Link: 22LL page
What is 22LL?
22LL is a subset of last layer algorithms. Each case consists of a 2-cycle of corners and a 2-cycle of edges. So, some PLLs such as T perm are 22LL cases. Z perm is not a 22LL case because while it has 2 2-cycles, they are both edge cycles.
How many cases are in 22LL?
22LL is only 56 cases. There are 5 sets, based on the corner 2-cycle, each of which represents a separate CLL case. The A and D sets contain adjacent and diagonal corner swap PLLs, respectively. The A set contains T, J, L, F, and both R perms, and the D set contains Y, V, and both N perms. The D set contains less algorithms than the other 4 sets because there were 4 cases which were reducible by AUF, which I didn’t notice until I made the case pictures for them. Additionally, there is no F set ( for the 2-cycle (UBR LUF) ) because it is identical to the I set with a U2 AUF.
Why make 22LL?
Originally, I planned to learn how to solve each 2x2-cycle for (UBR x) (DF y), where x is an edge and y is a corner, by using setups to PLLs or ZBLLs. I soon realised that it would be much simpler if I allowed edges to be unoriented.
You can read more about it on my blog.
If you have any better algs, or see any errors, post them here or send me a message.
Example Solves:
What is 22LL?
22LL is a subset of last layer algorithms. Each case consists of a 2-cycle of corners and a 2-cycle of edges. So, some PLLs such as T perm are 22LL cases. Z perm is not a 22LL case because while it has 2 2-cycles, they are both edge cycles.
How many cases are in 22LL?
22LL is only 56 cases. There are 5 sets, based on the corner 2-cycle, each of which represents a separate CLL case. The A and D sets contain adjacent and diagonal corner swap PLLs, respectively. The A set contains T, J, L, F, and both R perms, and the D set contains Y, V, and both N perms. The D set contains less algorithms than the other 4 sets because there were 4 cases which were reducible by AUF, which I didn’t notice until I made the case pictures for them. Additionally, there is no F set ( for the 2-cycle (UBR LUF) ) because it is identical to the I set with a U2 AUF.
Why make 22LL?
Originally, I planned to learn how to solve each 2x2-cycle for (UBR x) (DF y), where x is an edge and y is a corner, by using setups to PLLs or ZBLLs. I soon realised that it would be much simpler if I allowed edges to be unoriented.
You can read more about it on my blog.
If you have any better algs, or see any errors, post them here or send me a message.
Example Solves:
Speedsolve:
BLD:
U' B' R' D B R' D2 B2 L' F U' F2 U B2 D F2 U2 L2 U L2 D' F2
z2 y // inspection
U2 L R U R F B' D // cross
U' R' U' R // F2L #1
U' R U' R' U' L' U' L // F2L #2
R U' R' U y' R' U' R // F2L #3
y' R U2 R' // F2L #4
U F U R' U' R D' R2 U R' U' R2 D F' U' // 22LL
z2 y // inspection
U2 L R U R F B' D // cross
U' R' U' R // F2L #1
U' R U' R' U' L' U' L // F2L #2
R U' R' U y' R' U' R // F2L #3
y' R U2 R' // F2L #4
U F U R' U' R D' R2 U R' U' R2 D F' U' // 22LL
F' R' D' L D' F' R' L B D F2 U R2 U2 L2 U B2 F2 D2 F'
y // memo
// corners
D L' U2 L D' L' U2 L // UBR->LDB->FDL
R' U2 R' D' R U2 R' D R2 // UBR->URF->ULB
y' R2 D2 R U2 R' D2 R U2 R y // UBR->RBD->RFU
U2 L D' L' U2 L D L' // UBR->ULF->RDF
// edges
U2 M' U L U' M U L' U // DF->BU->LB
M U2 M U M' U2 M' U' // DF->UF->UR
R2 u M' U L U' M U L' U' u' R2 // DF->RD->FL
z L' U M' U' L U M U' z' // DF->DL->LU
(M2) R2 U R U R' U' R2 F' U F R' F' U' F R2 U' R2 (M2) // 22LL parity
z2 M' U M' U M' U2 M U M U M U2 // flip edges
y // memo
// corners
D L' U2 L D' L' U2 L // UBR->LDB->FDL
R' U2 R' D' R U2 R' D R2 // UBR->URF->ULB
y' R2 D2 R U2 R' D2 R U2 R y // UBR->RBD->RFU
U2 L D' L' U2 L D L' // UBR->ULF->RDF
// edges
U2 M' U L U' M U L' U // DF->BU->LB
M U2 M U M' U2 M' U' // DF->UF->UR
R2 u M' U L U' M U L' U' u' R2 // DF->RD->FL
z L' U M' U' L U M U' z' // DF->DL->LU
(M2) R2 U R U R' U' R2 F' U F R' F' U' F R2 U' R2 (M2) // 22LL parity
z2 M' U M' U M' U2 M U M U M U2 // flip edges
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