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Thread: Maneuver and its Inverse (question)

  1. #11
    Member Cielo's Avatar
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    S = B D2 R U B D' R D2 F U D' B2 D' L2 D' F2 R2 B2 D2 R2
    M1 = B2
    S M1 has two blocks: DB&DBL (now in RU&RUB), LB&LBU (now in UL&ULB).
    For the first one, X = z y', so after executing S M1 X, DB&DBL are "solved". As a result, DB&DBL are also "solved" if we execute X' M1' S'.
    In fact, DB&DBL block is in RU&RUB after X'. So M1' S' moves the RU&RUB block to DB&DBL.

    After writing down this, I find my explanation was too complicated -_-| But It still can answer your question in the first post.
    Just as qqwref posted: S moves a block from position A to B, then S' moves a block from position B to A.

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    Member siva.shanmukh's Avatar
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    Quote Originally Posted by Cielo View Post
    S = B D2 R U B D' R D2 F U D' B2 D' L2 D' F2 R2 B2 D2 R2
    M1 = B2
    S M1 has two blocks: DB&DBL (now in RU&RUB), LB&LBU (now in UL&ULB).
    For the first one, X = z y', so after executing S M1 X, DB&DBL are "solved". As a result, DB&DBL are also "solved" if we execute X' M1' S'.
    In fact, DB&DBL block is in RU&RUB after X'. So M1' S' moves the RU&RUB block to DB&DBL.

    After writing down this, I find my explanation was too complicated -_-| But It still can answer your question in the first post.
    Just as qqwref posted: S moves a block from position A to B, then S' moves a block from position B to A.
    Okay.. I got what you are trying to say. But the best answer would be your last line and qqwref's answer
    Success is how high you bounce when you hit bottom.

  3. #13
    Premium Member mariano.aquino's Avatar
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    Quote Originally Posted by siva.shanmukh View Post
    This I already knew. I mean if there is a set of solved cubies, they would be solved in the inverse maneuver as well. My question is regarding unsolved blocks. Anyways don't bother, I got my answer from qqwref's reply



    In FMC, if you are stuck at a point, say right after solving a 2x2x2 block (by saying stuck, I mean not getting any optimal maneuver from there on), you could just invert the scramble and the partial solution you have got so far on a solved cube and try to proceed from there. This not only gives a different unsolved state with the same solved block, but also lets you use it as your solution. for example if Ma S Mb renders the cube solved where S is your scramble, then Mb Ma is a solution for S. You can prove it in the following way.

    Ma S Mb = I
    Ma' Ma S Mb Ma = Ma' I Ma
    S Mb Ma = I

    Hence Mb Ma is a solution for S
    Thanks! I'll give it a try!

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