Round Robin Tournament Scheduling

### Catan league. 4 player matches, 6 matches each, 10 player league

ringcinema · 6 · 390

#### ringcinema

• Newbie
• Posts: 0
on: January 07, 2020, 03:35:15 PM
It seems that it should be solvable to have each of the 10 players meet every other player exactly twice in the course of 6 matches of four players each. That's 6 matches with 3 opponents per player, with 9 others in the league meeting each one twice. 9*2 = 6*3.

So far, I'm only coming close. Perhaps there's only one solution?
« Last Edit: January 07, 2020, 04:03:38 PM by ringcinema »

#### Ian Wakeling

• Forum Moderator
• God Member
• Posts: 1131
Reply #1 on: January 08, 2020, 05:57:53 AM
Yes, this is possible, but the 15 games all have to be played at different times.  For example:

` 5  3  2  7 8  4  2  9 3  1  6  8 9  7  4  1 8  5  4 10 6  2  3  4 1  5  9  6 3 10  1  4 9  8  3  5 1  2  8  7 4  6  7  5 7  8  6 10 2 10  5  110  7  9  3 6  9 10  2`

« Last Edit: January 20, 2020, 08:15:40 AM by Richard A. DeVenezias »

#### ringcinema

• Newbie
• Posts: 0
Reply #2 on: January 08, 2020, 10:34:01 AM
Yes, thank you. You made short work of it. Is it true that there is a unique solution for this particular form?

#### Ian Wakeling

• Forum Moderator
• God Member
• Posts: 1131
Reply #3 on: January 08, 2020, 10:55:53 AM
It depends on what you mean by unique.  If the order in which the 15 games are played is ignored, and the many possible permutations of player numbers is ignored, then there are 3 different solutions. They are given in full on p29 of the Handbook of Combinatorial Designs, 2nd edition.

#### ringcinema

• Newbie
• Posts: 0
Reply #4 on: January 08, 2020, 11:27:47 AM
I was unaware of this book. Many thanks.

#### ringcinema

• Newbie
• Posts: 0
Reply #5 on: January 11, 2020, 03:48:49 PM
I see what you were saying now. No two games have completely different sets of players, so no two games can be played at the same time. That's novel.