Sometimes you're left with a case that leaves two un-oriented corners. If you keep repeating the alg until the last D-layer corner is solved, then the U-layer will appear scrambled. The reason it happens is because the number of repetitions of the (R U R' U') alg you used to solve the pieces isn't exactly divisible by 6.
I didn't know this cycle was a multiple of 6. Interesting.
[*] When you get to the last corner don't watch the corner itself, but instead watch the rest of the cube. Repeat the R U R' U' alg until all but 2 unoriented corners are solved.
When I first read this several weeks ago I didn't understand what you meant at all. But it's been on my mind ever since and a few days ago something made sense and I did some experiments. Now I get it!
In terms of explaining this to someone who's new to 8355, it's practically impossible. I think my beginner's explanation is easier. But for someone experienced with the last 5 corners of 8355 this is a very good technique.
I'm tempted to write a long tutorial about it with examples and sample solves. For now I'm just going to throw out some key points:
1. As Cride says, the corner manipulation algorithms run in a cycle of 6. Any time you get to 6, or a multiple, or 0, the cube has the possibility of being solved, or only having two corners twisted out of place. I'm going to call this the "zero state" of the cube for lack of a better term.
2. The "normal" corner alg is R' D' R D. But, if you've done 8355 a bit, you'll know that D' R' D R also works, but runs the corner cycle in reverse. I.E., the FRU and FRD corners swap and twist in the opposite direction from R' D' R D. This can save you a ton of algs by twisting a corner clockwise (D' R' D R) instead of counter clockwise (R' D' R D), and of course makes your solves faster.
2A. So when you do R' D' R D, add 1 to the count. When you do D' R' D R, subtract 1 from the count. The count can be negative and negative multiples of 6 are also valid "zero states" for the cube.
2B. This same logic applies to R U R' U' and U R U' R' if you are doing the corner cycling with the 4 corners in the D layer instead of U. This is how Cride likes to do it and how the Wiki shows (since he wrote it!).
3. You may come to a multiple of 6 or zero and have FRD correct, but twisted, and one of the 4 U layer corners also correct but twisted. A beginner would do AUF to bring the twisted corner to FRU and then do a corner twisting alg until it was correct. Instead, do that same AUF, but then rotate the cube such that FRU and FRD become FLU and FRU (a z' rotation I think). Then do the 2 corner fix up routine. Finally undo your AUF turn and the cube is solved.
4. As in 3 above you may find other combinations where two corners (not FRU and FRD) are correctly permuted (in the right place), but not correctly oriented (twisted). A setup move or two will allow you to bring them to FRU and FRD and then the cube can be solved with the corner fixup routine. Of course you need to reverse the setup move(s) at the end.
Points 3 and 4 above were a revelation to me. But I seem to be the slow cuber around here so maybe others (like Cride) saw this themselves and it's no big deal. I thought it was interesting enough to write about. I hope this made some sense.
Brian.
