# 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,

 $\left(\begin{array}[]{cccc}a&b&c&d\\ c&d&a&b\\ d&c&b&a\\ b&a&d&c\end{array}\right)*\left(\begin{array}[]{cccc}1&2&3&4\\ 4&3&2&1\\ 2&1&4&3\\ 3&4&1&2\end{array}\right)=\left(\begin{array}[]{cccc}(a,1)&(b,2)&(c,3)&(d,4)\\ (c,4)&(d,3)&(a,2)&(b,1)\\ (d,2)&(c,1)&(b,4)&(a,3)\\ (b,3)&(a,4)&(d,1)&(c,2)\end{array}\right)$

In general, the combined square is not a Latin square unless the original two squares are equivalent     : $f_{1}(i,j)=f_{1}(k,\ell)$ iff $f_{2}(i,j)=f_{2}(k,\ell)$. Nevertheless, the more interesting aspect of pairing up two Latin squares (of the same order) lies in the function $f$:

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.

## References

• 1 H. J. Ryser, Combinatorial Mathematics, The Carus Mathematical Monographs, 1963
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