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Suppose there is a global set of $n$ elements, out of which I want to commit to $2n/3$ elements, i.e., anyone can take my commitment and test what $2n/3$ of the possible $n$ elements I committed to.

Is there any way to do so with a constant size commitment that doesn't require exponential calculations. I had Merkle tree root in mind, but of course that doesn't work without a bit-vector of the elements included in the root.

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  • $\begingroup$ "i.e., anyone can take my commitment and test what 2𝑛/3 of the possible 𝑛 elements I committed to." So there's no hiding? It doesn't sound you're looking for a commitment in the usual sense then. $\endgroup$ – Maeher Jul 30 '20 at 7:00
  • $\begingroup$ No hiding. In other words, can one party tell another party with a constant size message of the elements in a subset containing $2n/3$ elements from a total of $n$ globally known elements. $\endgroup$ – Od Na Jul 30 '20 at 9:51
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I will first restate the question as I understand it after the clarifying comment. If I'm misunderstanding something, please let me know.

Let $M$ be a set of size $n$. There are two parties $A$ and $B$. We are looking for a protocol that does the following. $A$ chooses a subset $N\subseteq M$ with $|N|=2n/3$ and computes some message $m$ with $|m|=\ell$, where $\ell$ is a constant, independent of $n$. Given $m$, party $B$ is able to reconstruct and output $N$ with probability $1$.

If that is a correct description of what you're looking for, then it's sadly impossible, since it would imply infinite lossless data compression.

There are $\binom{n}{\frac{2n}{3}}$ many possible subsets, so by sending $m$ you are transferring $\log_2 \binom{n}{\frac{2n}{3}}$ bits of information.

We have $$\log_2 \binom{n}{\frac{2n}{3}} \geq \log_2\left(\frac{n}{2n/3}\right)^{2n/3}=\log_2\left(\frac{9}{4}\right)^{n/3} > \log_2 2^{n/3} = \frac{n}{3}.$$

Since $m$ is only $\ell$ bits long, the pigeonhole principle tells us that you cannot transfer more than $\ell$ bits of information. For any $n>3\ell$¹ that's clearly being violated here.


¹This happens even earlier, I just can't be bothered to do a tight analysis here, we're using pretty loose lower bounds.

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  • $\begingroup$ Thank you! Does this answer also apply when the elements of $M$ are known to both $A$ and $B$? $\endgroup$ – Od Na Jul 30 '20 at 13:54
  • $\begingroup$ Yes. The information being transferred is which elements (think: the indices of an ordered representation) of $M$ are in $N$, not the elements of $M$. $\endgroup$ – Maeher Jul 30 '20 at 13:56
  • $\begingroup$ Thank you! Makes sense $\endgroup$ – Od Na Jul 30 '20 at 14:27

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