# Space Partition Puzzle

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Q: What is the maximal number of polytopes generated by the intersection of three cubes. (more specifically, I mean 3-polytopes or "cells") I'm usually pretty good at visualizing this stuff, but 3 cubes creates too many cells to keep track of. Is there a way to solve this problem in a general way - rather than just counting? Reminds me a bit of the zone theorem for line arrangements, but I don't know how to account for the cube constraints. This might sound like a homework problem - it's not; I've earned my degree. I took CS384 - Computational Geometry, and it's something of a brainworm.

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3-polytopes? I'm guessing you mean polyhedrons.

What exactly do you mean with "number of polytopes"? Do you mean number of faces (polygons) to build a polyhedron? i.e. for one cube: 6. Or do you mean number of polyhedrons? i.e. a cube is one polyhedron.

What about the shape of the polyhedrons? (Convex or concave?)
If concave... goodluck! I wouldn't have a clue. Well... perhaps then you could then split the concave polyhedron in to multiple convex polyhedrons and reason with that.

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If I understand the problem correctly, the answer is 7. But perhaps you should explain your question better.

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I thought it was 42. Douglas Adams said it was anyway. :)

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