Solutions 5. - University of St Andrews

UNIVERSITY OF ST ANDREWS
School of Mathematics and Statistics
MT4516 Finite Mathematics: Solutions 5.
1. Show that the number of n × n Latin squares is 1, 2, 12, 576 for
n = 1, 2, 3, 4 respectively.
The Latin squares of orders 1 and 2 are
1 ,
1 2
2 1 , 2 1
1 2 .
To count the Latin squares of order 3, first note that there are 3! = 6 ways of
permuting the values. Hence we may assume without loss of generality that the
first row is 123. The second can then be 231 or 312. In either case, the third row
is then fixed. So there are 6 · 2 = 12 Latin squares.
To count the Latin squares of order 4, first note that there are 4! = 24 choices for
the first row. Assume without loss of generality that the first row is 1234. Then
there are 9 possible choices for the second row:
2143, 2341, 2413, 3142, 3412, 3421, 4123, 4321, 4312.
If the second row is 2143 then there are four choices for the third row, namely 3412,
3421, 4312, 4321. The same is true if the second row is 3412 or 4321. However, if
the second row is one of 2341, 2413, 3142, 3421, 4123 or 4312 then there are only
two choices for the third row.
The fourth row is always uniquely determined by the previous three rows. Therefore
there are 24(3 · 4 + 6 · 2) = 576 Latin squares of order 4.
2. Prove that there is an 11 × 11 Latin square which cannot be obtained
from the multiplication table of a group by permuting the rows, columns
and entries. (Hint: You may use the fact that there is only one group
of order 11. )
Since 11 is prime, there is only one group of order 11. There are 11! ways to
permute the rows of its multiplication table, 11! ways to permute the columns,
and 11! ways to permute the symbols of the table. Thus in this way we obtain at
most (11!) 3 Latin squares.
We showed in lectures that there are at least 11! × 10! × · · · × 2! × 1 Latin squares
of order 11.
(11!) 3 = (11!) × (11 × 10!) × (11 × 10 × 9!)
= 11! × 10! × 9! × 1210
< 11! × 10! × 9! × 8! < 11! × 10! × 9! × 8! × 7! × · · · × 2! × 1.
So there are many Latin squares not arising from a group.
3. Let F be the finite field with four elements (see the lecture notes). Use
this field to construct three mutually orthogonal Latin squares of order
4.

Recall that the addition and multiplication tables for F are:
+ 0 1 x x + 1
0 0 1 x x + 1
1 1 0 x + 1 x
x x x + 1 0 1
x + 1 x + 1 x 1 0
× 0 1 x x + 1
0 0 0 0 0
1 0 1 x x + 1
x 0 x x + 1 1
x + 1 0 x + 1 1 x
Let f 1 = 1, f 2 = x, f 3 = x + 1 and f 4 = 0. The formula for the squares
A k = (a (k)
ij ) 4×4, for k = 1, 2, 3 is a (k)
ij
= f i f k + f j .
So a (1)
ij
= f i + f j .
a (2)
ij
= f i · x + f j .
a (3)
ij
= f i · (x + 1) + f j .
After renaming the symbols:
A 1 =
1 x 1 + x 0
1 0 x + 1 x 1
x x + 1 0 1 x
1 + x x 1 0 x + 1
0 1 x x + 1 0
f i xf i 1 x 1 + x 0
1 x 1 + x 0 1 x
x x + 1 x 1 0 x + 1
x + 1 1 0 x + 1 x 1
0 0 1 x x + 1 0
4 3 2 1
3 4 1 2
2 1 4 3
1 2 3 4
f i (x + 1)f i 1 x x + 1 0
1 x + 1 x 1 0 x + 1
x 1 0 x + 1 x 1
x + 1 x x + 1 0 1 x
0 0 1 x x + 1 0
, A 2 =
3 4 1 2
2 1 4 3
4 3 2 1
1 2 3 4
, A 3 =
2 1 4 3
4 3 2 1
3 4 1 2
1 2 3 4
4. Prove that there is no Latin square orthogonal to the square
A =
1 2 3 4
2 4 1 3
3 1 4 2
4 3 2 1
Assume that B = (b ij ) 4×4 is orthogonal to A. We may assume without loss of
generality that B is in standard form, so that b 11 = 1, b 12 = 2, b 13 = 3, b 14 = 4.
There are two possible choices for b 21 : either b 21 = 3 or b 21 = 4 since (a 12 , b 12 ) =
(2, 2). If b 21 = 3 then we deduce that b 22 = 1, b 24 = 2, b 23 = 4, b 31 = 4, b 32 = 3,
b 33 = 2, b 34 = 1. After this there is no possibility for b 41 .
If b 21 = 4 then b 31 = 2, b 41 = 3, b 23 = 2 and then there is no possible value for b 44 .
5. Construct two orthogonal Latin squares of order 12. (Hint: Use direct
products.)
.
.

