The Problem of the 36 Officers

Kalei Titcomb

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

1900: 6x6 case.

812,851,200 possiblereduced 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

2 1

1 2

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

1 2 3

3 1 2

2 3 1

c b a

b a c

a c b

1 2 3

3 1 2

2 3 1

c b a

b a c

a c b

Orthogonal

Orthogonal Latin squares

Two Latin squares of order nEvery 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}

c b a

b a c

a c b

1 2 3

3 1 2

2 3 1

Exampill n=2

Impossible

1 2

2 1

2 1

1 22 1

1 2

1 2

2 1

Have the pairs {1,2} and {2,1} twice

Don’t have {1,1} and {2,2}

M for n=3R C N L 1 2 3 4 5 6 7 8 9

r1 1 1 1 1

r2 1 1 1 1

r3 1 1 1 1

c1 1 1 1 1

c2 1 1 1 1

c3 1 1 1 1

1 1 1 1 1

2 1 1 1 1

3 1 1 1 1

a 1 1 1 1

b 1 1 1 1

c 1 1 1 1

1c 2b 3a

3b 1a 2c

2a 3c 1b

r1

r2

r3

c1 c2 c3

1 2 3

4 5 6

7 8 9

R: {r1,r2,r3} C: {c1,c2,c3}N: {1,2,3} L: {a,b,c}

M for n=6R C N L 1 … 36

r1:

r6

c1:

c6

1:6

a: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)=0And 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 MMust 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}

a 11

b 11

c 11

d 11

e 11

f 11

g 11

h 11

1 11 11 11 11

2 11 11 11 11

3 11 11 11 11

4 11 11 11 11

5 11 11

6 11 11

7 11 11

8 11 11

9 11 11

10

11 11

11

11 11

12

11 11

13

11 11

14

11 11

15

11 11

16

11 11

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 11

b 11

c 11

d 11

e 11

f 11

g 11

h 11

1 11 11 11 11

2 11 11 11 11

3 11 11 11 11

4 11 11 11 11

5 11 11

6 11 11

7 11 11

8 11 11

9 11 11

10

11 11

11

11 11

12

11 11

13

11 11

14

11 11

15

11 11

16

11 11

G and H

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

5,16}

Lemma III

Everyone needs to be friends with everyone else in H

Everyone is friends only once in HNeighbors in G can’t be a group of

friends in H

WLOG1 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 evenNamely, 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 examplesOrthogonal Latin Squares and

examplesInformed them of the 6 x 6 case

Magic SquaresConstant

Construct magic squares of odd order

Activity I

Latin squaresdefinition and examples2 x 2 2 cases3 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

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1 2 3

3 1 2

2 3 1

1 3 2

2 1 3

3 2 1

Activity II

Orthogonal Latin Squaresdefinition and examples2 x 2 exampill

Used playing cards3 x 3 case4 x 4 case

Mentioned the 6 x 6 exampill

Activity III

Magic Squaresdefinition and examples

Constant using 1+…+k=k(k+1)/2got a little lost

Gave an example using the formula

Activity IV

How to construct a magic square of odd ordermethod of up one and over one

Gave a 3 x 3 example5 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 CaughmanJoe EdigerKaren MarrongellePSU Mathematics Department Faculty

and Staff Eileen Mitchell-Babbitt