A * transversal* in a Latin Square is a collection of positions, one from each row and one from each column, so that the colors that are in these positions are all different.

Here is an example of a transversal in a 4×4 Latin square

Now look at the bottom squares of the Graeco-Latin square that correspond to this transversal of the top Latin square. Notice that all these squares have the same color!

If you now look at the small squares with the same color in the top Latin square, you'll notice that they are on a transversal of the big (bottom) Latin square. So, if you have a Graeco-Latin square, then the transversals of one correspond to the positions with the same color in the other, and vice-versa.

So, how can we use this idea? Suppose you are trying to make a Graeco-Latin square. If you start with a bottom square and **can** break it up into transversals which don't overlap each other, then you can make a second Latin square which will form a Graeco-Latin pair with the first. You form the second square by taking each of these transversals and putting the same color in each of the positions. Use a different color for each transversal. Here's an example. We first start by breaking up a Latin square into non-overlapping transversals.

Now we form a new square by coloring each transversal with the same color and put them back together.

Looking (carefully) at this final square and the one we started with, we see that they form a Graeco-Latin pair. In fact, this always works ... the problem is finding the non-overlapping transversals. If they don't exist, then you can't form a Graeco-Latin pair from your starting Latin square.

There are more transversals of the small square than just the 4 I used. Can you find them all? Can you break up the small square by using a different collection of non-overlapping transversals?

Look at the complete Latin square of order 5 that we have just constructed. We are trying to find non-overlapping transversals. Did you spot the transversal that goes down the diagonal? Now look at the positions just above the diagonal and also the bottom left corner. Yep, another transversal. In the picture below I've marked 5 non-overlapping transversals with lines of different colors.

You should build other cyclic Latin squares of different orders, both odd and even. Notice that only when the order is odd do you get the transversals in the same pattern as the order 5 example.

The picture below shows an unfinished Latin square with 4 transversals drawn in. You should find and record (you will need this information later) each of the 4 transversals.

If you would like to see more transversals, click on the number you want to see:

Jay's solution to the 10 × 10 Graeco-Latin Square.