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Subject:

Jerry Bryan wrote in his e-mail message of 1995/01/07

Note that this model does not include the face centers. That is, it

is G[C,E] rather than G[C,E,F]. 56 numbers would be required to

include the face centers. The distinction between 48 facelets and

56 facelets bears on the nitpicky question of whether the set C of

rotations is a subgroup of G or not.

Absolutely right. This part of the GAP documentation was written years

ago. These days I represent MG, CG, etc. as permutation groups on 54

points. I also changed the numbering, so that the [1..24] represent the

edges, [25..48] points represent the edges, and [49..54] represent the

centers.

Jerry continued

What I don't see is how to

model the Supergroup in GAP. It looks like you would have to

label each Face center with four numbers so you could see the

rotations of the Face centers, but that seems like overkill.

This is also correct. But GAP doesn't mind those 24 more points.

Jerry continued

When I write the model out to disk, I only write out 8 corner facelets

and 12 edge facelets. For example, I only write out the front and

back corner facelets. This saves space and converts the model from

a facelet model to a cubie model, with the twists implicitly encoded

rather than being explicitly encoded via multiplication tables. It

also automatically establishes a frame of reference by which a

proof of conservation of twist and flip can be accomplished.

In terms of computational group theory this sequence of 8 corner and

12 edgde facelets is called a *base* for the permutation group G.

That is, each element of the group is uniquely determined by the

images of those 20 facelets. Of course if you have already proved

that no single corner can be twisted and no single edge can be flipped,

you can reduce this to 7 corner and 11 edge facelets.

Mark Longridge wrote in his e-mail message of 1995/01/03

... I don't know how a normal 4x4x4 could be represented though.

Jerry answered

I fail to see the problem. Just number the facelets. The only

problem would then lie in deciding what the generators are -- i.e.,

which kind of slice moves do you accept. You would also have to

decide whether to model the invisible 2x2x2 inside, but again if you

did, just number the invisible facelets and include their movements

with your generators.

The problem is that many different positions all look solved. For

example, you can permute the 4 center facelets of one face or exchange

two adjacent edges, and the cube still looks solved (of course you cannot

do all this independently). So if we take the obvious permutation group

on the 6*16 points, then a whole subgroup would correspond to what a

puzzler would consider solved states. If by a model we mean a group

whose elements correspond to the different states a puzzler would see,

and whose identity corresponds to what a puzzler would consider solved,

then I have no good idea how to model the 4x4x4 cube as a permutation

group.

Have a nice day.

Martin.

-- .- .-. - .. -. .-.. --- ...- . ... .- -. -. .. -.- .- Martin Sch"onert, Martin.Schoenert@Math.RWTH-Aachen.DE, +49 241 804551 Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany