Algebraic Generation of Orthogonal Fractional Factorial Designs

Transcription

1 Algebraic Generation of Orthogonal Fractional Factorial Designs Roberto Fontana Dipartimento di Matematica, Politecnico di Torino Special Session on Advances in Algebraic Statistics AMS 2010 Spring Southeastern Sectional Meeting Lexington, March 27th-28th,2010 Joint paper with Giovanni Pistone, Politecnico di Torino [Fontana and Pistone(2010)]

2 Outline Aim of the work Full factorial design, fractions and counting functions Strata Generation of fractions Concluding remarks

3 Aim of the work To find a general method for fractional factorial design generation, where general means without any restriction for the number of factors/level

4 Full factorial design - complex coding of levels We refer to [Pistone and Rogantin(2008)] m factors, D 1,..., D m D j has n j levels coded with the n j -th roots of the unity: D j = {ω 0,..., ω nj 1} ( 1 2π ω k = exp n j D is the full factorial design with complex coding D = D 1 D j D m. ζ = (ζ 1,..., ζ m ) is a point of D #D is the cardinality of D, #D = m i=1 n j ) ( k exp i 2π n j ) k ;

5 Fractions and Counting functions A fraction F of D is a multiset (F, f ) whose underlying set of elements F is contained in D and f is the multiplicity function f : F N that for each element in F gives the number of times it belongs to the multiset F. Definition The counting function R of a fraction F is a response defined on D so that for each ζ D, R(ζ) equals the number of appearances of ζ in the fraction. We denote by c α the coefficients of the representation of R on D using the monomial basis {X α, α L = Z n1 Z nj Z nm }: R(ζ) = α L c α X α (ζ) ζ D c α C.

6 Full factorial design - complex coding of levels L is the full factorial design with integer coding α = (α 1,..., α m ) L = Z n1 Z nj Z nm, α j = 0,..., n j 1, j = 1,..., m is an element of L X j is the j-th component function, which maps a point to its i-th component: X j : D (ζ 1,..., ζ m ) ζ j D j ; the function X j is called simple term or, by abuse of terminology, factor. X α is the interaction term X α 1 1 Xm αm, i.e. the function X α : D (ζ 1,..., ζ m ) ζ α 1 1 ζαm m ;

7 Counting functions Proposition Let F be a fraction of a full factorial design D and R = α L c αx α be its counting function. 1 The coefficients c α are equal to 1 #D X α (ζ) ; ζ F 2 The term X α is centered on F, i.e. E F (X α ) = 1 if, and only if, c α = c [ α] = 0. #F ζ F X α (ζ) = 0, 3 The terms X α and X β are orthogonal on F, i.e. E F (X α X β ) = 0, if, and only if, c [α β] = 0.

8 Projectivity and Orthogonal Arrays Definition A fraction F factorially projects onto the I -factors, I {1,..., m}, if the projection is a multiple full factorial design, i.e. a full factorial design where each point appears equally often.

9 Projectivity and Orthogonal Arrays Definition A fraction F factorially projects onto the I -factors, I {1,..., m}, if the projection is a multiple full factorial design, i.e. a full factorial design where each point appears equally often. Definition A fraction F is a mixed orthogonal array of strength t if it factorially projects onto any I -factors with #I = t.

10 Projectivity and Orthogonal Arrays Definition A fraction F factorially projects onto the I -factors, I {1,..., m}, if the projection is a multiple full factorial design, i.e. a full factorial design where each point appears equally often. Definition A fraction F is a mixed orthogonal array of strength t if it factorially projects onto any I -factors with #I = t. Proposition (Projectivity)

11 Projectivity and Orthogonal Arrays Definition A fraction F factorially projects onto the I -factors, I {1,..., m}, if the projection is a multiple full factorial design, i.e. a full factorial design where each point appears equally often. Definition A fraction F is a mixed orthogonal array of strength t if it factorially projects onto any I -factors with #I = t. Proposition (Projectivity) 1 A fraction factorially projects onto the I -factors if, and only if, all the coefficients of the counting function involving only the I -factors are 0.

12 Projectivity and Orthogonal Arrays Definition A fraction F factorially projects onto the I -factors, I {1,..., m}, if the projection is a multiple full factorial design, i.e. a full factorial design where each point appears equally often. Definition A fraction F is a mixed orthogonal array of strength t if it factorially projects onto any I -factors with #I = t. Proposition (Projectivity) 1 A fraction factorially projects onto the I -factors if, and only if, all the coefficients of the counting function involving only the I -factors are 0. 2 A fraction is an orthogonal array of strength t if, and only if, all the coefficients of the counting function up to the order t are zero: c α = 0 for all α of order up to t, α (0, 0,..., 0).

