Low-discrepancy constructions in the triangle

Transcription

1 Low-discrepancy constructions in the triangle Department of Statistics Stanford University Joint work with Art B. Owen. ICERM 29th October 2014

2 Overview Introduction 1 Introduction The Problem Motivation 2 Discrepancy 3 4

3 Introduction The Problem Motivation Problem : Numerical Integration over triangular domain using quasi-monte Carlo (QMC) sampling. QMC in [0, 1] d. µ = f (x)dx ˆµ n = 1 [0,1] d n Koksma-Hlawka inequality n f (x i ) i=1 ˆµ n µ D n(x 1,..., x n ) V HK (f ) Recent work relating to general spaces by Aistleitner et al., (2012), Brandolini et. al (2013)

4 Motivation Introduction The Problem Motivation Need in computer graphics, genetic experimental studies, etc. Mapping by special functions/transformation from [0, 1] d Pillards and Cools (2005). Several notions of discrepancy on the triangle/simplex but no explicit constructions. Pillards and Cools (2005), Brandolini et. al (2013).

5 General Notions of Discrepancy Discrepancy The signed discrepancy of P at the measurable set S Ω R d is δ N (S; P, Ω) = vol(s Ω)/vol(Ω) A N (S; P)/N. The absolute discrepancy of points P for a class S of measurable subsets of Ω is where D N (S; P, Ω) = sup D N (S; P, Ω), S S D N (S; P, Ω) = δ N (S; P, Ω). Standard QMC works with Ω = [0, 1) d and takes for S the set of anchored boxes [0, a) with a [0, 1) d.

6 Discrepancy Discrepancy due to Brandolini et al. (2013) S C = {T a,b,c 0 < a < A C, 0 < b < B C } The parallelogram discrepancy of points P for Ω = (A, B, C) is Figure: The construction of the parallelogram T a,b,c = CDFE B D P N (P; Ω) = D N(S P ; P, Ω) for S P = S A S B S C. A F L b T a,b,c L a D a E b C

7 Discrepancy Discrepancy due to Pillards and Cools (2005) Ω = ((0, 0) T, (0, 1) T, (1, 1) T ) Their discrepancy where D PC N (P; Ω) = D N(S I, P, Ω) S I = {[0, a) a [0, 1) 2 }. Figure: Star Discrepancy on the Simplex (0, 1) (1, 1) ã a (0, 0)

8 Discrepancy Relationship between the discrepancies Lemma 1 Let T PC be the triangle from Pillards and Cools and for N 1, let P be the list of points x 1,..., x N T PC. Then Proof D PC N (P, T PC ) 2D P N (P, T PC ) [0, a 1 ) [0, a 2 ) = [0, a 1 ) [0, 1) [0, a 1 ) [a 2, 1) Taking C to be the vertex (0, 1) T of T PC, D PC N (P; T PC ) 2D N (S C, P, T PC ) 2D P N (P, T PC ).

9 Triangular van der Corput construction van der Corput sampling of [0, 1] the integer n = k 1 d kb k 1 in base b 2 is mapped to x n = k 1 d kb k. Points x 1,..., x n [0, 1) have a discrepancy of O(log(n)/n). Our situation : 4-ary expansion.

10 Triangular van der Corput construction n 0 in a base 4 representation n = k 1 d k4 k 1 where d k {0, 1, 2, 3} A B C Figure: A labeled subdivision of (A, B, C) into 4 and then 16 congruent subtriangles. Next are the first 32 triangular van der Corput points followed by the first 64. The integer labels come from the base 4 expansion.

11 Triangular van der Corput construction Computation : T = (A, B, C) ( B+C 2, A+C 2, A+B ) 2, d = 0 ( A, A+B 2 T (d) =, A+C ) 2, d = 1 ( ) B+A B+C 2, B, 2, d = 2 ( C+A 2, C+B 2, C ), d = 3. This construction defines an infinite sequence of f T (i) T for integers i 0. For an n point rule, take x i = f T (i 1) for i = 1,..., n.

12 Discrepancy Results Introduction Theorem 1 For an integer k 0 and non-degenerate triangle Ω = (A, B, C), let P consist of x i = f Ω (i 1) for i = 1,..., N = 4 k. Then DN P (P; Ω) = 7 9, N = N 1 9N, else.

13 Discrepancy Results Introduction Theorem 2 Let Ω be a nondegenerate triangle, and let P contain points x i = f Ω (s + i 1), i = 1,..., N = 4 k, for a starting integer s 1 and an integer k 0. Then DN P 2 (P; Ω) 1 N N.

14 Proof of Theorem 2 Introduction Proof δ N (S) = m j=0 δ N(S j ) where m is the number of subtriangles touching a boundary line of T a,b,c. 1/N δ N (S j ) 1/N. D N (S; P) m/n m 2 N 1 D N (S C ; P) (2 N 1)/N L a L b L b

15 Discrepancy Results Introduction Theorem 3 Let Ω be a non-degenerate triangle and, for integer N 1, let P = (x 1,..., x N ), where x i = f Ω (i 1). Then Proof: D P N (P; Ω) 12/ N. Let N = k j=0 a j4 j for some k, with a k 0. Let P l j denote a set of 4 j consecutive points from P, for l = 1,..., a j and j k. These P l j can be chosen to partition the N points x i. Fix any S S P.

