Round Robin Tournament Scheduling

Schedules - You must register to Post and Download => Requests => Topic started by: bez on June 06, 2013, 06:25:24 AM

Title: 6 Golfers, 4 rounds, playing in 3s
Post by: bez on June 06, 2013, 06:25:24 AM

Hi there,

I'm off on golf tour with 5 friends (6 in total), playing 4 rounds in groups of 3.

Can anyone help with the best schedule ?

I understand that a perfect schedule isn't possible based on the numbers above

Many thanks in advance

Title: Re: 6 Golfers, 4 rounds, playing in 3s
Post by: Ian Wakeling on June 06, 2013, 09:07:08 AM

It's actually impossible to arrange that everyone plays with each other at least once.  The best that can be done is:

(1 2 3) (4 5 6)
(1 2 6) (4 5 3)
(1 5 3) (4 2 6)
(1 5 6) (4 2 3)

where the pairs (1 4) (2 5) & (3 6) never play with each other and the other 12 pairs all play together twice.

Title: Re: 6 Golfers, 4 rounds, playing in 3s
Post by: bez on June 06, 2013, 09:25:45 AM

Hi Ian,

Thanks for the quick response....

I came up with this that sees everyone playing in the same group as everyone else at least once, but player 2 plays with player 3 every day :

123   456
156   234
145   236
146   235

I can see 10 possible groupings, so it should just be a question of combining the best 4 for maximum spread

round 1      123      456
round 2      124      356
round 3      125      346
round 4      126      345
              
round 5      134      256
round 6      135      246
round 7      136      245
              
round 8      145      236
round 9      146      235
              
round 10      156      234

If anyone can see a different solution I'd be very grateful

Title: Re: 6 Golfers, 4 rounds, playing in 3s
Post by: Ian Wakeling on June 06, 2013, 11:14:50 AM

Your are correct of course, I should have been more careful.  Thinking about the pairs of players who play together, some pairs will play together only m times (or not at all in the case of m=0), while other pairs will play together as many as n times, where m<=n.  Ideally we would like m=n for a fair tournament, but this is not possible.  My solution makes m and n as close together as possible, while your solution maximizes m.  They are two different ways of thinking about making it as fair as possible.  The cost of having  having m=1, is that n must be 4 for one pair, while the cost of having m and n as close as possible is that m=0 for three pairs.   I don't believe that there is any other useful schedule in between, so you should go for one or the other.  Hope that helps.

Title: Re: 6 Golfers, 4 rounds, playing in 3s
Post by: bez on June 06, 2013, 03:52:02 PM

Thanks again Ian, having run through a fair few manual permutations I'm fast coming to the same conclusion...