From: cliftonr@netcom.com (Pope Clifton)
Subject: Re: Prime Polyhedra
Date: Mon, 14 Sep 92 02:41:54 GMT
Organization: t.b, t.p.d, & a.d
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Warning: extreme math weenie-talk coming up!!  Hit 'n' now!

Gregory Nowak (greg@phoenix.Princeton.EDU) wrote:
: }billbill@wetware.com:
: }: Now I want more:
: 
: }: I want the term for polyhedra not bisectable into two (etc.)
: }: polyhedra, each with fewer vertices than the original.
: }: 
: }: I want an algorithm to generate these [{("prime")}] polyhedra,
: }: so I can encrypt 3space.
: }
: Here you go, billbill:
: 
: The concave span of three points is only two-dimensional, so you need
: at _least_ four points: a tetrahedron, but it may not be regular.
: Suppose you have n points (n>4), no one of which is contained in the
: concave span of the other n-1. Then, pick a vertex V. V is connected
: to k other vertices P1, ... , Pk by edges.  By renumbering we can
: choose P1, P2, P3 so they are the closest to V ( the edges V-P1, V-P2,
: V-P3 are the three shortest edges including V) . Slice this polyhedron
: with the (unique) plane passing through P1, P2, P3, and you end up
: with two new polyhedra, each with fewer vertices than the first: an
: irregular tetrahedron with vertices V, P1, P2, P3, and "the rest" with
: n-1 vertices, since V is gone but the cutting procedure added no new
: ones.  (you can see this easily in playing with the cube. Buy play
: Doh. Do your topology homework in 3-space! Dilute! Dilute! OK!)
: Lather, rinse, repeat, until you've reduced n to 4.
: 
: So the answer to your question is 
: 
: {prime polyhedra} = {tetrahedra} .
  Well, not exactly.  In your renumbering step above, you assume that 
the tetrahedron defined by V-P1-P2-P3 is in fact included in the initial 
polyhedron.  But billbill didn't stipulate a convex polyhedron (though
I believe the initial post did mention regularity.)  You did prove:
  {prime polyhedra} intersect {convex polyhedra} = {tetrahedra}

 A (minimal) concave counter-example:
  Grab that Play-Doh (or exercise your visualisation skills) and create an
equilateral tetrahedron.  Now indent one side in the shape of a shallow
tetrahedron, whose base is one face of the original ttrh and whose apex
is enclosed within the original ttrh.  There are several possible dissections
from this point -- but all have at least one of the sections containing
as many vertices as the original figure.

A counter-example invvolving a regular polyhedron:
  Take an icosahedron; indent one or more of the 12 pentagons where five
equilateral triangles come together.  (You can't do this operation with 
any of the Platonic polyhedra containing less faces.)  Now you have a 
concave polyhedron with equilateral triangular faces.  My visualisation is
not good enough to tell without building a model, but I think you can not 
reduce the number of vertices by removing any of the vertices of the indented 
pentagonal pyramids.  On the other hand, I believe you can slice off some
other vertices as ttrh, so it's not prime -- it can be reduced to a non-
regular concave prime polyhedron.  

  I think I will just leave this there and not try to formulate any rules
for identifying or constructing prime polyhedra -- I thought about it too
much this morning while I was trying to meditate and I need to get it 
off my mind.  At least billbill gets to have some interesting concave 
polyhedra in his encrypted universe, besides boring old tetrahedra. 

  -- Clifton (former algebraist wannabe, never much on geometry.)
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