- 145 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Discrepancy and SDPs' - gilead

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Discrepancy and SDPs

Nikhil Bansal (TU Eindhoven)

Outline

Discrepancy: definitions and applications

Basic results: upper/lower bounds

Partial Coloring method (non-constructive)

SDPs: basic method

Algorithmic Spencer’s Result

Lovett-Meka result

Lower bounds via SDP duality (Matousek)

Material

Classic: Geometric Discrepancy by J. Matousek

Papers:

Bansal. Constructive algorithms for discrepancy minimization, FOCS 2010

Matousek. The determinant lower bound is almost tight

Lovett, Meka. Discrepancy minimization by walking on the edges

Survey with fewer technical details:

Bansal. …

Discrepancy: What is it?

Study of gaps in approximating the continuous by the discrete.

Original motivation: Numerical Integration/ Sampling

Problem: How well can you approximate a region by discrete points

Discrepancy:

Max over intervals I

|(# points in I) – (length of I)|

Discrepancy: What is it?

Study of gaps in approximating the continuous by the discrete.

Problem: How uniformly can you distribute points in a grid.

“Uniform” : For every axis-parallel rectangle R

| (# points in R) - (Area of R) | should be low.

Discrepancy:

Max over rectangles R

|(# points in R) – (Area of R)|

n1/2

n1/2

Distributing points in a grid

Problem: How uniformly can you distribute points in a grid.

“Uniform” : For every axis-parallel rectangle R

| (# points in R) - (Area of R) | should be low.

n= 64

points

Van der Corput Set

Uniform

Random

n1/2 discrepancy

n1/2 (loglog n)1/2

O(log n) discrepancy!

Quasi-Monte Carlo Methods

With N random samples: Error \prop 1/\sqrt{n}

Quasi-Monte Carlo Methods: \prop Disc/n

Can discrepancy be O(1) for 2d grid?

No. \Omega(log n) [Schmidt …]

d-dimensions: O(log^{d-1} n) [Halton-Hammersely ]

\Omega(log^{(d-1)/2} n) [Roth ]

\Omega(log^{(d-1)/2 + \eta} n [Bilyk,Lacey,Vagharshakyan’08]

Discrepancy: Example 2

Input: n points placed arbitrarily in a grid.

Color them red/blue such that each rectangle is colored as evenly as possible

Discrepancy: max over rect. R ( | # red in R - # blue in R | )

Continuous: Color each element

1/2 red and 1/2 blue (0 discrepancy)

Discrete:

Random has about O(n1/2 log1/2 n)

Can achieve O(log2.5 n)

S3

S4

S1

S2

Combinatorial DiscrepancyUniverse: U= [1,…,n]

Subsets: S1,S2,…,Sm

Color elements red/blue so each

set is colored as evenly as possible.

Find : [n] ! {-1,+1} to

Minimize |(S)|1 = maxS | i 2 S(i) |

If A is m \times n incidence matrix.

Disc(A) = min_{x \in {-1,1}^n} |Ax|_\infty

Applications

CS: Computational Geometry, Comb. Optimization, Monte-Carlo simulation, Machine learning, Complexity, Pseudo-Randomness, …

Math: Dynamical Systems, Combinatorics, Mathematical Finance,

Number Theory, Ramsey Theory, Algebra, Measure Theory, …

Rounding

Lovasz-Spencer-Vesztermgombi’86

Given any matrix A, and x \in R^n

can round x to \tilde{x} \in Z^n s.t.

|Ax – A\tilde{x}|_\infty < Herdisc(A)

Proof: Round the bits one by one.

Nothing known until recently.

Thm [B’10]. Can efficiently round so that

Error \leq O(\sqrt{log m log n}) Herdisc(A)

Dynamic Data Structures

N points in a 2-d region.

Weights update over time.

Query: Given an axis-parallel rectangle R, determine the total weight on points in R.

Preprocess:

- Low query time
- Low update time (upon weight change)

Example

Line:

Query = O(n) Update = 1

Query = 1 Update = O(n^2)

Query = 2 Update = O(n)

Query = O(log n) Update = O(log n)

Recursively can get for 2-d.

What about other objects?

Query

Circles arbitrary rectangles aligned triangle

Turns out t_q t_u \geq n^{1/2}/log^2 n ?

Larsen-Green: t_q t_u \geq disc(S)^n/log^2 n

Sketch of idea

A good data structure implies

D = A P

A = row sparse P = Column sparse

(low query time) (low update time)

Best Known Algorithm

Random: Color each element i independently as

x(i) = +1 or -1 with probability ½ each.

Thm: Discrepancy = O (n log n)1/2

Pf: For each set, expect O(n1/2) discrepancy

Standard tail bounds: Pr[ | i 2 S x(i) | ¸c n1/2 ] ¼e-c2

Union bound + Choose c ¼ (log n)1/2

Analysis tight: Random actually incurs ((n log n)1/2).

