My understanding is that a typical neural network without any fancy activation functions can only solve problems that can be modelled as a continuous function. If this is the case, why can a standard neural network (1 hidden layer) solve xor? I'm not looking for a solution to xor here but an explanation as to why this is possible with a non-continuous function.

Any help on this matter would be greatly appreciated.


1 Answer 1


I guess the observation you need is that a neural network is typically a function $f \colon \mathbb{R}^2 \to \mathbb{R}$ or something along those lines, whereas the XOR function is a map from $\{0, 1\}^2$ to $\{0, 1\}$.

It's easy to construct a function that, when restricted to $\{0, 1\}^2$, is the XOR function, but is continuous on $\mathbb{R}^2$. Consider

$$\begin{align*} g \colon \mathbb{R}^2 &\to \mathbb{R}, \\ (x, y) &\mapsto x + y - 2xy.\end{align*}$$

Certainly $g(0, 0) = 0$, $g(1, 0) = 1$, $g(0, 1) = 1$ and $g(1, 1) = 0$, so $g|_{\{0, 1\}^2} = \operatorname{XOR}$. But we can see that $g$ is a continuous function (on $\mathbb{R}^2$). We don't care what it does on almost the entirety of $\mathbb{R}^2$; it just needs to agree on the four points we care about.

It's a fun exercise to try and explicitly construct a neural network which computes the XOR function. You should be able to do it fairly easily with ReLU activation; it might be slightly simpler to do it with a couple of layers. If it's still not clear, I can expand further.

  • $\begingroup$ Thank you for your response! Just to clarify, a true representation of xor would be discontinuous as they are 4 disconnected points? However, the neural net is learning a continuous function that must return g(0,0)=0, g(1,0)=1, g(0,1)=1 and g(1,1)=0? E.g. the neural net is learning a function similar to what I have drawn in blue ibb.co/RhZwvjT (sorry made this in paint) $\endgroup$
    – Bradley
    Jul 1, 2021 at 11:03
  • 1
    $\begingroup$ @Bradley Here's a question for you to think about: what does it actually mean for a function defined on four discrete points to be continuous? I dodged that question a bit because for most people, a continuous function is one where "if you change the input a little bit, the output changes correspondingly a little bit", and you can't do that for XOR. There are ways that you can define continuity in this situation (look up "point-set topology") but you don't need to care about that. I think you might have got a bit mixed up with your graph; the network takes two inputs and returns a (continued) $\endgroup$
    – htl
    Jul 1, 2021 at 11:44
  • $\begingroup$ ... single output, so you would most naturally draw it as a graph with three axes; I don't think yours makes sense for the situation. Does this help at all? $\endgroup$
    – htl
    Jul 1, 2021 at 11:45
  • $\begingroup$ With a discrete topology, every function is continuous (in the topological meaning). (I have no clue how this relates to neural networks, though.) $\endgroup$ Jul 1, 2021 at 23:12

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