Teams of 6 Players, playing in four Rounds of Foursomes

[Sifo-Dyas]

Hallo Board,
I need a schedule for a Skat (Trick-Card-Game) event of six Teams (A,B...) of six Players (A1,A2,...,B1 ...)

We warnt to play in foursome tables with following rules:
- No Player has a bye 
- Players of Same team never play at the same table 
- every Team should play every other Team nearly the same number (6 Players, 4 rounds, 3 opponents = 72 opponents for every team, absolut 14+14+14+15+15 meetings, nearly equal).
- two Players shouldn't meet twice

I figured this schedule for 2 and 3 Rounds, bug failed with a 4th round).
Have a nice Weekend 

[Ian Wakeling]

This is a tough problem, and the hardest part is trying to meet rule 3.  I know you are asking for 4 rounds, but my instinct was to try for 6 rounds with a cyclic solution.  I have found something potentially useful, but I am a long way from being optimal on rule 3.  As there are more rounds, the optimal solution would be 21+21+22+22+22 instead of 14+14+14+15+15 meetings, what I have achieved in the schedule below is 18+18+24+24+24.

(c1 b1 f3 d1) (a1 d4 f5 e6) (d3 e1 b5 c2) (e3 d6 a2 b3) (f4 a5 d5 b4) (a4 f1 b2 c3) (b6 a3 c4 e5) (d2 c5 e2 f2) (c6 f6 e4 a6)
(c2 b2 f4 d2) (a2 d5 f6 e1) (d4 e2 b6 c3) (e4 d1 a3 b4) (f5 a6 d6 b5) (a5 f2 b3 c4) (b1 a4 c5 e6) (d3 c6 e3 f3) (c1 f1 e5 a1)
(c3 b3 f5 d3) (a3 d6 f1 e2) (d5 e3 b1 c4) (e5 d2 a4 b5) (f6 a1 d1 b6) (a6 f3 b4 c5) (b2 a5 c6 e1) (d4 c1 e4 f4) (c2 f2 e6 a2)
(c4 b4 f6 d4) (a4 d1 f2 e3) (d6 e4 b2 c5) (e6 d3 a5 b6) (f1 a2 d2 b1) (a1 f4 b5 c6) (b3 a6 c1 e2) (d5 c2 e5 f5) (c3 f3 e1 a3)
(c5 b5 f1 d5) (a5 d2 f3 e4) (d1 e5 b3 c6) (e1 d4 a6 b1) (f2 a3 d3 b2) (a2 f5 b6 c1) (b4 a1 c2 e3) (d6 c3 e6 f6) (c4 f4 e2 a4)
(c6 b6 f2 d6) (a6 d3 f4 e5) (d2 e6 b4 c1) (e2 d5 a1 b2) (f3 a4 d4 b3) (a3 f6 b1 c2) (b5 a2 c3 e4) (d1 c4 e1 f1) (c5 f5 e3 a5)

At the team member level opposition is more balanced, as each player sees either 3 or 4 players from each of the other 5 teams.  Let me know if this will be useful or not.

[Ian Wakeling]

I thought some more about this and came up with the following 4 round schedule:

(a5 e3 c6 b2) (b3 c2 f1 d1) (e4 f4 d2 a2) (e1 b1 c5 d3) (a3 e2 b4 f3) (d5 a4 f5 c4) (e5 d4 a1 f2) (d6 b5 e6 c1) (b6 c3 f6 a6)
(a4 e6 b1 f6) (c4 d4 a2 b2) (c3 f2 d5 e2) (f4 a3 e3 b3) (b4 a1 c2 d6) (d1 e4 f3 c5) (d2 f5 b5 a5) (c1 e5 d3 b6) (c6 f1 a6 e1)
(f4 c1 e1 a5) (e2 b6 a2 d1) (f1 c3 d2 b1) (a6 f3 b3 d3) (e3 c5 d4 a4) (b4 f2 e6 c4) (d5 a1 b5 e4) (e5 b2 f5 c2) (c6 d6 a3 f6)
(a1 b1 c1 f3) (c2 d2 e2 a6) (f6 d5 e1 b2) (d3 a5 f1 e6) (b3 a4 c3 e4) (f4 d4 b4 c6) (f5 b6 d6 e3) (a3 e5 c4 d1) (b5 c5 a2 f2)

I made this in two steps.  First I chose the teams that participate in all 36 games - the advantage is that it is relatively easy to make this choice in such a way that it meets rule 3 with all team meetings nearly equal.  Then keeping the team pattern fixed, it is possible to look for optimal assignments of player 1 to 6 within each team.  I believe that each player should meet with at least 2, and at most 3, members of the other teams.

[Sifo-Dyas]

Thank you Ian,
this schedule-update is super.

I worked out a balanced schdedule by myself, working from a "Soccer-League"-6-Team-Schedule
and used it for the teams and for team members

A-E / B-D / C-F will be played with A1-E1-C5-F5 and B1-D1-C5-F5 and C1-F1-A5-E5, called (1,5)

Triple-Table / Round 1, Round 2, Round 3, Round 4
A-E / B-D / C-F (1,5) (3,4) (x,x) (2,6)
A-F / B-E / C-D (2,4) (x,x) (3,5) (x,x)
B-A / C-E / D-F (3,6) (x,x) (2,1) (x,x)
B-F / C-A / D-E (x,x) (1,6) (x,x) (4,5)
C-B / D-A / E-F (x,x) (2,5) (4,6) (3,1)
gives a posible solution :-)

[Ian Wakeling]

That's really clever, thanks for posting.

I think there is a typo above - shouldn't the first group of 4 in round 1 be A1-E1-B5-D5?

[Sifo-Dyas]

Yes, it was a typo.