firstname.lastname@example.org (Ronnie B. Kon) asks for hints for exchanging a pair
I tend to solve using commutators, but I don't see a way here....
The key is that commutators are always odd permutations. So do the
move that is an odd permutation of the edges, then use commutators.
Dik.Winter@cwi.nl (dik t. winter) shows a neat way of moving most of
the cubies affected by the odd permutation into the top slice, where
they can be cycled using Ronnie's commutator, which cycles the
TB(R), TR(F), and TF(L) cubies:
. . 0 . . . . . . . . 0 . 0 . .
(I'm naming them by their edge and their near side.) I suspect Ronnie
is using something like (F Ti F') T (F Ti' F) T (F Ti F') T^2 (F Ti' F).
(For this I'm using "i" to mark inside slabs).
But you can cycle the FL(T), FR(T), RB(T) cubies directly, using a
different commutator. With more effort, there is a commutator that
doesn't mess up face centers. We are getting to the part where it's
hard to distinguish between the hintable and the obvious, but if
people send me email about not being able to figure out what
commutators I'm talking about I'll answer, and post them if such
nobility is common.
My solution for the 4x4x4 always was: first corners, next edges and finally
centers. Because there are many identical pieces for the centers those are
reasonably simple. It would be much more difficult if each center had its
As I mentioned years ago, I've made places for mine by cutting corners
of to clusters of face centers and their neighboring edges on each
+----+----+----+----+ | | | | | | | | | | +----+---( )---+----+ | | | | | | | | | | +---( )---+----+----+ | | | | | | | | | | +----+----+----+----+ | | | | | | | | | | +----+----+----+----+
It's not that hard to fix the face centers, just time-consuming. It's
a good thing we do the edges first, though, or it would be hard to
figure where the cuts go.
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