The well known Decisional Diffie Hellman assumption (DDH) assert that for any $n = \log q$ and generator $g$ of $\mathbb{Z}_q$, for uniformly i.i.d $A, B, C \sim U(\mathbb{Z}_q)$, the following are indistinguishable for any PPT $M$: $g^A, g^B, g^C$ vs. $g^A, g^B, g^{AB}$. That is, up to negligible advantage: $$\epsilon = \left| \Pr[M(g^A, g^B, g^C) = 1] - \Pr [M(g^A, g^B, g^{AB}) = 1 \right| \leq 1/\omega(n) $$ I'm interested however how small the error can be taken? That is, is there any distinguisher for the extremely small $\epsilon = 2^{-O(n)}$? Is it known how ``negligible'' should $\epsilon$ be?

  • $\begingroup$ I think $\epsilon$ should be a function of $q$, not $n$, and therefore you need to specify a connection between $q$ and $n$ for this question to have an answer. $\endgroup$
    – Mark
    Mar 2, 2022 at 17:52
  • $\begingroup$ Thanks, I meant that $n = \log q$ $\endgroup$
    – Napoleon
    Mar 3, 2022 at 7:49

1 Answer 1


Negligible has a precise meaning in cryptography. It is really defined in terms of growth (or rather, decay), for example, with respect to the security parameter.

A function $\mu$ is negligible if it grows slower (or decays faster) than 1 over any polynomial function. Specifically, for any polynomial $\mathsf{poly}$, for some constant $N$, then for all $x \geq N$, we have: $$|\mu(x)| < \frac{1}{\mathsf{poly}(x)}.$$

An example of a negligible function is $\mu(x) = 2^{-x}$. This is because for any polynomial, we can always find an $N$ such that the previous inequality holds, since the decay is exponential. For example, using the polynomial $x^3$, the inequality doesn't hold at $x = 2$ (since $1/4 > 1/8$), $x = 3$ (since $1/8 > 1/27$), and so on. But when $x \geq 10$, then the inequality does hold (e.g. $2^{-10} < 1/10^3$). So in this specific eaxmple, we'd set $N = 10$.

In the specific example of DDH, suppose $M$ spends a polynomial amount of time computing random DDH triples itself, ($g^a,g^b,g^{ab}$). Then there is some tiny probability that the DDH challenge it is given was one that it computed, so it would win slightly more than $1/2$ the time (it wins half the time from a uniformly random guess). However, this advantage is negligible in the technical sense, because as $M$ is PPT, it can only compute polynomially many tuples, but the number of possible tuples grows exponentially with the security parameter. Therefore the advantage looks something like $\textsf{poly}(\kappa)/2^{\kappa}$, which is negligible in the formal sense above.

  • $\begingroup$ I agree that with the specific machine $M$ you mentioned, that simply guesses elements, has $1/2^{O(n)}$ advantage. Nevertheless, there might be other PPTs that do something different, and able to gain better advantage, say, $1/n^{\log n}$, which is still negligible. My question concerns more how much negligible the advantage can be: $n^{-\log \log n}, n^{-\log n}, 2^{-O(n)}$? $\endgroup$
    – Napoleon
    Mar 3, 2022 at 7:55
  • $\begingroup$ Theoretically, any of those things could exist, the security assumption says nothing about it. We just draw the line at "negligible" and leave it there. $\endgroup$ Mar 3, 2022 at 7:58
  • $\begingroup$ I agree that all of those error are theoretically possible, but practically -- are you aware for example about a distinguisher with advantage $n^{-\log \log n}$? $\endgroup$
    – Napoleon
    Mar 3, 2022 at 8:27
  • $\begingroup$ No, I'm not aware off the top of my head. It sounds like it would be a weird algorithm. $\endgroup$ Mar 3, 2022 at 9:45

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