diff --git a/docs/2025/puzzles/day10.md b/docs/2025/puzzles/day10.md
index 9bc04ba37..751d78d07 100644
--- a/docs/2025/puzzles/day10.md
+++ b/docs/2025/puzzles/day10.md
@@ -6,6 +6,396 @@ import Solver from "../../../../../website/src/components/Solver.js"
 
 https://adventofcode.com/2025/day/10
 
+## Solution Summary
+
+The input includes multiple lines, each of them can be handled independently, the result is the sum from the output of each scenario.
+
+For part 1, I applied the [Breadth-first Search](https://en.wikipedia.org/wiki/Breadth-first_search) algorithm (BFS), BFS allows finding the shortest-path from a state to another when every transition has the same cost, this was quick to implement and produced the correct output.
+
+For part 2, BFS wasn't adequate due to the number of potential states, there is an alternative Reduce-and-conquer approach (a variation of [Divide-and-conquer](https://en.wikipedia.org/wiki/Divide-and-conquer_algorithm)) which is fast enough.
+
+
+## Parsing the input
+
+Let's represent the input scenario as a case class:
+
+```scala
+case class InputCase(target: String, buttons: IndexedSeq[Set[Int]], joltage: IndexedSeq[Int])
+```
+
+For parsing the input, I'll mix pattern matching to extract the input pieces combined with `String` utilities (`replace` and `split`):
+
+```scala
+def parseInput(input: String): IndexedSeq[InputCase] = {
+  input.split("\n").collect { case s"[$lightDiagram]$buttonsStr{$joltageStr}" =>
+    val buttons = buttonsStr.split(" ").filter(_.nonEmpty)
+      .map(_.replace("(", "").replace(")", "").split(",").map(_.toInt).toSet)
+      .toIndexedSeq
+    val joltage = joltageStr.split(",").map(_.toInt).toIndexedSeq
+    InputCase(target = lightDiagram, buttons = buttons, joltage = joltage)
+  }
+}
+```
+
+## Part 1
+
+The BFS algorithm is adequate because in every step we can take a button to produce a new state, we have to avoid visiting the same state more than once.
+
+For example, given this input `[####] (0) (3) (0,1) (1,2)`:
+
+- The target state is `####` (all lights turned on).
+- The initial state is `....` (all lights off).
+- The buttons allow toggling light `0` or `3`, or, lights `0,1` or lights `1,2`.
+
+I have created a few helper functions:
+
+- `flip` receives the state from a light and toggles its value, `.` to `#`, and, `#` to `.`.
+- `transform` receives the lights state and toggles the lights covered by a button.
+
+```scala
+def flip(value: Char): Char = if (value == '#') '.' else '#'
+def transform(source: String, button: Set[Int]): String = {
+  source.toList.zipWithIndex.map { case (value, index) =>
+    if (button.contains(index)) flip(value)
+    else value
+  }.mkString("")
+}
+```
+
+The way to implement a BFS requires a `Queue`, a `State`, and, a way to track the visited values, for what I used a `Set`.
+
+The state for the BFS can de defined with the current light's state and the number of steps required to get there:
+
+```scala
+case class State(value: String, steps: Int)
+```
+
+Naturally, the initial state would have `value` to be only dots (`.`) with `0` steps.
+
+Given everything described until now, it is only required to define the BFS process, for this I used a tail-recursive function that does the following:
+
+- When the queue is empty, there is no solution.
+- When the current state has reached the goal, return the number of steps.
+- Otherwise, find the new states that can be visited, push them to the queue and repeat.
+
+```scala
+@scala.annotation.tailrec
+def loop(queue: Queue[State], touched: Set[String], inputCase: InputCase): Option[Int] = {
+  queue.dequeueOption match {
+    case None => None
+    case Some((current, _)) if current.value == inputCase.target => Some(current.steps)
+    case Some((current, nextQueue)) =>
+      val newValues = inputCase.buttons
+        .map(button => transform(current.value, button))
+        .filterNot(touched.contains)
+
+      val newTouched = touched ++ newValues
+      val newStates = newValues.map { newValue => State(newValue, current.steps + 1) }
+      val newQueue = nextQueue.enqueueAll(newStates)
+
+      loop(newQueue, newTouched, inputCase)
+  }
+}
+```
+
+The final piece is just wiring the existing functionality to cover each scenario and sum the results:
+
+```scala
+def part1(input: String): Unit = {
+  import scala.collection.immutable.Queue
+
+  def flip(value: Char): Char = ???
