## Suboptimal solvers for the 4x4 and 5x5 cubes?

Submitted by Werner Randelshofer on Thu, 06/29/2006 - 14:15.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.

## Using latex2html utility for posting

Submitted by cubex on Thu, 06/15/2006 - 08:59.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

## the search for a 21f, an idea for some candidates

Submitted by WarrenSmith on Sat, 05/20/2006 - 18:59.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

## A plan to settle the maximin distance problem so we can all go home

Submitted by WarrenSmith on Fri, 05/19/2006 - 20:50.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 =

## In search of: 21f*s and 20f*s; a four month odyssey.

Submitted by rokicki on Sun, 05/07/2006 - 12:21.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.

## Thistlethwaite's 52-move algorithm

Submitted by jaap on Wed, 04/19/2006 - 01:31.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 calculations for the 4x4x4 "squares set"

Submitted by Bruce Norskog on Mon, 04/03/2006 - 21:46.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.

## Rubik can be solved in 27f

Submitted by silviu on Sat, 04/01/2006 - 16:39.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.

## Two more classes with exacly 4 symmetries done - most 20f* are antisymmetric

Submitted by Herbert Kociemba on Thu, 03/30/2006 - 12:09.The definition of the classes D2(edge) and C2v(b) are explained on this page. Here you also can get some more information about these and other classes.

What is interesting, that from the 12 20f*-cubes of the class D2 (edge), 10 also have antisymmetry and from the 94 20f*-cubes of the class C2v(b) 92 also are antisymmetric.

## Results of two more cosets of the H group, this time face turn metric.

Submitted by rokicki on Fri, 03/24/2006 - 23:35.cosets, I decided to give it a shot in the face turn metric. So far

I've completed the identity coset and the flip8 (upper and lower edges)

cosets; the superflip coset and flip4 (middle edges) are still running.

The identity, flip4, flip8, and superflip are the four centers of the

H group. I've also shown that of the approximately 234,101,145,600

positions represented by these four cosets, none have a depth greater

than 21. This exploration covers more than 5/1,000,000,000 of the total

cube space.

The identity coset has the following depths. For comparison on the right