Mutually orthogonal latin squares (MOLS) and Orthogonal arrays (OA)

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1 and Orthogonal arrays (OA) Bimal Roy Indian Statistical Institute, Kolkata. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

2 Outline of the talk 1 Latin squares 2 3 Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

3 Outline of the talk 1 Latin squares 2 3 Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

4 Latin squares Definition A latin square of side Ò (or order Ò) is an Ò Ò array in which each cell contains a single symbol from an Ò-setS, such that each symbol occurs exactly once in each row and exactly once in each column. Example: Fact For every Ò there is a latin square of order Ò - addition table of Z Ò. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

5 Definition Two latin squares Ä and Ä of side Ò are isomorphic if there is a bijection Φ Ë Ë such that Φ(Ä(, )=Ä (Φ( ),Φ( )) for every, Ë, where Ë is not only the symbol set of each square, but also the indexing set for the rows and columns of each square. Enumeration Ò ½ ¾ Isomorphism classes ½ ½ ½ ½½ ½, ¼, ½ ½¾, ½,, Theorem Let Ä(Ò) denote the number of distinct latin squares of side Ò. Then(Ò!) ¾Ò Ò Ò¾ Ä(Ò) Ò =½ (!) Ò. Asymptotically Ä(Ò) Ò ½ Ò ¾ as Ò. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

6 Definition A Ò latin rectangle is a Ò array (where Ò) in which each cell contains a single symbol from an Ò-set Ë, such that each symbol occurs exactly once in each row and at most once in each column. Theorem A Ò latin rectangle, < Ò, can always be completed to a latin square of order Ò. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

7 Definition An Ò Ò array Ä with cells that are either empty or contain exactly one symbol is a partial latin square if no symbol occurs more than once in any row or column. The size of a partial latin square is its number of filled cells Theorem A partial latin square of order Ò and size at most Ò ½ can always be completed to a latin square of order Ò. Example: Non-completable latin square. ½ ¾ Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

8 Outline of the talk 1 Latin squares 2 3 Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

9 Definition Two latin squares Ä and Ä of the same order are orthogonal if Ä(, )=Ä (, ) and Ä (, )=Ä(, ), implies = and =. Equivalently, two latin squares of side Ò Ä=(, ) (on symbol set Ë) and Ä =(, ) (on symbol set Ë ) are orthogonal if every element in Ë Ë occurs exactly once among the Ò ¾ pairs (,,, ), ½, Ò. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

10 Definition A set of latin squares Ä ½,...,Ä Ñ is mutually orthogonal, or a set of MOLS, if for every ½ < Ñ, Ä and Ä are orthogonal. These are also referred to as POLS, pairwise orthogonal latin squares. Example: MOLS of side. ½ ¾ ¾ ½ ¾ ½ ½ ¾ ½ ¾ ½ ¾ ¾ ½ ¾ ½ ½ ¾ ¾ ½ ½ ¾ ¾ ½ Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

11 Æ(Ò) Maximum number of MOLS in a set of latin squares of side Ò. Convention: Æ(¼)=Æ(½)=. Æ(¾) =? ¼. Theorem For every Ò>½, ½ Æ(Ò) Ò ½. Theorem If Õ= Ô is a prime power, then Æ(Õ)=Õ ½. For each α F Õ {¼}, define the latin square Ä α (, )= +α, where, F Õ.The set of latin squares{ä α α F Õ {¼}} is a set of Õ ½ MOLS of side Õ. Æ( )=¾, Æ( )=, Æ( )=, Æ( )=, Æ( )=, Æ( )=. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

12 Æ(Ò) Maximum number of MOLS in a set of latin squares of side Ò. Convention: Æ(¼)=Æ(½)=. Æ(¾) =? ¼. Theorem For every Ò>½, ½ Æ(Ò) Ò ½. Theorem If Õ= Ô is a prime power, then Æ(Õ)=Õ ½. For each α F Õ {¼}, define the latin square Ä α (, )= +α, where, F Õ.The set of latin squares{ä α α F Õ {¼}} is a set of Õ ½ MOLS of side Õ. Æ( )=¾, Æ( )=, Æ( )=, Æ( )=, Æ( )=, Æ( )=. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

13 Given latin squares Ä ½ and Ä ¾, define Ä ½ Ä ¾ (Kronecker like product) (Ä ½ Ä ¾ )(Ü ½, Ü ¾ )(Ý ½, Ý ¾ ) =(Ä ½ (Ü ½, Ý ½ ), Ä ¾ (Ü ¾, Ý ¾ )). Example: Ä= ½ ¾ ¾ ½ ½ ¾, Å= ½ ¾ ¾ ½ Ä Å= (, ½) (½, ½) (¾, ½) (, ¾) (½, ¾) (¾, ¾) (¾, ½) (, ½) (½, ½) (¾, ¾) (, ¾) (½, ¾) (½, ½) (¾, ½) (, ½) (½, ¾) (¾, ¾) (, ¾) (, ¾) (½, ¾) (¾, ¾) (, ½) (½, ½) (¾, ½) (¾, ¾) (, ¾) (½, ¾) (¾, ½) (, ½) (½, ½) (½, ¾) (¾, ¾) (, ¾) (½, ½) (¾, ½) (, ½) Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

