Here are some methods of creating semi-Latin squares.
Inflation
Take an n×n Latin or semi-Latin
square with s letters per cell. Replace
each letter by r new letters. This gives
an (n×n)/(sr) semi-Latin
square. It is the r-fold inflation
of the original square.
Superposition
Take Latin or semi-Latin squares of size
(n×n)/ki
for i = 1, ..., r. The squares
must all have their rows labelled in the same
way, and also their columns, and the letters
in any one square must not appear in any other
square. The superimposition of these squares
has blocks of size k1 + ...
+ kr: in the block in row
a and column b put all the letters
that occur in this block of any of the squares.
Inflation is the special case of superposition
that arises when the original squares are the
same as each other apart from a renaming of
letters. Superimposition of mutually orthogonal
Latin squares gives a Trojan square.
Group method
Let G be a group of order nk.
Suppose that G has subsets H,
R and C of sizes k,
n and n respectively, such that
G = RH = HC. Label the
rows and columns of an n×n
square by the elements of R and C
respectively. Then the elements in the block
in row r and column c are all
those of the form rhc for h
in H.
If H is a subgroup of G then
this method gives an inflated Latin square.
Relaxed Product
Take an (n×n)/s
Latin or semi-Latin square M
on the numbers 1, ... ns,
and n2 further
squares Lij of size
(m×m)/r all on the same set of mr
letters. In row i and column j of M put the whole
of Lij and replace each of its letters by
subscripted forms, using as subscripts all the numbers in that cell of
the original M.
Example
|
A1 B1 |
C1 D1 |
A2 B2 |
C2 D2 |
A3 C3 |
B3 D3 |
|
C1 D1 |
A1 B1 |
C2 D2 |
A2 B2 |
B3 D3 |
A3 C3 |
|
A3 D3 |
B3 C3 |
A1 C1 |
B1 D1 |
A2 B2 |
C2 D2 |
|
B3 C3 |
A3 D3 |
B1 D1 |
A1 C1 |
C2 D2 |
A2 B2 |
|
A2 B2 |
C2 D2 |
A3 D3 |
B3 C3 |
A1 D1 |
B1 C1 |
|
C2 D2 |
A2 B2 |
B3 C3 |
A3 D3 |
B1 C1 |
A1 D1 |
This is a direct product if all the Lij are
the same.
Deletion-augmentation
Start with a semi-Latin square formed by the orthogonal superposition
of an n×n Latin square L and an
(n×n)/(n-2) semi-Latin square. Arrange it
so that a given letter A of L is on the diagonal. Add an extra
row and column, and ``letters'' 1, ..., n. In the ith
diagonal block, first replace A by i, then move that block to
the ith position in the new row and column, then put all of 1,
..., n except i on the diagonal. Finally, put all
letters of L except A in the cell in the new row and column.
Example
| 2 3 4 | D E F | G H I | J K L |
B C 1 |
| D K I | 1 3 4 | J E C | G B F |
H L 2 |
| G E L | J B I | 1 2 4 | D H C |
K F 3 |
| J H F | G K C | D B L | 1 2 3 |
E I 4 |
| B C 1 | H L 2 |
K F 3 | E I 4 |
D G J |
Page maintained by
R. A. Bailey
Page updated 4/8/00