Continuous trapdoor functions?

Question

Every trapdoor function I've seen has been a discrete function.

Do there exist continuous trapdoor functions?
If so, what's an example of a continuous trapdoor function? And if not, why not?

Answer 1

The purpose of a trapdoor function means it has to be discrete. After all, the purpose is that it will be done by computer. The typical ways you can talk about continuous functions and computability are largely irrelevant here - that is, saying that $f$ is computable if, given an oracle outputting digits of $x$ (or other equivalent formulation), it is possible to compute as many digits of $f(x)$ as is desired.

Except that in cryptography, we want to send the entire computation - always operating on things that we can be certain are subject to no error. The guarantee that computability gives - i.e. that we can always compute more digits, if we want - isn't sufficient here because that would mean we might end up continuously asking the other party, "Hey, I need more digits for my computation, send 'em over" (since they can't very well send the algorithm that generates the digits!) and not accomplishing anything quickly. It makes a lot more sense to say, "We are using this many bits. Compute them exactly" than to let fuzzy continuous stuff get in the way.

Moreover, if we allowed any amount of uncertainty in our calculation, the function would have to be reasonably well-behaved such that specifying that $x$ lies in a small interval suffices to specify $f(x)$ pretty well too. But that's a problem, since it means that, to keep the system secure, we have to enforce a certain distance between any $x$ and $y$ for which $f(x)$ and $f(y)$ might want to be computed - such that no information about $f(y)$ is gained by knowing $f(x)$, for instance. However, once we're enforcing that $x$ and $y$ be separated by some amount, we're back to a discrete function, since we can't have more than countably many $x$ and $y$, all adequately separated.

Answer 2

There are two reasons they are discrete. First, they operate on natural numbers, not reals, so they are by nature discrete. Second, you would like an adversary not to learn anything about $f(n)$ from $f(n-1)$ so they have to be very discontinuous