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?
  • $\begingroup$ 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, 2014 at 21:29
  • 3
    $\begingroup$ Excellent question! This will be my exemplar of an expert question about puzzles. $\endgroup$
    – xnor
    Dec 5, 2014 at 3:12

1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – user20
    Jun 14, 2015 at 18:59

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