Lower bounds for epsilon-nets

Epsilon-nets
Let R be a family of ranges. Say, R = {all discs in R 2 }.
Let P be a finite point set.
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.
P

Epsilon-nets
Let R be a family of ranges. Say, R = {all discs in R 2 }.
Let P be a finite point set.
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.
P
N
ε|P| = 4

Epsilon-nets
Let R be a family of ranges. Say, R = {all discs in R 2 }.
Let P be a finite point set.
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.
P
N
ε|P| = 4
Want N as small as possible.
Examples of families of ranges:
R = {all rectangles in R 2 }
R = {all ellipsoids in R d }
R = {all halfspaces in R d }
[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 R 2 }
Indep. of |P|.
Infinite
VC-dim

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 log d–1 1/ε) for
weak ε-nets w.r.t. convex sets in R d .
• [Alon ‘10]: Ω(1/ε α(1/ε)) for ε-nets w.r.t. lines in R 2 . (*)
• [Pach, Tardos ‘11]:
• Ω(1/ε log log 1/ε) for ε-nets w.r.t. axis-parallel
rectangles in R 2 .
(Tight [Aronov, Ezra, Sharir ‘10].)
• Ω(1/ε log 1/ε) for dual ε-nets for axis parallel rectanlges in R 2 .
• Ω(1/ε log 1/ε) for ε-nets w.r.t. axis-parallel
boxes and w.r.t. halfspaces in R 4 .

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 R 2 .
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 R 2 .
Proof strategy: Given n, we will construct an n-point set
P such that every ε-net N for P has size > n/2.
???
It should say
“Given ε”
For ε suitably chosen in terms of n.
(Spoiler: ε = (log log n) / n.)
???
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 R 2 .
Proof strategy: Given n, we will construct an n-point set
P such that every ε-net N for P has size > n/2.
???
It should say
“Given ε”
For ε suitably chosen in terms of n.
(Spoiler: ε = (log log n) / n.)
???
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
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, …, t max
, where t max
= log
2
1/(2ε).
N

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, …, t max
, where t max
= log
2
1/(2ε).
N
Just before step t, the points lie on
2 t–1 vertical lines:
…
2 t–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”
We want a rectangle to be heavy, and to be missed by N :
2 –t
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 t max
).
Proof:
# black orphans ≤ 2 t–1 *εn ≤ 1/(4ε)*εn = n/4.
t max
= log
2
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:
At most 4εn points
Pr[ given good rectangle is heavy and missed by N ] ≥ 2 –4εn
# good rectangles in first t max stages ≥ 1/(12ε) t max
≈ 1/ε log 1/ε t max
= log
2
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

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 ≤ 2 n
Pr[ some N is a good ε-net ] ≤ 2 n – 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:
2 n – n / (log log n) * (log n) * 1/√(log n) = 2 n – 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)
Finally:
Theorem: For every ε there exists an n
0 s.t. for every n ≥ n
0 there exists an n-point set P that requires ε-nets w.r.t. axisparallel 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.

Lower bound for lines
Theorem [Alon ‘10]: For every ε there exists an n
0 there exists an n
0 and
-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 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 m tic-tac-toe board, you will contain a complete line of size k
(maybe diagonal).
[Polymath ‘09]: Elementary proof that gives a bound on m:
It is enough to take m ≈ A k
(1/δ).

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 ≈ A k
(constant) ≈ A(k).
• Project the board into R 2 in “general position” (so that no point falls into a line it does not belong to).
3*3*3

Lower bound for lines
Number of points: n = k m ≈ k A(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