Round Robin Tournament Scheduling

12 golfers  3 foursomes 7 days/ balanced

mariano_torre · 7 · 5218

mariano_torre

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on: February 08, 2010, 07:58:06 PM
please  somebody can help?
thanks
« Last Edit: February 08, 2010, 07:58:36 PM by mariano_torre »


Ian Wakeling

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Reply #1 on: February 09, 2010, 03:19:15 AM
Please review some of the old threads from this message board here and here.
« Last Edit: February 09, 2010, 03:19:37 AM by Ian »


mariano_torre

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Reply #2 on: February 09, 2010, 06:15:24 AM
thank you very much!!!!


mariano_torre

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Reply #3 on: March 22, 2010, 04:58:37 PM
Dear Ian

After  a  carefull analysis of the proposed grid  we found out that I may have not specified clearly our needs

 

Please let me specify our needs so  that you may be able to  help us  find a more appropriate solution.

 

 I)We are looking for  a grid of  12 players ,playing 7 days.

 

Each  day there are 3 lines of 4 players each

 

The format is such that  each day  the players play simultaneously  3 formats, as follow

 

a)      Each player plays a medal play (individual score)

 

b)      fourball :  which means that  each day, each player plays in a twosome  with a different partner   ( thus,any player  plays in 7 days in twosomes with  other 6  players) choosing the best score of the two balls, hole by hole and adding up the  scores  for the 18 holes. Each team of two players play against the other 5 teams.

The lower score  give points to  each player of the winning team

 


c)      Each line formed  by four  players play as a team, choosing in each hole the best score adding up the  scores of the  18 holes.The lower  score give points to each player of the winning team.

 

At the end of each day, then each player get a number of points as a result of his own individual game (medal), plus  the points he  may have  earned as part of a twosome plus the points he may have earned as part of a line of  four.

 

With this format in mind what we need is to have each player play with a different partner each day in a twosome without repetition and to play with all the players in lines of four with the minimum possible repetitions.

 

 

 

II)  12 players ,  12 days

The same concept we would need  for 12 players and 12 days ( in this case a particular player will play  with the other 11 partners in a team of two at least once with  some repetitions on the 12th day ,but we would  need to minimize the repetitions with  the other players when forming the groups of  four players


Hope that is not to much and you can help us
thank you very much


Ian Wakeling

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Reply #4 on: March 24, 2010, 04:47:20 AM
I am not sure that I understand your requirements - and am confused by the three simultaneous formats.  Do you mean that you have three different ways of scoring the same matches?

If I have understood correctly that the twosomes are different on each day and they go around the course 2 twosomes at a time, then the following might help:

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


That's only 6 rounds so you would have to make up a 7th using twosomes  that have not yet occured.

I have something similar for 12 rounds:

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


All possible twosomes occur (either on the left or right of the 'v') at least once, six occur twice (1&12 2&10 3&11 6&7 8&4 9&5).


mariano_torre

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Reply #5 on: March 24, 2010, 01:31:07 PM
FIRST OF ALL thanks
I will try to phrase in another way
we are going to play 7 days 12 people in 4 somes( 2 *2somes)
we would like no 2somes repeats  (taht meas evry player will play (2somes)at least with 7 different guys)and maximun socilaize in 4somes
thanks again


Ian Wakeling

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Reply #6 on: March 25, 2010, 04:09:31 AM
I think the two schedules abouve will work for you.  The 'v' is slightly misleading, just consider the pair of players on the left of the 'v' as one twosome and the two players on the right as the second twosome and it should work out well.

Something like

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

added to the first scedule will make a 7th round without repeating any twosomes.

Hope that helps.