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Golf pairings solution required

# Golf pairings solution required

Posted in the Mathematics Forum

Mike
#123 Mar 15, 2013
I need to pair 12 golfers for 4 rounds with the goal of everyone playing with as many players as possible, but never more than twice.
mike zisman
#124 Mar 15, 2013
you do not neet a scheduler to do this. there are 15 ways to take 4 golfers out of 6. do that and repeat as necessary

1 2 3 4
1 2 3 5
1 2 3 6
1 2 4 5
1 2 4 6
1 2 5 6
1 3 4 5
1 3 4 6
1 3 5 6
1 4 5 6
2 3 4 5
2 3 4 6
2 3 5 6
2 4 5 6
3 4 5 6

everyone is in 10 of these 15 foursomes. just repeat throughout the season.
Birdie_numnum

UK

#125 Apr 11, 2013
terry mellis wrote:
<quoted text>
Hi
Did you ever get an answer to 8 guys over 4 rounds? I have organised a trip to Scottsdale Az from here in the UK in May and trying to sort it so that there is as little duplication in the guys being paired as possible is a nightmare. Thanks
terry mellis
Hi Terri
Are you the same Terri who was at Barclays Bank ? If so, I have been trying to find you for a long long time.
bill
#127 May 12, 2013
mike zisman wrote:
having lived with this problem for 15 years, and having a Ph.D. in operations research, i finally decided to deal with it. i have both an integer programming solution and an excellent algorithm implemented. input is number of golfers, number of rounds, who is missing from rounds, who must always play together, and who must be spread among groups (e.g., players 1,2,3 and 4 should always be in different groups). i would be happy to run some real world cases and send you the results - both a text file and an ecxel spreadsheet are created. contact me at [email protected] this is going to be part of a larger web site supporting buddy golf trips - paitings, marked scorecards, tournaments, group expenses, etc.
have 6 golfers playing 4 rounds. want everyone to play with each other at least once. Can you help? Thanks Bill Rivers
Gordon Cambridge UK

UK

#129 May 16, 2013
8 golfers playing 9 rounds

How can I sort pairings as equally as possible?
Jason

Bangkok, Thailand

#131 Jul 17, 2013
Try this iPhone app, specially designed for this.

http://isaacl.net/blog/2013/07/17/robinn-app-...
saravanan

Chennai, India

#132 Sep 10, 2013
12 players, 5 rounds.
12 players divided in to 2 halves of 6 each.
top half low handicappers.
bottom half high handicapper.
each team is one from top and other from bottom.

everybody has to play with each other
is it possible
Sam

United States

#134 Sep 18, 2013
We have nine players, playing four rounds and need to make threesomes, with everyone playing each other
lenny

#135 Sep 18, 2013
Need help with groupings. 20 players over 6 rounds.
RON
#136 Sep 21, 2013
Richard Croxford wrote:
12 players in 3 teams of 4. Playing 4 rounds of golf.... How do I arrange the pairings so that everybody plays each other????
There is an opportunity tp play 4 rounds with 3 different people (12 pesons)so there must be one repetition ??? There must be a formula for this but I am no mathematician. help would be greatly appreciated. Thank you. [email protected]
It cannot be done with out some folks playing together NUMEROUS times
RON
#137 Sep 21, 2013
mike zisman wrote:
you do not neet a scheduler to do this. there are 15 ways to take 4 golfers out of 6. do that and repeat as necessary
1 2 3 4
1 2 3 5
1 2 3 6
1 2 4 5
1 2 4 6
1 2 5 6
1 3 4 5
1 3 4 6
1 3 5 6
1 4 5 6
2 3 4 5
2 3 4 6
2 3 5 6
2 4 5 6
3 4 5 6
everyone is in 10 of these 15 foursomes. just repeat throughout the season.
That didn't answer the question at all...he only wants to play a couple rounds of golf...not a season.
Gary
#138 Oct 29, 2013
I am planning a golf trip with 8 golfers. We play 6 rounds with two man teams. Looking for a way to set up pairings where you play against different teams, in different groups.
Ashvin Patel

