ELL
Edges of Last Layer, or normally ELL, is a method that solves the last layer edges of the 3x3x3 in one algorithm as the last step in the solution, normally after performing CLL to solve the last layer corners.
ELL is also used for a method of solving the last layer edges on larger cubes. Since there are 8 of each type of wing, though, ELL for larger cubes requires more than one algorithm.
Intuitive:
If you like to find algs for ELL in the MU-group, then knowing these short "proto algs" will help a lot:
- O = M' U M ... Orient
- O' = M' U' M ... Orient inverse
- P = M' U2 M ... Permute
If you combine these and also insert U-turns you can find optimal or near optimal algs for most ELL's
Example, 4-flip: P U2 O U2 P U' ... P U2 sets up for O that flips four edges (O' works the same in this case, choose the fastets), U2 P is restoration of the setup and last U is AUF.
Deducing the number of cases
For orientation, there are 4 possibilities.
1. OLL #28 -- this OELL can be situated in 4 ways
2. OLL #57 -- this OELL can be situated in 2 ways
3. OLL #20 -- this OELL is symmetrical
4. All edges oriented correctly.
For permutation, there are also 4 possibilities.
1. Ua permutation -- this EPLL can be situated in 4 ways
2. Ub permutation -- this EPLL can be situated in 4 ways
3. Z permutation -- this EPLL can be situated in 2 ways
4. H permutation -- this EPLL can be situated in 1 way
Because OLL #20 always is symmetrical, for this orientation, there are 5 possible ELL cases. These are the two U-perms (one case for each), the Z-perm (1 case), the H-perm (which is 1 case anyways), and the solved EPLL.
OLL #28 has two orientations: The incorrectly oriented edges on UL and UR, and on UF and UB. This means that there are 8 ELL cases possible. These are the two U-perms (two cases for each), the Z-perm (two cases), the H-perm (one case), and the solved EPLL.
OLL #57 has four orientations. This means that there are 12 possible ELL cases.
We have 5 possibilities if all edges are correctly oriented, one of which is the solved position.
The cases for this follow the same principle of the cases for OLL #20 since solved OELL is symmetrical as well.
In total we have 30 ELL cases.
The ELLs wil be denoted by the cycles, beginning with the UF sticker. When there are 2 cycles, there will be started with UL. If that is not possible, they will start with UB. When an edge is correctly permuted, but incorrectly oriented, it will be denoted by the notation of that edge, between parenthesis.
ELL 1: UF-BU-UF UL-RU-UL: (R2' U2 Lw) D2 (x) U2' (R2 F) (R2 (U2'(x')) D2 (x) R' U2 R2 U
ELL 2: (UF, UB, UL, UR): U R2 U2' R' F2 U2' R2' F R Lw F2 D2 R F2 R2'
ELL 3: UF-LU-UR-FU (UB): F U R U' R' Bw' R2 (z') R' U' R' F R F' U R U2
ELL 4: UF-RU-UL-FU (UB): U R' U' R' F R F' y' R' U2 R2 U2' R2' U' R2 U' R' U y R U
ELL 5: UF-RU-UF UL-BU-UL: R2' U2 Lw D x U2' R2' U2 F2 R2 U2' R2' F R' U2' R2
ELL 6: (UF, UB): R F R U R' F' R' F U' y' R2' U2 R U' R' U' R
ELL 7: UF-BU-UF UL-UR-UL: R' U2 R U' R' U' R' F R2 U R' U' R' F' R2
ELL 8: UF-RU-FU UL-BU-FU: R2' U R' U' R F R2 U R U' F' U R2
ELL 9: UF-LU-RU-FU (UB): U' F R2 U' R' F U R2 U' R' F' U R U F' U
ELL 10: UF-RU-LU-FU (UB): R' F2 U F R' U' F2 U F R U' F' U' R
ELL 11: UF-RU-BU-UF: R U R' U' M' U R U' r'
ELL 12: UF-BU-UR-UF: L R' U' R U M U' R' U
To be continued.
See Also
External Links
- Erik Akkersdijk's ELL algs
- Kirjava's ELL algs
- Speedsolving.com: New ELL Algorithms!
- Speedsolving.com: Why Not Teach CLL/ELL?
ELL Algorithms
Note that all of these algorithms are written in the Western notation, where a lowercase letter means a double-layer turn and rotations are denoted by x, y, and z.
Click on an algorithm (not the camera icon) to watch an animation of it.
The naming convention here is; wich edges are fliped if they where permuted and which EPLL would it be if all was correctly oriented.
Pure flips
2-flip (Adjacent)
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2-flip (Opposite)
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4-flip
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Smile =) |
EPLL's
Please do not add algs to the EPLL section, the ones here are just examples, all EPLL's are at the PLL-page, put your contributions there too, else we will end up having to separate lists and we don't want that.
U-PLL a
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U-PLL b
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Z-PLL
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H-PLL
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B edge solved (3-cycles)
RF-flip U-PLL a
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LF-flip U-PLL b
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The following four cases are mirror + inverses of the first so you only need '1 alg' for all. |
Mirror to the side and inverse in diagonal. | ||||||||||||||||||||||||
RL-flip U-PLL a
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RL-flip U-PLL b
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LF-flip U-PLL a
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RF-flip U-PLL b
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B edge in position but fliped
The first six cases in this group are the worst of ELL, and that counts for both recognition and solving :-/
4-flip U-PLL a
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4-flip U-PLL b
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The following four cases are mirror + inverses of the first so you only need '1 alg' for all. |
Mirror to the side and inverse in diagonal. | ||||||||||||||||||||||||||||||||
BF-flip U-PLL a
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BF-flip U-PLL b
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BL-flip U-PLL a
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BR-flip U-PLL b
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BR-flip U-PLL a
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BL-flip U-PLL b
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No edges in position
RF-flip Z-PLL
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RB-flip Z-PLL
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RL-flip Z-PLL
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FB-flip Z-PLL
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4-flip Z-PLL
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4-flip H-PLL
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LB-flip H-PLL
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FB-flip H-PLL
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