# $n$ Tennis players took part in the one-round table tennis tournament $(n \geq 3)$. We say that player $A$ is better than player $B$, if …

$$n$$ Tennis players took part in the one-round table tennis tournament $$(n \geq 3)$$. We say that player $$A$$ is better than player $$B$$, if $$A$$ won $$B$$ or there is such a player $$C$$, that $$A$$ won $$C$$, and $$C$$ won $$B$$. For what $$n$$ in the tournament could it be that each player is better than everyone else? There are no draws in tennis.

I proved that $$n = 3k$$ is suitable, I also learned how to make an example for $$n = 5$$, I assume that $$n = 3k + 2$$ is suitable, but I cannot prove it, it is also not clear what to do if $$n = 3k + 1$$.

• What is a "one-round tournament"? Perhaps if you show the $n=5$ example I can better understand the tournament format? – antkam Mar 12 '19 at 13:20
• one-round tournament is when everyone has played exactly one time with each – Yaroslav Mar 12 '19 at 13:26
• Ah, thanks, what I usually call a "round-robin" tournament then. :) – antkam Mar 12 '19 at 13:28
• @MikeEarnest maybe I'm missing something obvious, but how does a Rock beat (directly or indirectly) another Rock? I can see this happening if, among the Rocks, you use something like my $n=odd$ solution, but then if each of the 3 groups (no need for equal size) are odd numbered, then the total number is odd and you could have just used my solution to begin with. – antkam Mar 12 '19 at 18:00
• @MikeEarnest - Your RPS idea does imply this: If some even $n$ is feasible, then any larger even $N > n$ is also feasible, because you can always divide $N-n$ into two odd groups and use my solution for each. – antkam Mar 12 '19 at 20:01

Partial solution... specifically:

Claim A: Any odd $$n$$ is feasible.

Claim B: $$n = 4$$ is infeasible.

Proof of A: Arrange the players in a circle and number them $$0, ..., n-1$$, and let $$i$$ beat $$i+1, i+3, ..., i+n-2$$. All arithmetic is modulo $$n$$.

First of all, this assignment is consistent: For any $$i \neq j$$, if $$j = i + odd$$ (i.e. $$i$$ beats $$j$$) then $$i = j + even$$ (i.e. $$j$$ does not beat $$i$$).

Next, clearly $$i$$ beats all the $$i+odd$$ directly, but since each $$j$$ beats $$j+1$$, $$i$$ also indirectly beats all the $$i+odd+1$$, i.e. all the $$i+even$$.

Proof of B: Among the $$n=4$$ players, clearly nobody can beat everyone or be beaten by everyone. Since each plays $$3$$ games, that means each must win only $$1$$ or $$2$$ games. Since there are $$6$$ games total, the only way to do this is if two players $$W,X$$ win twice each and two other players $$Y,Z$$ win once each. But consider the match between $$Y,Z$$ and without loss assume $$Y$$ beats $$Z$$. This is $$Y$$'s only win, and $$Z$$ beats only $$1$$ person (e.g. $$W$$), so $$Y$$ does not directly nor indirectly beat the other person (e.g. $$X$$).