[next] [prev] [up] Date: Fri, 13 Jan 95 02:35:45 +0100
[next] [prev] [up] From: Dik T. Winter <Dik.Winter@cwi.nl >
[next] [prev] [up] Subject: Re: superflip

i've also done some searching for short maneuvers for superflip,
although not to the extent that dik has. i was never really
satisfied with my efforts to exploit its symmetry and centrality.
however, i've recently had some new thoughts about this which
look promising.

I have indeed considered this, but have not yet come to a conclusion.

case 1:
suppose that there is a minimal sequence for superflip which
contains a half-turn. then, by applying R' U2 to superflip,
we've moved 3q (or 2f ) closer to start.

I do not know whether this is clear for all readers. My reasoning
was similar but the conclusion different, but someway equivalent:
If the minimal sequence contains a half-turn, we may just as well
assume that that half turn is the last, and F2. I do not know
whether the proof has been shown on this list, but it is simple.
Suppose M is a minimal sequence, and Z is some random sequence,
in that case Z M Z' is also superflip. Take Z the maximal
sequence at the end consisting of quarter-turns only, we end with
a sequence of equal length terminating with a half-turn.
Because of symmetry we may just as well consider it to be F2.

case 2:
otherwise, every minimal sequence contains only 90 degree turns.
then either R' U' gets us 2q (or 2f ) closer to start,
or R' U gets us 2q (or 2f ) closer to start. (and probably
both do.)
it would be nice to reduce this latter case to only one of R' U'
or R' U . can anyone do this?

This needs more than simple symmetry. There are 12*8 different
endings, and we have 48 symmetries (24 by rotation * 2 by inversion).
Leaving 2 cases. I considered this, but have not yet come to
conclusions. On the other hand I do not yet know what to conclude
from M M' = I for every superflip sequence.

when searching for superflip in the face turn metric, it's
sufficient to search through depth 17 in stage 1!

suppose we have a 19f sequence for superflip. then, by considering
parity, some turn must be a half-turn. now we may suppose (as above)
that the last two face turns are U R2 , which is in stage 2!

Yes, I had seen that. One of the major reasons I was not amused when
the system crashed doing depth 17 in stage 1! I will restart the
program doing depth 17, but I will first redo the counting so that
counts larger than 2^32 are correct.


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