There are 10 tables in the Conversing Club and 15 members.
Each day, 3 people sit together around each of 5 of 10 possible tables in the club talking to each other.
Every week (7 days) everyone is at the same table with everyone else exactly once. Also, nobody is at the same table twice in the course of a week to provide a change of scenery each time. The first day is as following:
ABC DEF GHI JKL MNO
(The second day A couldn't sit with B, or C; B couldn't sit with C; D couldn't sit with E or F, but could sit with A, B, or C.)
How could their schedule be configured?
(Based on Fifteen Schoolgirls)
(In reply to re: solution--computer used
That algorithm has been repaired so it no longer assigns the same person to the same table twice in the week, but now the minimum number of tables among the 84 solutions is 11, so I need a more efficient algorithm.
Posted by Charlie
on 2003-12-30 11:30:23