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I just wrote a quick program to count the number of QT to move from

the full cube group to the subgroup generated by <F^2,B^2,L,R,T,D>.

Thistlethwaite computed that this takes at least 7 HT in the worst

case. The surprisingly good result is that it also takes only 7 *QT*

in the worst case. This reduces the upper bound I posted Friday to 91

QT.

I had wondered if the worst cases could be reduced by choosing a

different pair of faces to restrict to half-twists. Unfortunately,

the all-edges-flipped position is one of those that requires at least

7 HT (and so 7 QT), and by symmetry it cannot be improved.

Allan C. Wechsler <ACW@YUKON.SCRC.Symbolics.COM> analyzed his own

cube-solving method, finding that:

For example, flipping two edges in place takes 22 qtw.

This can be done in 16 QT, though I don't know if that is the best

known. Any pair can be flipped with a conjugate of the 14 QT slice

mono-op FOTAROFATO-RAM TAFORATOFA-ROM (FT'RF'T L'R B'TR'BT' LR').

Adjacent and antipodal pairs require the introduction of a

non-cancelling QT in the conjugator.

Obviously a lot could be gained from tweaking the earlier part of

the algorithm to guarantee that I don't need to do this at the end.

Probably, but it's hard to make that guarantee. The problem is that

unless you flip edges in place with no other action (the very problem

you're trying to avoid) you may affect the later choices in the

algorithm, making the earlier tweaks wrong for that branch of the

algorithm.

For instance, the 7-QT method my program found solves the orientation

of all the edges (using a particular non-standard labeling of the

orientation that, when solved, is invariant under F^2, B^2, L, R, T,

and D). But it may permute edges, and permute and twist corners, so

it may not form a useful part of an arbitrary cube-solving algorithm.

It works in Thistlethwaite's only because he is careful in all

branches of the rest of the algorithm never to mix up the orientation

of those edges.

Dan Hoey

Hoey@AIC.NRL.Navy.Mil