I'm not even sure if it's possible to do this or not.

I have a hypothetical PLL-situation where I need to swap one opposite-corner pair of a single layer.

Like, a V-perm but without moving the edges:

enter image description here

I don't think this is possible, but hope it is.

Can someone either explain why it isn't possible or show the algorithm that would do this?

  • $\begingroup$ Does this solve your question? $\endgroup$ Commented Dec 10, 2017 at 17:47
  • $\begingroup$ Or possibly this? $\endgroup$ Commented Dec 10, 2017 at 17:48
  • $\begingroup$ @Randal'Thor neither work $\endgroup$
    – minseong
    Commented Dec 10, 2017 at 17:54

2 Answers 2


Apparently it's impossible because the Rubik's cube can never be one swap from being solved.

It's an even-parity puzzle, the number of swaps it is from being solved must always be even.

I found this helpful to understand why: https://www.reddit.com/r/Cubers/comments/3yg7q0/parity_claritypart_ii/

Every Face-turn on a 3x3 does an Odd-perm of Edges, and it also does an Odd-perm of Corners. If we combine those, then Odd + Odd = Even overall. So with every turn the parity of Edges change, and the parity of corners change, but the parity of the entire cube can never change no matter how many turns we make on it. A 3x3 cube is considered to be in Even parity when it is solved. (Zero swaps left.) So if we started Even, no amount of turns will ever change that. Even + Even = Even. It will always be solvable.


Looking through the "speedsolving the cube" by Dan Harris, I couldn't find anything in the rubik's revenge (4x4x4) section, or the 3x3x3 section either. It must be impossible.


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