Edge Movements Part II

Last U/D edges

Since EM1 affects FD, there will be one edge in U and one in D that cannot be settled by EM1. Solve them simultaneously as follows:

This assumes that at least one of the two required pieces is in the M slice. It does not work if FD and UF are occupying each others' location. In that case, we consider it as an edge exchange (FD,FU) and settle them later.

C edges

At this stage, all U and D edges (except perhaps the exchange required at the end of last section) have been settled. Physically rotate the cube sideways to view the remaining four M edges as C edges.

Now that both the the R and L faces (ie the original U and D faces) have been (largely) settled, we shift to a new paradigm. Regard these two faces as two metal plates with the C slice in between. Only this C slice have edges to be solved. They can always be solved with either of the following algorithms with suitable conjugates of Cⁿ.

Use two different ways to look at EM3:

1. In the subview of the R and L faces, C º I. Therefore EM3 = U²ÅU² = I.
2. In the C slices subview, U² exchanges UF and UB. C^-1 moves another pair of C edges to the U face for U² to exchange. Since the two exchanges are made on overlapping edges, the net change is the movement of three edges.

Formally, we can also say that U² is a mono-move on the F face. The commutator EM3 is thus analogous to CM3 and EM3R.

The (FD, FU) exchange required in the last section can be done by conjugates of EM4. After the physical rotation (ie making the four M edges as C edges), the FD, FU will lie on the R and L faces. Choose a suitable physical rotation so that those two edges stay on the top, that is, at UR and UL. If the C face has more than two edges incorrectly located, treat them with EM3 and EM4 above, ignoring UR and UL.