In the game of Sushi Go! you play one card out of your hand, after which you pass your hand to another player (receiving another hand from yet another player). The card you play can help yourself, or hinder the next player (by playing a card the next player needed).
In the official rules, you always pass to your left, which means you're always hindering the same player and always get hindered by the same player.
That got me thinking.
Suppose we have a game of Sushi Go! with four players. If the players first pass left, then across, then right, then across again, every player has had every hand once and every hand ends up with the original player again, just like when always passing left. But now every player has passed to every other player at least once: once to the player on their left, once to the player on their right, and twice to the player across.
With only three players, the only way of doing this would be doing a full cycle of passing left first, then doing a full cycle of passing right. However, this means that when switching between cycles, this means players pass their hands back to the player they received it from. I'd like to avoid that if possible.
So my question is for $N \in \Bbb N$ players where $N \ge 4$, how do I find a way of passing hands such that:
- Every player gets every hand just as often as any other hand
- No player ever passes their hand directly back to the player they just received it from
- Every player passes to every other player at least once
- Preferably, every player passes to every other player as often as to any other player
When $n$ is even, you can always do a zig-zag, and in particular your strategy for $n=4$ an instance of this. Assuming that $+1$ means a pass to the next person and $+(n-1)$ means a pass to the previous person, you can do this:
$$+1, +(n-2), +3, +(n-4), \ldots, +4, +(n-3), +2, +(n-1)$$
If you were to draw this on a regular $n$-gon, you would get a nice zig-zag ;-)
For an odd $n$ the best I could find right now is just repeating $n$ times $+1$'s, then $n$ times $+2$'s, and so on, up to $n$ times $+(n-1)$'s, at which point the cycle repeats.
However, please note that this does not work for $n=3$, because it violates the "no player ever passes their hand directly back to the player they just received it from". In fact that particular rule forces a stream of $+1$'s in the case of $n=3$, because it forbids following $+1$ with a $+2$, but that is the only alternative, there are no other options.
I hope this helps $\ddot\smile$