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From: roger.broadie@iclweb.com (Roger Broadie)
To:
Cc: "Tenie Remmel"
Subject: Re: Rubiks Revenge moves
Date: Sat, 22 Nov 1997 22:51:33 -0000
Message-Id: <19971122224927.AAA7296@home>
Tenie Remmel wrote (19 November 1997 )
> Is there an easy way to cycle three adjacent top edges on the
> Rubiks Revenge? I can't find one shorter than 62 moves, but if
> there was a short one I could simplify my solution greatly.
>
> . b c . . a b .
> a . . . => c . . .
> . . . . . . . .
> . . . . . . . .
Rather than just throw a few more solutions into the pot, I'd like to start
with some comments on the sort of process everyone, including me, seems to
use to deliver 3-cycles of edge pieces in the 4x4x4. It is of the general
form
[P, TQT']
where the square brackets are used to show a commutator, that is, [A,B]
means ABA'B'.
In this process P and Q are turns of layers that are parallel to one
another, and T is a turn of a layer transverse to P and Q. For instance, P
and Q could be L and r and T could be U (capitals for the outer layers,
lower case for the neighbouring inner layers, with sense parallel to the
corresponding outer layer). That gives
[L, UrU'] == L UrU'.L' Ur'U'
which is a (not especially appealing) 3-cycle of edges.
In fact any process of this form is a 3-cycle provided it takes one piece
from the layer Q into the layer P. That will happen if T is a quarter turn
in either sense - I haven't found anything useful with T as a half turn.
But P and Q can be any power. The reason that processes of this form are
3-cycles is simple. If two permutations intersect at only one element,
then their commutator is a 3-cycle. Thus if
A = (...a1, x, a2...) and B = (...b1, x, b2...)
then
[A,B] -> (a1, x, b1)
If you do just UrU' you will find there is a line of displaced pieces along
the intersection Ub, but no other displaced piece in any of the layers
parallel to r. Any of these pieces can be picked out to form part of a
3-cycle by selecting the layer that is parallel to r and contains the piece
and using a turn of that layer as the component P of the commutator, with
UrU' forming the component TQT'.
In general, if all of P, Q and T are outer layers we will have a 3-cycle of
corner pieces, if two are outer layers and one an inner layer we will have
a 3-cycle of edge pieces, if one is an outer layer and the other two inner
layers we will have a 3-cycle of centre pieces, and if all three are inner
layers we will have done nothing visible to the cube, but in fact there
will have been an invisible 3-cycle of the pieces of the imaginary internal
2x2x2.
We can derive the last of these cases from the first
quite neatly applying a fascinating concept called evisceration, which I
recently met trawling through the archives. It was first quoted from David
Singmaster's Cubic Circular by Stan Isaacs on 26 May 1983 and our present
acting moderator also discussed it on 1 June 1983. If you turn a cube
inside out by changing each outer layer in a process into an inner and
vice-versa (i.e. capitals to lower case), then, in the effect of the
process, you will interchange corner pieces with the pieces of the internal
2x2x2, and edge pieces with centre pieces. Making P, Q and T all to be
outer layers gives just a 3-cycle of corner pieces; therefore applying
evisceration takes that cycle into one on the pieces of the internal 2x2x2.
Singmaster's Notes on Rubik's Magic Cube, the fifth edition, interprets
processes of the type
[P, TQT']
as
[P,[T,Q]].
This expands to P TQT'Q'.P' QTQ'T', but the sequence Q'P'Q in the centre
reduces to just P', giving the same expansion as before. Of course, the two
components of the commutators TQT' and [T,Q] have different total effects,
but what they have in common is that they put the same single piece into P.
We can look at them both as being sort of like a mono-operation. Let's
call it a "monopop": each process pops a piece into P; you then turn P,
then reverse the pop operation, which extracts a different piece, and
finally restore P.
It's relatively straightforward to use this form to design specific
processes. Say we want to move an edge piece from ULf to FLd and keep the
third member of the 3-cycle in the top layer. Then we can take P to be L
to achieve the required part of the cycle. We now know that Q must be in r
or l. Let's take l. The transverse move in T has to take a piece from l
into the point of intersection of the two components of the commutator,
FLd. So it must be in F. Playing with F and l shows that the following
does the job.
[L, FlF'] == L FlF'. L' Fl'F' -> (ULf, FLd, UBl)
If we'd taken Q to be in r we'd have needed a bit more care to keep the
third piece of the 3-cycle in the top layer, but [L, F'r^2F] does, putting
it at UBr.
If we want to move a piece along a diagonal - from ULf to FLu, say - we
need to use the other component of the commutator, TQT'. Thus we can build
3-cycles which include ULf to FLu around the component U'FU. For instance
[f', U'FU] -> (ULf, LFu, RUf)
With a clear head and a good following wind it's possible to work out these
processes on the fly.
They also transform nicely into another process of the same type by cycling
the elements, which has the effect of conjugating the original process.
Thus the last process can be dealt with as follows
U[f', U'FU]U' = [Uf'U', F] -> (UFr, LFu, BUr)
This cycling procedure comes from Singmaster.
Let's now think about top-layer edge processes. I'll denote the pieces like
this.
X a1 a2 X
d2 o o b1
d1 o o b2
X c2 c1 X
The purpose of the numbering in pairs is to emphasize that the processes
come in pairs. Each process has a twin created by changing each inner
layer turn into its next-door inner neighbour. Thus the simplest U process
of the general type we're using is of this form:
[l, F'LF] -> (a1, c2, d1)
Its twin is
[r', F'LF] -> (a2, c1, d2)
In the twin process, each edge piece is changed into its next door
neighbour.
We want to capture this regularity. I will therefore represent this pair
by
[M', F'LF] -> (a*, c', d)
In this representation, M is either r or l', the asterisked piece defines
the layer that contains M and primed letters denote a piece with the
opposite suffix number to the asterisked piece. Obviously, these suffixes
are closely related to flip in a 3x3x3 and the assignment of the numbers is
arbitrary. Some assignments are more helpful than others in a particular
context, and the method used in the diagram above is the obvious one of
giving the same number ("flipperty"?) to the pieces in the positions that a
single piece moves into during a complete U turn.
Here, then is a complete set of top layer 3-cycles of edge pieces, to
within a reflection. It comes from a fairly systematic search I did for
processes of the type [P, TQT'] that can be conjugated by at most one turn
into a top-layer process. They are oriented in a way I find easy to do.
I will leave them as commutators, because it is very easy to perform the
full set of turns from them. The T/T' sequence remains constant for both
halves and the only adjustment needed is to invert P and Q the second time
around. Inverses are also easy to perform, since all one has to do is read
off the second component first.
(a*, c', d) [M', F'LF]
(a*, b, c) F2 [D R2 D', M2] F2
(c*, a, b') R2 [M' D' M, U2] R2
(d1, b2, b1) (Bb)' [U l U', R2] (Bb)
(d2, b1, b2) (Ff) [U r' U', R2] (Ff)'
(d2, c2, d1) (Bb) [L2, D l D'] (Bb)' (Tenie's question)
(d1, c1, d2) f' [L2, D r' D'] f
The last two pairs could have been left in the M form if say N was
introduced to represent either f or b'. But keeping them separate lets us
save a wrist movement for the first three by combining the inner and
neighbouring outer layers for a turn relative to the central cut. That
won't work for the final process, since the F layer is already included in
the 3-cycle and can't be amalgamated with the f layer.
All these process can be directly transferred to the 3x3x3, using the one
single central slice as M (or l' or r). The primes then correspond to
flipping the edge piece relative to the top surface.
Roger Broadie
22 November 1997