# CxLL T F

## T F

This case and the case U F are inverses of each other so the same algorithm (9 move commutator or mirror) can be used to solve both.

As for all T cases you look at two "bars" of stickers, one at Fur-Ful and one at Urb-Ulb. In this case the bar at Ub has got one colour and the bar at the F-side has got two opposite colours. |

### COLL

- (y2 x') U' R2 U' L2 U R2 U' L2 U2 (x)
- (y') l' U2 R' D2 R U2 R' D2 R2 (x')
- F U' r' F R2 F' r U R2 F'
- (U) F' R' F U2 F' R F R' U2 R
- (U2) F R U R' U' R U' R' U' R U R' F'

### CLL

- L' U2 R d' R' U' R d R' U2 L ... X Y -X (Conjugate)

### CMLL

- ...

### CLLEF

- y2 R B' R' B y U2 R U2 B2 L' B' L B' R'

### CF / 2x2x2 (Waterman)

- ...

### EG

**EG 2** --> 2x2x2

**EG 1**

- ...

**EG 0**

- ...

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 |