Croquet - 15 Players, 5 per game, 16 Events

[billmoors]

My question is similar to others I've seen, but I can't find the solution in the other examples. For my croquet league, there are 15 members. Each night, we play 3 games concurrently with 5 players on each course (game). I'm hoping to create a schedule so that everyone will play each other at least once over the course of 16 weeks. My hope is that the schedule can be created to maximize the number of times each player plays one another or alternatively minimize the number of times each player plays one another. If it would be easier to perform the calculation with 16 members, that would work too. Any help you can offer will be greatly appreciated!

[Ian Wakeling]

With 15 weeks rather than 16, then it is quite easy to find cyclic solutions where each round can be generated from the round before.  For example:

(1 7 3 14 2) (4 13 9 12 6) (8 10 11 5 15)
(2 8 4 15 3) (5 14 10 13 7) (9 11 12 6 1)
(3 9 5 1 4) (6 15 11 14 8) (10 12 13 7 2)
(4 10 6 2 5) (7 1 12 15 9) (11 13 14 8 3)
(5 11 7 3 6) (8 2 13 1 10) (12 14 15 9 4)
(6 12 8 4 7) (9 3 14 2 11) (13 15 1 10 5)
(7 13 9 5 8) (10 4 15 3 12) (14 1 2 11 6)
(8 14 10 6 9) (11 5 1 4 13) (15 2 3 12 7)
(9 15 11 7 10) (12 6 2 5 14) (1 3 4 13 8)
(10 1 12 8 11) (13 7 3 6 15) (2 4 5 14 9)
(11 2 13 9 12) (14 8 4 7 1) (3 5 6 15 10)
(12 3 14 10 13) (15 9 5 8 2) (4 6 7 1 11)
(13 4 15 11 14) (1 10 6 9 3) (5 7 8 2 12)
(14 5 1 12 15) (2 11 7 10 4) (6 8 9 3 13)
(15 6 2 13 1) (3 12 8 11 5) (7 9 10 4 14)

Above all pairs of players play together either 4 or 5 times and because of the cyclic structure it is easy to arrange that they play on each course exactly 5 times each.   However I think your suggestion of adding a 16th player is good, then it is possible to have a 16 week schedule where each player plays in 15 rounds, has one bye round, and plays against all other players exactly 4 times.

(3 16 8 11 12) (1 7 9 5 13) (15 14 6 4 10)
(9 4 1 12 16) (14 15 11 7 8) (2 13 10 6 5)
(11 16 6 2 4) (12 7 15 1 3) (8 10 9 13 14)
(6 13 11 3 7) (10 8 1 16 15) (5 4 14 9 2)
(13 16 2 8 7) (11 14 4 3 1) (12 10 5 9 15)
(6 15 7 2 9) (16 13 3 10 4) (5 8 12 1 14)
(4 7 5 15 16) (3 14 2 6 8) (1 10 13 12 11)
(7 4 12 14 13) (9 8 16 5 6) (2 15 3 11 10)
(8 5 10 7 3) (11 6 14 12 9) (13 2 1 15 4)
(6 11 15 5 1) (10 12 8 4 2) (16 3 9 14 7)
(4 5 11 8 13) (15 2 14 16 12) (1 9 10 3 6)
(9 1 8 2 11) (14 3 15 13 5) (7 12 6 10 16)
(3 2 16 1 5) (15 6 12 13 8) (4 9 7 10 11)
(2 3 13 9 12) (7 1 4 8 6) (14 11 5 16 10)
(13 6 16 14 1) (12 5 7 11 2) (8 9 3 4 15)
(10 1 2 7 14) (16 11 13 15 9) (5 12 4 6 3)

There is no course balance, but note that each player should appear 3 times in each of the 5 positions within a bracket.

Hope that helps.

[billmoors]

Thank You so much for your help last year, the schedule worked great! We're getting ready to kick off another season and I'm hoping you can help me again. This year, we have 18 players. Each night, we play 3 games concurrently with 6 players on each course (game). I'm hoping to create a schedule so that everyone will play each other at least once over the course of 18 weeks. Last year, you were able to arrange it so everyone played each other 4 or 5 times and if you could arrange something similiar, I would greatly appreciate it! Thank You - Josh Moors: Martian Mallet Club

[Ian Wakeling]

Josh,

Good to know that the schedule above worked for you.  18 players is special as it's possible to have all pairs playing each other exactly 5 times.  The schedule below will do this for you:

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

There are a few things to note. It does not have the cyclic structure, so it's not balanced for courses.  Either you need to randomize the course assignments for each of the 17 weeks or try to get best possible balance somehow.  Secondly rounds 1 and 17 of the schedule are the same, which is why I have placed them as far apart as possible.  If this is an issue then simply leave out the last round and you will have pairs playing together either 4 or 5 times.

Hope that helps.

[billmoors]

Thanks again for the quick response! I'll drop the values into my spreadsheet tonight to make sure it works out and let you know. Once again, THANK YOU!

[billmoors]

Ian - Thanks again for your help. I actually need 18 weeks instead of 17. Since 17 was a repeat of 1, do I just grab week 2 and repeat it for week 18? Or is there some other evaluation that needs to take place for 18 events? Thanks again - Josh

[Ian Wakeling]

Josh,

Since the 17 weeks are fully balanced with respect to pairs, then you are free to choose any 18th week you like (no need to repeat a previous round).  The 45 pairs that occur within the 3 games for 18th week will be the ones that are repeated 6 times, all other pairs will be repeated 5 times.

Ian.

[billmoors]

Great! Thanks again for your help!!!