I have been started to learn how to solve rubiks cube with Philip Marshall's ultimate solution. With this algorithm, I can easily solve cage and 6/8 corners, but however two corners comes up in an awkward position. Is there a special move for this ?


  • $\begingroup$ It's been too long since I last touched a Rubik's cube to write a decent answer, but this page has solution for all possible corner configurations. I believe you've got D1 there which is solved by applying the "Sune" sequence twice. There's also this but I'm not familiar with the notation there. $\endgroup$ – Martin Ender Feb 29 '16 at 13:03

What I always do in this situation is keep the blue face in this case on the left side. Notice that one of the corners needs to be rotated counter clock wise and the other clock wise.

Make it so the one that needs to be rotated clock wise is in the top left back position then do this

U R -U -R U R -U -R

then rotate the left face (just that one, DONT rotate the entire cube!) so the other is in the top back left corner and to the opposite:

R U -R -U R U -R -U

Finnally just rotate the left face

Additional information:

U R -U -R is a basic move that is used in a lot of other formulas. It's interesting to see what actually happens when doing this move:

  • top-back-left and top-back-right are switched
  • top-front-right and bottom-front-right are switched
  • middle-front-right, top-middle-right and top-back-middle are switched

With some interesting other properties:

  • Doing it twice puts the corner pieces involved back on their original position but rotated
  • Doing it three times puts the edge-pieces involved back in their original position AND rotation.
  • Doing it 6 times puts the cube in it's original position.

Why it works here

In the left-face only top-back-left is touched with this formula. Because we know that this move only rotates the top-back-left piece when done twice and we then rotate the left face and reverse the move twice we have essential rotated to different corners in the left face while leaving everything else intact.

In the case you actually have 3 corners that are rotated, it's actually always the case that all need to be rotated clockwise or all need to be rotated counter clockwise. You could of course perform this formula twice to solve it but there is a faster way. In the case of 3 corners that need to be rotated clockwise: Do the same as before by putting the corner you want rotated in the top-left-back and do U R -U -R U R -U -R. then rotate the left face and just repeat it two times for the other two corners. This works because we know that after 6 time U R -U -R the cube is in it's starting position while the top-left-back is rotated after every 2 moves.


You can use a special move for this, but it is not required. The situation you describe can be solved within the framework of Marshall's solution. Marshall has 2 corner piece series, one rotating three corners clockwise, the other one rotating three corners counter-clockwise.

Just apply any one of them so that it affects both of the misoriented corners. This will take them out (and another corner) out of position. You now have exactly 3 corners out of position, and you can always put them into correct position and orientation with a single corner piece series. This is what Marshall describes as 'The End Game'. You need to first find a corner that can slide into proper position along an edge, then figure out where the second one will go and how it will be oriented, apply face turns to bring the third corner into that space where the second will go, perform the corner piece series, then undo the face turns.

Yes, this sounds much more complicated than it really is to perform.

  • $\begingroup$ perhaps a diagram would be helpful? $\endgroup$ – Jason V Oct 24 '17 at 16:47

The method I use (because it's so easy to remember and hard to mess up) is to swap the piece with the diagonally opposing corner by using the left side, then swap it back using the front side. (six moves in total thus far, counting the bottom half turn as one move)

Start by aligning the cube so that the corner piece to be twisted clockwise is in the top-left-front corner, and the other unsolved corner piece is in the top layer, like you have in the picture. (blue is top, white is front.) Then do L-D2-L' and F'-D2-F.

This twists the corner piece, and leaves the top layer otherwise intact. You can then turn the top layer so that the other unsolved corner is at that same place (top-left-front), and do the sequence in reverse (F'-D2-F and L-D2-L') to twist the other corner and restore the lower layers.

After this, just turn the top layer back, and you are done.

PS. The corners can only be twisted in CW-CCW pairs, so hopefully your cube has another unsolved corner that isn't visible in the picture. If it doesn't, it's time to break up the cube and reassemble it so that it becomes solvable.

PPS. no, even another unsolved corner won't be enough, the other misaligned piece seems to be in the wrong place altogether. Maybe you just created the visible part of the picture without actually solving the rest of the cube?


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