## Solving the 4x4x4 in 68 turns

Submitted by Bruce Norskog on Wed, 02/14/2007 - 01:03.I have completed a five-stage analysis of the 4x4x4 cube showing that
it can always be solved using at most 68 turns.
The analysis used the same five stages that were used in my prior posts where I claimed the 4x4x4 cube
can be solved in 79 single-slice turns, or alternatively in 85 twists.
The difference in this analysis is that it allows any single layer turn or double layer turn
(where the two layers are any two adjacent layers and moved together) to be counted as a "turn."
In some prior posts, I referred to these turns as "block turns."
So the set of turns about the U-D axis that count as one turn are the following:

U,U',U^{2},u,u',u^{2},(Uu),(Uu)',(Uu)^{2},

D,D',D^{2},d,d',d^{2},(Dd),(Dd)',(Dd)^{2},

(ud'),(u'd),(ud')^{2}

## Lower bound using the Edges antipode

Submitted by mdlazreg on Mon, 01/29/2007 - 15:01.1) First solve all the edges [less than 18 moves as proved by Tom Rokicki ].

2) Take it to the cube identity [less than 22 moves as proved by Silviu Radu]. Which gives a maximum of 40 moves.

Is it possible to follow the same logic but use the Edges Antipode instead?. So the two phases become:

1) Solve all the edges by taking them to the Antipode edges instead of the identity edges. [this is guaranteed to be 18 moves].

2) From the edges antipode position take it to the cube identity. [Does anyone know the maximum moves needed for this phase?]. To find this maximum it would take the same computational effort to prove the 22 moves I guess. Let's call this maximum X.

## A Little More on Odd and Even Permutations

Submitted by Jerry Bryan on Thu, 01/11/2007 - 12:50.If you take a cube apart and put it back together, you can put it back together in twelve times as many ways as there are positions in the cube group. Twelve is because 3 * 2 * 2 = 12. Three of the ways are because the twist of the last corner cubie you put back together can be set in one of three different ways, only one of which is legal. Two of the ways are because the flip of the last edge cubie you put back together can be set in one of two different ways, only one of which is legal. The last two of the ways are because the corners and edges can by put back together either with the same odd/even parity (legal) or with the opposite odd/even parity (illegal).

## X1+X3+X5+...

Submitted by mdlazreg on Wed, 01/10/2007 - 18:02.Finding the diameter of God's Algorithm for the 3x3x3 cube is equivalent to finding the length of the sequence:

X0

X1

X2

X3

X4

X5

:

:

Xn

where X0=1; Xn is all positions reachable from the Xn-1 positions.

I have noticed and proved that for any rubik like puzzle , the following identity is true:

X1+X3+X5+....+X2n+1=X0+X2+X4+X6+....+X2n

This can be verified for the 2x2x2 cube sequence for example:

1

6

27

120

534

2,256

8,969

33,058

114,149

360,508

930,588

1,350,852

782,536

90,280

276

Though I have been reading many books and articles about rubik's like puzzles, I never came across the above identity. Did anyone see the above identity somewhere? Thanks.

## Starts-with and Ends-With

Submitted by Jerry Bryan on Thu, 12/21/2006 - 16:12.On the old Cube-Lovers list, the terms Starts-with and Ends-with were defined as follows. For a cube position x, StartsWith(x)=S(x) is the set of all moves with which a minimal maneuver can start and EndsWith(x)=E(x) is the set of all moves with which a minimal maneuver can end.

The concept is much older than Cube-Lovers, of course. It's obvious that from any position except for Start itself, there must be at least one move which takes the Cube closer to Start. The set of all such moves is simply the set of inverses of E(x).

## God's Algorithm, Face Turn Metric, Out to 11 Moves from Start

Submitted by Jerry Bryan on Thu, 11/30/2006 - 22:23.Distance Patterns Unique Positions from up to Symmetry Start 0 1 1 1 2 18 2 9 243 3 75 3240 4 934 43239 5 12077 574908 6 159131 7618438 7 2101575 100803036 8 27762103 1332343288 9 366611212 17596479795 10 4838564147 232248063316 11 63818720716 3063288809012

## Solving the 4x4x4 in 85 twists

Submitted by Bruce Norskog on Wed, 11/29/2006 - 00:45.In my posting titled "The 4x4x4 can be solved in 79 moves (STM)," I reported about an analysis I did where the 4x4x4 cube is solved in five stages. In that analysis, a move was considered to be any quarter-turn or half-turn of a single slice.

I have now completed a similar analysis of the 4x4x4 cube where a move is considered to be any quarter- or half-turn twist of the cube, and where a twist is considered to be one portion of the cube (a face layer, or a block consisting of a face layer and the adjacent inner layer) being turned with respect to the rest of the cube. The analysis indicates that any valid position of the 4x4x4 cube can be solved via these five stages using no more than 85 twists.

## Some Thoughts on Representing the Cube

Submitted by Jerry Bryan on Tue, 11/21/2006 - 00:34.I wanted to post a number of miscellaneous items about representing the cube, and I will also include a few other related items.

I'll start with the group S_{3} as an example.
I will treat the group S_{3} as acting on the
set {0, 1, 2}. As I have been
doing recently, I'll use the notation (a b c) to represent the
permutation 0→a, 1→b, 2→c.

In this notation, the entirety of S_{3} can be listed as
follows:

(0 1 2) (0 2 1) (1 0 2) (1 2 0) (2 0 1) (2 1 0)

This basic idea, or something very similar to it, is probably the way
most people represent the cube in a computer program.
Variations on the
theme could include an S_{54} model, an S_{48} model,
an S_{24} × S_{24} model, and some sort of wreath
product model. The most common wreath product model would probably
be something like (S_{8} wr C_{3}) ×
(S_{12} wr C_{2}). In the wreath product model,
S_{8} and S_{12} represent the corner cubies and the
edge cubies, respectively, and C_{3} and C_{2}
represent the twists of the corner cubies and the flips of the edge cubies,
respectively. Of course, none of these various models are isomorphic
to the cube group. Rather, the cube group is a subgroup of
whatever group is chosen as the computer model.

## 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.