in reply to Re^2: Unique Combos with Math::Combinatorics
in thread Unique Combos with Math::Combinatorics
Unfortunately, that would mean that W only gets to play on 4 teams while everybody else gets to play on more and six players (A B C D F O) get to play on 7 teams. I doubt your players will consider that fair.
Update: You can get pretty fair with this subset of teams:
A, E, F, G A, H, I, J A, N, O, P A, Q, R, S A, T, U, V B, E, H, K B, F, I, L B, N, Q, T B, O, R, U B, P, S, V C, E, I, M C, G, K, O C, J, L, P C, Q, U, W C, R, T, X D, E, J, N D, G, H, L D, M, R, V D, S, T, W E, P, Q, X F, M, S, U F, V, W, X G, I, N, R H, M, O, W I, K, S, X J, K, Q, V L, N, U, X
Which has half of the players on 5 teams (A B C E I N Q R S U V X) and half of them on 4 teams (D F G H J K L M O P T W).
Update: And if you add the following teams:
T, K, G, M D, F, H, P W, J, L, O
Then every player gets to play on 5 teams and there are only 4 pairings that ever get repeated (DH KG WO JL).
- tye
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Re^4: Unique Combos with Math::Combinatorics (games)
by ketema (Scribe) on Apr 03, 2008 at 09:04 UTC |