Date: Thu, 13 Nov 86 13:16:44 -0500 (EST)
From: Dan Hoey <hoey@nrl-aic.ARPA >
~~~ Subject: Magic

Date: Thu, 6 Nov 86 18:13 EST
From: Allan C. Wechsler <acw@WAIKATO.S4CC.Symbolics.COM>

... Here is my model. It might be wrong.

>From what I have seen of the puzzle, your model seems to correctly
specify necessary conditions for puzzle movement. I have some

Each pair of squares is bound together by two loops of string (nylon
fishing wire, actually).

This is acceptable for a model, though the physical puzzles I have seen
are have twice as much wire on half of the squares as on the others. I
don't know why, and I am nearly ready to take one apart just to verify
my several curiosities about the construction.

... Only the square on the right (the rotor) will move. Keep the left
square (the stator) fixed. Flip the rotor ....
You have moved the rotor all the way around the stator. At some point,
each edge of both served as the hinge. This amazing orbit is the basis
for the bewilderingness of the puzzle. The path is a bizarre
three-dimensional cloverleaf.

This corresponds with my understanding of two-square interaction.
Thanks for nailing it down.

I think that I have now given all the "laws of motion" of the puzzle.

There is something more, because each square has *two* neighbors.
Consider a stator with two rotors. There are two ``bizarre three-
dimensional cloverleaves'' about the stator, but both rotors move in
the same b3dc. Furthermore, they cannot pass each other in their
common b3dc. Thus if one goes around more than a complete revolution,
the other must make a net advance in the same direction.

Since the laws are all local, governing the motion of one adjacent
pair of squares, there is no obvious invariant that forbids the
doughnut.

I think I can demonstrate one, given the fact that the pieces are
joined in a cycle. Let us tie each pair of adjacent squares together
as in your model. We notice that in the starting position, each of the
occupied channels is occupied by two wires, one for each neighbor of
the square. Suppose we dyke out a loop from the left neighbor, and
splice the left neighbor's connections into the right neighbor's loop.
Then do this all the other duplicate loops. I believe we end up with a
puzzle held together by four pieces of wire, each (* 8 (sqrt 2)) units
long. One of them is drawn below, assuming transparent squares. I
believe that this puzzle will move in the same way as the original one.

---- ---- ---- ----
|/\  |    |/\  |    |
|\ \ |   /|  \ |    |
| \ \|  / |   \|    |
|  \ |\/  |    |\   |
---- ---- ---- ----
|   \|    |  /\| \  |
|    |\   | /  |\ \ |
|    | \  |/   | \ \|
|    |  \/|    |  \/|
---- ---- ---- ----

Now put away the dykes and try to fold the puzzle into a doughnut. If
you succeed, you will have formed two (* 10 (sqrt 2)) wires and two
(* 6 (sqrt 2)) wires:

---- ---- ----          ---- ---- ----
|/\  |    |/\  |        |    |/\  |    |
|\ \ |   /|  \ |        |   /|  \ |    |
| \ \|  / |   \|        |  / |   \|    |
|  \ |\/  |   /|        | /  |    |\   |
---- ---- ----          ---- ---- ----
|   \|    |  / |        |/   |    | \  |
|   /|    | /  |        |\   |    |  \ |
|  / |    |/   |        | \  |    |   \|
| /  |    |\   |        |  \ |    |   /|
---- ---- ----          ---- ---- ----
|/   |  /\| \  |        |   \|    |  / |
|\   | /  |\ \ |        |    |\   | /  |
| \  |/   | \ \|        |    | \  |/   |
|  \/|    |  \/|        |    |  \/|    |
---- ---- ----          ---- ---- ----

Changing the string lengths is a little more magic than I expect from
Rubik.

My thanks to John Robinson for encouraging me to believe that the
string-length criterion has merit. He tells me Andy Latto has a
similar proof.

Dan