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Golf foursomes
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Harold Druss
science forum beginner


Joined: 15 Feb 2006
Posts: 2

PostPosted: Wed Feb 15, 2006 9:15 pm    Post subject: Golf foursomes Reply with 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?
Thanks
Harold
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Proginoskes
science forum Guru


Joined: 29 Apr 2005
Posts: 2593

PostPosted: Thu Feb 16, 2006 1:19 am    Post subject: Re: Golf foursomes Reply with quote

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
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Rob
science forum beginner


Joined: 20 Jun 2005
Posts: 44

PostPosted: Wed Feb 22, 2006 5:17 am    Post subject: Re: Golf foursomes Reply with 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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Harold Druss
science forum beginner


Joined: 15 Feb 2006
Posts: 2

PostPosted: Thu Feb 23, 2006 10:14 am    Post subject: Re: Golf foursomes Reply with quote

Hi Rob
Thanks for your reply.

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
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Dana DeLouis
science forum beginner


Joined: 06 Mar 2006
Posts: 37

PostPosted: Sat Mar 11, 2006 3:57 am    Post subject: Re: Golf foursomes Reply with quote

Thanks for this info. I've never understood the additional info at this
link...
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
news:1140585474.423234.27620@g43g2000cwa.googlegroups.com...
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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