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Assume that Alice and Bob are respectively given two strings $x \in \{0,1\}^n$ and $y \in \{0,1\}^n$ such that the hamming distance between $x$ and $y$ is either $> n/2+\sqrt{n}$ or $< n/2-\sqrt{n}$.

Alice picks a set $A \subseteq \{1,...,n\}$ of size $k$ uniformly at random, and sends to Bob the set $\{(i,x_i) \mid i \in A\}$. Bob compares $x_i$ with $y_i$ for every $i \in A$ and accepts iff the majority of these indices agree. What is (asymptotically) the minimal $k$ for which this protocol succeeds (i.e. Bob accepts iff the distance is $< n/2-\sqrt{n}$) with probability at least $2/3$?

I guess that $k$ should be $\Omega(n)$, and that in order to show this we should bound the tail of the distribution (e.g. by using the Central limit theorem or something similar), but I don't see how this can be done.

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  • $\begingroup$ Are you familiar with the Chernoff bound? This should give you one direction. $\endgroup$ – Yuval Filmus Apr 21 at 17:34
  • $\begingroup$ In the other direction you could use the reverse Chernoff bound, but in your case there might be a simpler way. $\endgroup$ – Yuval Filmus Apr 21 at 17:35
  • $\begingroup$ @YuvalFilmus What do you mean by two directions? also $A$ is picked uniformly at random. How do we change this to binomial distribution? $\endgroup$ – user91015 Apr 21 at 17:54
  • $\begingroup$ You need both a lower bound and an upper bound on $k$. These are the two directions. $\endgroup$ – Yuval Filmus Apr 21 at 18:00
  • $\begingroup$ As to relating it to the binomial distribution, you're right that it's more like the hypergeometric distribution (though Chernoff's bound works in this case as well). I suggest first considering what happens when instead of a set of size $k$ you draw $k$ integers from $\{1,\ldots,n\}$ with repeats. $\endgroup$ – Yuval Filmus Apr 21 at 18:01
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Suppose that the Hamming distance between $x$ and $y$ is $n/2 - \sqrt{n}$. Suppose that you sample $m$ indices with replacement, and let $S$ be the number of indices on which $x$ and $y$ disagree. Then $S$ has expectation $mp$ and variance $mp(1-p)$, where $p = 1/2 - 1/\sqrt{n}$. According to the central limit theorem, for every fixed $t$ we have $$ \Pr\left[ \frac{S-mp}{\sqrt{mp(1-p)}} \leq t \right] \longrightarrow \Pr[N(0,1) \leq t]. $$ We are interested in the event $S \leq m/2$, which is the same as $S - mp \leq m(1/2-p) = m/\sqrt{n}$, or equivalently $$ \frac{S-m/p}{\sqrt{mp(1-p)}} \leq \frac{1/2-p}{\sqrt{p(1-p)}} \sqrt{m} \approx \sqrt{\frac{m}{4n}}. $$ In particular, if $m = cn$ then (following a short argument) the probability of error tends to $$ \Pr[N(0,1) > \sqrt{c/4}]. $$ For small enough $c$, this would be more than $1/3$, giving an $\Omega(n)$ lower bound for this version of your algorithm.

In your case, you choose indices without replacement. A similar argument works, using the central limit theorem for hypergeometric random variables, see for example Lahiri and Chatterjee, A Berry–Esseen theorem for hypergeometric random probabilities under minimal conditions, Theorem 2.1.

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  • $\begingroup$ If you are assuming that the distance between $x$ and $y$ is $n/2-\sqrt{n}$, then why $p = 1/2- 1/\sqrt{n}$? $\endgroup$ – user12 Apr 22 at 14:12
  • $\begingroup$ Thanks, changed agree to disagree. $\endgroup$ – Yuval Filmus Apr 22 at 14:13
  • $\begingroup$ You'll have to choose the correct version of "agreement" and "disagreement" for the argument to make sense. The idea is to show that the answer is wrong with probability more than 1/3. $\endgroup$ – Yuval Filmus Apr 22 at 14:38

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