# 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)

• ...

### CLLEF

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

• ...

EG 2 --> 2x2x2

EG 1

• ...

EG 0

• ...

## Navigator

 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