CxLL U F
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U F
This case and the case T F are inverses of each other so the same algorithm (9 move commutator or mirror) can be used to solve both.
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As for all U cases you look at two "bars" of stickers, one at Fur-Ful and one at Urb-Ulb. In this case both bars has got opposite colours (chess board pattern). |
COLL
- (y x') U2 L2 U R2 U' L2 U R2 U (x)
- (y2) l' R' D2 R U2 R' D2 R U2 R (x')
- (y2 x) R2 D2 R U2 R' D2 R U2 R (x') ... same in diffrent notation
- (U) F R2 U' r' F R2 F' r U F'
CLL
- ...
CMLL
- ...
CF / 2x2x2 (Waterman)
- ...
EG
EG 2 --> 2x2x2
EG 1
- ...
EG 0
- ...
| 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 |
