Groupings- one for all you mathematicians.

[John_Findlay]

Morning all.

Help, please. I can't get my head around this one.

I'm organising an annual golf trip. 12 players. 3 days golf. First two days we're in 3 groups of fourballs. 3rd day we're being joined by a further 4 members to sign us on.... so our group of 12 will be in four groups of threeballs with an added member in each group.

I'm trying to work it so that everyone plays with as many different people as possible. Is it mathematically possible to avoid playing with the same person over the 3 days? I don't think so. Here's the best formula I could come up with.

The various colours and numbers represent the 10 tee times. So, for example, Sid & Gerr play together on both days one and two. I can't find a better solution. At least this way I'm avoiding someone playing with the same person 3 times in 3 days but the best I can do is 6 of the players playing with the same person on days 1 and 2.

Any logical help appreciated. It's causing my head to hurt.

 

[madandra]

FFS John, its not rocket science .....




You need a sachet of brown sauce, tomato sauce, mayonaise and salad cream and draw them out of a hat

 

[John_Findlay]

Ahhh, I forgot about the "Wardrop Condimental Theorum of Randomtivity!!

Psssst....it disnae work!

 

[vig]

????

Let me get this right John.
You have 3 days golf = 3 rounds?
There are 12 players + on day 3 you are joined by an additional 4 = 16?
You would like the groups mixed so that no one plays with the same player twice?

Is it possible? Of course it is if that is what you want above

 

[John_Findlay]

Yip, Vig. That's the scenario. Can you tell me how every player avoids playing with anyone else twice, given that there are only 10 tee times? I've tried it various ways and none of them work. Bear in mind that the 4 groups of three on the last day have to be joined by one member.

 

[nomadpaul]

only possible i think if you played 4x3 balls as opposed to 3x4 balls ???

 

[John_Findlay]

That's what I thought, Paul. Problem is we only have three tee times booked on days 1 & 2 and we want to play fourballs.

 

[vig]

Just looked again and my head hurts. Nearest I can get at the mo is 1 doubling up but leave it with me
Here goes.

Day 3 1 new member in each group.

Players No 1,2,3 - 4,5,6 - 7,8,9 - 10,11,12

Day 2
Players No 1,4,7,10 - 2,5,8,11 - 3,6,9,12

Day 1
Players No 1,6,8,11 - 2,4,9,10 - 3,5,7,12

I'll come back to you if i can solve it

 

[JustOne]

It's not possible. Using letters as the players this is the best I could come up with #I think#
People that have played together are in bold. It might be the same as what you already have.... I didn't check as the colours confused me <wink> hehe

DAY1 abcd - efgh - ijkl
DAY2 abei - cfjk - ghdl
DAY3 afl+1 bgj+1 chi+1 dek+1

 

[JustOne]

Just looked again and my head hurts. Nearest I can get at the mo is 1 doubling up but leave it with me
Here goes.

Day 3 1 new member in each group.

Players No 1,2,3 - 4,5,6 - 7,8,9 - 10,11,12

Day 2
Players No 1,4,7,10 - 2,5,8,11 - 3,6,9,12

Day 1
Players No 1,6,8,11 - 2,4,9,10 - 3,5,7,12

I'll come back to you if i can solve it

You have 3 doubles... (4,10) (8,11) and (3,12) are in the same groupings on days 1 and 2

It can't be done as it's impossible to choose 4 unique players from only 3 groups, there'd HAVE to be two players from one group in there somewhere.

 

[John_Findlay]

I think the number of groups per day (in this case 3 on days 1 & 2) has to exceed the number of other golfers you play with on any given day by 1 for it to be possible.

Vig's solution is no nearer the mark than mine.... so no banana. 3 played with 12 twice, 4 played with 10 twice, 8 played with 11 twice. So six people play with the same golfer twice. Same as in the scenario by me and JustOneUK.

This takes me back to SYS Calculus! Where are those wee sachets of brown sauce.....

Do you think it would make a difference if group A set off from Edinburgh a 8am travelling at 60mph and group B started off at 9am at 75mph?

 

[mono217]

Just draw cards

 

[tigerj]

Hi
Have a look at this:
http://www.cs.brown.edu/~sello/uniques.html
It's the nearest you'll get!

 

[Herbie]

I think mono's answer would save everyone getting a headache and give you less to answer for on the days in question

 

[Dodger]

Jesus John,I had a wee think but gave up as my heed was stotting!!
I have tried the same on our tours and if it is possible we havn't found out how either!

 

[John_Findlay]

Typical! After all that, it's been decided that on day 3 we're going out in traditional leading order after all (player in last position goes out first, etc). Still, it was a useful exercise for the brain, eh, and I got to use my crayons.

 

[Herbie]

Give every player a number in each group.
First group of three would all have the nuber 1 against their names. Next group of three would all have number 2. etc.

Later when you add players to a group give them the prefix number and a letter eg. 1a

If you have it written down you will know simply by looking if you have placed them with same folk twice.

(little more needed but Im sure you will see it as soon as you plan it on paper)