4 Mathematical Programming tools Mutually Orthogonal Latin Squares (MOLS) G Appa, R Euler, A Kouvela, D Magos, Y Mourtos, & A Tran

L1 = L2 = L1 & L2 SUPERIMPOSED A 3x3 LS and its Orthogonal mate In any LS Each element exactly once in each row & col 2 LS R orthogonal mates if, when superimposed, All n 2 pairs of n elements occur exactly once

Summary We use LP+IP+CP+Structural Properties of MOLS to show that a)a 6x6 LS cannot have an orthogonal mate b)for the open 10x10 triple problem: I. it is sufficient to consider orthogonal pairs with the northwest corner 5x5 sub-squares having a full house, i.e., having all 10 symbols II.Maximum no. of types of full house pairs is PhDs to solve the triple problem?

R there 3 MOLS of size 10? Best work: McKay, Meynert & Myrvold (2006) Showed: approximately10 15 distinct pairs Concluded: Finding all pairs 1 st not viable MMM used transversals We have been using LP+IP+CP combo CP has ‘All Diff’ & symmetry breaking rules Excellent for n < 10 but hopeless for n=10? Based on structural properties of orthogonality We provide a new tool – at least for CP

Extending an old theorem provides a dating agency for finding Orthogonal Mates i.e., Compatible Pairs

Outline Mann’s theorem (1944) extended A new proof for n=6 requiring analysis of only four 3x3 sub-matrices The Full House lemma Dating agency results for n=10 pairs Dating agency results for 10x10 tripls

Mann’s results (1944) Best for LS that can’t have a mate A Latin Square (LS) of size 4n+2 can’t have an orthogonal mate if it is too fat A remarkable result identifying properties of a LS that cannot have an orthogonal mate Euler (1782) had identified simple ones e.g., a circular matrix Mann did much more with a beautiful proof That I could not easily understand

Mann Extended We provide a constructive proof that delivers mush more than Mann’s results Take any 2 even dimensional LS If their characteristics don’t match they can’t be orthogonal mates Characteristics: Freq dist of symbols in sub-matrix I

Mann: Mateless LS of Dim 4n+2 < n+1 cells with symbols other than 1 to 2n+1 in the North West sub-sq I ? Toooo fat!Can’t have a mate!

Basics of Mann Illustrated Take a 6x6 LS L1 Divide it into: 4 sub-squares I to IV I in rows & cols 1 to 3; IV in rows & cols 4 to 6 Let k denote the value in a cell of L1; k=1,…,6 Each k value occurs equal no. of times in I & IV So the freq of k is even in I+IV If < 4 symbols in I+IV diff from k1 to k3, L1 is TOO FAT; it can’t have an orthogonal mate L1 = I II III IV

What does Mann eliminate for 4n+2=6? For n=1, i.e., 6x6 case, the following 2 cases of frequency distribution (D) of values in I+IV are eliminated. D1 = [ ] &D2 =[ ] Here we represent values in terms of decreasing frequency. So k1 occurs 6 times in I+IV, k2 also 6 times etc. Why no mate for L1 with [ ]?

Why no mate for [ ]? A constructive proof When only one symbol in I is diff from 1, 2, 3 we get the [ ] case of frequencies of 1 to 6 in I+IV in L1 This implies for I+IV in L2: a)All six l values must occur twice (for k = 1,2) b)No l value more than 4 times (0 freq for k = 5,6) b) Four l values to be paired with k=3 c) Two l values to be paired with k=4 Is such an assignment of k and l values possible? To check this systematically, we can solve an assignment problem

Why no mate for [ ]? Assigning a Freq Dist in L2 for a given FD in L1 We can do this systematically with IP L2 L Given Freq in L Freq L2 even odd even

IP to find all Compatible FDs in L2 for a given FD in L1of dim 2n? Let k & l = 1 to 2n be elements in L1 & L2 (respectively) Let x k l = 1 if in I+IV k of L1 assigned to l of L2; Let x k l = 0 otherwise Let b l €N + represent the freq. of element l in I of L2 Let a k represent the given freq. of element k in I of L1 (1) ∑ l x kl = 2a k for all k (Assign 2a k for each k) (2) ∑ k x kl - 2b l = 0 for all l (Assign each l even times) (3) ∑ l 2b l = 2n 2 (2n 2 elements in I+IV of L2) All distinct feasible solutions give compatible pairs If no feasible sol to (1), (2) & (3) L1 has no mate

What else can happen for D=6? only 6 more types of freq Dist of k values in I+IV (a)[ ] &(b)[ ] (c) [ ] (d) [ ] &(e)[ ] (f) [ ] They cannot be ruled out by Mann but we can rule out many combinations as incompatible Solve an IP to check if k & l values for a pair of distributions can be assigned orthogonally Easy to see that IP will rule out (a) & (b); (a) & (c) and (a) & (e) as incompatible

IP for checking compatibility of 2 LS of dim 2n Let k & l = 1 to 2n be elements in L1 & L2 Let x k l = 1 if in I+IV k of L1 assigned to l of L2 Let x k l = 0 otherwise b l represents the given freq. of element l in I of L2 a k represents the given freq. of element k in I of L1 ∑ l x kl = 2a k for all k (Assign 2a k for each k) (1) ∑ k x kl = 2b l for all l (Assign 2b l for each l) (2) If (1) and (2) satisfied, freq dist. compatible Only compatible pairs can be ortho mates Let them date!

