# CxLL L B

## 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

CxLLedit |
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 CLLedit |
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 |