Take two orthogonal Latin squares of order 4, say A 1 and A 2 from Q3, and two
orthogonal Latin squares of order 3, say
The direct products are:
B =
1 2 3
2 3 1
3 1 2
, C =
1 2 3
3 1 2
2 3 1
.
A 1 ×B =
B 4 B 3 B 2 B 1
B 3 B 4 B 1 B 2
=
B 2 B 1 B 4 B 3
B 1 B 2 B 3 B 4
10 11 12 7 8 9 4 5 6 1 2 3
11 12 10 8 9 7 5 6 4 2 3 1
12 10 11 9 7 8 6 4 5 3 1 2
7 8 9 10 11 12 1 2 3 4 5 6
8 9 7 11 12 10 2 3 1 5 6 4
9 7 8 12 10 11 3 1 2 6 4 5
4 5 6 1 2 3 10 11 12 7 8 9
5 6 4 2 3 1 11 12 10 8 9 7
6 4 5 3 1 2 12 10 11 9 7 8
1 2 3 4 5 6 7 8 9 10 11 12
2 3 1 5 6 4 8 9 7 11 12 10
3 1 2 6 4 5 9 7 8 12 10 11
A 2 ×C =
C 3 C 4 C 1 C 2
C 2 C 1 C 4 C 3
=
C 4 C 3 C 2 C 1
C 1 C 2 C 3 C 4
These are orthogonal Latin squares of order 12.
7 8 9 10 11 12 1 2 3 4 5 6
9 7 8 12 10 11 3 1 2 6 4 5
8 9 7 11 12 10 2 3 1 5 6 4
4 5 6 1 2 3 10 11 12 7 8 9
6 4 5 3 1 2 12 10 11 9 7 8
5 6 4 2 3 1 11 12 10 8 9 7
10 11 12 7 8 9 4 5 6 1 2 3
12 10 11 9 7 8 6 4 5 3 1 2
11 12 10 8 9 7 5 6 4 2 3 1
1 2 3 4 5 6 7 8 9 10 11 12
3 1 2 6 4 5 9 7 8 12 10 11
2 3 1 5 6 4 8 9 7 11 12 10
6. A Latin square is said to be self-orthogonal if it is orthogonal to its
own transpose.
We let A T = (b ij ) n×n where b ij = a ji .
(i) Prove that if A = (a ij ) n×n is a self-orthogonal Latin square then all
the entries a ii , 1 ≤ i ≤ n, are distinct.
If a ii = a jj then (a ii , b ii ) = (a ii , a ii ) = (a jj , a jj ) = (a jj , b jj ), which contradicts
self-orthogonality.
(ii) Prove that there are no self-orthogonal Latin squares of order 3;
find a self orthogonal Latin square of order 4.
Assume that A = (a ij ) 3×3 is self-orthogonal. By permuting the symbols if
necessary, we may assume without loss of generality that a 11 = 1, a 12 = 2
.

and a 13 = 3. From part (i) it then follows that a 22 = 3 and a 33 = 2. But then
there is no possibility for a 21 , since a 11 = 1 so a 21 ≠ 1, a 12 = 2 so a 21 ≠ 2
and a 22 = 3 so a 21 ≠ 3.
We use similar arguments to construct the following self-orthogonal Latin
square of order 4:
1 2 3 4
4 3 2 1
.
2 1 4 3
3 4 1 2
(iii) Prove that the direct product of two self-orthogonal Latin squares is
again a self-orthogonal Latin square. (Hint: You may use (A × B) T = A T × B T ).
If A is orthogonal to A T and if B is orthogonal to B T then A×B is orthogonal
to A T × B T = (A × B) T .
(iv) Let n > 3 be a prime power, and let F = {f 1 , f 2 , . . . , f n } be the finite
field with n elements. For an arbitrary t ∈ F, with t ≠ 0, 1 and
t + t ≠ 1, define an array A t = (a ij ) n×n by a (t)
ij
= tf i + (1 − t)f j . Prove
that A t is a self orthogonal Latin square.
First we show that A t is a Latin square. If a (t)
ij
= a (t)
ik
then
tf i + (1 − t)f j = tf i + (1 − t)f k
so (1 − t)f j = (1 − t)f k and hence f j = f k and so j = k. So there are no
repeats in any row.
If a (t)
ij
= a (t)
kj then
tf i + (1 − t)f j = tf k + (1 − t)f j
so tf i = tf k and hence f i = f k and i = k. So there are no repeats in any
column and A t is a Latin square.
Now we show that A t is self-orthogonal. Suppose that (a (t)
ij , a(t) ji ) = (a(t)
Then
Equation (1) implies that
Substituting into (2) we get:
tf i + (1 − t)f j = tf k + (1 − t)f l (1)
tf j + (1 − t)f i = tf l + (1 − t)f k (2)
f i = t −1 (tf k + (1 − t)f l + (t − 1)f j )
= f k + t −1 (1 − t)f l + t −1 (t − 1)f j .
kl , a(t) lk ).
tf j + f k + t −1 (1 − t)f l + t −1 (t − 1)f j − tf k − (1 − t)f l − (t − 1)f j = tf l + (1 − t)f k
⇒ f j (t + 1 − t −1 − t + 1) = f l (1 − t −1 − t + 1 + t)
⇒ f j (2 − t −1 ) = f l (2 − t −1 )
⇒ f j (2t − 1) = f l (2t − 1).
Since 2t ≠ 1 we deduce that f j = f l . Then (1) ⇒ f i = f k . So i = k and j = l.