Math Problem - 12

Discussion in 'Puzzle Theory' started by Lucas Garron, Aug 15, 2008.

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  1. Lucas Garron

    Lucas Garron Super-Duper Moderator Staff Member

    Sune as a single commutator

    I think I should post a problem sometime. This, however, will be a cube problem; nevertheless, it's a lot like the others presuming some cube knowledge.

    Many cubers know knows that the Sune (RUR'URU2R') can be written as a two commutators ([R, U][U2, R]) or as a conjugated commutator ([R U2; U', R']). But even Johannes didn't know it could be written as a single commutator. In fact, I wasn't even sure until I tried it, and the task turned out to be pretty fun.

    So, the problem: Write the Sune as a commutator in the pure form [moves, moves].
    (Doesn't have to be pretty unless you want it to be.)

    If you've found it or already know it, please use spoiler tags, as seeing a solution even for a glance may be distracting for people who haven't solved it yet.


    And if you think that's too easy: It's well-known that commutators generate all the even permutations. Show that any even permutation can be written as a single commutator, or show that it can't be done.
     
  2. DavidWoner

    DavidWoner The Punchmaster

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    i found the ones for U-perm, still working on the sune one.

    EDIT: im pretty sure these are actually wrong. yes. they are wrong. i will remove them now.
     
    Last edited: Aug 17, 2008
  3. Tim Reynolds

    Tim Reynolds Premium Member

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    That was fun to work on...I can't really explain my steps very well, but I got
    [R U R2, R U2 R2] = R U R2 R U2 R2 R2 U' R' R2 U2 R' = R U R' U2 U' R U2 R'= sune
    now working on the general case...
     
  4. nitrocan

    nitrocan Member

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    I don't know commutators that well, but this doesn't require so much commutator knowledge I guess. I will give it a try:
    R U R' U R U2 R' as A B A' B'
    I'm guessing A will start with R U and B will start with R U2 and they should be 3 or 4 moves most. I will call the 3rd (and if it's 4 moves the 3rd and the 4th) move x(A) and y(B)
    Our commutator can be written as : R U x R U2 y x' U' R' y' U2 R'
    so x R U2 y x' U' R' y' must be R' U R
    then our x becomes R2 and then we can see that:
    R U R2 R U2 y R2 U' R' y' U2 R' = R U R' U R U2 R'
    then U2 y R2 U' R' y' must be U R
    then y is R2 so that in cancels the R2 and U2 U' makes U and R' R2 makes R
    so R U x is R U R2 and R U2 R2, our commutator becomes:
    [R U R2, R U2 R2]
    this wasn't so easy so I'm not going to answer the 2nd question :D
     
  5. Stefan

    Stefan Member

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    Do you know the answer?
     
  6. cuBerBruce

    cuBerBruce Member

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    Actually, every legal position of Rubik's cube is an even permutation, and half of them can't be generated by commutators (of legal Rubik's cube moves). Apparently by "even permutation," you really meant that the edge permutation is even and the corner permutation is even. :)

    Assuming that's what you really meant, Joyner's book Adventures in Group Theory has the answer, but his proof "cheats" by using another theorem.
     

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