orthogonal Latin squares

Given two Latin squares $L_1=(A,B,C_1,f_1)$ and $L_2=(A,B,C_2,f_2)$ of the same order $n$, we can combine them coordinate-wise to form a single square, whose cells are ordered pairs of elements from $C_1$ and $C_2$ respectively. Formally, we can form a function $f:A\times B\to C_1\times C_2$ given by

$$f(i,j)=(f_1(i,j),f_2(i,j)).$$

This function $f$ says that we have created a new square $A\times B$, whose cell $(i,j)$ contains the ordered pair of values, the first coordinate of which corresponds to the value in cell $(i,j)$ of $L_1$, and the second to the value in cell $(i,j)$ of $L_2$. We may write the combined square $L_1*L_2$.

For example,

In general, the combined square is not a Latin square unless the original two squares are equivalent: $f_1(i,j)=f_1(k,\mathrm{\ell })$ iff $f_2(i,j)=f_2(k,\mathrm{\ell })$. Nevertheless, the more interesting aspect of pairing up two Latin squares (of the same order) lies in the function $f$:

Definition. We say that two Latin squares are orthogonal if $f$ is a bijection.

Since there are $n^2$ cells in the combined square, and $|C_1\times C_2|=n^2$, the function $f$ is a bijection if it is either one-to-one or onto. It is therefore easy to see that the two Latin squares in the example above are orthogonal.

Remarks.

• The combined square is usually known as a Graeco-Latin square, originated from statisticians Fischer and Yates.

• It can be shown that if $L_1,\mathrm{\dots },L_m$ are Latin squares of order $n\ge 3$ such that each pair of them are orthogonal, then $m\le n-1$. If the equality occurs, then the set of Latin squares are said to be complete.

• (Bose) If $n\ge 3$, then $L_1,\mathrm{\dots },L_m$ form a complete set of pairwise orthogonal Latin squares of order $n$ iff there exists a finite projective plane of order $n$.