Last Layer Cross
Place the yellow edges correctly
Step 5: Yellow cross will be finished.
Our goal in this step
We can turn the roof and the whole cube freely and are looking for a situation where exactly one edge matches the F wall centre and the other three edges needs to be rotated clockwise. In the example no edge is matching.
We do a U turn, turn the whole cube clockwise and have the intended situation: The yellow/red edge matches with the red F wall, the other three edges need to move clockwise: yellow/green from UL to UB, yellow/green from UB to UR and yellow/blue from UR to UL.
The arrows in the picture show, why I’m talking about a clockwise rotation. We need a third algorithm for this:
R, U2, R', U', R, U', R’ (algorithm III)
We analyze this algorithm:
· We count 7 moves.
· Only the R wall and the roof U are involved. We start with a clockwise R and connect this with intended clockwise rotation. We alternate between U and R and the R turns alternate between clockwise and anticlockwise.
Here are the 7 moves executed on a solved Rubik’s Cube:
U2
R’
R
U’
U’
R’
R
A good way for memorizing the sequence is the following:, We concentrate on the corner in the cellar at our right (FDR) and the adjacent edge (in the example the red/green edge at FR). They always stay together.By the first two moves, this corner edge pair emigrates to the left wall (L= blue wall). The third move restores the R wall, except of the slot that our pair has left. Now the pair returns by four more moves and in a different way.
(If the pair would return by exactly the same three moves reversed, nothing is gained at all. The sequence R, U2, R', R, U2, R', would be a no-operation.)
Therefore the pair returns by a first anticlockwise turn of the roof, The following R move brings the slot for the pair to the roof again, the second anticlockwise roof turn pushes the pair back to its slot and the last R’ turn restores the R wall completely.
When we look at the result on our example above, we recognize that we have achieved our goal, the yellow cross is completed:
We have still to look at some special cases: all edges can be turned to their correct place by turning the roof. That’s easy, just turn U until you recognize that the cross is finished.
In this case (picture right), we have two matching edges at UF and UL.
We turn the roof by U and turn the whole cube clockwise
We recognize that we have one matching edge in the F wall, but the other three edges need an anticlockwise rotation.
We do a second U turn and see that no edges match at all. We do a third U turn, see one matching yellow/orange edge
and turn the whole cube that orange becomes our F wall.
We have reached the wanted situation, where exactly one edge matches the F wall centre and the other three edges needs to be rotated clockwise.
We use our algorithm
R, U2, R', U', R, U', R'U;
We see that this was a simple case, where two matching edges could be changed by simple U turns to the situation we want. Furthermore, we have seen that a situation with one matching edge and the three others needing an anticlockwise rotation can be handled by simple U turns as well.
There is another case of two matching edges that needs to be covered: Two matching edges in a line. In the example the red and orange (yellow) edges build such a line. Simple U turns do not change the situation: Either two edges match or none at all. We just do randomly our algorithm; in this case with F = red
Now, we have two matching edges, but in an L shape. We handle this as in the case above, by U turns until we get the situation we want, where exactly one edge matches the F wall centre and the other three edges needs to be rotated clockwise.
