Ronnie Kon <email@example.com> writes: >I write: >>The state of the cube is not: >> >>X|O|X|X|X X|A|X|C|X >>X|X|X|X|X X|X|X|X|X >>X|X|X|X|X But rather: X|X|X|X|X >>X|X|X|X|X X|X|X|X|X >>X|X|X|O|X X|X|X|B|X >> >>Where cubie "C" just "looks" like it's in the right place. >> >>You need an operator that rotates A->B->C->A. [...] >> >>This will very likely leave an inconvenient number of edges flipped. For >>the answer to _this_ problem, see my last post. ;) > >I think you must be wrong here (but would love to be proved wrong--I'm no >mathematician so group theory is very much beyond me). > > [Proofs deleted.]
I think we're in complete agreement, at least up to here. (I particularly
enjoyed your "proof by hardware ;).
I didn't mean to imply that the A->B->C->A operator preserved flipped-ness
of the Non-Central-Edge[NCE] Cubies. Moreover, I was being imprecise
where I said "a NCE cubie is simply flipped"; rather "the cubie *appears*
as if it were in the right place (i.e. judged by its colors) and flipped".
As you point out, *really* means that it is in the slot of its "twin".
To recap more succinctly, what I was proposing was a rather pedestrian,
two-step solution to the original problem. Starting from the initial
state in FIG1 (where the cube is completely "solved" except that the
cubies marked "O" are swapped. Also they are swapped in such a way that
the visible face is all a single color).
FIG1: X|O|X|X|X FIG2: X|Q|X|Q|X X|X|X|X|X X|X|X|X|X X|X|X|X|X A->B->C->A gives: X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|O|X X|X|X|X|X
STEP1] If we then apply the A->B->C->A operator, we end up with the state
in FIG2, where the cube is completely "solved" except that the cubies
marked "Q" "appear" to be "simply" flipped.
STEP2] We can then solve this problem, which (imo) is easier. For example
see the method that I described in an earlier post; this involves turning
the non-central plane (containing the flipped cubie) through a quarter turn.
Of course, now that I say it, it seems that the correct course would be to
*start* with the quarter turn of the non-central plane. This would leave
five NCE cubies out of place, but the cube would be in the right orbit.
>From there the solution should be straightforward (e.g. two intersecting
Finally, it seems clear that this entire problem --and all the subsequent
discussion-- maps directly onto a virtually identical problem on the 4by
cube (i.e. simply be removing the center planes).
Note that you get an apparant parity reversal by flipping the cubies, but
this does not actually move anything. In other words, no amount of
flipping and moving will allow you to end up moving A->B->C->A. That's
why I solve edges first.
Again, perhaps I'm missing the point, but if you don't care about how
the flipping comes out, the A->B->C->A 3-cycle is certainly doable:
[WARNING: EVEN MORE BORING STUFF AHEAD!! ;]
(I have no idea how to show this notationally, so I'll try pictorially.)
| 1] 2] V 3] X|A|X|C|X X|Y|X|C|X ->Z|Z|Z|Z|Z X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|B|X X|A|X|B|X X|A|X|B|X 4] 5] 6] Z|A|Z|Z|Z X|B|X|X|X X|B|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|B|X Z|Z|Z|A|Z ->?|?|?|?|? ^ | [Rotate Face one-half turn] >--- 7] \ 8] 9] X|B|X|X|X \ X|B|X|X|X Z|C|Z|Z|Z X|X|X|X|X | X|X|X|X|X X|X|X|X|X X|X|X|X|X | X|X|X|X|X X|X|X|X|X X|X|X|X|X V X|X|X|X|X X|X|X|X|X ?|?|?|?|? Z|Z|Z|C|Z<- X|X|X|B|X
[Rotate next- [Rotate Face to-bottom one-half turn] plane 1/4 Turn]
<---- 10] 11] \ 12] Z|C|Z|Z|Z Z|C|Z|Z|Z \ Z|C|Z|Z|Z X|X|X|X|X X|X|X|X|X | X|X|X|X|X X|X|X|X|X X|X|X|X|X | X|X|X|X|X X|X|X|X|X X|X|X|X|X ^ X|X|X|X|X ->?|?|?|?|? ?|?|?|?|? X|X|X|A|X<-
plane 1/4 turn]
| 13] V 14] 15] Z|Z|Z|Z|Z X|Y|X|B|X<- X|C|X|B|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|C|X|A|X X|C|X|A|X X|X|X|A|X ^ |