CxLL L B
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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.
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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 |