Arnold, UK

#139 Dec 16, 2013
12 players heading to Thailand for golf. 3 teams of 4 playing 6 rounds of golf.... How do I arrange the pairings so that everybody plays each other?
Janeman

Thornlie, Australia

#140 Jan 1, 2014
Can you help. We have 16 players playing each week for the year and i am having problems with the pairing.
Janeman

Darlinghurst, Australia

#141 Jan 6, 2014
Steve Page

London, UK

#143 Apr 27, 2014
I have 2 teams of six (Team A & B)- playing each other 4 times.

Is it possible to have a player from Team A play with a different partner from Team A for each of the 4 rounds, and likewise for Team B.

Whilst no player from either team plays an opponent more than twice during the 4 rounds?
mike zisman
#144 Apr 27, 2014
the best possible solution has every player playing with another member of their team once. as to the other team, of which there are 6 players and 8 "slots" (4 rounds in which there are two opposing players in each round), team A player will play with 3 team B players twice, 2 team B players once and one team B player not at all. all 12 players will have this profile so all are treated equally.

GolfTripGenius.com .\$48 and you can experiment as much as you want, as well as make changes to what is proposed and immediately see impact.
Steve Page

London, UK

#145 Apr 28, 2014
Hi Mike

Hi

I sent you a question a couple of days ago, but when I re-read it I'm not sure that I've explained it very well.
So I will try and explain clearly this time.

I have 2 teams of six (Team A & B)- playing each other 4 times.
They will be in 3 x 4ball matches with a pair from each team in each 4 ball match

Is it possible to have a player from Team A play with a different partner from Team A for each of the 4 rounds,
and likewise for Team B.

In addition (this is the hard bit)-Whilst no player from either team plays an opponent more than twice during the 4 rounds?
The best I could come up with was the following
yet there were still 4 players playing 3 times (F&5 B&4)
A Team has players 1-6
B Team has players A-F

round 1
1&3 vs AB
2&4 vs CD
5&6 vs EF
round 2
1&4 vs BC
2&5 vs DE
3&6 vs AF
round 3
1&2 vs AE
4&6 vs BD
3&5 vs CF
round 4
1&5 vs DF
2&6 vs AC
3&4 vs BE

Thanks again
Steve
mike zisman
#146 Apr 28, 2014
Steve - you did a fine job explaining the problem the first time. and the answer i provided is still correct :) here is a matrix of how many times each player is with each other player. players 1 to 6 are on one team and 7 to 12 on other team:

1: 0 1 1 1 1 0 2 0 2 1 1 2
2: 1 0 0 1 1 1 2 1 1 2 2 0
3: 1 0 0 1 1 1 0 1 1 2 2 2
4: 1 1 1 0 0 1 1 2 2 0 2 1
5: 1 1 1 0 0 1 1 2 2 2 0 1
6: 0 1 1 1 1 0 2 2 0 1 1 2
7: 2 2 0 1 1 2 0 1 1 1 1 0
8: 0 1 1 2 2 2 1 0 0 1 1 1
9: 2 1 1 2 2 0 1 0 0 1 1 1
10: 1 2 2 0 2 1 1 1 1 0 0 1
11: 1 2 2 2 0 1 1 1 1 0 0 1
12: 2 0 2 1 1 2 0 1 1 1 1 0

table entry [x,y] is the same as table entry [y,x] and is the number of times that players x and y are together. the diagonal (player [x,x] is 0) as it is meaningless.

the scheduler generated 13.7 million possible schedules in less than 0.5 seconds on my mac. i doubt that you looked at 13.7 million schedules :). people do not realize the size of the "search space" - in your case the number of ways to place 2 players from team A and two players from team B into 3 foursomes over 4 rounds. it is a mathematical programming problem (integer programming specifically) just like scheduling airplanes or crews of kids in classes.
macgolf
#148 Oct 7, 2014
Pairing 20 guys in foursomes for 6 days and to play with as many of the other 19 players as possible

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