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##### Requests / Re: Threesome-League with 8 or 10 Players

« Last post by**Sifo-Dyas**on

*April 05, 2020, 05:15:39 AM*»

Thank you very much :-)

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Thank you very much :-)

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With 8 players there is no balanced schedule. There are 28 possible pairs of players, while each threesome covers 3 pairs, however 28 (or 56 for play each other twice) is not divisible by 3.

For 10 players there is a balanced solution as follows:

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

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

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

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

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

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

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

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

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

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

Where the byes are in the last column.

Hope that helps.

For 10 players there is a balanced solution as follows:

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

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

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

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

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

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

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

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

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

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

Where the byes are in the last column.

Hope that helps.

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Hallo Board,

I Need help because of the corona-virus.

We Play in a Skat (german Card game) club and normaly, we meet in real life and can Play threesomes and Foursomes. Because of the corona-virus, we must not meet in bar and have to Play online.

The Problem is, online you only can Play threesomes.

Because of this, I Need a round Robin schedule for 8 or 10 Players of threesomes, everybody Meeting any other Player the same number: (e.g. with 10 Players and a bye, every Player should Play every Opponent twice in 10 rounds).

Hope you can help me.

Thanks

I Need help because of the corona-virus.

We Play in a Skat (german Card game) club and normaly, we meet in real life and can Play threesomes and Foursomes. Because of the corona-virus, we must not meet in bar and have to Play online.

The Problem is, online you only can Play threesomes.

Because of this, I Need a round Robin schedule for 8 or 10 Players of threesomes, everybody Meeting any other Player the same number: (e.g. with 10 Players and a bye, every Player should Play every Opponent twice in 10 rounds).

Hope you can help me.

Thanks

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I don't believe that this is possible. If you meet the criteria regarding the pros (don't play together, and play 10 members each), then you would have to have pairs of members who play each other twice. I think there would be as many as 8 of these pairs. The best solutions will have some asymmetry regarding the pros and the distribution of players over the threesomes and foursomes. For example consider the following:

( 3 12 6) ( 9 8 11) (4 13 1) (5 2 7 10)

(10 1 6) ( 9 12 7) (4 8 2) (5 13 3 11)

(11 2 6) (13 9 10) (4 7 3) (5 8 1 12)

( 2 1 3) ( 7 13 8) (5 9 6) (4 10 11 12)

Where players 4 and 5 are the two pros. Pro 4 meets 9 different members, while Pro 5 meets all 11 members. No pair of players meeet twice. The members have different numbers of opponents too, 10, 11 and 12 have two foursomes, while 6 and 9 are never in a foursome. Hope that helps.

Ian

( 3 12 6) ( 9 8 11) (4 13 1) (5 2 7 10)

(10 1 6) ( 9 12 7) (4 8 2) (5 13 3 11)

(11 2 6) (13 9 10) (4 7 3) (5 8 1 12)

( 2 1 3) ( 7 13 8) (5 9 6) (4 10 11 12)

Where players 4 and 5 are the two pros. Pro 4 meets 9 different members, while Pro 5 meets all 11 members. No pair of players meeet twice. The members have different numbers of opponents too, 10, 11 and 12 have two foursomes, while 6 and 9 are never in a foursome. Hope that helps.

Ian

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This does not work for me either. An alternative would be to use the Excel generator here, but note that you would need to create an account and log on, in order to see the download link.

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This topic has been moved to Requests.

https://www.devenezia.com/round-robin/forum/index.php?topic=1614.0

https://www.devenezia.com/round-robin/forum/index.php?topic=1614.0

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Hi Ian,

I had trouble implementing your idea in a manner which required less computational time than the way I was approaching (probably due to my limited programming skills / the inefficiency of my algorithm), but I am happy to say that I just stumbled across a solution I was looking for after a couple of months of systematically checking permutations using my method. I am posting it below in case anyone else has a need for it, either as is or as a starting point for further investigations.