16 Proof of Theorem 3 Introduction Now, δ N (S; P) = 1 N a k j 4 j δ(s; Pj l ). j=0 l=1 Therefore from Theorem 2, D N (S; P) = δ N (S; P) 1 N a k j j=0 l=1 4 j( 2 2 j 1 4 j ) 1 N k a j (2 j+1 1) j=0 3 ( 2(2 k+1 1) (k + 1) ) N 12 2k N and then k log 4 (N), gives D N (S; P) 12/ N. Taking the supremum over S S P yields the result.

17 We use Theorem 1 of Chen and Travaglini (2007) This construction yields parallel discrepancy of O(log N/N) Definition 1 A real number θ is said to be badly approximable if there exists a constant c > 0 such that n nθ > c for every natural number n N and denotes the distance from the nearest integer. Definition 2 Let a, b, c and d be integers with b 0, d 0 and c > 0, where c is not a perfect square. Then θ = (a + b c)/d is a quadratic irrational number.

18 Let Θ = {θ 1,..., θ k } be a set of k 1 angles in [0, 2π). Then let A(Θ) be the set of convex polygonal subsets of [0, 1] 2 whose sides make an angle of θ i with respect to the horizontal axis. Theorem 1 (Chen and Travaglini (2007)) There exists a constant C Θ < such that for any integer N > 1 there exists a list P = (x 1,..., x N ) of points in [0, 1] 2 with D N (A(Θ); P, [0, 1] 2 ) < C Θ log(n)/n.

19 Lemma 2 (Davenport) Suppose that the angles θ 1,..., θ k [0, 2π) are fixed. Then there exists α [0, 2π) such that tan(α),tan(α π/2), tan(α θ 1 ),... tan(α θ k ) are all finite and badly approximable.

20 Figure: Set of Angles for Kronecker (0, 1) R = ((0, 0) T, (0, 1) T, (1, 0) T ). Θ = {0, π/2, 3π/4} 3π 4 0 π 2 0 π 2 3π 4 (0, 0) (1, 0)

21 Lemma 3 Let α be an angle for which tan(α) is a quadratic irrational number. Then tan(α), tan(α π/2) and tan(α 3π/4) are all finite and badly approximable. tan(3π/8) = tan(5π/8) = 1 2.

22 Theorem 4 Let N > 1 be an integer and let R defined above be the triangle. Let α (0, 2π) be an angle for which tan(α) is a quadratic irrational. Let P 1 be the points of the lattice (2N) 1/2 Z 2 rotated anticlockwise by angle α. Let P 2 be the points of P 1 that lie in R. If P 2 has more than N points, let P 3 be any N points from P 2, or if P 2 has fewer than N points, let P 3 be a list of N points in R including all those of P 2. Then there is a constant C with D P (P 3 ; R) < C log(n)/n.

23 Triangular Lattice Points 3π 8 5π 8 π 4 π 2 Figure: Triangular lattice points for target N = 64. Domain is an equilateral triangle. Angles 3π/8 and 5π/8 have badly approximable tangents. Angles π/4 and π/2 have integer and infinite tangents respectively and do not satisfy the conditions for discrepancy O(log(N)/N).

24 Algorithm Introduction Given a target sample size N, an angle α such as 3π/8 satisfying Lemma 3. and a target triangle (A, B, C), Take integer grid Z 2

25 Algorithm Introduction Given a target sample size N, an angle α such as 3π/8 satisfying Lemma 3. and a target triangle (A, B, C), Take integer grid Z 2 Rotate anti clockwise by α

26 Algorithm Introduction Given a target sample size N, an angle α such as 3π/8 satisfying Lemma 3. and a target triangle (A, B, C), Take integer grid Z 2 Rotate anti clockwise by α Shrink by 2N

27 Algorithm Introduction Given a target sample size N, an angle α such as 3π/8 satisfying Lemma 3. and a target triangle (A, B, C), Take integer grid Z 2 Rotate anti clockwise by α Shrink by 2N Remove points not in the triangle.

28 Algorithm Introduction Given a target sample size N, an angle α such as 3π/8 satisfying Lemma 3. and a target triangle (A, B, C), Take integer grid Z 2 Rotate anti clockwise by α Shrink by 2N Remove points not in the triangle. (Optionally) add/subtract points to get exactly N points

29 Algorithm Given a target sample size N, an angle α such as 3π/8 satisfying Lemma 3. and a target triangle (A, B, C), Take integer grid Z 2 Rotate anti clockwise by α Shrink by 2N Remove points not in the triangle. (Optionally) add/subtract points to get exactly N points Linearly map R onto the desired triangle (A, B, C) Angle 3pi/8 Angle 5pi/8 Triangular

30 Introduction Parallel discrepancy of triangular lattice points for angle α = 3π/8 and various targets N. The number of points was always N or N + 1. The dashed reference line is 1/N. The solid line is log(n)/n. Discrepancy Grid rotated by 3pi/8 radians Sample size N

31 Introduction The Kronecker construction attains a lower discrepancy than the van der Corput construction. van der Corput construction is extensible and the digits in it can be randomized. If f is continuously differentiable, then for N = 4 k, the randomization in Owen (1995) will give root mean square error O(1/N) Future Work Generalization to higher dimensional simplex. in tensor product spaces.

32 Thank you. Questions?