A telephone game problem Let's say you are playing a telephone game
Everyone is in a circle.
First turn you come up with a secret sentence then you tell it to your left neighbour and listen to the secret from your right neighbour.
Next turns are basically the same but you don't come up with a new secret, you just pass on the last one you heard.
It ends when you hear back your first secret coming back to you (and it has gone full circle).
(Fun lays in the fact that there usually is something that deforms the message when it is passed, like loud noise or whatnot and you end up with nonsensical sentences)
Now let's imagine that I want secrets be passed on from person to person just as much as in a circle like previously defined without anyone hearing a same one twice, but I want to have as many different person to person interaction.
eg: For a group of 4 I could go that

*

*turn 1 : you tell your secret to your left neighbour

*turn 2 : you tell the secret you heard last turn to your left neighbour's left neighbour

*turn 3 : you tell the secret you heard last turn to your left neighbour's left neighbour's left neighbour
And everyone has been in contact with every secret once and once only and you never told a secret twice to the same person.

But for 5 ...
Assuming that you are numbered 1, your left neighbour 2 his left neighbour 3 and so on

*

*turn 1 : (next neighbour) you know secrets 1 and 5

*turn 2 : (nextnext neighbour) you know secrets 1 and 5 and 3

*turn 3 : (nextnextnext neighbour) you know secrets 1 and 5 and 3 and hear again from 5 before hearing about secret 4 or 2

So my question is : is there a simple method for N participants to find the interaction patterns that ensure non repetition before everyone has heard every secret and minimal interaction duplication (and that does not require the generation of every possibility and checking those one by one)?
 A: Let $n$ be the number of people in the group.
When $n$ is even...
You can arrange it so you hear a secret from a different person each time.

In each turn, pass the message to the person, one spot to your left, then two spots to your right, then three spots to your left, then four spots to your right, and so on, ending with passing $n-1$ spots to your left.

When $n=2m+1$, and $m$ is odd...
With this method, there will be only one instance where you here a secret from the same person twice.

For the first $m$ turns, you will pass one to the left, two to the right, three to the left, and so on, ending with passing $m$ spots to the left.
Next, pass one spot to the left (this is the repeat). Then, pass $(m-1)$ spots left, then $(m-2)$ right, $(m-3)$ left, $(m-4)$ right, and so on down to one spot right.

When $n=2m+1$, and $m$ is even...
With this method, there will be only one instance where you here a secret from the same person twice.

For the first $m$ turns, you will pass one to the left, two to the right, three to the left, and so on, ending with passing $m$ spots to the left.
Next, pass one spot to the right. Then, pass $(m-1)$ spots right, then $(m-2)$ left, $(m-3)$ right, $(m-4)$ left, and so on down to one spot right (this is the repeated pass).

When $n$ is odd, do we need the repeated pass?
This is a tricky question. When $n$ is not prime, the answer is yes, but the construction is not easy to describe. It is described in this paper by Higgam. When $n$ is prime and odd, then it is an open problem to determine if this is possible. We have confirmed it is impossible for $n=3,5$ and $7$. See my other answer on this topic for more details.
