Introduction to Discrepancy Theory

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David Oswin Wheeler
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1 Introduction to Discrepancy Theory Aleksandar (Sasho) Nikolov University of Toronto Sasho Nikolov (U of T) Discrepancy 1 / 23

2 Outline 1 Monte Carlo 2 Discrepancy and Quasi MC 3 Combinatorial Discrepancy Sasho Nikolov (U of T) Discrepancy 2 / 23

3 Discrepancy Theory How well can a discrete object approximation a continuous object? How well can a small object approximation a big object? Sasho Nikolov (U of T) Discrepancy 3 / 23

4 How to Compute an Integral? A fundamental problem in sciences: How to approximate the integral of a function? 1 0 f (x)dx =? f(x) 1 0 f(x)dx 0 1 Sasho Nikolov (U of T) Discrepancy 4 / 23

5 The Issues Didn t we learn this in calculus? xe x dx = [xe x ] 1 0 e x dx = [(x 1)e x ] Sasho Nikolov (U of T) Discrepancy 5 / 23

6 The Issues Didn t we learn this in calculus? xe x dx = [xe x ] 1 0 e x dx = [(x 1)e x ] Integration is hard! Many interesting functions do not have a closed form integral at all. The function f may be very complicated! Sasho Nikolov (U of T) Discrepancy 5 / 23

7 The Issues Continue Or we may not even know what f really is! f (x) may be: The speed of a particle at time x The price of a stock at time x The amount of energy released when we burn x grams of some chemical substance Sasho Nikolov (U of T) Discrepancy 6 / 23

8 The Issues Continue Or we may not even know what f really is! f (x) may be: The speed of a particle at time x The price of a stock at time x The amount of energy released when we burn x grams of some chemical substance Sometimes all we can do is treat f as a black box: we can query f (x) at a specific x each query may require a new experiment: expensive! Sasho Nikolov (U of T) Discrepancy 6 / 23

9 The Issues Continue Or we may not even know what f really is! f (x) may be: The speed of a particle at time x The price of a stock at time x The amount of energy released when we burn x grams of some chemical substance Sometimes all we can do is treat f as a black box: we can query f (x) at a specific x each query may require a new experiment: expensive! Can we compute 1 0 f (x)dx with a black box f, under minimal assumptions, with few queries? Sasho Nikolov (U of T) Discrepancy 6 / 23

10 The Monte Carlo Idea During the Manhattan Project, von Neumann and Ulam had an idea (inspired by Ulam s uncle s gambling habbit): 1 Pick n random points in [0, 1] 2 Estimate integral by average 0 1 Sasho Nikolov (U of T) Discrepancy 7 / 23

11 The Monte Carlo Idea During the Manhattan Project, von Neumann and Ulam had an idea (inspired by Ulam s uncle s gambling habbit): 1 Pick n random points in [0, 1] 2 Estimate integral by average f(x 1 )f(x 2 ) f(x3 ) f(x 12 ) 0 x 1 x 2 x 3 x 12 1 Sasho Nikolov (U of T) Discrepancy 7 / 23

12 The Monte Carlo Idea During the Manhattan Project, von Neumann and Ulam had an idea (inspired by Ulam s uncle s gambling habbit): 1 Pick n random points in [0, 1] 2 Estimate integral by average f(x 1 )f(x 2 ) f(x3 ) f(x 12 ) 1 f(x)dx i=1 f(x i) 0 x 1 x 2 x 3 x 12 1 Sasho Nikolov (U of T) Discrepancy 7 / 23

13 Monte Carlo: Convergence If we pick n random points x 1,..., x n [0, 1] then 1 f (x)dx 1 n f (x i ) n E(f ), n 0 i=1 where E(f ) is a measure of the energy of f. ( 1 E(f ) = 0 ) 1/2 f (x) 2 dx Sasho Nikolov (U of T) Discrepancy 8 / 23

14 Can we do better? We have a sequence x = (x 1, x 2, x 3,...) that we want to use to estimate integrals, using an average. Sasho Nikolov (U of T) Discrepancy 9 / 23

15 Can we do better? We have a sequence x = (x 1, x 2, x 3,...) that we want to use to estimate integrals, using an average. When we want a more accurate estimate, we evaluate f at a few more points from the sequence. Sasho Nikolov (U of T) Discrepancy 9 / 23

