22LL - a last layer subset

Discussion in 'General Speedcubing Discussion' started by aronpm, Jan 5, 2012.

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  1. aronpm

    aronpm Member

    2,010
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    Sep 9, 2009
    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:
    Speedsolve:
    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
    BLD:
    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
     
    Last edited: Jan 5, 2012
  2. AJ Blair

    AJ Blair Member

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    0
    Feb 7, 2010
    Lakewood, Colorado
    WCA:
    2009BLAI01
    YouTube:
    Platymoo
    IAB - y' U' R' U' R U R' F' R U R' U' R' F R2

    I like these...I use a few already, but I could always use a few more tricks for last layer!
     
  3. aronpm

    aronpm Member

    2,010
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    Sep 9, 2009
    Thanks, that's a nice cancellation/cyclic shift. I've added it.
     
  4. PandaCuber

    PandaCuber Member

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    Jun 24, 2011
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    YouTube:
    brusinque
    Can I see an example solve? Im a visual person...
     
  5. irontwig

    irontwig Member

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    Apr 6, 2009
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    WCA:
    2010JERN01
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    Tbh I'm kinda surprised that someone hasn't done this earlier, especially considering that a lot of BLD prås seem to be pretty insane.
     
  6. AJ Blair

    AJ Blair Member

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    Being the noob I am...I really have no clue what that means...it's just an Old Pochman alg with an R cancellation...if that's what you said...then oops...
     
  7. irontwig

    irontwig Member

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    Apr 6, 2009
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    WCA:
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    Cyclic shifts is e.g: ABC => BCA => CAB.
     
  8. AJ Blair

    AJ Blair Member

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    Feb 7, 2010
    Lakewood, Colorado
    WCA:
    2009BLAI01
    YouTube:
    Platymoo
    Gotcha!



    ABI - M' U R U R' F' R U R' U' R' F R2 U' R' U2 M
    EAE - B' U R' U' R' F R2 U' R' U' R U R' F' R B

    Two more...just turned them into a J perm or a T perm...not sure if that's what we're looking for here...but it's fast
     
  9. aronpm

    aronpm Member

    2,010
    2
    Sep 9, 2009
    Sure. (I made the speedsolve scramble using Cube Explorer, based on F2L from a qqtimer scramble and a random 22LL case):

    Speedsolve:
    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
    BLD:
    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
    Excuse my terribly inefficient CFOP...

    It's not like ZBLL, where you use ZBF2L to orient edges; there's no extra step before 22LL to setup a case. You just do it whenever it comes up.

    Cyclic shift of an algorithm is when you take moves from one side of an algorithm and moving them to the other. J perm is a cyclic shift of T perm, by moving R U R' F' from the end of T perm to the start. The alg you gave moves the U' R' U' from the end of J perm and takes it to the front.

    Originally I had that for ABI, but I still think there might be a direct alg that faster than both of them. That's a good EAE though, I'll add that one.
     
    Last edited: Jan 5, 2012
  10. PandaCuber

    PandaCuber Member

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    brusinque
    thats really cool.
     
  11. gyc6001

    gyc6001 Member

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    example solve ftw.
     
  12. cubersmith

    cubersmith Member

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  13. StachuK1992

    StachuK1992 statue

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    I already gave this to him, but
    F2 r U r' F R U R' F' U' F might interest some others.
     
  14. Athefre

    Athefre Member

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    Jul 25, 2006
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    I once considered this as a direct solve LL method. The first step being to solve two corners and two edges. I stopped when I realized that the first step has the same "problem" as CLL+1 and Snyder LL. Start mixing corners with edges in a direct solve and the two piece types' orientations and positions relative to each other create a large number of cases.

    I didn't consider if the second step has potential for BLD.
     
  15. Egide

    Egide Premium Member

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    Dec 27, 2009
    Belgium
    WCA:
    2011HIRW01
    YouTube:
    ZZmethod
    an alternative for AAD that l find faster is y L U' R' U L' U2 R U' R' U2 R y'
     
  16. y235

    y235 Member

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    Jan 29, 2011
    Israel
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    YouTube:
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    If using it in a solve, how do you recognize if this a 22LL case or not?
    I think that start checking the cycles isn't good, because if it's not a 22LL case you just wasted a lot of time.
     
  17. Kirjava

    Kirjava Colourful

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    22LL isn't an LL system.
     
  18. bamilan

    bamilan Member

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    But it can easily be. If reducing all other LL cases to 22LL takes only low-number-move algorithms(6-9) we should give it a try.
    Just think about that. For every LL case you should solve "only" any 2 corners and 2 edges.
     
  19. Kirjava

    Kirjava Colourful

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    ahahahaha the best part was when you said 'easily'
     
  20. riffz

    riffz Member

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    Awesome. So you'll have a LL system with terrible recognition and step 2/2 will require one more algorithm than simply OLL/PLL.
     

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