Note: since asking this question I've purchased several other twisty puzzles and now regret taking the "look up algorithms online" approach to solving - in the end I get much more enjoyment from the puzzles where I worked out to solve them myself than the ones where I didn't. If you're stuck at this point I would recommend seeking resources to understand the concept of parity and then try to come up with a solution yourself. You won't end up with a super efficient algorithm, but you will understand it, which is ultimately more satisfying. Or it is for me at least. That said, answers to this question are still welcome.

I'm solving my 4x4 Rubik's cube (aka Rubik's revenge) for the first time, and I have the "OLL parity" case:

enter image description here

There are plenty of algorithms for this available online, but different people use different notation and nobody ever says which one they're using, so I risk messing up my cube if I guess wrongly.

So, please give me algorithms for solving the OLL parity, using the following notation:

  • R: turn just the rightmost face clockwise
  • Rw (R wide): turn the rightmost half of the cube clockwise, i.e. the rightmost face and the slice adjacent to it
  • r: turn the rightmost slice clockwise, but not the rightmost face
  • R2, Rw2, r2: as above but turning it twice
  • R', Rw', r': as above but turning it anticlockwise
  • L, F, B, U, D: left, front, back, up, down. (All clockwise by default, can be modified as above)
  • x: if you use this, please explain what it means, as I have no clue.
  • - (hyphen): if you use this, please explain what it means, as I have no clue.
  • (, ) (parentheses): if you use these, please explain what they mean, as I have no clue.

If you use anything else at all in your notation, please explain what it means. If you prefer a different notation from the above that's fine, but please explain it.

I'm holding my cube with the unsolved cubies facing me at the top. (i.e. the red face would be the front in the diagram above.) If I should start in some other orientation, please say so.

It would be helpful to have an algorithm that leaves all other pieces alone, as my cube is in exactly the almost-solved state above. But if algorithms that don't do this are significantly shorter and/or easier to remember then that would be fine too. (Please tell me which I can expect in your answer!)

  • 3
    $\begingroup$ Has a useful answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$
    – Rubio
    Apr 28, 2018 at 13:32

4 Answers 4


Here is the algorithm I use (from http://www.rubiksplace.com/cubes/4x4/), but using your notation:

Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'

So it's basically a bunch of half cube turns and top face 180 degree turns.

The "x" means you rotate the cube clockwise along the x-axis. In other words, if the Red face was facing you, it will move to the top while the sides stay the same color.

  • $\begingroup$ It doesn't matter how carefully I perform this move, whether directly from this answer or elsewhere, it always completely messes up the cube. (Or to be precise, it does remove the parity, but it scrambles it as a 3x3 so I have to solve it again using 3x3 moves.) I must be making a mistake, but I've no idea what. Sorry, I thought asking for more precise notation would help. $\endgroup$
    – N. Virgo
    May 13, 2018 at 9:12
  • $\begingroup$ I just tried it again to make sure it works. Maybe watching someone do it will help: youtu.be/DX5aGS5k2o8 $\endgroup$ May 14, 2018 at 5:27

I use the following that leaves all the other cubes alone:

r U2 r U2 r' U2 r U2 l' U2 r U2 r' U2 lr' r' U2 r'

If you look at it this way, the net effect is just an r' :

r U2 r
U2 r' U2 r U2 l'
U2 r U2 r' U2 l
r' U2 r'

  • $\begingroup$ "If you look at it this way, the net effect is just an r':" That is not correct. On the other hand, the following algorithm for double parity IS just the result of r'. (Lw' Uw2 Lw' Rw' Uw2 Lw2 U2 Lw' U Rw' U2 F' R2 U' F U) r' (U' F' U R2 F U2 Rw U' Lw U2 Lw2 Uw2 Rw Lw Uw2 Lw). (Where Rw = (Rr), Uw2 = (Uu)2, etc. $\endgroup$ May 16, 2019 at 20:21

I can never remember any clean parity fix for this case, but there is a very easy to remember dirty one, which leaves only about 6 edge pieces out of place. Of course this is no good for speed solving, but it is so simple that I'll never forget it.

It is: R2 u R2 u R2 u R2 u R2 u R2

What this does is move the u layer, while the R2 moves will swap out the centre pieces from that layer and put them back in a different place. The net effect is that the u layer has been given one quarter turn, but the centre pieces have been restored. This is an odd permutation on the affected edge pieces.

If you start with the flipped pair at the FR location, you then only have to perform a few 3-cycles to fix the last 6 edges.


Not technically an answer, but here is the meaning of the symbols:

x, y, z :

These are notation for rotation of the entire cube. they mean to rotate following the R, U, and F moves, respectively.


I don't know what the hyphen means. Can you tell me where you saw the hyphen?

( )

These don't change what to do, they just tell you what moves are triggers, or sets of moves that repeat through many algorithms.You usually see them around sets of moves like (R U R' U') (R' F R F') and (R U2 R')


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