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Thread: Z --> ZxZ Function

  1. #1
    Vee is offline
    Apr 2013

    Z --> ZxZ Function


    I would like to define a function from Z --> ZxZ that is both onto and one to one. I think it is going to have to be a piecewise function, but I am totally lost on how to get started. What I do know is that the function can be represented graphically by a spiral where the inputs and outputs look something like this:

    f(0) = (0,0)
    f(1) = (1,0)
    f(2) = (1,1)
    f(3) = (0,1)
    f(4) = (-1,1)
    f(5) = (-1,0)
    f(6) = (-1,-1)
    f(7) = (0,-1)... so on

    I know this is only for positive Z, and this is part of the reason I am stuck. I am also not sure how to represent this info as a function. Any help would be appreciated. This is bugging the crap out of me. Thank you.
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  2. #2
    Super Member
    Jan 2008

    Re: Z --> ZxZ Function

    It may be easier to write a bijection $\displaystyle f:\mathbb{Z}\to \mathbb{N}$, so that $\displaystyle g=(f,f):\mathbb{Z}\times \mathbb{Z} \to \mathbb{N}\times \mathbb{N}$ is a bijection. For example, take

    $\displaystyle f(x)=\begin{cases} 2x, & \text{if }x\geq 0 \\ -2x+1, & \text{if }x<0\end{cases}$

    It remains to find a bijection $\displaystyle \varphi: \mathbb{N}\times \mathbb{N} \to \mathbb{N}$. The standard one is the Cantor pairing function

    $\displaystyle \varphi(x,y)= \frac{(x+y+1)(x+y)}{2}+x$

    This last function makes precise the usual snake-like enumeration diagram for $\displaystyle \mathbb{N}\times \mathbb{N}$.

    The function you want is $\displaystyle g^{-1} \circ \varphi^{-1} \circ f$. Since each function in the composition is a bijection, the result is also a bijection.
    Thanks from topsquark and Vee
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