Game Draw Brainteaser

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psychman

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Oct 13, 2005
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I am designing a tournament.
There are 22 teams.
Each team must play each other team once, and only once (that's 21 games per team.)
Each game involves 4 teams.
I need ideas about how to create the draw, so that the combination of teams ends up with the above conditions being met.
Call them Team 1, Team 2 and so on. I imagine I will end up with a list of team combinations like that below
Game 1: 2, 12, 16, 21
Game 2: 3, 7, 14, 19

Thanks
 
Since each game involves 4 teams, then each team plays 7 games:
1,2,3,4
1,5,6,7
1,8,9,10
1,11,12,13
1,14,15,16
1,17,18,19
1,20,21,22

Then, something like:
2,5,8,11
2,6,9,12
2,7,10,13
2,14,17,20
2,15,18,21
2,16,19,22

So we now have 1 and 2 completed...
Am I understanding it correctly?
 
One grows confused.
Game 1: 2, 12, 16, 21
Does that mean you have satisfied
team 2 playing teams 12, 16 and 21
team 16 playing teams 2, 12 and 21
etc.
all in game 1 so there will be only 7 games played by team 2, not 21?
Help
------------------
Gene :!:
P.S. Ok, overlapping again.
 
I got as far as what Gene and Denis got, and thought I could just keep creating 4 digit combinations, but there came a point where I ran out of places to go - kept getting doubleups. Would appreciate anybody who can come up with a magic formula for working it out. (Or just doing it all for me and posting the entire list!)
Psychman
 
2,5,8,11
2,6,9,12
2,7,10,13
2,14,17,20
2,15,18,21
2,16,19,22

So, carrying on using "2" above as a guide, we can go diagonally down, starting with "5,9,13":

3,5,9,13
3,6,10,20
3,7,17,21
3,14,18,22 ; now we're left with top right corner (8,11,12) and bottom left corner(15,16,19):
3,11,15,19
3,8,12,16

This takes care of team #3.
If you're able to continue this to team#7, you're done.

I guess you'd need to keep a running tab of the teams played so far by a certain team;
like, 8 has played 1,2,3,5,9,10,11,12,16

I'm too lazy to continue...plus don't wanna headache :shock:
 
Thanks for the reply Denis. Unfortunately I could not make it work.

From
3, 5,9,13
3,6,10,20
3,7,17,21
3,14,18,22
3,11,15,19
3,8,12,16

Using the diagonal method, I got to
4,5,10,21
4,6,17,22
4,7,18,19

Then realised that 18 and 19 have already met back in Game 1.
Not sure if I have understood your instructions properly or not, any tips greatly appreciated!

Psychman
 
I would, and I have been cogitating, but I spent a GREAT deal of time on a similar problem a few years ago and bombed out. (It seems to have been deleted.) I strongly suspect it is impossible but I can't prove that either.

PS. I posted one that had a hole in the theory so big I couldn't see it, so I'm deleting it now.
 
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