I actually got currious about this and tried it. This answer is not quite
correct. This returns all of the combinations of players (like lottery
numbers) when actually you want the combinations where no 2 players are ever
together. I think this going in the right direction, but you need to
eliminate more rows w/ the where clause.
===== Original Message from [EMAIL PROTECTED] at 8/29/01 9:43 pm
>Troy,
>Thanks for this information. I will give it a try and see if it works. I
>thought however that there would be some mathematical representation of it.
>But I have never tried this. I attempted a cursor and drove down from that
>but the code became extensive.
>Phil
>
>-----Original Message-----
>From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On
>Behalf Of Troy Sosamon
>Sent: Tuesday, August 28, 2001 10:04 PM
>To: [EMAIL PROTECTED]
>Subject: RE: Tough Puzzle
>
>
>A few years ago there was a discussion on the boards about computing lottery
>combinations using SQL, and if my memory serves, our master golfer (Bill
>Downall) figured out the solution using a cartesion join. Now your
>situation
>is very simelar. If you had 28 golfers, I think this will work.
>
>Say you had table golfers with id and name
>
>select t1.name, t2.name, t3.name, t4.name from golfers t1, golfers t2,
>golfers
>t3, golfers t4 where t1.id not in (t2.id, t3.id, t4.id) and t2.id not in
>(t1.id, t3.id, t4.id) and t3.id not in (t1.id, t2.id, t4.id) and t4.id not
>in
>(t1.id, t2.id, t3.id)
>
>I think this would give you all of the combinations of 4 possible. Since
>you
>only have 25 golfers, you would need to add 3 dummy players to fill it out
>so
>it would work.
>
>Troy Sosamon
>Denver, Co.
>
>
>
>
>>===== Original Message From [EMAIL PROTECTED] =====
>>To the guru, sage, developers of RBase:
>>
>>Every year I have gone with my business friends to Myrtle Beach for a "golf
>>happening".
>>Every year I try to come up with a solution.
>>I have been going for 5 years now and am ready to consult the guru's.
>>
>>Here's the issue that I am requesting help with:
>>
>>Known facts:
>>1. We have approximately 25 golfers that show up.
>>2. We play for six days all 25 golfers.
>>3. We take the total number of players and compute into foursomes and
>>threesomes.
>>4. The first day paring is done by unit assignments (usually four to a
>>unit)
>>5. The second day starts the problem. Everything works great for the
>first
>>day but I start to have problems with the second and thereafter days
>because
>>of this rule: a player must play with a different set of golfers the next
>>days. So if player one played with 2,3, 4 on day 1, any combination of 2,
>>3, 4 in any of the next five days is less than desired. We have never been
>>able to get it to work.
>>
>>I created a matrix in to try to drive the combinations but usually run out
>>of good combinations by day 4. Is it mathematically possible?
>>
>>Reward for solution.
>>
>>Phil Nolette
>>[EMAIL PROTECTED]
>
>Troy Sosamon
>Denver Co
>[EMAIL PROTECTED]