+  def transform(source: String, button: Set[Int]): String = ???
+
+  case class State(value: String, steps: Int)
+
+  @scala.annotation.tailrec
+  def loop(queue: Queue[State], touched: Set[String], inputCase: InputCase): Option[Int] = ???
+
+  def resolveCase(inputCase: InputCase): Int = {
+    val initial = inputCase.target.map(_ => '.')
+    loop(Queue(State(initial, 0)), Set(initial), inputCase)
+      .getOrElse(throw new RuntimeException("Answer not found"))
+  }
+
+  val total = parseInput(input).map(resolveCase).sum
+  println(total)
+}
+```
+
+
+There are potential alternatives to deal with this but given the input size, they are not necessary, for example:
+
+- Use a `BitSet` data structure instead of the `String`.
+- Applying the same button twice does not make sense because the effect gets reverted which simplifies the problem to either use a button or not.
+
+
+## Part 2
+
+My initial reaction was that resolving this might be trivial to do by reusing the BFS implementation by changing a few operations:
+
+- State is now the joltage.
+- Instead of going from the empty state to the goal, let's go from the given joltage to `0` values.
+- Filter out invalid states (`joltage[k] < 0`)
+
+```diff
+diff --git a/Main.scala b/Main.scala
+index 21ccc48..b98c9e4 100644
+--- a/Main.scala
++++ b/Main.scala
+@@ -19,25 +19,26 @@ object Main extends App {
+   def part1(input: String): Unit = {
+     import scala.collection.immutable.Queue
+
+-    def flip(value: Char): Char = if (value == '#') '.' else '#'
+-    def transform(source: String, button: Set[Int]): String = {
+-      source.toList.zipWithIndex.map { case (value, index) =>
+-        if (button.contains(index)) flip(value)
++    def isValid(source: IndexedSeq[Int]): Boolean = source.forall(_ >= 0)
++    def transform(source: IndexedSeq[Int], button: Set[Int]): IndexedSeq[Int] = {
++      source.zipWithIndex.map { case (value, index) =>
++        if (button.contains(index)) value - 1
+         else value
+-      }.mkString("")
++      }
+     }
+
+-    case class State(value: String, steps: Int)
++    case class State(value: IndexedSeq[Int], steps: Int)
+
+     @scala.annotation.tailrec
+-    def loop(queue: Queue[State], touched: Set[String], inputCase: InputCase): Option[Int] = {
++    def loop(queue: Queue[State], touched: Set[IndexedSeq[Int]], inputCase: InputCase): Option[Int] = {
+       queue.dequeueOption match {
+         case None => None
+-        case Some((current, _)) if current.value == inputCase.target => Some(current.steps)
++        case Some((current, _)) if current.value.forall(_ == 0) => Some(current.steps)
+         case Some((current, nextQueue)) =>
+           val newValues = inputCase.buttons
+             .map(button => transform(current.value, button))
+             .filterNot(touched.contains)
++            .filter(isValid)
+
+           val newTouched = touched ++ newValues
+           val newStates = newValues.map { newValue => State(newValue, current.steps + 1) }
+@@ -48,7 +49,7 @@ object Main extends App {
+     }
+
+     def resolveCase(inputCase: InputCase): Int = {
+-      val initial = inputCase.target.map(_ => '.')
++      val initial = inputCase.joltage
+       loop(Queue(State(initial, 0)), Set(initial), inputCase)
+         .getOrElse(throw new RuntimeException("Answer not found"))
+     }
+```
+
+This resolved the example input but it was too slow with the actual test scenarios, I tried a few heuristics to trim unnecessary paths, I also tried `DFS` with prunning, `A*`, and, I was close to implement a [bidirectional BFS](https://en.wikipedia.org/wiki/Bidirectional_search).
+
+Eventually, I got an idea from [reddit](https://old.reddit.com/r/adventofcode/comments/1pk87hl/2025_day_10_part_2_bifurcate_your_way_to_victory/) about using the Reduce-and-conquer approach instead, it goes like this; if the path from `0` to `T` takes `N` steps, the path from `0` to `2T` takes `2N` steps, with this:
+
+- We can try to get convert the `joltage` into even numbers.
