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.

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).

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 

 

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.

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.

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.

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