From:

~~~ Subject:

Jerry asks, in his message of Sun, 6 Nov 1994 09:15:37 (EST):

This is not a shift invariance question, but rather two

questions about your searches. One question is, do you perform

separate searches for q-turns and h-turns, or only for h-turns?

The <U, R> group searches are q-turns only, but if I see a way to

compress it a bit by eye I do so. For example:

UR2 = U3 R3 U3 R2 U1 R1 U1 R3 U3 R1 U1 (R1 U3)^2 R3 U3

...was reduced to

UR2 = U2 R3 U3 R2 U1 R1 U1 R3 U3 R1 U1 (R1 U3)^2 R3

The second question is, how on earth do you keep track

of all those processes in your searches?

With the computer hardware I'm using (486-DX40 with 4 megs) and

my current algorithm I created a file "ur.dat" which is a flat

ascii file. It contains all the processes which generate distinct

positions up to 12 q turns. I also have the file "ursum.is" which

contains all the `cubesums' for each element of <U, R>.

My program "x3bin.exe" loads the "ursum.is" database into memory

and it does a binary search on the cubesums to try and find a

match for the current position. If not found, it turns the

current cube and tries again, with longer and longer sequences

until a match is found.

Using this method I have found a process for all <U, R> positions

with the longest being 22 q turns (so far). The hardest positions

can take as long as a couple of hours. I have no idea what the

antipodes look like at this time, but I'll probably try the

random approach soon.

Jerry continues:

>p183 6 Twist R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3 > (18 q or 16 h moves) ^^^^^^^^^^^^^^^^^^^^^

This process was found using the Kociemba algorithm in a program

written by Dik Winter, which I ran on a Sun4. This program uses

h turns in it's searches and uses all 6 generators.

After inspecting the original process found by the program,

I was able to manually reduce it somewhat, resulting in p183 above.

The original process....

Solution (13+ 4=17): R1 U1 D1 R3 U1 R1 D2 R1 U1 R3 U1 D1 R3 U1 R2 U1 R2 That is (or as Dan would say i.e.) 13 h turns in phase 1 and 4 turns in phase 2. Hmmm, looks like I used the inverse of the original.

Jerry Continues:

I have been asked how I search

so many positions. I have answered the question before, but I guess

another part of the answer that I haven't mentioned is that I don't

keep up with processes at all, only positions. If I am asked to provide

processes, I can do so, but it is a very painful task. I have thought

about keeping up with processes, but I am quite sure that if I did

so it would reduce the number of positions I could search.

Well, that explains the fact you counted the number of antipodes but

had no processes for them, but do you know what they look like?

If you can tell me what one of the 87 positions at 25 q turns

looks like, I should be able to generate a sequence for it.

-> Mark <-

Email: mark.longridge@canrem.com