Round Robin Tournament Scheduling

More petanque, 12 players and below

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bakdal

  • Guest
on: May 20, 2007, 10:56:28 AM
Hi there.

I wrote a week ago, with a question of a 13 players petanque tournament. I've got a solution through this page, and thanks a lot for that.

I figured though, that I should be prepared for one player cancellation. Embarrased to write again, I have tried and tried and tried to make it work, but I can't.

The thing is this:

12 players, where each player should play once with all other players.

I can see a structure of 10 tripple matches and 3 doubles should do the trick. Preferably each player should have the others as opponent at least once and and a maximum of two times.

And, are there any suggestions for a good sceme for 11, 10 or 9 players? I arrange tournaments a number of times a year, and I might as well have the schemes in hand, in stead of writing every other week ;-)

The criteria for any number of players in the tournament is, that everybody play on the same team with everybody once. A mix of doubles and tripple are fine, and - in case of more than once solution - as few games as possible.

Thanks in advance,

Klaus


Ian Wakeling

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Reply #1 on: May 21, 2007, 03:00:47 AM
Klaus,

12 players is a much harder problem as the cyclic method I used for 13 players no longer works.  I am aware of the solution below for 10 triple matches and 3 double matches, however it only balances the partnerships, so everyone gets to play on the same team as everyone else exactly once.

(A D K) v (H J L)
(A G I) v (B D H)
(C F I) v (D E J)
(A C J) v (D I L)
(A E L) v (B G J)
(B C K) v (D F G)
(C E H) v (F J K)
(B F L) v (H I K)
(B E I) v (C G L)
(A F H) v (E G K)

(A B) v (C D)
(E F) v (G H)
(I J) v (K L)


I haven't checked but there will be some inbalance in the opponents. You could try to move things around, keeping the same brackets above but making different matches, some arrangements will no doubt have better opponent balance than others.

Hope that helps,

Ian.
« Last Edit: May 21, 2007, 03:03:21 AM by Ian »


bakdal

  • Guest
Reply #2 on: June 03, 2007, 02:21:54 PM
Dear Ian.

Thank you very much for your solutions. It has litterally made life easier. I used to get stuck in groups, winner halfs/looser halfs and the like, but now it's all different.

I had my first tournament this weekend after finding this site. 13 announced their participation, 12 showed up! Now, the solution you made for 13 players was brilliant and well balanced. The solution for 12 players (10 doubles - 3 triples) only balanced who you played with, not against. I haven't rearranged the scheme, so what to do? Then it strucked me: The 13 player scheme was all triples. I don't mind a pair playing a triple in one game, so I simply took out the player number 13, and voila!

So, for all eternity I have nice and balanced schemes for 13 and 12 players.

Given the new opening on the criterias, is it possible to make schemes for 10 and 11 players as well, and I would be home free.

Excact criterias are:

- All play on the same side with all once.
- All play against all as balanced as possible.
- The above two criterias can be switched.
- Triples games should be the standard, but 2 versus 3 games a allowed, as are 2 versus 2 (but then prefered balance between number of times each player plays in doubles and in triples).
- Not more than 16 games in total (prefered a bit lower if possible).

Thanks once more for help and aid already given. If this last request if too little math and too much....something else, it's perfectly ok.

In awe,

Klaus