Pairings

OK here’s a problem I’m trying to solve for pairings.

12 players, 9 rounds. 4 teams of 3 player each. Team members do not play each other but play everyone else.

Team 1 = 1,2,3
Team 2 = 4,5,6
Team 3 = 7,8,9
Team 4 = 10,11,12

Can you find a pairing method where no one will have more 2 colors in a row, and half of the field retains 5Wvs4B while the other have retains 4Wvs5B.

I’ve gotten it down to where only 2 players are screwy with 3W for one of them and 3B for the other one.

1-4, 5-2, 3-6, 7-10, 11-8, 9-12
4-2, 3-5, 6-1, 10-8, 9-11, 12-7
5-1, 2-6, 4-3, 7-11, 8-12, 10-9

1-7, 8-2, 3-9, 11-4, 12-5, 6-10
9-1, 2-7, 3-8, 4-12, 5-10, 11-6
1-8, 2-9, 7-3, 10-4, 5-11, 6-12

12-1, 11-2, 3-10, 4-7, 8-5, 9-6
1-11, 2-10, 12-3, 8-4, 5-9, 6-7
10-1, 12-2, 11-3, 9-4, 7-5, 6-8

You can try manipulating this option.

Close - we’re down to just 1 person with the 3W (player #12).

But if we switch the final round of 12-2 to be 2-12 I think that problem is solved, no?

We still have 7 players with 5W and 4B, and 5 player with 4W and 5B. Is there a way to solve this without creating the three in a row color problem?

The 2-12 switch would be good. There are 6 5W4B and 6 4W5B. Also there are 2 14W13B teams and 2 13W14B teams.

Ok I just recounted. This looks like this is good. Thanks Jeff!!

–Sevan

Here’s another one provided by FM/IO Eric Schiller:

1-12,2-10,3-11;9-4,7-5,8-6

10-1, 11-2,12-3;7-4,8-5,9-6

1-11,2-12,3-10;4-8,5-9,6-7

6-1,4-2,5-3;12-7,10-8,11-9

1-4,2-5,3-6;7-10,8-11,9-12

5-1,6-2,4-3;11-7,12-8,10-9

1-9,2-7,3-8;4-12,5-10,6-11

7-1,8-2,9-3;10-4,11-5,12-6

1-8,2-9,3-7;4-11,5-12,6-10

It actually follows RR concepts more closely - everyone in the top half has 5W and 4B and everyone in the bottom half has 4B and 5W. Less number of people have double color (BB or WW) also.

But nonetheless - thanks for the assist Jeff and all others that emailed me.

You may want to modify that depending on whether or not your team concept has team prizes. the 1/2/3 and 4/5/6 teams have 15W12B while the other two teams are 12W15B. Finishing with 7-3 and 10-6 would add double colors but change the overall team colors to 14W13B and 13W14B.