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 The Conversing Club 2 (Posted on 2004-02-06)
Unfortunately the Conversing Club has lost members so now they are down to 12 members. Many of the members felt talking with 3 members about things ended up with 2-against-1 situations. To address this pain, the members decided to have only 2 people sit at a table each day. They also decided that with the decline in members, they only needed 6 tables and didn't need "move every time" rule. (In other words, there aren't any extra tables once two people are placed at each of 6 tables and a person can sit at a particular table more than once each "cycle") The first day they sit as following:

AB CD EF GH IJ KL

(So the second day A could sit with C or D (and others too), but not B. C could sit with A, B or E (and others too) but not D)

After 11 days, they wouldn't have anyone left to talk with, so would repeat their schedule. Their schedule was still constructed such that everyone would talk to everyone else exactly once. How was it constructed?

 See The Solution Submitted by Gamer Rating: 3.0000 (2 votes)

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 re: solution -- the simple way | Comment 3 of 5 |
(In reply to solution by Charlie)

Now, looking up "round robin" on the internet, it seems that the simple way is the equivalent of

2 11 ( 9); 3 10 ( 7); 4 9 ( 5); 5 8 ( 3); 6 7 ( 1),

where the chords cut straight across the axis formed by the center and position 1, and which is always available automatically giving all lengths of chord so long as there are an even number of people.
 Posted by Charlie on 2004-02-06 13:42:18

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