Golf Teams

bill simpson · 2011-08-07 23:41:31

I am leading a group of 8 golfers on a 5 day tour. We play in teams of 4. How can I make sure that each player plays with each of the other players on at least 2 of the 5 days?

bobbym · 2011-08-08 00:28:41

Hi bill;

I am not sure what you are asking. These progressive dinner type problems are very difficult and do not always have known solutions. I need to know more about the structure you want.

John E. Franklin · 2011-08-08 13:21:55

The best option I could come up with involves playing 2 games on Monday, Tuesday, Thursday, and Friday, and playing 3 games on Wednesday.

Here's the layout, players are numbered 0 to 7.
Monday AM 0 1 2 4
Monday PM 1 2 3 5
Tuesday AM 2 3 4 6
Tuesday PM 3 4 5 7
Wednesday EARLY AM 4 5 6 0
Wednesday Noontime 5 6 7 1
Wednesday LATE PM 6 7 0 2
Thursday AM 7 0 1 3
Thursday PM 0 3 5 6
Friday AM 1 4 6 7
Friday PM 2 5 7 0
Hope this helps you. With these 11 games, each player will see each other 2 or more times, 2 minimum.
(This fun project took me 35 minutes.) If you require fewer games, please write back, it might be possible with 10 games,
or even nine, I don't really know. (over and out)

John E. Franklin · 2011-08-08 13:26:30

Note on Wednesday, player 6 will get tired, playing 3 games, so you might want to
take all my 11 games and mix up the order differently, this should be easy to do.
Good Luck!!

John E. Franklin · 2011-08-08 14:03:35

You might consider exchanging monday AM with wednesday late PM, so player 6
is less busy on wednesday and everyone will play the first day except #4, so that's not bad.
Then #4 player can start tuesday morning.

TMorgan · 2011-08-08 14:51:10

I assumed that the team plays together all day, regardless of how many games are played in a day. Otherwise why state the 5 day limit. The first test I thought of was to consider golfer zero (actually I had called him golfer 1 but I like your system better). He has 5 days to play with the other 7 players twice. Each day he plays with 3 other players, so he has 15 slots available. Each of the other 7 players has to fill 2 of those slots, which requires 14 slots. This of course does not prove that it is possible, as Bobby said it is quite complex. But this first simple test did not prove it impossible.

bill simpson · 2011-08-09 04:33:16

Sorry, everyone seems to have misunderstood my problem. There are eight golfers who are to be split into teams of 4 each day. Everyone plays only 1 game each day and, over the 5 days we are playing I want to make sure that the teams are mixed up so that, ideally, everyone plays with everyone else at least twice. I do not want anyone to play with anyone else more than 3 times in the 5 days and I do not want anyone to play any less than 2 times with each player. It should be possible. Player number 1 for example will play 5 games with 3 people each time making 15. As there are another 7 players if he plays with 6 of them 2 times (=12) and one of them 3 times that equals the 15. Help

bobbym · 2011-08-09 04:40:17

Hi bill simpson;

I think I understand what you are asking but I can find no block design ( social golfer schedule ) that fits exactly what you want. For some numbers of players and number of days no answers are known or the answers have not been found. I will continue to look/work on the problem, I just thought you should know I might fail.

John E. Franklin · 2011-08-09 10:11:43

Hi bill simpson:

I have found an inelegant partial solution that addresses about 75% of the conditions necessary,
in case you have to go with something not perfect.

In this layout, everyone meets at least once, but also two pairs happen four times.
There are 28 total meetings of pairs of players in this problem.
This solution has 6-one-meetings, 14-two-meetings, 6-three-meetings, and 2-four-meetings,
which adds up to 28 total meetings, which is a triangular 7-row bowling pin Pascal number.
Here are the five days:
Day 1: 3 0 4 5 and 7 6 2 1
Day 2: 5 1 2 3 and 4 0 6 7
Day 3: 4 6 7 2 and 0 1 3 5
Day 4: 0 2 5 6 and 1 3 4 7
Day 5: 3 5 6 7 and 0 1 2 4
At least now you have something to try to improve upon.
We can have a contest to see who can get the closest to the ideal conditions!!