14 Lemma If{ } are MOLS of order Ò, =½ and{ } =½ Ñ, the{ } =½ are MOLS of order ÑÒ. are MOLS of order Theorem (McNeish) Æ(Ò Ñ) Ñ Ò{Æ(Ò), Æ(Ñ)}. Corollary If Ò=Ô ½ ½ Ô ¾ ¾...Ô Æ(Ò) Ñ Ò{Ô, where each Ô is a prime ½, then ½ = ½, ¾,..., }. Æ(Ò)>½, for Ò, odd and Ò=¼ ÑÓ ( ). Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

15 Problem Æ(Ò)=? for Ò=¾ ÑÓ ( ). In particular, Æ( )=? Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

16 Problem Æ(Ò)=? for Ò=¾ Euler s officers problem ÑÓ ( ). In particular, Æ( )=? Arrange officers from different regiments and different ranks in a array so that each row and each column contains one officer of each rank and one officer of each regiment. Conjecture (Euler) Æ( )=¼. Theorem (G. Tarry (exhaustive search), Fisher and Yates, Mann, Stinson) Æ( )=¼. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

17 Conjecture (Euler) Æ(Ò)=¼, for Ò=¾ ÑÓ ( ). Was open for ½ years. Bose and Shrikhande (1958) constructed MOLS of order ¾¾. Independently, Parker constructed MOLS of order ½¼. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

18 MOLS (contd..) Figure: Mutually orthogonal latin sqaures of order ½¼. (Source: Wolfram Mathworld) Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

19 MOLS (contd..) Theorem (Bose, Parker, Shrikhande) Æ(Ò) ¾, for all Ò ¾,. Theorem (Chowla, Erdős, Straus) Æ(Ò) for Ò. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

20 Definition (Transversal design) A transversal design of order or groupsize Ò, blocksize, and index λ, denoted TD λ (, Ò), is a triple(î,g,b), where 1 Î is a set of Ò elements; 2 G is a partition of Î into classes (the groups), each of size Ò; 3 B is a collection of -subsets of Î (the blocks); 4 every unordered pair of elements from Î is contained either in exactly one group or in exactly λ blocks, but not both. When λ=½, one writes simply TD(, Ò). A TD(, Ò) is a uniform -GDD of group size Ò. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

21 Outline of the talk 1 Latin squares 2 3 Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

22 Definition An orthogonal array of size Æ, with constraints (or of degree ), levels (or of order ), and strength Ø, denoted OA(Æ,,, Ø), is a Æ array with entries from a set of ¾ symbols, having the property that in every Ø Æ submatrix, every Ø ½ column vector appears the same number λ= Æ times. The parameter λ is Ø the index of the orthogonal array. An OA(Æ,,, Ø) is also denoted by OA λ (Ø,, ). If Ø is omitted, it is understood to be ¾. If λ is omitted, it is understood to be ½. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

23 Example: An OA ½ (,, ) Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

24 Theorem For Õ a prime power and ¾ Õ, there exists an OA(, Õ). Let ½,..., F Õ. Define two vectors in F Õ Ú ½ =(½,...,½) Ì Ú ¾ =( ½,..., ) Ì. and Next, construct a Õ ¾ array whose columns are indexed by F Õ F Õ, and(, )-th column is given by Ú ½ + Ú ¾ ; is OA(, Õ). Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

25 Theorem Suppose that Ò ¾ and. Then the existence of any one of the following designs implies the existence of the other two designs: 1 ( ¾) MOLS(Ò). 2 an OA(, Ò), 3 a TD(, Ò). (Ã ¾) MOLS(Ò) OA(, Ò) Let Ä ½, Ä ¾,...Ä ¾ be latin squares of order Ò on set Ë. An array with columns[,, Ä ½ (, ), Ä ¾ (, ),...,Ä ¾ (, )] Ì,, Ë is an OA(, Ò). Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

26 Example: =, Ò=. Ü Ý Þ Þ Ü Ý Ý Þ Ü Ü Ý Þ Ý Þ Ü Þ Ü Ý Ü= ½, Ý= ¾, Þ= Ü Ü Ü Ý Ý Ý Þ Þ Þ Ü Ý Þ Ü Ý Þ Ü Ý Þ Ü Ý Þ Þ Ü Ý Ý Þ Ü Ü Ý Þ Ý Þ Ü Þ Ü Ý Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

