Using some new ideas, techniques, and computer programs, we
have successfully found optimal solutions to all symmetric positions
of Rubik's cube in the face turn metric (FTM). Furthermore we have
maneuvers for 1,091,994 20f* (positions whose optimal solutions
have 20 face turns) cubes and proven that there are no symmetric
21f* cubes. So if there are any cubes at depth 21 then these must
be unsymmetrical. To the best of our knowledge, at the start of this
investigation in January, only a few such positions were known (less
than a dozen). Expressions for all these cubes can be found on
Rokicki's home page http://tomas.rokicki.com/all20.txt.gz.

I have done a five-stage analysis of the 4x4x4 cube. My analysis considers the four centers for each face to be indistinguishable. It also assumes that there is no inner 2x2x2 cube in the middle of the cube.

Like Morwen Thistlethwaite's well-known four-stage 3x3x3 analysis, my five-stage procedure consists of multiple stages where each successive stage only allows use of a subset of the moves allowed in the previous stage, with the final stage only allowing half turns. So far, I have completed analyses of the five stages using the slice turn metric (STM). Use of other metrics is possible. (In fact I have done some other metrics for some of the stages.) My analyses for each individual stage are optimal with respect to the specified move restrictions for each stage. The results indicate that the 4x4x4 can be solved using a maximum of 79 slice turns.

I am looking for a suboptimal solution algorithm for the 4x4 and 5x5 cubes.

I am pondering about prepending a third phase to Kociemba's two-phase algorithm for the 3x3 cube. The initial phase performs two-layer twists on a 4x4 cube or a 5x5 cube until the stickers on the edge parts and on the side parts line up to form a 3x3 cube. Then Kociemba's two phase algorithms takes over and solves the 3x3 cube.

Does anyone have experience with such an algorithm? I currently don't know how to create the pruning tables for the initial phase. Also I am not sure, if my approach will work at all.

Silviu Radu has suggested using the latex2html utility to generate html for use in messages posted to the forum. Let's try a test:

Test

The files would have to be hosted somewhere but this would provide a way of handling all those mathematical symbols although no doubt some minor hand-turning would be necessary.

Mark

There is a unique Rubik cube position maximally far from Start *PROVIDED*
you only look at edges and ignore corners - it was found by J.Bryan and
is all edges flipped in place, composed with a mirror reflection of the whole
cube.

That suggests, taking this one edge position and exploring all possible configurations
(there are about 3 million) of the 8 corners to get 3 million cube positions.
If you are seeking a cube configuration with 21f or more distance to start,
these 3 million candidates seem tolerably likely to include a winner.

Because 3 million is a lot of searching, you might try a cheaper approach like just

I outline an approach which may be able to determine the maximin depth of the Rubik cube R - may be able to prove the answer is 20 face turns - with a feasible amount of computer time.

Because R has 4.3*10^19 configurations, exhaustive search is not feasible.

At present at least 10000 configurations are known (including the "superflip") that require 20 face-turns (20f) to solve.

Silviu Radu has a proof at at most 27f are necessary.

So the answer is somewhere in [20, 27]. What is it?

H.Kociemba's "two phase solver" works by first getting into the H = subgroup (which is known to be possible in at most 12f because of an exhaustive search of R/H) and then solving (which is known to be possible in at most 18f because of an exhaustive search of H) thus proving an upper bound of 30f.

In January of this year I set out to find a 21f* position---or, at the very least, extend the set of known 20f* positions. At that time I knew of only three 20f* positions, despite having exhaustively solved several collections of pretty patterns and performed months of optimal cube solutions.

At this point, I have found no 21f* positions, but with Herbert Kociemba and Silviu Radu, have found 11,313 (mod M+inv) 20f* positions. This set represents 16,510 mod M positions, and 428,982 overall cube positions. The majority of the positions were found by Silviu using a spectacular coset solver that he will write about soon.

I have put Thistlethwaite's 52-move algorithm on my site. This might be of historic interest to those of you who haven't seen a complete copy of it before.

David Singmaster had a copy that he scanned in, and put on his Singmaster CD6. That is a cd with all his notes and research on all kinds of recreational mathematics, which he makes available to anyone who is interested. I have converted those scans to text and put it all on my site.

Jaap
Jaap's Puzzle Page

God's algorithm for 4x4x4 Squares Set

I have computed God's algorithm for the set of Rubik's Revenge (4x4x4 cube) positions that are reachable by the following set of moves: { U^2, u^2, d^2, D^2, L^2, l^2, r^2, R^2, F^2, f^2, b^2, B^2 }. Actually, not all of the above moves need to be included in order to generate the whole set. Since these moves are expressed as squares of other basic moves in the group theory notation, the set of positions reachable by these moves is referred to as the Squares Group. In my analysis thus far, I have considered the four centers of a given face to be indistinguishable. That is, I am considering only the "plain" 4x4x4, not the 4x4x4 supercube (where all centers are taken to be distinguishable from the others). With this simplification, this set of positions does not actually form a mathematical group, so I will refer to it as the Squares Set here. 19 slice turns was found to be sufficient to solve any of the positions in this set.

In this paper we give a proof that Rubiks cube can be solved in 27f.
The idea is to eliminate the 476 cosets at distance 12 in the group H=< U,D,L2,F2,R2,B2 >.
In this way we never have to consider in the 2 phase algorithm that a coset is at distance 12.
So we only solve cosets at distance 11. Together with my earlier result of 28 this gives a proof of 27.
The same idea was used by Bruce Norskog in his 38q proof.




However we do not really need to compute all 476 cosets. In fact we only need to compute 7 cosets of the group
T = Intersection ( < U,D,L2,F2,R2,B2 > , < F,B,L2,U2,R2,D2 > , < L,R,F2,B2,U2,D2 > )
The group H is not invariant under all symmetries. But the group T is invariant under all 48.