2025 day 11 blog post (#925)

merlinorg · web-flow · commit 9a376ac94869 · 2025-12-13T07:17:19.000-08:00
diff --git a/docs/2025/puzzles/day11.md b/docs/2025/puzzles/day11.md
@@ -2,14 +2,306 @@ import Solver from "../../../../../website/src/components/Solver.js"
 
 # Day 11: Reactor
 
+by [@merlinorg](https://github.com/merlinorg)
+
 ## Puzzle description
 
 https://adventofcode.com/2025/day/11
 
+## Solution Summary
+
+We use a simple recursive algorithm to count the number of paths in
+part 1. We add memoization to make part 2 tractable.
+
+### Part 1
+
+Part 1 challenges us to count the number of paths from a start node
+to an end node in a
+[directed acyclic graph](https://en.wikipedia.org/wiki/Directed_acyclic_graph) (DAG).
+
+#### Input Modeling
+
+We will model the input as an
+[adjacency list](https://en.wikipedia.org/wiki/Adjacency_list)
+from each device to those to which it is connected:
+
+```scala 3
+type AdjacencyList = Map[String, List[String]]
+```
+
+We can then add an extension method to `String` to parse the input.
+One line at a time, we parse the start device and its connections,
+and then split the connections into a list:
+
+```scala 3
+extension (self: String)
+  def parse: AdjacencyList = self.linesIterator
+    .collect:
+      case s"$a: $b" => a -> b.split(' ').toList
+    .toMap
+```
+
+For example:
+
+```scala 3
+val a = "aaa: you hhh".parse
+a: Map("aaa" -> List("you", "hhh"))
+```
+
+#### Counting Paths
+
+Graph traversal is typically done with either
+[breadth-first search](https://en.wikipedia.org/wiki/Breadth-first_search)
+or [depth-first search](https://en.wikipedia.org/wiki/Depth-first_search).
+Breadth-first is appropriate for finding shortest paths and other
+such minima, typically with a short-cut exit, but can have significant
+[space complexity](https://en.wikipedia.org/wiki/Space_complexity)
+(memory usage). In this case, we want to traverse all paths, so depth-first
+is more appropriate, having _O(log(N))_ space complexity.
+
+To count the paths we will use a recursive loop:
+
+- The number of paths from **B** to **B** is 1.
+- The number of paths from **A** to **B** is equal to the sum of the
+  number of paths from all nodes adjacent to **A** to **B**.
+
+We will implement this as an extension method on our `AdjacencyList`
+type:
+
+```scala 3
+extension (adjacency: AdjacencyList)
+  def countPaths(from: String, to: String): Long =
+    def loop(loc: String): Long =
+      if loc == to then 1L else adjacency(loc).map(loop).sum
+    loop(from)
+```
+
+We use `Long` as our result type; AoC is notorious for
+problem that overflow the size of an `Int`. This is not a
+tail-recursive loop because the recursive call is not the last statement.
+
+#### Solution
+
+With this framework in place, we can now easily solve part 1:
+
+```scala 3
+def part1(input: String): Long = input.parse.countPaths("you", "out")
+```
+
+### Part 2
+
+Part 2 asks us to again count the number of paths through a graph,
+but this time from a different start location, and counting only
+paths that pass through two particular nodes (in any order).
+
+One might be tempted to approach this by keeping track of all the
+nodes through which we have traversed using a `Set[String]`, and
+checking for the two named nodes when we reach the end; or, indeed,
+by simply keeping track of whether we have encountered the two
+specific nodes using two `Boolean`s.
+This extra work is unnecessary, however, if you make the following
+observation:
+
+- If there are **n** paths from **A** to **B**, and **m** paths
+  from **B** to **C**, then there are **n**×**m** paths from
+  **A** to **C**.
+
+To solve the problem, we then just need to consider two routes
+through the system:
+
+- **svr** →⋯→ **fft** →⋯→ **dac** →⋯→ **out**
+- **svr** →⋯→ **dac** →⋯→ **fft** →⋯→ **out**
+
+For each option, we count the number of paths between each pair
+of nodes, take the product of those intermediate results, and then
+sum the two results. (Because this graph is acyclic, there in fact
+cannot be both a path **fft** ⇢ **dac** and a path **dac** ⇢ **fft**,
+so one of these values will be zero.)
+
+#### Initial Solution
+
+Our initial solution is to sum the solutions through these these two routes;
+we split each route into pairs of nodes using `sliding(2)`, count
+the paths between these pairs, and take the product of those values:
+
+```scala 3
+def part2(input: String): Long =
+  val adjacency = input.parse.withDefaultValue(Nil)
+  List("svr-dac-fft-out", "svr-fft-dac-out")
+    .map: route =>
+      route
+        .split('-')
+        .sliding(2)
+        .map: pair =>
+          adjacency.countPaths(pair(0), pair(1))
+        .product
+    .sum
+```
+
+Note one small addition; we add a default value of `Nil`
+(the empty list) to our adjacency list, to easily accommodate
+routes that do not terminate at the **out** node.
+
+##### The Problem
+
+It rapidly becomes clear that this solution will not complete
+within any reasonable time. If **A** is connected to both **B** and **C**,
+and both of those are connected to both **D** and **E**, and so on,
+then the number of paths we have to traverse will grow exponentially.
+We did not encounter this in part 1 because the problem was constructed
+such that there was no exponential growth from that other starting point.
+
+#### Memoization
+
+The solution to this problem is
+[memoization](https://en.wikipedia.org/wiki/Memoization). In the problem
+described just above, when we are counting the number of paths from **B**, we
+have to count number of paths from **D** and **E**. When we are then looking
