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Thread: Void Parity Resolution/Algs

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    Super-Duper Moderator Lucas Garron's Avatar
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    Default Void Parity Resolution/Algs

    The "Void Parity" is the issue that occurs when solving anything equivalent to a 3x3x3, particularly a void cube, where you have odd total parity with your current assignment of centers.
    Layer methods hit this (i.e. you can recognize the parity) at PLL, and I think with CF or Roux, you can try to avoid it.

    It's well-known that a reassigning centers with a quarter-slice twist + resolving pieces will resolve this. My approach used to be something like an M-fix with a 4-flip and two 3-cycles.
    But for speedsolving, you'd want something efficient. Tomas once asked us for a fix in #rubik, and so I got kinda interested.

    The most practical is to find a fix that preserves OLL (which can be applied before PLL). It's not hard to imagine that the best fixes are URrML-ish and preserve CPLL, which is also good for CF and Roux, depending on how you do it (and what kind of mistakes you make).

    After a while, I found l U r' U M U' M' r U' R' U M U'.
    Then, qqwref found the very nice M' U M' U M' U' M' U' M' U2' M' U' M'.

    So, does anyone have any other algs, or better approaches?
    Last edited by Lucas Garron; 09-04-2008 at 05:18 PM.
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    brah blah's Avatar
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    Y'know what? I used to have a number of these algs, back when I first proposed my "new" 5x5x5 BLD method, but then I never went back to solving a 5x5x5 BLD again, so I think I may have deleted that file (can't find it right now)
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    I've been using the following to swap UF and UR: (M' U')4 (R2 d' M' d2 M' d' R2) (F2 M' F2 M2).
    Essentially a 4-flip, an Allan (U-Perm), and an M-layer 3-cycle + M move. But qqwref's alg (which swaps UR and UB after adding a missing U' on the end ) is shorter.
    For swapping UFL and UBR, I've been using (R' U L' U2 R U' L)2 U' F R F' (M' U')4 F R' F' B2 M B2 M2.
    Not great algs, but I didn't have to memorize anything new.

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    Member Meep's Avatar
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    My approach to that was to simply do M', then insert the two cross pieces again via the M layer, then orient any edges that were misoriented from doing that.

    Ex. M' + U' M' U M U M U2 (M' + M <-- These cancel out) U M U M U2 M' U M' U M' U'

    A really inefficient intuitive approach. =\

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    M' U M' U' M U' M U2 M' R U R U' M' U R U' r'
    edit: joey again

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    Quote Originally Posted by coinman View Post
    M' U M' U' M U' M U2 M' R U R U' M' U R U' r'
    edit: joey again
    joey, your alg seems to move a lot of pieces. Did you make a mistake?

    BTW, there is this old thread that is somewhat related to this one.

    http://www.speedsolving.com/forum/showthread.php?t=2516

    (And joey, your alg in that thread doesn't work for me either. Are there people who actually post algs without using an alg checker first? )

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    Super-Duper Moderator Lucas Garron's Avatar
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    joey's alg is M' U M' U' M U' M U2 (M' R) U R' U' M' U R U' r'. Missed a '.

    (And thanks, I know about that thread. I made this thread specifically to discuss approaches to centerless speedsolving, though. )
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    Member Derrick Eide17's Avatar
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    i came up with a really good one for the first opposite swap awhile ago but totally forgot it

    Nice alg for 2nd one though from Michael!

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    Member Ethan Rosen's Avatar
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    On the last layer make it so all you have is two edges opposite each other on the layer that need to be swapped. Then do cube rotations so those two pieces are BR and FR. Do E' L2 E L2 E'. The problem with this is that it then leaves you with four edges to orient.

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    M' (D' R2 D) (U M' U' M) (D' R2 D)
    The M' changes parity, the (conjugated) 5-edges-cycle repairs the damage.

    Now with ACube and this input: FD ? -? ? BD DR -? DL FR FL BR BL ? ? ? ? DRF DFL DLB DBR
    13 qtm: M' U' B D B U S' U' F' R' B'
    9 stm, 12 htm: M' U B2 U2 M' U M U B2
    Last edited by Stefan; 09-09-2008 at 07:29 AM.

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