mightypuzzle wrote:

No, this is not a regular parity case

Do you solve using Fridrich F2L? If so then this is a standard OLL algorithm both unoriented corners at the front (R2, D, R', U2, R, D', R', U2) and then a single edge OLL parity, and then the application of a PLL parity for both corners.

If you solve using a different method, I'd be happy to see what else I can do.

You're welcome if I helped, I'm sorry if I didn't.

No, I use beginners method

lol

I did solve it however, I just thought there might be a special (simple) algorithm for this particular case.

Reason why I didn't know how to solve this parity is that the only parity alg. that is in the booklet that came with the cube is r2 B2 U2 l2 U2 r' U2 r U2 F2 r F2 l' B2 r2 and there are picture above it that show 2 adjacent dedges swapped, 2 opposite corners swapped, 2 adjacent corners swapped, and 1 flipped dedge.

There IS one other alg. in the booklet that flips 2 opposite dedges: r2 U2 r2 Uw2 r2 u2

I use that for the adjacent dedge swap too with 2 setup moves.

Thanks for helping!

EMI94100 wrote:

1NSAN3 wrote:

Is this a regular parity case?

How do I solve this state?

Thanks!

Yes and using an OLL parity alg. (as explained in every 4x4 tutorial, btw)

Rw2 B2 Rw' U2 Rw' U2 x' U2 Rw' U2 Rw U2 Rw' U2 Rw2 U2

I tried this algorithm on the solved cube, and it gives me the exact same case! Thanks!!!