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Scala Advent of Code 2021 byScala Center

Day 1Day 2Day 3Day 4Day 5Day 6Day 7Day 8Day 9Day 10Day 11Day 12Day 13Day 14Day 15Day 16Day 17Day 18Day 19Day 20Day 21Day 22Day 23Day 24Day 25
Credit to https://github.com/OlegIlyenko/scala-icon

Learn Scala 3

A simpler, safer and more concise version of Scala, the famous object-oriented and functional programming language.

Solve Advent of Code puzzles

Challenge your programming skills by solving Advent of Code puzzles.

Share with the community

Get or give support to the community. Share your solutions with the community.

- + \ No newline at end of file diff --git a/2022/index.html b/2022/index.html index 5df1eb88d..ce97f7ca2 100644 --- a/2022/index.html +++ b/2022/index.html @@ -5,13 +5,13 @@ Scala Center Advent of Code | Scala Center Advent of Code - +
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Scala Advent of Code 2022 byScala Center

Day 1Day 2Day 3Day 4Day 5Day 6Day 7Day 8Day 9Day 10Day 11Day 12Day 13Day 14Day 15Day 16Day 17Day 18Day 19Day 20Day 21Day 22Day 23Day 24Day 25
Credit to https://github.com/OlegIlyenko/scala-icon

Learn Scala 3

A simpler, safer and more concise version of Scala, the famous object-oriented and functional programming language.

Solve Advent of Code puzzles

Challenge your programming skills by solving Advent of Code puzzles.

Share with the community

Get or give support to the community. Share your solutions with the community.

- + \ No newline at end of file diff --git a/2022/puzzles/day01/index.html b/2022/puzzles/day01/index.html index e1bee4b08..f331a45c5 100644 --- a/2022/puzzles/day01/index.html +++ b/2022/puzzles/day01/index.html @@ -5,14 +5,14 @@ Day 1: Calorie Counting | Scala Center Advent of Code - +
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Day 1: Calorie Counting

by @bishabosha

Puzzle description​

https://adventofcode.com/2022/day/1

Solution Summary​

First transform the input into a List of Inventory, each Inventory is a list of Int, representing the calorie count of an item in the inventory, this is handled in scanInventories.

Part 1​

Given the List of Inventory, we must first find the total calorie count of each inventory.

For a single Inventory, we do this using the sum method on its items property (found in the List class). e.g. inventory.items.sum.

Then use the map method on the List class, to transform each Inventory to its total calorie count with an anonymous function.

Then sort the resulting list of total calorie counts in descending order, this is provided by scala.math.Ordering.Int.reverse.

The maxInventories method handles the above, returning the top n total calorie counts.

For part 1, use maxInventories with n == 1 to create a singleton list of the largest calorie count.

Part 2​

As in part 1, construct the list of sorted total calorie counts with maxInventories. But instead, we need the first 3 elements. We then need to sum the resulting list.

Final Code​

import scala.math.Ordering

def part1(input: String): Int =
  maxInventories(scanInventories(input), 1).head

def part2(input: String): Int =
  maxInventories(scanInventories(input), 3).sum

case class Inventory(items: List[Int])

def scanInventories(input: String): List[Inventory] =
  val inventories = List.newBuilder[Inventory]
  var items = List.newBuilder[Int]
  for line <- input.linesIterator do
    if line.isEmpty then
      inventories += Inventory(items.result())
      items = List.newBuilder
    else items += line.toInt
  inventories.result()

def maxInventories(inventories: List[Inventory], n: Int): List[Int] =
  inventories
    .map(inventory => inventory.items.sum)
    .sorted(using Ordering.Int.reverse)
    .take(n)

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Part 1​

Part 2​

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- + \ No newline at end of file diff --git a/2022/puzzles/day02/index.html b/2022/puzzles/day02/index.html index 6dcfb4896..3d47a8979 100644 --- a/2022/puzzles/day02/index.html +++ b/2022/puzzles/day02/index.html @@ -5,13 +5,13 @@ Day 2: Rock Paper Scissors | Scala Center Advent of Code - +
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Day 2: Rock Paper Scissors

by @bishabosha

Puzzle description​

https://adventofcode.com/2022/day/2

Final Code​

import Position.*

def part1(input: String): Int =
  scores(input, pickPosition).sum

def part2(input: String): Int =
  scores(input, winLoseOrDraw).sum

enum Position:
  case Rock, Paper, Scissors

  // two positions after this one, wrapping around
  def winsAgainst: Position = fromOrdinal((ordinal + 2) % 3)

