The Problem of the 36 Officers Kalei Titcomb

1780: No mutual pair of orthogonal Latin squares of order n=4k+2. k=0, : 6x6 case. 812,851,200 possible reduced to 9,408 pairs 1984: short, four page, noncomputer proof.

Latin Square of order n n x n array of n different symbols Each symbol occurs once in each row and once in each column Example n=2 {1,2}

Example: n=3 {1,2,3}{a,b,c} cba bac acb cba bac acb Orthogonal

Orthogonal Latin squares Two Latin squares of order n Every ordered pair of symbols occurs once when superimposed Example: n=3 {1,a} {1,b} {1,c} {2,a} {2,b} {2,c} {3,a} {3,b} {3,c} cba bac acb

Exampill n=2 Impossible Have the pairs {1,2} and {2,1} twice Don’t have {1,1} and {2,2}

M for n=3 RCNL r11111 r21111 r31111 c11111 c21111 c a1111 b1111 c1111 1c2b3a 3b1a2c 2a3c1b r1 r2 r3 c1c2c R: {r1,r2,r3}C: {c1,c2,c3} N: {1,2,3}L: {a,b,c}

M for n=6 RCNL1 … 36 r1 : r6 c1 : c6 1:61:6 a:fa:f 7 ones per row 6 or 4 ones per column

Lemma I Our 24 x 40 matrix must have dependencies in our rows (r1,…,r6)+(c1,…,c6)=0 (r1,…,r6)+(1,…,6)=0 (r1,…,r6)+(a,…,f)=0 And one more in addition to these! (proof uses properties of the Latin square) Call that set of rows Y.

Lemma II Must have submatrices Y and Y’ of M Must still have same amount of ones in each row (and if you sum them, each column) By a counting argument, Y must have 8 or 12 rows. So… Y’ has 16 or 12 rows

Y=8 and Y’=16 By some more counting, we find that - Y has two from each group {r1, r2, c1, c2, 1, 2, a, b} So Y’ has four from each group {r3, r4, r5, r6, c3, c4, c5, c6, 3, 4, 5, 6, c, d, e, f}

a1 b1 c1 d1 e1 f1 g1 h

Y=8 and Y’=16 By some more counting, we find that - Y has 28 columns with 2 ones and the rest have no ones. This splits our matrix into 6 parts….

a1 b1 c1 d1 e1 f1 g1 h

G and H Y={a,b,c,d,e,f,g,h} Q={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}

Lemma III Everyone needs to be friends with everyone else in H Everyone is friends only once in H Neighbors in G can’t be a group of friends in H

WLOG 1 is friends with 5,9,13 {1,5,9,13} are neighbors Blocks {1,6,10,14} {1,7,11,15} {1,8,12,16}

Lemma IV These graphs show Y cannot be of size 8. Therefore Y must be of size 12 according to Lemma II. We have five cases of how these 12 rows are spread among the 4 groups: {6,6,0,0} {6,4,2,0} {6,2,2,2} {4,4,4,0} {4,4,2,2}

Lemma V We can’t have Y=12! This is because: {6,6,0,0} is already a dependency {6,4,2,0}, {4,4,4,0}, {6,2,2,2}, {4,4,2,2} Sum of two even subsets (mod 2) is even Namely, of size 4 and 8, which is not possible

Lemma II through V gives us our non- existence of a pair of 6 x 6 orthogonal Latin squares!

So beware the puzzle 36 Cube by Thinkfun!

Curriculum Latin Squares and examples Orthogonal Latin Squares and examples Informed them of the 6 x 6 case Magic Squares Constant Construct magic squares of odd order

Activity I Latin squares definition and examples 2 x 2 2 cases 3 x 3 12 cases Had them find all 12, or as many as they could Gave an example of how to prove all 12 had been found

Two of each case

Activity II Orthogonal Latin Squares definition and examples 2 x 2 exampill Used playing cards 3 x 3 case 4 x 4 case Mentioned the 6 x 6 exampill

Activity III Magic Squares definition and examples Constant using 1+…+k=k(k+1)/2 got a little lost Gave an example using the formula

Activity IV How to construct a magic square of odd order method of up one and over one Gave a 3 x 3 example 5 x 5 example in groups got a little lost Ran out of time for construction of even order magic square for multiples of 4.

Reference Stinson, D.R.. “A Short Proof of the Nonexistence of a Pair of Orthogonal Latin Squares of Order Six”. Journal of Combinatorial Theory, Series A, Volume 36, pg , 1984.

Thank You John Caughman Joe Ediger Karen Marrongelle PSU Mathematics Department Faculty and Staff Eileen Mitchell-Babbitt