The other day I wrote about orthogonal Latin squares. Two Latin squares are orthogonal if the list of pairs of corresponding elements in the two squares contains no repeats.

A self-orthogonal Latin square (SOLS) is a Latin square that is orthogonal to its transpose.

Here’s an example of a self-orthogonal Latin square:

1 7 6 5 4 3 2 3 2 1 7 6 5 4 5 4 3 2 1 7 6 7 6 5 4 3 2 1 2 1 7 6 5 4 3 4 3 2 1 7 6 5 6 5 4 3 2 1 7

To see that this Latin square is orthogonal to itself, we’ll concatenate the square and its transpose.

11 73 65 57 42 34 26 37 22 14 76 61 53 45 56 41 33 25 17 72 64 75 67 52 44 36 21 13 24 16 71 63 55 47 32 43 35 27 12 74 66 51 62 54 46 31 23 15 77

Each pair of digits is unique.

## Magic squares

As we discussed in the earlier post, you can make a magic square out of a pair of orthogonal Latin squares. If we interpret the pairs of digits above as base 10 numbers, we have a magic square because each row, column, and diagonal has the digits 1 through 7 in the 1s place and the same set of digits in the 10s place.

Typically an *n* × *n* magic square is filled with the numbers 1 through *n*². We can make such a magic square with a few adjustments.

First, we subtract 1 from every element of our original square before we combine it with its transpose. This gives us the following.

00 62 54 46 31 23 15 26 11 03 65 50 42 34 45 30 22 14 06 61 53 64 56 41 33 25 10 02 13 05 60 52 44 36 21 32 24 16 01 63 55 40 51 43 35 20 12 04 66

If we interpret the elements above as base 7 numbers, then the entries are the numbers 0 through 66_{seven} which equals 48_{ten}. If we add 1 to every entry we get all the numbers from 1 through 100_{seven} = 49_{ten}. Written in base 10, this is the following.

1 45 40 35 23 18 13 21 9 4 48 36 31 26 34 22 17 12 7 44 39 47 42 30 25 20 8 3 11 6 43 38 33 28 16 24 19 14 2 46 41 29 37 32 27 15 10 5 49

## Possible sizes

It’s surprising that orthogonal Latin squares exist as often as they do. Self-orthogonal Latin squares are more rare, and yet they exist for all sizes except *n* = 2, 3, or 6. (Source: The CRC Handbook of Combinatorial Designs.)

It’s hard to prove that self-orthogonal Latin squares exist, but it’s easy to verify the sizes where they *don’t* exist. There is no self-orthogonal Latin square of size 6 because there is no pair of orthogonal Latin squares of size 6. The latter is a hard theorem to prove, but given it, the former is trivial.

There are orthogonal Latin squares of size 3, but no self-orthogonal Latin squares of size 3. And it’s not hard to see why. Suppose there were such a square. Without loss of generality we can label the top row with 1, 2, and 3. Let *x* be the entry directly below 1.

1 2 3 x * * * * *

What could *x* be? Not 1, because elements in a column of a Latin square are unique. It can’t be 2 either, because otherwise when you join the square and its transpose would have two 22s. So *x* must be 3, and the element below *x* must be 2. But then when you join the square and its transpose you have two 32s. There’s not enough wiggle room for a 3 × 3 Latin square to be self-orthogonal.