3360 LECTURES. R. Craigen. October 15, 2016

Transcription

1 3360 LECTURES R. Craigen October 15, 2016

2 Introduction to designs Chapter 9 In combinatorics, a design consists of: 1. A set V elements called points (or: varieties, treatments) 2. A collection B of subsets of V, called blocks (or lines) 3. Generally we impose further conditions or structures. What sort of conditions and structures? Some possibilities: regularity: all points appear in the same number of blocks balance: any two points appear together in the same number λ of blocks (λ is the index) constant block size Blocks might be ordered lists of points A block might be all of V (i.e., complete). If all blocks are complete it is a complete design; else it is incomplete there may be repeated blocks; or these might be forbidden

3 Example A simple design EG: V = {1, 2, 3}, B = {B 1, B 2, B 3, B 4 }, where B 1 = {1, 2, 3}, B 2 = {1, 2}, B 3 = {1, 3}, B 4 = {2, 3} What can we say about this design? it is regular (each point is in r = 3 blocks) it is balanced (index λ = 2) there are no repeated blocks there is a complete block. But the design is not complete it does not have constant block size Note that eliminating the complete block B 1 from B leaves a (r = 2), balanced (λ = 1) design with constant block size (k = 2). The replication number r of a regular design is the number of blocks in which each point occurs.

4 Latin squares In these terms a Latin square is a complete design where, for some k (the order), V = {1, 2,..., k} (symbols) and k (complete) blocks, in which elements are listed in some order, so that no point appears in the same position in two blocks EG A Latin square of order 5 (blocks are columns): NOTES: equivalently, every point appears once in every position the rows also form a Latin square So we could think of this as two complete designs with blocks ordered a certain way... a pretty complex idea...

5 Latin squares Abstract definition A Latin square of order n is an n n array with entries from an n symbol set (by default V = {1, 2,..., n}); each symbol appears exactly once in each row and once in each column. Easy construction: List the symbols to make the first row. Circulate the row one position to the right for the next row. Repeat this until there are n rows. EG: the 5 5 square on the last slide: (row 1) (circulate row 1) (circulate row 2) (circulate row 3) (circulate row 4) Such an array is called circulant. This construction gives a Latin square of every order there is always a circulant one!

6 More examples of Latin squares Not all Latin squares are circulant Any n symbols may be used, not just {1, 2,..., n}. Some small Latin squares: α β β α a b c d d a b c c d a b b c d a a b c d c d a b b a d c d c b a The 3 3 square above illustrates a back-circulant array. EG 9.1 a simple experimental use of Latin squares: Test 7 medical treatments (EG different drugs or in this case dentures) on 7 patients. Each patient tries each treatment for one month so the experiment takes 7 months. Experimental design: - Rows months - Columns patients - Symbols treatment (kind of dentures used) - Arrange these into a Latin square.

7 Orthogonal latin squares - an example Another experiment. This time we test two kinds of things (a) 4 tire brands 1,2,3,4; (b) 4 brake linings a,b,c,d: - one tire on each of four cars (= columns) - one tire in each of four positions (= rows) using a Latin square for the design. Same setup for brake linings (... also use a Latin square) Tires car: A B C D L. front R. front L. rear R. rear Brakes car: A B C D LF a b c d RF b a d c LR c d a b RR d c b a Why might this be a poor arrangement for such an experiment? A design to test every tire with every brake lining: (Table 9.10) & d a b c c b a d a d c b b c d a = (1, d) (2, a) (3, b) (4, c) (2, c) (1, b) (4, a) (3, d) (3, a) (4, d) (1, c) (2, b) (4, b) (3, c) (2, d) (1, a)

8 Orthogonal latin squares (and MOLS) Two Latin squares A = [a ij ] and B = [b ij ] of order n are orthogonal if the combined array C(A, B) = [(a ij, b ij )] has each of the n 2 possible symbol pairs (a, b) in exactly one position. EG: circulant and back-circulant Latin squares of order 3: & = (1, 1) (2, 2) (3, 3) (3, 2) (1, 3) (2, 1) (2, 3) (3, 1) (1, 2) EG: Latin squares of order 2. Why not? There is (essentially) only one Latin square of that size! If several Latin squares are each orthogonal to all the others together they are called mutually orthogonal Latin squares (MOLS). (RT uses less-common term orthogonal family).

