The Problem of the 36 Officers Kalei Titcomb

1780: No mutual pair of orthogonal Latin squares of order n=4 k+2. k=0, 1. 1900: 6 x 6 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} 1 2 2 1 1 2

Example: n=3 {1, 2, 3} {a, b, c} 1 2 3 1 c 2 b 3 ac b a 3 1 2 3 b 1 a 2 cb a c 2 3 1 2 a 3 c 1 ba c b 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} 1 3 2 c b a 2 1 3 b a c 3 2 1 a c b

Exampill n=2 Impossible 1 2 2 1 12 21 21 12 2 1 1 2 Have the pairs {1, 2} and {2, 1} twice Don’t have {1, 1} and {2, 2}

M for n=3 R r 1 1 r 2 1 r 3 1 r 1 c 2 c 1 N 1 c r 2 3 b 2 3 1 1 1 c 3 3 a r 3 5 6 1 1 1 2 c 4 1 1 1 2 a 1 3 c 3 1 c 1 1 1 b c 2 1 a 1 2 a L 2 b 1 c 3 1 C 1 b 1 1 R: {r 1, r 2, r 3} 1 1 1 N: {1, 2, 3} 1 1 8 1 1 1 5 3 1 8 9 1 1 1 6 1 1 7 1 2 7 1 9 1 1 1 C: 1{c 1, c 2, c 3} L: {a, b, c} 1 1 1

M for n=6 R r 1 : r 6 c 1 : c 6 1 : 6 a : f C N L 1 … 36 7 ones per row 6 or 4 ones per column

Lemma I Our 24 x 40 matrix must have dependencies in our rows (r 1, …, r 6)+(c 1, …, c 6)=0 (r 1, …, r 6)+(1, …, 6)=0 (r 1, …, r 6)+(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 {r 1, r 2, c 1, c 2, 1, 2, a, b} So Y’ has four from each group {r 3, r 4, r 5, r 6, c 3, c 4, c 5, c 6, 3, 4, 5, 6, c, d, e, f}

a 1 b 1 c 1 d 1 e 1 f 1 g 1 h 1 1 1 2 1 3 1 4 1 1 1 1 5 1 6 1 7 1 8 1 1 1 9 1 10 1 11 1 12 1 1 13 1 14 1 15 1 16 1 1 1

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….

a 1 b 1 c 1 d 1 e 1 f 1 g 1 h 1 1 1 2 1 3 1 4 1 1 1 1 5 1 6 1 7 1 8 1 1 1 9 1 10 1 11 1 12 1 1 13 1 14 1 15 1 16 1 1 1

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 nonexistence 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 1 2 3 3 2 1 1 1 1

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. 373 -376, 1984.

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