146 lines
4.7 KiB
Markdown
146 lines
4.7 KiB
Markdown
---
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title: "Reservoir Sampling"
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date: 2024-08-02T18:30:56+01:00
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draft: false # I don't care for draft mode, git has branches for that
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description: "Elegantly sampling a stream"
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tags:
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- algorithms
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- python
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categories:
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- programming
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series:
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- Cool algorithms
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favorite: false
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disable_feed: false
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mathjax: true
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---
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[_Reservoir Sampling_][reservoir] is an [online][online], probabilistic
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algorithm to uniformly sample $k$ random elements out of a stream of values.
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It's a particularly elegant and small algorithm, only requiring $\Theta(k)$
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amount of space and a single pass through the stream.
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[reservoir]: https://en.wikipedia.org/wiki/Reservoir_sampling
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[online]: https://en.wikipedia.org/wiki/Online_algorithm
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<!--more-->
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## Sampling one element
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As an introduction, we'll first focus on fairly sampling one element from the
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stream.
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```python
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def sample_one[T](stream: Iterable[T]) -> T:
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stream_iter = iter(stream)
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# Sample the first element
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res = next(stream_iter)
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for i, val in enumerate(stream_iter, start=1):
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j = random.randint(0, i)
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# Replace the sampled element with probability 1/(i + 1)
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if j == 0:
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res = val
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# Return the randomly sampled element
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return res
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```
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### Proof
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Let's now prove that this algorithm leads to a fair sampling of the stream.
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We'll be doing proof by induction.
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#### Hypothesis $H_N$
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After iterating through the first $N$ items in the stream,
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each of them has had an equal $\frac{1}{N}$ probability of being selected as
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`res`.
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#### Base Case $H_1$
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We can trivially observe that the first element is always assigned to `res`,
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$\frac{1}{1} = 1$, the hypothesis has been verified.
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#### Inductive Case
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For a given $N$, let us assume that $H_N$ holds. Let us now look at the events
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of loop iteration where `i = N` (i.e: observation of the $N + 1$-th item in the
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stream).
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`j = random.randint(0, i)` uniformly selects a value in the range $[0, i]$,
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a.k.a $[0, N]$. We then have two cases:
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* `j == 0`, with probability $\frac{1}{N + 1}$: we select `val` as the new
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reservoir element `res`.
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* `j != 0`, with probability $\frac{N}{N + 1}$: we keep the previous value of
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`res`. By $H_N$, any of the first $N$ elements had a $\frac{1}{N}$ probability
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of being `res` before at the start of the loop, each element now has a
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probability $\frac{1}{N} \cdot \frac{N}{N + 1} = \frac{1}{N + 1}$ of being the
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element.
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And thus, we have proven $H_{N + 1}$ at the end of the loop.
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## Sampling $k$ element
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The code for sampling $k$ elements is very similar to the one-element case.
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```python
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def sample[T](stream: Iterable[T], k: int = 1) -> list[T]:
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stream_iter = iter(stream)
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# Retain the first 'k' elements in the reservoir
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res = list(itertools.islice(stream_iter, k))
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for i, val in enumerate(stream_iter, start=k):
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j = random.randint(0, i)
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# Replace one element at random with probability k/(i + 1)
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if j < k:
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res[j] = val
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# Return 'k' randomly sampled elements
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return res
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```
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### Proof
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Let us once again do a proof by induction, assuming the stream contains at least
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$k$ items.
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#### Hypothesis $H_N$
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After iterating through the first $N$ items in the stream, each of them has had
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an equal $\frac{k}{N}$ probability of being sampled from the stream.
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#### Base Case $H_k$
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We can trivially observe that the first $k$ element are sampled at the start of
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the algorithm, $\frac{k}{k} = 1$, the hypothesis has been verified.
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#### Inductive Case
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For a given $N$, let us assume that $H_N$ holds. Let us now look at the events
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of the loop iteration where `i = N`, in order to prove $H_{N + 1}$.
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`j = random.randint(0, i)` uniformly selects a value in the range $[0, i]$,
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a.k.a $[0, N]$. We then have three cases:
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* `j >= k`, with probability $1 - \frac{k}{N + 1}$: we do not modify the
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sampled reservoir at all.
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* `j < k`, with probability $\frac{k}{N + 1}$: we sample the new element to
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replace the `j`-th element of the reservoir. Therefore for any element
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$e \in [0, k[$ we can either have:
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* $j = e$: the element _is_ replaced, probability $\frac{1}{k}$.
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* $j \neq e$: the element is _not_ replaced, probability $\frac{k - 1}{k}$.
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We can now compute the probability that a previously sampled element is kept in
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the reservoir:
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$1 - \frac{k}{N + 1} + \frac{k}{N + 1} \cdot \frac{k - 1}{k} = \frac{N}{N + 1}$.
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By $H_N$, any of the first $N$ elements had a $\frac{k}{N}$ probability
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of being sampled before at the start of the loop, each element now has a
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probability $\frac{k}{N} \cdot \frac{N}{N + 1} = \frac{k}{N + 1}$ of being the
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element.
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We have now proven that all elements have a probability $\frac{k}{N + 1}$ of
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being sampled at the end of the loop, therefore $H_{N + 1}$ has been verified.
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