Beyond 3-D, Chapter 8

David Akers

Slicing the cube

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Coordinates of hypercubes . . .

The combinatorial explanation for the number of edges in a hypercube makes perfect sense to me now. Obviously the number of points in an n-cube will be 2^n, since we have exactly two choices for each coordinate (1 or 0). For some reason I hadn't thought of it in this way before.

The discussion of hypercube slicing was a bit more mystifying, unfortunately. I guess I just can't seem to visualize it very well without that movie in front of me. I sometimes wished that the book had a few more of the mathematical details. I found myself forced to make a lot of assumptions and (in some cases) guesses about the math. Here's some of what I came up with (please correct me if I am wrong on any of this).

Regarding slices perpendicular to the longest diagonal: In three space, the slice which goes through (1, 0, 0), (0, 1, 0), and (0, 0, 1) gives us the coordinates of a two-simplex which lies on the barrier between slices which are triangles and slices which are truncated triangles (6 sided). Similarly, in four space, the hyperslice which contains (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1) should be on the boundary between hyperslices which are tetrahedrons and hyperslices which are truncated tetrahedrons (8 sided). If, then, we could find the point which lies on the longest diagonal of the hypercube, and is contained by this hyperslice, we would then know the distance we had travelled from the vertex, right? I am going on the assumption that there is only one hyperplane which slices the hypercube perpendicular to its longest diagonal and goes through a particular point on this diagonal. This would seem logical, but I'm not quite sure. Does this make sense?

In a cube, the division between tetrahedra and truncated tetrahedra occurs exactly 1/3 of the way down the object. Is this also true for the hypercube? Maybe calculating this result comes down to finding the center of the tetrahedron formed by the four points, then applying the extended version of the pythagorean theorem to find the distance from this center to the vertex.