Subfield: Combinatorics


Question: Let $n \geq 1$ be an integer and let $t_{1}<t_{2}<\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:

i) The number of games played by each person is one of $t_{1}, t_{2}, \ldots, t_{n}$,

ii) For every $i$ with $1 \leq i \leq n$, there is someone who has played exactly $t_{i}$ games of chess.

Solution:
Let $\mathcal{T}=\left\{t_{1}, \ldots, t_{n}\right\}$. The proof proceeds by induction on $n=|\mathcal{T}|$. If $n=1$ and $\mathcal{T}=\{t\}$, choose a group of $t+1$ people and let every pair of two persons play against each other. Then every person has played $t$ games and the conditions of the problem are satisfied.

In the inductive step, suppose that $\mathcal{T}$ has $n \geq 2$ elements $t_{1}<t_{2}<\cdots<t_{n}$. Consider the set

$$
\mathcal{T}^{\prime}=\left\{t_{n}-t_{n-1}, t_{n}-t_{n-2}, \ldots, t_{n}-t_{1}\right\}
$$

By the inductive hypothesis, there exists a group $G^{\prime}$ of $t_{n}-t_{1}+1$ people that satisfies the conditions of the problem for $T^{\prime}$.

Next construct a group $G^{\prime \prime}$ of $t_{n}+1$ people by adding $t_{1}$ people who do not know any of the other $t_{n}-t_{1}+1$ people in $G^{\prime}$. Finally, construct a group $G$ by complementing the knowledge relation in $G^{\prime \prime}$ : two persons play against each other in $G$ if and only if they do not play against each other in $G^{\prime \prime}$.

By construction $t \in \mathcal{T}$ if and only if there exists a person in $G^{\prime \prime}$ that played against exactly $t_{n}-t$ other people (if $t=t_{n}$, choose one of the $t_{1}$ people added to $\left.G^{\prime}\right)$. That person knows $t_{n}-\left(t_{n}-t\right)=t$ other students in $G$, completing the proof.
Let $\mathcal{T}=\left\{t_{1}, \ldots, t_{n}\right\}$. The proof proceeds by induction on $n=|\mathcal{T}|$. If $n=1$ and $\mathcal{T}=\{t\}$, we choose a group of $t+1$ people such that everyone plays with everyone else. If $n=2$ and $\mathcal{T}=\left\{t_{1}, t_{2}\right\}$ with $t_{1}<t_{2}$, divide the $t_{2}+1$ people into groups $A$ resp. $B$ of size $t_{1}$ resp. $t_{2}-t_{1}+1$ such that everyone from group $A$ played with everyone else whereas people from group $B$ only played with the people from group $A$. Then the people from group $A$ resp. $B$ played with exactly $t_{2}$ resp. $t_{1}$ other people. In the inductive step, suppose that $T$ has $n>2$ elements $t_{1}<\ldots<t_{n}$. Consider the set

$$
\mathcal{T}^{\prime}=\left(\mathcal{T} \backslash\left\{t_{1}, t_{n}\right\}\right)-t_{1}=\left\{t_{n-1}-t_{1}, t_{n-1}-t_{1}, \ldots, t_{2}-t_{1}\right\}
$$

By the induction hypothesis there exists a group $C$ of $t_{n-1}-t_{1}+1$ people that satisfies the conditions of the problem for $T^{\prime}$. Next add groups $D$ resp. $E$ of $t_{1}$ resp. $t_{n}-t_{n-1}$ people such that people from group $D$ played with everyone else whereas people from group $E$ only played with the people from group $D$. Then the people from group $C$ played with $t$ other people if and only if they played with $t-t_{1}$ many people among $C$, i.e. if and only if $t \in\left\{t_{2}, \ldots, t_{n-1}\right\}=$ $\mathcal{T} \backslash\left\{t_{1}, t_{n}\right\}$. People from group $D$ resp. $E$ played with $t_{n}$ resp $t_{1}$ peoples, which completes the proof.
The proof proceeds by induction on $\left|t_{n}\right|$. If $t_{n}=1$ we have $n=1$ and we can consider two persons that play against each other. Then every player has played 1 game and the conditions of the problem are satisfied. If $t_{n}>1$ we distinguish the two cases $t_{1}>1$ and $t_{1}=1$. If $t_{1}>1$ there exists, by the induction hypothesis, a group $A$ of size $t_{n}$ that satisfies the conditions of the problem for $t_{1}^{\prime}=t_{1}-1, \ldots, t_{n}^{\prime}=t_{n}-1$. Now add a new person to $A$ and let him/her play against everyone from $A$. The new group will be of size $t_{n}+1$ and there exists a person which has played $t$ games if and only if there exists a person that has played $t-1$ games within $A$, i.e. if and only if $t \in\left\{t_{1}, \ldots, t_{n}\right\}$. Hence the conditions of the problem are satisfied.

If $t_{1}=1$ there exists, by the induction hypothesis, a group $B$ of size $t_{n-1}$ that satisfies the conditions of the problem for $t_{1}^{\prime}=t_{2}-1, \ldots, t_{n-2}^{\prime}=t_{n-1}-1$. Now add a new person $P$ and let him/her play with everyone from group $B$ and a group $C$ of size $t_{n}-t_{n-1}>0$ and let them play with $P$. The new group will be of size $t_{n-1}+1+\left(t_{n}-t_{n-1}\right)+1=t_{n}+1$. Since person $P$ has played against everyone he will have played $t_{n}$ games. The people in $C$ will have played $1=t_{1}$ games. There exists a person in $B$ that has played $t$ games if and only if there exist a person in $B$ that has played $t-1$ games within $B$, i.e. if and only if $t \in\left\{t_{2}, \ldots, t_{n-1}\right\}$. Hence the conditions of the problem are satisfied.