Lower bounds for epsilonnets Gabriel Nivasch Presenting work by: Noga Alon, János Pach, Gábor...

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Transcript of Lower bounds for epsilonnets Gabriel Nivasch Presenting work by: Noga Alon, János Pach, Gábor...
Lower bounds for epsilonnets
Gabriel NivaschPresenting work by: Noga Alon,
János Pach, Gábor Tardos
EpsilonnetsLet 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 εnet w.r.t. R if, for every r in R that contains at least εP points of P, r contains a point of N.
Want N as small as possible.
EpsilonnetsLet 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 εnet w.r.t. R if, for every r in R that contains at least εP points of P, r contains a point of N.
εP = 4
Want N as small as possible.
N
EpsilonnetsLet 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 εnet w.r.t. R if, for every r in R that contains at least εP points of P, r contains a point of N.
εP = 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 VCdimension then there exists N of size O(1/ε log 1/ε).
R = {all convex sets in R2} Infinite VCdim
N
Indep. of P.
EpsilonnetsGeneral belief until recently: In geometric settings, there always exist εnets of size O(1/ε).But:• [Bukh, Matoušek, N. ‘09]: Ω(1/ε logd–1 1/ε) for weak εnets w.r.t. convex sets in Rd.
• [Alon ‘10]: Ω(1/ε α(1/ε)) for εnets w.r.t. lines in R2. (*)
• [Pach, Tardos ‘11]:• Ω(1/ε log log 1/ε) for εnets w.r.t. axisparallel
rectangles in R2.• Ω(1/ε log 1/ε) for dual εnets for axis parallel
rectanlges in R2.• Ω(1/ε log 1/ε) for εnets w.r.t. axisparallel boxes
and w.r.t. halfspaces in R4.
(Tight [Aronov, Ezra, Sharir ‘10].) We
will
show
Lower bound for axisparallel rectangles
[Pach, Tardos ‘11]: Ω(1/ε log log 1/ε) for εnets w.r.t. axisparallel rectangles in R2.
Proof strategy: Given n, we will construct an npoint set P such that every εnet N for P has size > n/2.
For ε suitably chosen in terms of n.(Spoiler: ε = (log log n) / n.)
Lower bound for axisparallel rectangles
[Pach, Tardos ‘11]: Ω(1/ε log log 1/ε) for εnets w.r.t. axisparallel rectangles in R2.
Proof strategy: Given n, we will construct an npoint set P such that every εnet 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 axisparallel rectangles
[Pach, Tardos ‘11]: Ω(1/ε log log 1/ε) for εnets w.r.t. axisparallel rectangles in R2.
Proof strategy: Given n, we will construct an npoint set P such that every εnet 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 axisparallel rectangles
Given n, we will construct an npoint set P such that every εnet 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 axisparallel rectangle.
Rectangle that contains εn points.
Lower bound for axisparallel rectangles
Choosing n random points in the unit square:• Step 0: Choose the ycoordinates, 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 axisparallel 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 axisparallel 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 axisparallel rectangles
For each vertical line, partition the points into intervals, where each interval contains exactly εn black points.
“orphans”
Lower bound for axisparallel 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 axisparallel 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 axisparallel 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 axisparallel 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 εnet ] ≤ (1 – 2–4εn )1/ε log 1/ε ≤ e–1/ε log 1/ε * 2^(–4εn)
1 – x ≤ e–x
Lower bound for axisparallel rectangles
Pr[ our N is a good εnet ] ≤ e–1/ε log 1/ε * 2^(–4εn)
# subsets N of size n/2 ≤ 2n
Pr[ some N is a good εnet ] ≤ 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 axisparallel rectangles
To summarize, an εnet 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 npoint set P that requires εnets w.r.t. axisparallel 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 n0point set P that requires εnets 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 HalesJewett theorem, which is about playing highdimensional tictactoe:
Density HalesJewett theoremDensity HalesJewett 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
tictactoe 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 εnets 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 εnet 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))