13 Orthogonal array of strength t and indicator functions Let F = α L b αx α an indicator function of a fraction F of D Let C = {α L : 0 < α t}; F is an indicator function of an Orthogonal Array of strengt t if it is a solution of the following system { b α = β L b βb [α β] b α = 0, α C Remark F is an indicator function F (F 1) = 0 b α = β L b βb [α β]

14 Level set of X α Proposition (level set 1) Let X j the simple term with level set Ω nj = {ω 0,..., ω nj 1}. Let s consider the term Xj r and let s define { 1 r = 0 s j,r = n j / gcd(r, n j ) r > 0 Over D, the term X r j takes all the values of Ω sj,r equally often. Example n j = 4, D j = {ω 0, ω 1, ω 2, ω 3 } {1, i, 1, i} ζ j Xj 0 Xj 1 Xj 2 Xj ω 1 1 ω 1 ω 2 ω 3 ω 2 1 ω 2 1 ω 2 ω 3 1 ω 3 ω 2 ω 1

15 Level set of X α Proposition (level set 2) Let X α = X α 1 1 Xm αm an interaction. X α i i takes values in Ω si,αi where s i,αi is determined according to the previous Proposition level set 1. Let s define s α = lcm(s 1,α1,..., s m,αm ). Over D, the term X α takes all the values of Ω sα equally often. Example n j = n k = 4, X 2 j X 2 k. From Proposition level set 1 we have s j = s k = 2. We obtain s = 2. Indeed X 2 j (ζ j) {1, 1}, X 2 k (ζ k) {1, 1} X 2 j X 2 k (ζ j, ζ k ) {1, 1}

16 Strata Definition Given a term X α, α L = Z n1... Z nm the full design D is partitioned into the the following strata { } Dh α = ζ D : X α (ζ) = ω h where ω h Ω sα and s α is determined according to the previous Propositions level set 1 and level set 2. We use n α,h to denote the number of points of the fraction F that are in the stratum D α h, with h = 0,..., s α 1, n α,h = ζ D α h y ζ

17 Counting function as linear combination of indicator functions Let s consider the indicator functions 1 ζ of all the single points of D { 1 ζ = (ζ 1,..., ζ m ) 1 ζ : D (ζ 1,..., ζ m ) 0 ζ (ζ 1,..., ζ m ) The counting function R of a fraction F can be written as ζ D y ζ1 ζ with y ζ R(ζ) {0, 1,..., n,...}. The particular case in which R is an indicator function corresponds to y ζ {0, 1}. Proposition Let F be a fraction of D. Its counting fraction R can be expressed both as R = α c αx α and R = ζ D y ζ1 ζ. The relation between the coefficients c α and y ζ is c α = 1 y ζ X #D α (ζ) ζ D

18 c α and n α,h Proposition Let F be a fraction of D with counting fraction R = α L c αx α. Each c α, α L, depends on n α,h, h = 0,..., s α 1, as c α = 1 s α 1 n α,h ω h #D h=0 where s α is determined by X α (see Proposition level set 1 and Proposition level set 2 ). Viceversa, each n α,h, h = 0,..., s α 1, depends on c [ kα], k = 0,..., s 1 as n α,h = #D s α s α 1 k=0 c [ kα] ω [hk]

19 Conditions on n α,h to have X α centered Let F be a fraction of D with counting fraction R = 1 sα 1 #D α L h=0 n α,hω h X α. Proposition (*) Let X α be a term with level set Ω sα on full design D. s α is prime The term X α is centered on the fraction F if, and only if, its s α levels appear equally often: n α,0 = n α,1 =... = n α,s 1 = λ α s α is not prime Let P α (ζ) = s α 1 h=0 n α,hζ h and let s denote by Φ sα the cyclotomic polynomial of the s α -roots of the unity. The term X α is centered on the fraction F if, and only if, the remainder H α (ζ) = P α (ζ)modφ sα (ζ) whose coefficients are integer linear combinations of n α,h, h = 0,..., s α 1, is identically zero.

20 Conditions on R(ζ) to have X α centered n α,h are related to the values of the counting function R of a fraction F by the following relation n α,h = y ζ, ζ D α h Proposition * allows to express the condition X α is centered on F ( c α = 0) as integer linear combinations of the values y ζ of the counting function over the full design D.

21 Orthogonal Arrays For Orthogonal Arrays, using Proposition projectivity we have c α = 0 α C where C = {α L : 0 < α t} and α is the number of non null elements of α.

