How to arrange a conversation group

Question

Imagine a group of 8 people, initially arranged in two rows like so:

1 2 3 4

8 7 6 5

How do I proceed to make sure that each person is able to have conversation with the person immediately across from them (1-8,2-7, etc) for a fixed period of time and then rotate the arrangement in such a way that eventually each person is able to have a conversation with each of the others (with no duplicates)?

Does the same answer apply for odd numbers of people (one person out per arrangement)?

Answer 1

First, you can't have pairs of people if you have an odd number of people.

With your arrangement, or any even number $n$ of people, you need $n-1$ periods since a given person will speak with every other person. For your example, we can rotate the bottom row until all have conversed, then swap half of each row and rotate the halves, and continue this halving and rotating the $n-1$ times. Here, the permutations are

Rotating the bottom row: $$ \color{red}{1234} \\ \color{blue}{8765} $$ $$ \color{red}{1234} \\ \color{blue}{7658} $$ $$ \color{red}{1234} \\ \color{blue}{6587} $$ $$ \color{red}{1234} \\ \color{blue}{5876} $$ Now swap half of each row and rotate: $$ \color{red}{12}\color{blue}{56} \\ \color{red}{34}\color{blue}{78} $$ $$ \color{red}{12}\color{blue}{56} \\ \color{red}{43}\color{blue}{87} $$ And the final permutation: $$ \color{red}{13}\color{blue}{57} \\ \color{red}{24}\color{blue}{68} $$

Answer 2

I figured out that this is actually the "Round Robin Scheduling Problem" and has a simple solution. For an even number of participants you freeze the position of one participant in the first or last column and shift the others past them.

For an odd number you add a dummy participant in a frozen position as above, and shift the others past them (this results in each participant sitting out one conversation 'slot' when paired with the dummy).