  // one position after this one, wrapping around
  def losesAgainst: Position = fromOrdinal((ordinal + 1) % 3)

end Position

def readCode(opponent: String) = opponent match
  case "A" => Rock
  case "B" => Paper
  case "C" => Scissors

def scores(input: String, strategy: (Position, String) => Position) =
  for case s"$x $y" <- input.linesIterator yield
    val opponent = readCode(x)
    score(opponent, strategy(opponent, y))

def winLoseOrDraw(opponent: Position, code: String): Position = code match
  case "X" => opponent.winsAgainst // we need to lose
  case "Y" => opponent // we need to tie
  case "Z" => opponent.losesAgainst // we need to win

def pickPosition(opponent: Position, code: String): Position = code match
  case "X" => Rock
  case "Y" => Paper
  case "Z" => Scissors

def score(opponent: Position, player: Position): Int =
  val pointsOutcome =
    if opponent == player then 3 // tie
    else if player.winsAgainst == opponent then 6 // win
    else 0 // lose

  // Rock = 1, Paper = 2, Scissors = 3
  val pointsPlay = player.ordinal + 1

  pointsPlay + pointsOutcome
end score

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Part 1​

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- + \ No newline at end of file diff --git a/2022/puzzles/day03/index.html b/2022/puzzles/day03/index.html index f7234b4ab..16f58c17c 100644 --- a/2022/puzzles/day03/index.html +++ b/2022/puzzles/day03/index.html @@ -5,13 +5,13 @@ Day 3: Rucksack Reorganization | Scala Center Advent of Code - +
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Day 3: Rucksack Reorganization

by @bishabosha

Puzzle description​

https://adventofcode.com/2022/day/3

Final Code​

def part1(input: String): Int =
  val intersections =
    for line <- input.linesIterator yield
      val (left, right) = line.splitAt(line.length / 2)
      (priorities(left) & priorities(right)).head
  intersections.sum

def part2(input: String): Int =
  val badges =
    for case Seq(a, b, c) <- input.linesIterator.grouped(3) yield
      (priorities(a) & priorities(b) & priorities(c)).head
  badges.sum

def priorities(str: String) = str.foldLeft(Priorities.emptySet)(_ add _)

object Priorities:
  opaque type Set = Long // can fit all 52 priorities in a bitset

  // encode priorities as a random access lookup
  private val lookup =
    val arr = new Array[Int](128) // max key is `'z'.toInt == 122`
    for (c, i) <- (('a' to 'z') ++ ('A' to 'Z')).zipWithIndex do
      arr(c.toInt) = i + 1
    IArray.unsafeFromArray(arr)

  val emptySet: Set = 0L

  extension (priorities: Set)
    infix def add(c: Char): Set = priorities | (1L << lookup(c.toInt))
    infix def &(that: Set): Set = priorities & that
    def head: Int = java.lang.Long.numberOfTrailingZeros(priorities)

end Priorities

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- + \ No newline at end of file diff --git a/2022/puzzles/day04/index.html b/2022/puzzles/day04/index.html index 9a582419d..aa7637db9 100644 --- a/2022/puzzles/day04/index.html +++ b/2022/puzzles/day04/index.html @@ -5,13 +5,13 @@ Day 4: Camp Cleanup | Scala Center Advent of Code - +
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Day 4: Camp Cleanup

by @bishabosha

Puzzle description​

https://adventofcode.com/2022/day/4

Final Code​

def part1(input: String): Int =
  foldPairs(input, subsumes)

def part2(input: String): Int =
  foldPairs(input, overlaps)

def subsumes(x: Int, y: Int)(a: Int, b: Int): Boolean = x <= a && y >= b
def overlaps(x: Int, y: Int)(a: Int, b: Int): Boolean = x <= a && y >= a || x <= b && y >= b

def foldPairs(input: String, hasOverlap: (Int, Int) => (Int, Int) => Boolean): Int =
  val matches =
    for line <- input.linesIterator yield
      val Array(x,y,a,b) = line.split("[,-]").map(_.toInt): @unchecked
      hasOverlap(x,y)(a,b) || hasOverlap(a,b)(x,y)
  matches.count(identity)

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- + \ No newline at end of file diff --git a/2022/puzzles/day05/index.html b/2022/puzzles/day05/index.html index 965dce033..526283987 100644 --- a/2022/puzzles/day05/index.html +++ b/2022/puzzles/day05/index.html @@ -5,13 +5,13 @@ Day 5: Supply Stacks | Scala Center Advent of Code - +
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Day 5: Supply Stacks

by @bishabosha

Puzzle description​

https://adventofcode.com/2022/day/5

Final Code​

def part1(input: String): String =
  moveAllCrates(input, _ reverse_::: _) // concat in reverse order

def part2(input: String): String =
  moveAllCrates(input, _ ::: _) // concat in normal order

/** each column is 4 chars wide (or 3 if terminal) */
def parseRow(row: String) =
  for i <- 0 to row.length by 4 yield
    if row(i) == '[' then
      row(i + 1) // the crate id
    else
      '#' // empty slot

def parseColumns(header: IndexedSeq[String]): IndexedSeq[List[Char]] =
  val crates :+ colsStr = header: @unchecked
  val columns = colsStr.split(" ").filter(_.nonEmpty).length

  val rows = crates.map(parseRow(_).padTo(columns, '#')) // pad empty slots at the end

