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Suppose $𝐿_1,…,𝐿_𝑘$ are lists with $n$ elements each. We use a fully independent hash function ℎ to compute a value for each element of each list. (We suppose the hash function returns a value uniformly at random from $[𝑛^c]$ for some large constant $c$. Note also that two elements $x$, and $y$ where $x = y$ has $ℎ(𝑥)=ℎ(𝑦)$). Let $𝑀_1,\ldots,𝑀_𝑘$ be sublists with the smallest $𝑋$ elements by computed hash value in each list and let $𝐶_1 = L_1$. Define more $𝐶$s by: $$ 𝐶_{𝑖+1}= \begin{cases} 𝐶_𝑖\cup𝐿_𝑖 & \text{if }|𝐶_𝑖\cap 𝑀_𝑖|<𝑋/𝑟,\\ 𝐶_𝑖 & \text{otherwise}. \end{cases} $$

What size of $𝑋$ will ensure in expectation that $c_1\left(\frac{|C_i \cap M_i|}{|M_i|}\right) \leq \frac{|𝐶_𝑖\cap𝐿_𝑖|}{|L_i|} \leq c_2\left(\frac{|C_i \cap M_i|}{|M_i|}\right)$ for some constants $c_1$ and $c_2$; I am happy if this works for any constants (e.g. $c_1 = 0.5$ and $c_2 = 2$).

I am looking to minimize the size of $X$.

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    $\begingroup$ I can't quite parse what you are asking. What does it mean to ensure "in expectation" that $V<r$ if $A$, where $V$ is a random variable and $A$ is an event? (Also, just to check, I guess you mean that $C_1=L_1$, not just that $C_1$ contains $L_1$?) $\endgroup$ Commented Nov 20, 2020 at 10:42
  • $\begingroup$ @JamesMartin I fixed the problem statement. Does it make more sense now? $\endgroup$ Commented Nov 24, 2020 at 17:04

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