Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "Lower bounds for epsilon-nets Gabriel Nivasch Presenting work by: Noga Alon, János Pach, Gábor Tardos."— Presentation transcript:

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

2 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. 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.

3 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. 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

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. 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 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 } Infinite VC-dim N Indep. of |P|.

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

6 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.)

7 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.) ??? It should say “Given ε” ??? Normally |N| does not depend on |P|

8 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.) ??? It should say “Given ε” ??? Normally |N| does not depend on |P| We’ll fix both problems later on.

9 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.

10 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 ? …

11 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, …, t max, where t max = log 2 1/(2ε).

12 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

13 Lower bound for axis-parallel rectangles For each vertical line, partition the points into intervals, where each interval contains exactly εn black points. “orphans”

14 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–t2–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.

15 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).

16 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)

17 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 t max stages ≥ 1/(12ε) t max ≈ 1/ε log 1/ε t max = log 2 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

18 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

19 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 n 0 s.t. for every n ≥ n 0 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.

20 Lower bound for lines Theorem [Alon ‘10]: For every ε there exists an n 0 and there exists an n 0 -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:

21 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 ≈ A k (1/δ).

22 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

23 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) QED 1/ε = n/k ≈ A(k) / k ≈ A(k)  k ≈ α(1/ε) |N| = Ω(1/ε α(1/ε))


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

Similar presentations


Ads by Google