Volleyball 4 v 4 king of the beach

[rsswan]

I am running a 4 v 4 tournament, king of the beach whist style with 8 players, playing equal rounds with and against each other.  I can find a lot of doubles tournaments but not teams of 4.  Any help is appreciated.   Thanks!

[Ian Wakeling]

You can have a schedule in 7 rounds where pairs of players are on the same team three times and oppose four times:

(8 5 1 6) : (2 3 4 7)
(8 6 2 7) : (3 4 5 1)
(8 7 3 1) : (4 5 6 2)
(8 1 4 2) : (5 6 7 3)
(8 2 5 3) : (6 7 1 4)
(8 3 6 4) : (7 1 2 5)
(8 4 7 5) : (1 2 3 6)

[rsswan]

Thanks Ian.  That's exactly what I needed.  You are awesome!!!

[robert_hume]

You can have a schedule in 7 rounds where pairs of players are on the same team three times and oppose four times:

(8 5 1 6) : (2 3 4 7)
(8 6 2 7) : (3 4 5 1)
(8 7 3 1) : (4 5 6 2)
(8 1 4 2) : (5 6 7 3)
(8 2 5 3) : (6 7 1 4)
(8 3 6 4) : (7 1 2 5)
(8 4 7 5) : (1 2 3 6)

I was looking to do just this! And perhaps a 10 person (5 x 5) version. Are you hand calculating these through trial and error or is there a formula / algorithm you use similar to whatever bjerring did for 7 and 9 person triples?

[Ian Wakeling]

There is a 'formula', but it is probably not what you are expecting.  Have a look at this 2003 paper by Abel et al, which is freely downloadable.  Theorem 14 gives a way to make the 8 player schedule above (with p=2, n=2, m=1 & s=0), unfortuately Theorem 7 says that the 10 person schedule that you are looking for, specifically a (5,10) GWhD(10), does not exist, in other words no balanced solution is possible.

I am not sure what you mean by the 7 and 9 person triples, can you provide a link?

[Ian Wakeling]

To clarify, the non-existence of the 10 player scedule refers to one in which there are 9 rounds of play.  However I can find an 18 round scedule where all partners occur 8 times and all opponents occur 10 times.  For example:

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

To find this I am using a special program that searches for cyclic schedules.

[robert_hume]

There is a 'formula', but it is probably not what you are expecting.  Have a look at this 2003 paper by Abel et al, which is freely downloadable.  Theorem 14 gives a way to make the 8 player schedule above (with p=2, n=2, m=1 & s=0), unfortuately Theorem 7 says that the 10 person schedule that you are looking for, specifically a (5,10) GWhD(10), does not exist, in other words no balanced solution is possible.

I am not sure what you mean by the 7 and 9 person triples, can you provide a link?

Hi Ian, thanks for the response. I was referring to the Bjerring tournament designs: https://vbc.cyburi.com/resource/LinearRankingTournament.pdf

See page 11 of this document.

[Ian Wakeling]

Many thanks for the link.  The two schedules on p11 look to me like they have been made with the "Balanced Incomplete Blocks" method that is hinted at on page 29.  The way the 9 player schedule is made, the same teams of 3 players are used twice each, once on the left of the court and once on the right.  It might be possible to have better mixing, so all teams of 3 are different - I will think some more.