Doubles pairings with five people

[HeuristicBisc]

Hi,

I am looking to determine an appropriate permutation for table tennis. Generally, there are five of us that play, but we only require four players for doubles. To find a permutation of the five players (where two players can only play once in a round of five games and each person only sits off once) is fairly straightforward, for example:

1 (2-3) (4-5)
2 (1-4) (3-5)
3 (2-4) (1-5)
4 (2-5) (1-3)
5 (1-2) (3-4)

where the left hand column sits out, and the two in brackets play together. However, we would like to determine a permutation so that, in future, the pairs are swapped in such a way that no two pairs match. It is clear that we would have three 'rounds', so in the second round, the first game would involve 1 sitting out, and, for example, (2-4) (3-5). In round three, the first game would then involve 1 sitting out again, with the pair (2-5) (3-4) playing.

Note that with the example round given above, there is no round two that works. So it seems as though round 1 needs to be changed??

Can anyone help to provide a permutation? We believe that there is one, and possibly only one, that works.

HeuristicBisc

 

[Muphrid]

Take the existing list of games that you have, switch the 3 and the 4 everywhere, reorder so that the rotation of sitting is the same, and you have your second round, I believe.

 

[KroneckerD]

I think that is close to being a second round Muphrid, though game 5 remains the same by just swapping 3 and 4?

For anybody interested...

My friend posted the original question, and after quite a lot (lots++) of work I think I have found that there are too many constraints to be satisfied. One of the constraints is that in a 'round' every pair has to play with each other exactly once. I believe, if my calculations are correct, there are only 6 such rounds (listed below, hopefully without errors)

(a)
1 (2-3) (4-5)
2 (1-4) (3-5)
3 (2-4) (1-5)
4 (1-3) (2-5)
5 (1-2) (3-4)

(b)
1 (2-3) (4-5)
2 (1-5) (3-4)
3 (1-4) (2-5)
4 (1-2) (3-5)
5 (1-3) (2-4)

(c)
1 (2-4) (3-5)
2 (1-3) (4-5)
3 (1-4) (2-5)
4 (1-5) (2-3)
5 (1-2) (3-4)

(d)
1 (2-4) (3-5)
2 (1-5) (3-4)
3 (1-2) (4-5)
4 (1-3) (2-5)
5 (1-4) (2-3)

(e)
1 (2-5) (3-4)
2 (1-4) (3-5)
3 (1-2) (4-5)
4 (1-5) (2-3)
5 (1-3) (2-4)

(f)
1 (2-5) (3-4)
2 (1-3) (4-5)
3 (2-4) (1-5)
4 (1-2) (3-5)
5 (1-4) (2-3)

(I think round (c) is the same as Muphrid's suggestion)

What you notice, if you look very closely, is that every 'round' listed above shares exactly one game with every other round. For example in game 1 of round (a) is the same as game 1 of round (b), game 3 of round (d) is the same as game 3 of round (e), game 5 of round (b) is the same as game 5 of round (e), etc.

Therefore, there is no permutation (if my calculations are correct) of the rounds listed above that can you give you three rounds which satisfy the conditions

1) Players must sit off in sequential order (1,2,3,4,5) until person 5 has sat off, then person 1 sits off etc.
2) Everybody must play with everybody else exactly once in a 'round'.
3) Every pair must play every other pair exactly once over the course of 15 games.

Therefore, we have an underdetermined system (too many constraints and no solution exists), therefore, we need to remove one of the constraints (to achieve a square system or an overdetermined system (which will give us at least one solution)).

However, using any 3 of the rounds (a)-(f) will result in conditions 1 and 2 being satisfied, but 3 not being satisfied on two occasions. (Which for anyone interested might be sufficient).

Removing conditions 1 and 2 and keeping 3 gives an obvious solution, player 1 sits off three times, the other 4 people play 3 times with every other person, then player 2 sits off etc. However, this seems unfair as a game of table tennis can last 3-5 minutes up to 11, therefore, 9-15 minutes over the course of three games, so condition 1, I think, should stay.

Maybe the best solution is to keep conditions 1 and 3 and weaken 2 and write a new condition

2a) Two players cannot play together two times in a row.

Then the following (I believe) three rounds will give a solution to conditions 1 2a and 3.

Round 1

1 (2-5) (3-4)
2 (1-3) (4-5)
3 (1-4) (2-5)
4 (1-5) (2-3)
5 (1-3) (2-4)

Round 2

1 (2-3) (4-5)
2 (1-4) (3-5)
3 (2-4) (1-5)
4 (1-2) (3-5)
5 (1-4) (2-3)

Round 3

1 (2-4) (3-5)
2 (1-5) (3-4)
3 (1-2) (4-5)
4 (1-3) (2-5)
5 (1-2) (3-4)

(Of course condition 2a is not satisfied going from Round 3 back to round 1 again, but after quite a bit of searching I haven't been able to fix this).

I would appreciate any feedback or comments about this (or any errors).

 

[Diffy]

I agree that the first condition can not be solved. And I didn't check your six groups but it is absolute true that there are only six.

Perhaps the problem of Round 3 back to round 1 can be solved by playing

Round 1
Round 2
Round 3
Round 2
Round 1

 

[KroneckerD]

Yeah that would work Diffy, I think we would struggle to get through 3 rounds in an hour but even if we do, then starting round 2 isn't unfair, it's still player 1's turn to sit off and the last time (2-3) or (4-5) played each other in round 3 was game 3. Thanks for the suggestion.