  // transpose the rows to get the columns, then remove the terminal empty slots from each column
  rows.transpose.map(_.toList.filterNot(_ == '#'))
end parseColumns

def moveAllCrates(input: String, moveCrates: (List[Char], List[Char]) => List[Char]): String =
  val (headerLines, rest0) = input.linesIterator.span(_.nonEmpty)
  val instructions = rest0.drop(1) // drop the empty line after the header

  def move(cols: IndexedSeq[List[Char]], n: Int, idxA: Int, idxB: Int) =
    val (toMove, aRest) = cols(idxA).splitAt(n)
    val b2 = moveCrates(toMove, cols(idxB))
    cols.updated(idxA, aRest).updated(idxB, b2)

  val columns = parseColumns(headerLines.to(IndexedSeq))

  val columns1 = instructions.foldLeft(columns) { case (columns, s"move $n from $a to $b") =>
    move(columns, n.toInt, a.toInt - 1, b.toInt - 1)
  }
  columns1.map(_.head).mkString
end moveAllCrates

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- + \ No newline at end of file diff --git a/2022/puzzles/day06/index.html b/2022/puzzles/day06/index.html index 1f93298d9..2b037e512 100644 --- a/2022/puzzles/day06/index.html +++ b/2022/puzzles/day06/index.html @@ -5,7 +5,7 @@ Day 6: Tuning Trouble | Scala Center Advent of Code - + @@ -21,7 +21,7 @@ the multiset described above, you only care about the first and last element of each window, which can be represented by two indexes into the string.

The final optimisation is to only update the set when the last element of the window is different to the first element of the previous window.

The final optimised code is presented below, including an implementation of the multiset:

def part1(input: String): Int =
  findIndexOptimal(input, n = 4)

def part2(input: String): Int =
  findIndexOptimal(input, n = 14)

class MultiSet:
  private val counts = new Array[Int](26)
  private var uniqueElems = 0

  def size = uniqueElems

  def add(c: Char) =
    val count = counts(c - 'a')
    if count == 0 then
      uniqueElems += 1
    counts(c - 'a') += 1

  def remove(c: Char) =
    val count = counts(c - 'a')
    if count > 0 then
      if count == 1 then
        uniqueElems -= 1
      counts(c - 'a') -= 1
end MultiSet

def findIndexOptimal(input: String, n: Int): Int =
  val counts = MultiSet()
  def loop(i: Int, j: Int): Int =
    if counts.size == n then
      i + n // found the index
    else if j >= input.length then
      -1 // window went beyond the end
    else
      val previous = input(i)
      val last = input(j)
      if previous != last then
        counts.remove(previous)
        counts.add(last)
      loop(i = i + 1, j = j + 1)
  end loop
  input.iterator.take(n).foreach(counts.add) // add up-to the first `n` elements
  loop(i = 0, j = n)

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- + \ No newline at end of file diff --git a/2022/puzzles/day07/index.html b/2022/puzzles/day07/index.html index 8699c9c2c..f251424b1 100644 --- a/2022/puzzles/day07/index.html +++ b/2022/puzzles/day07/index.html @@ -5,13 +5,13 @@ Day 7: No Space Left On Device | Scala Center Advent of Code - +
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Day 7: No Space Left On Device

code by Jan Boerman

Puzzle description​

https://adventofcode.com/2022/day/7

Solution​

First of all, we need to create types for commands, to differentiate the input:

enum Command:
  case ChangeDirectory(directory: String)
  case ListFiles

enum TerminalOutput:
  case Cmd(cmd: Command)
  case Directory(name: String)
  case File(size: Int, name: String)

Let's make a directory structure, in which we will define files as mutable.Map, that can contain name (String) and size (Integer), will have reference to parent directory, and will be able to contain subdirectories:

class DirectoryStructure(val name: String,
                         val subDirectories: mutable.Map[String, DirectoryStructure],
                         val files: mutable.Map[String, Int],
                         val parent: DirectoryStructure | Null)

And now we need to come up with a way to parse out input code:

def input (str: String) = str.linesIterator.map {
  case s"$$ cd $directory" => Cmd(ChangeDirectory(directory))
  case s"$$ ls" => Cmd(ListFiles)
  case s"dir $directory" => Directory(directory)
  case s"$size $file" => File(size.toInt, file)
}.toList

We have to come up with a way to calculate directory size -- we can use sum for the size of all files in directory and define size of all of the following subdirectories recursively, which will take care of problem:

def directorySize(dir: DirectoryStructure): Int =
  dir.files.values.sum + dir.subDirectories.values.map(directorySize).sum

Now we need to create a function to build the directory structure from the input. For that we can use match and separate input, -- for that we can use cases and recursion will do the rest for us:

def buildState(input: List[TerminalOutput], currentDir: DirectoryStructure | Null, rootDir: DirectoryStructure): Unit = input match
  case Cmd(ChangeDirectory("/")) :: t => buildState(t, rootDir, rootDir)
  case Cmd(ChangeDirectory("..")) :: t => buildState(t, currentDir.parent, rootDir)
  case Cmd(ChangeDirectory(name)) :: t => buildState(t, currentDir.subDirectories(name), rootDir)
  case Cmd(ListFiles) :: t => buildState(t, currentDir, rootDir)
  case File(size, name) :: t =>
    currentDir.files.put(name, size)
    buildState(t, currentDir, rootDir)
  case Directory(name) :: t =>
    currentDir.subDirectories.put(name, DirectoryStructure(name, mutable.Map.empty, mutable.Map.empty, currentDir))
    buildState(t, currentDir, rootDir)
  case Nil => ()

And now, we need to assemble the program, in part one, we will search for all directories with size smaller 100000, and calculate the sum of their sizes.

def part1(output: String): Int =
  val rootDir = buildData(output)
  collectSizes(rootDir, _ < 100000).sum

In part two, we are looking for the smallest directory, which size is big enough to free up enough space on the filesystem to install update (30,000,00). We have to find out how much space is required for update, considering our available unused space:

def part2(output: String): Int =
  val rootDir = buildData(output)
  val totalUsed = directorySize(rootDir)
  val totalUnused = 70_000_000 - totalUsed
  val required = 30_000_000 - totalUnused
  collectSizes(rootDir, _ >= required).min

Final Code​

import scala.annotation.tailrec
import scala.collection.mutable

import TerminalOutput.*
import Command.*

def input (str: String) = str.linesIterator.map {
  case s"$$ cd $directory" => Cmd(ChangeDirectory(directory))
  case s"$$ ls" => Cmd(ListFiles)
  case s"dir $directory" => Directory(directory)
  case s"$size $file" => File(size.toInt, file)
}.toList

enum Command:
  case ChangeDirectory(directory: String)
  case ListFiles

enum TerminalOutput:
  case Cmd(cmd: Command)
  case Directory(name: String)
  case File(size: Int, name: String)

class DirectoryStructure(val name: String,
                         val subDirectories: mutable.Map[String, DirectoryStructure],
                         val files: mutable.Map[String, Int],
                         val parent: DirectoryStructure | Null)

def buildState(input: List[TerminalOutput], currentDir: DirectoryStructure | Null, rootDir: DirectoryStructure): Unit = input match
  case Cmd(ChangeDirectory("/")) :: t => buildState(t, rootDir, rootDir)
  case Cmd(ChangeDirectory("..")) :: t => buildState(t, currentDir.parent, rootDir)
  case Cmd(ChangeDirectory(name)) :: t => buildState(t, currentDir.subDirectories(name), rootDir)
  case Cmd(ListFiles) :: t => buildState(t, currentDir, rootDir)
  case File(size, name) :: t =>
    currentDir.files.put(name, size)
    buildState(t, currentDir, rootDir)
  case Directory(name) :: t =>
    currentDir.subDirectories.put(name, DirectoryStructure(name, mutable.Map.empty, mutable.Map.empty, currentDir))
    buildState(t, currentDir, rootDir)
  case Nil => ()

def directorySize(dir: DirectoryStructure): Int =
  dir.files.values.sum + dir.subDirectories.values.map(directorySize).sum

def collectSizes(dir: DirectoryStructure, criterion: Int => Boolean): Iterable[Int] =
  val mySize = directorySize(dir)
  val children = dir.subDirectories.values.flatMap(collectSizes(_, criterion))
  if criterion(mySize) then mySize :: children.toList else children

def buildData(output: String) =
  val rootDir = new DirectoryStructure("/", mutable.Map.empty, mutable.Map.empty, null)
  buildState(input(output), null, rootDir)
  rootDir


def part1(output: String): Int =
  val rootDir = buildData(output)
  collectSizes(rootDir, _ < 100000).sum

def part2(output: String): Int =
  val rootDir = buildData(output)
  val totalUsed = directorySize(rootDir)
  val totalUnused = 70_000_000 - totalUsed
  val required = 30_000_000 - totalUnused
  collectSizes(rootDir, _ >= required).min

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- + \ No newline at end of file diff --git a/2022/puzzles/day08/index.html b/2022/puzzles/day08/index.html index 671a1f179..be8aabd1f 100644 --- a/2022/puzzles/day08/index.html +++ b/2022/puzzles/day08/index.html @@ -5,7 +5,7 @@ Day 8: Treetop Tree House | Scala Center Advent of Code - + @@ -28,7 +28,7 @@ For example trees of height 3 can see lengths(3) trees. And we update this list with each new tree we see, if it's x big, all trees at least x small will only see that tree, and all other trees will see one more: at index i of value v: if i <= x then 1 else v+1.