22 OA(n, s m, t), s prime, as a system of linear equations Let s consider OA(n, s m, t), s prime. We consider α C and, using Proposition *, we write the condition c α = 0 using strata as ζ D0 α y ζ = λ y ζ = λ Varying α C we get Considering λ as a variable, ζ D α 1... ζ Ds 1 α y ζ = λ where à = [A, 1] and Ỹ = (Y, λ). AY = λ1 ÃỸ = 0

23 Computation of generators for OA(n, s m, t), s prime The sum of two Orthogonal Arrays, Y 1 OA(n 1, s m, t) and Y 2 OA(n 2, s m, t), is an Orthogonal Array Y 1 + Y 2 OA(n 1 + n 2, s m, t). The Hilbert Basis [Schrijver(1986)] is a minimal set of generators such that any OA(n, s m, t) becomes a linear combination of the generators with positive or null integer coefficients. This approach extends that of [Carlini and Pistone(2007)] where homogeneous 2-level fractions were considered and the conditions ζ F X α (ζ) = 0 were used. c α = 1 #D The advantage of using strata is that we avoid computations with complex numbers (X α (ζ)). We give some examples. For the computation we use 4ti2 [4ti2 team(2008)].

24 Some examples of OA(n, s m, t), s prime Some classes of OA s: A(n, 2 5, 2) The matrix à has 30 rows and 33 columns. We find 26, 142 solutions A(n, 3 3, 2) The matrix à has has 54 rows and 28 columns. We find 66 solutions: 12 have 9 points, all different 54 have 18 points, 17 different. A(n, 3 4, 3) The matrix à has has 192 rows and 82 columns. We find 131, 892 solutions

25 One example of OA(18, 3 3, 2) Classification of OA(n, 3 4, 3) # support # of points max rep N , , , , , , , , , , , , , , ,184 Total 131,892

26 OA(n, s n sn k, t) as a system of linear equations k Let s consider mixed level orthogonal arrays. OA(n, s m, t), without the condition s prime, become a particular case. Using Proposition *, we express the condition c α = 0, α C using strata. Let s consider, as an example, OA(n, 2 3 3, 3). Let s take α (1, 1, 0, 0) C = {α Z 2 Z 3 Z 3 Z 3 : 1 α 3}. We get s α = 6. The condition c 1,1,0,0 = 0 is equivalent to X 1,1,0,0 is centered that is equivalent to H α (ζ) P α (ζ)modφ 6 (ζ) is identically zero. We have and therefore P α (ζ) = n (1,1,0,0),0 + n (1,1,0,0),1 ζ n (1,1,0,0),5 ζ 5 Φ 6 (ζ) = 1 ζ + ζ 2 H α (ζ) P α (ζ)modφ 6 (ζ) = n (1,1,0,0),0 n (1,1,0,0),2 n (1,1,0,0),3 + n (1,1,0,0),6 + (n (1,1,0,0),1 + n (1,1,0,0),2 n (1,1,0,0),5 n (1,1,0,0),6 )ζ

27 OA(n, s n sn k, t) as a system of linear equations k Varying α C and reminding that n α,h = ζ Dh α y ζ we get ÃỸ = 0. Some classes of OA s: The matrix à has 89 rows and 54 columns. We find 403 solutions OA(n, 2 3 3, 3) OA(n, 4 2, 2) OA(n, 6 2, 2) The matrix à has 10 rows and 16 columns. We find 24 solutions corresponding to all the LHD. The matrix à has 18 rows and 36 columns. We find 620 solutions corresponding to all the LHD. Classification of OA(n, 2 3 3, 3) # support # of points max rep N Total 403

28 Sampling - example find one single replicate OA(9, 3 3, 2);

29 Sampling - example find one single replicate OA(9, 3 3, 2); find one solution of AY = 1 with y ζ {0, 1}

30 Sampling - example find one single replicate OA(9, 3 3, 2); find one solution of AY = 1 with y ζ {0, 1} we used standard Simulated Annealing with objective function V (Y ) defined as the number of equations of AY = 1 that are satisfied; we implemented this algorithm using SAS/IML.

31 Sampling - example find one single replicate OA(9, 3 3, 2); find one solution of AY = 1 with y ζ {0, 1} we used standard Simulated Annealing with objective function V (Y ) defined as the number of equations of AY = 1 that are satisfied; we implemented this algorithm using SAS/IML. we found a solution in 2, 702 iterations (a couple of seconds on a standard laptop).

32 Sampling - example find one single replicate OA(9, 3 3, 2); find one solution of AY = 1 with y ζ {0, 1} we used standard Simulated Annealing with objective function V (Y ) defined as the number of equations of AY = 1 that are satisfied; we implemented this algorithm using SAS/IML. we found a solution in 2, 702 iterations (a couple of seconds on a standard laptop). we have also experimented the algorithm on

33 Sampling - example find one single replicate OA(9, 3 3, 2); find one solution of AY = 1 with y ζ {0, 1} we used standard Simulated Annealing with objective function V (Y ) defined as the number of equations of AY = 1 that are satisfied; we implemented this algorithm using SAS/IML. we found a solution in 2, 702 iterations (a couple of seconds on a standard laptop). we have also experimented the algorithm on OA(9, 3 4 ; 3); we found one solution in 2, 895 iterations;