+- Resolve `joltage / 2`.
+- The answer for the current step would be `f(joltage / 2)*2 + currentCost`.
+- There are many overlaping sub-problems but we can use a cache to avoid unnecessary recomputation.
+
+**DISCLAIMER** I have no proof that this handles every possible scenario but with the test cases I prepared, the BFS result leads to the same from this, and, the result has been accepted by Advent of Code.
+
+There is an important detail to resolve part 1, applying a button more than once does not make sense because it invalidates the previous action, in the case of part 2 which uses integer values, we can focus on the value parity instead, for this, the same statement holds, applying the same button twice invalidates the previous action.
+
+For example, applying the button `0, 3` to joltages `3, 4, 5, 6` lead to `2, 4, 4, 6` (all even) but applying the same button again will revert the parity back.
+
+When all joltages are even (like `2, 4, 4, 6`), we can divide each value by 2, leaving us with `1, 2, 2, 3`, then, we can apply the same process recursively.
+
+Having said this, let's generate all possible transitions from the available buttons, this is, all [subsets](https://en.wikipedia.org/wiki/Subset) from the given buttons, leveraging the powerful Scala stdlib, we can call the `Set#subsets` function:
+
+```scala
+scala> List(1, 2, 3).toSet.subsets.foreach(println)
+Set()
+Set(1)
+Set(2)
+Set(3)
+Set(1, 2)
+Set(1, 3)
+Set(2, 3)
+Set(1, 2, 3)
+```
+
+**NOTE**: The empty subset is important because that allow us to take the existing joltages which could be already divisible by 2.
+
+The code to generate the moves generates the transformation vector for every button, for example, in the case of a button `0,2`, with `4` joltages, the transformation vector becomes `1,0,1,0`, this is the delta to apply to the joltages with the cost equal to the number of buttons required to get this combination:
+
+```scala
+val allMoves = inputCase.buttons.toSet.subsets().toList.map { set =>
+  set.foldLeft(IndexedSeq.fill(inputCase.joltage.size)(0) -> 0) { case ((acc, cost), button) =>
+    val newVector = acc.zipWithIndex.map { case (x, index) =>
+      if (button.contains(index)) x + 1
+      else x
+    }
+    newVector -> (cost + 1)
+  }
+}
+```
+
+Essentially, `allMoves` have the transformation options associated with the cost to apply them.
+
+We'll require a cache to avoid recomputing the same value twice:
+
+```scala
+// min moves to switch the given joltages to 0
+var cache = Map.empty[IndexedSeq[Int], Option[Int]]
+```
+
+At last, the actual function to compute the cost required to transform the given joltages into 0s:
+
+- When joltages is composed by only 0s, we have the answer (no cost).
+- When the answer for the given joltages is already cached, reuse it.
+- Otherwise, apply any transformations that lead to even-joltages, resolve the smaller problem and compute the answer, out of those options, keep the minimum cost and put it into the cache.
+
+```scala
+def f(joltage: IndexedSeq[Int]): Option[Int] = {
+  if (joltage.forall(_ == 0)) Option(0)
+  else if (cache.contains(joltage)) cache(joltage)
+  else {
+    val choices = allMoves.flatMap { case (delta, cost) =>
+      val newJoltage = joltage.zip(delta).map( (goal, diff) => goal - diff)
+
+      if (newJoltage.forall(_ >= 0) && newJoltage.forall(_ % 2 == 0))
+        f(newJoltage.map(_ / 2)).map(res => cost + (res * 2))
+      else
+        None
+    }
+
+    val best = choices.minOption
+    cache = cache + (joltage -> best)
+
+    best
+  }
+}
+```
+
+Now, the whole code becomes:
+
+```scala
+def part2(input: String): Unit = {
+  def resolveCase(inputCase: InputCase): Int = {
+    val allMoves = ???
+
+    // min moves to switch the given joltages to 0
+    var cache = Map.empty[IndexedSeq[Int], Option[Int]]
+
+    def f(joltage: IndexedSeq[Int]): Option[Int] = ???