We can then use this in a similar way to what we did with max and rollingMax before:

val rollingLengths = line.scanRight( List.fill(10)(0) ){
  case (curr, lengths) =>
    lengths.zipWithIndex.map{ case (v, i) => if i <= curr then 1 else v+1 }
}.init

We then get the score by reading lengths at the appropriate point, again as was done with rollingMax:

rollingLengths.zip(line).map{ case (lengths, curr) => lengths(curr) }

By combining everything, noticing once again our calculation is the same for each line, we get:

def computeScore(ls: HeightField): ScoreField = ls.map{ line =>
  val rollingLengths = line.scanRight( List.fill(10)(0) ){
    case (curr, lengths) =>
      lengths.zipWithIndex.map{ case (v, i) => if i <= curr then 1 else v+1 }
  }.init
  rollingLengths.zip(line).map{ case (lengths, curr) => lengths(curr) }
}

Where ScoreField is identical to HeightField, but serves to make the code more readable:

type ScoreField = Field[Int]

We can use the same trick as before to get all the other directions for free:

val scoreFields: List[ScoreField] = computeInAllDirections(parsed, computeScore)

This time instead of or-ing, we need to multiply "A tree's scenic score is found by multiplying together its viewing distance in each of the four directions.":

val scoreField: ScoreField = scoreFields.reduce(combine(_ * _))

And this time the last step is to get the heighest value instead of the sum:

scoreField.megaReduce(_ max _)

Final Code​

def part1(input: String): Int =
  val parsed = parse(input)
  val visibilityFields: List[VisibilityField] = computeInAllDirections(parsed, computeVisibility)
  val visibilityField: VisibilityField = visibilityFields.reduce(combine(_ | _))
  visibilityField.megaMap(if _ then 1 else 0).megaReduce(_ + _)

def part2(input: String): Int =
  val parsed = parse(input)
  val scoreFields: List[ScoreField] = computeInAllDirections(parsed, computeScore)
  val scoreField: ScoreField = scoreFields.reduce(combine(_ * _))
  scoreField.megaReduce(_ max _)

type Field[A] = List[List[A]]

extension [A](xss: Field[A])
  def megaZip[B](yss: Field[B]): Field[(A, B)] = (xss zip yss).map( (xs, ys) => xs zip ys )
  def megaMap[B](f: A => B): Field[B] = xss.map(_.map(f))
  def megaReduce(f: (A,A) => A): A = xss.map(_.reduce(f)).reduce(f)

def combine[A](op: ((A,A)) => A)(f1: Field[A], f2: Field[A]): Field[A] = f1.megaZip(f2).megaMap(op)

def computeInAllDirections[A, B](xss: Field[A], f: Field[A] => Field[B]): List[Field[B]] =
  for
    transpose <- List(false, true)
    reverse <- List(false, true)
  yield
    val t = if transpose then xss.transpose else xss
    val in = if reverse then t.map(_.reverse) else t
    val res = f(in)
    val r = if reverse then res.map(_.reverse) else res
    val out = if transpose then r.transpose else r
    out

type HeightField = Field[Int]
type ScoreField = Field[Int]

type VisibilityField = Field[Boolean]

def parse(input: String): HeightField = input.split('\n').toList.map(line => line.map(char => char.asDigit).toList)

def computeVisibility(ls: HeightField): VisibilityField = ls.map{ line =>
  val rollingMax = line.scanLeft(-1){ case (max, curr) => Math.max(max, curr) }.init
  rollingMax.zip(line).map{ case (max, curr) => max < curr) }
}

def computeScore(ls: HeightField): ScoreField = ls.map{ line =>
  val rollingLengths = line.scanRight( List.fill(10)(0) ){
    case (curr, lengths) =>
      lengths.zipWithIndex.map{ case (v, i) => if i <= curr then 1 else v+1 }
  }.init
  rollingLengths.zip(line).map{ case (lengths, curr) => lengths(curr) }
}

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Part 2​

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- + \ No newline at end of file diff --git a/2022/puzzles/day09/index.html b/2022/puzzles/day09/index.html index 1e6fcbc31..d65bab2e5 100644 --- a/2022/puzzles/day09/index.html +++ b/2022/puzzles/day09/index.html @@ -5,7 +5,7 @@ Day 9: Rope Bridge | Scala Center Advent of Code - + @@ -27,7 +27,7 @@ Each line you can extract the direction and steps count with a pattern binding val (s"$dir $n") = line, then use Direction.valueOf to lookup the direction, and .toInt to convert n to the number of steps.

