Solve 5 last corners in 4x4x4

Question

OK, so I'm trying to solve this 4x4x4 cube with least possible memorized algorithms. So first I follow the basic conversión to a 3x3x3 configuration, and then I used some Pitrus method, so I build a 2x2 block que then expanded to a 2x2x3. From here, I dont know if this will be a dead end, but I want to get to a commutator and conjugates situation (Ryan Heiss).

So What I do, is create a second 2x2x3 connected to the first one, and now my bottom layer is completely solved except one full corner column. I use the edge in this column as a socket to solve the top layer's edges. This ends up with the top layer's 4 corners unsolved and the unsolved column's bottom corner unsolved as well. So just 5 corners left to complete the puzzle.

Is this situation documented for a 4x4x4? Can I run into parity issues that will not happen in 3x3x3 cube? Are there known algorithms for this situation? Will I be able to use commutators and conjugates un this situation? I would love some examples.

Maybe little variation in the method I'm using will make things easier?

Thanks for the ideas!

Answer 1

Is this situation documented for a 4x4x4?

If it is, it'd be documented for 3x3x3 as well.

Can I run into parity issues that will not happen in 3x3x3 cube?

Yes, since if you've converted to a 3x3x3 and don't make two-layer moves, you'll still be in the same parity. Unless of course I'm misunderstanding what you're saying.

Are there known algorithms for this situation?

Yes; just apply any corner permutation algorithm; e.g. L' U R U' L U R' U' will permute the BLU, BRU, and FLU corners.

If you meant "for some parity situation", see here.

Will I be able to use commutators and conjugates un this situation?

This question seems pretty vague. Plenty of steps when solving end up being commutators and conjugates; the last-layer corner orientation step, in some beginner algorithms, for instance, is a commutator.