Discussions on the mathematics of the cube

Lower bound using the Edges antipode

One way of solving the cube is using two phases:

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

One of the basic tenents of cube theory (a highly specialized branch of group theory - grin!) is that odd permutations of the corners can only occur with odd permutations of the edges, and that even permutations of the corners can only occur with even permutations of the edges. This fact is one of the three factors that causes the order of the cube group to be smaller than the order of what is sometimes the illegal cube group.

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+...

Hi,

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

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

```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

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

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 S3 as an example.  I will treat the group S3 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 S3 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 S54 model, an S48 model, an S24 × S24 model, and some sort of wreath product model.  The most common wreath product model would probably be something like (S8 wr C3) × (S12 wr C2).  In the wreath product model, S8 and S12 represent the corner cubies and the edge cubies, respectively, and C3 and C2 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!

Ok, this method is called Simul Block, or Shotgun. It's a very powerful bld method, using 8 new algs, and putting some old algs to new uses, or using uncommon variations of a common algorithm.

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

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