Date: Mon, 17 May 82 17:30:00 -0400 (EDT)
From: Allan C. Wechsler <ACW@MIT-AI >
~~~ ~~~ Subject: Reactions to 4^3.

Bernie and I have divided the double-takers into three major
classes:

```1. "Oh, a four-by-four cube!"

2. "Oh, a four-sided cube!"
```

(and the best of all)

3. "Oh, a four-dimensional cube!"

error #3 by calling the 4x4x4 a "C^4". Which brings me to
an interesting question. What would a four-dimensional
would have sixteen hypercubies, all of the "corner" type.
Each hypercubie presents to the outside world four
three-dimensional hyperstickers.

The 2^4 has eight three-dimensional hyperfaces, presumably
each its own color. I like the idea of using black and
violet along with the traditional red, yellow, orange,
blue, green, and white. We can call the hyperfaces Back,
Front, Up, Down, Left, Right, In, and Out. In is across
the puzzle from Out. I doesn't touch O, but does touch
all the other hyperfaces.

The 2^3 can be seen as two square slabs stuck together. Of
course, they aren't really "square" since they have
thickness. A move consists of rotating one of these
squares with respect to the other.

Similarly, the 2^4 is two cubical hyperslabs stuck
together. Of course, they aren't really cubical since they
have hyperthickness. A move consists of rotating one of
these cubes with respect to the other.

While the two slices of a 2^3 can only have four relative
positions, the hyperslices of a 2^4 can have twenty-four
different alignments. There are twenty-four corresponding
twists. One is the null twist, which consists of just
sitting and looking at the thing. (In this case, this is
not such a trivial operation!) Then there are six
different quarter-twists, as opposed to two in the 3D case.
Then there are, get this, eight "third-twists", which when
repeated three times bring the slice home. These have no
half-twists, three of one kind and six of another.

As in the three-dimensional case, the half-twists and
regard one of the sixteen hypercubies as fixed (without
loss of generality, if you'll believe that) then there are
twenty-four different quarter-twists in all. Twelve of
these are inverses of the other twelve, but selecting the
twelve "clockwise" ones is a lot harder than it is in the
three-dimensional case. My intuition fails me.

I haven't tried to apply the Furst-Hopcroft-Luks algorithm
to this monster. At most there are 6^15*15! =
4.7*10^11*1.3*10^12 = (very roughly) 6*10^22. I suspect
that some kind of parity, trinity, or quaternity argument
will reduce this by a factor of two, three, or four.

```--- Allan