Round Robin Tournament Scheduling

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Title: Complicated euchre
Post by: mldarby77 on January 14, 2007, 05:47:04 PM

Hi. First of all, thanks for taking the time to possibly think about this.

I'm trying to put together a euchre league. I want it to have 2 divisions (A and B) of 8 people each. I want a schedule so that you play as a teammate with everyone in your own division, but play against different combinations of player from the other division.

So, a short verion for one player would be..

Round 1: 1A & 8A vs 1B & 8B
Round 2: 1A & 7A vs 2B & 4B
Round 3: 1A & 6A vs 3B & 5B
etc.

Thanks for any help, and feel free to ask me any questions.
-Matt

Title: Re: Complicated euchre
Post by: Ian Wakeling on January 16, 2007, 12:34:08 PM

An interesting problem Matt.  I think it makes sense to look for a schedule with 7 rounds where players are partnered once with each of the other 7 players from their own division.  This means that a player sees 14 opponents, which creates some difficulty as 14 is not divisible by 8.  So how about the following where each player opposes 7 of the players from the other division exactly twice, but never opposes the remaining player?

Round
  1   (1A 2A) v (4B 7B)   (3A 4A) v (1B 5B)   (5A 6A) v (3B 8B)   (7A 8A) v (6B 2B)
  2   (1A 3A) v (7B 6B)   (2A 7A) v (5B 3B)   (4A 5A) v (2B 8B)   (6A 8A) v (1B 4B)
  3   (1A 4A) v (6B 8B)   (2A 8A) v (4B 5B)   (3A 5A) v (7B 2B)   (6A 7A) v (3B 1B)
  4   (1A 5A) v (3B 4B)   (2A 6A) v (8B 7B)   (3A 8A) v (5B 6B)   (4A 7A) v (1B 2B)
  5   (1A 6A) v (5B 2B)   (2A 3A) v (1B 8B)   (4A 8A) v (3B 7B)   (5A 7A) v (6B 4B)
  6   (1A 7A) v (8B 5B)   (2A 4A) v (3B 6B)   (3A 6A) v (2B 4B)   (5A 8A) v (7B 1B)
  7   (1A 8A) v (2B 3B)   (2A 5A) v (6B 1B)   (3A 7A) v (4B 8B)   (4A 6A) v (7B 5B)

1A never opposes 1B, 2A never opposes 2B etc..

Ian.