Ruling out non-isomorphic pairs using characteristics of FDs No need to solve IPs for all possible pairs Tricks available to rule out many pairs If found incompatible (by tricks or IP) Don’t let them date for being mates Cuts out a lot… a lot… a lot… of pairs

Orthogonal Mating Agency Trick 1 Ignore Complements Take (a)=[ ] & (b)=[ ] If I+IV has (a) in L1, II+III has its complement (b) (d) & (e) also complements if a pair with (a) in L1 and (d) in L2 compared, No need to worry about (b) in L1 and (e) in L2 My Ortho Mates R your Ortho Mates

Orthogonal Mating Agency Trick2 Look 4 Hidden structures Hidden Mann structure Consider |1 2 3| |2 3 4| |3 4 5| Hidden Mann in Rows & Cols 1, 3, 5 C1C3C5 R R R Freq of 1 to 6 [ ] So D = [ ] D = [ ]

Orthogonal Mating Agency Trick 3 Check Upper & Lower Bounds If I+IV in L1 has [ ] For values l in I+IV of L2: 2 ≤ l ≤ 4 So for 6x6 (a) (b) & (c) only compatible with (f) Incompatible Pair of Lower & Upper Bounds means Can’t B Ortho Mates

Dating Agency Results for n=6 Given L1 and L2, if I+IV are incompatible, they cannot be mates Leaves only 1 case of L1 for 6x6 pairs So what is a case? STAY AWAKE!

Remember the 6 types of FDs? Of course you do Let me remind you anyway only 6 more types of freq Dist of k values in I+IV (a)[ ] &(b)[ ] (c) [ ] (d) [ ] &(e)[ ] (f) [ ]

A new proof for n=6 There are 8 freq distributions Mann rules out [ ] & [ ] Left with 6 listed as (a) to (f) earlier (f) = [ ] is the only mate for 3 of them – (a), (b) and (c) (f) is also a mate for (d) & (e) but not the only mate However, (d) and (e) are complements So 2 cases of L1 left (d) or (f)

A new proof for n=6 (continued) every case of (d) has a hidden (f) in it Only 4 cases of I with FD (f) LP solution for all 4 infeasible Q E D

More from Trick2 Hidden (f) in the first 3 rows of every (d) Consider (d) = [ ] in normalised form |1 2 3| |2 3 4| |3 5 6| whichever way the cells with ? are filled (f) = [ ] emerges; e.g., Col 1, 4, give give 3 2 4

Can every date lead to a mate? Compatible but not necessarily mates

Orthogonal Mating Agency Trick 4 Only FH cases matter A new Strategy for finding even dim. MOLS LOOK ONLY AT LS WITH A FULL HOUSE

A new lemma for 2nx2n MOLS Def. A sub-matrix has a full house (FH) if all 2n values are present in it Suppose there are m MOLS of size 2n For any nxn sub-matrices I and II in the m MOLS at least m-1 must have a FH in I or in II Or both We illustrate with 2n=4

Full House Lemma Sketch of a proof Suppose I does not have a full house in L1 So there is a k* with freq 0 in I of L1 But then in II of L1 k* has freq n And in II+III k* occurs 2n times So in L2 sub-matrix II and III must have a full house As m-1 LS are orthogonal to L1 all of them must have a full house in I or II

Dating Results for LS with right character For n=6 Leads to a new proof For n=10 ? Only 43 types of LS have a full house Or 1832 pairs qualify to date and be mates 10x10 Triples Only 716 pairs with a full house qualify to have a 3 rd mate Roughly 6,000+ triples qualify

Some details for n=10? Mann rules out 6 out of 141 FDs where the most frequent k 1 to k 5 add up to 46 or more [ ] to [ ] So what about (44|6) split to (26|24)? There are 135 freq. dist. not ruled out by Mann 43 have a FULL HOUSE (all 10 values present) There are 716 compatible pairs with FH in both

10x10 triple? The best one could say is: Complete search remains hopeless but not all tools have been employed yet Don’t give up on solving the 10x10 triple MOLS