We have a tournament in which 13 players participate, and matches are played in groups of 4 players (i.e. it's a 4-player game).
There are 13 rounds in total, each player skips one round, and the other 12 are divided into three matches of 4 competitors.
So, for example, if we denote players by 1-13:
in round 1, player 1 skips that round, 2-3-4-5 is one match, 6-7-8-9 the other one, and 10-11-12-13 the last one.
In the end, every player would play in 12 (out of 13) rounds.
Is it possible to make a bracket so that every player plays every opponent exactly three times in total across the duration of 13 rounds?
(In the example above, in Round 1, player 2 played once against players 3, 4 and 5, and zero times against all of the others.)
Best Answer
How about this:
Round $1$: $(2, 3, 4, 5), (6, 7, 8, 9), (10, 11, 12, 13)$
Round $2$: $(1, 4, 5, 7), (3, 9, 12, 13), (6, 8, 10, 11)$
Round $3$: $(1, 5, 6, 11), (2, 9, 10, 13), (4, 7, 8, 12)$
Round $4$: $(1, 3, 8, 12), (2, 7, 11, 13), (5, 6, 9, 10)$
Round $5$: $(1, 2, 4, 9), (3, 7, 10, 11), (6, 8, 12, 13)$
Round $6$: $(1, 3, 9, 11), (2, 7, 10, 12), (4, 5, 8, 13)$
Round $7$: $(1, 2, 6, 12), (3, 5, 8, 10), (4, 9, 11, 13)$
Round $8$: $(1, 7, 9, 10), (2, 5, 6, 13), (3, 4, 11, 12)$
Round $9$: $(1, 7, 8, 13), (2, 5, 11, 12), (3, 4, 6, 10)$
Round $10$: $(1, 2, 8, 11), (3, 5, 7, 13), (4, 6, 9, 12)$
Round $11$: $(1, 3, 6, 13), (2, 4, 8, 10), (5, 7, 9, 12)$
Round $12$: $(1, 4, 10, 13), (2, 3, 6, 7), (5, 8, 9, 11)$
Round $13$: $(1, 5, 10, 12), (2, 3, 8, 9), (4, 6, 7, 11)$
Edit #1:
To see how to come up with this solution, consider the image below:
This image depicts the first round: The white node labelled "1" skips the first round while the three teams are made up of the blue nodes $2,3,4,5$, the green nodes $6,7,8,9$ and the red ones $10,11,12,13$.
In the second round, all quadrilaterals are rotated clockwise by $1/13$ turn (or, equivalently, the node labels are rotated counter-clockwise), so that node "2" gets to skip the round. The new blue team will then consist of nodes $11$, $6$, $8$, and $10$ and so on.
The reason why this construction works is that the $3\cdot {4\choose2}=18$ distances of players of the same team are evenly distributed among the numbers $1$ through $6$:
The 3 node pairs $(6,7)$, $(11,12)$ as well as $(12,13)$ each have distance $1$, the 3 node pairs $(4,5)$, $(7,8)$ and $(11,13)$ have distance $2$ etc - each distance is obtained exactly three times, so each pair of nodes will be included in a team in exactly three rotations.
Edit #2:
I just noticed that an equivalent scenario (even distribution of the distances of pairs of team members) can be achieved in a more symmetric way: