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Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu> has some more questions.

Take a standard 3x3x3 Rubik's cube, and remove the corner and

center labels to make an Edges Cube.... Scramble..., how will

the cubemeister distinguish Start from Pons Asinorum?

... if you identify the identity with Start, you are in the

disquieting situation of having a group with two distinct identities

(grin!).

The problem is that we would not be dealing with a _group_ then, but a

collection of cosets of M. Just as in the edge `group', we deal with

either 1) a less-symmetric group in which one of the edges never

moves, or 2) a larger group in which we distinguish positions that

differ by rigid motions of the cube, or

3) a non-group in which we consider cosets--equivalence

classes of group #2, where group elements that differ by rigid motions

are equivalent. You have got a lot of mileage out of working with

group #2 to save duplication among symmetries, then reducing to

non-group #3. But what you lose is the group structure of the object

you are studying. Instead, you have to work in the large group and

then deduce information about the cosets. All in all, though, I'm

very glad of it, for the lost symmetries of group #1 were sorely

missed.

For most of the other questions, mouse@collatz.mcrcim.mcgill.edu

provides satisfactory answers. However, strictly speaking we should

not call an equivalence class to be a group element (unless it is a

coset of a normal subgroup, and neither C nor M is normal in the large

group). I'll admit I've also abused the term when considering

distances in the ``edge group'', as if all 24 rotations of a position

were the same element of some group. But when we start dealing with

the distinction between fixed and movable cubes I think we need to

start being more careful.

[ mouse also mentions that quarter-turn ``usually doesn't include

slice turns on the 3-Cube, but on the 4-Cube and higher, they must

of necessity be included.'' I'll take that as an argument for

eccentric slabism: a QT rotates any 1xNxN slab except a central slab

of an odd-edged cube. As opposed to cutism, where a QT consists of

a rotation of part of the cube with respect to the other. ]

Other questions:

> ...since Start and Pons Asinorum differ only by a simple

> reflection, why had not my version of M-conjugation declared them

> to be equivalent?

Your versino treats positions X,Y for which m'Xmc=Y (m in M, c in C)

as equivalent. If you instead determine when m'Xmn=Y (m,n in M) you

would find them equivalent. This is equivalent to changing the loop

in your version of M-conjugacy.

For j = 1 to 24 for k = 1 to 24 for m = 1 to 2 for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i)))))

so that the two occurrences of Qm need not be the same.

(I speak of "my version of M-conjugation", but the question is no

different if you look at Dan Hoey's original M-conjugation).

No, I didn't use M-conjugation except for a cube with a fixed

orientation in space [or equivalently, with face centers]. So in the

original concept of M-conjugation that Jim Saxe and I put together,

Start and Pons Asinorum don't just differ by a reflection.

I found Dan Hoey's postings about the four special states of the

Edge Group to be delightful.... However, [without the results on

distances] if we identified the group as being rectangular, would we

be led to saying which of the four special states were diagonally

opposed without the computer search? Without the search, I might be

tempted to say that Start and Pons Asinorum were diagonally opposed.

Well, really the `group' is in the shape of a sphenoid, a word I

learned yesterday for a tetrahedron whose three pairs of opposite

edges are equal. [Or equivalently, a tetrahedron whose edges are face

diagonals of a rectangular prism.] But it might be more accurate to

consider it as a large ball of string with a bunch of symmetries.

Calling it a rectangle or sphenoid may lead us to ignore the structure

that is not representable in Euclidean space.

Dan Hoey

Hoey@AIC.NRL.Navy.Mil