Difference between revisions of "ELL"
m (→Intuitive:) |
m (→Intuitive:) |
||
Line 6: | Line 6: | ||
If you like to find algs for ELL in the MU-group, then knowing these short "proto algs" will help a lot: | 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 | ||
* O' = M' U' M ... Orient mirror | * O' = M' U' M ... Orient mirror | ||
+ | * P = M' U2 M ... Permute | ||
If you combine these and also inserts U-turns you can find optimal or near optimal algs for most ELL's | If you combine these and also inserts U-turns you can find optimal or near optimal algs for most ELL's |
Revision as of 10:01, 28 July 2010
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 mirror
- P = M' U2 M ... Permute
If you combine these and also inserts U-turns you can find optimal or near optimal algs for most ELL's
Example, 4-flip: P U2 O U2 P U'
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?
|