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I Multivariable function that is injective?

  1. Jun 26, 2016 #1
    Hey all, is it possible to find a function that for $$ a,b,c.. \in \mathbb{R} $$ $$ y= f(a,b,c,..) , \hspace{5mm} y= \rho , \rho \in \mathbb{R} \hspace{2mm} for \hspace{2mm} only \hspace{2mm} 1 \hspace{2mm} set \hspace{2mm} of \hspace{2mm} a,b,c.. $$
    Any help appreciated
     
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  3. Jun 26, 2016 #2

    micromass

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    What's your domain and codomain?
     
  4. Jun 26, 2016 #3
    I'm fine with any that isn't completely trivial (if there are any trivial solutions), will try adapt to whatever I can get, basically this is for a program that has a graph, and a vertex has to have a single number input as a function of the labels of the vertices already in the path. Inputing the entire path so far will take up way too many resources.
     
  5. Jun 26, 2016 #4

    mfb

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    There are injective functions ##g: \mathbb{R} \times \mathbb{R} \to \mathbb{R}##, you can use f(a,b,c)=g(a,g(b,c)). Those functions are messy, and need infinite precision to be truly injective, I'm quite sure you don't want to use them.

    More context would help, but I guess there is an easier solution. Why can't you just use the set of three numbers as label? Expressed as string or whatever if the data format is an issue.
     
  6. Jun 26, 2016 #5
    Don't want to pass entire label as worsens the time complexity of the program by at the minimum of increasing the power by 1. /: Could you please list an example of one of the functions which satisfies $$ \mathbb{R} \times \mathbb{R} \to \mathbb{R} $$ ?
     
  7. Jun 26, 2016 #6
    The sets ##\mathbb R^m## and ##\mathbb R^n## have the same cardinality for all ##m,n##
     
  8. Jun 26, 2016 #7

    mfb

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    Here is an example
    Your label will have to be longer than the length of a single coordinate, in a suitable format. If you have N possible values for the single coordinate, you need N3 possible labels.

    I don't see how concatenating strings would increase the time complexity of anything.
     
  9. Jun 26, 2016 #8

    micromass

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    Maybe you're interested in a function ##\mathbb{Q}\times \mathbb{Q}\rightarrow \mathbb{Q}## instead? That is much easier to give, but I'm sure it's not going to be useful in a computational context.

    In any case, given ##m/n## and ##m'/n'## in reduced form (meaning that ##m## and ##n## have no common divisors and ##n>0## and likewise for ##m'## and ##n'##), you can send this to ##2^m 3^n 5^{m'} 7^{n'}##.
     
  10. Jun 26, 2016 #9
    The processing required for what those strings will be processed as it will
     
  11. Jun 26, 2016 #10
    What's $$ m' $$ and $$n'$$?
     
  12. Jun 26, 2016 #11

    mfb

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    It is linear in the number of vertices, which is as good as it can get.

    m' and n' are the numerator and denominator of the second fraction.
     
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