G.l.p., optimal coefficients, rank-1 lattice rules, ... Dirk - PowerPoint PPT Presentation

Lattice rules Lattice sequences Cos space + tent transform Conclusions G.l.p., optimal coefficients, rank-1 lattice rules, ... Dirk Nuyens Department of Computer Science KU Leuven, Belgium RICAM Special Semester on Algebra and Number Theory Workshop 1: Uniform distribution and quasi-Monte Carlo methods Austrian Academy of Sciences, Linz October 14–18, 2013 Lattice rules — Dirk Nuyens (KU Leuven) 1/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Definition Rank-1 lattice rules For a given generating vector z ∈ Z d N take � N − 1 � z k mod N � � Q N ( f ; z ) := 1 [ 0 , 1 ) d f ( x ) d x . f ≈ N N k = 0 Korobov (1959), Sloan, Joe, Niederreiter, Kuo, Dick, ... Components z j are taken relatively prime to N ( → permutations of k / N ). Here: N = 17, z = ( 1 , 5 ) . Looks good. ♣ how to generate these points: N=17; z=[1; 5]; P = mod(z*(0:N-1), N)/N; But, how to pick “good” z ? ♣ goodandbad.m; try z = [1; 1]; Lattice rules — Dirk Nuyens (KU Leuven) 2/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Definition Rank-1 lattice rules For a given generating vector z ∈ Z d N take � N − 1 � z k mod N � � Q N ( f ; z ) := 1 [ 0 , 1 ) d f ( x ) d x . f ≈ N N k = 0 Korobov (1959), Sloan, Joe, Niederreiter, Kuo, Dick, ... Components z j are taken relatively prime to N ( → permutations of k / N ). Here: N = 17, z = ( 1 , 5 ) . Looks good. ♣ how to generate these points: N=17; z=[1; 2]; P = mod(z*(0:N-1), N)/N; But, how to pick “good” z ? ♣ goodandbad.m; try z = [1; 1]; Lattice rules — Dirk Nuyens (KU Leuven) 2/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Definition Rank-1 lattice rules For a given generating vector z ∈ Z d N take � N − 1 � z k mod N � � Q N ( f ; z ) := 1 [ 0 , 1 ) d f ( x ) d x . f ≈ N N k = 0 Korobov (1959), Sloan, Joe, Niederreiter, Kuo, Dick, ... Components z j are taken relatively prime to N ( → permutations of k / N ). Here: N = 17, z = ( 1 , 5 ) . Looks good. ♣ how to generate these points: N=17; z=[1; 5]; P = mod(z*(0:N-1), N)/N; But, how to pick “good” z ? ♣ goodandbad.m; try z = [1; 1]; Lattice rules — Dirk Nuyens (KU Leuven) 2/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Worst-case error A Koskma–Hlawka error bound Suppose f can be represented as � � ˆ | ˆ f ( h ) exp ( 2 π i h · x ) where f ( x ) = f ( h ) | < ∞ , h ∈ Z d h ∈ Z d then N − 1 � � Q N ( f ; z ) − I ( f ) = 1 ˆ f ( h ) exp ( 2 π i h · z k / N ) − ˆ f ( 0 ) N k = 0 h ∈ Z d Lattice rules — Dirk Nuyens (KU Leuven) 3/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Worst-case error A Koskma–Hlawka error bound Suppose f can be represented as � � ˆ | ˆ f ( h ) exp ( 2 π i h · x ) where f ( x ) = f ( h ) | < ∞ , h ∈ Z d h ∈ Z d then � � N − 1 � � 1 ˆ exp ( 2 π i ( h · z ) k / N ) Q N ( f ; z ) − I ( f ) = f ( h ) N k = 0 0 � = h ∈ Z d Lattice rules — Dirk Nuyens (KU Leuven) 3/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Worst-case error A Koskma–Hlawka error bound Suppose f can be represented as � � ˆ | ˆ f ( h ) exp ( 2 π i h · x ) where f ( x ) = f ( h ) | < ∞ , h ∈ Z d h ∈ Z d then � ˆ Q N ( f ; z ) − I ( f ) = f ( h ) 1 h · z ≡ 0 0 � = h ∈ Z d Lattice rules — Dirk Nuyens (KU Leuven) 3/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Worst-case error A Koskma–Hlawka error bound Suppose f can be represented as � � ˆ | ˆ f ( h ) exp ( 2 π i h · x ) where f ( x ) = f ( h ) | < ∞ , h ∈ Z d h ∈ Z d then � f ( h ) r α, γ ( h ) r − 1 ˆ Q N ( f ; z ) − I ( f ) = α, γ ( h ) 1 h · z ≡ 0 � �� � 0 � = h ∈ Z d = 1 Lattice rules — Dirk Nuyens (KU Leuven) 3/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Worst-case error A Koskma–Hlawka error bound Suppose f can be represented as � � ˆ | ˆ f ( h ) exp ( 2 π i h · x ) where f ( x ) = f ( h ) | < ∞ , h ∈ Z d h ∈ Z d then � | ˆ f ( h ) | r α, γ ( h ) r − 1 | Q N ( f ; z ) − I ( f ) | ≤ α, γ ( h ) 1 h · z ≡ 0 0 � = h ∈ Z d Lattice rules — Dirk Nuyens (KU Leuven) 3/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Worst-case error A Koskma–Hlawka error bound Suppose f can be represented as � � ˆ | ˆ f ( h ) exp ( 2 π i h · x ) where f ( x ) = f ( h ) | < ∞ , h ∈ Z d h ∈ Z d then   