Sets of Orthogonal Latin Squares

Transcription

1 Sets of Ortogonal Latin Squares To obtain sets of 1mutually ortogonal Latin Squares (MOLS) of side were is rime or a ower of a rime, we associate eac of te treaments wit an element of te Galois Field of s n elements (i.e.gfs n in a 1 to 1 corresondence. Galois Fields Te setg o, g 1,...,g 1of elements is a finite field of order if: 1. Addition: a. g i g j g j g i b. g i g j g k g i g j g k c. Given g i and g k! g j g i g j g k d. te element aving tre additive rooerty of zero is g o g j g 0 g j j 2. Multilication: a. g i g j g j g i b. g ig j g k g i g jg k c. g ig j g k g i g j g i g k d. Given any g i g 0 and any g k! g j g i g j g k and g 0 as te multilicative roerty of zero i.e.g 0 g i g 0 e. Te element aving te multilicative roerty of unity is g 1. Case 1: If is a rime. Te finite field of elements is reresented by g 0 0,g 1 1,g i i, i 2,..., 1. Addition and multilication are ordinary aritmetic oerations, excet te resulting number is reduced mod. Mills

2 Case 2: Galois Field of s n elements were s is a rime Let Px be an irreducible olynomial of degree n wit integer coefficients. i.e. Px P 1 xp 2 x sp 3 x were P 1, P 2 and P 3 are olynomials (wit integer coefficients) of degrees less tan n. For any olynomial Fx, a olynomial in x wit integer coefficients). ten Fx fx mod s,px i.e. tis means we can write and Fx sqx PxQx fx fx a 0 a 1 x 1 a 2 x 2... a n1 x n1 is te residue of Fx mods,px and a o,...,a n1 0, 1,...,s1. If s and Px are fixed and fx varies, we get s n classes formed since a i takes s values. (Note: In order tat division be unique, s must be rime and Px irreducible mod s). Te finite field formed by te s n classes of residues fx is called GFs n (i.e. Galois Field of s n ) and te s n classes are te same regardless of te coice of Px, as long as Px is irreducible. GFs n exists if s is rime and n is a ositive integer. Te classes of residues may be reresented by te different ossible f i x. We denote tem by g 0, g 1,..., g 1. We generally reresent te elements of GF as owers of an element y (called te rimitive mark or P.M.) of te field suc tat y 1 1 and tis is te smallest ower for wic tis is true. i.e. elements are g o 0,g 1, g 2 y,..., g 1 y 2 Ten te addition table forms a L.S.D. and oter squares are obtained by cyclically rotating all rows but te first. e.g. 4 s n so s 2,n 2) GF s n is GF4 s 2 ere. Its elements are g 0 0, g 1 1, g 2 xte rimitive mark y and g 3 y 2 1x Aritmetic is carried out mod 2 and y x is P.M. Te irreducible olynomial Px of degree 2 in te field is Px x 2 x1( it is irreducible mod s 2). Now we can write tat f 1 x a 0 a 1 x were a 0, a 1 0,,1 so Mills

3 fx 0 1 x 1x Te P.M. is suc tat y 1 fx 1 1 and 4 Note: If y x: ten we ave y 3 x 3 1 mod Px,s so y x If we ad set y 1x: ten we ave tat y 3 1x 3 and one lower ower 1 y 2 y i.e. x 2 x 2 x1 1 x 1 x mods, Px Note: g 4 y 3 y 2 y 1xx x 2 x x 2 x mods,px In te addition table, te first row consists of te elements 0, 1,x, 1xand te table becomes 0 1 x 1x ten add (1) 1 0 1x x ten add (x) x 1x 0 1 ten add (1x) 1x x 1 0 Writing A,B,C,D for te elements 0,1, x,1xresectively, and rotating all but te first row cyclically we get A B C D A B C D A B C D B A D C C D A B C D A B D C B A D C B A B A D C i.e. 3 MOLS of side 4. D C B A B A D C C D A B Mills

4 Construction of L.S. and MOLS Teorem:aset of -1 mutually ortogonal Latin Squares (MOLS) of side were s n ( is rime or a ower of a rime). We associate eac of te treatments wit an element of GF s n in a 1-1 corresondence. Elements of te field are ordered as g 0 0,g 1 1,g 2 y,..., g 1 y 2 were y is a rimitive mark (P.M.) of te field. (i.e.y 1 1 and no lower ower y q 1 for 0 q ). Te additive table forms a L.S. and oter squares are obtained by rotating cyclically all rows but te first. Teorem: Te i t square of a set of 1MOLS as g i x i1 as te first element of te second row. Te first element of te (m 1) st row is g i y m1 g i g m Terefore te i t square is g 0 g 1 g 1 g i g 1 g i g 1 g i g i g 2 g i g i g 2 g 1 g i g 2 g i g 1 g i g i g 1 g 1i g i g 1 Note:A tyical element is g i g j g l were i 1,...,1;j 0,1,..., 1;l 0, 1,..., 1 Teorem: Suc a square is a Latin Square Proof: (by contradiction) Suose it is not a Latin square and terefore tat 2 elements in (q1) t row (say) are identical. i.e. For some air t, u, (t u g i g q g t g i g q g u Ten g t g u (t u Contradiction! (since te elements of te field are distinct) Do te same for columns: so wic imlies g q g l since g ii 0. Contradiction! g i g q g t g i g l g t g i g q g l 0 q l Mills

5 Teorem: Te squares in te set are mutually ortogonal. Proof: (by contradiction) Suose squares i and j are not ortogonal. Terefore wen j t is suerimosed on i t, at least 2 cells are te same. (i.e. one air of elements occurs togeter in two of te cells) Suose (w.l.o.g) in (q 1 t row, (t1) t column and (r1) t row, (y1) t column, were q r,t y Elements of i t square coincide:latin letters equal Elements of j t square coincide:greek letters equal g i g q g t g i g r g u or or Contradiction! g j g q g t g j g r g t g i g j g q g i g j g r g i g j g q g r Table of Px s and P.M. s s n Px P.M. 2 2 x 2 x1 x 2 3 x 3 x 2 1 x 2 4 x 4 x1 1x 3 2 x 2 1 1x 3 3 x 3 2x1 x 5 2 x 2 x1 2x Mills

6 Analysis of Several Latin Squares Running MOLS allows us to get more d.f. for error and to conduct more yotesis tests. Consider te model were square, i row, j column, k treatment y ijk i j k k ijk Imose side conditions: 0; k 0; and for eac i 0; j 0; k 0; and for eac k k 0 k i j k Suose we ave squares eac of side terefore i, j,k 1,..., ; 1,..., s Te solutions to te N.E. s are: y y y k y k y Holding fixed, i y i y ; j y j y ; k y k y y y y k y y k y y k y Mills

7 ANOVA table Source of Variation d.f. S.S. Total s 1 Squares (S) Treatments (T) T x S interaction Rows in squares s1 1 s1 1 s1 Columns in squares s1 Error 2 y ijk C.. T.S.S. i j k s 1 k k1 s k 1 k1 s1 C.M. S S C.M. S T i j j1 s 1 2 by subtraction C.M. S T S S Randomization for Latin Squares: Select a random square. Assign rows, columns, and treatments at random in eac square. Do tis for eac of te s squares. Mills