I've recently acquired a Geared Mixup puzzle, and have spent several hours trying to solve this particular problem.
I've pretty much solved the puzzle. All centers, edges, and corners are in the right positions. I just have one small problem: I have three individual edges rotated on the Geared Mixup puzzle. Two are shown in the picture below, and a (hidden) one is opposite the green-white edge.
I'd like to find a move-efficient way to rotate the edges back on this puzzle, but I'm struggling. I have two major thoughts:
- It's definitely possible to rotate a single edge individually. I know two centers can be rotated, and I know that edges can be swapped with centers, so I'd just swap and edge with a center and rotate them both.
- Since it's possible to rotate two centers, it should, in theory, be possible to rotate two edges at once: the front right and back left (which is also rotated). I'm not sure how to do this, though it seems like it might (?) be more move-efficient.
I'm looking to construct something move-efficient, too.
I'm starting with a base sequence
[A] = (R4 U' R4 U)3. Sequence
[A] does a few things. First, it swaps the
FL edge with the
BR edge. Second, it swaps the
F center with the
B center. Third, it rotates the
If we construct a reversible sequence
[B] that sets up the puzzle such that a center is opposite the edge that needs to be rotated, both are in center-slot positions, and we can easily reverse the edge swap and center swap, then this problem is solved.
The problem is that last bit: I can construct a sequence that does this, but I can't easily reverse the side-effect center and edge swap. If the edges are opposing edge pairs, they can be easily reversed by
[A]. (I'm not worried about centers, because pairs of centers can be swapped easily.) For the life of me, I can't get it that way.
Here's an example. Let's say that we construct a
[B] = B' D R (s.t.
[B'] = R' D' B). We've moved the green-yellow edge to the opposite of the white center in positions where they'll both be rotated. However, as a side effect of the way this was constructed, executing
[B A B'] will result in a swap of the orange enter and blue-red edge, and the yellow center and an orange-red edge. This is a swap that's not easily reversible.
I'm looking for a sequence
[C] that will put reversible swaps in those positions, such that I can execute
[B C A C' B'] (or, in other words, use
[B C] as a setup sequence) without having to worry later about struggling to swap those pieces back. (To swap them back is pretty simple: just execute
[A] on the right edge pair - but they need to be effecacious for this change.)
My two questions are:
- What is a sequence
[C]that completes this algorithm?; or:
- Is there an easy way to do this that I've totally missed?