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Date: Sun, 8 Jan 95 13:47 WET
From: "Martin Schoenert"
To: BRYAN@wvnvm.wvnet.edu
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To: "Jerry Bryan"'s message of Sat, 7 Jan 1995 10:15:52 EST <9501071609.AA16758@life.ai.mit.edu>
Subject: Re: Re: Cube with GAP
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