Judith,
Thanks for bringing your tournament question here, as this is an interesting problem. If I understand correctly your "couples" are permanent bridge parnternships who always play together against another couple. At the top level the problem is to find a schedule for 24 couples, that I have labelled A to X, and then arrange them in groups of 4, over 6 sites and 7 rounds, in such a way the number of times the same two couples meet is minimised.
Round Site 1 Site 2 Site 3 Site 4 Site 5 Site 6
1 (P K C N) (B R X H) (T A Q J) (U G E L) (M F W O) (V D S I)
2 (O H D E) (P I F T) (L B C J) (N G R A) (V U X M) (Q S K W)
3 (G Q M C) (N H V F) (O I B K) (U T R D) (L X A S) (J P E W)
4 (T N O L) (C D X W) (A U K F) (S P R M) (I J H G) (V B E Q)
5 (L K H M) (X Q I N) (W V G T) (F E R C) (U J O S) (A P B D)
6 (C O V A) (M J N D) (F S B G) (R L W I) (U P H Q) (X E K T)
7 (W N B U) (T S H C) (R J V K) (Q L F D) (O X P G) (M E I A)
In fact this is the same problem as arranging golf foursomes. So the schedule above could be used for arranging 24 golf players for rounds of 6 social foursomes, or for arranging your 24 bridge couples at 6 sites. The important property of the schedule is that no two couples meet together more than once, so every couple meets with exactly 21 out of the 23 other couples over the course of the 7 rounds.
Now consider what happens when 4 couples meet at a site, you are free to play as many or as few of the three possible bridge matches as you like. For example if the four couples at a site are 1,2,3 and 4. Then you could play upto three mini-rounds from the following list:
Mini-round Table 1 Table 2
1 ( 1 vs 2) ( 3 vs 4)
2 ( 1 vs 3) ( 2 vs 4)
3 ( 1 vs 4) ( 2 vs 3)
None of these bridge matches can occur again in another main-round since no two of these 4 couples will meet again at the same site.
In the schedule the assignment of groups of 4 couples to sites is purely at random, so some couples will play more than once at a site and others not at all. I believe it is not possible to have a schedule with the additional property that all couples play at least once at a site.
I hope that's clear.
Ian.
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Note added August 15th.
The following set of triples contain the couples who never meet in the schedule above.
(A H W) (B M T) (C I U) (D G K) (E N S) (F J X) (L P V) (O Q R)
for example H and W are the couples that A never meets. If the schedule is being used for golf, this could be used for a final round of 8 threesomes and would ensure that all possible pairs of players occur exactly once. The triples could also be used to divide the 24 couples/players into 8 teams of 3, then no matches with couples/players from the same team would be scheduled.