9 Existence of MOLS BASIC Q: For given r and n, do there exist r MOLS of order n? EG: 3 MOLS of order 4 (Table 9.11): Theorem 9.1: If there exist r MOLS of order n > 1 then r n 1. Proof: Let A (1), A (2),..., A (r) be MOLS of order n Relabelling the symbols of one array in any way (say {1, 2,..., n} {x 1,..., x n }) does not affect orthogonality Now, relabel the entries of each array so they begin: 1 2 n x n x n x r x k {2,..., n}; for p q, x p x q. By PHP, r n 1.

10 MOLS: More existence results A set of n 1 MOLS of order n is said to be complete (or a complete orthogonal family of Latin squares) EG: We have already seen 2 MOLS of order 3 (thus complete) and also 3 MOLS of order 4 (also complete) Theorem 9.2: Suppose n = p k, where k Z + and p is prime (that is, n is a prime power). Then there is a complete set of MOLS of order n. Proof: We shall see soon after we develop some machinery EG: A complete set of MOLS of order 5 (Table 9.18)

11 The state of the art for small MOLS Euler s 36 Officer Problem Theorem 9.2 gives complete sets of MOLS in orders 2, 3, 4 (= 2 2 ), 5, 7, 9 (= 3 2 ), 11, 13,... But the theorem does not say anything about orders that are not prime powers. What is known of order 6? Euler s 36 Officer Problem (1779): 6 regiments each have 6 officers of 6 different ranks. Can all 36 officers be arranged in a square formation so that each row and each column contains one officer of each rank and also one of each regiment? The array formed by ranks of officers in such a formation is a Latin square of order 6 The array formed by the regiments to which they belong also forms a Latin square of order 6 Since no two officers of any regiment have the same rank all 36 (rank, regiment) combinations must occur Therefore the two Latin squares must be orthogonal

12 MOLS of orders 6, 10 and 12 Euler believed there s no solution to his 36 Officer Problem But he was unable to prove this G. Tarry (1900) proved that Euler was right there do not exist 2 MOLS of order by reducing the over 800 million possible pairs of Latin squares of order 6 to 9408 and systematically eliminating Euler conjectured (1782): no MOLS of order 2 (mod 4) Shown false (Bose, Shrikhande, Parker 1960): Theorem 9.4: If n > 6, n 2 (mod 4) there are MOLS of order n Lam, Thiel, Swiercz (1989): no complete set of (9) MOLS of order 10 (massive computer search taking several years) It is still not known if there are 4 MOLS of order 10 Not known if a complete set of MOLS of order 12 exists Best so far: 5 (Johnson, Dulmage, Mendelsohn, 1961)

13 MOLS with holes of order 10 A partial Latin square is a k k array with entries from among k symbols and possibly some entries left blank (holes), in which no symbol appears twice in any row or column. Partial Latin squares are orthogonal if pairing corresponding entries gives no pair twice. If orthogonal partial Latin squares share the same holes they are called MOLS with holes. Brouwer (1984) found 4 MOLS of order 10 sharing a 2 2 hole Fig. 1. Four MOLS of order 10 with a hole of size 2

14 prehensive enumerations of the sort undertaken in the previous sections. However, giv tremendous interest in the existence otherwise of a triple of MOLS of order 10 ( ] and Thethe latest references on therein), MOLS we ofdid order use our 10programs to investigate the latin squa h autoparatopism groups of order at least 3. A catalogue of these squares was produc the authors of [24]. It was already established in [24] that none of these squares is Judith Egan and Ian Wanless (Monash University, 2014) found the first set of 3 MOLS of order 10: triple of MOLS. However, some of them come much closer than any previously kno mples, as we discovered. Consider the following three squares A = , B = , C = are A is orthogonal to both B and C. When B and C are overlayed, 91 different pa produced out of a possible 100. Moreover, the only duplicated pairs involve symb (Unknown at the time of writing of RT (2009) text reports,9 in C. We conclude that A and B have 7 common transversals. The previously b blished best resultknown showed so a pair far : of 2MOLS by Bose, of order Shrikhande 10 with 4and common Parker transversals in 1960) [2]. Note that A is semisymmetric and A, B and C all have the automorphi (123)(456)(789). Constructing MOLES in an ad-hoc manner is not easy! knowledgments. This research was supported in part by an Australian Mathemati

