CxLL L L
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L L
The four cases, U R, U L, L L and L F are mirrors and inverses of each other so the same algorithm (9 move commutator) can be used to solve them all.
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First look in the U-face, there are two stickers of the same colour, next find the two stickers of the opposite colour. Case L F looks the same but mirrored. |
COLL
- (y) R U2 R D R' U2 R D' R2
CLL
- ...
CMLL
- R U2 L' U' L U2 R' U R' F2 R
CLLEF
- y' R U R' U R' F R F' U2 R' F R F'
CF / 2x2x2 (Waterman)
- ...
EG
EG 2 --> 2x2x2
EG 1
- ...
EG 0
- y R'UR'F'U'FUF2R2
| CxLL edit |
 U |
 D |
 R |
 L |
 F |
 B |
 U |
 U U |
 U D |
 U R |
 U L |
 U F |
 U B |
 T |
 T U |
 T D |
 T R |
 T L |
 T F |
 T B |
 L |
 L U |
 L D |
 L R |
L L |
 L F |
 L B |
 S |
 S U |
 S D |
 S R |
 S L |
 S F |
 S B |
 -S |
 -S U |
 -S D |
 -S R |
 -S L |
 -S F |
 -S B |
 Pi |
 Pi U |
 Pi D |
 Pi R |
 Pi L |
 Pi F |
 Pi B |
 H |
 H U |
 H D |
 H R |
 H L |
 H F |
 H B |
| Hyper CLL edit |
U |
D |
R |
L |
F |
B |
 U |
 U U |
 U D |
 U R |
 U L |
 U F |
 U B |
 T |
 T U |
 T D |
 T R |
 T L |
 T F |
 T B |
 L |
 L U |
 L D |
 L R |
 L L |
 L F |
 L B |
 S |
 S U |
 S D |
 S R |
 S L |
 S F |
 S B |
 -S |
 -S U |
 -S D |
 -S R |
 -S L |
 -S F |
 -S B |
 Pi |
 Pi U |
 Pi D |
 Pi R |
 Pi L |
 Pi F |
 Pi B |
 H |
 H U |
 H D |
 H R |
 H L |
 H F |
 H B |
