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**Lower bounds for epsilon-nets**

Gabriel Nivasch Presenting work by: Noga Alon, János Pach, Gábor Tardos

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**Epsilon-nets Let R be a family of ranges. Say, R = {all discs in R2}.**

Let P be a finite point set. P Let 0 < ε < 1 b a parameter. A subset N of P is an ε-net w.r.t. R if, for every r in R that contains at least ε|P| points of P, r contains a point of N. Want N as small as possible.

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**Epsilon-nets Let R be a family of ranges. Say, R = {all discs in R2}.**

Let P be a finite point set. P Let 0 < ε < 1 b a parameter. N A subset N of P is an ε-net w.r.t. R if, for every r in R that contains at least ε|P| points of P, r contains a point of N. ε|P| = 4 Want N as small as possible.

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**Epsilon-nets Let R be a family of ranges. Say, R = {all discs in R2}.**

Let P be a finite point set. P Let 0 < ε < 1 b a parameter. N A subset N of P is an ε-net w.r.t. R if, for every r in R that contains at least ε|P| points of P, r contains a point of N. ε|P| = 4 Want N as small as possible. [Haussler, Welzl, ‘87]: If R has finite VC-dimension then there exists N of size O(1/ε log 1/ε). Examples of families of ranges: R = {all rectangles in R2} R = {all ellipsoids in Rd} R = {all halfspaces in Rd} Indep. of |P|. Infinite VC-dim R = {all convex sets in R2}

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Epsilon-nets General belief until recently: In geometric settings, there always exist ε-nets of size O(1/ε). But: [Bukh, Matoušek, N. ‘09]: Ω(1/ε logd–1 1/ε) for weak ε-nets w.r.t. convex sets in Rd. [Alon ‘10]: Ω(1/ε α(1/ε)) for ε-nets w.r.t. lines in R2. (*) [Pach, Tardos ‘11]: Ω(1/ε log log 1/ε) for ε-nets w.r.t. axis-parallel rectangles in R2. Ω(1/ε log 1/ε) for dual ε-nets for axis parallel rectanlges in R2. Ω(1/ε log 1/ε) for ε-nets w.r.t. axis-parallel boxes and w.r.t. halfspaces in R4. We will show (Tight [Aronov, Ezra, Sharir ‘10].)

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**Lower bound for axis-parallel rectangles**

[Pach, Tardos ‘11]: Ω(1/ε log log 1/ε) for ε-nets w.r.t. axis-parallel rectangles in R2. Proof strategy: Given n, we will construct an n-point set P such that every ε-net N for P has size > n/2. For ε suitably chosen in terms of n. (Spoiler: ε = (log log n) / n.)

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**Lower bound for axis-parallel rectangles**

[Pach, Tardos ‘11]: Ω(1/ε log log 1/ε) for ε-nets w.r.t. axis-parallel rectangles in R2. Proof strategy: Given n, we will construct an n-point set P such that every ε-net N for P has size > n/2. ??? Normally |N| does not depend on |P| For ε suitably chosen in terms of n. (Spoiler: ε = (log log n) / n.) ??? It should say “Given ε”

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**Lower bound for axis-parallel rectangles**

[Pach, Tardos ‘11]: Ω(1/ε log log 1/ε) for ε-nets w.r.t. axis-parallel rectangles in R2. Proof strategy: Given n, we will construct an n-point set P such that every ε-net N for P has size > n/2. ??? Normally |N| does not depend on |P| For ε suitably chosen in terms of n. (Spoiler: ε = (log log n) / n.) ??? It should say “Given ε” We’ll fix both problems later on.

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**Lower bound for axis-parallel rectangles**

Given n, we will construct an n-point set P such that every ε-net N for P has size > n/2 (for ε suitably chosen). Construction: P is a set of n random points in the unit square. We will show: With high probability, every subset N of n/2 points misses some heavy axis-parallel rectangle. Rectangle that contains εn points.

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**Lower bound for axis-parallel rectangles**

Choosing n random points in the unit square: Step 0: Choose the y-coordinates, and place all points at the left side. Step 1: Jump right by 1/2? Step 2: Jump right by 1/4? … Step t: Jump right by 2–t?

