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Only these two edges are misplaced. Any algorithm to solve this?

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    $\begingroup$ If blue is the front face... F B L F2 L F' B D' F2 D2 L2 B2 R2 U' F2 R2. $\endgroup$ May 10, 2017 at 13:05
  • $\begingroup$ Thank you!! if you can post this as an answer I can accept :) $\endgroup$
    – Dasun
    May 10, 2017 at 13:15
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    $\begingroup$ I like to use F R' F' R' F2 L D R D' L' R' F2 R2 for your exact orientation where orange is F, Blue is R, and yellow is U $\endgroup$
    – spex
    Sep 10, 2018 at 2:25

6 Answers 6

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Why so complicated? You can use even a simpler combo. First, move the other flipped edge to the opposite side. If BLUE is Front and ORANGE is Left do:

  1. Moving the orange edge to the opposite side: L', B'
  2. Combo to remember: 4x(M' U), 4x(U M')
  3. Moving back the orange edge to where it was: B, L

Basically you can flip sides on any two edges anywhere using this method.

P.S. M' - that's middle layer turning 90º in the same direction as R

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  • $\begingroup$ Good answer for a rubik's cube, although if you're in a cube where the center piece rotation matters, like a ghost cube or a photo cube, this will also rotate two opposite centers on the equator 180 degrees. $\endgroup$
    – John Smith
    May 19 at 4:06
  • $\begingroup$ Update: I just worked this out. This will reorient two opposite edges in place without upsetting the centers: ((U M)4 L (U M)4) L' ((U M)4 L M (U M)4) L' M'. So on cubes where center rotation matters, you can move one adjacent edge to the opposite as described in this answer, and then do that. $\endgroup$
    – John Smith
    May 20 at 0:21
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For this example, an algorithm that is more simple to remember, and can be altered for the other case (opposite edges), is this: F' E F2 E2 F' U' F E2 F2 E' F and U. Replace U and U' with U2 for the case of flipping two opposite edges on top face.

(E is the equatorial slice, viewed from face U)

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    $\begingroup$ This used to confuse me, too, but Equatorial turns follow the the D layer. So if you were to put this algorithm into a cube solver as you wrote it, it wouldn't work. It should be F' E' F2 E2 F' U' F E2 F2 E F U and F' E' F2 E2 F' U2 F E2 F2 E F U2 (Slice Turns, Solver) I still gave you an up vote since you did explain your atypical notation and gave a simple, reusable algorithm that covers both adjacent and opposite edge flips. $\endgroup$
    – spex
    Sep 10, 2018 at 11:40
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You should learn how to use commutators as briefly sketched in this post about general permutation puzzles. Then you would immediately know how to solve such states.

In particular, it is trivial to find a sequence A to first flip one edge without affecting the top face and then make a single turn B of the top face to move the other edge into that same position before undoing A and undoing B. This is, of course, a commutator A B A' B'.

Using this general approach you will find numerous solutions. One solution is as follows, assuming the two edges are at the front and right. First shift the edge to the middle horizontal slice (R) and then slice it away (E) before rotating the right face (R2) to be able to receive the edge via another slice (E2) before putting it back to the top face (R). That is, A = R E R2 E2 R and B = U'.

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I did M U M’ U M U M’ U M U M’ U

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r U R’ U’ r’ U2 R U R U’ R2 U2 R

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R U R' U' M' U R U' r'. This a standard CFOP OLL Algorithm

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