1

##### Comments and Thanks / Printing

« Last post by**Angelo**on

*January 16, 2020, 08:16:53 AM*»

How can I print a round robing chart

1

How can I print a round robing chart

2

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.

3

There is a schedule here for 7 rounds and it links to another with 14 rounds. I think one of those should work for you.

4

I am trying to figure out a 14 player Euchre Round Robin Rotation. Everyone plays with a different partner. Everyone sits twice. Is this possible?

5

I was unaware of this book. Many thanks.

6

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.

7

Yes, thank you. You made short work of it. Is it true that there is a unique solution for this particular form?

8

It would be much better to have 10 weeks, then each player can have exactly 2 bye rounds, and exactly 8 games. For example:

( 5 6 v 8 9) (4 3 v 2 10) (1 7)

( 1 7 v 9 10) (5 4 v 3 6) (2 8)

( 2 8 v 10 6) (1 5 v 4 7) (3 9)

( 3 9 v 6 7) (2 1 v 5 8) (4 10)

( 4 10 v 7 8) (3 2 v 1 9) (5 6)

( 8 10 v 9 5) (1 6 v 4 2) (3 7)

( 9 6 v 10 1) (2 7 v 5 3) (4 8)

(10 7 v 6 2) (3 8 v 1 4) (5 9)

( 6 8 v 7 3) (4 9 v 2 5) (1 10)

( 7 9 v 8 4) (5 10 v 3 1) (2 6)

Leave out the last round if 9 rounds is a hard limit for you, however only players 2 and 6 will have 8 games, the rest of the players will only have 7 games.

( 5 6 v 8 9) (4 3 v 2 10) (1 7)

( 1 7 v 9 10) (5 4 v 3 6) (2 8)

( 2 8 v 10 6) (1 5 v 4 7) (3 9)

( 3 9 v 6 7) (2 1 v 5 8) (4 10)

( 4 10 v 7 8) (3 2 v 1 9) (5 6)

( 8 10 v 9 5) (1 6 v 4 2) (3 7)

( 9 6 v 10 1) (2 7 v 5 3) (4 8)

(10 7 v 6 2) (3 8 v 1 4) (5 9)

( 6 8 v 7 3) (4 9 v 2 5) (1 10)

( 7 9 v 8 4) (5 10 v 3 1) (2 6)

Leave out the last round if 9 rounds is a hard limit for you, however only players 2 and 6 will have 8 games, the rest of the players will only have 7 games.

9

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 1

10 7 9 3

6 9 10 2

10

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?

So far, I'm only coming close. Perhaps there's only one solution?