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Let’s have two Latin Squares 6X6 superimposed on each other, represented here by marking one with 1s and one with 0s. The two Latin Squares contain twelve 5,2 pairs, a total of twelve pairs 6,4 and 4,3 there is, and also a total of twelve pairs 3,1 and 1,6

Does any one know which two Latin Squares satisfy the above requirements?

NOTE: It is self-evident that twelve pairs is the maximum number of a given pair, e.g. 5,2.

superimposed Latin Squares

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  • $\begingroup$ What is meant by pair? They are adjacent? $\endgroup$ – Moti Jun 30 '19 at 4:06
  • $\begingroup$ They are in the same column, as in the figure, with 1 above, 0 below. Together they form a pair. $\endgroup$ – Vassilis Parassidis Jun 30 '19 at 5:46
  • $\begingroup$ If I interpret your input - all the 5 and 2 in the two squares are at the same locations. Right? $\endgroup$ – Moti Jun 30 '19 at 5:59
  • $\begingroup$ The numbers can be anywhere in the Latin Squares. A pair comprises numbers from each of the two superimposed Latin Squares, one above the other. 5 up, 2 down or 2 up, 5 down. $\endgroup$ – Vassilis Parassidis Jun 30 '19 at 6:12
  • $\begingroup$ I am not sure how you can construct the 5 pair combination simultaneously unless you ask to construct any? $\endgroup$ – Moti Jun 30 '19 at 6:16
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If I'm understanding the question correctly, any 6x6 Latin square can be paired with another using the substitutions (2<->5)(->6->4->3->1->). For example:

123456
234561
345612
456123
561234
612345

651324
513246
132465
324651
246513
465132

In fact, I suspect all solutions will be of this form, meaning there are exactly 1,625,702,400 solutions, two for each 6x6 Latin square, reversing the arrows of the above substitution for the second solution (or if you prefer, swapping the squares labeled 0 and 1 in your table)

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  • $\begingroup$ The example you have provided gives solutions to one of the fifteen combinations, namely 2,5. Does your system give twelve pairs for the remaining fourteen combinations? Thanks for your correct answer. $\endgroup$ – Vassilis Parassidis Jul 1 '19 at 22:13
  • $\begingroup$ Yes, there are six kinds of every pair because the permutation (1643)(25) - or its inverse (1346)(25) - sends each number to its required "mate". $\endgroup$ – AxiomaticSystem Jul 2 '19 at 1:32
  • $\begingroup$ Could you elaborate, please? By "every pair", do you mean the fifteen combinations? The combinations are the following (1,2) (1,3) (1,4) (1,5) (1,6) (2,3) (2,4) (2,5) (2,6) (3,4) (3,5) (3,6) (4,5) (4,6) (5,6). Do you suggest we get twelve pairs of each combination by superimposing two Latin Squares? Specific examples would make everything clear. $\endgroup$ – Vassilis Parassidis Jul 2 '19 at 3:20
  • $\begingroup$ Only the combinations (1,3),(1,6),(2,5),(3,4),(4,6) are present. (2,5) is present 12x, and the other four are present 6x, for a total of 36 pairs, one for each cell of the 6x6 square. For a literal reading of your initial conditions, no 6x6 square has 60 distinct cells, so you'd need to count most of the pairs 'forwards and backwards' $\endgroup$ – Zomulgustar Jul 2 '19 at 3:45
  • $\begingroup$ Thank you. I wish there was a way to carry on our discussion of the subject. $\endgroup$ – Vassilis Parassidis Jul 2 '19 at 5:21

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