7
$\begingroup$

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.

enter image description here

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 U and D centers.

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?
$\endgroup$
  • $\begingroup$ Is there any chance somebody without the device at hand could help you? i.e. are there any rules which can be abstracted to pure text and used to solve it? $\endgroup$ – BmyGuest Dec 4 '14 at 21:29
  • 3
    $\begingroup$ Excellent question! This will be my exemplar of an expert question about puzzles. $\endgroup$ – xnor Dec 5 '14 at 3:12
4
$\begingroup$

That's funny, I just finished finding my own set of algorithms.

My base algorithm is effective in select situations.

I found a 2-2 swap from [A]=RU'RURURU'R'U'R2' swapping right-up edge and back center in additon to the left center and left-front edge. The key comes in the setup move [B]=R2F2x. (reminder: x is rotating the entire cube as if you were doing an R turn). This setup puts the originally swapped pairs into a new configuration to be swapped back. However, in swapping back, the original left-front edge will be rotated 90 degrees counter-clockwise. Thus the full algorithm I use is [A][B][A][B]'.

This is efficient if I just have one gear that needs to go in the right direction, but I found some others for other cases.

  1. Mess with more than one gear: Let [C]=FU'[A]UF'. This does the 1-1 swap on only edges. Add on isolating and rotating move [D]=UFnBnU' with Fn and Bn being n moves in that direction. Now we have algorithm [E]=[C][D][C][D]' which will rotate the left-bottom and left-back edges 90*n degrees.

  2. Mess with more than two gears: Let [F]=[R2U2]x3. Now use U[F]U'[F] to rotate all top four gears 180 degrees. I'm assuming you also know that [F] will also flip all gears not on the S slice 180 degrees.

  3. Finally, operate one gear with precision. Namely, we wish to rotate just one gear on a 90, 180, or 270. We do this with the following long monster. F[E]F' will rotate JUST the left-back gear 90*n degrees counter-clockwise.

The main advantage that I see with my approach over yours is that your base sequence manipulates pieces that are opposite each other. This adds a lot of symmetry, which when attempting to isolate a single edge to rotate, isn't exactly what helps. My initial technique was an accident I stumbled upon, [1] was going to be my intended solution as it is flexible and the [D] setup isolates the gear to be rotated on the Back face and then proceeds to rotate it (but it had a side effect), and [3] was finally settled upon as my preferred rotator as it exchanged one of the rotated gears of [1] for a center, so it goes on unnoticed.

$\endgroup$
  • $\begingroup$ Sorry for taking so long to get back to you! I only recently remembered I have this thing. And... your solution works! Thanks so much! $\endgroup$ – Aza Jun 14 '15 at 18:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.