Then to run n steps, create the steps iterator, then drop n elements to advance the state n steps, then take the next() element:

def uniquePositions(input: String, knots: Int): Int =
  val end = input.linesIterator.foldLeft(initialState(knots)) { case (state, line) =>
    val (s"$dir $n") = line: @unchecked
    steps(state, Direction.valueOf(dir)).drop(n.toInt).next()
  }
  end.uniques.size

Part 1 needs 2 knots, and part 2 needs 10 knots, they can be implemented as such:

def part1(input: String): Int =
  uniquePositions(input, knots = 2)

def part2(input: String): Int =
  uniquePositions(input, knots = 10)

Final Code​

import Direction.*

def part1(input: String): Int =
  uniquePositions(input, knots = 2)

def part2(input: String): Int =
  uniquePositions(input, knots = 10)

case class Position(x: Int, y: Int):
  def moveOne(dir: Direction): Position = dir match
    case U => Position(x, y + 1)
    case D => Position(x, y - 1)
    case L => Position(x - 1, y)
    case R => Position(x + 1, y)

  def follow(head: Position): Position =
    val dx = head.x - x
    val dy = head.y - y
    if dx.abs > 1 || dy.abs > 1 then Position(x + dx.sign, y + dy.sign) // follow the head
    else this // stay put

case class State(uniques: Set[Position], head: Position, knots: List[Position])

enum Direction:
  case U, D, L, R

def followAll(head: Position, knots: List[Position]) =
  var prev = head // head was already moved with `moveOne`
  val buf = List.newBuilder[Position]
  for knot <- knots do
    val next = knot.follow(prev)
    buf += next
    prev = next
  (prev, buf.result())
end followAll

def step(dir: Direction, state: State) =
  val head1 = state.head.moveOne(dir)
  val (last, knots1) = followAll(head1, state.knots)
  State(state.uniques + last, head1, knots1)

def steps(state: State, dir: Direction): Iterator[State] =
  Iterator.iterate(state)(state => step(dir, state))

def initialState(knots: Int) =
  val zero = Position(0, 0)
  State(Set(zero), zero, List.fill(knots - 1)(zero))

def uniquePositions(input: String, knots: Int): Int =
  val end = input.linesIterator.foldLeft(initialState(knots)) { case (state, line) =>
    val (s"$dir $n") = line: @unchecked
    steps(state, Direction.valueOf(dir)).drop(n.toInt).next()
  }
  end.uniques.size

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- + \ No newline at end of file diff --git a/2022/puzzles/day10/index.html b/2022/puzzles/day10/index.html index 8f3b38e95..6ddf2637e 100644 --- a/2022/puzzles/day10/index.html +++ b/2022/puzzles/day10/index.html @@ -5,14 +5,14 @@ Day 10: Cathode-Ray Tube | Scala Center Advent of Code - +
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Day 10: Cathode-Ray Tube

code and article by Mewen Crespo (reviewed by Jamie Thompson)

Puzzle description​

https://adventofcode.com/2022/day/10

Solution​

Today's goal is to simulate the register's values over time. Once this is done, the rest falls in place rather quickly. From the puzzle description, we know there are two commands availaible: noop and addx. This can be implemented with a enum:

enum Command:
  case Noop
  case Addx(x: Int)

Now, we need to parse this commands from the string. This can be done using a for loop to match each line of the input:

import Command.*

def commandsIterator(input: String): Iterator[Command] =
  for line <- input.linesIterator yield line match
    case "noop" => Noop
    case s"addx $x" if x.toIntOption.isDefined => Addx(x.toInt)
    case _ => throw IllegalArgumentException(s"Invalid command '$line''")

Here you can use linesIterator to retrieve the lines (it returns an Iterator[String]) and mapped every line using a for .. yield comprehension with a match body. Note the use of the string interpolator s for a simple way to parse strings.

tip

Error checking: Althought not necessary in this puzzle, it is a good practice to check the validity of the input. Here, we checked that the string matched with $x is a valid integer string before entering the second case and throw an exception if none of the first cases were matched.

Now we are ready to compute the registers values. We choose to implement it as an Iterator[Int] which will return the register's value each cycle at a time. For this, we need to loop throught the commands. If the command is a noop, then the next cycle will have the same value. If the command is a addx x then the next cycle will be the same value and the cycle afterward will be x more. There is an issue here: the addx command generates two cycles whereas the noop command generates only one.

To circumvent this issue, generate an Iterator[List[Int]] first which will be flattened afterwards. The first iterator is constructed using the scanLeft method to yield the following code:

val RegisterStartValue = 1

def registerValuesIterator(input: String): Iterator[Int] =
  val steps = commandsIterator(input).scanLeft(RegisterStartValue :: Nil) { (values, cmd) =>
    val value = values.last
    cmd match
      case Noop => value :: Nil
      case Addx(x) => value :: value + x :: Nil
  }
  steps.flatten

Notice that at each step we call .last on the accumulated List[Int] value which, in this case, is the register's value at the start of the last cycle.

