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I would like to know what $O(v(n))$ really means in detailed and simple words please. I found it everywhere in the literature I am reviewing but I cannot find what the intuition of it (especially if it means the Big O Notation, then what it has to do with the Probability here; also what is $v$, it is never defined except in another paper as a constant -but maybe not related directly to v in this formula-).

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Many Thanks.

References 1 , 2

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    $\begingroup$ can you give a reference to where you got the quote from and what [1] is? after all, others may be better than you in tracing references. took a quick look at two references, I did not see any $\nu(n)$. $\endgroup$ – kodlu Aug 31 at 22:24
  • $\begingroup$ @kodlu Just added the references. This (1) is the equation number whereas $s{_1}^{i-1}$ means the subsequence starting at the first bit and ends at $i-1$. $\endgroup$ – Mike Aug 31 at 22:43
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But this notation is defined (informally) in the first paper.

The notation $O(\nu(n))$ is used for any function, $f(n)$, that vanishes faster than the inverse of any polynomial, that is for every polynomial, $\mathrm{poly} (n)$, and $n$ large enough, $f(n) \leq 1/\mathrm{poly}(n)$

Therefore, what it means is no probabilistic polynomial time (PPT) algorithm $A$ can guess the next bit at inverse polynomially decreasing error rate.

Given any PPT algorithm $A$ this error probability decays super-polynomially, for example at a rate $\exp(-\log^2 n)$ which goes to zero faster than any polynomial.

Claim: $\exp(-\log^2 n)$ is less than $n^{-c}$ for any constant $c$ for $n$ large enough.

Proof: Look at the reciprocal. $$\exp(\log^2 n) > n^c$$ if and only if $$\log^2 n > c \log n$$ which will clearly happen as soon as $$\frac{\log^2 n}{\log n}>c$$ i.e., as soon as $\log n>c$.

Note that $\exp(\log^{1+\epsilon} n)$ is superpolynomial for all $\epsilon>0.$

Edit: The superpolynomial convergence is essentially what is referred to as negligibility.

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  • $\begingroup$ Thanks and +1. So a successful predictor should predict with a probability greater than or equal $1-O(ν(n))$, How is the latter evaluated numerically. That's from where should I derive the $O(ν(n))$ numerically? A simple example of a simple algorithm will be very much appreciated. $\endgroup$ – Mike Sep 1 at 9:57
  • $\begingroup$ Is O(v(n)) the error rate? And how is it derived? Practically from the experiment/simulation? Thanks. $\endgroup$ – Mike Sep 1 at 10:38
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    $\begingroup$ It is implicitly negligibility. Adding this into answer will increase the awareness. $\endgroup$ – kelalaka Sep 1 at 18:17
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First, that's not the Next Bit Test, which is purely theoretical, but rather an attempt at a practical approximation of it.

From your first citation:

The notation $O(\nu(n))$ is used for any function, $f(n)$, that vanishes faster than the inverse of any polynomial, that is for every polynomial, $poly(n)$, and $n$ large enough, $f (n) < 1/poly (n)$.

So you can think of it as a function that asymptotically approaches 0, very, very quickly. Overall, this means that the probability in question must then be extremely small. Since that probability is |(the chance that an attacker guesses the next bit)-50%| it means the attacker has extremely small advantage.

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  • $\begingroup$ Thanks for your answer. +1 for explaining the intuition of the notation. $\endgroup$ – Mike Sep 2 at 9:17
  • $\begingroup$ Just to mention, the paper is a second attempt of the NBT, and based on improving a previous attempt and the Shamir's paper (which proves that the NBT proposed by Yao is not universal where Shamir provides a better version of the NBT) $\endgroup$ – Mike Sep 2 at 10:05

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