13 players, triple games (petanque)

[bakdal]

Hi there.

Great page, but I haven't found a solution for my problem. I'm sure it's laughable easy for some one here.
 
The thing is: I'm hosting a petanque tournament for 13 people, playing at my one home court. I figure, that if 13 games of triples (3 players vs. 3 players) are played, everybody have played in 6 games, and so it should be possible to balance, that everybody is on the same side with all other 12 players once. Of course, the number of times you play the others as opponent is unbalanced.
 
Can anybody help? Criteria are:
 
1) Balanced game plan in terms of all players having all other 12 players as partners once
2) As little spread as possible is terms of number of times every player has the other 12 players as opponent.
 
Thanks in advance,
 
Klaus

[Ian Wakeling]

I think a simple cyclic schedule will work for you.  Number the 13 players as 0 to 12.

( 0  1  4)  v  ( 3  5 10)
( 1  2  5)  v  ( 4  6 11)
( 2  3  6)  v  ( 5  7 12)
( 3  4  7)  v  ( 6  8  0)
( 4  5  8)  v  ( 7  9  1)
( 5  6  9)  v  ( 8 10  2)
( 6  7 10)  v  ( 9 11  3)
( 7  8 11)  v  (10 12  4)
( 8  9 12)  v  (11  0  5)
( 9 10  0)  v  (12  1  6)
(10 11  1)  v  ( 0  2  7)
(11 12  2)  v  ( 1  3  8)
(12  0  3)  v  ( 2  4  9)

As you ask, all 12 possible partners occur exactly once over the 6 matches for any one player.  The opposing players should occur at least once, and at most twice.

[bakdal]

Thanks a lot.

Then, you write "a simple cyclic schedule", indicating that the work to produce the actual scheme was easy. As a mathematical novice, I would be interested to know, how to attend such a simple cyclic schedule modulator? I've checked on this page, and it seems that possibilities are limited. For instance, I only found schedules for even numbers, and not including the triple preference.

Are there other pages I could attend, for more parametres, and for the sake of self-helpness (so I don't need to ask here every time I set up a tournament).

All the best,

Klaus