Golf - 17 players/16 in tourney over 6 rounds-??

[Mass]

Tough problem to solve, but the forum folks here seem brilliant...

Parameters:

17 golfers/16 playing in tourney
8 persons on each team - "Ryder Cup" format
6 total rounds over 3 days
Playing format - three foursomes, 1 threesome and 1 twosome (total 17 golfers)

Problem :

In order for each player on the two teams to play against each other once, two rounds will need to be "double matches" - i.e. for two rounds, each player in a foursome has to play 2 matches at once.
(Player 1 on Team A vs. players 3 and 4, both on team B..... PLayer 2 on Team B vs. players 1 and 3.... etc. --> all 4 players in a single foursome here)
This will obviously not work for the threesome or the twosome, but I canot seem to generate an algorithym to get the 7 matches needed for each player (in 6 rounds)  :-/

We also want people playing with different/varied partners over these rounds...  8-)

Whist cannot generate these odd numbers.... !!

Thanks for your help !

Mass

[Ian Wakeling]

Mass,

I am struggling to understand exactly what you are looking for, in particular the 17th player who is not on either team is a little confusing.  Perhaps if you explained, from the point of view of one player on one of the teams of 8, how the schedule would look from their perspective, i.e. how many foursomes/threesomes/twosomes and how many opponents and partners, that would help a lot.

There is one schedule that it may be possible to adapt, so please have a look here.

Ian.

[Mass]

Ian,

Thanks for responding....

Hightlights :

1 guys has opted out of the "tourney" but will still play with us - this has really screwed up our pairings, etc.
Now we cannot go with 4 foursomes (total 16).

- So we have opted to play with 3 foursomes, 1 threesome and 1 twosome.
- We have 2 "teams" (team A AND TEAM B), each with 8 players who will be playing individual matches against the other team's players (i.e. 1A vs. 1B, 1A v. 2B, 1A vs. 2C, etc. and conversely, 1B vs. 1A, 1B vs. 2A, 1B vs. 3A, etc.) Obviously some of these matches are the same/overlap (1A vs. 1B) = ( 1B vs. 1A).
- We are playing a total of 6 rounds of 18 holes to complete these matches in.
- I cannot envision us getting in all the matches in 6 rounds with that many to play (each player needs 8 matches against the 8 players of the opposing team), so I am thinking
we could play 2 matches at once in each foresome - 1st foresome could be players 1A, 2A, 1B, 2B. Each player could have 2 matches (i.e. 1A vs. 1B, 1A vs. 2B....... 2A vs. 1B, 2A vs. 2B)
- But the threesome and twosomes would only have 1 match - (the threesome would have the "lone" 17th player who is not in the tourney every round)

I told you this was baffling to me..... I only have 7 days to figure it out...

Thanks again -

Mass
  

[Ian Wakeling]

So we can ignore player 17 and just consider it a problem of 3 foursomes and 2 twosomes?  I don't have any software that would help with this problem, so I can only really suggest that you use the schedule in the link above.  You need to take one foursome from each round and break it in to 2 twosomes - this will give you the first four rounds.  Finally you need to construct 2 more rounds that cover the pairings that were lost when the foursomes were broken.  Here I am thinking that all foursomes are of the 'two matches at once' kind.

[Mass]

THANKS Much Ian !
Will work on this straight away !

JM