the problem of the 36 officers kalei titcomb. 1780: no mutual pair of orthogonal latin squares of...
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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