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Given a subroup G of permutations, we may define a cube position P to be

G-symmetric if every permutation in G preserves P up to color

relabelling. Explicitly, if x and y are two facelets which have the

same color in position P, then g(x) and g(y) also have the same color

in P.

Consider, for instance, the group R of whole-cube moves. The Pons

Asinorum is R-symmetric. Any whole-cube move moves the set of blue

facelets to positions occupied (before the move) by facelets of some

single other color. In fact, even if the cube were reflected, this

would be true. This can be verified by looking at one blue facelet of

the cube and realizing that you know from that alone where all the

other blue facelets are. Letting the M denote the group composed of R

augmented by the set of whole-cube mirror reflections. The Pons

Asinorum is also M-symmetric.

McKeeman has stated that any R-symmetric position (with the exception

of the solved state) is a local maximum. I do not know if this is true,

but I can show that any M-symmetric position (with the same exception)

is a local maximum.

Consider a robot cubenik which knows how to do whole-cube moves, but

can only QTW the "up" face. Clearly, this robot has no problems,

because it can move any face to the top, perform U or U', and move it

back. The robot could get along without U' by doing U^3 instead, but if

we're counting QTWs it's not going to win any races. Solution? Build a

transdimensional robot which can perform whole-cube reflections. This

robot performs U' by reflecting the cube, performing U, and reflecting

it back. The point of this parable is that for any two QTWs x and y,

there is an element m in M for which y = (m x m'). Note also that for

any m in M, the permutation (m x m') is a QTW.

Let P be an M-symmetric position, and let (x1 x2 ... xn) = P' be the

shortest solution of P in QTWs. Assume that P is not the solved

position, so n > 0. For any QTW y, I will demonstrate an n-step

solution of P which begins with y. Write y = (m x1 m'). Since P is

M-symmetric, (P m) is a relabelling of P. This implies that (P m P')

and therefore (P m P' m') are relabellings of I. Therefore the sequence

((m x1 m') (m x2 m') ... (m xn m')) = (m P' m') will essentially solve

P, up to a whole-cube move. The existance of an n-step solution

starting with an arbitrary QTW y implies that P is a local maximum.

There is a twelve-element subgroup T of M which will suffice instead of

M for this argument. Representing elements as permutations of faces, T

is generated by the permutations (represented as cycles):

(F L U)(R D B) -- Rotating the cube about the FLU-RBD axis

(F B)(U R)(L D) -- Rotation exchanging corners FLU and RBD

(L U)(R B) -- Reflection in the LU-RB plane

Question: Does there exist a position other than the solved position

and the Pons Asinorum which is T-symmetric or R-symmetric?