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Lower bounds for epsilon-nets Gabriel Nivasch Presenting work by: Noga Alon, János Pach, Gábor Tardos

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Lower bounds for epsilon-nets

Gabriel NivaschPresenting work by: Noga Alon,

János Pach, Gábor Tardos

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

Let P be a finite point set.

Let 0 < ε < 1 b a parameter.P

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.

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

Let P be a finite point set.

Let 0 < ε < 1 b a parameter.P

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.

N

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

Let P be a finite point set.

Let 0 < ε < 1 b a parameter.P

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.

Examples of families of ranges:R = {all rectangles in R2}R = {all ellipsoids in Rd}R = {all halfspaces in Rd}

[Haussler, Welzl, ‘87]: If R has finite VC-dimension then there exists N of size O(1/ε log 1/ε).

R = {all convex sets in R2} Infinite VC-dim

N

Indep. of |P|.

Epsilon-netsGeneral 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.

(Tight [Aronov, Ezra, Sharir ‘10].) We

will

show

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.)

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.)???

It should say “Given ε”

???Normally |N|

does not depend on |P|

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.)???

It should say “Given ε”

???Normally |N|

does not depend on |P|

We’ll fix both problems later on.

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.

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? …

Lower bound for axis-parallel rectangles

N

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).

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 Just before step t, the points lie on 2t–1 vertical lines:

2t–1

Lower bound for axis-parallel rectangles

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

“orphans”

Lower bound for axis-parallel rectangles

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

“orphans”

2–t

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.

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).

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)

Lower bound for axis-parallel rectangles

Focus on good rectangles – rectangles of good intervals:

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)

At most 4εn points

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

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

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)

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).

Finally:

Proof: To get larger sets P, replace every point by a tiny cloud of points.

Lower bound for linesTheorem [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:

Density Hales-Jewett theoremDensity 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).

Lower bound for linesBack 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

Lower bound for linesNumber 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)

QED

1/ε = n/k ≈ A(k) / k ≈ A(k) k ≈ α(1/ε)

|N| = Ω(1/ε α(1/ε))