Last edited by John E. Franklin (2011-08-09 11:54:24)

John E. Franklin · 2011-08-09 11:52:40

Hi bill simpson:

I have found a slightly better solution to my #9 post previous an hour ago or so.
This solution has 22 meeting in the 2 or 3 meetings times range out of a possible 28.
There are no 4 or higher meetings, but there are still 6 single meetings of some players.
Specifically there are 12-two-meetings, and 10-three-meetings, making this slightly
superior to my last post. Here is the data:

Day 1: 1 2 3 7 and 0 4 5 6
Day 2: 1 2 5 6 and 0 3 4 7
Day 3: 1 3 4 6 and 0 2 5 7
Day 4: 2 3 4 5 and 0 1 6 7
Day 5: 2 3 6 7 and 0 1 4 5

Hope that helps a little more. Try to beat it!!

bill simpson · 2011-08-09 21:10:38

We are getting nearer. I am sure though that there is a perfect answer. Keep trying everyone

John E. Franklin · 2011-08-09 21:41:22

Hello Bill!!
Not perfect, but here's the best one yet!!
You won't believe how close it is!!
There are 24 out of 28 meetings of 2 players that are in the twice or three times league.
There are only four remaining single meetings of 2 players.
Here is the data:
Day A: 0 1 2 3 and 4 5 6 7
Day B: 0 1 4 5 and 2 3 6 7
Day C: 0 2 5 7 and 1 3 4 6
Day D: 1 2 4 7 and 0 3 5 6
Day E: 1 2 5 6 and 0 3 4 7

The first four days A to D I generated by a BASIC program I wrote, but the Eth day I chose by hand.

bobbym · 2011-08-09 21:57:11

Hi bill simpson;

I am sure though that there is a perfect answer.

Why are you sure there is a perfect answer? Have you seen a matchup like this in practice?

John E. Franklin · 2011-08-09 23:50:39

I have found you (bill simpson) a solution that you might not be able to use because you will have to change the five day tour to a six day tour. In this schedule, there are twelve double meetings of partners and sixteen triple meetings of partners, and nothing else like ones or fours or zeros or fives or sixes. 12 + 16 = 28, as expected.
(12 * 2) + (16 * 3) = 72, as expected, due to 72 / 6 equals 12 games of 4 people in 6 days. (6 combinations of pairs in 4 people)
Here is the six day data:
Day 1: 0 1 2 5 and 3 4 6 7
Day 2: 2 4 5 7 and 0 1 3 6
Day 3: 0 2 3 4 and 1 5 6 7
Day 4: 2 3 5 6 and 0 1 4 7
Day 5: 0 3 5 7 and 1 2 4 6
Day 6: 1 3 4 5 and 0 2 6 7
Sorry it's not the required five days though, but I wanted to succeed in some way or another!?!

bill simpson · 2011-08-10 03:22:23

Hi John

Thanks for all your help. You are getting closer but I cannot make it 6 days. I could, however make it 4 days with the 5th day to be set up with the best 4 performers iin the first group and the rest in the next group irrespective of what it does to the initial objective.

Will your programme give the perfect answer for 4 days?

Bill

John E. Franklin · 2011-08-10 10:45:50

Allo bill

Le meilleur (the best) my BASIC program could find was twelve-one-time-meetings,
twelve-twice-meetings, and four-thrice-meetings.
Here are the daily layouts:
dayA: 1246 and 0357
dayB: 2345 and 0167
dayC: 0247 and 1356
dayD: 0123 and 4567
Sorry I couln't find better.
I let the program run under
tighter constraints for 30 minutes,
but it found nothing better.
(over and out)

Math Is Fun Forum

#1 2011-08-07 23:41:31

Golf Teams

#2 2011-08-08 00:28:41

Re: Golf Teams

#3 2011-08-08 13:21:55

Re: Golf Teams

#4 2011-08-08 13:26:30

Re: Golf Teams

#5 2011-08-08 14:03:35

Re: Golf Teams

#6 2011-08-08 14:51:10

Re: Golf Teams

#7 2011-08-09 04:33:16

Re: Golf Teams

#8 2011-08-09 04:40:17

Re: Golf Teams

#9 2011-08-09 10:11:43

Re: Golf Teams

#10 2011-08-09 11:52:40

Re: Golf Teams

#11 2011-08-09 21:10:38

Re: Golf Teams

#12 2011-08-09 21:41:22

Re: Golf Teams

#13 2011-08-09 21:57:11

Re: Golf Teams

#14 2011-08-09 23:50:39

Re: Golf Teams

#15 2011-08-10 03:22:23

Re: Golf Teams

#16 2011-08-10 10:45:50

Re: Golf Teams

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