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Date: Thu, 6 Nov 86 18:13 EST
From: Allan C. Wechsler
Subject: Magic
To: hoey@NRL-AIC.ARPA, Cube-lovers@MIT-MC.ARPA
In-Reply-To: <531686509/hoey@nrl-aic>
Message-ID: <861106181326.6.ACW@ROCKY-MOUNTAINS.S4CC.Symbolics.COM>
Date: 6 Nov 1986 13:41:48 EST (Thu)
From: Dan Hoey
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.