CxLL L B
From Speedsolving.com Wiki
L B
The four cases, T R, T L, L R and L B are mirrors and inverses of each other so the same algorithm (8 move commutator) can be used to solve them all.
|
The two stickers in the U-face has got opposite colours, find the other two. The case L R looks the same but mirrored. |
COLL
- (y' x') U L' U' R U L U' l'
CLL
- ...
CMLL
- ...
CLLEF
- y F R' F' R2 r' U R U' R' U' M'
CF / 2x2x2 (Waterman)
- ...
EG
EG 2 --> 2x2x2
EG 1
- ...
EG 0
- (y) R' U2 F U' F' U2 F' R
| 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 |
