Step six: Positioning the Last Layer-corners
In this step we will put the corners of the Last Layer (LL) in the correct positions relative to each other. The method presented here is a bit hard for me to explain in words, so for now, just try to understand what I mean. If my story is hard to follow, just study the moves yourself by doing them a couple of times. I am sure you’ll get the idea.
The method we will use to solve the corners will not affect the relative positions of the edges. This means that we can’t swap two corners. (In general, it’s impossible to swap two corners, without affecting the edges). However, we can cycle three corners, or swap URF with URB and ULF with ULB (case number 3). Take a look at the images below to see the permutations that can be solved with this method
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| 1. Three cycle anti clockwise | 2. Three cycle clockwise | 3. Two transpositions |
If the permutation on your cube looks like you have to swap URF with ULF, then do the move U. Now, you will have the first three-cycle. (Or you can choose to do the move U' to get the second three-cycle)
To solve the second three-cycle (the one that cycles three corners in clockwise direction), first look at the applet here:
35. R'D2R - U2 - R'D2R |
Before starting the applet: the corner in the ULF position (Red Green Yellow) is the solved (!) corner. We can use this corner as reference to solve the other ones. The corner in the URF position has to move to UBL. Now, we do the move [C] = R'D2R. Notice that this move is it’s own inverse. This move temporarily places one of the Last Layer corners in the D layer. That corner had to go to the UBL position, so we do the move U2, to move the UBL corner to the UFR position. Now, after doing [C] again, the first corner is correctly positioned relative to the Red-Green-Yellow corner. Run applet 35 to see this.
36. U' - R'D2R |
Next, we want to solve the other piece that is temporarily stored in the DBL position. Notice that corner has an orange sticker, just like the corner we just solved. Thus, we want the DBL corner, to end up at the left side of the corner we just solved. To do this, we do the move U', and then [C]. Run applet 36 to see this.
37 |
Now, there are three solved corners in the U layer, so there is only one possibility for the position of the last corner. Move the ‘wrong’ corner to the URF position by doing U'. Then do [C] again. This will solve the last corner, and restore the F2L! Run applet 37 below to see this. Thus, the whole algorithm for this case is:
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| R’D2R - U2 - R'D2R - U' - R'D2R - U' - R'D2R | |
| Similarly the algorithms for the other cases are: | |
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| R'D2R - U - R'D2R - U - R'D2R - U2 - R'D2R | |
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| R'D2R - U - R'D2R - U' - R'D2R - U' - R'D2R - U' - R'D2R – U - R'D2R |
These algorithms should not be memorized! If you use these algorithms, you should study the moves first: make sure you understand what the moves do!