SEARCH HOME
 Math Central Quandaries & Queries
 The problem is that I have 12 couples, 1 through 12, in a dinner club, and each couple is host once with 3 of the other couples. he dinners will take place over 4 months. I am trying to not repeat any of the couples. We had the formula for 16 and 20 couples but not for 12. Are you able to help with the groupings for 12. Thanks very much. Irv

Hi Irv,

There are 66 pairs of couples, each of whom has to dine together once. Each night on the schedule accounts for 18 pairs (each foursome accounts for 6 of them). Since 66 is not divisible by 18, the schedule you want does not exist. Scheduling is funny that way.

It is possible to come close, I think. Would you prefer to repeat couples, or have some couples not dine together? Is there some flexibility in the last week (say, not dining in groups of 4 couples)?

There is a half-decent schedule where three couples dine together at a time for the first 4 weeks, and then four couples dine together at a time in week 5. What you can do is start with the schedule for 16 couples, choose any one of the "foursomes" on week 5, and delete this foursome from week 5 and the couples in it from the foursomes where they occur in the first four weeks.

If this isn't quite what you want, then please write back.

Victoria

Irv wrote back

Hi Again,

I have tried finding the answer as per the last email sent to me, but I can only pick up a previous reply
to Cheryl which does show some combinations.  Example: using letters numbers 1 -2 rather than letters A-B.   Whether you combine 1 -2 or 2-1 that would still be a repeat.  So. using numbers 1 through 12 is it possible to follow through with three groups of 4 each every month for four months so that none of the
numbers repeat at any time.  If not possible can the numbers be set up with a minimum of repeats.
Thank you.
Irv Snyder

Irv,

It is remarkable that something so small can prove to be so difficult.

First, the non-repeating schedule you seek does not exist. The easiest way to convince yourself of that is to recognize that there are 66 pairs of couples who should dine together once, and every club night accounts for 18 of them. Since 66 is not a multiple of 18, it can't be done (if everyone participates in each of the 4 nights).

The smallest possible amount of duplication would be if there were a schedule with four club nights so that every couple dines together at least once. There would be 4 x 18 = 80 pairs of couples who dine together
in all. If such a schedule existed (and I don't know if it does), the duplication would be 14 pairs.

Here is the best suggestion I have at the moment. Start with your schedule for 16 couples. Suppose that one of the foursomes in the final week (week 5) is {13, 14, 15, 16}. Delete this foursome, and delete these four people from every other foursome where they occur. Now, in each of weeks 1, 2, 3, and 4, pick one of the resulting threesomes and assign each of its members to one of the other three groups. (You may want to experiment with ways of doing this to balance the duplication a bit.).

What you now have is a 5 event schedule where everyone dines on 4 of the nights and 8 couples dine on the remaining night. Every couple dines together at least once. There are 92 pairs of couples that dine together in all, so the duplication is 26 pairs (which is a lot!).

I realize this is probably not exactly what you want, and if I see how to do better then I will email again.

Happy dining.

Victoria

Hi Again,

I have tried finding the answer as per the last email sent to me, but I can only pick up a previous reply
to Cheryl which does show some combinations.  Example: using letters numbers 1 -2 rather than letters A-B.   Whether you combine 1 -2 or 2-1 that would still be a repeat.  So. using numbers 1 through 12 is it possible to follow through with three groups of 4 each every month for four months so that none of the
numbers repeat at any time.  If not possible can the numbers be set up with a minimum of repeats.
Thank you.
Irv Snyder

Hi again Irv,

I have thought about the dinner club problem some more, One can prove that if only four dinner nights are held, then not all couples can dine together over the course of the four months. The best case scenario for four couples at a time and all pairs meeting at least once would be five dinner nights. I don't know if this schedule is possible, but speculate that it is.

This suggests that it is time to investigate alternatives. What flexibility is there to accommodate different group sizes or different numbers of nights?

Victoria

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.