Set of Unique pairs

[wolle]

There are more than 2 sets of solutions for 6 numbers

Yes I've realized that so I removed my failed solution.

[wolle]

Yes I've realized that so I removed my failed solution.

Well, it turns out I was on the right track after all so I restore the solution and continue.

I had the idea that a pair (i,j) can be viewed as a square on a 6 by 6 chess board. We place 6 rocks on the board so that no rock threats any other rock. It is called a valid 6-rock configuration. We make sure that no rock is placed in the diagonal going from upper left to lower right. This is because the pair (i,j) to describe this position will have i=j and such a pair can never be part of a valid set (because if the members of one pair are equal then the set cannot go from 1 to 6).

A 6-rock configuration will represent 6 pairs and it turns out that they can always be used to form 2 sets of 3 pairs where both set abide by the 1 to 6 rule. This is extraordinary and at this point I don't know exactly why it works (well it seems it doesn't always work. I'll have to check it out. Or maybe I leave it as an exercise for the OP ).

The solution strategy is to fill the whole chess board with as many valid 6-rock configurations as possible. To do it systematically we start with the diagonal going from lower left to upper right. We then take the diagonal below, and then the one below again etcetera until the whole board is filled with rocks. I name the configurations 1 to 5 and this is how the board turns out:

x 3 5 2 4 1
3 x 4 5 1 2
5 4 x 1 2 3
2 5 1 x 3 4
4 1 2 3 x 5
1 2 3 4 5 x

Note that no rock in any configuration threats another rock within the same configuration.

If we turn the rock configurations into pair equivalents we have

Conf. 1 : (6,1), (5,2), (4,3), (3,4), (2,5), (1,6)
Conf. 2 : (6,2), (5,3), (3,5), (2,6), (4,1), (1,4)
Conf. 3 : (6,3), (5,4), (4,5), (3,6), (2,1), (1,2)
Conf. 4 : (6,4), (4,6), (5,1), (3,2), (2,3), (1,5)
Conf. 5 : (6,5), (5,6), (4,2), (3,1), (2,4), (1,3)

We re-organize these configurations into 2 sets of 3 pairs making sure both sets abide by the 1 to 6 rule

Conf. 1 : {(6,1), (5,2), (4,3)}, {(3,4), (2,5), (1,6)}
Conf. 2 : {(6,2), (5,3), (4,1)}, {(2,6), (3,5), (1,4)}
Conf. 3 : {(6,3), (5,4), (2,1)}, {(3,6), (4,5), (1,2)}
Conf. 4 : {(6,4), (3,2), (5,1)}, {(4,6), (2,3), (1,5)}
Conf. 5 : {(6,5), (3,1), (4,2)}, {(5,6), (2,4), (1,3)}

This, ladies and gentlemen, is an optimal solution. All possible pairs are present exactly once. It means there are no repeated pairs and there are no pairs left to form more sets. There are 10 sets in total, all abiding with the 1 to 6 rule.

[2kaud]