Solving the 2x2 with 8355 - The '35' Method

[Cride5]

At a first glance, the idea of solving the 2x2 using the 8355 concept is as simple as just solving 3/4 of a layer intuitively, and then repeating the sexy-move to solve the remaining 5 pieces - easy! If you've ever tried solving a 2x2 like this, you'll probably have noticed that it sometimes just doesn't work. When returning the buffer layer to solved, the cube sometimes ends up with just two corners swapped. No matter how you place the last corner, it seems impossible to do it with the buffer layer returning to the solved state.

The reason for this is that the method uses a series of corner swaps to solve pieces. The problem is that in order to return the 'buffer' layer to its original state 6x repetitions of the sexy move are required (or if the inverse is used, the number of repetitions mod-6 must be 0). Unlike the 3x3, the 2x2 can be in an odd parity permutation. If solving with a series of swaps, an odd parity permutation must use an odd number of swaps.

So how can an odd parity permutation be solved with '35'? It can be done using a 'pure' 8355 approach, simply by re-solving the last layer with all pieces rotated around by 1 quarter turn. Simply insert the final cubie from the buffer position, so that it replaces one of the ll cubies adjacent to the final unsolved position. Then solve all the remaining ll cubies relative to that one. This will require an odd number of swaps and will eventually allow the buffer layer to return to the solved state with all pieces in place.


On my search for this being spoken about before all I could find was this in Chinese:
http://www.youtube.com/watch?v=rFNirQ1kEDs

Unfortunately my chinese is not that good, so the majority of it flew way over my head ... apart from maybe the bit at 3:23

 

[cubacca1972]

Not sure that I understand what you are asking, but in the event that you can't decipher what to do in the case of the U layer being solved but the D layer being mixed up, just keep doing R'D'RD until only UFR and DFR need to be solved.

Next, regrip your cube so that the 2 unsolved cubes are in the U layer at UFR and UBR. Solve one of them with the R'D'RD move, then solve the other the same way. That should solve it. I don't know why this works.

 

[Cride5]

OK, suppose you have this scramble: R' F2 R' F R' F U' R' U2

... and you solve all-but-two pieces by doing this:

3/4 of D-layer: F R2
Flip cube: x2 y
3/4 of opposite layer: D (RUR'U')*3 D (URU'R') D2 (URU'R')*2
​

That leaves exactly two unsolved corners. The trick you described above only works for two pieces which are twisted, but in place. In this case they are twisted, but also swapped, so that won't work. Try it yourself.

The solution is to re-solve the final layer, placing the pieces a quarter turn round from where they originally were, like so:

D' (RUR'U') D' (URU'R') D' (RUR'U') D' (URU'R') D2​