There are $\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} \cdot (1 - o(1))$ possible $n$-element subsets of a $2n$-element set. Therefore, any data structure storing such a set must use at least $2n - O(\log n)$ bits.

It’s easy to store such a set in $2n$ bits by using a bitvector while getting $O(1)$ membership queries. Is there a way to store such a set in $2n - \omega(1)$ bits, so that membership queries are “fast” for some reasonable definition of “fast?” (E.g. $O(1)$ time, polylog $n$, etc.)

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    $\begingroup$ This might not be relevent, but if I'm not mistaken, there is a datastructe that uses $n+o(n)$ space, and answers queries in constant time: en.wikipedia.org/wiki/…. If you allow one sided error, this is also known as static approximate membership query data structure, or a static filter. When the set of elements is not known in advance, and you need to support insertions as well this is called an incremental filter. A classical example of such is Bloom filter. $\endgroup$ Commented Feb 27 at 8:39
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    $\begingroup$ @TheHolyJoker Actually, this part of the Wikipedia article seems to directly answer the question: en.wikipedia.org/wiki/… . It references doi.org/10.1137/S0097539795294165 and (for support of more operations) doi.org/10.1145/1290672.1290680 . $\endgroup$ Commented Feb 27 at 9:28
  • $\begingroup$ No, wait, it doesn’t work: the error margins are too large. For the case at hand, $B(n,2n)+o(B(n,2n))$ just means the same as $2n+o(n)$, and is likely even larger than $2n$. The bounds in doi.org/10.1145/3357713.3384274 are much better, but still not good enough: $\log\binom{2n}n+\log n+O(\log\log\log\log\log n)$ is about $2n+\frac12\log n$. $\endgroup$ Commented Feb 27 at 10:34
  • $\begingroup$ Further references are in cstheory.stackexchange.com/a/19317 , but I haven’t checked if they are any good. $\endgroup$ Commented Feb 27 at 10:47
  • $\begingroup$ For the record, any algorithm/data structure for this is completely galactic. You'd need on the order of $2^{64}$ bits to save a single 32-bit register compared to the naive full storage. $\endgroup$
    – orlp
    Commented Feb 27 at 22:57


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