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Constructive Algorithms for Discrepancy Minimization

Nikhil Bansal (IBM)

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Combinatorial Discrepancy

Universe: U= [1,…,n] Subsets: S1,S2,…,Sm

Color elements red/blue so each set is colored as evenly as possible.

S1

S2

S3

S4

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ApplicationsCS: 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, …

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General Set System

Universe: U= [1,…,n] Subsets: S1,S2,…,Sm

Find : [n] ! {-1,+1} toMinimize |(S)|1 = maxS | i 2 S (i) |

For simplicity consider m=n henceforth.

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

Union bound + Choose c ¼ (log n)1/2

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

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Better Colorings Exist![Spencer 85]: (Six standard deviations suffice) Always exists coloring with discrepancy · 6n1/2

(In general for arbitrary m, discrepancy = O(n1/2 log(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.

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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 discrepancyWidely used: TU set systems, Geometry, …

S1

S2

…

S’1

S’2

…

1 2 … n 1’ 2’ … n’

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Our Results

Thm 1: Can get Spencer’s bound constructively. That is, O(n1/2) discrepancy for m=n sets.

Thm 2: For any set system, can findDiscrepancy · O(log (mn)) Hereditary discrepancy.

General Technique: Constructive bounds for geometric problems, Beck Fiala setting, k-permutation problem, …

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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 (vector coloring) | 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 the major tool.

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Key Point

The SDP gap example does not work if discrepancy << n1/2

As we will see, SDPs are very useful in that regime.

But seems useless for Spencer’s problem

Idea: “Tighter” bounds for some sets. |i 2 S vi |2 · 2 n | i 2 S’ vi|2 · n/log n |vi|2 = 1

Why can one do this: Entropy Method.

Tighter bound for S’

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Talk Outline

Introduction

The Method Low Hereditary discrepancy -> Good coloring

Additional IdeasSpencer’s O(n1/2) bound

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Algorithm (at high level)

Cube: {-1,+1}n

Analysis: Few steps to reach a vertex (walk has high variance) Disc(Si) does a random walk (with low variance)

start

finish

Algorithm: “Sticky” random walk Each step generated by rounding a suitable SDP Move in various dimensions correlated, e.g. t

1 + t2 ¼ 0

Each dimension: An ElementEach vertex: A Coloring

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An SDP Hereditary disc. ) the following SDP is feasible

SDP: Low discrepancy: |i 2 Sj

vi |2 · 2

|vi|2 = 1

Rounding: Pick random Gaussian g = (g1,g2,…,gn) each coordinate gi is iid N(0,1)

For each i, consider i = g ¢ vi

Obtain vi 2 Rn

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Properties of RoundingLemma: 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)

1. Each i » N(0,)

2. For each set S, i 2 S i = g ¢ (i2 S vi) » N(0, · 2)

(std deviation ·)

SDP:|vi|2 = 1|i2 S

vi|2 ·2

Recall: i = g ¢ vi

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

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Algorithm Overview Construct coloring iteratively.Initially: Start with coloring x0 = (0,0,0, …,0) at t = 0.At Time t: Update coloring as xt = xt-1 + (t

1,…,tn)

( tiny: 1/n suffices)

x(i)

xt(i) = (1i + 2

i + … + ti)

Color of element i: Does random walk over time with step size ¼

Fixed if reaches -1 or +1.

time

-1

+1

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

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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: Walks for various elements are correlated

Consider basic walk x(t+1) = x(t) 1 with prob ½Define Energy (t) = x(t)2 if never reached n1/2 = (t-1) + 1 otherwise E[(t)] = ½ (x(t-1)+1)2 + ½ (x(t-1)-1)2 = x(t-1)2 + 1 = (t-1)+1

So, E[(10 n) ] = 10 n. Moreover, (10 n) · 11 n So only a small fraction of walks can have < n.

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AnalysisConsider time T = O(1/2)

Claim 2: Each set has O() discrepancy in expectation.Pf: For each S, xt(S) doing random walk with step size ¼

Define a round as T = O(1/2) time steps. Claim 1: ) Everything colored in O(log n) rounds.

Claim 2: ) Expected discrepancy of a set at end = O( log n)

+ By Chernoff Bounds, discrepancy = O( log mn) whp over all sets.

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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

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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.

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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)

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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 positionsConsider X = (X1 – X2)/2

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

X1 = ( 1,-1, 1 , …,1,-1,-1)X2 = (-1,-1,-1, …,1, 1, 1) X = ( 1, 0, 1 , …,0,-1,-1)

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A useful generalization

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

Provided S = ( n1/2) in some average sense

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An SDP Suppose there exists partial coloring X:1. On ¸ n/2 elements2. 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 vi 2 Rn

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Algorithm

Initially write SDP with S = c n1/2

Each set S does random walk and expects to reach discrepancy of O(S) = O(n1/2)

Some sets will become problematic. Reduce their S on the fly.Not many problematic sets, and entropy penalty low.

0 20n1/2 30n1/2 35n1/2 …

Danger 1 Danger 2 Danger 3 …

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Concluding RemarksProbable right answer: O( (log m)1/2) for her. Disc.

Can be derandomized [Bansal-Spencer] (add new constraints to SDP)

Gives derandomization for prob. inequalities stronger than Chernoff bounds.(Exponential moment technique for derandomizing Chernoff loses (log n)1/2)

Other non-constructive problems: Lattices (Minkowski Thm) Fixed-Point Based (Nash, Sperner’s Lemma) Topological (Hypergraph matching), …

Discrepancy problems: Beck Fiala Conjecture (more generally Komlos conjecture) Erdos Discrepancy problem 3-permutation conjecture, …

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Thank You!