Newsgroups: sci.math
From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
Date: 9 Sep 1994 21:07:47 GMT
Subject: Unfolding polyhedra and polytopes (was Re: How many hexominoes...)

ad...@super.org (Jeffrey P. Adams) showed 11 ways of unfolding the
surface of a cube into hexominoes (20 if you count reflections of the
chiral ones).

> Now, I'm thinking about octohedra, dodecahedra, and icosahedra.  I
> wonder, is there an easier way to go about this?  Does anyone know
> whether this has already been done?

I did it for the octahedron a while ago.  There is a nice
correspondence between the unfoldings of the cube and the octahedron.
Note that an unfolding of any polyhedron is completely determined
by the set of edges that are cut to leave a simply-connected set of
faces.  If two polyhedra are dual, then the set of edges cut for an
unfolding of the first is dual to the set of edges _not_ cut for an
unfolding of the other.  This forms a chirality-preserving bijection
between the sets of unfoldings.  So the duality between cube and
octahedron induces a bijection between their unfoldings.  Some
bijection will occur between the dodecahedron and icosahedron.  The
tetrahedron is self-dual and duality induces the identity bijection
between its two distinct unfoldings (three if you count reflections).

Notice that when you unfold a non-convex polyhedron and flatten it
out, the surface may overlap itself.  For instance, the following
depicts one form of the unfolded surface of the L-tricube:

              +--+-----+-----+
              |  |     |     |
  +--+     +--+--+-----+-----+
  |  |     |XX|  |
  |  +--+--+--+  |
  |     |  |     |
  +-----+--+-----+

The XX square is actually two overlapping squares, one joined at the
bottom and the other joined at the right.  Can an overlapping
unfolding be found for _every_ nonconvex polyedron?  Can one ever
occur with a convex polyhedron?  If so, can it occur on the
dodecahedron or icosahedron?

I do know it doesn't happen on the tesseract, and I think it doesn't
happen on a hypercube of any dimension, but I haven't got a proof.

d...@mirage.esl.com (Dana Albrecht) asked:

> Related question: If one "unfolded" n-dimensional cubes, how many
> "different" unfoldings are there? ...

I wrote on this subject in 1992.  There are 261 octocubes you can
get from unfolding the surface of a tesseract, or 355 if you count
reflections of the chiral ones.  Then I found a paper by Peter Turney
in J Rec Math V17(1), 1984, in which he shows the problem to be
equivalent to counting the number of 8 vertex graphs that consist of
a tree together with a differently-colored pairing of the 8 vertices,
disjoint from the tree (the tree represents face adjacency in the
unfolding, and the pairing represents opposite pairs of faces in of
the solid.  This would plainly extend to the problem of unfolding the
surface of a 5-dimensional hypercube into a dekatesseract, but I don't
know if it's been done.  Also, I don't know how to figure out which
ones are chiral.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil