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I manually arranged two complete latin squares from ordered pairs of 4 and 6 like so:

1 2 3 4
2 4 1 3
3 1 4 2
4 3 2 1

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

I like this kind of sequence, but can tell it is going to be more difficult to manually arrange as the number of pairs increase.

In a related question about calculating row-complete latin squares, the following method generated a slightly different arrangement:

For any even n, say n = 2m, a row complete latin square of order n can be formed by writing down

0, 1, 2m - 1, 2, 2m - 2, 3,..., m + 1, m

as the first row and then developing subsequent rows by adding 1 modulo n.

0 1 3 2
1 2 0 3
2 3 1 0
3 0 2 1

which can be made column-complete by rearranging the rows (incremented for comparison with the first example)...

1 2 4 3
2 3 1 4
4 1 3 2
3 4 2 1

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

What I'm trying to figure out is how to generate the first pattern with a method similar to that row-complete calculation, or a way to convert the above squares into the first pattern by swapping some values.

It looks like they both have their values reversed in rows and columns in the same way, yet their sequences are different.

That makes it seem like a simple matter, but I'm not finding a simple (enough) pattern between them.

For example putting one of each square together and looking at it:

11 22 63 34 55 46
22 34 11 46 63 55
63 11 55 22 46 34
34 46 22 55 11 63
55 63 46 11 34 22
46 55 34 63 22 11

      63 34    46
   34    46 63   
63          46 34
34 46          63
   63 46    34   
46    34 63 

Does anyone see a way of transposing one to the other, or modifiying the row-complete calculation to match the other row sequence?

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What about when you try this method for constructing your latin square: Arrange the n elements in any order for the first column. Successive columns are constructed using the difference set: 1, n-2, 3, n-4, 5, n-6, ......., n-1 ? –  Djaian Oct 28 '10 at 11:43

1 Answer 1

I'm sorry I didn't see your question earlier. The answer is in fact quite easy: Just permute the symbols with 3->6, 4->3 and 6->4 (or its inverse if you want to go in the other direction).

In general, you can permute the symbols arbitrarily in a Latin square and it will preserve row-completeness and column-completeness (or lack thereof).

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PS. Annoyed at missing this question, I tagged the relevant questions latin-square. –  Douglas S. Stones Nov 19 '10 at 10:49

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