Subfield: Combinatorics


Question: An $(n, k)$-tournament is a contest with $n$ players held in $k$ rounds such that:

(i) Each player plays in each round, and every two players meet at most once.

(ii) If player $A$ meets player $B$ in round $i$, player $C$ meets player $D$ in round $i$, and player $A$ meets player $C$ in round $j$, then player $B$ meets player $D$ in round $j$.

Determine all pairs $(n, k)$ for which there exists an $(n, k)$-tournament.

Solution:
For each $k$, denote by $t_{k}$ the unique integer such that $2^{t_{k}-1}<k+1 \leq 2^{t_{k}}$. We show that an $(n, k)$-tournament exists if and only if $2^{t_{k}}$ divides $n$.

First we prove that if $n=2^{t}$ for some $t$ then there is an $(n, k)$-tournament for all $k \leq 2^{t}-1$. Let $S$ be the set of $0-1$ sequences with length $t$. We label the $2^{t}$ players with the elements of $S$ in an arbitrary fashion (which is possible as there are exactly $2^{t}$ sequences in $S$ ). Players are identified with their labels in the construction below. If $\alpha, \beta \in S$, let $\alpha+\beta \in S$ be the result of the modulo 2 term-by-term addition of $\alpha$ and $\beta$ (with rules $0+0=0,0+1=1+0=1$, $1+1=0$; there is no carryover). For each $i=1, \ldots, 2^{t}-1$ let $\omega(i) \in S$ be the sequence of base 2 digits of $i$, completed with leading zeros if necessary to achieve length $t$.

Now define a tournament with $n=2^{t}$ players in $k \leq 2^{t}-1$ rounds as follows: For all $i=1, \ldots, k$, let player $\alpha$ meet player $\alpha+\omega(i)$ in round $i$. The tournament is well-defined as $\alpha+\omega(i) \in S$ and $\alpha+\omega(i)=\beta+\omega(i)$ implies $\alpha=\beta$; also $[\alpha+\omega(i)]+\omega(i)=\alpha$ for each $\alpha \in S$ (meaning that player $\alpha+\omega(i)$ meets player $\alpha$ in round $i$, as needed). Each player plays in each round. Next, every two players meet at most once (exactly once if $k=2^{t}-1$ ), since $\omega(i) \neq \omega(j)$ if $i \neq j$. Thus condition (i) holds true, and condition (ii) is also easy to check.

Let player $\alpha$ meet player $\beta$ in round $i$, player $\gamma$ meet player $\delta$ in round $i$, and player $\alpha$ meet player $\gamma$ in round $j$. Then $\beta=\alpha+\omega(i), \delta=\gamma+\omega(i)$ and $\gamma=\alpha+\omega(j)$. By definition, $\beta$ will play in round $j$ with

$$
\beta+\omega(j)=[\alpha+\omega(i)]+\omega(j)=[\alpha+\omega(j)]+\omega(i)=\gamma+\omega(i)=\delta
$$

as required by (ii).

So there exists an $(n, k)$-tournament for pairs $(n, k)$ such that $n=2^{t}$ and $k \leq 2^{t}-1$. The same conclusion is straightforward for $n$ of the form $n=2^{t} s$ and $k \leq 2^{t}-1$. Indeed, consider $s$ different $\left(2^{t}, k\right)$-tournaments $T_{1}, \ldots, T_{s}$, no two of them having players in common. Their union can be regarded as a $\left(2^{t} s, k\right)$-tournament $T$ where each round is the union of the respective rounds in $T_{1}, \ldots, T_{s}$.

In summary, the condition that $2^{t_{k}}$ divides $n$ is sufficient for an $(n, k)$-tournament to exist. We prove that it is also necessary.

Consider an arbitrary $(n, k)$-tournament. Represent each player by a point and after each round, join by an edge every two players who played in this round. Thus to a round $i=1, \ldots, k$ there corresponds a graph $G_{i}$. We say that player $Q$ is an $i$-neighbour of player $P$ if there is a path of edges in $G_{i}$ from $P$ to $Q$; in other words, if there are players $P=X_{1}, X_{2}, \ldots, X_{m}=Q$ such that player $X_{j}$ meets player $X_{j+1}$ in one of the first $i$ rounds, $j=1,2 \ldots, m-1$. The set of $i$-neighbours of a player will be called its $i$-component. Clearly two $i$-components are either disjoint or coincide.

Hence after each round $i$ the set of players is partitioned into pairwise disjoint $i$-components. So, to achieve our goal, it suffices to show that all $k$-components have size divisible by $2^{t_{k}}$.

To this end, let us see how the $i$-component $\Gamma$ of a player $A$ changes after round $i+1$. Suppose that $A$ meets player $B$ with $i$-component $\Delta$ in round $i+1$ (components $\Gamma$ and $\Delta$ are



not necessarily distinct). We claim that then in round $i+1$ each player from $\Gamma$ meets a player from $\Delta$, and vice versa.

Indeed, let $C$ be any player in $\Gamma$, and let $C$ meet $D$ in round $i+1$. Since $C$ is an $i$-neighbour of $A$, there is a sequence of players $A=X_{1}, X_{2}, \ldots, X_{m}=C$ such that $X_{j}$ meets $X_{j+1}$ in one of the first $i$ rounds, $j=1,2 \ldots, m-1$. Let $X_{j}$ meet $Y_{j}$ in round $i+1$, for $j=1,2 \ldots, m$; in particular $Y_{1}=B$ and $Y_{m}=D$. Players $Y_{j}$ exists in view of condition (i). Suppose that $X_{j}$ and $X_{j+1}$ met in round $r$, where $r \leq i$. Then condition (ii) implies that and $Y_{j}$ and $Y_{j+1}$ met in round $r$, too. Hence $B=Y_{1}, Y_{2}, \ldots, Y_{m}=D$ is a path in $G_{i}$ from $B$ to $D$. This is to say, $D$ is in the $i$-component $\Delta$ of $B$, as claimed. By symmetry, each player from $\Delta$ meets a player from $\Gamma$ in round $i+1$. It follows in particular that $\Gamma$ and $\Delta$ have the same cardinality.

It is straightforward now that the $(i+1)$-component of $A$ is $\Gamma \cup \Delta$, the union of two sets with the same size. Since $\Gamma$ and $\Delta$ are either disjoint or coincide, we have either $|\Gamma \cup \Delta|=2|\Gamma|$ or $|\Gamma \cup \Delta|=|\Gamma|$; as usual, $|\cdots|$ denotes the cardinality of a finite set.

Let $\Gamma_{1}, \ldots, \Gamma_{k}$ be the consecutive components of a given player $A$. We obtained that either $\left|\Gamma_{i+1}\right|=2\left|\Gamma_{i}\right|$ or $\left|\Gamma_{i+1}\right|=\left|\Gamma_{i}\right|$ for $i=1, \ldots, k-1$. Because $\left|\Gamma_{1}\right|=2$, each $\left|\Gamma_{i}\right|$ is a power of 2 , $i=1, \ldots, k-1$. In particular $\left|\Gamma_{k}\right|=2^{u}$ for some $u$.

On the other hand, player $A$ has played with $k$ different opponents by (i). All of them belong to $\Gamma_{k}$, therefore $\left|\Gamma_{k}\right| \geq k+1$.

Thus $2^{u} \geq k+1$, and since $t_{k}$ is the least integer satisfying $2^{t_{k}} \geq k+1$, we conclude that $u \geq t_{k}$. So the size of each $k$-component is divisible by $2^{t_{k}}$, which completes the argument.