = 0 }, for some parameters α 1, …, α d+1. So we need to prove that f(t) = α 1 c 1 t α d c d t + α d+1 has at most d zeros. Enough to prove that f'(t) = β 1 c 1 t β d c d t has at most d–1 zeros. = c 1 t ( β 1 + β 2 (c 2 /c 1 ) t β d (c d /c 1 ) t ) QED By induction.

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(m 3 / (n 6 log 2 n)) [Eppstein ’93, corrected by NS’08] Current upper bound: O(m 2 / n 3 ) [Eppstein ’93] (Actually for every S there is a T that achieves this.) We show: Upper bound of O(m 2 / (n 3 log n)), taking S = stretched grid.

Other results 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. We showed: 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.)

Open problems For weak epsilon-nets: Close the gap between Ω(r log r) and O(r 2 ) in the plane. Close the gap between Ω(r) and O(r α(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!