My first thought was to create a circle with each players name around the circle and draw lines which indicate who'd they'd be playing with.
But this creates problems because as you can see that means there are 364 different ways of grouping 3 people together amongst the 14 so that every players plays with another player at least once. (or so I think, I took Combinatorics too many years ago). I thought to myself, there must be some mistake, perhaps this model doesn't take into account duplicates. So I worked out an easier case: 2 person groups amongst a pool of 4 people.
Since order doesn't matter, all we are worried about is the fact that player 1 plays with player 2 at least once, duplicate entries do exist. moreover, this seems to be a horrid way of grouping folks together.
Perhaps a better method would be to have the 14 players count off to 5's, and group the 3 1's together, the 3 2's together, and so on. But how would you mix the players after they've played one set so that every player plays with every other player. The problem still stands. Any thoughts, ideas or solutions?
4 comments:
my face just melted.
i think i've got a method. i'll try and post this afternoon when i get off. - Matt
take 14 index cards and writes names on them, then number them 1 to 14. arrange in 3 rows as follows:
1 2 3 4
9 8 7 6 5
10 11 12 13 14
group by columns. (5 and 14 get byes)
then shift as follows:
14 1 2 3
8 7 6 5 4
9 10 11 12 13
iterate.
conjecture: after 14 iterations each group composition will have been attained with no repeats.
i think i've got a proof that all compositions are attained. haven't worked out the no repeats part tho.
- Matt
Matt,
That is very clever. nice job. I'd be interested in seeing the proof. email me at thebigrokh@gmail.com
Do you have any formal mathematical education?
-S
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