Posted on 08/11/2004 12:26:49 PM PDT by weave09
Forgive me if I have posted this in the wrong area, I am not new to FR but haven't posted much. A group of my friends are going on a golf trip. There are 12 golfers playing 4 rounds of golf as foursomes. I have been asked to come up with a schedule that has each golfer playing at least one round of golf with the other eleven golfers. Any help would be greatly appreciated. Thanks.
This seems quite easy (which means I'm sure I'll get it WRONG!). Each golfer is a number, 1 through 12. Assign these 12 golfers the following group numbers:
1. A, a, a, a
2. a, b, c, d
3. a, c, d, b
4. a, d, b, c
5. b, b, b, b
6. b, a, c, d
7. b, d, a, c
8. b, c, d, a
9. c, c, c, c
10.c, a, b, d
11.c, b, d, a
12.c, d, a, c
What did I win? Everyone plays with different person, each day.
No, they want to be assured that each one would play at least a round with the other eleven.
Sad to say, but you did. Golfers 1, 5, and 9 never play each other, for example.
CRAP! An error. I'm reviewing it, I think I am close.
It would work with a 5th round...
I suggest that after the first round you always pair-up the winning golfers, second place finishers, third place and fourth place from the previous round.
That way every group is broken up each round but you will start with 4 threesomes and end up with 3 foursomes.
Not really.
Ironically, I was installing a foreign language disk at the same time.
I meant you would start with 3 foursomes and end with 4 threesomes.
Hrmph.
1. a, b, c
2. a, c, b
3. b, c, a
4. b, a, c
5. c, a, b
6. c, b, a
7. a, b, c
8. a, c, b
9. b, c, a
10. b, a, c
11. c, a, b
12. c, b, a
YEEEHAWWWWW!! Got it.
See above, I think that works.
My mistake ma'am, no offense intended.
N/M, it's closer but no cigar. DAMN.
I still don't get 1, 5, and 9 playing each other at any time. Of course, you still have one round to work with.
sorry, tisn't possible
(please don't ask for a proof)
OK, my proof is based in logic, not a formula.
4 rounds equals 12 playing partners for each player (3 each round).
You want each to play the other eleven, which means you can only duplicate one playing partner over all 4 rounds.
All told, 12 players must have exactly 1 duplicate each, spread over 4 rounds.
However, you can't even have two rounds of totally unique foursomes (in fact, at least half of the players must repeat a playing partner from the previous day).
Proof:
there are 3 groups each day - A, B, C
players 1,2,3,4 play in group A first day
where will they play day 2?
put 1 in A, 2 in B, 3 in C ... player 4 must go into a group with someone he played the previous round with.
Go ahead and try to design 2 totally unique rounds.
btw, you could make this work with 16 players, I think.
What this means is there are at least 6 duplicates each round after the first, for a minmum of 18 duplicates. Since your premise only allows for 12 duplicates, you can't win.
I'll take payment in beer ...
Show your work.
In conjunction with pythagoreans theorum in conjunction with virgo it appears that chi square for N=12 in a four round rotation is inverse to the sum of the square root of 3.
Proof is in the pudding.
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