+
+    f(inputCase.joltage).getOrElse(throw new RuntimeException("Answer not found"));
+  }
+
+  val total = parseInput(input).map(resolveCase).sum
+  println(total)
+}
+```
+
+
+## Final code
+
+```scala
+case class InputCase(target: String, buttons: IndexedSeq[Set[Int]], joltage: IndexedSeq[Int])
+
+def parseInput(input: String): IndexedSeq[InputCase] = {
+  input.split("\n").collect { case s"[$lightDiagram]$buttonsStr{$joltageStr}" =>
+    val buttons = buttonsStr.split(" ").filter(_.nonEmpty)
+      .map(_.replace("(", "").replace(")", "").split(",").map(_.toInt).toSet)
+      .toIndexedSeq
+    val joltage = joltageStr.split(",").map(_.toInt).toIndexedSeq
+    InputCase(target = lightDiagram, buttons = buttons, joltage = joltage)
+  }
+}
+
+def part1(input: String): Unit = {
+  import scala.collection.immutable.Queue
+
+  def flip(value: Char): Char = if (value == '#') '.' else '#'
+  def transform(source: String, button: Set[Int]): String = {
+    source.toList.zipWithIndex.map { case (value, index) =>
+      if (button.contains(index)) flip(value)
+      else value
+    }.mkString("")
+  }
+
+  case class State(value: String, steps: Int)
+
+  @scala.annotation.tailrec
+  def loop(queue: Queue[State], touched: Set[String], inputCase: InputCase): Option[Int] = {
+    queue.dequeueOption match {
+      case None => None
+      case Some((current, _)) if current.value == inputCase.target => Some(current.steps)
+      case Some((current, nextQueue)) =>
+        val newValues = inputCase.buttons
+          .map(button => transform(current.value, button))
+          .filterNot(touched.contains)
+
+        val newTouched = touched ++ newValues
+        val newStates = newValues.map { newValue => State(newValue, current.steps + 1) }
+        val newQueue = nextQueue.enqueueAll(newStates)
+
+        loop(newQueue, newTouched, inputCase)
+    }
+  }
+
+  def resolveCase(inputCase: InputCase): Int = {
+    val initial = inputCase.target.map(_ => '.')
+    loop(Queue(State(initial, 0)), Set(initial), inputCase)
+      .getOrElse(throw new RuntimeException("Answer not found"))
+  }
+
+  val total = parseInput(input).map(resolveCase).sum
+  println(total)
+}
+
+def part2(input: String): Unit = {
+  def resolveCase(inputCase: InputCase): Int = {
+    val allMoves = inputCase.buttons.toSet.subsets().toList.map { set =>
+      set.foldLeft(IndexedSeq.fill(inputCase.joltage.size)(0) -> 0) { case ((acc, cost), button) =>
+        val newVector = acc.zipWithIndex.map { case (x, index) =>
+          if (button.contains(index)) x + 1
+          else x
+        }
+        newVector -> (cost + 1)
+      }
+    }
+
+    // min moves to switch the given joltages to 0
+    var cache = Map.empty[IndexedSeq[Int], Option[Int]]
+
+    def f(joltage: IndexedSeq[Int]): Option[Int] = {
+      if (joltage.forall(_ == 0)) Option(0)
+      else if (cache.contains(joltage)) cache(joltage)
+      else {
+        val choices = allMoves.flatMap { case (delta, cost) =>
+          val newJoltage = joltage.zip(delta).map( (goal, diff) => goal - diff)
+
+          if (newJoltage.forall(_ >= 0) && newJoltage.forall(_ % 2 == 0))
+            f(newJoltage.map(_ / 2)).map(res => cost + (res * 2))
+          else
+            None
+        }
+
+        val best = choices.minOption
+        cache = cache + (joltage -> best)
+
+        best
+      }
+    }
+
+    f(inputCase.joltage).getOrElse(throw new RuntimeException("Answer not found"));
+  }
+
+  val total = parseInput(input).map(resolveCase).sum
+  println(total)
+}
+
+val input = readInput()
+part1(input)
+part2(input)
+```
 ## Solutions from the community
 
 - [Solution](https://github.com/merlinorg/advent-of-code/blob/main/src/main/scala/year2025/day10.scala) (and