6-couple 10-round SAMDRR

[Richard B]

I know that the 6-couple 5-round SAMDRR is impossible. However, I have a symmetrical 5-round schedule in which each man plays plays with one woman twice and one woman zero times, and against the first woman zero times and the second one twice. Is there a way to find some sort of inverse schedule where the positions of the women are reversed, such that when the two schedules are combined to make a 10-round competition each man plays with each woman twice and against each woman twice?

[Ian Wakeling]

Your question is interesting, and I am not sure if this is going to be possible or not.  Can you post your 5 round schedule please, as I would be interested in taking a look?

From what you say I assume that you are only interested in scenarios where all 6 couples play simultaneously in full rounds.  So schedules with rounds of 2 games and 4 byes per round would be of no interest.

[Richard B]

I tried to attach an .xlsx file... here's a zipped version

[Richard B]

Thanks for getting back to me and sorry, I didn't realise that the message with the unzipped Excel file was not sent at all, even without the file. I hope you can read the zipped version. And yes, you are correct that I'm looking for simultaneous full rounds.

[Ian Wakeling]

Thanks for sharing the Excel file.  I have been experimenting and it seems it is possible to have a double length 6 couple SAMDRR.   For example:

 (H1 W2 v H3 W4) (H6 W1 v H5 W3) (H2 W5 v H4 W6)
 (H1 W3 v H4 W5) (H2 W1 v H6 W4) (H3 W6 v H5 W2)
 (H1 W4 v H5 W6) (H3 W1 v H2 W5) (H4 W2 v H6 W3)
 (H1 W5 v H6 W2) (H4 W1 v H3 W6) (H5 W3 v H2 W4)
 (H1 W6 v H2 W3) (H5 W1 v H4 W2) (H6 W4 v H3 W5)

 (H1 W2 v H4 W6) (H3 W1 v H5 W4) (H2 W3 v H6 W5)
 (H1 W3 v H5 W2) (H4 W1 v H6 W5) (H3 W4 v H2 W6)
 (H1 W4 v H6 W3) (H5 W1 v H2 W6) (H4 W5 v H3 W2)
 (H1 W5 v H2 W4) (H6 W1 v H3 W2) (H5 W6 v H4 W3)
 (H1 W6 v H3 W5) (H2 W1 v H4 W3) (H6 W2 v H5 W4)

Where H1W1, H2W2, H3W3,... are the spouse pairs who never meet.

From the perspective of players H1 and W1, the first 5 rounds and last 5 rounds appear to be a SAMDRR.  For everyone else, there is an imbalance in the opposite sex pairs in both halves, which cancels out when all 10 rounds are considered together.  For the latter players, there is always one player with who they partner twice, and one opposite sex opponent who they oppose twice, in both halves of the tournament.  For example, in the first 5 rounds, H2 plays twice with W5, and plays against W6 twice.  Conversely, in the last 5 rounds, H2 plays twice with W6, and plays against W5 twice.

Note that same sex opponents are balanced in each half.

[Richard B]

Brilliant... thanks very much! I had also found, and was working on, the case where everything works perfectly for one couple to see whether it was more straightforward than the symmetrical case I sent you, but obviously hadn't managed to do it although I felt sure that it was possible. May I ask if you just worked it out with a pencil, so to speak, or was the approach more analytic? Although I have a first in Maths it's a long time ago, and my interest is now more in putting together amusing golf tournaments. I did buy the Anderson book on Combinatorial Designs but much of it is beyond me now!

[Ian Wakeling]

I did start out with a pencil and the idea of a cyclic design on a set with two infinite elements (i.e. H1 and W1), which of course reduces the problem to finding just the first and sixth rounds.  I didn't get this to work, but I found a few that were encouragingly close to a double SAMDRR.  In then end, I wrote a program that enumerated all possible first rounds.  There are 480 of these that have spouse avoidance and same-sex opposition pair balance, but which are imbalanced in the opposite-sex pairs.  Finally there are 64 ways of combining 2 from the 480 to make a double SAMDRR, of which I have shown one example above.  If you wish to send me an e-mail by clicking on my name, then I can send a file with all 64 solutions.