From:

~~~ Subject:

On 01/21/94 at 18:32:15 hoey@AIC.NRL.Navy.Mil said:

>I welcome Jerry Bryan's <BRYAN%WVNVM.BITNET@mitvma.mit.edu> efforts to

>improve the terminology of the groups associated with Rubik's cube.

>But there is some additional clarification I think is necessary.

Let G\C be the corners with edges without centers group. I intend

for the notation to indicate G reduced by C, where C is the rotation

group for the cube....

Let GC\C be the corners without edges without centers group....

Let GE\C be the edges without corners without centers group....First, these are not, strictly speaking, groups. Well, you can make

them groups, by defining what the group operation is. But I don't

know any way of doing that without losing the symmetrical nature of

the problem.Second, I would suggest that G/C, GC/C, and GE/C are more standard

names for these objects. The elements are nominally 24-element sets,

each of which is an equivalence class when two positions are

considered equivalent when they differ by their position with respect

to the corners. The classes are called the cosets of C in G, GC, and

GE, respectively.

Dan Hoey's criticism's are quite valid. I will attempt to repair

the damage as follows: 1) accept the Gx/C notation in lieu

of Gx\C, 2) define an operation within Gx/C such that

Gx/C is a group, and 3) use Gx/C as a model for cubes without

centers in such a way that the symmetrical nature of the problem

is retained.

Let C be the set of twenty-four whole cube rotations of the cube, and

let G be the standard 3x3x3 cube group. We observe that if X is a cube

in G, then c'Xc is also a cube in G for every c in C. We could call

this operation C-conjugancy. However, there is seldom (if ever) any

reason to speak of C-conjugancy. That is, C is just a subset of M,

the set of forty-eight whole cube rotations and reflections. Indeed,

C is half of M, and the other half of M is the reflection of C. Hence,

M-conjugancy of the form m'Xm is more powerful than C-conjugancy, and

there is normally no reason to speak of C-conjugancy. I only bring it

up to emphasize that if X is in G, then c'Xc is in G.

On the other hand, if I understand correctly the model most people use

for G, elements of the form Xc or cX are not in G except for the

trivial case where c=I. The problem is that C is considered to move

the centers, but G is generated by Q, the set of quarter-turns of the

faces, and Q does not move the centers. For example, there is a

c in C such that F=c'Rc, but there is not a c in C such that

F=Rc or F=cR. And indeed, neither Rc nor cR are in G at all unless

c=I.

As we said, G is generated as G=<Q>, where Q is the set of

quarter-turns Q={F,B,U,D,L,R,F',B',U',D',L',R'}. Elements of Q

move the corners and edges, but Q is the identity on the centers.

C, on the other hand, is generally considered to move the centers.

Hence, the group generated as <Q,C> is a supergroup of G, and there

are elements of the supergroup which are not in G. (This

supergroup, by the way, is not The Supergroup. The Supergroup is

generated by Q alone, but with orientations of the (otherwise fixed)

centers considered.) Therefore, our first order of business is to

make C into a sub-group of G.

We observe that since the elements of Q are the identity on the

centers, the primary function of the centers is to provide a

frame of reference. But we can provide a frame of reference

without the centers actually being there.

For example, consider the group GC consisting of cube centers

and corners. You can model this group by removing the edge

labels from a physical cube. Establish the cube at Start and

perform RL'. The corners will be rotated forward, and will be

positioned properly with respect to each other, but the cube is

clearly not solved. You can tell that the cube is not at Start

because the corners are not aligned properly with the centers.

Now, do the same thing except remove both the edge and center

labels. If you perform RL' at Start, the cube "looks"

solved but rotated forward. However, we can adopt the convention

that the cube is solved only if the Up color is Up, the Front

color is Front, etc. With this convention in place, RL' is

clearly seen not to be solved; it is two moves from Start.

The convention provides the fixed frame of reference.

Furthermore, RL' (which is in GC) is equal to an element of C, and

indeed all elements of C are in GC, as are all elements of the

form Xc or cX for c in C and X in GC. Hence, we have <Q>=<Q,C>.

Similar comments apply to GE, the group of edges and centers, except

that processes composed from elements of Q to accomplish rotations

in C are not quite so short in GE as they are in GC.

G, the full 3x3x3 cube group consisting of corners, edges, and centers

is a bit more difficult. The problem is that if X is in G, then

objects of the form Xc or cX are in G only if c is even. Twelve

elements of C are even and twelve are odd. Indeed, C[even] is

a sub-group of C, but C[odd] is not.

We will deal with this situation (as circumstances require) in two

different ways. One is simply to restrict ourselves to C[even]

when dealing with G. The other is to define a new group we will

call GS. In our model for G in which the centers are implied by

a frame of reference convention rather than by actual physical

centers, we can easily add slice moves to the standard face moves.

If the centers were physically present, then the slice moves would

move the centers, but without the physical centers there is

no problem.

If S is the set of slice moves, then GS is generated as <Q,S>.

GS is essentially G with parity restrictions removed. Hence we

observe that |G|=|GC|*|GE|/2, |GS|=|GC|*|GE|, and |GS|=|G|*2.

Also, if X is in G or in GS, then elements of the form cX or

Xc are in GS for all c in C. In those occasions where we are

willing to think of GS rather than G, we can use C rather than

C[even].

At this point, we can say that GS/C, G/C[even], GC/C, and GE/C are

cosets of C in GS, C[even] in G, C in GC, and C in GE, respectively.

To be a little more conformant with standard coset notation, we will

write cube elements as lower case letters for the remainder of this

note, and hence for a particular cube x a coset of C is denoted as

Cx={y: y=cx} or xC={y: y=xc}.

Now, we propose a group operator for the cosets: Cx Cy = C(xy) and

xC yC = (xy)C. Showing that we have a group is easy. I originally

included a proof in this note, but there is a proof in Chapter 8 of

Frey and Singmaster's _Handbook of Cubik Math_. Hence, I will defer

to their proof instead.

According to Frey and Singmaster, G/C is called the factor group of

C in G, or the quotient group of G by C. Of most significance to

us right now is the fact that the identity of the factor group is

Ci or iC, where i is the identity of G. But Ci or iC is just C.

Hence, the identity of the factor group is C. This justifies our

identification of G/C with a centerless cube. In English, it means

that we can rotate a centerless cube in space without changing

anything. I think this would comply with most people's intuitive

sense of what it means for a cube to be centerless.

Finally, as to whether this model retains the "symmetrical nature of

the problem", I will have to leave that as an open question,

depending on precisely what we mean by "symmetrical". It seems to

me that this model does a better job of being "symmetrical" than

a model which includes only seven corner cubies or only eleven

edge cubies, but maybe not. What does "symmetrical" mean when

it comes to centerless cubes?

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are

going to have time to do it over again tomorrow?