Suppose that $\mu(n)$ is a negligible function, which means that for every $c>0$ there is some $N$ such that for all $n>N$ it holds that $\mu(n)\leq n^{-c}$.

Now, imagine that some encryption scheme, signature scheme, or some cryptographic primitive in general has an "error" (distinguishing error, forging error, etc.) of $\mu(n)$. Imagine that now I want to set up my primitive so that this error is no larger than $2^{-80}$. If I had the explicit expression for $\mu(n)$ (which is likely to be the case if I read the security proof of the given primitive) then I would solve the inequality $\mu(n)\leq 2^{-80}$ for $n$. In practice (even though this is not necessarily the case), negligible functions look pretty much like exponentials $2^{-n}$, so we know that this $n$ will be polynomial in $80$ and therefore it won't be too large.

Now think of the general case. Suppose I want to find some $n$ such that $\mu(n)\leq 2^{-\kappa}$ for some $\kappa$, let's denote the minimum of such $n$'s by $n(\kappa)$. We know that such value must exist simply because $\mu$ tends to zero as $n$ approaches infinity. Moreover, since $\mu$ is negligible, intuition on the definition says that $n(\kappa)$ shouldn't be too large in terms of $\kappa$ (e.g. if $\mu(n) = 2^{-n}$ then $n(\kappa) = \kappa$). My question is: how do we actually prove this?

Here's my question, extracted from the motivation above:

Let $\mu$ be a negligible function, and let $n(\kappa)$ be the smallest $n$ such that $\mu(n)\leq 2^{-\kappa}$. Prove that $n(\kappa)$ does not grow too fast as $\kappa\to\infty$.

Part of the question requires to specify precisely what is meant by "too fast". I would love some statement like "$n(\kappa)$ is polynomial in $\kappa$", but I'm not sure how to prove this.

I believe this property is important as this is one of the reasons why we choose negligible functions (besides being closed on additions and multiplication by polynomials): it allows us to get small error probabilities without increasing the security parameter too much.

Any insights on this are welcome.

  • $\begingroup$ I don't think you can get something very explicit. The definition of negligible itself is not very explicit, so I would expect that the best you can get here is something like "for every $c$ there is a $\kappa$ such that $n(\kappa)\leq \ldots$". I need to work the math harder to figure out the exact expression, but this is my intuition. $\endgroup$ – Cristina Mar 10 '18 at 15:22
  • $\begingroup$ Hmmm, yes, I was also expecting that. I just thought this question might have been studied before, and a more satisfactory solution could exist. $\endgroup$ – Daniel Mar 10 '18 at 15:23

The function $n^{-\log(n)}=2^{-\log(n)^2}$ is negligible, but $n(\kappa)$ for it is $2^{\sqrt{\kappa}}$ which is not polynomial in $\kappa$.

If we look at the functions $\mu_c(n)=n^c$, their $n_c(\kappa)$ is $2^\frac{\kappa}{c}$, so if $\mu$ is negligible, its $n(\kappa)$ must be smaller than those (at least for big enough $\kappa$, this is ok since you asked for the asymptotics). This means that $n(\kappa)$ is subexponential, i.e., $n(\kappa)=2^{o(\kappa)}$.

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  • $\begingroup$ True. But then what can we really say about $n(\kappa)$? Sounds like there is nothing really interesting we can say about it. $\endgroup$ – Daniel Mar 12 '18 at 18:42
  • $\begingroup$ @Daniel I edited my answer to include the asymptotics of $n(\kappa)$ $\endgroup$ – Florian Bourse Mar 13 '18 at 9:05

Functions like $f(n) = 2^{-n^{0.01}}$ are negligible, but if you want to get $f(n) < 2^{-\kappa}$ for this function then you need to set $n > \kappa^{100}$. So in general you can make negligible functions that require $n(\kappa) = n^c$ for any $c$ you like.

I'm fairly sure $n(\kappa)$ should always be polynomial, but don't have a proof.

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  • $\begingroup$ Thanks! That makes sense. Actually, you made me notice an error I had in my question (which I just fixed). $n$ is not logarithmic in $\kappa$, but polynomial (at least it should). It would be great to get a proof of this fact. $\endgroup$ – Daniel Mar 10 '18 at 17:04
  • $\begingroup$ Thinking about this... it seems very hard to ensure this property! After all, why does it have to be the case that $n(\kappa)$ is polynomial? If one were to argue by contradiction, using the negation of being polynomial doesn't fit that much with the properties of $\mu$ $\endgroup$ – Daniel Mar 11 '18 at 0:42

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