I got curious about this when making groups for a tournament. We have have 11 players, and the tournament consists of 4 rounds. In every round, there will be three groups. Two groups of 4 people, and one group of 3 people. My questions are:
Is there a way to set up the groups so that everyone plays with everyone at least once? If not, what algorithm can be used to minimize repetition in the groups?
And further, is ensuring that everyone is in the group of three at least once also possible?
Thanks!
Sorry for having to bring bad news, but the setup you are looking for does not exist.
We can prove this by contradiction, assuming that there exists an arrangement of eleven people, labeled $a$ through $k$, into four rounds (I - IV) with three groups $X$, $Y$, and $Z$ each such that $|X| = |Y| = 4$ and $|Z| = 3$ and such that every person plays with everyone else at least once.
By counting the possible numbers of players in the groups, we can make the following observations
$\tag{1}\text{No player may be part of group $Z$ more than twice.}$ $\tag{2}\text{A player who plays in group $Z$ twice plays with everyone else exactly once.}$
On the other hand, since there is a total of $4 \cdot |Z| = 12$ players in group $Z$, the pigeonhole principle tells us that there is at least one person to play in group $Z$ twice, and by observations $(1)$ and $(2)$, we know that she must play in a group other than $Z$ twice, and that she plays with everyone else exactly once. Therefore, without loss of generality, we can assume the arrangement to look something like this:
$$ \begin{array}{c|cccc|cccc|ccc} & & & X & & & & Y & & & Z & \\ \hline \text{I} & . & . & . & . & . & . & . & . & a & b & c \\ \text{II} & . & . & . & . & . & . & . & . & a & d & e \\ \text{III} & a & f & g & h & . & . & . & . & . & . & . \\ \text{IV} & a & i & j & k & . & . & . & . & . & . & . \\ \end{array} $$
We now look at the set $\mathcal{S }= \{b, c, d, e\}$ of players sharing group $Z$ with player $a$ in rounds I and II. First we observe $\tag{3}\text{In rounds III and IV, players $b$ and $c$ never play together in group $Z$.}$
Otherwise, both of them would be in $Z$ twice, so by $(2)$ they can only play with anyone else once, but since they already play together in round I, they couldn't play together again at all, a contradiction.
Moreover, $\tag{4}\text{In rounds III and IV, players b and c never play together at all.}$
Otherwise, by $(3)$ they would need to be in group $Y$ both in rounds III and IV, contradicting $(1)$.
Of course, the same holds for players $d$ and $e$, so $(4)$ implies that the four pairs $\{b, d\}$, $\{b, e\}$, $\{c, d\}$, and $\{c, e\}$ must be placed in groups $Y$ and $Z$ in rounds III and IV in such a way that no two of these pairs ever appear in the same group within the same round. However, since disjoint pairs must be in the same rounds, there is no way disjoint pairs can be in different groups, so inparticular at least one element of $\mathcal{S}$ must be in group $Z$ both in round III and round IV which contradicts $(1)$.