From:

~~~ Subject:

I've been working on a new algorithm to find move sequences to reach

certain positions on the 3x3x3 cube. The basic idea is to find a

sequence such that:

(S1 S2 S3... SX) ^N = Goal State

where X is the number of moves in the fragment and

N is the number of times the fragment is repeated.

I call such a process to be "Cyclicly Decomposable".

Certain states, such as the 12-flip, require quite a few moves, in

fact more moves than possible to search using brute force even when

using high-order computers. The best results using the Kociemba

algorithm need 28 q turns or 20 q+h turns for the 12-flip.

While the cyclic decomposition algorithm (henceforth the CD algorithm

for short) usually requires more moves than the Kociemba algorithm

it does have an mnemonic advantage. The following is the best

result using the CD algorithm to date:

p192 2 Flip in face (F1 B1 L1 T1 D1)^6 (30 q) p195 12 Flip (T1 B3 T1 L1 F3 R3)^6 (36 q)

Note that both of these processes use 5 of the 6 generators.

Some cube positions are extremely resistant to CD but flip and

twist patterns are no problem. In particular, the 6 X order 3

pattern does not yield to CD with values of X up to 7.

Naturally with N = 1 we can always find one of the optimal paths

but I am more interested in cases where N > 1. One may note that

with N > 1 using CD processes we are still free to use any number

of q turns except a prime number, for which N must be 1, but this

should not be too constraining.

By relaxing the conditions somewhat we can conceive of sequences

which are near-CD, that is CD with a small suffix or prefix:

p169 4 Y's Rot + 2 X (F2 B2 D1 L2 R2) ^2 + T1 (11 q+h)

By looking at the best sequence the Kociemba algorithm can

find for a position, we can count the number of q turns and

use this as a starting point for an attempt with CD...

p1 6 X order 3 R2 L3 D1 F2 R3 D3 F1 B3 U1 D3 F1 L1 D2 F3 R1 L2 (20 q) Looking at p1 we can infer that possibly X=5 and N=4 may lead to finding a CD process or X=4, N=5 X=10, N=2 X=20, N=2 etc.

At the very least we can discount X * N = odd number.

-> Mark <-

Email: mark.longridge@canrem.com