Speed Networking

Question

For a speed networking session, I would like to have 20 people at 4 tables (5 people per table) move four times during the session to ultimately meet everyone in the session. Is there a formula for seating and moving so that everyone gets to meet everyone else during the four rotations?

Answer 1

The answer is impossible.

In four moves, i.e. 5 rounds, each one have a total of 20 opponents, but there are only 19 other participants. That means, if such scheme existed, everyone had to meet with exactly one other participant twice.

Say participants $A,B,C,D$ and $E$ are in the same table in Round 1. For Round 2, two of them, say $A$ and $B$, are in the same table by pigeonhole principle, as there are 5 participants but only 4 tables. This is still fine, as long as $A$ will not meet the remaining 4 participants again, and also $B$ will not meet the other 4 participants again.

For Round 3, again 2 of the 5 participants are in the same table. These two participants either contains $A$ or $B$, which is forbidden, or does not contain $A$ or $B$. Say these two participants are $C$ and $D$. Then $C$ cannot meet with any of $A,B,D,E$ again, and $D$ cannot meet with any of $A,B,C,E$ again.

Then what about the Rounds 4 and 5? In each round, at least two participants in $A,B,C,D,E$ will meet again, but any pair of participants will involve one of $A,B,C,D$.

So for some participant in the 5 participants, he/she will meet someone he/she had met in the first round again in two of the later rounds. But then, there are not enough seats for him/her to meet the remaining $20-5=15$ participants anymore.

Answer 2

These problem are similar to "tournament problems", for example, a chess clubb pays blitz a night, 18 emberspresent, how to arrange in a practical (and fast!) way that everybody meets everybody once? There is a large literature on such tournament problems, and is is associated to theories of designs. So you can use that for a search.

For your specific problem, it seems impractical to have boards with 5 people each, since you want to make pairs, that way each round one person is unpaired (on each table).

One idea: Put your tables in a row, with people seeted at both sides (and not at the ends). First round, everybody meets the person opposite. After that round, on the one side people move one chair left, on the other side, one chair left. When reaching the end , move cyclically to the other side. That way everybody will see everybody once. This is the way it is done in chess clubs.

For the usual speed-dating session, you will presumably have two disjoint groups of people (men/women) which are only interested in the other group (for homosexual speed dating, you can forget this and use the setup above!). Then we need some alternative method. But this is easy: Put the boards in a row as above, men on the one side, women the other. Now, one group must be immobile, the other only moving, linearly. When coming out at one end, just reenter at the other end.