When you have seen Rubik's Revenge, have you seen everything?
There is some confusion about the meaning of the term Supergroup
for the larger cubes. There are two issues at stake:
(1) In the 4^3 and larger cubes, there are pieces that may
be permuted but are colored the same. Are positions
that differ only in the permutation of identically-
colored facets to be considered distinct?
(2) In all odd-sized cubes, each face has a center facet
that may be twisted. Are positions that differ only in
the twist of face centers to be considered distinct?
If the answer to (1) is `no', the puzzle is not a group, but a
collection of cosets. I find this interesting only in that it is
understood by the masses, and then the answer to (2) is `no'.
If the answer to both (1) and (2) is `yes', we have what I would call the Supergroup. I mark my cubes so that I can distinguish patterns in this group. I discussed a way of doing this for the 3^3 in my message of 9 January 81 0551-EST. For Rubik's Revenge, I use a similar procedure, except now two spots must be cut out on each face: +-----------------------+ | | | | | | | *|* | | |-----+-----+-----+-----| | | *|* | | | | | *|* | |-----+-----+-----+-----| | | | *|* | | | | | | |-----+-----+-----+-----| | | | | | | | | | | +-----------------------+ Also, I now arrange the faces T-symmetrically, with the spots toward the girdle.
Jim Saxe first brought it to my attention that we may answer `yes'
to (1) and `no' to (2). We get a group that ignores face center
orientation. This is probably what people mean when they say there
is no Supergroup for Rubik's Revenge: This is the group of the
puzzle, and the Supergroup (as I have defined it) is the same for
the 4^3. But in the 5^3 this group is distinct both from the
Supergroup and from the Color Cosets.
The 57th Piece
Alan Bawden, in his message of 5 June 82, despaired of explaining
the mechanical workings of Rubik's Revenge. I will now rush in.
If you take your Rubik's Revenge apart (as described in Ronald B.
Harvey's message of 19 May 82, not recommended by Alan Bawden), you
will find that the cubies ride around on a sphere: the mysterious
57th piece. Only the face centers are connected to the sphere;
they have flanges that hold the edges in place, and the centers and
edges have flanges that hold the corners in place. The tricky part
is the linkage between the sphere and the face center cubies.
The four center cubies of a face have extensions that together form
a mushroom-shaped plug. Each plug extends from the center of the
face inward and is cut in quarters lengthwise. There are six
sockets on the sphere in which these plugs will fit and may rotate,
but may not be pulled out. This is sufficient to implement a
puzzle isomorphic to the 3^3, namely the 4^3 where we allow only
The other necessity for a 4^3 is to be able to twist the six center
slices, which we name after their adjacent faces. To accomplish
this, there are grooves in the sphere that form three orthogonal
great circles. Each groove has the cross section of half of a
socket, and includes the four half-sockets that correspond to a
center slice. When we twist one of those center slices, the
half-plugs formed by pairs of face centers in that slice ride
around the sphere in the grooves. Of course there is an adjacent
center slice; when it is moved, it takes the sphere with it. The
reason the grooves cannot have the cross-section of a socket is
that then, when we twisted half the cube with respect to the other,
the sphere might turn forty-five degrees, preventing center slice
twists along the other great circles.
When you twist the U center-slice of your cube, does the sphere
move or stay fixed? To find out without taking the cube apart,
hold the cube by the D center slice and repeatedly twist the U
center slice clockwise. Don't touch the U or D faces while doing
this. Mostly, the U face will turn and the D face will stay fixed,
because the cubie-cubie friction is greater than the cubie-sphere
friction. Eventually, however, you will either see the U face lag
behind the U center slice, or the D face move to follow the U
center slice. If the U face lags, the sphere is not moving; if the
D face moves, the sphere is moving.
I will take this opportunity to mention another feature of the
interior sphere: It has screws in it. I took my screws out, but
the sphere didn't come apart. Then I put them back in, on the
`don't screw with it' principle. Perhaps they are there so Rubik's
Revenge won't float? This issue was raised by Tom Davis (12 August
80) back when people were interested in solving the 3^3 underwater.
The last I heard, Richard Pavelle (25 July 1980) was able to solve
the cube with only five gulps of air. I imagine some of the
30-second whiz kids can solve Rubik's Cube while completely
submerged. But Rubik's Revenge? Don't hold your breath.
The Mechanical Invisible Group
Alan Bawden's message posed an interesting question about the 57th
piece, which I will state somewhat differently. Suppose we paint
the sockets of the sphere according to the colors of the face
centers that inhabit them. Then we mix up Rubik's Revenge and
solve it. Must the sockets still match their face centers?
The answer is no. In fact, the sphere may be in any of the
twenty-four positions consistent with it having once matched the
face centers. To show this, we will show how to perform a
ninety-degree whole-cube move of the outside without moving the
sphere. This is equivalent to turning the sphere ninety degrees,
and we can repeat the procedure and its conjugates to move the
sphere to any of its twenty-four positions.
In order to twist Rubik's Revenge without moving the sphere, we
place the cube in a position such that the U, F, and R center
slices do not move the sphere and restrict ourselves to the moves:
U1 Clockwise quarter twist of the U face U2 Clockwise quarter twist of the U half (face and center slice together) D1 Clockwise quarter twist of the D face U1',U2',D1' Counterclockwise quarter twists R1,R2,L1,R1',R2',L1' Likewise for R F1,F2,B1,F1',F2',B1' Likewise for F
The move is actually fairly simple. The tricky part is moving the
L center slice cubies without moving the sphere. To do this,
remember the eight-flip
X = (R1 L1 U1 D1 F1 B1)^2
from the 3^3. This exchanges the L and R center slices in the 4^3,
allowing us to cycle the L center slice cubies in the R center
R2 L1' X R2 R1' X
is a sphere-fixing whole-cube move taking 24 qtw after cancellation.
The Theoretical Invisible Group
So much for tawdry reality. As Allan C. Wechsler pointed out on 6
August 82, we can imagine a 4^3 puzzle that contains a 2^3 on the
inside. If we solve the outside, must the inside be solved? The
process shown above for the 57th piece implies not, for that
process performs the RL antislice on the 2^3 with respect to the
outside. Can any move of the 2^3 be accomplished? Elementary
group theory says no, for odd permutations of the Rubik's Revenge
edge cubies are also odd permutations of the 2^3's cubies.
I ran the Furst, Hopcroft, and Luks algorithm on the problem. It
turns out that the permutation parity is the only restriction on
what can be done with the 2^3. Thus if we solve the outside, the
inside may be in any one of the (8! 3^7)/2 positions obtainable in
an even number of quarter twists on a 2^3 puzzle. This in
particular includes all whole-cube moves of the 2^3.
Unfortunately, I don't know any simple processes for whole-cube
moves or for turning two adjacent faces.