Better Colorings Exist!

[Spencer 85]: (Six standard deviations suffice)

Always exists coloring with discrepancy ·6n1/2

(In general for arbitrary m, discrepancy = O(n1/2log(m/n)1/2)

Tight: For m=n, cannot beat 0.5 n1/2 (Hadamard Matrix, “orthogonal” sets)

Inherently non-constructive proof

(pigeonhole principle on exponentially large universe)

Challenge: Can we find it algorithmically ?

Certain algorithms do not work [Spencer]

Conjecture[Alon-Spencer]: May not be possible.

S3

S4

S1

S2

Beck Fiala ThmU = [1,…,n] Sets: S1,S2,…,Sm

Suppose each element lies in at most t sets (t << n).

[Beck Fiala’ 81]: Discrepancy 2t -1.

(elegant linear algebraic argument, algorithmic result)

Beck Fiala Conjecture: O(t1/2) discrepancy possible

Other results: O( t1/2 log t log n ) [Beck]

O( t1/2 log n ) [Srinivasan]

O( t1/2 log1/2 n ) [Banaszczyk]

Non-constructive

1’ 2’ … n’

S1

S2

…

S’1

S’2

…

Approximating DiscrepancyQuestion: If a set system has low discrepancy (say << n1/2)

Can we find a good discrepancy coloring ?

[Charikar, Newman, Nikolov 11]:

Even 0 vs. O (n1/2) is NP-Hard

(Matousek): What if system has low Hereditary discrepancy?

herdisc (U,S) = maxU’ ½ U disc (U’, S|U’)

Robust measure of discrepancy (often same as discrepancy)

Widely used: TU set systems, Geomety, …

Our Results

Thm 1: Can get Spencer’s bound constructively.

That is, O(n1/2) discrepancy for m=n sets.

Thm 2: If each element lies in at most t sets, get bound of O(t1/2 log n) constructively (Srinivasan’s bound)

Thm 3: For any set system, can find

Discrepancy ·O(log (mn))Hereditary discrepancy.

Other Problems: Constructive bounds (matching current best)

k-permutation problem [Spencer, Srinivasan,Tetali]

Geometric problems , …

Relaxations: LPs and SDPs

Not clear how to use.

Linear Program is useless. Can color each element ½ red and ½ blue. Discrepancy of each set = 0!

SDPs(LP on vi¢ vj, cannot control dimension of v’s)

| i 2 S vi |2· n 8 S

|vi|2 = 1

Intended solution vi = (+1,0,…,0) or (-1,0,…,0).

Trivially feasible: vi = ei (all vi’s orthogonal)

Yet, SDPs will be a major tool.

Punch line

SDP very helpful if “tighter” bounds needed for some sets.

|i 2 S vi |2· 2 n

| i 2 S’ vi|2· n/log n

|vi|2· 1

Not apriori clear why one can do this.

Entropy Method.

Algorithm will construct coloring over time and

use several SDPs in the process.

Tighter bound for S’

Talk Outline

Introduction

The Method

Low Hereditary discrepancy -> Good coloring

Additional Ideas

Spencer’s O(n1/2) bound

Slight improvement

Can be improved to O(\sqrt{n})/2^n

If you pick a random {-1,1} coloring s

w.p. say >= ½ |a \cdot s| \leq c \sqrt{n}

2^{n-1} colorings s, with |a\cdot s| \leq c \sqrt{n}

Algorithmically

Easy: 1/poly(n) (How?)

Answer: Pick any poly(n) colorings.

[Karmarkar-Karp’81]: \approx 1/n^log n

Huge gap: Major open question

Remark: {-1,+1} not enough. Really need color 0 also.

E.g. a_1 = 1, a_2=…=a_n = 1/(2n)

Yet another enhancement

There is a {-1,0,1} coloring with at least

n/2 {-1,1}’s s.t. \sum_i a_i s_i \leq n/2^{n/5}

Make buckets of size 2n/2^{n/5}

At least 2^{4n/5} sums fall in same bucket

Claim: Some two s’ and s’’ in same bucket and differ in at least n/2 coordinates

Again consider s = (s’-s’’)/2

Proof of Claim

Claim: Any set of 2^{4n/5} vertices of the

boolean cube has

[Kleitman’66] Isoperimetry for cube.

