I wrote "it might be some other number", but that's wrong. If the average doesn't depend on the choice of P, Q, and R, then it must be 4, since that's what you get when they are all rectangles with their axes aligned. Jim On Monday, November 20, 2017, James Propp <jamespropp@gmail.com> wrote:
Anyone know of software (preferably in Mathematica) that computes the intersection of polytopes, or at least polygons?
I'd like to explore some more random-nonempty-intersection problems, and it'd be very helpful to be able to do numerical experiments in a flexible setting.
For instance, how many sides on average does the intersection of P, Q+v, and R+w have, conditioned on the event that the intersection is nonempty, where P, Q, and R are fixed parallelograms, and v and w are random translation vectors? (The usual caveats and clarifications about "random translation vectors" apply.) The answer might be that it depends on P, Q, and R, or it might be 4 (no matter what), or it might be some other number; maybe one of you can see a non-experimental approach, but I don't, so doing experiments is the next step for me. But I don't trust myself to write correct infrastructural code for handling polygons.
Jim Propp
PS: I'm still mulling over Veit's two puzzles about averages. I'm guessing that a key tool for solving his first puzzle is "V-E+F=2", or rather "V-E+F is negligible when V, E, and F are large", but I don't see what the answer is.