(1 2 v 3 4) (5 6 v 7 8 ) (9 10 v 11 12) (13 14 v 15 16) (17 18 v 19 20)

(1 3 v 9 17) (2 15 v 11 14) (4 12 v 18 20) (5 13 v 8 16) (6 19 v 7 10)

(1 4 v 8 14) (2 18 v 6 16) (3 19 v 5 15) (7 13 v 17 20) (9 11 v 10 12)

(1 5 v 9 13) (2 6 v 10 17) (3 7 v 14 18 ) (4 11 v 15 19) (8 12 v 16 20)

(1 6 v 15 20) (2 5 v 12 18 ) (3 9 v 16 19) (4 7 v 10 13) (8 11 v 14 17)

(1 7 v 11 16) (2 4 v 17 19) (3 6 v 12 13) (5 10 v 14 20) (8 9 v 15 18 )

(1 10 v 16 18 ) (2 8 v 13 19) (3 5 v 11 20) (4 6 v 9 14) (7 12 v 15 17)

(1 12 v 14 19) (2 7 v 9 20) (3 8 v 10 15) (4 5 v 16 17) (6 11 v 13 18 )

To reiterate, this is a schedule for 20 players, 8 rounds, where every possible pair of 2 players meets at least once, doesn't meet more than twice, and doesn't play as partners twice.

Cheers,

Ray

I had trouble implementing your idea in a manner which required less computational time than the way I was approaching (probably due to my limited programming skills / the inefficiency of my algorithm), but I am happy to say that I just stumbled across a solution I was looking for after a couple of months of systematically checking permutations using my method. I am posting it below in case anyone else has a need for it, either as is or as a starting point for further investigations.

(1 2 v 3 4) (5 6 v 7 8 ) (9 10 v 11 12) (13 14 v 15 16) (17 18 v 19 20)

(1 3 v 9 17) (2 15 v 11 14) (4 12 v 18 20) (5 13 v 8 16) (6 19 v 7 10)

(1 4 v 8 14) (2 18 v 6 16) (3 19 v 5 15) (7 13 v 17 20) (9 11 v 10 12)

(1 5 v 9 13) (2 6 v 10 17) (3 7 v 14 18 ) (4 11 v 15 19) (8 12 v 16 20)

(1 6 v 15 20) (2 5 v 12 18 ) (3 9 v 16 19) (4 7 v 10 13) (8 11 v 14 17)

(1 7 v 11 16) (2 4 v 17 19) (3 6 v 12 13) (5 10 v 14 20) (8 9 v 15 18 )

(1 10 v 16 18 ) (2 8 v 13 19) (3 5 v 11 20) (4 6 v 9 14) (7 12 v 15 17)

(1 12 v 14 19) (2 7 v 9 20) (3 8 v 10 15) (4 5 v 16 17) (6 11 v 13 18 )

To reiterate, this is a schedule for 20 players, 8 rounds, where every possible pair of 2 players meets at least once, doesn't meet more than twice, and doesn't play as partners twice.

Cheers,

Ray

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Good afternoon

I have two golf professionals and 11 members playing 4 rounds of golf.

The end result (hopefully)…..each pro will play in a 4-some twice and a 3-some twice.....therefore, playing with 10 of the 11 members.

Hope this makes some sense......I have got the result close, but just missing something.

Thank you

Chris

I have two golf professionals and 11 members playing 4 rounds of golf.

- each round will be 1 4some & 3 3somes.
- perfect scenario is no one plays with another player more than once AND the two professionals CANNOT play with each other at all.

The end result (hopefully)…..each pro will play in a 4-some twice and a 3-some twice.....therefore, playing with 10 of the 11 members.

Hope this makes some sense......I have got the result close, but just missing something.

Thank you

Chris

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First of all, the "Rounds viewer" has been invaluable in helping me create schedules for my cribbage league. Thank you for making it available. I used the "Table/Cyclic format to initially create a 12-player schedule. As the league grew to 16, 20, 24 and currently 28 players, I was able to easily create new schedules to accommodate the growth.

However, the scheduler does not allow me to create a schedule for greater than 32 players, ie. 36 or 40. Not sure if I'm doing something wrong or if it's my computer or something else.

Any help is greatly appreciated.

Thanks,

Dick

However, the scheduler does not allow me to create a schedule for greater than 32 players, ie. 36 or 40. Not sure if I'm doing something wrong or if it's my computer or something else.

Any help is greatly appreciated.

Thanks,

Dick

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The Excel macros only deal with singles play, so you will find it very hard to achieve any balance for the pairs of players who oppose each other. I think your best strategy is to mould the whist schedules into something that will work with 2 courts. For example if you use the generator that you have linked to, then you need to cut each round down by removing one game, and then pair up the removed games into new rounds - for example with 12 players [2 8 v 3 6] & [9 5 v 10 1] can form a new round, as can the two games underneath and so on. 16 players is really easy, simply partition each round of 4 games into 2 rounds of 2 games. There is a Whist schedule for 13 players (see half way down the Durango Bill page) and you could probably do something similar. 14 or 15 players is problematic, there is no way to balance partners and opponents fairly.

Ian

Ian