How do you solve Alexander's Star?

Alexander's Star is actually a very simple puzzle, in a certain way.
There are undoubtedly quicker and more clever solutions, but it can
actually be solved with repeated applications of just one four-turn
"operation".

The only way I can think to describe the move is the way it resides in
my head: "in, in, out, out".  Hold one hand on each of two adjacent
vertices.  (A vertex is a corner where three of the edge pieces meet.
Adjacent vertices are ones that share an edge.)  Turn one of the
vertices towards the other (120 degrees).  That's "in".  Then turn the
other vertex towards the first (the other "in").  Next turn the first
one back "out", and finally turn the second one back "out".

The result is that three edge pieces have been rotated: the three that,
at various times during the move, were temporarily located between the
two vertices.  These three edges form a triangle, either pointing
toward you or away from you (depending on your interpretation of
"towards" in "turn one vertex toward the other").  Whether they were
rotated clockwise or counterclockwise depends on which vertex you move
first.  (Two clockwises is the same as one counterclockwise, so if they
move the wrong direction, you just do the move again.)

Judicious, prolonged, and painful application of this operation will
get you where you want to be.  (It's made painful by how difficult it
is to make turns on the stupid thing.)  My approach is to start by
building one "face" of the dodecahedron, and work around to the other
side, putting pieces in their places and then leaving them alone.  By
paying careful attention to the colors on the pieces you've already put
in place, you can figure out which ones you need to put in place next
to them.

It's worth noting that there are duplicates of all the edge-color
combinations.  That is, there are two red/yellow edges, two red/blue
ones, and so on.  The result of this is that (with my solution
technique) you can end up with a situation where it appears that only
two edge pieces are disturbed.  In this case, you need to swap one of
the two disturbed pieces with its mate of the same color (which is
somewhere around on the other side of the cube).  Annoying, but still
doable with but repeated applications of the same move.

One advantage of a solution like this is that, when you finally get
done, onlookers will admire you not only for your cleverness in solving
it, but also your persistence.  :-)

-- Tom Magliery <mag@ncsa.uiuc.edu>