Round Robin Tournament Scheduling

10 or 11 players petanque tournament

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bakdal

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on: June 02, 2008, 04:07:29 AM
Hi there.

I wrote this site last year, and got a very usable solution. This time it's actually the same prolem, but with fewer players.

I host a petanque tournament, and the basics and criterias are as follows:
- One court and one day tournament
- 10 or 11 participants
- Games played by two team of three or two participants (games with one side of three the other side of two is perfectly ok).
- Every participant should play exactly one game on same side with each and every of the other participants, and once or twice oppose every other participant.

I've got 11 players announced to show up, but I'd like to be prepared for a "no show".

Can something be worked out? Any help would be recieved with gratitude. Suggestions to alternate tournament formats are welcomed as well. The thing simply is: these 10 or 11 players, one court, and no early elimination (we are in for the fun, and some come from a far distance).

Best regards,

Klaus


Ian Wakeling

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Reply #1 on: June 06, 2008, 03:00:20 AM
Klaus,

Here is a cyclic schedule for 11 players (numbered 0 to 10) similar to the schedule that I made last year.  Each player is matched with all others at least once as a partner and at least once as an opponent.

( 0  7 10)  ( 2  6  8)
( 1  8  0)  ( 3  7  9)
( 2  9  1)  ( 4  8 10)
( 3 10  2)  ( 5  9  0)
( 4  0  3)  ( 6 10  1)
( 5  1  4)  ( 7  0  2)
( 6  2  5)  ( 8  1  3)
( 7  3  6)  ( 9  2  4)
( 8  4  7)  (10  3  5)
( 9  5  8)  ( 0  4  6)
(10  6  9)  ( 1  5  7)


For 10 players and 10 matches, I have not found such a good solution. Below, each player gets exactly 6 matches and is partnered with all other players at least once and at most twice.  The opponent balance is perhaps not as good as it might be with some pairs of players never playing in opposition, however these are the same pairs that partner each other twice, so in general the social mixing is good.

 R   team 1      team 2
 1 ( 1, 5, 4)  ( 7, 6, 9)
 2 ( 4, 3, 7)  ( 6, 9, 8)
 3 ( 1, 9,10)  ( 6, 3, 2)
 4 (10, 7, 1)  ( 8, 9, 3)
 5 ( 6, 5, 1)  ( 7, 8, 2)
 6 ( 3, 1, 4)  (10, 2, 6)
 7 ( 5, 8,10)  ( 4, 2, 9)
 8 ( 5,10, 3)  ( 4, 6, 8)
 9 ( 5, 7, 3)  ( 2, 1, 8)
10 ( 7, 4,10)  ( 9, 5, 2)


Hope that helps.
« Last Edit: June 06, 2008, 04:38:03 AM by Ian »


Ian Wakeling

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Reply #2 on: June 06, 2008, 04:54:23 AM
Klaus,

Ignore the 10 player schedule above, the one below is much better.  Now all players oppose each other exactly twice.

(0 1 4)  (3 5 8)
(1 2 5)  (4 6 9)
(2 3 6)  (5 7 0)
(3 4 7)  (6 8 1)
(4 5 8)  (7 9 2)
(5 6 9)  (8 0 3)
(6 7 0)  (9 1 4)
(7 8 1)  (0 2 5)
(8 9 2)  (1 3 6)
(9 0 3)  (2 4 7)


bakdal

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Reply #3 on: June 06, 2008, 05:11:51 AM
Dear Ian.

Thanks alot. I'm garded now for 10-13 players. Below there's hardly a tournament, and above is not suitable for just one court, so I'm home free.

It saves me many hours, and a lot of respect from the participants  ;)

Klaus