How do I construct two 4 by 4 orthogonal Latin Squares?

Question

I am constructing two, 4 by 4, orthogonal Latin Squares from the alphabet {$a,b,c,d$}. I have already created one Latin Square. Is there a method for constructing the other Latin Square or is it just trial and error?

Answer 1

How comfortable are you with finite field arithmetic? For latin squares of size a prime power $q$ there is a very nice construction for producing $q-1$ mutually orthogonal latin squares. Here it is for $q = 4$:

First identify the $4 \times 4$ grids with $\mathbb{F}_4^2$ in the obvious way. For each nonzero element $a$ of $\mathbb{F}_4$ we construct a latin square by labeling the points of $\mathbb{F}_4^2$ as follows: For a point $p \in \mathbb{F}_4^2$ take the line through $p$ of slope $a$ and label $p$ by where this line intersects the $x$-axis.

Since two points in $\mathbb{F}_4^2$ define a unique line it follows that these are mutually orthogonal latin squares.

Answer 2

According to Brendan McKay's data, there is only one main class of $4 \times 4$ orthogonal Latin squares up to:

• permutations of the rows, columns, and symbols in the first square,
• permutations of the rows, columns, and symbols in the second square, and
• permutations of the roles (row, column, symbol-1, symbol-2).

Here's the representative given:

$$\begin{bmatrix} a & d & b & c\\ c & b & d & a\\ d & a & c & b\\ b & c & a & d\\ \end{bmatrix}, \begin{bmatrix} a & d & c & b\\ d & a & b & c\\ c & b & a & d\\ b & c & d & a\\ \end{bmatrix}.$$

Both Latin squares will be isotopic to the Cayley Table of $\mathbb{Z}_2 \times \mathbb{Z}_2$, as the other non-isotopic Latin square of order $4$ (namely, the Cayley table of $\mathbb{Z}_4$) has no orthogonal mate.

Answer 3

Graeco Latin Squares are also referred to as Orthogonal Squares, in that each cell is comprised of two sets of symbols, each in a different Latin Square arrangement. Read more about it on my blog: http://www.glennwestmore.com.au/category/latin-squares/.