- #1
HeuristicBisc
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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
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