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Subject:

On 10/07/95 at 23:54:00 mark.longridge@canrem.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

was minimal.

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