Speed Meeting Pairings, part 2

Question

I'm an elementary grade teacher and I want to organize a speed meeting event at the start of the school year for the students. Think "speed dating", but for the purpose of getting to know each other instead of dating. There would be 10 times 3 minutes rounds during which six persons from each of the 6 classes would meet 5 other students they hadn't previously met in the speed meeting.

The important thing is of course that nobody sees somebody else two times during those rounds. The 6 classes each have 20 students. Thus, we would make 20 groups of 6 students each round. We could add a rule that a student of a class should not meet others from their own class, but that is optional.

Do you think you could help me figure out how to achieve this? I've tried with trial and error, but I quickly hit a block and I'm afraid it might be long before I find a solution.

Thanks.

Answer 1

Number the tables $0$ to $19$, and break the students up into six groups of $20$. In each group, every student is assigned a different starting table.

Start with this system: - Group A never moves, - Group B moves 1 table up cyclicly each round (the student at $19$ moves to $0$), - Group C moves 1 table down each round (the student at table $0$ moves to $19$), - Group D moves 2 tables up each round, - Group E moves 2 tables down each round, - Group F moves 3 tables up each round.

This isn't perfect. Make out the Progression Table of the students starting at table $0$: $$\begin{array}{c|c|c|c|c|c|c}\text{Round}&\text{A}&\text{B}&\text{C}&\text{D}&\text{E}&\text{F}\\\hline 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 1 & 19 & 2 & 18 & 3 \\ 3 & 0 & 2 & 18 & 4 & 16 & 6 \\ 4 & 0 & 3 & 17 & 6 & 14 & 9 \\ 5 & 0 & 4 & 16 & 8 & \color{red}{12} & \color{red}{12}\\ 6 & 0 & 5 & \color{red}{15} & \color{red}{10} & \color{red}{10} & \color{red}{15}\\ 7 & 0 & 6 & 14 & 12 & 8 & 18\\ 8 & 0 & 7 & 13 & 14 & 6 & 1\\ 9 & 0 & 8 & 12 & 16 & \color{red}4 & \color{red}4\\ 10 & 0 & 9 & 11 & 18 & 2 & 7\end{array}$$

But now it is a fairly simple matter to make 4 arbitrary changes to get unique groupings: $$\begin{array}{c|c|c|c|c|c|c}\text{Round}&\text{A}&\text{B}&\text{C}&\text{D}&\text{E}&\text{F}\\\hline 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 1 & 19 & 2 & 18 & 3 \\ 3 & 0 & 2 & 18 & 4 & 16 & 6 \\ 4 & 0 & 3 & 17 & 6 & 14 & 9 \\ 5 & 0 & 4 & 16 & 8 & 12 & \color{green}{13}\\ 6 & 0 & 5 & 15 & 10 & \color{green}{11} & \color{green}{16}\\ 7 & 0 & 6 & 14 & 12 & 8 & 18\\ 8 & 0 & 7 & 13 & 14 & 6 & 1\\ 9 & 0 & 8 & 12 & 16 & 4 & \color{green}5\\ 10 & 0 & 9 & 11 & 18 & 2 & 7\end{array}$$

Make this modified table your assignments for table $0$. Students starting at other tables add their starting table number cyclicly to the entries in this table to get their assignments. I believe that should be sufficient to assure unique pairings for everyone. But it is very late, so I am going to bed without checking it.

Answer 2

I think I found a solution that works! Here's the summary of it. I hope I downloaded the image right. Basically, each round, I tried moving everybody up a certain amount of position from the initial round. That's the number we see in each column A to F. When the new match was in conflict with another pairing, I manually found a pairing that would work. This table shows only one person from each class, but you simply have to add 1 to the number shown everytime for the next person, and so on.

I managed to make it to the 10th round. Of course, this is by no means a "resolution" of the problem, but only a solution that happens to work.

Thanks for everybody who contributed.