This post is rather technical in nature but if you do understand the
technicalities the
details are not difficult. It deals with solving sizeable cubes in a canonical
way. To alert
you, this post is about infinite cubes. The scrambling is assumed to be a
function from
some ordinal into the slice space (including face turns and inner slices). In
particular the
scramble might not be finite (so it's not part of the group generated by the
slices, in
general). The solution described will also often be infinite even if the
scramble is not. The
aim is to solve the cube in type alpha where the cube length is of type
1+alpha+n (+1)
+n+alpha*+1.
Consider a cube of length type 1+alpha + n (+1 if odd) + n +alpha* +1 (where
alpha is
some limit ordinal - I'm not going to try to consider more complicated than
that - and
alpha* is the reverse order type of alpha (so n+alpha* is actually the reverse
order type of
n) - so the centres are a square of length type alpha + n (+1 if odd) + n +
alpha*.
If alpha is 0, this is just a regular cube, but what if it is not? That's the
case under
discussion here.
We can fix an initial orientation of the scrambled cube by either (in the case
of an even
cube) assuming the UFR corner is solved or (in the case of an odd cube)
assuming the very
centre squares are solved placed in whatever way we usually would place them -
i.e. U at
U, F at F etc.
It is clear that each centre can be decomposed into pieces 4 of which are
easily type
(2n)*alpha=alpha or (2n+1)*alpha=alpha (depending on n being odd or even), 4 of
which
are easily typed alpha^2 (going in from the corners) and then canonically
reordered in type
alpha and (in the odd case) 4 of type alpha and either (2n)^2 (even case) or
(2n+1)^2 (odd
case).
Moreover the edges can be typed either as 2*alpha or 1+2*alpha (i.e. alpha).
Since m+alpha=alpha for all finite m, we can put the centres (2n)^2 or (2n+1)^2
first
without increasing the type and then it is easy to arrange the 8 (or 12) alphas
to get
12*alpha=alpha, so that we can type each centre in type alpha and hence the
whole in
type 6*alpha=alpha.
Moreover the edges can be typed in 24*alpha or 12=24*alpha=alpha and the
corners in
type 8, giving a total typing of 8+2*alpha=alpha. So we can type the entire
cube in type
alpha. We can also do it in such a way that the corners all come first and this
for each
centre position all the edges that can be reached by producing from its orbit
(either 12 or
24 of them) come before that centre position.
We can assume that centre postions (in a particular coordinate) are all
consecutive and
ordered according to a particular face dictionary (e.g. U,D,F,B,R,L). e.g. the
position in the
U face diagonally next to the UBL piece comes directly before the corresponding
position
in the D face (adjacent DFL) which is directly before the one in the F face
(adjacent FLU)
etc. - using a net diagram of the cube surface.
Memorize the cube position in type alpha. (You could go for type card(alpha)
but the
mapping might not be as easy and some of the canonical aspects may not be
preserved.)
Now we proceed to solve the cube by transfinite induction. We are solving
ordering on the
positions not the pieces (so the fact that we have lots of pieces that could
conceivably go
in the same place is irrelevant).
We start by making sure that the corners are evenly permuted (by using D if
necessary).
We can then solve the corners with 3 cycles and orientation moves. If the cube
is odd we
can then also do the 12 central most edges in 3 cycles and orientation moves.
So assume
all that is done. It takes finite type.
Given a position, we assume all lower positions have been solved. There are two
cases:
1. the current position is solved - and we are done with the stage.
2. the current position is not solved.
If the current position is not solved then there must be another element in its
orbit (by
which I mean the orbit of the pieces that can go into this position) which is
also not solved
(a later position).
If there are two such other positions then we can certainly apply one of the
standard
algorithms for a finite cube (e.g. a 5x5x5) to cycle the 3 pieces in the
earliest such
positions, solving the earliest of those positions without changing any of the
rest of the
cube and we are done with the stage.
So next we must consider the possibility that there are only two unsolved
positions in the
cycle.
If the position is an edge then all the centre positions in the slice
containing that edge are
later positions.
Slicing the edge quarter of a turn will disturb either 2 or 3 of the already
solved positions
but will reduce things to an even permutation for that orbit so that we can go
back to the
earlier elements of the orbit and solve again (this time the whole bunch) in a
couple of 3
cycles. We can choose to slice the edge so that the current position goes into
the earlier
position, for definiteness.
So now the problem comes - what if the current position is a centre position.
Since the corners are solved and so are all the edges produced from its
orbit.this case
cannot in fact arise. The evenness of the other two implying the eveness of the
permutation of this orbit, so we can't possibly arrive in this situation.
Therefore we can always solve the position without disturbing the lower
positions (except
for edge slicing and here there is only once choice of slice) and either no
higher positions
or a canonical choice of higher positions, using canonical permutations (either
2 earliest
unsolved pieces in a 3-cycle or all the D centre on the first move (in case the
corner
permutation is odd) but that is a canonical permutation or of 2-3 higher edges
and a
bunch of centres in a slice but again the slice direction has been chosen in a
canonical way
and since in the original ordering the centre pieces are arranged in blocks of
6 (each
corresponding to the same position but in each face according to a face
dictionary - e.g.
U,D,F,B,R,L) the slice even reorders the centre pieces in a canonical way.
Note each step (plus, if necessary the fixing of the finite number of displaced
earlier
positions - in edge slicing - though not in such step the fixing of the later
centre
positions) takes only finitely many moves.
Therefore this system will solve the cube in type alpha and because all the
distortions
along the way are all canonical, it could even be in principle be used as a
useful method
for blindfold cubing.
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