# Invert the softmax function

Is it possible to revert the softmax function in order to obtain the original values $$x_i$$?

$$S_i=\frac{e^{x_i}}{\sum e^{x_i}}$$

In case of 3 input variables this problem boils down to finding $$a$$, $$b$$, $$c$$ given $$x$$, $$y$$ and $$z$$:

$$\begin{cases} \frac{a}{a+b+c} &= x \\ \frac{b}{a+b+c} &= y \\ \frac{c}{a+b+c} &= z \end{cases}$$

Is this problem solvable?

## 2 Answers

Note that in your three equations you must have $x+y+z=1$. The general solution to your three equations are $a=kx$, $b=ky$, and $c=kz$ where $k$ is any scalar.

So if you want to recover $x_i$ from $S_i$, you would note $\sum_i S_i = 1$ which gives the solution $x_i = \log (S_i) + c$ for all $i$, for some constant $c$.

• So it’s solvable up to a constant. Thank you! – jojek May 18 '18 at 17:39
• Which c constant should I use? There is any way of calculating it? – Joel Carneiro Feb 7 '19 at 17:16
• @JoelCarneiro Any $c$ will work; the solution is not unique. – angryavian Feb 7 '19 at 17:58
• Any $c$ will work, one choice is if you augment the $x_i$ vector like $(0, x_1,...,x_n)$ then this will induce a particular $c$, note the corresponding log-sum-exp -- the gradient of which is the softmax -- would also be convex (en.wikipedia.org/wiki/LogSumExp). – Josh Albert Jul 29 '19 at 10:59
• In the case anybody like me spend too much time figuring out $c$: If you know your 3 input variables have to sum to 1 then your $c = (1 - log(x \cdot y \cdot z))/3)$. – Rasmus Ø. Pedersen Sep 14 '20 at 13:04

A similar question was asked in a post of reddit. The answer below is adapted from that post:

$$S_{i}$$ = $$\exp(x_{i})/(\sum_{i} \exp(x_{i}))$$

Taking ln on both sides:

$$\ln(S_{i}) = x_{i} - \ln(\sum_{i} \exp(x_{i}))$$

Changing sides:

$$x_{i} = \ln(S_{i}) + \ln(\sum_{i} \exp(x_{i}))$$

The second term of the right hand side is constant for a particular $$i$$ and can be written as $$C_{i}$$. Therefore, we can write:

$$x_{i} = \ln(S_{i}) + C_{i}$$

• better use j in the summation – eyaler Jan 22 at 18:26