diff --git a/docs/2025/puzzles/day08.md b/docs/2025/puzzles/day08.md
index 494818045..eb023414c 100644
--- a/docs/2025/puzzles/day08.md
+++ b/docs/2025/puzzles/day08.md
@@ -2,10 +2,126 @@ import Solver from "../../../../../website/src/components/Solver.js"
 
 # Day 8: Playground
 
+by [@mbovel](https://github.com/mbovel)
+
 ## Puzzle description
 
 https://adventofcode.com/2025/day/8
 
+## Data Model
+
+To solve this puzzle, we use a class to represent junction boxes:
+
+```scala
+/** A junction box in 3D space with an associated circuit ID. */
+case class Box(val x: Long, val y: Long, val z: Long, var circuit: Int):
+  def distanceSquare(other: Box): Long =
+    (x - other.x) * (x - other.x) + (y - other.y) * (y - other.y) + (z - other.z) * (z - other.z)
+```
+
+Each `Box` has:
+- Three coordinates (`x`, `y`, `z`) representing its position in 3D space
+- A mutable `circuit` field to track which circuit the box belongs to (each circuit is identified by a distinct integer)
+- A `distanceSquare` method that computes the squared Euclidean distance to another box (we use squared distance to avoid computing square roots, since we only need to compare distances)
+
+## Data Loading
+
+The following functions parse the input into a sequence of boxes and compute all unique pairs sorted by distance:
+
+```scala
+/** Parses comma-separated coordinates from the given `line` into a `Box`
+  * with the given `circuit` ID.
+  */
+def parseBox(line: String, circuit: Int): Box =
+  val parts = line.split(",")
+  Box(parts(0).toLong, parts(1).toLong, parts(2).toLong, circuit)
+
+/** Parses the input, returning a sequence of `Box`es and all unique pairs
+  * of boxes sorted by distance.
+  */
+def load(input: String): (Seq[Box], Seq[(Box, Box)]) =
+  val lines = input.linesIterator.filter(_.nonEmpty)
+  val boxes = lines.zipWithIndex.map(parseBox).toSeq
+  val pairsByDistance = boxes.pairs.toSeq.sortBy((b1, b2) => b1.distanceSquare(b2))
+  (boxes, pairsByDistance)
+```
+
+The `pairs` extension method generates all unique pairs from a sequence:
+
+```scala
+extension [T](self: Seq[T])
+  /** Generates all unique pairs (combinations of 2) from the sequence. */
+  def pairs: Iterator[(T, T)] =
+    self.combinations(2).map(pair => (pair(0), pair(1)))
+```
+
+## Part 1
+
+For Part 1, we process the 1000 closest pairs of boxes and merge their circuits. The algorithm iterates through pairs in order of increasing distance; when two boxes belong to different circuits, we merge them into one. Finally, we find the three largest circuits and return the product of their sizes.
+
+```scala
+def part1(input: String): Int =
+  val (boxes, pairsByDistance) = load(input)
+  for (b1, b2) <- pairsByDistance.take(1000) if b1.circuit != b2.circuit do
+    merge(b1.circuit, b2.circuit, boxes)
+  val sizes = boxes.groupBy(_.circuit).values.map(_.size).toSeq.sortBy(-_)
+  sizes.take(3).product
+```
+
+The `merge` function updates all boxes in one circuit to belong to another:
+
+```scala
+/** Sets all boxes with circuit `c2` to circuit `c1`. */
+def merge(c1: Int, c2: Int, boxes: Seq[Box]): Unit =
+  for b <- boxes if b.circuit == c2 do b.circuit = c1
+```
+
+
+## Part 2
+
+For Part 2, we continue merging circuits until only one remains. We track the number of distinct circuits and return the product of the x-coordinates of the two boxes in the final merge.
+
+```scala
+def part2(input: String): Long =
+  val (boxes, pairsByDistance) = load(input)
+  var n = boxes.length
+  boundary:
+    for (b1, b2) <- pairsByDistance if b1.circuit != b2.circuit do
+      merge(b1.circuit, b2.circuit, boxes)
+      n -= 1
+      if n <= 1 then
+        break(b1.x * b2.x)
+    throw Exception("Should not reach here")
+```
+
+[`boundary` and `break`](https://www.scala-lang.org/api/3.x/scala/util/boundary$.html) provide a way to exit the loop early and return a value when only one circuit remains.
+
+## Potential Optimizations
+
+On my machine, both parts run in under two seconds, which is acceptable for this puzzle. Still, several optimizations are possible.
+
+What we implemented is essentially the [Euclidean minimum spanning tree (EMST)](https://en.wikipedia.org/wiki/Euclidean_minimum_spanning_tree) computed via [Kruskal’s algorithm](https://en.wikipedia.org/wiki/Kruskal%27s_algorithm), after generating all pairwise distances.
+
+This algorithm is normally paired with a [union–find](https://en.wikipedia.org/wiki/Disjoint-set_data_structure) structure to maintain connected components efficiently. Using it would speed up our current `merge` step, which is $\mathcal{O}(n)$ per merge, and reduce it to near-constant time.
+
+We could also improve how we find the $k$ closest pairs. Computing all pairs and sorting them has $\mathcal{O}(n^2 \log n)$ complexity. A spatial index such as a [k-d tree](https://en.wikipedia.org/wiki/K-d_tree) would avoid generating all pairs and, in the average case, remove the quadratic blow-up. Another option is to restrict candidates to a geometric graph such as the [relative neighborhood graph](https://en.wikipedia.org/wiki/Relative_neighborhood_graph) or the 3D [Delaunay triangulation](https://en.wikipedia.org/wiki/Delaunay_triangulation), both of which contain the EMST and are much sparser than the complete graph.
+
+## Final Code
+
+See the complete code on [GitHub](https://github.com/scalacenter/scala-advent-of-code/blob/main/2025/src/day08.scala).
+
+## Run it in the browser
+
+Thanks to the [Scala.js](https://www.scala-js.org/) build, you can also experiment with this code directly in the browser.
+
+### Part 1
+
+<Solver puzzle="day08-part1" year="2025"/>
+
+### Part 2
+
+<Solver puzzle="day08-part2" year="2025"/>
+
 ## Solutions from the community
 
 Share your solution to the Scala community by editing this page.