15 Multiplication of arrays Notation. Let B = [b ij ] n n be any array. Let a be a symbol. Define (a, B) to be the array whose (i, j) entry is (a, b ij ). That is, (a, B) = [(a, b ij )] n n. EG: If B is the circulant Latin square with first row (1 2 3) then (a, 1) (a, 2) (a, d) (a, B) = a, = (a, 3) (a, 1) (a, 2) (a, 2) (a, 3) (a, 1) By extension we define the product of two arrays as follows. If A = [a ij ] m m then (A, B) is the mn mn partitioned array (a 11, B) (a 1m, B).... (a m1, B) (a mm, B) = (a 11, b 11 ) (a 11, b 1n ) (a 1m, b 11 ) (a 1m, b 1n ) (a 11, b n1 ) (a 11, b nn) (a 1m, b n1 ) (a 1m, b nn) (a m1, b 11 ) (a m1, b 1n ) (a mm, b 11 ) (a mm, b 1n ) (a m1, b n1 ) (a m1, b nn) (a mm, b n1 ) (a mm, b nn)

16 Multiplication of Latin squares A and B: Latin squares of orders m and n (A, B) is an array of symbols corresponding to pairs (a, b) (a runs over the symbols in A; b runs over the symbols in B) EG: x y y x, = x1 x2 x3 y1 y2 y3 x3 x1 x2 y3 y1 y2 x2 x3 x1 y2 y3 y1 y1 y2 y3 x1 x2 x3 y3 y1 y2 x3 x1 x2 y2 y3 y1 x2 x3 x1 Entries of (A, B) are the mn pair symbols (a, b) ab each symbol appears once in every row each symbol appears once per column so (A, B) is also a Latin square of order mn.

17 Multiplication of MOLS (a ij, b uv ) (u, v)-entry of the (i, j)-block of (A, B) Quadruple (i, j, u, v) specifies a unique position in (A, B) Let A, C: MOLS of order m and B, D: MOLS of order n Suppose (a ij, b uv ), (c ij, d uv ) = (a hk, b st ), (c hk, d st ) It follows that (a ij, c ij ) = (a hk, c hk ) A, C are orthogonal so we must have (i, j) = (h, k) Similarly (b uv, d uv ) = (b st, d st ) so (u, v) = (s, t) It follows that (i, j, u, v) = (h, k, s, t) Thus pairs of corresponding entries in (A, C) and (B, D) must all be distinct. So (A, C) is orthogonal to (B, D). Theorem 9.6 (MacNeish, 1922): If there exist r MOLS of both orders m and n then there also exist r MOLS of order mn Proof: Reason as above with products (A 1, B 1 ),... (A r, B r )

18 Existence of MOLS of non-prime power orders Recall: Theorem 9.2: Suppose n = p k, k Z + and p prime. Then there is a complete set of MOLS of order n. What of non-prime power orders? Theorem 9.3 Suppose n Z + has prime power decomposition p t 1 1 pt 2 2 pts s. Let r = min { p t 1 1 1, pt 2 2 1,..., pts s 1 }. Then there is a set of r MOLS of order n. Proof: Apply Theorem 9.6 recursively among orders p t i i. OBS: This value is nowhere near n 1 except for prime powers! EG: For n = 6, r = 1. For n = 10, r = 1. For n = 15, r = 2. Corollary 9.3.1: If n Z + has prime power decomposition 2 t 1 p t 2 2 pts s and t 1 1 then there is a pair of MOLS of order n. Bose, Shrikhande, Parker construct pairs for all orders n = 2q, q odd except n = 2, 6. (Theorem 9.4).

19 Balanced Incomplete Block Designs (BIBDs) 9.4 Recall: A Design is a set V of points (or varieties ) together with a collection B = {B 1,..., B b } of subsets ( blocks ) of V We focus on a particular kind, BIBDs, which satisfy All blocks have the same number, k, of points (0 < k < v) All points have the same replication number, r > 0 Each pair of points appear together in the same number (the index, λ) of blocks Notation: BIBD(v, b, r, k, λ), where V = v 2, B = b > 0 EG: A BIBD(7, 7, 3, 3, 1): V = {0, 1, 2, 3, 4, 5, 6}, B = {013, 124, 235, 346, 450, 561, 602} NOTE: Text uses (b, v, r,...) This is wrong!