Round Robin Tournament Scheduling

20 golfer schedule recap

don8564 · 3 · 7295

don8564

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on: October 08, 2011, 06:14:06 PM
This may seem like a re-hash of several postings.  We take an annual golf trip with a group of 16 that will most likely grow to 20 next year.  During the trip we play 7 rounds at 7 different courses.  Our goal is to have everyone play with everyone once at least once in a foursome, and once we start doubling up pairs we try not to triple any pair.  For 16 it's fairly easy - we use a social square for the first 5 rounds which puts everyone with everyone else exactly once.  For the next two rounds we repeat any two of the first 5 groupings, modifying them somewhat to get some specific doubling of pairs.  To be clear, in a foursome together is all that matters, regardless of whether they are partners (same cart) or opponents (other team/cart).

For next year, and most likely 20 golfers, still over 7 rounds, we want to do the same.  If I've read thru the various posts on this site correctly, it appears we can't do what I want, specifically, have each of the 20 golfers play with everyone else at least once and not triple up any pair.  The first 5 rounds are fairly straightforward, using the 16 social square and adding 4 more golfers.  After 5 rounds, there are no doubles but each golfer has played with 15 other golfers once, but each golfer has yet to play with the remaining 4 golfers (and the 4 golfers are not the same for any of the 16).  So far, so good, and then it gets hit/miss.

The best I've been able to do in 7 rounds leaves 6 different pairs of players that have not played with each other, but every other pair has play together at least once and several have played twice, and there are no triples, which is good.  I went to one site from a link here and got the Whist schedule for 20 over 7 rounds.  That schedule takes 19 rounds to complete and has more conditions than we want, i.e., playing "against" everyone else twice, our only requirement is being in the same foursome.  I was hopeful that I could use 7 of the Whist schedule pairing lines to create our schedule but I've not been able to find seven that give any better result than what I've done with the social square and hit/miss process.

So the first question I have is does anyone have a schedule for 20 golfers over 7 rounds that puts everyone in a foursome with everyone else at least once but no more than twice?  The second or followon quetstion, assuming the first answer is no, is does anyone have a schedule that improves on mine, i.e., fewer than 6 pairs that have not played together?
« Last Edit: October 08, 2011, 08:53:36 PM by dlinnell »


Ian Wakeling

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Reply #1 on: October 09, 2011, 02:43:52 PM
If you allow fivesomes for one round then there is a solution in 6 rounds where all pairs play together exactly once.

However, the answer to your principal question is, yes.  For example the following schedule covers all pairs once and no more than twice:

10   3   8  16
13   1  20   4
14   6  15   9
 2   5  17  12
11  18  19   7

 8  20  12  14
 7  15  17   3
16  18   1   2
 4  10  19   6
 9   5  11  13

15  20   1  18
13   5   8   4
 7  16   6  12
 2   9   3  19
17  11  10  14

20   5   3   6
10   9  18  12
15  16  11   4
 8   1  19  17
14  13   7   2

11   1   3  12
 8   7   9   4
10  15   2  20
 6  18  13  17
19   5  16  14

13  19  12  15
 6   8   2  11
 1   7  10   5
14   3   4  18
17   9  16  20

14   1   6   9
20  11   7  19
 8   5  15  18
 4  17   2  12
10  13  16   3


don8564

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Reply #2 on: October 15, 2011, 09:32:36 PM
Thanks.  Sorry I took a while getting back, been traveling.  The numbers work well, appreciate the help, just what we were looking for.
« Last Edit: October 15, 2011, 09:33:00 PM by dlinnell »