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Very often I get situation like this:

Four flipped edges in the equator layer

I am interested in manual solving, so automatically generated solution is not useful (like with Kociemba algorithm). Currently I use twice the two edges flip (E R' E R' E R2 E' R' E' R' E' R2), but this is too long solution with total of 28 moves.

Is there a shorter algorithm to solve this situation?

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  • $\begingroup$ Hmm, what solving method do you use that you get into this situation? (When you solve layer by layer you would never get this case.) Do you use a corners-first method, or a middle-layer last method? $\endgroup$ – Kevin Cruijssen Oct 28 '16 at 22:06
  • $\begingroup$ I do not know, probably corners first seems likely. I have figured out 2x2x2 first, and now apply it on 3x3x3. $\endgroup$ – Evil Oct 28 '16 at 22:11
  • $\begingroup$ @Evil. You counted both quarter turns and slice turns as 1 move. But in slice turn metric (stm), any layer is only counted once whether a quarter turn or a half turn, and R, R2, E, and E2 are all one turn. - So, with the slice turn metric, your formula is really length 24. Jonathan's formula is of length 12, and Jaap's formula is of length 14 but easier. - Yes, this occurs with the corners first method. Have you checked out the Varasano method? It solves 8 corners in 3 steps: 1. orient first face (intuitive), 2. orient opposite face (7 cases), 3. pbl (position both layers, 5 cases). Done. $\endgroup$ – Cuc Apr 3 '17 at 5:12
  • $\begingroup$ Wow. Thank you @Cuc. I had apparently counted in QTM. I never heard about Varasano method, but honestly I have never learned any known solving system, as I tried to generalize from 2^3 to 5^3 on my own, then picked some algorithms to speed up solving, trained fingers for speed. Now it is probably good moment to learn something, so I will definitely learn Varsano method. It seems far faster with only 12 cases, because right now I use more with slow 2-3 swaps. Thank you again for pointing this out. $\endgroup$ – Evil Apr 3 '17 at 18:31
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The shortest number of moves in face turns (HTM) from here to solved would be

$16$, for example: U R F' D L2 B' R U' D B' L F2 D' R B' D'

The shortest number of moves in slice turns (STM) from here to solved would be

$12$ (matching yours), for example R U2 R2 S R' E2 R' U2 R2 S R E2

I don't do speed solving so maybe others can give some advice, but I think using some look-ahead or a different approach that avoids this may be worth looking into. Also note that a longer sequence might actually be faster to perform.

A method that is probably a good start might be the one made by Lars Petrus (also a good start for fewest moves), which may be found at his website.

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One simple 4-flip sequence that I like a lot is (E R)4. It doesn't quite flip the four edges you want (it does FL BL BR and UR). A small conjugation to to bring the FR edge to UR then gives the following sequence:

F' U' F (E R)4 F' U F.

It is not quite as short as Jonathan's, but it is easy to remember and understand.

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  • $\begingroup$ Wow, thank you this one is also very cool and intuitive. $\endgroup$ – Evil Oct 30 '16 at 19:24
  • $\begingroup$ Oh, so "the basic case" is for this, with additional moves to put top edge to the equator. $\endgroup$ – Evil Oct 30 '16 at 19:43

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