From:

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[ Cube-Lovers,

There has recently been a discussion on Usenet group rec.puzzles

about some cube topics. There were a few pieces of new information,

such as that you can now get Ishi's 5^3 cubes in a lot of places (I

got mine in Learningsmith's) for about $35. Here's a message I sent

that's relevant to some Cube-Lovers topics.

By the way, I'm still working through Jerry Bryan's articles on his

brute-force program and his approach to symmetry. I hope to get a

reply out soon.

]

eric@gsb002.cs.ualberta.ca (Holleman Eric) wrote:

By the way, I found the Revenge somewhat easier than the Cube, and I

don't think that it was because of my familiarity with the earlier

puzzle.

x87bennett@gw.wmich.edu (Joe) wrote:

From my experience, if you can solve a Rubik's Revenge, you can

solve the Cube very easily. Once you get each of the middle 2 cubes

on each edge to match, and all 4 center cubes on each face to match,

it works exactly like a Rubik's cube.

and alan@saturn.cs.swin.oz.au (Alan Christiansen) wrote:

I have both. I solved both. The 4x4x4 is a superset of the 3x3x3.

ie by fixing all the face centres and then pairing all edges you are

left with a 3x3x3 cube, except that when you have solved this 3x3x3

there may be a single pair of edges flipped. This is impossible

on a real 3x3x3. Fixing this requires a middle layer to be rotated

1/4 revolution and then all the bits put back.I cant see how it can be [any] easier than a 3x3x3.

I, too, found the 3^3 easier than the 4^3. But I can imagine ways in

which a solver could find the 4^3 easier. Let us first consider a 4^3

with the faces fixed, the edges together, and the correct simulated

edge flip parity. I would solve this as if it were a 3^3, and a lot

of people do. But another solver might find it easier to take

advantage of the extra moves that are not possible on a 3^3. To take

a concrete example, it could be that the solver has a hard time with

flipping edges by pairs, as is needed to solve the 3^3. On the 4^3

you can flip one edge at a time. So the solver would find the 4^3

position easier than the corresponding position on a 3^3. If the

solver finds this so much easier that it overcomes the difficulty of

putting the faces and edges together--or in fact puts the faces and

edges together in the course of solving the corners and the edge

positions--then the 4^3 could be easier. It depends on the solution

procedure.

alan@saturn.cs.swin.oz.au (Alan Christiansen) continues:

ANyway the real reason I am writing this is that I have written

a cube simulator.

It can simulate 3x3x3 4x4x4 5x5x5 .... cubes.

I am working on 4x4x4x4 cube simulation.

This is interesting, as there is more than one way to model the

four-dimensional cube problem. Consider the 3^4 cube. It has eight

hyper-faces, each in the shape of a cube. One model of this puzzle is

that you could turn any face of any hyper-face as if it were a face of

a 3^3 Rubik's cube. In a second model, you cannot move part of a

hyper-face, but can turn each hyper-face as if it were a solid cube in

space. A third model allows either kind of move.

These models are different from each other. The second model permits

the face centers of the hyper-faces to move around, whereas in the

first model only edges and corners move. In the first model, odd

permutations of corners are possible, which is not true in the second

model. Of course, the third model is the closure of the first two.

According to Hofstatder's column reprinted in _Metamagical_Themas_,

there is an unpublished 1982 manuscript by H J Kamack and T R Keane

entitled ``The Rubik Tesseract''. They calculated the size of the

group of the 3^4 puzzle, but I don't know which model was used.

Alan Christiansen indicates he has gone directly to the 4^4 puzzle. I

don't know which model he plans, or if the models become more similar

with the extra possibilities inherent in the larger cube. I don't

even know whether he plans to figure out how big the groups are or

whether they are identical.

Dan Hoey

Hoey@AIC.NRL.Navy.Mil