Good, low degree, Rank-1 Lattice rules in High Dimensions Tor Srevik - - PowerPoint PPT Presentation

Good, low degree, Rank-1 Lattice rules in High Dimensions

Tor Sørevik1

1joint work with James N. Lyness

MCQCM 2012, Sydney

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Basic definitions

An s−dimensional simple rank 1 lattice rule Qf = 1 N

N−1

• j=0

f

• j (1, x1, · · · , xs−1)

N

• (1)

x = (1, x1, · · · , xs−1) ∈ I Ns and N ∈ I N A quadrature rule is of trigonometric degree d iff it integrates exactly all trigonometric polynomials of degree less or equal d. For lattice rules the enhanced trigonometric degree, δ = d + 1, can be computed as δ(Q) = min

p∈Λ⊥\{0} ||p||1

(2) where Λ⊥ is the dual lattice defined as Λ⊥ = {p | pT z ∈ Z for all z ∈ Λ} (3)

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Computing the enhanced degree

For rank 1 lattice rules: p ∈ Λ⊥ ⇔ ∃λ ∈ Z Zs such that: pT =  

s−1

• j=1

λjxj + λsN, λ1, · · · , λs−1   (4) Plugging (4) into (2) we get δ(Q) = min

λ∈Zs\{0} | s−1

• j=1

λjxj + λsN| +

s−1

• j=1

|λj| (5) Restriction to the subset of λ’s with λs = 0 gives δ(Q) ≤ min

λ∈Zs−1\{0} | s−1

• j=1

λjxj| +

s−1

• j=1

|λj| (BC)

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A simple idea

Devide the search into two steps: Basic algorithm

• 1. For fixed δ, find candidate x-vectors satisfying (BC). (”δ− sequences”)
• 2. Find the smallest N for x-vectors satisfying (BC).

To make this a finite search we need to restrict the set of admissible x-vectors.

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Limiting the search:

Theorem

Every rank 1 simple lattice have a symmetric copy which satisfy: 1 < x1 < · · · < xs ≤ N/2; (Low sequence) (6) and another satisfying: N/2 < x1 < · · · < xs < N; (High sequence) (7)

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..and a little refinement

Lemma

A necessary condition for a low sequence to produce a lattice rule of enhanced trigonometric degree larger or equal to δ is that: x1 ≥ δ − 1 xj − xj−1 ≥ δ − 2; 1 < j < s − 1 xs−1 ≤ (N − δ)/2 + 1 Since N ≥ 2(xs−1 − 1) + δ we seek δ−sequences with small xs−1.

Lemma

A necessary condition for x = (x1, · · · , xk) to be a δ− sequence is that (x1, · · · , xk−1) is a δ− sequence.

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Extension Algorithm for finding δ−sequences

extend sequence (s, δ, k, x(1 : k − 1)) for xk = xk−1 + δ − 2, upper

• k = check if δ−sequence (δ, k, x(1 : k))

if ok then if (k = s) then find minimal N, write result and return else extend sequence (s, δ, k, x(1 : k)) endif endif end for This computation took us to δ = 5 and s ≤ 10, δ = 6 and s ≤ 7. Not good enough!

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The case δ = 5

James’ last theorem:

Theorem

Let xs < 2x1 − 5 (”high-sequence”) and εi,j = xi − xj; i > j Then x1, x2, · · · , xs is a δ− sequence with δ ≥ 5 if and only if the set {εi,j; 1 ≤ j < i ≤ s} (8) contains only distinct elements and εi,j ≥ 3.

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Golomb rulers

Definition

A set of integers A = {a1, a2, · · · , an} a1 < a2 < · · · < an (9) is called a Golomb ruler if ak − al = aj − ai iff {k, l} = {j, i} (10) If, in addition there exist an p such that that the above hold for all differences modulo p, the Golomb ruler is said to be cyclic modulo p. G(n) = an − a1 is said to be the length of the Golomb ruler (”Golomb rulers” are equivalent to ”Sidon sets” or ”B2 sequences”)

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Properties of Golomb rulers

Theorem

If the set A = {a1, a2, · · · , an} is a Golomb ruler then so is the sets: Translation: A1 = {a1 + x, a2 + x, · · · , an + x} Multiplication: A2 = {ca1, ca2, · · · , can} Mirroring: A3 = {an − an, an − an−1, · · · , an − a1} and for cyclic rulers: Cyclic shift+ Translation: A4 = {p − an, a1 + p − an, · · · , an−1 + p − an}

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Length of Golomb rulers

G(n) > n2 − 2n√n + √n − 2 Proved by Lindstr¨

• m -69.

