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Constructive Algorithms for Discrepancy Minimization

Constructive Algorithms for Discrepancy Minimization. Nikhil Bansal (IBM). S 3. S 4. S 1. S 2. Combinatorial Discrepancy. Universe: U= [1,…,n] Subsets: S 1 ,S 2 ,…,S m Color elements red/blue so each set is colored as evenly as possible. Applications.

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Constructive Algorithms for Discrepancy Minimization

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  1. Constructive Algorithms for Discrepancy Minimization Nikhil Bansal (IBM)

  2. S3 S4 S1 S2 Combinatorial Discrepancy Universe: U= [1,…,n] Subsets: S1,S2,…,Sm Color elements red/blue so each set is colored as evenly as possible.

  3. 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, …

  4. General Set System Universe: U= [1,…,n] Subsets: S1,S2,…,Sm Find : [n] ! {-1,+1} to Minimize |(S)|1 = maxS | i 2 S(i) | For simplicity consider m=n henceforth.

  5. 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).

  6. 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.

  7. 1 2 … n 1’ 2’ … n’ S1 S2 … S’1 S’2 … Approximating Discrepancy Question: 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 Widely used: TU set systems, Geometry, …

  8. 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 find Discrepancy ·O(log (mn))Hereditary discrepancy. General Technique: Constructive bounds for geometric problems, Beck Fiala setting, k-permutation problem, …

  9. 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.

  10. 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’

  11. Talk Outline Introduction The Method Low Hereditary discrepancy -> Good coloring Additional Ideas Spencer’s O(n1/2) bound

  12. start 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)

  13. 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

  14. 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 Si = 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})

  15. +1 time -1 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 +  (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)

  16. 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 ½ Define 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.

  17. Analysis Consider 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.

  18. 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

  19. 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.

  20. 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)

  21. X1 = ( 1,-1, 1 , …,1,-1,-1) X2 = (-1,-1,-1, …,1, 1, 1) X = ( 1, 0, 1 , …,0,-1,-1) Proving Partial Coloring Lemma Beautiful 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]

  22. 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

  23. 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

  24. 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

  25. Concluding Remarks Probable 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, …

  26. Thank You!

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