Solved!

After RUF' RU2R'

So solve to these and do

RU2R' FU'R' and its done!

Last night I discovered a way to twist 4 corners

*! Yay!*

**without cycling edges at the same time**( ( R' F' ( R U R' U' ) F R ) (R' F2 y' ( R U2 R' U' R U' R' ) y ) ) x 2

I think of it as doing the RU corner swapper followed by an antisune, and doing it twice.

It twists 2 and 8 anti, and 4 and 6 clock. 2 is what I call UFR, 4 is DFL, 6 is DBR, and 8 is UBL.

Then later there was cause for more rejoicing. Whereas the first discovery was the result of deliberate documentation, the second one was stumbled upon while scrambling. After a few twists I realized that all three 2x1x1 posts were planted in the F2L. After solving it, further experimentation uncovered the way to get there from the place where the 2x1x1 pieces are solved—RUF' RU2R'. This state allows use of F RUR'U' F' and its buddy, for flipping edges. It also allows for the Sune family of cycling edges and twisting corners, which may make the previously mentioned discovery obsolete.

While on the topic of Sune, I'd like to mention a very nice little Word document that Burgo put online that has a way to use Sune to do a pure 3-cycle of edges. Right-handed Sune, y, Left-handed Sune, y' cycles UF > UL > UB. After undoing the setup with RU2R' FU'R' it translates to FD > UB > UL.

In the previous post I included some algorithms. I don't know exactly how many I will actually use now, but here is a more complete list:

**Solving the 2x1x1 Pieces**

Something that can help in some situations to get the 3 2x1x1 pieces is R ( U R U' R' ) or ( R U R' U' ) R'.

**3-cycle of Corners**

3>4>2 followed by 3>6>2 followed by 4>3>2 cycles 2>6>4 and returns the 3 2x1x1 pieces to their solved state.

**Corner Swappers and Twisters**

R' F' ( R U R' U' ) F R swaps 2 and 4, and 6 and 8, and cycles edges.

R' F' ( R U R' U' ) x 2 F R twists 2 and 4 clock, and 6 and 8 anti, and cycles edges.

R' F' ( R U R' U' ) x 4 F R twists 2 and 4 anti, and 6 and 8 clock, and cycles edges.

F R ( F' U' F U ) R U R2 F' swaps only 2 and 4, and cycles edges.

( ( R' F' ( R U R' U' ) F R ) (R' F2 y' ( R U2 R' U' R U' R' ) y ) ) x 2 twists 2 and 8, and 4 and 6

**.**

*without cycling edges***3-cycle of Edges**

R' F' ( R U R' U' ) F R swaps some corners and does FD > RB > UB.

R' F2 y' ( R U R' U R U2 R' ) y F2 R swaps some corners and does FD > RB > UB.

( RUF' RU2R' ), Right-handed Sune, y, Left-handed Sune, y' ( RU2R' FU'R' )

cycles FD > UB > UL

*with no corners affected.***Edge Flippers**

( R' F' ( R U R' U' ) F R ) (R' F2 y' ( R U2 R' U' R U' R' ) y F2 R )

flips FD and UB and swaps corners.

( R' F' ( R U R' U' ) F R ) (R' F2 y' ( R U R' U R U2 R' ) y F2 R )

flips FD and RB and swaps corners.

Alternately, one could just solve to the above picture using familiar algorithms, and be 6 twists from solved! :D

**Strategy:**

- Solve the 2x1x1 pieces.
- Check corners and permute if necessary.
- With white up, solve the two middle layer edges.
- RUF' RU2R'
- Solve the edges.
- Solve the corners.
- RU2R' FU'R'

**Conclusion:**

Although only a small handful of algorithms are actually needed, it was fun coming up with all of these and sorting through them to come up with a strategy.