34 Sampling - example find one single replicate OA(9, 3 3, 2); find one solution of AY = 1 with y ζ {0, 1} we used standard Simulated Annealing with objective function V (Y ) defined as the number of equations of AY = 1 that are satisfied; we implemented this algorithm using SAS/IML. we found a solution in 2, 702 iterations (a couple of seconds on a standard laptop). we have also experimented the algorithm on OA(9, 3 4 ; 3); we found one solution in 2, 895 iterations; 4 4 sudoku; we found one solution in 2, 852 iterations;

35 Sampling - example find one single replicate OA(9, 3 3, 2); find one solution of AY = 1 with y ζ {0, 1} we used standard Simulated Annealing with objective function V (Y ) defined as the number of equations of AY = 1 that are satisfied; we implemented this algorithm using SAS/IML. we found a solution in 2, 702 iterations (a couple of seconds on a standard laptop). we have also experimented the algorithm on OA(9, 3 4 ; 3); we found one solution in 2, 895 iterations; 4 4 sudoku; we found one solution in 2, 852 iterations; 9 9 sudoku; we did not find any solution in 100, 000 iterations;

36 Sampling example move from one single replicate OA(9, 3 3, 2) to another single replicate OA(9, 3 3, 2) ;

37 Sampling example move from one single replicate OA(9, 3 3, 2) to another single replicate OA(9, 3 3, 2) ; move from one solution of AY = 1 with y ζ {0, 1} to another solution of AY = 1 with y ζ {0, 1};

38 Sampling example move from one single replicate OA(9, 3 3, 2) to another single replicate OA(9, 3 3, 2) ; move from one solution of AY = 1 with y ζ {0, 1} to another solution of AY = 1 with y ζ {0, 1}; we used 4ti2 to generate M, the Markov Basis corresponding to the homogeneous system AX = 0.

39 Sampling example move from one single replicate OA(9, 3 3, 2) to another single replicate OA(9, 3 3, 2) ; move from one solution of AY = 1 with y ζ {0, 1} to another solution of AY = 1 with y ζ {0, 1}; we used 4ti2 to generate M, the Markov Basis corresponding to the homogeneous system AX = 0. M contains 81 different moves;

40 Sampling example move from one single replicate OA(9, 3 3, 2) to another single replicate OA(9, 3 3, 2) ; move from one solution of AY = 1 with y ζ {0, 1} to another solution of AY = 1 with y ζ {0, 1}; we used 4ti2 to generate M, the Markov Basis corresponding to the homogeneous system AX = 0. M contains 81 different moves; as an initial fraction we considered the nine-run regular fraction F 0 whose indicator function R 0 is R 0 = 1 3 (1 + X 1X 2 X 3 + X1 2X 2 2X 3 2 ). We run simulations generating R i as R i = R i 1 + ɛ MRi 1 M Ri 1 (and moving to F i if R i 0);

41 Sampling example move from one single replicate OA(9, 3 3, 2) to another single replicate OA(9, 3 3, 2) ; move from one solution of AY = 1 with y ζ {0, 1} to another solution of AY = 1 with y ζ {0, 1}; we used 4ti2 to generate M, the Markov Basis corresponding to the homogeneous system AX = 0. M contains 81 different moves; as an initial fraction we considered the nine-run regular fraction F 0 whose indicator function R 0 is R 0 = 1 3 (1 + X 1X 2 X 3 + X1 2X 2 2X 3 2 ). We run simulations generating R i as R i = R i 1 + ɛ MRi 1 M Ri 1 (and moving to F i if R i 0); we obtained all the 12 different 9-run fractions, each one with 9 different points as known in the literature and as shown previously.

42 Concluding remarks + no restrictions on the number of levels

43 Concluding remarks + no restrictions on the number of levels + no computation with complex numbers

44 Concluding remarks + no restrictions on the number of levels + no computation with complex numbers + Hilbert basis, Markov basis, MCMC available

45 Concluding remarks + no restrictions on the number of levels + no computation with complex numbers + Hilbert basis, Markov basis, MCMC available computational effort

46 Some references 4ti2 team, ti2 a software package for algebraic, geometric and combinatorial problems on linear spaces. Available at Carlini, E., Pistone, G., Hibert bases for orthogonal arrays. Journal of Statistical Theory and practice, Fontana, R., Pistone, G., Algebraic strata for non symmetrical orthogonal fractional factorial designs and application. La matematica e le sue applicazioni 2010/1, Politecnico di Torino DIMAT. Pistone, G., Rogantin, M., Indicator function and complex coding for mixed fractional factorial designs. J. Statist. Plann. Inference 138 (3), Schrijver, A., Theory of linear and integer programming. Wiley-Interscience Series in Discrete Mathematics. John Wiley & Sons Ltd., Chichester, a Wiley-Interscience Publication.

47 Thanks for your attention!