The Gilles Roux method
The right block 
The right block (from now on: RB) is also 1x2x3 sized block, parallel to
the left one. Thus it means the following pieces: BR, DR, FR edges and DBR, DFR
corners. How fortunate that the center is in it’s place alreadyJ! The two blocks are shown on the hereunder picture.
It’s solution is a bit tougher than the LB, since we don’t have that
much room to perform moves. (Actually this is the slowest part of the whole
method.) R or r (or M) and U will be rotated most times, on rare occasions F
and B. Practically speaking, D will never be used, only in reasonable
cases.
However, if you are apprehensive of disassembling the LB, you can limit
yourself to R, r and M. The RB can be solved this way, too, although in a bit
more steps. But it has it’s advantages:
- It leaves the LB alone, so you “won’t lose scent”.
- Your left hand remains in the same state all along -> faster
movements. Therefore, fingertricks can be used effectively.
- Less thinking.
The solution is similar to the one of the LB: at first, solve a proper
corner-edge pair, then insert the corresponding edge to the right side (1x2x2).
In the end, solve the remaining corner-edge pair, and put it to the 1x2x2
block.
A simple solution of RB follows.
Using M turns, only edges are flipped, we’ll use this fact, e.g. how
could the Green-Yellow-Red corner and the Green-Yellow edge be put together?
F’U2F is a good idea, but beware, LB is temporally disassembled. Another
solution can be that you move the edge up from under. To perform it, you first
have to move it down: U’ (so it is located on M layer), then M’ moves it to the
D layer. You move the corner to the right place with U2 (yeah, U2 rulez! J), bring the edge back with M (or even r). A third possible variation:
After U’, you don’t put the edge aside, but the corner: R’U2R. Although it is
the same as our first idea (F’U2F), but here U were turned to “not disturb” the
LB.
But what happens when the corner and the edge are standing quite
chaotically? It is practical to do the following: Get the corner be on U and
the edge on D. It’s not too difficult, since using U/R/r/M you have quite wide
freedom to move one single edge and a corner. When it’s done, use R and U to
move and turn the corner, so that it matches the edge when moved up with M/M’.
A corner can be turned at places FRU and BRU, using R and R’. U simply moves
it.
When you have assembled a corner-edge pair, it is probably located on U
layer. Turn it if necessary, so none of it’s parts are above the LB. Then, turn
it using r, so that the part that belongs to R side, will be on B or F. (see
picture, here, yellow goes to R, so after an r turn the yellow part will be
located on B).
After that an U/U’ puts the corner-edge pair in their place, but before
doing so, find the corresponding edge. Try to move it behind the corner-edge
pair, in a way that the color going to the right side is on F or B. (see
picture).
As for now a U’RU or UR’U’ finishes the 1x2x2! (The later is the
mirrored case of the above picture).
If the edge is not there, it needs to be moved there.
Move the corner-edge pair to the R side (before that, move the edge far away,
if possible, so they don’t interfere), and turn R to get the edge on U, and no
parts of the corner-edge pair on U. Then the edge can be moved to the right
place with U/U’, and the corner with R.
For example, taken the following case (the edge should be flipped):
First disjoin the corner-edge pair and the edge: URU’. It is nearly OK,
only the edge needs to be flipped.
After r'U2R:
And right away it’s behind the corner-edge pair. URU’, the 1x2x2 is
finished.
Of course there are many algorithms to solve the 1x2x2. Let’s see the
following picture.
It could be solved the following way: A Red-Yellow edge will be put to
it’s place by disassembling the small 1x1x2: R’UMU’RU’. (Not recommended, it’s
hard to find out while playing). It could also be r'U'rURU2, which is the same
as the strategy discussed afore. r'U'rURU2 is similar to this, but it takes
care of the edge remained alone. You can do the following, too: move all parts
to the right side and then assemble: U'rUR'U'R2U. Or another chance: U'Rr2U'MU.
Essentially you pass the corner part of the corner-edge pair to the other edge
(this one is not an easy case though). To choose the shortest way: BUB'U2,
which sets the edge rapidly, while disassembling the LB (temporally). Choose
the one that you can recognize first.
Now, only the second corner and edge are to be assembled, and attached
to the 1x2x2. Chances are a bit
restricted: Try to keep the 1x2x2 under (no parts touch U). So you have U, M
and (restrictedly) R and r turns (R ’n’ r, get it?). Strategy: Bring the corner
to U with R, and move the edge down with M (Similarly to the first corner-edge
pair). After that, use R and U until the corner can be put together with the
edge.
E.g. the following case:
After Ur’, the the corner is up, the edge is down.
If the green side of the edge was up, then after UM’, if the yellow,
then after M2 the second corner-edge pair was together. In this case however,
the corner must be flipped: UR.
If the block was “underfoot”, the turn U2 to shelve the corner, R/R’ for
the block, and bring the corner back with U2. (But for now, it isn’t
underfoot).
Now only a UM’ and the second corner-edge pair is finished.
Of course there are other ways to solve the second corner-edge pair. An
example is shown in the following picture.
Instead of niggling with corner turning after UM, a R’U2R is well enough
and it’s finished.
Assembling is not difficult from now on. Move the second corner-edge
pair to U so that it’s color belonging to R is on F or B, and the 1x2x2 doesn’t
touch it. Then U/U’ and R/R’ and you are finished.
For example the following figure:
With an r turn the yellow side of the corner-edge pair gets to B.
However, it moves the 1x2x2 up, making it to touch the U (it is “underfoot”).
Shelve it! Using U’ (or U2) shelve the corner-edge pair, then use R’ to shelve
the 1x2x2. U (or U2) brings the corner-edge pair back.
Now the r is free. Only a UR’ and the RB is finished. Instead of rUR’ a
FR’F’R is also suitable, although it disassembles the LB temporally.
It was just a short guide. If you practice much and solve the blocks
slowly, you can discover many simplifying turns – even disassembling the LB.
Getting into the way of doing them the first two blocks can be solved in a few
seconds! That needs a lot of practice though…
More information is in the Right block building techniques section.
Remark (color neutrality): The RB can be turned 180° related to the LB
(by color, see the pictures). This is the color neutrality for RB.
The first one shows where the RB goes, the first one show it during the
solution. Of course in the end it needs to be corrugated (CMLL, 6E4C), and in
the very end a R2 is needed. Sometimes it is faster. Use this technique only if
you really know how many beans make five, it can really mess up the solution…
See also Learning advices and Full solution examples sections.
Right block building techniques
CMLL fast recognition technique
Fridrich method vs. Roux method