16 Can we do better? We have a sequence x = (x 1, x 2, x 3,...) that we want to use to estimate integrals, using an average. When we want a more accurate estimate, we evaluate f at a few more points from the sequence. We want the error Err(f, x, n) := to be as small as possible. 1 0 f (x)dx 1 n n f (x i ) We know that if x is random, Err(f, x, n) 1/ n? for f with constant energy. i=1 Sasho Nikolov (U of T) Discrepancy 9 / 23

17 Can we do better? We have a sequence x = (x 1, x 2, x 3,...) that we want to use to estimate integrals, using an average. When we want a more accurate estimate, we evaluate f at a few more points from the sequence. We want the error Err(f, x, n) := to be as small as possible. 1 0 f (x)dx 1 n n f (x i ) We know that if x is random, Err(f, x, n) 1/ n? for f with constant energy. i=1 Can we achieve Err(f, x, n) 1/n for all nice f? Sasho Nikolov (U of T) Discrepancy 9 / 23

18 Outline 1 Monte Carlo 2 Discrepancy and Quasi MC 3 Combinatorial Discrepancy Sasho Nikolov (U of T) Discrepancy 10 / 23

19 Intervals Are Enough If a sequence x has small error for all intervals, then it has small error for all smooth functions. δ( x, n) = max a b 1 n {i : a x i b}. a,b [0,1] Sasho Nikolov (U of T) Discrepancy 11 / 23

20 Intervals Are Enough If a sequence x has small error for all intervals, then it has small error for all smooth functions. δ( x, n) = max a b 1 n {i : a x i b}. a,b [0,1] Koksma-Hlawka inequality: Err(f, x, n) V (f )δ( x, n), where V (f ) is a measure of the smoothness of f (total variation). Sasho Nikolov (U of T) Discrepancy 11 / 23

21 Intervals Are Enough If a sequence x has small error for all intervals, then it has small error for all smooth functions. δ( x, n) = max a b 1 n {i : a x i b}. a,b [0,1] Koksma-Hlawka inequality: Err(f, x, n) V (f )δ( x, n), where V (f ) is a measure of the smoothness of f (total variation). van der Corput (1934): Can δ( x, n) = O(1/n) for some sequence x? Sasho Nikolov (U of T) Discrepancy 11 / 23

22 From Intervals to Rectangles Roth showed that studying δ( x, n) is equivalent to placing n points uniformly in a unit square. (Think of one dimension as the index.) n = 20 Sasho Nikolov (U of T) Discrepancy 12 / 23

23 From Intervals to Rectangles Roth showed that studying δ( x, n) is equivalent to placing n points uniformly in a unit square. (Think of one dimension as the index.) n = R 1 2 area(r) - 1 n #{points in R} = = 1 10 Sasho Nikolov (U of T) Discrepancy 12 / 23

24 Discrepancy of Rectangles For a set P of n points in [0, 1] 2 and a rectangle R = [a, b] [c, d] P R d(p, R) = area(r) n = (b a)(d c) d(p) = max D(P, R) R P R n Sasho Nikolov (U of T) Discrepancy 13 / 23

25 Discrepancy of Rectangles For a set P of n points in [0, 1] 2 and a rectangle R = [a, b] [c, d] P R d(p, R) = area(r) n = (b a)(d c) d(p) = max D(P, R) R P R n We can construct a sequence x with δ( x, n) = O(f (n)). For any n, we can construct a set P of n points s.t. d(p) = O(f (n)). Sasho Nikolov (U of T) Discrepancy 13 / 23

26 Grid Can we construct P s.t. d(p) = O(1/n)? Sasho Nikolov (U of T) Discrepancy 14 / 23

27 Grid Can we construct P s.t. d(p) = O(1/n)? Grid: d(p) 1 n : area(r) 0 and P R = n Sasho Nikolov (U of T) Discrepancy 14 / 23

28 Irrational Lattice P = { (i/n, {i 2}) : i = 0,..., n 1 } : d(p) = Θ( log n n ) {x} = fracional part of x = x x Sasho Nikolov (U of T) Discrepancy 15 / 23

