Math puzzle for tournament setup

Cephyr

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Jun 8, 2015
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I have been trying to set up a foosball tournament which has the following criteria:

8 players, say A, B, C, D, E, F, G, H, need to form rotating, unique teams (gives 28 combinations). This gives 2 matches each round and 7 rounds. Pretty easy to solve, if not for the last criteria: Everyone has to play against each individual player exactly twice. But how do you get it all to add up? I'm fairly good at math and fairly good at sudoku, but this nut feels more like seppuku :p Failing to put the pieces into programming too.
 
I have been trying to set up a foosball tournament which has the following criteria:

8 players, say A, B, C, D, E, F, G, H, need to form rotating, unique teams (gives 28 combinations). This gives 2 matches each round and 7 rounds. Pretty easy to solve, if not for the last criteria: Everyone has to play against each individual player exactly twice. But how do you get it all to add up? I'm fairly good at math and fairly good at sudoku, but this nut feels more like seppuku :p Failing to put the pieces into programming too.

Probably the easiest way is just to make it 1 match, 7 rounds and duplicate it for 2 matches a round, i.e. in Round 1 A plays B twice, C plays D twice, E plays F twice, and G plays H twice.

However, if you want to make it 'random', put the 28 matches on individual pieces of paper and draw four (or more to exclude duplicates) for the first match. Then do the same for the second match of the round. That would make 1 round. Make up 7 rounds. Then choose a round randomly.
 
8 players, say A, B, C, D, E, F, G, H, need to form rotating, unique teams (gives 28 combinations). This gives 2 matches each round and 7 rounds. Pretty easy to solve, if not for the last criteria: Everyone has to play against each individual player exactly twice. But how do you get it all to add up? I'm fairly good at math and fairly good at sudoku, but this nut feels more like seppuku :p Failing to put the pieces into programming too.
Hostilely, I do no know what the puzzle actually says. The only way to meet the "exactly twice" requirement" is the play in pairs. There are twenty-eight pairs. Double that to fifty-six pairs; there we have it: each two players together in a pair exactly twice.
\(\displaystyle \begin{array}{*{20}{c}}& {A,B}&& {}&& {}&& {}&& {}&& {}&& {}&& {}\\\hline& {A,C}&& {B,C}&& {}&& {}&& {}&& {}&& {}&& {}\\\hline& {A,D}&& {B,D}&& {C,D}&& {}&& {}&& {}&& {}&& {}\\
\hline& {A,E}&& {B,E}&& {C,E}&& {D,E}&& {}&& {}&& {}&& {}\\
\hline& {A,F}&& {B,F}&& {C,F}&& {D,F}&& {E,F}&& {}&& {}&& {}\\
\hline& {A,G}&& {B,G}&& {C,G}&& {D,G}&& {E,G}&& {F,G}&& {}&& {}\\\hline& {A,H}&& {B,H}&& {C,H}&& {D,H}&& {E,H}&& {F,H}&& {G,H}&& {}\end{array}\)
 
Probably the easiest way is just to make it 1 match, 7 rounds and duplicate it for 2 matches a round, i.e. in Round 1 A plays B twice, C plays D twice, E plays F twice, and G plays H twice.

However, if you want to make it 'random', put the 28 matches on individual pieces of paper and draw four (or more to exclude duplicates) for the first match. Then do the same for the second match of the round. That would make 1 round. Make up 7 rounds. Then choose a round randomly.

Ah, but the matches are played with two-player teams that are also rotating. So the first round is probably easy, AB plays against CD and EF plays against GH. Then A can not play with B again, and can only play once against C or D in the next six rounds. It is no problem at all setting up a tournament setup without the "X can play against individual players exactly twice" criterion.
 
Hostilely, I do no know what the puzzle actually says. The only way to meet the "exactly twice" requirement" is the play in pairs. There are twenty-eight pairs. Double that to fifty-six pairs; there we have it: each two players together in a pair exactly twice.
\(\displaystyle \begin{array}{*{20}{c}}& {A,B}&& {}&& {}&& {}&& {}&& {}&& {}&& {}\\\hline& {A,C}&& {B,C}&& {}&& {}&& {}&& {}&& {}&& {}\\\hline& {A,D}&& {B,D}&& {C,D}&& {}&& {}&& {}&& {}&& {}\\
\hline& {A,E}&& {B,E}&& {C,E}&& {D,E}&& {}&& {}&& {}&& {}\\
\hline& {A,F}&& {B,F}&& {C,F}&& {D,F}&& {E,F}&& {}&& {}&& {}\\
\hline& {A,G}&& {B,G}&& {C,G}&& {D,G}&& {E,G}&& {F,G}&& {}&& {}\\\hline& {A,H}&& {B,H}&& {C,H}&& {D,H}&& {E,H}&& {F,H}&& {G,H}&& {}\end{array}\)

Well, the teams are pairs. First round A and B play against C and D. That means both player A and player B have played once against both C and D and can only meet either of them once more in the next 6 rounds. Your table shows the 28 different teams. 4 teams play each round. 7 rounds. 7 rounds means every player will have a different team mate each round, but it is the "play individual opponents exactly twice" criterion that is difficult :)
 
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