+at **C**, we have already calculated the results for **D** and **E**
+so we don't need to repeat those calculations. If we store every
+intermediate result, we can avoid the exponential growth.
+
+To apply this fix, we can use a `mutable.Map` as a memo to store these
+intermediate values inside our `countPaths` function. The logic we want
+basically looks like the following:
+
+```scala 3
+  def countPaths(from: String, to: String): Long =
+    val memo = mutable.Map.empty[String, Long]
+    def loop(loc: String): Long =
+      if memo.contains(loc) then memo(loc)
+      else
+        val count = if loc == to then 1L else adjacency(loc).map(loop).sum
+        memo.update(loc, count)
+        count
+    loop(from)
+```
+
+The `mutable.Map` class provides a `getOrElseUpdate` method that
+allows us to efficiently and cleanly express this:
+
+```scala 3
+  def countPaths(from: String, to: String): Long =
+    val memo                    = mutable.Map.empty[String, Long]
+    def loop(loc: String): Long =
+      lazy val count if loc == to then 1L else adjacency(loc).map(loop).sum
+      memo.getOrElseUpdate(loc, count)
+    loop(from)
+```
+
+We use a `lazy val` just for clarity here. The second parameter to
+`getOrElseUpdate` is a
+[by-name parameter](https://docs.scala-lang.org/tour/by-name-parameters.html),
+so the `count` computation will only be evaluated if the key is not already
+present in the map.
+
+With this enhancement, the computation completes in moments and the
+result is very large – almost a quadrillion in my case. Without
+memoization, the universe would be a distant memory before the initial
+solution would complete.
+
+## On Mutability
+
+As someone (although apparently not on the Internet) once said:
+
+> “Abjure mutability, embrace purity and constancy”
+
+Mutability is absolutely necessary in order to efficiently solve
+many problems. But, like children, it is best kept in a small
+room under the stairs.
+
+Oftentimes we seek to encapsulate mutability
+in library helper functions, to avoid sullying our day to day code.
+We can do just the same here.
+
+If you consider the `loop` function inside `countPaths`: It takes an
+input parameter of type `A` (`String` in this case) and produces an
+output of type `Result` (`Long` in this case), and is called with an
+initial value (`from`). As part of its operation, it needs to be able
+to recursively call itself.
+
+Consider now the following `memoized` function signature: It has the same two
+type parameters, `A` and `Result`, an initial value `init`,
+and a function from `A` to `Result` – but this function has a second
+parameter, a function from `A` to `Result` (the recursion call) –
+and it returns a value of type `Result`.
+
+```scala 3
+def memoized[A, Result](init: A)(f: (A, A => Result) => Result): Result
+```
+
+With something like this, we can rewrite `countPaths` just so:
+
+```scala 3
+  def countPaths(from: String, to: String): Long =
+    memoized[Result = Long](from): (loc, loop) =>
+      if loc == to then 1L else adjacency(loc).map(loop).sum
+```
+
+If you squint, this is now almost identical to the non-memoized
+code. Our shameful mutability is hidden.
+
+This uses [named type arguments](https://docs.scala-lang.org/scala3/reference/experimental/named-typeargs.html),
+an experimental language feature that lets us avoid specifying
+both types. In this case, the result type can't be automatically inferred.
+
+And what does `memoized` look like? It creates a `Function1`
+class that encapsulates the memo and allows the recursive function
+to be called, like the [Y combinator](https://en.wikipedia.org/wiki/Fixed-point_combinator#Y_combinator).
+
+```scala 3
+def memoized[A, Result](init: A)(f: (A, A => Result) => Result): Result =
+  class Memoize extends (A => B):
+    val memo = mutable.Map.empty[A, B]
+    def apply(a: A): B = memo.getOrElseUpdate(a, f(a, this))
+  Memoize()(init)
+```
+
+## Final Code
+
+```scala 3
+def part1(input: String): Long = input.parse.countPaths("you", "out")
+
+def part2(input: String): Long =
+  val adjacency = input.parse.withDefaultValue(Nil)
+  List("svr-dac-fft-out", "svr-fft-dac-out")
+    .map: route =>
+      route
+        .split('-')
+        .sliding(2)
+        .map: pair =>
+          adjacency.countPaths(pair(0), pair(1))
+        .product
+    .sum
+
+type AdjacencyList = Map[String, List[String]]
+
+import scala.language.experimental.namedTypeArguments
+
+extension (adjacency: AdjacencyList)
+  def countPaths(from: String, to: String): Long =
+    memoized[Result = Long](from): (loc, loop) =>
+      if loc == to then 1L else adjacency(loc).map(loop).sum
+
+def memoized[A, Result](init: A)(f: (A, A => Result) => Result): Result =
+  class Memoize extends (A => B):
+    val memo = mutable.Map.empty[A, B]
+    def apply(a: A): B = memo.getOrElseUpdate(a, f(a, this))
+  Memoize()(init)
+
+extension (self: String)
+  def parse: AdjacencyList = self.linesIterator
+    .collect:
+      case s"$a: $b" => a -> b.split(' ').toList
+    .toMap
+```
+
 ## Solutions from the community
 - [Solution](https://codeberg.org/nichobi/adventofcode/src/branch/main/2025/11/solution.scala) by [nichobi](https://nichobi.com)
 
 - [Solution](https://github.com/Philippus/adventofcode/blob/main/src/main/scala/adventofcode2025/Day11.scala) by [Philippus Baalman](https://github.com/philippus)
 
+- [Solution](https://github.com/merlinorg/advent-of-code/blob/main/src/main/scala/year2025/day11.scala) by [merlin](https://github.com/merlinorg/)
+
 Share your solution to the Scala community by editing this page.
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