G(n) < n2 Conjectured by Erd¨

• s and Turan -41.

This conjecture is proved for n prime and computational verified for n ≤ 65000. A crude and pessimistic upper bound for lattice rules with δ = 5 based on Golomb rulers are then N < 4(2s − 3)2. Compared to the lower bound based on the moment equations: N ≥ NME = 2s2 + 2s + 1 (Mysovskikh -87).

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Construction of good Golomb rulers

Theorem

Let p be a prime number and g a primitive element of the multiplicative group Z∗

• p. The following sequence is a cyclic Golomb ruler.

R(p, g) = pk + (p − 1)gk mod (p(p − 1)); for 1 ≤ k ≤ p − 1 (11) [Rusza -93]

Theorem

Let q = pm be a prime power and θ a primitive element in the Galois field GF(q2). Then the q integers d1, · · · , dq = {a|1 ≤ a < q2and θa − θ ∈ GF(q)} (12) have distinct pairwise modulo differences modulo q2 − 1. [Bose and Chowla -62]

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Creating δ−sequences from a Golomb ruler

Remove element(s) to satisfy the εij ≥ 3 condition. If shorter δ− sequences are wanted, remove more elements. Let A = {0, a1, ..., as−1} be the standard form of the Golomb ruler. For a cyclic GR modulo p we can do cyclic shifts + translations to get new GRs in standard form: C(A − a1) = {0, a2 − a1, ..., as−1 − a1, p − a1}. For each GR in standard form we may use the translation property A = A + c; c ∈ I N to produce multiple candidates. Bottom line: 1 Golomb ruler makes many related δ−sequences.

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Computing the optimal N for a fixed Delta-sequence

For fixed x find minimal N of the set: Ω(N) = {N : |

s−1

• j=1

λjxj + λsN| +

s−1

• j=1

|λj| ≥ δ; λ ∈ Zs\{0}} The second sum implies that we only need to consider λ where s−1

j=1 |λj| ≤ δ − 1.

For fixed λ1, · · · λs−1 the λs which minimize the expression is: λs = nint − s−1

j=1 λjxj

N (13)

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Computing the minimal N (cont.)

Let H ∈ Zr×(s−1) contain all s − 1-dim λ’s such that s−1

j=1 |λj| < δ then

for all admissible λ we compute q = Hx and l ∈ I Nr; li =

s−1

• j=1

|hij| then (x, N) of degree δ ⇔ min

i=1,...,r |qi − nint

qi N

• N| + li ≥ δ

Note: To check for multiple N, only the last step needs to be repeated.

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Computing the minimal N for related x’s

Creating related xn by translation xn = x0 + ne; eT = (1, 1, · · · , 1) (14) we have qn = Hxn = q0 + n∆q; q0 = Hx0; ∆q = He (15) Thus for multiple vectors created by translation only the first need a full matrix-vector multiply.

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Computational complexity for δ = 5

The number of rows in H is r ≈ 23 s − 2 + δ s − 1

• (16)

For δ = 5 and s ≫ 5 we have r ≈ (s + 2)4/3 N is bounded above by a N-best-so-far and from below by max(xs−1 + 3, NME). In typical computation we test ∼ 2s2 values of N for each x. The cyclic shift gives us s possible x0 for each Golomb ruler, and each of these have ∼ s2 translates.

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Computational results

By the exhaustive search we were able to compute optimal rank-1 lattice for δ = 5 s ≤ 10 and for δ = 6 s ≤ 7 Using Golomb rulers we have (so far) computed good rank-1 lattices of enhanced trig. degree up to s ≤ 27.

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Quality of results

5 10 15 20 25 30 10 20 30 40 50 60 70 80

Number of lattice points, δ = 5

s, Dimension N1/2

Lower bound Optimal rank−1 Best Golumb ruler Upper bound

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Conclusions

The exhaustive search by the extension algorithm produce to many delta sequences, thus we have to many candidates to examine. For δ = 5 Golomb rulers produce good δ− sequences almost free of cost. There do exist rank 1 lattice rules with δ = 5 having N = O(s2). The bottleneck is the computation of optimal N for fixed x. Preliminary results: New rank 1 lattice rules for δ = 5 and s < 28 (so far...).

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