Golf 10 Players / 2- fivesomes / 6 rounds

[tomc]

Golf,

Can anyone help me identify a table or schedule that would provide minimum overlap of playing partners?  Or put another way...  Our group  of 10 golfers want to mix up the rounds; not playing with the same person more than 3-4 times?  We will be playing in groups of 5 for 6 rounds total.

Thanks for the help!

tomc

[Ian Wakeling]

I believe the best solution you can have to this problem is the following:

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

30 of the 45 possible pairings occur within 2 of the fivesomes, the remaining 15 pairings occur within 4 of the fivesomes.

[tomc]

IAN,

Thanks for the quick response!  Now we have a road map to avoid all the confrontations about playing with one person to much!

Tom

[tomc]

Ian,

I put together a quick "crude" spreadsheet showing all the combinations with totals.  I came up with the following:
1-2 plays together 3 times (1-2;3) & 1-3 plays together 2 times (1-3;2) 1-4;3  1-5;4  1-8;4  2-4;4  2-10;4  3-5;4  3-9;4  3-10;4  4-7;4  4-9;3  4-10;3  5-7;4  6-7;4  6-8;4  6-10;4  8-9;4  all other combinations play together 2 times!

Can you give us a table if we play eight rounds instead of six?

Would adding these two rounds spread out the possible combinations better?  Or will adding two more rounds bring a five times playing together into the mix?  In a perfect world there wouldn't be anyone playing together more than 4 times!

Thanks for your brain power!

Tom

[Ian Wakeling]

Tom,

Have another look at the schedule above, as I think 1-2 does play 4 times, and 1-4 only twice.

The 8 round problem has an interesting solution that you might want to go for:

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

Here 40 of the 45 possible pairings occur in exactly 4 of the fivesomes.  The remaining pairings 1-2, 3-4, 5-6, 7-8, 9-10 never occur.  On the face of it, not so good, but you could assign the 10 players to 5 teams of 2 and then you have a round-robin type tournament where players play 4 times with all members of opposing teams.  Is that of any use?

Ian.

[tomc]

Ian,

We are really trying to just spread out the eight rounds of golf to maximize the number each person will play together...  

Can you provide a table that provides the best scenario in a eight round, five-some event?  Even if some players play together no more than five times; but at least all play at least three times together?  Or is this asking to much for the numbers?

Thanks again!

Tom