Part 1​

In the first part, the challenge asks you to compute the strength at the 20th cycle and then every 40th cycle. This can be done using a combination of drop (to skip the first 19 cycles), grouped (to group the cycles by 40) and map(_.head) (to only take the first cycle of each group of 40). The computation of the strengths is, on the other hand, done using the zipWithIndex method and a for ... yield comprehension. This leads to the following code:

def registerStrengthsIterator(input: String): Iterator[Int] =
  val it = for (reg, i) <- registerValuesIterator(input).zipWithIndex yield (i + 1) * reg
  it.drop(19).grouped(40).map(_.head)

The result of Part 1 is the sum of this iterator:

def part1(input: String): Int = registerStrengthsIterator(input).sum

Part 2​

In the second part, we are asked to draw a CRT output. As stated in the puzzle description, the register is interpreted as the position of a the sprite ###. The CRT iterates throught each line and, if the sprites touches the touches the current position, draws a #. Otherwise the CRT draws a .. The register's cycles are stepped in synced with the CRT.

First, the CRT's position is just the cycle's index modulo the CRT's width (40 in our puzzle). Then, the CRT draw the sprite if and only if the register's value is the CRT's position, one more or one less. In other words, if (reg_value - (cycle_id % 40)).abs <= 1. Using the zipWithIndex method to obtain the cycles' indexes we end up with the following code:

val CRTWidth: Int = 40

def CRTCharIterator(input: String): Iterator[Char] =
  for (reg, crtPos) <- registerValuesIterator(input).zipWithIndex yield
    if (reg - (crtPos % CRTWidth)).abs <= 1 then
      '#'
    else
      '.'

Now, concatenate the chars and add new lines at the required places. This is done using the mkString methods:

def part2(input: String): String =
  CRTCharIterator(input).grouped(CRTWidth).map(_.mkString).mkString("\n")

Final Code​

import Command.*

def part1(input: String): Int =
  registerStrengthsIterator(input).sum

def part2(input: String): String =
  CRTCharIterator(input).grouped(CRTWidth).map(_.mkString).mkString("\n")

enum Command:
  case Noop
  case Addx(x: Int)

def commandsIterator(input: String): Iterator[Command] =
  for line <- input.linesIterator yield line match
    case "noop" => Noop
    case s"addx $x" if x.toIntOption.isDefined => Addx(x.toInt)
    case _ => throw IllegalArgumentException(s"Invalid command '$line''")

val RegisterStartValue = 1

def registerValuesIterator(input: String): Iterator[Int] =
  val steps = commandsIterator(input).scanLeft(RegisterStartValue :: Nil) { (values, cmd) =>
    val value = values.last
    cmd match
      case Noop => value :: Nil
      case Addx(x) => value :: value + x :: Nil
  }
  steps.flatten

def registerStrengthsIterator(input: String): Iterator[Int] =
  val it = for (reg, i) <- registerValuesIterator(input).zipWithIndex yield (i + 1) * reg
  it.drop(19).grouped(40).map(_.head)

val CRTWidth: Int = 40

def CRTCharIterator(input: String): Iterator[Char] =
  for (reg, crtPos) <- registerValuesIterator(input).zipWithIndex yield
    if (reg - (crtPos % CRTWidth)).abs <= 1 then
      '#'
    else
      '.'

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- + \ No newline at end of file diff --git a/2022/puzzles/day11/index.html b/2022/puzzles/day11/index.html index 7f95ca02c..d516a8fc3 100644 --- a/2022/puzzles/day11/index.html +++ b/2022/puzzles/day11/index.html @@ -5,13 +5,13 @@ Day 11: Monkey in the Middle | Scala Center Advent of Code - +
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Day 11: Monkey in the Middle

Puzzle description​

https://adventofcode.com/2022/day/11

Final Code​

import scala.collection.immutable.Queue

def part1(input: String): Long =
  run(initial = parseInput(input), times = 20, adjust = _ / 3)

def part2(input: String): Long =
  run(initial = parseInput(input), times = 10_000, adjust = identity)

type Worry = Long
type Op = Worry => Worry
type Monkeys = IndexedSeq[Monkey]

case class Monkey(
  items: Queue[Worry],
  divisibleBy: Int,
  ifTrue: Int,
  ifFalse: Int,
  op: Op,
  inspected: Int
)

def iterate[Z](times: Int)(op: Z => Z)(z: Z): Z =
  (0 until times).foldLeft(z) { (z, _) => op(z) }

def run(initial: Monkeys, times: Int, adjust: Op): Long =
  val lcm = initial.map(_.divisibleBy.toLong).product
  val monkeys = iterate(times)(round(adjust, lcm))(initial)
  monkeys.map(_.inspected.toLong).sorted.reverseIterator.take(2).product

