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I'm trying to understand the statement of the Goldwasser-Sipser Set Low Bound Protocol as presented in "Computational Complexity: A Modern Approach" by Arora and Barak.

In particular, I'm trying to understand why the following claims yield what we would want:

$S \subseteq \{0,1\}^m$ is a set such that members can be certified. Both parties know a number $K$. Let $k$ be an integer such that $2^{k-2} < K \le 2^{k-1}$.

Claim: Let $S \subseteq \{0,1\}^m$ satisfy $|S| \le 2^{k}/2.$ Then for $p=|S|/2^k$ we have $p \ge \mathbf{P}_{h \in_{_R} \mathcal{H}_{m,k}, y \in_{_R} \{0,1\}^k}\left[\exists x \in S: h(x)=y \right] \ge \frac{3p}{4}-\frac{p}{2^k}.$

My understanding is that we want:

  • If $|S| \ge K$ then the verifier should accept (corresponding to the prover finding such an $x$ in the claim) with "high" probability;
  • If $|S| \le K/2$ then the verifier will accept with probability at most 1/2.

I'm getting thrown off by a few things here:

  • the claim assumes that $|S| \le 2^k/2 = 2^{k-1} = 2\times2^{k-2} < 2 K$, so how does this place us in one of the relevant cases?
  • How do we actually use those left ($p$) and right ($3p/4 - p/2^k$) bounds to conclude what we want?

Overall, I understand bits and pieces of this setup, but I'm having trouble seeing how it all fits together. Can someone walk through the logic/algebra here?

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To succeed in distinguishing the case of $|S|\ge K$ from $|S|\le\frac{K}{2}$, you just need a constant gap (independent of $|S|$) in the acceptance probability in both cases. If you managed to achieve such a gap, then applying Chernoff to multiple executions of the protocol would achieve high success probability.

Observe that the interesting case is $k\le m$, since otherwise $K\ge 2^{k-2}\ge2^{m-2}$, and the verifier can convince himself that $|S|\ge K$ simply by sampling strings in $\{0,1\}^m$ uniformly at random and checking whether they belong to $S$ (in this case $|S|\ge K$ implies that the probability of falling in $S$ is at least $1/4$, so Chernoff with a polynomial number of samples yields the desired result). From now on assume $m\ge k$, and thus the hash functions are $h_{a,b}:GF(2^m)\rightarrow GF(2^k)$ given by $h_{a,b}=(ax+b)_{1...k}$, where $x_{1...k}$ in the truncation of a string obtained by taking the first $k$ bits.

For uniformly distributed $a,b\in GF(2^m)$, and every $x\in GF(2^m)$, $ax+b$ is uniformly distributed. This means that $\forall x\in GF(2^m): \Pr\limits_{\substack{a,b\in GF(2^m)\\y\in GF(2^k)}}\left[(ax+b)_{1...k}=y\right]=\frac{1}{2^k}$. We conclude that $\Pr\limits_{a,b,y}[\exists x\in S: h_{a,b}(x)=y]\le \frac{|S|}{2^k}$. On the other hand, we can lower bound the same probability by:

$$ \Pr\limits_{a,b,y}[\exists x\in S: h_{a,b}(x)=y]\ge\\ \sum\limits_{x\in S}\Pr\limits_{a,b,y}\left[h_{a,b}(x)=y\right]-\sum\limits_{x\neq x'\in S}\Pr\limits_{a,b,y}\left[h_{a,b}(x)=y \land h_{a,b}(x')=y\right]=\\\frac{|S|}{2^k}-\frac{}{}\binom{|S|}{2}\frac{1}{2^{2k}}\ge \frac{|S|}{2^k}\left(1-\frac{|S|}{2^{k+1}}\right). $$

So far there were no restrictions on the size of $S$. Now, if $|S|\le K\le 2^{k-1}$ then $\frac{|S|}{2^k}\le\frac{1}{2}$, and we obtain $\Pr\limits_{a,b,y}[\exists x\in S: h_{a,b}(x)=y]\ge \frac{3}{4}\frac{|S|}{2^k}$. We conclude that if $|S|=K$ then the probability of acceptance is at least $\frac{3}{4}\frac{K}{2^k}$ (the bound trivially holds for larger $|S|$, just by taking a subset of size $K$). If on the other hand $|S|\le K/2$ then the acceptance probability is at most $\frac{1}{2}\frac{K}{2^K}$ and we have our constant gap $\ge\frac{1}{4}\frac{K}{2^k}$, and Chernoff completes the proof.

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  • $\begingroup$ Thanks for the detailed write-up, this helps clear up some of my confusion. I think what's holding me back from really seeing the final picture is understanding how the Chernoff bounds actually get applied here. Even in the first case you described ($k \le m$) I'm having trouble actually writing down and using Chernoff. Could you help me see how they specifically work in these cases? $\endgroup$
    – theQman
    Jan 9 at 23:21
  • $\begingroup$ Suppose we are trying to learn the bias of a coin, which is either $p$ or $p'$ such that $|p-p'|\ge\epsilon>0$. Toss the coin $n$ times, and denote $\hat{p}=\frac{1}{n}\sum\limits_{i=1}^n X_i$. Your algorithm then declares $p$ if $|\hat{p}-p|\le\epsilon/2$, otherwise it replies $p'$. If the bias is $p$ then your algorithm answers incorrectly with probability at most $2e^{-\frac{n\epsilon^2}{4}}$. If however the bias is $p'$, then again your algorithm errs with probability at most $2e^{-\frac{n\epsilon^2}{4}}$. $\endgroup$
    – Ariel
    Jan 10 at 6:47
  • $\begingroup$ In our case we don't really know $p,p'$, but only know a lower bound $p\ge q$ which also satisfies $p'\le q-\epsilon$ (the constant gap), so your algorithm can check whether $\hat{p}\ge q-\frac{\epsilon}{2}$, i.e. it suffices to know $q,\epsilon$, which in our case is $\frac{3}{4}\frac{K}{2^k}$ and $\epsilon=\frac{1}{4}\frac{K}{2^K}$. $\endgroup$
    – Ariel
    Jan 10 at 7:01

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