Hamming ball B(v,r) has the smallest diameter for a given number of vertices.

|B(v,n/4)| < 2^{4n/5}

finish

Algorithm (at high level)Each dimension: An Element

Each vertex: A Coloring

Cube: {-1,+1}n

Algorithm: “Sticky” random walk

Each step generated by rounding a suitable SDP Move in various dimensions correlated, e.g. t1 + t2¼ 0

Analysis: Few steps to reach a vertex (walk has high variance)

Disc(Si) does a random walk (with low variance)

An SDP

Hereditary disc. ) the following SDP is feasible

SDP:

Low discrepancy: |i 2 Sj vi |2 ·2

|vi|2 = 1

Obtain vi2 Rn

Rounding:

Pick random Gaussian g = (g1,g2,…,gn)

each coordinate gi is iid N(0,1)

For each i, consider i = g¢ vi

Properties of Rounding

Lemma: If g 2 Rn is random Gaussian. For any v 2 Rn,

g ¢ v is distributed as N(0, |v|2)

Pf: N(0,a2) + N(0,b2) = N(0,a2+b2) g¢ v = i v(i) gi» N(0, i v(i)2)

Recall: i = g ¢ vi

- Each i» N(0,1)
- For each set S,
- i 2 Si = g ¢ (i2 S vi) » N(0, ·2)
- (std deviation ·)

SDP:

|vi|2 = 1

|i2S vi|2·2

’s mimics a low discrepancy coloring (but is not {-1,+1})

time

-1

Algorithm OverviewConstruct coloring iteratively.

Initially: Start with coloring x0 = (0,0,0, …,0) at t = 0.

At Time t: Update coloring as xt = xt-1 + (t1,…,tn)

( tiny: 1/n suffices)

xt(i) = (1i + 2i + … + ti)

Color of element i: Does random walk

over time with step size ¼ N(0,1)

x(i)

Fixed if reaches -1 or +1.

Set S: xt(S) = i 2 S xt(i) does a random walk w/ step N(0,·2)

Analysis

Consider time T = O(1/2)

Claim 1: With prob. ½, at least n/2 elements reach -1 or +1.

Pf: Each element doing random walk with size ¼.

Recall: Random walk with step 1, is ¼ O(t1/2) away in t steps.

A Trouble: Various element updates are correlated

Consider basic walk x(t+1) = x(t) 1 with prob ½

Define Energy (t) = x(t)2

E[(t+1)] = ½ (x(t)+1)2 + ½ (x(t)-1)2 = x(t)2 + 1 = (t)+1

Expected energy = n at t= n.

Claim 2: Each set has O() discrepancy in expectation.

Pf: For each S, xt(S) doing random walk with step size ¼

Analysis

Consider time T = O(1/2)

Claim 1: With prob. ½, at least n/2 variables reach -1 or +1.

) Everything colored in O(log n) rounds.

Claim 2: Each set has O() discrepancy in expectation per round.

) Expected discrepancy of a set at end = O( log n)

Thm: Obtain a coloring with discrepancy O( log (mn))

Pf: By Chernoff, Prob. that disc(S) >= 2 Expectation + O( log m)

= O( log (mn))

is tiny (poly(1/m)).

Recap

At each step of walk, formulate SDP on unfixed variables.

Use some (existential) property to argue SDP is feasible

Rounding SDP solution -> Step of walk

Properties of walk:

High Variance -> Quick convergence

Low variance for discrepancy on sets -> Low discrepancy

Refinements

Spencer’s six std deviations result:

Goal: Obtain O(n1/2) discrepancy for any set system on m = O(n) sets.

Random coloring has n1/2(log n)1/2 discrepancy

Previous approach seems useless:

Expected discrepancy for a set O(n1/2),

but some random walks will deviate by up to (log n)1/2 factor

Need an additional idea to prevent this.

Spencer’s O(n1/2) result

Partial Coloring Lemma: For any system with m sets, there exists a coloring on ¸ n/2 elements with discrepancy O(n1/2 log1/2 (2m/n))

[For m=n, disc = O(n1/2)]

Algorithm for total coloring:

Repeatedly apply partial coloring lemma

Total discrepancy

O( n1/2 log1/2 2 ) [Phase 1]

+ O( (n/2)1/2 log1/2 4 ) [Phase 2]

+ O((n/4)1/2 log1/2 8 ) [Phase 3]

+ … = O(n1/2)

X1 = ( 1,-1, 1 , …,1,-1,-1)

X2 = (-1,-1,-1, …,1, 1, 1)

X = ( 1, 0, 1 , …,0,-1,-1)

Proving Partial Coloring LemmaBeautiful Counting argument (entropy method + pigeonhole)

Idea: Too many colorings (2n), but few “discrepancy profiles”

Key Lemma: There exist k=24n/5 colorings X1,…,Xk such that

every two Xi, Xj are “similar” for every set S1,…,Sn.

