5
$\begingroup$

I have solved the bottom white layer and the second layer. I have almost solved the top layer also, but I have not able to solve the corner pieces.

Normally I can solve this with the below sequence:
R U2 R' U' R U' R' L' U2 L U L' U L

But I have to this twice.

Can anyone please help me to solve this with more efficient way?

enter image description here enter image description here

$\endgroup$
4
$\begingroup$

Personally I use the beginner's method when I solve my 3x3x3 Cube. In which case you can use a combination of R' D' R D (2x) sequences with setup moves to solve this.

So for your case above, put Green at the front and Yellow at the top, and use the following algorithm:

R' D' R D (2x)
U2
R' D' R D (4x)
U2

This sequence of R' D' R D (2x) you will always have to do three or six times in order to orient all the corners.

And here is a video I made about a year ago explaining the entire Beginner's Method, including orienting the corners of the last layer.


EDIT: Hmm, noticed you already know how to solve the corners, but are looking for a more efficient algorithm.. My bad.
Will get back on that in a moment.

Perhaps holding it the same as above (Green front, Yellow top), and then use the following algorithm:

L U L2 D L2 U' L F2 U' F2 U L2 D'

I don't think it's particularly easy to remember. But it's shorter and therefore more efficient..

Personally I would just stick with that you already know, by using your algorithm (R U2 R' U' R U' R' L' U2 L U L' U L) twice. But maybe that's just me..

$\endgroup$
  • 1
    $\begingroup$ I tend to use (F D' F' R' D R) U2 (R' D' R F D F') U2. This is reasonably efficient in the number of moves, but still easy to remember and understand. It is not quick to perform though as you need to regrip a lot. It is quicker to do something similar to the beginner's method you listed, namely (F' R F R')2 U2 (R F' R' F)2 U2 which can be fingertricked quite well. $\endgroup$ – Jaap Scherphuis Jun 10 '17 at 21:25
4
$\begingroup$

I'd like to explain a somewhat intuitive approach of turning these corners.

Start with blue facing front and yellow on top, as in the pictures. Let's first attempt to just turn the yellow-blue-red corner in it's place without worrying about any of the other pieces. This can be done by F R. We notice that the corner turns into it's correct spot but obviously many other pieces are disturbed. How do we isolate the turning of just the corner we want?

The simplest way to undo something that was done to the cube is to retrace the same moves backwards. However, if we want to make some changes along the way, we can choose to not undo one of the moves, which can return many pieces back but not all. So to negate the unwanted effect of the first F turn, we want to incorporate a F' turn in our algorithm, but in a way that the corner we are manipulating doesn't go back to its starting spot.

Let's start with F D2 F'. Now the front side is almost as it started and we've "hidden" our corner "behind" the cube. To get the corner back we now have a few options. A short way is just to continue D2 R, undoing the hiding move and lifting the corner into its spot. The cube is still messed up, but compared to just F R, some things are more to our liking, specifically the two pieces next to the corner that have yellow and blue on them. Maybe there is a better way to fetch the corner from its hiding spot. If we consider how it got there and imagine doing the same moves as a mirror image on R instead of F, we see that it can be fetched by R' D2 R as well. This is even better as it anticipates the effect of the R and in a sense undoes it ahead of time.

Where are we now? We've found that using the moves F D2 F' R' D2 R we can turn the corner correctly in its place and not disturb quite as many pieces as just simply doing F R. Unfortunately, many pieces are still all over the place. We do however notice, that the top layer is looking better and it's only the first two layers that are in need of repair. Once again the simplest way to fix these two layers is to undo every move we've done so far: R' D2 R F D2 F', but of course this also restores the corner which we worked so hard to twist. So before we undo any of those moves, let's move that corner out of the way again. Similarly to how we previously did D2 and hid the corner, we now use U2 to swap the corner that is being turned. This works because the rest of the top layer is unaffected by the planned moves. Finally, when we fix the first two layers, we still need to turn U2 to undo the swap and the cube should be solved.

Putting it all together, the sequence ( F D2 F' R' D2 R ) U2 ( R' D2 R F D2 F' ) U2 does what we want. This algorithm has a lot of symmetry to it. We essentially take the corner down from the top layer, hide it away, undo the first move; then we fetch this back symmetrically to reach the half-way point of the algorithm by swapping the corners on the top layer, finishing off with doing all this backwards.

If you have three corners to twist, you'll have to perform this twice, but be careful not to turn all the corners in the same direction, but rather in pairs of two in opposite directions.

The idea is easily modified to twist the corners in the other direction:
( R' D2 R F D2 F' ) U2 ( F D2 F' R' D2 R ) U2
Or to just turn two adjacent corners instead of opposite ones:
( F D2 F' R' D2 R ) U ( R' D2 R F D2 F' ) U'
Or turning the corner a bit less intuitively but with fewer half-turns:
( R' D R F D F') U2 ( F D' F' R' D' R ) U2
Or to turn edges instead of corners (though this is not very efficient):
( M D M' D' M D2 M') U2 ( M D2 M' D M D' M) U2

Do comment if you come up with other cool variations :-)

$\endgroup$
  • $\begingroup$ Like an essay but well explained $\endgroup$ – VortexYT Jul 27 '17 at 13:15
2
$\begingroup$

By the beginners method You can use this algorithm R U R' U R U U R' By this algorithm two to three times by keeping green as your face you can solve your corner piece. enter image description here

$\endgroup$
2
$\begingroup$

This sequence twists the front two corners on the top face:

(R'DRFDF') U' (FD'F'R'D'R) U

We can write this

A U' A' U (14 turns)

where A is (R'DRFDF').

To swap the front right corner and the back left corner, both on the top face, use

A U2 A' U2 (also 14 turns).

Written out in full, this is

(R'DRFDF') U2 (FD'F'R'D'R) U2

$\endgroup$
2
$\begingroup$

You can use an OLL (CFOP Fridrich) algorithm to complete the yellow side. For this case: Position the cube as seen in the pic with yellow as the top face and carry out the algorithm.

enter image description here

You can then use either the beginner's method or any other layer-based solving method to arrange the top layer into the correct positions (PLL in CFOP).

image source: https://ruwix.com/the-rubiks-cube/advanced-cfop-fridrich/orient-the-last-layer-oll/

$\endgroup$
  • $\begingroup$ Welcome to the site! Is the image yours, or taken from somewhere else? If the latter is true, then I suggest adding attribution. $\endgroup$ – EKons Jul 9 '18 at 18:57
2
$\begingroup$

The most efficient and easy to remember is a meson discovered by Erno Rubik himself:

(U2 F L' D2 L F')2

Michael Keller Puzzle Laboratory

$\endgroup$
1
$\begingroup$

You can solve this with a slightly modified version of your algorithm. With green in front and yellow on top do:

B' R U2 R' U' R U' R' L' U2 L U L' U L B

You'll notice that all I did was add a B' and a B. This works because the algorithm only moves two pieces, allowing you to do set up moves to change which two pieces are changed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.