27 OA(, Ò) TD(, Ò) LetAbe a Ò ¾ array OA(, Ò). Define, Î={½,...,Ò} {½,..., } G={ ½ }, where ={½,...,Ò} { } B={ Ö ½ Ö Ò ¾ }, where Ö ={(A(, Ö), ) ½ }.. (Î,G,B) is a TD(, Ò) Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

28 Example (contd..): =, Ò=. Ü Ü Ü Ý Ý Ý Þ Þ Þ Ü Ý Þ Ü Ý Þ Ü Ý Þ Ü Ý Þ Þ Ü Ý Ý Þ Ü Ü Ý Þ Ý Þ Ü Þ Ü Ý Î={(Ü, ½),(Ý, ½),(Þ, ½),(Ü, ¾),(Ý, ¾),(Þ, ¾),(Ü, ),(Ý, ),(Þ, ), (Ü, ),(Ý, ),(Þ, )} G={{(Ü, ½),(Ý, ½),(Þ, ½)}, {(Ü, ¾),(Ý, ¾),(Þ, ¾)}, {(Ü, ),(Ý, ),(Þ, )}, {(Ü, ),(Ý, ),(Þ, )}} Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

29 ={{(Ü, ½),(Ü, ¾),(Ü, ),(Ü, )}, {(Ü, ½),(Ý, ¾),(Ý, ),(Ý, )}, {(Ü, ½),(Þ, ¾),(Þ, ),(Þ, )}, {(Ý, ½),(Ü, ¾),(Þ, ),(Ý, )}, {(Ý, ½),(Ý, ¾),(Ü, ),(Þ, )}, {(Ý, ½),(Þ, ¾),(Ý, ),(Ü, )}, {(Þ, ½),(Ü, ¾),(Ý, ),(Þ, )}, {(Þ, ½),(Ý, ¾),(Þ, ),(Ü, )}, {(Þ, ½),(Þ, ¾),(Ü, ),(Ý, )}} Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

30 Theorem (Bose, Bush) An OA λ (, Ú) exists only if λú ¾ ½. Moreover,if λ ½= Ú ½ ÑÓ (Ú ½), and ½< < Ú ½, then λú ¾ ½ Ú ½+ Ú(Ú ½ ) (¾Ú ¾ ½) ½ ½ ¾ Example: For λ=½, Ú= ½¼, there is no OA(, ½¼) for ½¾. For λ=, Ú= ½¼, there is no OA (, ½¼) for. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

31 Theorem Let Õ be a prime power and Ò ¼ be an integer. Then there exists an OA ¾Õ Ò(¾ ÕÒ+½ ½ ½, Õ). Õ ½ Theorem If an OA λ ( ½, ) and an OA µ ( ¾, ) both exist, then an OA λµ ¾(( ½ + ½)( ¾ + ½) ½, ) exists. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

32 Theorem An OA λ (Ø,, ) is an OA λ (Ø ½,, ). Theorem (Rao) An OA λ (Ø,, ) exists only if λ Ø ½+ Ø ½ ¾ Ø ¾ =½ ( )( ½) if Ø even, =½ ( )( ½) + ½ Ø ½ ¾ ( ½) Ø+½ ¾ if Ø odd. Example: For λ=½, Ø= ¾, = ¾¼, there is no OA(¾,, ¾¼) for ¾¾. For λ=½, Ø=, = ¾¼, there is no OA(,, ¾¼) for ¾½. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

33 Theorem (Bush) An OA ½ (Ø,, ) with Ø> ½ exists only if + Ø ½ if even and Ø, + Ø ¾ if odd and Ø, Ø+ ½ if Ø. Theorem (Bose, Bush) An OA λ (,, ) exists only if λ ¾ ½ +½. Moreover,if λ ½= ½ ÑÓ ( ½), and ½ < ½, then λ ¾ ½ ½+ ( ½ ) (¾ ¾ ½). ½ ¾ Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

34 Example: For λ=½, = ½¼, there is no OA(,, ½¼) for ½. For λ=, = ½¼, there is no OA (,, ½¼) for. Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

35 References C. j. Colbourn and J. H. Dinitz (Ed.), Handbook of Combinatorial Designs, Discrete Mathematics and its Applications, Second Edition, CRC Press, D. R. Stinson, Combinatorial Designs: Constructions and Analysis, Springer, J. H. van Lint and R. M. Wilson, A Course in Combinatorics, Second Edition, Cambridge University Press, Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

36 Thank You Bimal Roy, Indian Statistical Institute, Kolkata. and Orthogonal arrays (O

Overview of some Combinatorial Designs

Overview of some Combinatorial Designs Bimal Roy Indian Statistical Institute, Kolkata. Outline of the talk 1 Introduction 2 3 4 5 6 7 Outline of the talk 1 Introduction 2 3 4 5 6 7 Introduction Design theory: Study of combinatorial objects