Re^2: Unique Combos with Math::Combinatorics (fair)

Note that none of your solutions are fair because they all have some players playing on more teams than other players.

Update: Also note that a likely scenario for applying this algorithm is a gathering of 24 people where they break up into teams. This requires that the list of possible teams can be divided into "rounds" where each round has every player on exactly one team. I'm curious what the maximal solutions are with and without the requirement of rounds (but with the requirement of being "fair", of course). Even the maximal number of teams without a requirement of fairness would be interesting.

But from a practical standpoint, it may be even more valuable to come up with solutions that are more fair but don't require the number of repeated pairings to be zero, just to be minimized.

- tye

Comment on Re^2: Unique Combos with Math::Combinatorics (fair)

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Re^3: Unique Combos with Math::Combinatorics (fair) by ambrus (Abbot) on Mar 03, 2008 at 20:39 UTC
Someone directed me to the social golfer problem which is somewhat similar to what you ask.	[reply]
Re^4: Unique Combos with Math::Combinatorics (perfect) by tye (Sage) on Mar 03, 2008 at 23:03 UTC
Interestingly, it says that the maximum can be achieved, however, the given solution: 7 rounds play in 6 groups of 4 golfers [ 1 2 11 21 \| 9 10 19 5 \| 17 18 3 13 \| 4 7 6 24 \| 8 12 14 15 \| 16 20 2 +2 23 ] [ 1 3 12 22 \| 9 11 20 6 \| 17 19 4 14 \| 5 8 7 18 \| 2 13 15 16 \| 10 21 2 +3 24 ] [ 1 4 13 23 \| 9 12 21 7 \| 17 20 5 15 \| 6 2 8 19 \| 3 10 14 16 \| 11 18 2 +2 24 ] [ 1 5 14 24 \| 9 13 22 8 \| 17 21 6 16 \| 7 3 2 20 \| 4 10 11 15 \| 12 18 1 +9 23 ] [ 1 6 15 18 \| 9 14 23 2 \| 17 22 7 10 \| 8 4 3 21 \| 5 11 12 16 \| 13 19 2 +0 24 ] [ 1 5 16 19 \| 9 15 24 3 \| 17 23 8 11 \| 2 5 4 22 \| 6 10 12 13 \| 14 18 2 +0 21 ] [ 1 8 10 20 \| 9 16 18 4 \| 17 24 2 12 \| 3 6 5 23 \| 7 11 13 14 \| 15 19 2 +1 22 ] [download] Obviously doesn't work since 1 is paired with 5 in both the 4th and 6th round. But that appars to be a typo and should be a 7 in round 6. Spot checking didn't turn up any other problem but I didn't fully verify the solution, which using our A..X notation would be: `ABKU IJSE QRCM DGFX HLNO PTVW ACLV IKTF QSDN EHGR BMOP JUWX ADMW ILUG QTEO FBHS CJNP KRVX AENX IMVH QUFP GCBT DJKO LRSW AFOR INWB QVGJ HDCU EKLP MSTX AGPS IOXC QWHK BEDV FJLM NRTU AHJT IPRD QXBL CFEW GKMN OSUV` [download] - tye	[reply] [d/l] [select]