Some X1,X2 differ on ¸ n/2 positions

Consider X = (X1 – X2)/2

Pf: X(S) = (X1(S) – X2(S))/2 2 [-10 n1/2 , 10 n1/2]

A useful generalization

There exists a partial coloring with non-uniform discrepancy bound S for set S

Even if S = ( n1/2) in some average sense

An SDP

Suppose there exists partial coloring X:

1. On ¸ n/2 elements

2. Each set S has |X(S)| ·S

SDP:

Low discrepancy: |i 2 Sj vi |2·S2

Many colors:i |vi|2¸ n/2

|vi|2· 1

Pick random Gaussian g = (g1,g2,…,gn)

each coordinate gi is iid N(0,1)

For each i, consider i = g ¢ vi

Obtain vi2 Rn

Algorithm

Initially write SDP with S = c n1/2

Each set S does random walk and expects to reach

discrepancy of O(DS) = O(n1/2)

Some sets will become problematic.

Reduce their S on the fly.

Not many problematic sets, and entropy penalty low.

Danger 3 …

Danger 1

Danger 2

…

35n1/2

0

30n1/2

20n1/2

Concluding Remarks

Construct coloring over time by solving sequence of SDPs (guided by existence results)

Works quite generally

Can be derandomized[Bansal-Spencer]

(use entropy method itself for derandomizing + usual tech.)

E.g. Deterministic six standard deviations can be viewed as a way to derandomize something stronger than Chernoff bounds.

Rest of the talk

- How to generate i with required properties.
- How to update S over time.
Show n1/2 (log log log n)1/2 bound.

Why so few algorithms?

- Often algorithms rely on continuous relaxations.
- Linear Program is useless. Can color each element ½ red and ½ blue.

- Improved results of Spencer, Beck, Srinivasan, … based on clever counting (entropy method).
- Pigeonhole Principle on exponentially large systems (seems inherently non-constructive)

Partial Coloring Lemma

Suppose we have discrepancy bound S for set S.

Consider 2n possible colorings

Signature of a coloring X: (b(S1), b(S2),…, b(Sm))

Want partial coloring with signature (0,0,0,…,0)

Progress Condition

Energy increases at each step:

E(t) = \sum_i x_i(t)^2

Initially energy =0, can be at most n.

Expected value of E(t) = E(t-1) + \sum_i \gamma_i(t)^2

Markov’s inequality.

Missing Steps

- How to generate the \eta_i
- How to update \Delta_S over time

Partial Coloring

X1 = (1,-1, 1 , …, 1,-1,-1)

X2 = (-1,-1,-1, …, 1,1, 1)

If exist two colorings X1,X2

1. Same signature (b1,b2,…,bm)

2. Differ in at least n/2 positions.

Consider X = (X1 –X2)/2

- -1 or 1 on at least n/2 positions, i.e. partial coloring
- Has signature (0,0,0,…,0)
X(S) = (X1(S) – X2(S)) / 2, so |X(S)| ·S for all S.

Can show that there are 24n/5 colorings with same signature.

So, some two will differ on > n/2 positions. (Pigeon Hole)

Spencer’s O(n1/2) result

Partial Coloring Lemma: For any system with m sets,

there exists a coloring on ¸ n/2 elements with discrepancy O(n1/2 log1/2 (2m/n))

[For m=n, disc = O(n1/2)]

Algorithm for total coloring:

Repeatedly apply partial coloring lemma

Total discrepancy

O( n1/2 log1/2 2 ) [Phase 1]

+ O( (n/2)1/2 log1/2 4 ) [Phase 2]

+ O((n/4)1/2 log1/2 8 ) [Phase 3]

+ … = O(n1/2)

Let us prove the lemma for m = n

Ent(b1) · 1/5

Proving Partial Coloring Lemma-30 n1/2

-10 n1/2

10 n1/2

30 n1/2

-2

-1

0

1

2

Pf: Associate with coloring X, signature = (b1,b2,…,bn)

(bi = bucket in which X(Si) lies )

Wish to show: There exist 24n/5 colorings with same signature

Choose X randomly: Induces distribution on signatures.

Entropy () · n/5 implies some signature has prob. ¸ 2-n/5.

Entropy ( ) ·i Entropy( bi) [Subadditivity of Entropy]

bi = 0 w.p. ¼ 1- 2 e-50,

= 1 w.p. ¼ e-50

= 2 w.p. ¼ e-450

….

For each set S, consider the “bucketing”

-2

-1

2

0

1

S

-3S

-S

3S

5S

Bucket of n1/2/100

has penalty ¼ ln(100)

A useful generalizationPartial coloring with non-uniform discrepancy S for set S

Suffices to have s Ent (bs) · n/5

Or, if S = s n1/2 , then s g(s) · n/5

g() ¼ e-2/2 > 1

¼ln(1/) < 1

Recap

Partial Coloring:S¼ 10 n1/2 gives low entropy

) 24n/5 colorings exist with same signature.

) some X1,X2 with large hamming distance.

(X1 – X2) /2 gives the desired partial coloring.

Trouble: 24n/5/2n is an exponentially small fraction.

Only if we could find the partial coloring efficiently…

Download Presentation

Connecting to Server..