# Last 2 pieces in Rubik

I followed a guide in how to solve the Rubik Cube however from time ti time I incurr in a case that is not covered by guide and I obtain an almost solved Cube except for 2 pieces. Seems also the online solver of that website is bugged for this particolar status, so how do I get out of this stuck state?

• Note that this is a trivial commutator once you understand the general method for solving permutation puzzles. In particular, this is a 2-cycle commutator as described in that post, where you have a sequence A of moves that twists one corner but preserves everything else on the top face, and then you perform a single turn B of the top face to bring the other corner to that same position before you undo A and then undo B. Commented May 10, 2021 at 15:11

The case you are showing can be solved by doing a pure algorithm to solve corner orientation. Luckily the algorithm for solving this case quickly is very easy. You simply place the unsolved corners on the left and do sune U2 bsune (U2). I personally use this in both my speed solves and in my blind solves (though conjugated most of the time).

In longhand, this algorithm is R U R' U R U2 R' U2 R' U' R U' R' U2 R U2

If you are unfamiliar here is a short description of sune and bsune.

sune: The name of the algorithm R U R' U R U2 R' in which you take out one of the slots and then put it back in another way.

backsune: Also known as bsune, this is doing a sune from the back. The algorithm for this is R' U' R U' R' U2 R

I have no cube at hand, but one of the online solvers provides the solution.
https://rubiks-cube-solver.com/solution.php?cube=0111111214223222222131333333344444444555555555666666666

Added 07 May 2018 per request by Rand al'Thor.

The procedure devised by the site under the link above is as follows:
R' U F2 U' F2 U' R' U R2 D' F2 D F2

Please note I made a mistake in the problem definition: I defined the left-hand upper corner rotated clockwise and the right-hand upper corner rotated counter-clockwise, opposite to what you show on photographs. Luckily you can just rotate the whole cube so that the yellow and green faces swap their positions, and the solution would fit exactly your problem

...or just apply the solution twice ;-)

• Thanks you. I never tried this solver in particular.. it worked <3 Commented May 6, 2018 at 17:56
• @GameDeveloper I never used it, either. :) Just googled it out among others to find some, which allows to set up an arbitrary starting position to be solved. If I understand it correctly, the starting setup is encoded in that long string of digits. Commented May 6, 2018 at 18:15
• Link-only answers are very much discouraged on Stack Exchange, since if the link goes dead the answer becomes essentially useless. Please could you edit to include a description of the solution in the text of the post itself? :-) Commented May 6, 2018 at 20:06
• @Randal'Thor Done. Commented May 7, 2018 at 7:45

I'm not sure what guide you've followed, because there are loads of different beginner methods. Personally I always orient the top two, three or four corners in the same way, using multiples of R' D' R D (2x), since it's very easy to remember (although a bit longer to execute and counterproductive if you plan to transition into speedcubing in the future).

For your particular case it would mean the following algorithm (starting with yellow at the top, and red at the front):

R' D' R D (4x)
U
R' D' R D (2x)
U'


Or completely written out:

R' D' R D R' D' R D R' D' R D R' D' R D U R' D' R D R' D' R D U'


Here is a video I created a while back with the Beginner's Method I personally learned. Starting at the linked 30:40 it covers orienting the top layer corners. I showed it with three corners, four corners, or two corners unsolved. Yours is the same as the one with two corners that I show at 33:37, although your corners are next to each other instead of kitty-corner of each other.

• I've been cubing for about one week now. I've achieved the "fully solved" position 5 times. On 2 other occasions (including the last), my final "unsolved" configuration is as shown in the OP. All 7 final algos were: R' D' R D, applied in one or more multiples of 2. Question: If I arrive at the "unsolved" cofiguration, how many further repetitions of R' D' R D do I need to do in order to arrive back at the same "unsolved" config? (Of course, I might have solved it before that - but it's a useful number to know.) Thanks Commented Aug 24, 2023 at 11:24
• @OldGrantonian If you repeat R' D' R D (2x) three times on a solved cube, the top-right corner will have rotated three times 120 degrees, and it would then be back at its initial state. The U and U' are just there to put another corner in that top-right position, so after repeating it three times you can have one corner rotated once 120 degrees and another corner twice (e.g. 240 degrees). So unless I misunderstand your question, the answer would be six (in pairs of 2), aka 3x R' D' R D R' D' R D. Commented Aug 24, 2023 at 12:10
• Six pairs sounds about right. When I reach the config where only 2 adjacent corners reamain to be solved, I want to photograph the config, then start counting the pairs. I will then know if I have an "unsolvable" config. For example, it's possible that my first "unsolvable" config might actually have been solvable if I had continued for a few more interations. Thanks Commented Aug 24, 2023 at 16:19