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**Lower bound for axis-parallel rectangles**

Let N be a subset of n/2 points of P. We will calculate the probability that N misses some heavy rectangle. We will look at steps t = 1, 2, …, tmax, where tmax = log2 1/(2ε). N

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**Lower bound for axis-parallel rectangles**

Let N be a subset of n/2 points of P. We will calculate the probability that N misses some heavy rectangle. We will look at steps t = 1, 2, …, tmax, where tmax = log2 1/(2ε). Just before step t, the points lie on 2t–1 vertical lines: N … 2t–1

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**Lower bound for axis-parallel rectangles**

For each vertical line, partition the points into intervals, where each interval contains exactly εn black points. “orphans”

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**Lower bound for axis-parallel rectangles**

For each vertical line, partition the points into intervals, where each interval contains exactly εn black points. “orphans” We want a rectangle to be heavy, and to be missed by N: At step t: none of the black points should jump, all the red points should jump. 2–t

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**Lower bound for axis-parallel rectangles**

For each vertical line, partition the points into intervals, where each interval contains exactly εn black points. “orphans” Claim 1: There are at most n/4 black orphans (by choice of tmax). Proof: # black orphans ≤ 2t–1*εn ≤ 1/(4ε)*εn = n/4. tmax = log2 1/(2ε) Claim 2: There are at least 1/(4ε) intervals. Proof: # intervals ≥ (n/4) / (εn) = 1/(4ε).

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**Lower bound for axis-parallel rectangles**

Call an interval good if it contains at most 3εn red points; otherwise bad. Claim: There are at least 1/(12ε) good intervals. Proof: total # intervals ≥ 1/(4ε) # bad intervals ≤ (n/2) / (3εn) = 1/(6ε) # good intervals ≥ 1/(4ε) – 1/(6ε) = 1/(12ε)

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**Lower bound for axis-parallel rectangles**

Focus on good rectangles – rectangles of good intervals: At most 4εn points Pr[ given good rectangle is heavy and missed by N ] ≥ 2–4εn # good rectangles in first tmax stages ≥ 1/(12ε) tmax ≈ 1/ε log 1/ε tmax = log2 1/(2ε) Pr[ our N is a good ε-net ] ≤ (1 – 2–4εn )1/ε log 1/ε ≤ e–1/ε log 1/ε * 2^(–4εn) 1 – x ≤ e–x

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**Lower bound for axis-parallel rectangles**

Pr[ our N is a good ε-net ] ≤ e–1/ε log 1/ε * 2^(–4εn) # subsets N of size n/2 ≤ 2n Pr[ some N is a good ε-net ] ≤ 2n – 1/ε log 1/ε * 2^(–4εn) union bound Want to choose ε in terms of n so that this probability tends to 0. Let ε = (log log n) / (8n). Get: 2n – n / (log log n) * (log n) * 1/√(log n) = 2n – n (√ log n)/ (log log n) 0

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**Lower bound for axis-parallel rectangles**

To summarize, an ε-net N for P must have size at least n/2, for ε = (log log n) / n. Express |N| in terms of ε: |N| ≥ Ω(1/ε log log 1/ε) Finally: Theorem: For every ε there exists an n0 s.t. for every n ≥ n0 there exists an n-point set P that requires ε-nets w.r.t. axis-parallel rectangles of size Ω(1/ε log log 1/ε). Proof: To get larger sets P, replace every point by a tiny cloud of points.

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Lower bound for lines Theorem [Alon ‘10]: For every ε there exists an n0 and there exists an n0-point set P that requires ε-nets w.r.t. lines of size Ω(1/ε α(1/ε)). Note: No arbitrarily large n. “Cloud” method does not work, because lines are infinitely thin. Proof uses the density Hales-Jewett theorem, which is about playing high-dimensional tic-tac-toe:

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**Density Hales-Jewett theorem**

Density Hales-Jewett Theorem [Furstenberg, Katznelson ‘91]: For every integer k and every real δ > 0 there exists an m such that, no matter how you select a δ fraction of the points of a k * k * k * … * k tic-tac-toe board, you will contain a complete line of size k (maybe diagonal). m [Polymath ‘09]: Elementary proof that gives a bound on m: It is enough to take m ≈ Ak(1/δ).

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**Lower bound for lines Back to ε-nets w.r.t. lines in the plane:**

Let δ = 1/2 (density). Given k (board side), let m be the dimension guaranteed by DHJ. So m ≈ Ak(constant) ≈ A(k). Project the board into R2 in “general position” (so that no point falls into a line it does not belong to). 3*3*3

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**Lower bound for lines Number of points: n = km ≈ kA(k) ≈ A(k)**

Choose ε so that εn = k, namely ε = k/n. Then, every ε-net w.r.t. lines N must have more than n/2 points, since otherwise, the missing points would be more than a δ fraction, so a whole line would be completely missed. |N| ≥ n/2 = k/(2ε) ≈ 1/ε α(1/ε) 1/ε = n/k ≈ A(k) / k ≈ A(k) k ≈ α(1/ε) |N| = Ω(1/ε α(1/ε)) QED

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