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Date: 6 Nov 1986 13:41:48 EST (Thu) From: Dan Hoey <hoey@nrl-aic.ARPA>I have found 32 different 2x4 rectangles. I think that is all of them,

but I haven't got any proofs, nor even a decent mathematical model for

deciding when a flip is possible.

I think I have a proof. Wait a few paragraphs.

If a good model of the string interactions can be developed, we may be

able to make an attack on the doughnut problem based on the length of

string channels. In the doughnut, there are four ten-pair channels and

four six-pair channels. We may be able to show that the string wouldn't

reach one, and would exceed the other. More likely, the model will

prohibit the doughnut more directly.

Here is my model. It might be wrong.

The puzzle is a cycle of eight squares. Their underlying adjacency

relationships never change.

Each pair of squares is bound together by two loops of string (nylon

fishing wire, actually). Those two loops are dedicated to holding those

two squares together -- they never migrate to other squares, although

parts of a loop may sometimes lie on one of the pair, sometimes on the

other.

I need a diagram of the channels:

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

I offer the usual apologies about aspect ratio. Now, ignore the two

sided nature of the puzzle. Imagine that the two squares form a 1x2

unit billiard table, and think of the channels as trajectories of

billiard balls, and you will see that the channels form two disjoint

"orbits", each (* 4 (sqrt 2)) units long. The two loops of string

follow these two orbits, crossing from the obverse to the reverse sides

of the puzzle and back again at every chance they get. If you work it

out, you see that half the channels are empty at any given time.

The details are too mindbending, but the result is clear: those two

squares are bound together. They don't depend on synergy from the rest

of the puzzle to bind them.

Now close the pair of squares, putting the right on top of the left as

if you were finishing reading a book. You can open the pair again

vertically, but only in one direction. The front square can flip up, or

it can flip down, but not both. Without loss of generality, let's say

it can flip up. If you had folded the right square behind the left

instead of in front, it would be able to flip down.

Now gedankenfollow these gendankendirections.

Put the squares back as above. Only the square on the right (the rotor)

will move. Keep the left square (the stator) fixed.

Fold the rotor in front of the stator. Flip the rotor up, and over, and

behind the stator. Now it will open to the left. Flip the rotor around

the left edge of the stator until it is in front again. Then it will

flip down. Do so, until it is in back again. Now it will open to the

right.

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.

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

Since the laws are all local, governing the motion of one adjacent pair

of squares, there is no obvious invariant that forbids the doughnut.

There is another string-related question I am wondering about. I have

noticed some of the string-pairs getting twisted. I wonder how bad

this can get. Does anyone have an operation that can be repeated to

make the twists tighter and tighter? Are these puzzles built for

obsolescence?

I am convinced not.

I have been considering Magic metrics, but it's a difficult problem.

Counting flips is easy enough, but how do you count a move that skews a

parallelogram? Are such skew moves necessary?Dan

Your 32 configurations are characterized as follows: Turn the puzzle so

the unlinked rings are in front. Rotate it so Rubik's signature is in

the bottom row. The signature square could be in any of four positions.

The "matchbox" square could be clockwise from Rubik, or

counterclockwise. And Rubik himself could be in any of four

orientations. This determines the orientations of all the other

squares. There are no more degrees of freedom.

It is more instructive to consider the "supergroup", which includes

180-deg rotations but excludes turning the puzzle over. This group has

sixty-four elements in thirty-two similar pairs. I have a table.