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*)/*k*_{i}
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 *k*_{1} + ...
+ *k*_{r}: 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 *n*^{2} further
squares *L*_{ij} 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 *L*_{ij} and replace each of its letters by
subscripted forms, using as subscripts all the numbers in that cell of
the original *M*.
### Example

A_{1} B_{1} |
C_{1} D_{1} |
A_{2} B_{2} |
C_{2} D_{2} |
A_{3} C_{3} |
B_{3} D_{3} |

C_{1} D_{1} |
A_{1} B_{1} |
C_{2} D_{2} |
A_{2} B_{2} |
B_{3} D_{3} |
A_{3} C_{3} |

A_{3} D_{3} |
B_{3} C_{3} |
A_{1} C_{1} |
B_{1} D_{1} |
A_{2} B_{2} |
C_{2} D_{2} |

B_{3} C_{3} |
A_{3} D_{3} |
B_{1} D_{1} |
A_{1} C_{1} |
C_{2} D_{2} |
A_{2} B_{2} |

A_{2} B_{2} |
C_{2} D_{2} |
A_{3} D_{3} |
B_{3} C_{3} |
A_{1} D_{1} |
B_{1} C_{1} |

C_{2} D_{2} |
A_{2} B_{2} |
B_{3} C_{3} |
A_{3} D_{3} |
B_{1} C_{1} |
A_{1} D_{1} |

This is a direct product if all the *L*_{ij} 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 *i*th
diagonal block, first replace A by *i*, then move that block to
the *i*th 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