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Harold Druss
science forum beginner

Joined: 15 Feb 2006
Posts: 2

Posted: Wed Feb 15, 2006 9:15 pm    Post subject: Golf foursomes

Hi All
I have been asked by a group of golfers (20) if it would be possible
for every golfer to play at least once with every other golfer on a 7 day
golf trip.
They will play in foursomes.
If not 7 days, then how many days would it take?
Thanks
Harold
Proginoskes
science forum Guru

Joined: 29 Apr 2005
Posts: 2593

Posted: Thu Feb 16, 2006 1:19 am    Post subject: Re: Golf foursomes

Harold Druss wrote:
 Quote: Hi All I have been asked by a group of golfers (20) if it would be possible for every golfer to play at least once with every other golfer on a 7 day golf trip. They will play in foursomes. If not 7 days, then how many days would it take?

Google for "whist tournaments". This problem is well-studied, and
you'll find lots of useful links (or so I've heard).

--- Christopher Heckman
Rob
science forum beginner

Joined: 20 Jun 2005
Posts: 44

Posted: Wed Feb 22, 2006 5:17 am    Post subject: Re: Golf foursomes

(Note first that time has no relavence on the math: you would need to
determine how many games can be played in a day, and then work out an
ideal schedule for the correct number of games...I am concerned here
only with determining the number of games necessary)

Idealy, you are looking for a S(2,4,20) steiner system. A S(l,k,v) is
a collection of k-tuples of a set with v elements such that each
l-tuple appears in exactly 1 of the k-tuples.

EG: S(2,3,7) is a collection of triples of a set of 7 elements (say 1,
2, 3, ..., 7), such that each pair occures in exactly one triple:
1,2,4
2,3,5
3,4,6
4,5,7
5,6,1
6,7,2
7,1,3

However there can't be a S(2,4,20), since by a well known theorem on
general block designs (see Design Theory: Volume 1 by Thomas Beth, H
Lenz, D Jungnickel, pg11, corollary 2.11, 2.11.a), it must be that

v-1 \equiv 0 (mod k-1)

and here that would mean that 20-1 \equiv 0 (mod 4-1), or 19 \equiv 0
(mod 3), which is not true.

Thus, there is no possible way that everyone could play each other
exactly once (assuming you stick to only foursoms)--to answer this
question, you will need to allow some people to play others more than
once (for example, Peter, Paul, Jane, and Wendy may play one game, and
later, it may be necessary for Peter, James, Shaun, and Wendy to play
each other (Peter and Wendy repeat).

There is a maximum number of games you need: just (20 choose 4) = 4845
games. This would cover every possible foursome. The question
remains...how many of those games do we not have to play because
everyone in them has played everyone else at least once...

(It was recently showed that for 25 people, it would work out
perfectly: see http://www.maths.gla.ac.uk/~es/steiner.html, and I think
16 people works too, though I cant find a reference at this moment)

Thats about all I can post right now timewise... Maybe I'll post more
later.

Rob
Harold Druss
science forum beginner

Joined: 15 Feb 2006
Posts: 2

Posted: Thu Feb 23, 2006 10:14 am    Post subject: Re: Golf foursomes

Hi Rob

 Quote: Thus, there is no possible way that everyone could play each other exactly once (assuming you stick to only foursoms)--to answer this question, you will need to allow some people to play others more than once (for example, Peter, Paul, Jane, and Wendy may play one game, and later, it may be necessary for Peter, James, Shaun, and Wendy to play each other (Peter and Wendy repeat).

This is fine. The object is to try and have each golfer play at least once
with
every other golfer on the trip. The fact that they play more than once with
someone
is OK. This is a social thing, not something for competition.
Thanks
Harold
Dana DeLouis
science forum beginner

Joined: 06 Mar 2006
Posts: 37

Posted: Sat Mar 11, 2006 3:57 am    Post subject: Re: Golf foursomes

Thanks for this info. I've never understood the additional info at this
http://mathworld.wolfram.com/SocialGolferProblem.html
I've tried in the past to write a program, but never had any luck. :>(
--
Dana

"Rob" <robertborgersen@gmail.com> wrote in message
 Quote: (Note first that time has no relavence on the math: you would need to determine how many games can be played in a day, and then work out an ideal schedule for the correct number of games...I am concerned here only with determining the number of games necessary) Idealy, you are looking for a S(2,4,20) steiner system. A S(l,k,v) is a collection of k-tuples of a set with v elements such that each l-tuple appears in exactly 1 of the k-tuples. EG: S(2,3,7) is a collection of triples of a set of 7 elements (say 1, 2, 3, ..., 7), such that each pair occures in exactly one triple: 1,2,4 2,3,5 3,4,6 4,5,7 5,6,1 6,7,2 7,1,3 However there can't be a S(2,4,20), since by a well known theorem on general block designs (see Design Theory: Volume 1 by Thomas Beth, H Lenz, D Jungnickel, pg11, corollary 2.11, 2.11.a), it must be that v-1 \equiv 0 (mod k-1) and here that would mean that 20-1 \equiv 0 (mod 4-1), or 19 \equiv 0 (mod 3), which is not true. Thus, there is no possible way that everyone could play each other exactly once (assuming you stick to only foursoms)--to answer this question, you will need to allow some people to play others more than once (for example, Peter, Paul, Jane, and Wendy may play one game, and later, it may be necessary for Peter, James, Shaun, and Wendy to play each other (Peter and Wendy repeat). There is a maximum number of games you need: just (20 choose 4) = 4845 games. This would cover every possible foursome. The question remains...how many of those games do we not have to play because everyone in them has played everyone else at least once... (It was recently showed that for 25 people, it would work out perfectly: see http://www.maths.gla.ac.uk/~es/steiner.html, and I think 16 people works too, though I cant find a reference at this moment) Thats about all I can post right now timewise... Maybe I'll post more later. Rob

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