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I would like to map inputs to $(0, 1)$ with sum to one constraint.
At first, I came up with softmax function, but it's not invertible.
Is there any function that map inputs to (0, 1) with sum to one constraint?

For example, Let $x_i,y_i \in \mathbb{R}$ and $(_1,_2,_3)=(_1,_2,_3)$, then $Σ_=1$ with $0<_<1$.
I only consider a dimension of input is greater than 2.

alryosha
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  • What does "map inputs to (0,1) with sum to one constraint." mean? – lulu Jul 21 '20 at 00:18
  • If $f(x_1, x_2, x_3) = (y_1, y_2, y_3)$, then $\Sigma y_i = 1$ with $0 \lt y_i \lt 1$ – alryosha Jul 21 '20 at 00:20
  • And what are the inputs? Any number of real numbers? Only finite collections of real numbers? More broadly, please edit your post to include all the information you have in mind. – lulu Jul 21 '20 at 00:21
  • Worth noting: unless I misunderstand, any single input must be mapped to $1$, yes? So it could hardly be invertible. Or are you excluding single inputs? – lulu Jul 21 '20 at 00:22
  • I'm sorry. I only consider a dimension of input is greater than 2 – alryosha Jul 21 '20 at 00:26
  • Well, there are bijections between $\mathbb R^3$ and the unit cube, or whatever, but they aren't pretty. So, I'm not sure what you are hoping for... For instance, most functions of the form you want would take, say, $f(x,x,x)=\left(\frac 13, \frac 13, \frac 13\right)$ and, of course, lose invertibility. – lulu Jul 21 '20 at 00:29
  • I'm sorry but would you elaborate why $f(x_1,x_2,x_3) = (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$ leads to losing invertibility? – alryosha Jul 21 '20 at 00:37
  • What I said was $f(x,x,x)=\left(\frac 13, \frac 13, \frac 13\right)$. So if all three inputs coincide, so do all three outputs. That's sensible, but it destroys invertibility. – lulu Jul 21 '20 at 00:44
  • Then, it could be possible if function does not take the same components? I mean, there exists $i, j$ such that $x_i \neq x_j$ – alryosha Jul 21 '20 at 00:49
  • As I said, there are set theoretic bijections between the relevant sets, but I'd expect them to be quite ugly and hard to implement. Is that really the sort of thing you wanted? Very far cry from something like softmax. – lulu Jul 21 '20 at 00:55
  • I'm sorry but i can't imagine what set theoretic bijections look like, so I don't know that it is what I want.. Thank you for the detailed answer. – alryosha Jul 21 '20 at 01:05
  • here is a construction of a bijection between $\mathbb R^3$ and $\mathbb R$. That's what set theoretic bijections look like. – lulu Jul 21 '20 at 01:14

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