29 van der Corput set P = {(i/n, rev(i)) : i = 0,..., n 1}: D(P) = Θ( log n n ) rev(b k b k 1... b 1 b 0 ) = 0.b 1 b 2... b k Sasho Nikolov (U of T) Discrepancy 16 / 23

30 Roth s Lower Bound, and Questions D(P) = O( log n n ) is possible we can estimate integrals with error O But what about d(p) = O(1/n)? ( ) log n n Sasho Nikolov (U of T) Discrepancy 17 / 23

31 Roth s Lower Bound, and Questions D(P) = O( log n n ) is possible we can estimate integrals with error O But what about d(p) = O(1/n)? Theorem (Roth, 1954; Schmidt 1972) For any n-point set P, D(P) = Ω( log n n ). ( ) log n n Sasho Nikolov (U of T) Discrepancy 17 / 23

32 Roth s Lower Bound, and Questions D(P) = O( log n n ) is possible we can estimate integrals with error O But what about d(p) = O(1/n)? Theorem (Roth, 1954; Schmidt 1972) For any n-point set P, D(P) = Ω( log n n ). ( ) log n n What about boxes in dimension 3? In dimension k? for η k 0 as k. (log n) (k 1)/2+η k n d(p) (log n)k 1 n Sasho Nikolov (U of T) Discrepancy 17 / 23

33 Outline 1 Monte Carlo 2 Discrepancy and Quasi MC 3 Combinatorial Discrepancy Sasho Nikolov (U of T) Discrepancy 18 / 23

34 Tusnády s Problem Given: Set Q of n points in the unit square Goal: Color each point p Q red or blue so that each rectangle R is as balanced as possible. Sasho Nikolov (U of T) Discrepancy 19 / 23

35 Tusnády s Problem Given: Set Q of n points in the unit square Goal: Color each point p Q red or blue so that each rectangle R is as balanced as possible. Sasho Nikolov (U of T) Discrepancy 19 / 23

36 Tusnády s Problem Given: Set Q of n points in the unit square Goal: Color each point p Q red or blue so that each rectangle R is as balanced as possible. disc(q) := min χ where χ: P { 1, 1} is a coloring. max R p R P χ(p), Sasho Nikolov (U of T) Discrepancy 19 / 23

37 From Combinatorial to Geometric Discrepancy For any n there exists an n-point set P s.t. d(p) 1 n max disc(q), Q where Q ranges over n-point sets in the unit square. Sasho Nikolov (U of T) Discrepancy 20 / 23

38 From Combinatorial to Geometric Discrepancy For any n there exists an n-point set P s.t. d(p) 1 n max disc(q), Q where Q ranges over n-point sets in the unit square. Sasho Nikolov (U of T) Discrepancy 20 / 23

39 From Combinatorial to Geometric Discrepancy For any n there exists an n-point set P s.t. d(p) 1 n max disc(q), Q where Q ranges over n-point sets in the unit square. Sasho Nikolov (U of T) Discrepancy 20 / 23

40 From Combinatorial to Geometric Discrepancy For any n there exists an n-point set P s.t. d(p) 1 n max disc(q), Q where Q ranges over n-point sets in the unit square. Sasho Nikolov (U of T) Discrepancy 20 / 23

41 From Combinatorial to Geometric Discrepancy For any n there exists an n-point set P s.t. d(p) 1 n max disc(q), Q where Q ranges over n-point sets in the unit square. Sasho Nikolov (U of T) Discrepancy 20 / 23

42 Bounds for Tusnády Theorem (Nikolov, Matoušek, Talwar, 2014) log(n) d 1 max disc(q) log(n)d+1/2 Q The proof uses (the analysis of) an algorithm to estimate discrepancy. Sasho Nikolov (U of T) Discrepancy 21 / 23

43 Computational Questions How can we efficiently (i.e. fast) find balanced colorings? Can we compute disc(q)? This kind of balanced colorings problem has many other applications: computational geometry data structures approximation algorithms private data analysis. Sasho Nikolov (U of T) Discrepancy 22 / 23

44 Sasho Nikolov (U of T) Discrepancy 23 / 23

The Discrepancy Function and the Small Ball Inequality in Higher Dimensions

The Discrepancy Function and the Small Ball Inequality in Higher Dimensions Dmitriy Georgia Institute of Technology (joint work with M. Lacey and A. Vagharshakyan) 2007 Fall AMS Western Section Meeting