The Futurama Theorem

Discussion in 'Physics & Math' started by MacGyver1968, Feb 4, 2011.

  1. MacGyver1968 Fixin' Shit that Ain't Broke Valued Senior Member

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    7,028
    I was watching a re-run of an episode of Futurama last night, and problem is presented. The professor stares into the camera and declares "This can only be solved with MATH!"

    Here's the problem (from wiki)

    I tried to figure it out on my own...but quickly realized it well beyond me. With a little google-fu, I discovered that one of the writers for the show has a PhD. in Mathmatics and actually created a new theorem just for this show. I thought that was kind of cool.

    Here's the theorem if you want to check it out.
    The formal proof is shown in the episode for a brief moment while 'sweet' Clyde of the "globe trouders" manages to prove the theorem. The proof is as follows:

    First let π be some k-cycle on [n] = {1 ... n} WLOG [without loss of generality] write:

    π = 1 2 ... k k+1 ... n
    2 3 ... 1 k+1 ... n

    Let <a,b> represent the transposition that switches the contents of a and b. By hypothesis π is generated by DISTINCT switches on [n]. Introduce two "new bodies" {x,y} and write

    π* = 1 2 ... k k+1 ... n x y
    2 3 ... 1 k+1 ... n x y

    For any i=1 ... k-1 let σ be the (l-to-r) series of switches

    σ = (<x,1> <x,2> ... <x,i>) (<y,i+1> <y,i+2> ... <y,k>) (<x,i+1>) (<y,1>)

    Note each switch exchanges an element of [n] with one of {x,y} so they are all distinct from the switches within [n] that generated π and also from <x,y>. By routine verification

    π* σ = 1 2 ... n x y
    1 2 ... n y x

    i. e. σ reverts the k-cycle and leaves x and y switched (without performing <x,y>).

    NOW let π be an ARBITRARY permutation on [n]. It consists of disjoint (nontrivial) cycles and each can be inverted as above in sequence after which x and y can be switched if necessary via <x,y>, as was desired.

    http://en.wikipedia.org/wiki/Futurama_theorem#The_proof
     
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    The entire length of my PhD I had has my desktop a screen grab of Prof Farnsworth's blackboard during the lecture where he 'proved' the neutrino tastes of gravy. While I can't say I'd have been able to do it myself I'm surprised such a theorem wasn't already known, the structure of permutation groups is a popular area in group theory. Actually a guy I work with did his PhD in that so I might ask him this tomorrow

    Please Register or Log in to view the hidden image!

    (/edit Or rather Monday, today being Friday!)
     
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  5. rpenner Fully Wired Valued Senior Member

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    How would you phrase the theorem in a textbook or paper?

    "Every non-trivial element of the \(S_n\) subgroup of \(S_{n+2}\) has a non trivial inverse in \(S_{n+2}\) which may be factored into unique transpositions not part of \(S_n\) ?"

    "Every element of the \(S_n\) subgroup of \(S_{n+2}\) may be factored into zero or more unique tranpositions which are not part of \(S_n\)."

    "If G is a symmetric group, and H is a symmetric subgroup of G, and degree(G) >= 2 + degree(H) then for every element, h, of H there exists a finite sequence of unique elements of G - H where the product equals h."
     
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  7. alephnull you can count on me Registered Senior Member

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    The mathematics of wanton burrito meals?
    heh, maybe if I one day do a PhD it will be on superdupersymmetric string theory.


    I'm also surprised such a theorem didn't already exist. The proof itself is not difficult and looks like it could be one of those "optional" questions at the end of a first year abstract algebra problem set.

    It does make me happy that a "Futurama Theorem" exists though.
     

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