There's a pretty common riddle floating around which goes something like this:
An ant starts at one corner of a unit cube. They wish to reach the opposite corner. If they can traverse along any face, but cannot fly, what is the length of the shortest path they can take?
Pretty obviously the answer to this is $\sqrt{5}$ (from turning it into a net). But I was wondering if there was some generalization of this riddle for other shapes? It seems like the math gets quite messy if there are asymmetries, so for this question, I'd like to assume that every side length is 1.
For any n-simplex, I'm fairly sure that the length is 1. Though it doesn't really make sense to define an "opposite", any vertex is 1 away from another, so it also doesn't really matter. I guess I can define opposition as any pair of vertices which have the longest traversal length. But I'm having trouble with generalizing it for other polytopes. An octahedron would have $\sqrt{3}$, an icosahedron would have $\sqrt{7}$, a dodecahedron would have $\sqrt{\frac{7 \sqrt{5} + 17}{2}}$, and a tesseract would have $2 \sqrt{2}$ (the diagonals of 2 squares). But I'm not really sure how to compute this value for things like 120-cells, 600-cells, n-cubes, and n-orthoplexes.
I'm also interested in generalizing it so that traversal can also be higher dimensional. For an ant with 3-dimensional traversal, they would be able to traverse the cube in $\sqrt{3}$ and the octahedron in $\sqrt{2}$ (as these are the diameters of the circumscribed spheres). For a tesseract, they would be able to traverse along every diagonal within a cell, rather than just the faces, though I haven't figured out the length yet.
I would appreciate any help on this generalization, even some Mathematica code would be valuable to me.
Here are the nets I've used:
- Cube:
- Octahedron:
- Dodecahedron:
- Icosahedron:
