diff --git a/.markdownlint.yaml b/.markdownlint.yaml
new file mode 100644
index 0000000..419c334
--- /dev/null
+++ b/.markdownlint.yaml
@@ -0,0 +1,3 @@
+# MD024/no-duplicate-heading/no-duplicate-header
+MD024:
+  siblings_only: true
diff --git a/config.yaml b/config.yaml
index c0927fc..b0aadf8 100644
--- a/config.yaml
+++ b/config.yaml
@@ -67,6 +67,7 @@ params:
   webmentions:
     login: belanyi.fr
     pingback: true
+  mathjax: true
 
 taxonomies:
   category: "categories"
diff --git a/content/posts/2024-08-02-reservoir-sampling/index.md b/content/posts/2024-08-02-reservoir-sampling/index.md
new file mode 100644
index 0000000..270c02c
--- /dev/null
+++ b/content/posts/2024-08-02-reservoir-sampling/index.md
@@ -0,0 +1,145 @@
+---
+title: "Reservoir Sampling"
+date: 2024-08-02T18:30:56+01:00
+draft: false # I don't care for draft mode, git has branches for that
+description: "Elegantly sampling a stream"
+tags:
+  - algorithms
+  - python
+categories:
+  - programming
+series:
+  - Cool algorithms
+favorite: false
+disable_feed: false
+mathjax: true
+---
+
+[_Reservoir Sampling_][reservoir] is an [online][online], probabilistic
+algorithm to uniformly sample $k$ random elements out of a stream of values.
+
+It's a particularly elegant and small algorithm, only requiring $\Theta(k)$
+amount of space and a single pass through the stream.
+
+[reservoir]: https://en.wikipedia.org/wiki/Reservoir_sampling
+[online]: https://en.wikipedia.org/wiki/Online_algorithm
+
+<!--more-->
+
+## Sampling one element
+
+As an introduction, we'll first focus on fairly sampling one element from the
+stream.
+
+```python
+def sample_one[T](stream: Iterable[T]) -> T:
+    stream_iter = iter(stream)
+    # Sample the first element
+    res = next(stream_iter)
+    for i, val in enumerate(stream_iter, start=1):
+        j = random.randint(0, i)
+        # Replace the sampled element with probability 1/(i + 1)
+        if j == 0:
+            res = val
+    # Return the randomly sampled element
+    return res
+```
+
+### Proof
+
+Let's now prove that this algorithm leads to a fair sampling of the stream.
+
+We'll be doing proof by induction.
+
+#### Hypothesis $H_N$
+
+After iterating through the first $N$ items in the stream,
+each of them has had an equal $\frac{1}{N}$ probability of being selected as
+`res`.
+
+#### Base Case $H_1$
+
+We can trivially observe that the first element is always assigned to `res`,
+$\frac{1}{1} = 1$, the hypothesis has been verified.
+
+#### Inductive Case
+
+For a given $N$, let us assume that $H_N$ holds. Let us now look at the events
+of loop iteration where `i = N` (i.e: observation of the $N + 1$-th item in the
+stream).
+
+`j = random.randint(0, i)` uniformly selects a value in the range $[0, i]$,
+a.k.a $[0, N]$. We then have two cases:
+
+* `j == 0`, with probability $\frac{1}{N + 1}$: we select `val` as the new
+reservoir element `res`.
+
+* `j != 0`, with probability $\frac{N}{N + 1}$: we keep the previous value of
+`res`. By $H_N$, any of the first $N$ elements had a $\frac{1}{N}$ probability
+of being `res` before at the start of the loop, each element now has a
+probability $\frac{1}{N} \cdot \frac{N}{N + 1} = \frac{1}{N + 1}$ of being the
+element.
+
+And thus, we have proven $H_{N + 1}$ at the end of the loop.
+
+## Sampling $k$ element
+
+The code for sampling $k$ elements is very similar to the one-element case.
+
+```python
+def sample[T](stream: Iterable[T], k: int = 1) -> list[T]:
+    stream_iter = iter(stream)
+    # Retain the first 'k' elements in the reservoir
+    res = list(itertools.islice(stream_iter, k))
+    for i, val in enumerate(stream_iter, start=k):
+        j = random.randint(0, i)
+        # Replace one element at random with probability k/(i + 1)
+        if j < k:
+            res[j] = val
+    # Return 'k' randomly sampled elements
+    return res
+```
+
+### Proof
+
+Let us once again do a proof by induction, assuming the stream contains at least
+$k$ items.
+
+#### Hypothesis $H_N$
+
+After iterating through the first $N$ items in the stream, each of them has had
+an equal $\frac{k}{N}$ probability of being sampled from the stream.
+
+#### Base Case $H_k$
+
+We can trivially observe that the first $k$ element are sampled at the start of
+the algorithm, $\frac{k}{k} = 1$, the hypothesis has been verified.
+
+#### Inductive Case
+
+For a given $N$, let us assume that $H_N$ holds. Let us now look at the events
+of the loop iteration where `i = N`, in order to prove $H_{N + 1}$.
+
+`j = random.randint(0, i)` uniformly selects a value in the range $[0, i]$,
+a.k.a $[0, N]$. We then have three cases:
+
+* `j >= k`, with probability $1 - \frac{k}{N + 1}$: we do not modify the
+sampled reservoir at all.
+
+* `j < k`, with probability $\frac{k}{N + 1}$: we sample the new element to
+replace the `j`-th element of the reservoir. Therefore for any element
+$e \in [0, k[$ we can either have:
+  * $j = e$: the element _is_ replaced, probability $\frac{1}{k}$.
+  * $j \neq e$: the element is _not_ replaced, probability $\frac{k - 1}{k}$.
+
+We can now compute the probability that a previously sampled element is kept in
+the reservoir:
+$1 - \frac{k}{N + 1} + \frac{k}{N + 1} \cdot \frac{k - 1}{k} = \frac{N}{N + 1}$.
+
+By $H_N$, any of the first $N$ elements had a $\frac{k}{N}$ probability
+of being sampled before at the start of the loop, each element now has a
+probability $\frac{k}{N} \cdot \frac{N}{N + 1} = \frac{k}{N + 1}$ of being the
+element.
+
+We have now proven that all elements have a probability $\frac{k}{N + 1}$ of
+being sampled at the end of the loop, therefore $H_{N + 1}$ has been verified.