On 10/07/95 at 23:54:00 email@example.com said:
This problem was solved by David Benson in Oct. 1979, who was one of
the earliest cube pioneers. Dr. Singmaster reports on this in his
2nd Addendum of "Notes".Let A = R1 L3 F2 B2 R1 L3, then AUA = D1 AUA = R1 L3 F2 B2 R1 L3 U1 R1 L3 F2 B2 R1 L3 (17 q, 13 q+h)
Perhaps Jerry will find something shorter.
Well, without seeing the original article, I am not sure if I would
agree that the problem was solved or not back in 1979. By that I
mean that the problem I was proposing to solve was finding a minimal
solution. I don't know if the original article claimed that 17q
I can now confirm that 17q is indeed a minimal solution. With my
data base of positions through level 8, I was able to perform half-depth
searches to confirm that there are no solutions through 15q. Given the
existence of a 17q solution, that is sufficient to show that 17q is
minimal. But just to be sure, I extended my data base of positions
through level 9, and with my extended data base I was able to find
several solutions of length 17q.
I am not quite sure what we should count as a unique solution. But I can
report that my search found 16 unique (*not* unique up to conjugacy)
half-way positions. I use the term "half-way" advisedly. The "half-way"
positions are 9q from Start and 8q from B or vice versa. I guess you
could say that the vice versa gives you a total of 32=16+16 half-way
positions, but the whole concept of "half-way" is pretty slippery
in this case anyway.
Just because we know that 17q is minimal does not mean that we know
that 13 q+h is minimal. I have not done any searches of the q+h case.
With my extended data base, the results of the search with five generators
are as follows:
Level Number of Local Branching Positions Max Factor0 1 0 1 10 0 10.000 2 77 0 7.700 3 584 0 7.584 4 4,434 0 7.592 5 33,664 0 7.592 6 255,320 0 7.584 7 1,933,936 0 7.575 8 14,635,503 7.568 9 110,685,344 7.562
One more thing, and perhaps this particular problem can be put to rest.
I have mentioned several times that I could not figure out how to
use conjugacy for this particular problem. Well now I have, although
it is too late for doing the search. You certainly cannot use
M-conjugacy, but you can use a subgroup of M. The subgroup includes
four rotations and four reflections.
The four rotations are i, b, bb, and bbb, where we use lower case
letters to simulate Frey and Singmaster's script notation for rotations.
For example, b is the whole cube rotation consisting of grasping
the Back face and rotating the whole cube (not just the Back face)
clockwise by 90 degrees. For the reflections, we use Dan Hoey's
"permutations of face centers" notation. The four reflections are
(UL)(DR), (UR)(DL), (UD), and (LR). To tie the two notations
together, we could write the rotations as i=(), b=(ULDR),
bb=(UD)(RL), and bbb=(URDL). These eight rotations and reflections
form the subgroup of M which is called Q2 in Dan's taxonomy of
the 98 subgroups of M. Hence, we could have used Q2-conjugacy, which
would have reduced the size of the problem by about eight times.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU