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Discussions on the mathematics of the cube
The 4x4x4 can be solved in 77 single-slice turnsSubmitted by Bruce Norskog on Thu, 07/26/2007 - 00:17.Previously I announced that the 4x4x4 cube could be solved in 79 single-slice turns
by solving it in five stages,
in a manner similar to the Thistlethwaite 4-stage solution for the 3x3x3.
(See
The 4x4x4 can be solved in 79 moves (STM).)
However, I have now realized my solution to the 2nd stage could have allowed
the use of more basic turns than I used.
I have realized that: » 6 comments | read more
Rubik's cube can be solved in 34 quarter turnsSubmitted by silviu on Mon, 07/02/2007 - 10:24.I have proven that Rubik's cube can be solved in 34 quarter turns. The details can be found at:
http://www.risc.uni-linz.ac.at/publications/download/risc_3122/uppernew3.ps UFR / UF Coset SpaceSubmitted by B MacKenzie on Sat, 06/30/2007 - 16:52.Following up on an earlier thread I have explored the UFR/UF coset space. The number of UF cosets in the UFR group is given by: The first three factors are the number of ways the corner cubie position and the two edge cubie positions which are not on the UF faces may be configured. The factor of 6 is for the corner position permutation. Of the 720 corner position permutations on the UF faces achievable using the UFR face turns only 120 are achievable using the UF face turns. Thus the corner permutation may be one of six cosets. The flip of the seven UF edge cubies is constrained under the UF face turns, one cannot perform a double edge flip in this group. This gives rise to a factor of two to the sixth for the edge flip (flip parity determines the flip of the seventh edge cubie). As Bruce Norskrog pointed out in the earlier thread, the above number is the same as the order of the UFR group divided by the order of the UF group, so things check out. » 4 comments | read more
Antisymmetry, Corners of the 3x3x3 Cube, quarter turn metricSubmitted by Jerry Bryan on Thu, 06/21/2007 - 12:18.Distance Positions Positions Positions from reduced reduced Start by by Symmetry Symmetry and Anti-Symmetry 0 1 1 1 1 12 1 1 2 114 5 5 3 924 24 17 4 6539 149 96 5 39528 850 469 6 199926 4257 2289 7 806136 16937 8768 8 2761740 57848 29603 9 8656152 180787 91688 10 22334112 466220 235710 11 32420448 676786 342593 12 18780864 392342 199610 13 2166720 45600 23818 14 6624 163 110 Total 88179840 1841970 934778 As I have written before, my programs have seldom worked with positions. They have nearly always worked with representative elements of M-conjugate classes. In the table above, the summary of representative elements is labeled "Positions Reduced by Symmetry". The goal of this approach is to obtain a 48 times speedup in processing time, and also to obtain a 48 times reduction in storage requirements. » 19 comments | read more
Representation of edge permutations and move tableSubmitted by Herbert Kociemba on Mon, 06/18/2007 - 15:02.Stimulated by the last thread, where the representations of permutations and orientations was dealt I want to ask, what would be a good representation for the 12! edge permutations on the coordinate level, if the right multiplication of the edge permutation by any of the generators U,L.... should also be done with a MoveTable on the coordinate level like newcoordinate = MoveTable[oldcoodinate][generator].
This MoveTable would have 12!*18 4 Byte entries when we take the coordinate from 0..12!-1 and of course is far too big. Of course we could reduce this by 48 symmetries, but then we still would have a very large table. » 15 comments | read more
Two Face GroupSubmitted by B MacKenzie on Wed, 06/06/2007 - 09:47.Has the Rubik cube subgroup generated by the turns of two orthogonal faces been exhaustively expanded? My computer runs out of physical memory and bogs down after 18 q turns: Shell Classes Elements 0 1 1 1 1 4 2 3 10 3 6 24 4 15 58 5 35 140 6 85 338 7 204 816 8 493 1970 9 1189 4756 10 2863 11448 11 6862 27448 12 16324 65260 13 38550 154192 14 90192 360692 15 206898 827540 16 462893 1851345 17 992268 3968840 18 1973209 7891990 Totals 3792091 15166872 » 18 comments | read more
26f now claimed proven sufficientSubmitted by Bruce Norskog on Thu, 05/31/2007 - 20:52.It is now claimed proven that 26 face turns is sufficient to solve Rubik's Cube. I heard about this on the Yahoo speedsolving group forum. There is an article about it at the following link, which also contains a link for downloading the paper. That paper was written by two people at Northeastern University.
http://www.physorg.com/news99843195.html More on Branching FactorsSubmitted by Jerry Bryan on Mon, 05/21/2007 - 23:32.I think it's useful to define branching factors for some other situations than are normally considered. For the most part, I'll speak to the quarter turn metric, but generalizations to the face turn metric are not hard to come by.
I was pretty sure that I posted an article to this site about Starts-With and Ends-With, but if so I can't find it. In any case, for a position x we define Starts-With(x) to be the set of moves with which a minimal process for x can start, and Ends-With(x) to be the set of moves with which a minimal process for x can end. If Ends-With(x)=Q (the set of quarter turns), then x is a local maximum. A similar formulation of the same idea is that if |Ends-With(x)|=12, then x is a local maximum. » 5 comments | read more
Number of maneuvers for Rubik's CubeSubmitted by Herbert Kociemba on Sat, 05/19/2007 - 10:35.By accident I just ran across a formula I developed many years ago for the number of maneuvers in FTM which cannot be shortened in a trivial way. I did not see it anywhere else, so maybe it is of some interest.
Let r = Sqrt(6) and k the maneuver length, then we have N(k) = [(3+r)(6+3r)^n + (3-r)(6-3r)^n]/4 which gives 1, 18, 243, 3240, 43254, ... Round[(3+r)(6+3r)^n] is a good approximation even for small n. and we see that 6+3r = 13.348... is the asymtotic branching factor. Disjoint Cycles and Twist/Flip Parity RulesSubmitted by B MacKenzie on Wed, 05/16/2007 - 00:13.I've been fooling around writing a virtual Rubik's cube program, initially simply as an exercise for teaching myself the Open GL 3D rendering API. In representing the puzzle I was led to assign each cublet an X,Y,Z coordinate specifying its starting position in the cube.
(-1,-1,-1) being the coordinate for the (left,down,back) cublet through to (1,1,1) being the coordinate for the (right,up,front) cublet. The transformed position and orientation of each cublet is then specified as an element of the O symmetry point group, there being a one to one correspondence between the 24 elements of the O point group and the 24 states a cublet may assume via Rubik cube face turns: 12 edge positions with two flip states each or 8 corner positions with 3 twist states each. » 5 comments | read more
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