Semi-random team assignment: 36 players will play in 7 rounds of 6 games

[animals]

I'd like to set up a competition where a total of 36 players will play in 7 rounds of 6 games, each of which involves 2 teams of 3 players. Ideally, each player will play in a single game (with or against) each other player. I can make it work for 16 players, playing 5 rounds of 4 games, each involving 2 teams of 2 players. Can anyone come up with a solution for 36 players?

 

[Romsek]

Since you don't care whether players play for or against each other in a given game I'd simplify the problem statement
by saying that in each game there are 6 players w/o the need for considering teams.

If you can solve that problem, partitioning the game solutions into teams is trivial.

 

[animals]

Yes, there is no need to consider the composition of the individual teams; however, partitioning into teams is certainly not simple as, after a few rounds, it is difficult to create groups of 6 where all the players haven't played in a game with each of the others before.

 

[pka]

I'd like to set up a competition where a total of 36 players will play in 7 rounds of 6 games, each of which involves 2 teams of 3 players. Ideally, each player will play in a single game (with or against) each other player. I can make it work for 16 players, playing 5 rounds of 4 games, each involving 2 teams of 2 players. Can anyone come up with a solution for 36 players?

I would like to help with this. But I simply do not understand the setup. What does "will play in 7 rounds of 6 games" mean?
If there are thirty-six players to be placed on teams of three players each then that is twelve teams and there are \(\displaystyle \dfrac{36!}{(3!)^{12}(12!)}=356765771022012352000000\) ways to do that. These are known as unordered partitions because the team are not named.
Please post some your own work by way of telling us the setup.

 

[animals]

If there are 4 players:

games_per_round = sqrt(players) = 2​

players_per_game = sqrt(players) = 2​

rounds = sqrt(players) + 1 = 3​

​

Round 1: Game 1 - {1,2} / Game 2 - {3,4}​

Round 2: Game 1 - {1,3} / Game 2 - {2,4}​

Round 3: Game 1 - {1,4} / Game 2 - {2,3}​

​

Each player has played in a game with every other player, but only once

If there are 16 players:

games_per_round = sqrt(players) = 4​

players_per_game = sqrt(players) = 4​

rounds = sqrt(players) + 1 = 5​

​

Round 1: {1,11,5,15} / {6,13,4,7} / {3,14,12,9} / {10,8,16,2}​

Round 2: {11,7,9,16} / {5,6,12,8} / {1,4,14,10} / {3,13,2,15}​

Round 3: {7,14,15,8} / {4,9,5,2} / {13,12,10,11} / {16,6,1,3}​

Round 4: {11,6,2,14} / {9,1,8,13} / {3,5,7,10} / {16,15,12,4}​

Round 5: {15,9,10,6} / {1,12,2,7} / {8,11,3,4} / {5,14,16,13}​

If there are 36 players:

games_per_round = sqrt(players) = 6​

players_per_game = sqrt(players) = 6​

rounds = sqrt(players) + 1 = 7​

​

I can set up about 4/5 rounds, but then keep getting the situation where player A hasn't played in a group with player B, but player B has played against too many other players that player A has already played against to put them together in a group of 6.​