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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!"

One of the contributors to this list inadvertently made

error #3 by calling the 4x4x4 a "C^4". Which brings me to

an interesting question. What would a four-dimensional

cube really be like? Let's just start with a 2x2x2x2. It

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

analog in the 3D puzzle. The remaining nine twists are

half-twists, three of one kind and six of another.

As in the three-dimensional case, the half-twists and

third-twists are all products of quarter-twists. If we

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.

Yours with a headache,

--- Allan