For better parties
The other day I was at a dinner party. We were a total of six people seated around a rectangular table, two at each of the long sides and two at the head of the table. What struck me was that I was seated far away from people I wanted to talk to and close to people I knew too much about.
Please suggest a way to seat six people so the distance between any two is constant. If it can’t be done, suggest another reasonable criteria and find its solution.
Please think fast! I am invited for another dinner party this Saturday.
April 20th, 2010 at 2:02 pm
If there are 3 of you, you must sit at the corners of an equilateral triangle. If there are 4, you can still do it, e.g. one, like the Dormouse, in the teapot in the centre of the table, and the other 3 on the ground so that the 4 of you are at the corners of a regular tetrahedron. If there are 6 of you, though, your host will need to hold the party somewhere where there are at least 5 dimensions of space.
April 20th, 2010 at 5:56 pm
I will ask my host to move into 5 dimensions, but if he doesn’t listen: “If it can’t be done, suggest another reasonable criteria and find its solution.”
April 21st, 2010 at 1:07 am
Perhaps you could model your conversational preferences as a sort of distance–shorter for those you want to talk to, longer for those you rather not–and then arrange the locations to match the preference distances. With only 6 locations, you can use an exhaustive search to find an optimal solution in short time.
Alternatively, you could relax the “constant distance” criterion to just TWO distances (call them Close and Far) and seat folks at the corners of a regular octahedron–of course your host would need a dining room straight off the set of Auntie Mame. Still that would give everyone 4 close neighbors and only one Far one.
April 30th, 2010 at 12:02 am
[...] I am sure this is a famous problem, but I don’t know its name or any results. I came to think of it when I was a dinner party as explained here. [...]