Volume of n-cube truncates

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Volume of n-cube truncates

Postby quickfur » Fri Dec 20, 2013 1:24 am

Given a CD diagram of an n-cube truncate (_4_3_3_..._3_, with some combination of ringed nodes) and either an edge length or the size of the bounding box, how do you go about computing the volume of the polytope? I can sorta invent ad hoc solutions to specific cases, but how do you compute the volume from a CD diagram in general? Or how do you compute the volume given the coordinates of the polytope (which are trivially derived from the CD diagram)?
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Re: Volume of n-cube truncates

Postby wendy » Fri Dec 20, 2013 7:49 am

I've had several bashes at this. It's quite messy.

In essence, you have to evaluate each flag separately, where that section gives an orthoscheme. The process amounts to something like this.

a_1 = p_0 - p_1, a_2 = p_1-p_2 etc right down to p_n. Then you multiply up all of the a_n, to get v_1. You proceed through each of the flags and add the volumes together. Then you multiply by G/n! . 2^n, to get the final volume. The stott-matrix can be used for the dot-matrix product, ie A dot B = S_ij A_i B_j. It's a sheer system, so addition and subtraction works as normal.

The next trick is to evaluate the flags. You do this from the dynkin symbol, supposing that 0 is the vertex, and the remainder of the nodes are 1-n. A path is any connected set, including connections to the vertex-node.

The flags of x3o5x are (1x)3(2o)5(3x), it has four in number. I have not figured out yet (ie it is not bleedingly obvious), on how one might do this from the dynkin symbol, but it will come.
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