1 / q   1 / p  � �   | ˆ f ( h ) | p r p r − q    | Q N ( f ; z ) − I ( f ) | ≤ α, γ ( h ) α, γ ( h )   h ∈ Z d 0 � = h ∈ Z d � �� � h · z ≡ 0 � �� � =: � f � p ,α, γ =: e ( z , N ; �·� p ,α, γ ) for some positive function r α, γ ( h ) . Lattice rules — Dirk Nuyens (KU Leuven) 3/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Worst-case error The worst-case error The worst-case error in using Q for f ∈ F to approximate I ( f ) � � � Q ( f ) − I ( f ) � sup e ( Q ; F ) := ⇒ | Q ( f ) − I ( f ) | ≤ � f � F e ( Q ; F ) . f ∈F � f � F ≤ 1 For the Korobov space, with weights γ u ,   p � � � f h | p r p � f � p | ˆ  γ − 1 | ˆ f h | p | h j | α  p ,α, γ = = α, γ ( h ) < ∞ . u ( h ) h ∈ Z d j ∈ u ( h ) h ∈ Z d Riesz representer of the error for lattice rule when p > 1: � r − q α, γ ( h ) exp ( 2 π i h · x ) . ξ ( x ) = 0 � = h ∈ Z d Then e ( z , N ; � · � p ,α, γ ) = Q N ( ξ ; z ) 1 / q . Lattice rules — Dirk Nuyens (KU Leuven) 4/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Worst-case error Calculating the worst-case error ◮ Korobov space with product weights γ u = � j ∈ u γ j : d � � � � exp ( 2 π i hx ) 1 + γ q χ ( x ) = − 1 + where j ω ( x j ) , ω ( x ) = . | h | α q j = 1 0 � = h ∈ Z Need α > 1 / q . ◮ The R -criterion → star-discrepancy: � exp ( 2 π i hx ) where χ ( x ) = · · · ω ( x ) = . | h | α q − N 2 , N � � 0 � = h ∈ 2 Need α > 0. This corresponds to truncated Fourier series. Lattice rules — Dirk Nuyens (KU Leuven) 5/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Component-by-component The component-by-component algorithm Brute force search for good z ∈ Z d N would cost exponential in d . Therefore (Sloan, Joe, Kuo, ... and also Korobov) : for s = 1 to d do for all z s ∈ Z × N do calculate e s ( z 1 , . . . , z s − 1 , z s ) end for z s = argmin e s ( z 1 , . . . , z s − 1 , z ) z ∈ Z × N end for Cost of O ( d 2 N 2 ) . ◮ Rules with optimal rate of convergence in weighted Korobov space, i.e., O ( N − α + δ ) , δ > 0. ◮ Rules with optimal rate of convergence in weighted Sobolev space, i.e., O ( N − 1 + δ ) , δ > 0. Kuo (2003), Dick (2004) Lattice rules — Dirk Nuyens (KU Leuven) 6/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Fast component-by-component construction Fast component-by-component construction Component-by-component by using the recursion e s (( z ∗ 1 , . . . , z ∗ s − 1 , z s ) , N ; || · || p ,α, γ ) q s − 1 ) , N ; || · || p ,α, γ ) q + θ s ( z s ) . = e s − 1 (( z ∗ 1 , . . . , z ∗ Write ω ( k · z j ) := ω ( x k , j ) = ω ( { kz j / N } ) = ω (( kz j mod N ) / N ) � r − q α ( h ) exp ( 2 π i hkz j / N ) , = 0 � = h ∈ Z Lattice rules — Dirk Nuyens (KU Leuven) 7/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Fast component-by-component construction For prime N ... where N − 1 N − 1 � � � 1 Y u ( k ) ω ( k · z s ) = 1 θ s ( z s ) = γ u ∪{ s } Y s ( k ) ω ( k · z s ) N N u ⊆{ 1 : s − 1 } k = 0 k = 0 N − 1 � = 1 N Y s ( 0 ) ω ( 0 ) + 1 Y s ( k ) ω ( k · z s ) N k = 1 � �� � convolution with � � � r − q α ( h j ) exp ( 2 π i h j z ∗ ω ( k · z ∗ Y u ( k ) = j k / N ) = j ) , h u ∈ Z | u | j ∈ u j ∈ u ∗ � Y s ( k ) = γ u ∪{ s } Y u ( k ) . u ⊆{ 1 : s − 1 } Lattice rules — Dirk Nuyens (KU Leuven) 8/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Fast component-by-component construction Cyclic structure → circulant matrix When N is prime Z N = Z × N ∪ { 0 } ... and Ω n is basically just homomorphic to the multiplication table of the group Z × N . E.g., for N = 7: 1 2 3 4 5 6 · 1 1 2 3 4 5 6 2 2 4 6 1 3 5 3 3 6 2 5 1 4 4 4 1 5 2 6 3 5 5 3 1 6 4 2 6 6 5 4 3 2 1 Lattice rules — Dirk Nuyens (KU Leuven) 9/41

Lattice rules Lattice sequences Cos space + tent transform Conclusions Fast component-by-component construction Cyclic structure → circulant matrix When N is prime Z N = Z × N ∪ { 0 } ... and Ω n is basically just homomorphic to the multiplication table of the group Z × N . E.g., for N = 7: 1 2 3 4 5 6 · 1 1 2 3 4 5 6 2 2 4 6 1 3 5 3 3 6 2 5 1 4 4 4 1 5 2 6 3 5 5 3 1 6 4 2 6 6 5 4 3 2 1 Multiplication modulo N can be much easier using a representation in powers of a generator g ... g α g β ≡ g α + β ( mod ϕ ( N )) ( mod N ) . Lattice rules — Dirk Nuyens (KU Leuven) 9/41