# Estimating the number of distinct elements

Need to understand "intuition" part. It does not make sense to me why $$log(d)$$ is a good approximation.

We have a stream $$\sigma = \{a_1, ..., a_n\}$$, with each $$n \in [n]$$, and this implicitly defines a frequency vector $$F = {f_1, ..., f_n}$$. Let $$d = |\{j : f_j > 0$$}| be the number of distinct elements that appear in $$\sigma$$.

For an integer $$p > 0$$, let $$zeros(p)$$ denote the number of zeros that the binary representation of $$p$$ ends with. Formally,

$$zeros(p) = max\{i : 2^{i}$$ divides $$p\}$$

Algorithm:

Initialize:

1. Choose a random hash function $$h$$: $$[n] \mapsto [n]$$ from a 2-universal family;

2. z $$\leftarrow$$ 0;

Process $$j$$: 3. if $$zeros(h(j)) > z$$ then $$z \leftarrow zeros(h(j))$$

Output: $$2^{z+\frac{1}{2}}$$

The basic intuition here is that we expect $$1$$ out of $$d$$ distinct tokens to hit $$zeros(h(j)) \geq log(d)$$, and we don't expect any tokens to hit $$zeros(h(j)) >> log(d)$$. Thus the max value of $$zeros(h(j))$$ over the stream - which is what we maintain in $$z$$ - should give us a good approximation to $$log(d)$$.

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• 1. Please attribute your sources. You give a very long quotation. Is that a quote from some source? If so, where? Copying without attribution amounts to plagiarism, which isn't cool. 2. What is your question? I don't see any question here, merely a list of declarative statements. Can you articulate a specific, answerable question? – D.W. Mar 9 '16 at 4:58

The idea is that the hash function $h$ maps elements to random elements. Now think about the likelihood that the last $k$ digits of a random number are $0$. There's a 50/50 chance that the number ends with $0$, but it becomes more and more unlikely as you increase the number of zeroes at the end. For $000$, the probability is already down to $\frac{1}{8}$, and for ten zeroes it's around one in one thousand. So it's very unlikely that any one random number has a lot of zeroes at the end.
So what's the maximum number of zeroes can you expect when you take $d$ random numbers? It's unlikely that one of the $d$ numbers ends in loads of zeroes, but the probabilities are so that it's likely that at least one has $\log{d}$ zeroes.