4 Mathematical Programming tools Mutually Orthogonal Latin Squares(MOLS)

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

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

716

 716 PhDs to solve the triple problem? 

R there 3 MOLS of size 10?

• Best work: McKay, Meynert & Myrvold (2006)

• Showed: approximately10

distinct pairs

• Concluded: Finding all pairs 1

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 = [6 6 6 0 0 0] & D2 = [6 6 4 2 0 0]

• 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 [6 6 4 2 0 0]?

Why no mate for [6 6 4 2 0 0]?

A constructive proof

• When only one symbol in I is diff from 1, 2, 3 we get the [6 6 4 2 0 0] 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 [6 6 4 2 0 0]?

Assigning a Freq Dist in L2 for a given FD in L1

We can do this systematically with IP

L2

L1 1 2 3 4 5 6 Given

Freqin L1

1       6

2       6

3     4

4   2

5 0

6 0

Freq

L2 even even odd odd even 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_kl = 1 if in I+IV k of L1 assigned to l of L2;

• Let x_kl = 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 = 2n² (2n² 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

only 6 more types of freq Dist of k values in D=6?

I+IV

(a)[6 6 2 2 2 0] & (b) [6 4 4 4 0 0]

(c) [6 4 4 2 2 0]

(d) [6 4 2 2 2 2] & (e) [4 4 4 4 2 0]

(f) [4 4 4 2 2 2]

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_kl = 1 if in I+IV k of L1 assigned to l of L2

• Let x_kl = 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)=[6 6 2 2 2 0] & (b)=[6 4 4 4 0 0]

• 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| 4 5 6

• |2 3 4| 5 6 1

• |3 4 5| 6 1 2

• 4 5 6 1 2 3

• 5 6 1 2 3 4

• 6 1 2 3 4 5

Hidden Mann in Rows & Cols 1, 3, 5

C1 C3 C5 R1 1 3 5

R2 3 5 1 R3 5 1 3

Freq of 1 to 6 [3 0 3 0 3 0]

So D = [6 6 6 0 0 0]

D = [6 4 4 2 2 0]

Orthogonal Mating Agency Trick 3 Check Upper & Lower Bounds

If I+IV in L1 has [6 6 2 2 2 0]

• 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)[6 6 2 2 2 0] & (b) [6 4 4 4 0 0]

(c) [6 4 4 2 2 0]

(d) [6 4 2 2 2 2] & (e) [4 4 4 4 2 0]

(f) [4 4 4 2 2 2]

A new proof for n=6

There are 8 freq distributions

Mann rules out [6 6 6 0 0 0] & [6 6 4 2 0 0]

Left with 6 listed as (a) to (f) earlier

(f) = [4 4 4 2 2 2] 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) = [6 4 2 2 2 2] in normalised form

• |1 2 3| 4 5 6

• |2 3 4| <1,5,6 ?> <1,6 ?> <1,5 ?>

• |3 5 6| <1,2 ?> <1,2,4 ?> <1,2,4 ?>

• whichever way the cells with ? are filled

• (f) = [4 4 4 2 2 2] emerges; e.g.,

• 1 2 3 4 5 6 Col 1, 4, 5 1 4 5

• 2 3 4 1 6 5 give 2 1 6

• 3 5 6 2 4 1 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

1 2 3 4

2 1 4 3

3 4 1 2

4 3 2 1

1 2 3 4

3 4 1 2

4 3 2 1

2 1 4 3

1 2 3 4

4 3 2 1

2 1 4 3

3 4 1 2

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

mate

Roughly 6,000+ triples qualify

Some details for n=10?

• Mann rules out 6 out of 141 FDs where

the most frequent k

to k

add up to 46 or more

• [10 10 10 10 10] to [10 10 10 8 8]

• 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