![]() ![]() So, for example, my potential square root algorithm for R U R' U' is: That is, the sign of the piece (after the 4-cycle is executed the first time) which was originally in the slot (in the solved state before the 4-cycle was executed at all) they move to on the second iteration). (sign of the slot they move to when the 4-cycle is executed for the second time. (their sign after the 4-cycle is executed the first time) The signs of each of the corner pieces 1,2,3 and 5 (after we have executed this 4-cycle to the solved state) are: Let ufr = 1, ubr = 2, ubl = 3, ufl = 4, and dfr = 5 and let. Before I can describe the puzzle, I have to give you my numbering system and other needed information so that I can construct a step by step transition from the problem to my "puzzle". I made a model of the problem, and it basically came down to a simple puzzle. ![]() So I believe you can't have a square root for any position having two misoriented 4-cycles of corners.Ĭlick to expand.I apologize for posting material that you probably already know, but just in case you were asking about this and didn't catch everything I have in this post. But since all 8 corners are used in the two 4-cycles, you have no remaining corners to fix the net twist. Similarly for two misoriented 4-cycles, you would need both cycles to be twisted the same direction. This means there must be another piece or cycle that is misoriented so that the net twist is zero (mod 3). However, that always seems to generate two misoriented 2-cycles that are both twisted in the same direction as the 4-cycle. Basically odd-length misoriented cycles don't seem to be a problem.įor misoriented 2-cycles, you need to have a 4-cycle that when squared gives two 2-cycles. It essentially only applies to the situation of having two misoriented 2-cycles. You'll need at least another twisted corner (or pair of twisted corner cycles) to balance the twists. I think to have a square root for a position with twisted corner cycles, you need two same-length cycles with the same twist direction. While my program does find as many roots as you want it to, it does not point out the ones with the briefest solutions since it does not itself find solutions.įor any programmers who are interested, the source to my root taker is part of my Rubiks solver project. (R2 U2)3 (L2 D2)3 = (U2 F L2 B' L2 D2 L2 B' U2 B D2 R2)3Įxcuse the lengths of some of those root algorithms. Sqrt(E Perm) = B L2 F2 U B' U F2 U R2 F U2 B2 U' L2 B2 D2 F2 D With this program, I have calculated a few roots as a proof of concept: Of course, the program checks parity and validates the orientations of the edges and corners. The program itself figures out the positions and orientations of the edges and corners for the root rather than attempting to represent the square root with the standard group basis (i.e. So I finally sat down and wrote a program to take nth roots of elements in the cube group. Has anybody fully implemented or developed a method for doing such a thing? Is there even a use for finding such roots of cube group elements? With this in mind, I can see myself perhaps finding the square root of a cube group element by representing it's corners and edges as matrices and diagonalizing them to find the square roots. If such a matrix, let's call it A, is the square root of this permutation matrix, then A would need a complex determinant in order for det(A)^2 = -1. Then, doing an R turn will result in a permutation matrix with a negative determinant. Consider representing just the position of the corners in an 8x8 permutation matrix. In fact, with the following reasoning, I do not believe that there exists a square root for an R turn: However, when you look at something simple like R, it seems that a square root is not quite as obvious. What I found is clearly not the only such square root for the super flip. I recently came up with a Rubik's configuration which, when applied twice to a 3x3x3 cube, will put it in the super flip (R' F' D R D B2 D F L F' L' F2 R2 U D L2 F2 ). ![]()
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