round robin mens doubles

[colmegan]

Hi
I am trying to organise a mens doubles tennis round robin tournament with 10 men. each week 2 doubles matches and 1 singles match. everyone gets to play singles once in the tournament and no-one plays with the same partner in doubles more than once - any ideas please??  colm

[Ian Wakeling]

I think you are looking for just 5 rounds of play as you mention each player having just one singles match.  It is possible to arrange 5 rounds so that all possible opponents (combined over the doubles and singles matches) are seen exactly once each.  Also a player's four doubles partners are all different.

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

This schedule on the Wiseman website which is almost balanced for partners, may also interest you.

Ian.

[colmegan]

Hi Ian

Thank you so much for replying. I did not describe it properly though so here is what I want to do:
run a tournament with 10 men over 9 weeks playing once a week. each week there will be 2 doubles matches and 1 singles match - this way everyone plays singles once in the tournament and everyone plays doubles with the same partner only once but against everyone else twice throughout the tournament - now have I confused you more??  colm

[colmegan]

sorry - got it wrong again! everyone will play singles more than once as there are 9 weeks!!

[Ian Wakeling]

9 rounds is a much harder problem and the singles matches can never be balanced - most players will have 2 singles matches, but 2 players will only have one.  I can extend the idea from the schedule above to 10 rounds and get the following:

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

where player 10 is represented as "X".  But like the 11 round tournament (see Wiseman link) a few pairs of players oppose just once while another few pairs oppose three times. If you leave out the 10th round above, I think you may have a workable solution, at least all the partnerships will be different, so one of your criteria above is met.