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MathGroup Archive 1999

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polytopes and cross products

  • To: mathgroup at smc.vnet.net
  • Subject: [mg19290] polytopes and cross products
  • From: Russell Towle <rustybel at foothill.net>
  • Date: Thu, 12 Aug 1999 22:34:46 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

Hi all,

In prologue, I am using Mathematica to construct various regular and
semi-regular polytopes in n-dimensional space, especially in four
dimensions. I have spent many hundreds of hours on a notebook which
constructs these 4-polytopes, makes solid shadows of them, and projections
into various sub-spaces, and even hidden-detail-removed projections of
4-polytopes into 3-spaces.

In order to bring this notebook to a point where it can be placed on
MathSource,  I must solve one last nagging problem. First, a 4-polytope is
bounded by polyhedra. Take the regular 120-cell, {5,3,3}. It is bounded by
120 Platonic pentagonal dodecahedra. In the 4-space, supposing the
dodecahedra are opaque, an observer would only see half, or somewhat less
than half, of them. In order to find which ones are visible from an
arbitrary viewpoint in the 4-space, I must find the cross product for each
dodecahedron: the vector perpendicular to its own particular 3-space. My
cross product function "Cross4" requires four 4-vectors as input. So I will
write, for instance, for a dodecahedron named "dodeca",

Cross4[ dodeca[[ {1,5,11,17} ]] ].

I loop through all 120 dodecahedra in such a way. I find by trial-and-error
that, say, the 1st, 5th, 11th, and 17th vertices of any of the 120
dodecahedra span their respective 3-spaces. And I find by trial-and-error
that, say, the 1st, 2nd, 3rd, and 4th vertices of the 120 dodecahedra *do
not always* span the 120 3-spaces.

As it stands now, once I have picked a good set of four indices, I can find
the visible dodecahedra in a second or two. What I need, though, is a fast
and reliable way to pick out subsets of four vertices from *any* polyhedron
in a 4-space, such that the four vertices *span the 3-space of the
polyhedron*. For it may well happen, as in the case of the dodecahedra,
that some four of the 20 vertices of a polyhedron are all in a single plane
(a single polygon) and thus do not span the 3-space of the polyhedron.

In summary, how can I use Mathematica to select a set of four 4-vectors
which span a 3-space in the 4-space?

Russell Towle
Box 141
Dutch Flat, CA 95714
(530) 389-2872




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