Talk:Latin Squares in reduced form

From Rosetta Code

Four Reduced Latin Squares Of Order 4

I must be missing something here because everyone's answer is just the same Latin square shown four times. How is that four squares and not just one? --Puppydrum64 (talk) 21:04, 22 July 2022 (UTC)

Please inspect the answers more carefully. F# First square line 2 is 2 3 4 1. The second square is 2 1 4 3.--Nigel Galloway (talk) 08:35, 23 July 2022 (UTC)

OEIS A000315

The step of multiplying by n! * (n-1)! to compare to A002860 looks unnecessary. A002860 says (in its "Formula" section) that it is "n!*(n-1)!*A000315(n)". So, it would be more direct to just have us compare the number of reduced Latin Squares against OEIS A000315. Util (talk) 00:46, 13 July 2019 (UTC)

I hope that on completion of this task the contributer has appreciated that this is a key to generating all Latin Squares (A002860). I could add taking the set of order 4, creating for all 4! permutations of the columns all 3! permutations of rows 2 to 4 and verifying that each Latin Square so produced is unique, thus proving that it does so, if the consensus is that such is needed. Certainly anyone doubting it should do so. Adding OEIS A00031 as a reference not a problem.--Nigel Galloway (talk) 19:32, 13 July 2019 (UTC)