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Discussions on the mathematics of the cube
the search for a 21f, an idea for some candidatesSubmitted by WarrenSmith on Sat, 05/20/2006 - 18:59.There is a unique Rubik cube position maximally far from Start *PROVIDED*
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 » 6 comments | read more
A plan to settle the maximin distance problem so we can all go homeSubmitted by WarrenSmith on Fri, 05/19/2006 - 20:50.I outline an approach which may be able to determine the maximin depth of the
Rubik cube R - may be able to prove the answer is 20 face turns - with
a feasible amount of computer time.
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 = » 8 comments | read more
In search of: 21f*s and 20f*s; a four month odyssey.Submitted by rokicki on Sun, 05/07/2006 - 12:21.In January of this year I set out to find a 21f* position---or, at the
very least, extend the set of known 20f* positions. At that time I knew
of only three 20f* positions, despite having exhaustively solved several
collections of pretty patterns and performed months of optimal cube
solutions.
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. » 10 comments | read more
Thistlethwaite's 52-move algorithmSubmitted by jaap on Wed, 04/19/2006 - 01:31.I have put Thistlethwaite's 52-move algorithm on my site. This might be of historic interest to those of you who haven't seen a complete copy of it before.
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.God's algorithm for 4x4x4 Squares Set
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. » 17 comments | read more
Rubik can be solved in 27fSubmitted by silviu on Sat, 04/01/2006 - 16:39.In this paper we give a proof that Rubiks cube can be solved in 27f.
The idea is to eliminate the 476 cosets at distance 12 in the group H=< U,D,L2,F2,R2,B2 >. In this way we never have to consider in the 2 phase algorithm that a coset is at distance 12. So we only solve cosets at distance 11. Together with my earlier result of 28 this gives a proof of 27. The same idea was used by Bruce Norskog in his 38q proof. However we do not really need to compute all 476 cosets. In fact we only need to compute 7 cosets of the group T = Intersection ( < U,D,L2,F2,R2,B2 > , < F,B,L2,U2,R2,D2 > , < L,R,F2,B2,U2,D2 > ) The group H is not invariant under all symmetries. But the group T is invariant under all 48. » 10 comments | read more
Two more classes with exacly 4 symmetries done - most 20f* are antisymmetricSubmitted by Herbert Kociemba on Thu, 03/30/2006 - 12:09.I finished the analysis of two more classes with 4 symmetries now. The computation took more than two weeks. All can be solved within 20 moves.
The definition of the classes D2(edge) and C2v(b) are explained on this page. Here you also can get some more information about these and other classes. What is interesting, that from the 12 20f*-cubes of the class D2 (edge), 10 also have antisymmetry and from the 94 20f*-cubes of the class C2v(b) 92 also are antisymmetric. » 8 comments | read more
Results of two more cosets of the H group, this time face turn metric.Submitted by rokicki on Fri, 03/24/2006 - 23:35.After seeing Silviu have such success with H group (U, D, F2, B2, R2, L2)
cosets, I decided to give it a shot in the face turn metric. So far I've completed the identity coset and the flip8 (upper and lower edges) cosets; the superflip coset and flip4 (middle edges) are still running. The identity, flip4, flip8, and superflip are the four centers of the H group. I've also shown that of the approximately 234,101,145,600 positions represented by these four cosets, none have a depth greater than 21. This exploration covers more than 5/1,000,000,000 of the total cube space. The identity coset has the following depths. For comparison on the right » 20 comments | read more
Rubik can be solved in 35qSubmitted by silviu on Wed, 03/22/2006 - 10:19.Let H be the group < U,D,L2,F2,B2,R2 > and let N be the subgroup of H that contains
all even elements in H.
I have run an exhaustive search on the coset space G/N and got the following table:
0q 1 1q 9 2q 68 3q 624 4q 5544 5q 49992 6q 451898 7q 4034156 8q 35109780 9q 278265460 10q 1516294722 11q 2364757036 12q 235188806 13q 28144The group N contains no elements of odd length and the maximum length is 24. » 3 comments | read more
New optimal solutions for an important groupSubmitted by silviu on Wed, 03/15/2006 - 03:32.I have computed optimal solutions for every element in the group <U,D,L^2,R^2,F^2,B^2>.
For this task I made some minor modifications to Reid's solver. I would like to thank him for sharing it.
0q 1 1q 4 2q 10 3q 36 4q 123 5q 368 6q 1,320 7q 4,800 8q 15,495 9q 54,016 10q 194,334 11q 656,752 12q 2,222,295 13q 7,814,000 14q 26,402,962 15q 89,183,776 » 13 comments | read more
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