Recent content by cuBerBruce

I have done a 2-phase analysis of the Curvy Copter puzzle (sans jumbling). The first phase brings the puzzle into the <UF, UR, UB, UL, FR, BR> subgroup. The 2nd phase solves the puzzle using the 6 generators of that subgroup. This analysis gives an upper bound of 48 moves to solve the Curvy...

This is about solving the Rubik's Cube using only J-perms. Both the Ja and Jb versions are allowed, but only the normal AUF case that swaps two adjacent corner and two adjacent edges are allowed. Of course, the cube can be rotated into any of the 24 orientations before applying each J-perm. So...

In fact, all last layer positions can be solved in 16 moves (FTM) or less, including AUF. I gave the distribution here.

As for the first question, the answer can be found at https://oeis.org/A005452 (up to 18 moves). Allowing half turns (face turn metric), then the symmetry-reduced number of positions at each depth is known up to 15 moves deep as given here. I note that this link uses the term "mod M" to indicate...

The biggest order within the standard (fixed centers) cube group is 1260. Example: R U2 D' B D'. If you allow wide turns, inner layer turns or whole cube rotations, then the maximum order becomes 2520. Of course, the maneuver "R L2 U' F' d" contains a wide turn.

In my GAP example, I have 12 permutations defined, named U, u, d, etc. They all represent bijective mappings of a set unto itself - a set of 56 elements. (The permutations are specified in cycle notation.) You would seem to claim these permutations are not valid permutations because the set...

If you read my post, my point was that there are different possible interpretations of what it would mean for a cube state to be "odd" (vs. "even"). My point is that a mathematician would make it clear precisely what he means by "odd" or "even" first, before applying it to a given cube state...

What two sets? I have 1 set of elements of all the pieces of the puzzle. See my GAP example. Inside the parentheses, you needlessly and subtly impose additional conditions on what you consider a permutation to be. As a result, your definition of permutation differs from what mathematicians...

Any two 4-cycles that are disjoint make two 4-cycles, by definition. Since the two 4-cycles here are not even in the same orbit, they are obviously disjoint. A U move on the 3x3x3 is a 4-cycle of edge pieces, and a 4-cycle of corner pieces. The total number of 4-cycles is 2. What part of that do...

Yes, you are obviously misuing the word permutation. You are trying to define the "puzzle permutation" (singular term) as multiple separate permutations (plural). Mathematicians are happy to talk about a permutation where the elements are constrained to certain orbits, but such a permutation...

I count essentially 11 types of corner 3-cycles that have 8-move commutators. There are 48 symmetry variations of each type for 11*48 = 528 positions. Of the 11 types, 7 have two separate 8-move commutators. That makes a total of (11+7)*48 = 864 8-move maneuvers, if you're counting maneuvers...

In the block turn metric, there are 6-move maneuvers, but I don't think the OP meant to allow that metric. For example (SiGN notation): U2 2-4r2 2-4f2 D2 2-4r2 2-4f2 (same algorithm to scramble or solve) Using only single-layer turns, I can't think of any positions having only corners...

I would call [a : b] a "conjugation," not a "conjugate." Unfortunately, the speedcubing community tends to mangle mathematical terminology. Of course, unlike the commutator notation, the notation [a : b] is not used in mathematics, at least not for this purpose. It is purely a "cubing"...

I think maybe you should just list the order you that the graph vertices are visited, rather than your "LFL..." notation for describing the Hamiltonian circuits. It's like you're saying you are using a 6-vertex graph, but each vertex has 4 different ways of labeling the edges, depending how you...

I've thought about trying to create a function that would take a cube state and return the index of the position within my Hamiltonian circuit, but have never gotten around to it. It would probably involve some degree of searching, and for that you would have to be using some other way of...