## Solving the Rubik's Cube in Sub 13 Algs, BLD!

Submitted by Dbeyer on Sat, 07/29/2006 - 06:08.This is an advanced version of Pochmann, the method has three key steps.

Solve the F/B face + 1 S Edge (the UL as Default)

Roux Cycle the last three edges

Parity Fix.

The parity is something common in bld, so most of you will laugh at this. My method never encounters the 2 Corner 2 Edge swap parity, because that's what my system is based on.

It's something that 4 Step Solvers encounter, and they deal w/ it first.

## How to Compute Optimal Solutions for All 164,604,041,664 Symmetric Positions of Rubik's Cube

Submitted by silviu on Thu, 07/27/2006 - 17:07.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.

## The 4x4x4 can be solved in 79 moves (STM)

Submitted by Bruce Norskog on Sun, 07/09/2006 - 18:38.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.

## 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.