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
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
For most of the other questions, email@example.com
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. ]
> ...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.