Board game tournament

[1000rpm]

Hi all

My first post on here.  I've searched for a solution to this and can't find one.  Its seems similar to Whist scenarios, but without partners.  So in theory it should be simpler, but I can't find an answer anywhere.

I'm going to run a boardgame tournament.  There will be 28 or 40 participants - not sure yet - I've picked those numbers for a good reason - it makes it easier to set up the schedule.
The game is a 4 player game.  I want everyone to play against everyone else once and once only.
So player A might play against B, C and D in their first game.  Player A will not play any of those players again.

I figure that with 28 participants it needs each player to play 9 games to play each other player once (e.g. A plays BCD in game 1, EFG in game 2, HIJ in game 3, etc.  27 other players divided by 3 = 9 games for player A).
40 participants means 39 other people to play, at 3 per game = 13 games for each player.

I'm looking for the solution schedule.  Can anyone help?

Thanks
Tony

[Ian Wakeling]

Hi Tony,

You can find the first schedule on Ed Pegg's page, look for "28 golfers can play in foursomes for 9 days".  Some years ago now, I gave an example of the second schedule here.

While it is true that 12i + 4 players are necessary for the perfect tournament, it is possible to come very close for some other multiples of 4, for example the 32 golfer schedule from the first link, and the 36 player example that I gave here.

I hope that helps the tournament go well.

Ian.

[1000rpm]

Many thanks!

[1000rpm]

I have kicked my tournament off.  You can see the organisation thread here:

http://boardgamegeek.com/geeklist/149758/stone-age-online-tournament