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

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  • 1
    $\begingroup$ If blue is the front face... F B L F2 L F' B D' F2 D2 L2 B2 R2 U' F2 R2. $\endgroup$ – Ian MacDonald May 10 '17 at 13:05
  • $\begingroup$ Thank you!! if you can post this as an answer I can accept :) $\endgroup$ – Dasun May 10 '17 at 13:15
  • $\begingroup$ 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 '18 at 2:25

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 '18 at 11:40

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), U, 3x(M'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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