# Rubik's cube: Last edge problems (beginners method)

When solving the Rubik's cube using the beginners method, the last step which solves the last layer edges only requires 1 step of the algorithm if one face is already completed.

My problem is that when I have a complete face, I have to do the algorithm twice to solve.

What am I missing? • Not sure what tutorial you got there, because those hand-moves are kinda pointless compared to just letter-notations or small icons, but let's ignore that for now.. :S Also, there are loads of different layer-by-layer Beginner's methods for the 3x3x3 Cube, and the algorithm you show there I've personally never seen before.. For the tutorial I use the last step is solved in this order: orient top edges; place top edges; place top corner; orient top corners. You can find a video I made of that here. Aug 5 '17 at 14:59

I don't know how to read that algorithm, but from the pictures it looks like it does a 3-cycle of edges.

In the first case where one of the edges is already correct, you have a 50% chance that the given algorithm cycles them in the opposite way to what you need, so a 50% chance that you need to apply it twice instead of once.

In the second case where every edge is wrong, doing the algorithm once will solve one edge. You are then in the first case situation where you have an even chance of needing to do it twice instead of once.

• Yes, pretty sure this is a 3-cycle. Sexy, Lexy, 5 x Sexy, 5 x Lexy Aug 5 '17 at 21:47
• @ComputingCorn if you only want to do the algorithm once, then you can either Rx1, Lx1, Rx5, Lx5 or Lx1, Rx1, Lx5, Rx5 in the first case, depending on which direction you need to cycle those three edges. If you want them to go counter-clockwise, use the first one, otherwise the second one. Aug 5 '17 at 21:52
• @user38034 I think you may be on to something. Let me give it a try. Aug 6 '17 at 5:30
• @user38034 Thank you, this is the answer! I didn't realise how the algorithm worked but now I understand that it moves the edge piece counter/ clockwise. Aug 6 '17 at 5:39

There is no need to handle such a case, because if one edge is in the right place, then a rotation of the layer will reduce it to the case where two adjacent faces are correct.

If the edge pieces are all in the cross orientation, there are only two cases:

A:

  Y
N   Y
N


B:

  Y
N   N
Y


Any other situation reduces to these by rotating the face. (A) is permuted using: R U R' U R 2U R' U.

For (B), you can do "cross-to-line", rotate 180, "line-to-cross", rotate: (L U F U' F' L') 180 (L F U F' U' L') U.

That's my own algorithm; various online resources say to apply the (A) algorithm twice in the (B) case, which results in more moves.

You can eliminate the 180 flip if you can learn the "line-to-cross" from the other side. Plus, if you ever get a "line" originally, you can learn to predict when it will go to (B), and do it from the other side instead.