def round(adjust: Op, lcm: Worry)(monkeys: Monkeys): Monkeys =
  monkeys.indices.foldLeft(monkeys) { (monkeys, index) =>
    turn(index, monkeys, adjust, lcm)
  }

def turn(index: Int, monkeys: Monkeys, adjust: Op, lcm: Worry): Monkeys =
  val monkey = monkeys(index)
  val Monkey(items, divisibleBy, ifTrue, ifFalse, op, inspected) = monkey

  val monkeys1 = items.foldLeft(monkeys) { (monkeys, item) =>
    val inspected = op(item)
    val nextWorry = adjust(inspected) % lcm
    val thrownTo =
      if nextWorry % divisibleBy == 0 then ifTrue
      else ifFalse
    val thrownToMonkey =
      val m = monkeys(thrownTo)
      m.copy(items = m.items :+ nextWorry)
    monkeys.updated(thrownTo, thrownToMonkey)
  }
  val monkey1 = monkey.copy(
    items = Queue.empty,
    inspected = inspected + items.size
  )
  monkeys1.updated(index, monkey1)
end turn

def parseInput(input: String): Monkeys =

  def eval(by: String): Op =
    if by == "old" then identity
    else Function.const(by.toInt)

  def parseOperator(op: String, left: Op, right: Op): Op =
    op match
      case "+" => old => left(old) + right(old)
      case "*" => old => left(old) * right(old)

  IArray.from(
    for
      case Seq(
        s"Monkey $n:",
        s"  Starting items: $items",
        s"  Operation: new = $left $operator $right",
        s"  Test: divisible by $div",
        s"    If true: throw to monkey $ifTrue",
        s"    If false: throw to monkey $ifFalse",
        _*
      ) <- input.linesIterator.grouped(7)
    yield
      val op = parseOperator(operator, eval(left), eval(right))
      val itemsQueue = items.split(", ").map(_.toLong).to(Queue)
      Monkey(itemsQueue, div.toInt, ifTrue.toInt, ifFalse.toInt, op, inspected = 0)
  )
end parseInput

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- + \ No newline at end of file diff --git a/2022/puzzles/day12/index.html b/2022/puzzles/day12/index.html index d8c2ffdfe..5fb00396f 100644 --- a/2022/puzzles/day12/index.html +++ b/2022/puzzles/day12/index.html @@ -5,13 +5,13 @@ Day 12: Hill Climbing Algorithm | Scala Center Advent of Code - +
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Day 12: Hill Climbing Algorithm

Puzzle description​

https://adventofcode.com/2022/day/12

Solution​

Today's challenge is to simulate the breadth-first search over a graph. First, let's create a standard Point class and define addition on it:

case class Point(x: Int, y: Int):
  def move(dx: Int, dy: Int):
    Point = Point(x + dx, y + dy)
  override def toString: String =
    s"($x, $y)"
end Point

Now we need a representation that will serve as a substitute for moves:

val up    = (0, 1)
val down  = (0, -1)
val left  = (-1, 0)
val right = (1, 0)
val possibleMoves = List(up, down, left, right)

Let's make a path function that will help us to calculate the length of our path to the point, based on our moves, that we defined before:

def path(point: Point, net: Map[Point, Char]): Seq[Point] =
  possibleMoves.map(point.move).filter(net.contains)

A function that fulfills our need to match an entry with the point we are searching for:

def matching(point: Point, net: Map[Point, Char]): Char =
  net(point) match
    case 'S' => 'a'
    case 'E' => 'z'
    case other => other

Now we just need to put the program together. First of all, let's map out our indices to the source, so we can create a queue for path representation. After that we need to create a map, to keep track the length of our path. For that we will need to map E entry to zero. The last part is the implementation of bfs on a Queue.

def solution(source: IndexedSeq[String], srchChar: Char): Int =
  // create a sequence of Point objects and their corresponding character in source
  val points =
    for
      y <- source.indices
      x <- source.head.indices
    yield
      Point(x, y) -> source(y)(x)
  val p = points.toMap
  val initial = p.map(_.swap)('E')
  val queue = collection.mutable.Queue(initial)
  val length = collection.mutable.Map(initial -> 0)
  //bfs
  while queue.nonEmpty do
    val visited = queue.dequeue()
    if p(visited) == srchChar then
      return length(visited)
    for visited1 <- path(visited, p) do
      val shouldAdd =
        !length.contains(visited1)
        && matching(visited, p) - matching(visited1, p) <= 1
      if shouldAdd then
        queue.enqueue(visited1)
        length(visited1) = length(visited) + 1
    end for
  end while
  throw IllegalStateException("unexpected end of search area")
end solution

In part one srchChar is 'S', but since our method in non-exhaustive, we may apply the same function for 'a'

def part1(data: String): Int =
  solution(IndexedSeq.from(data.linesIterator), 'S')
def part2(data: String): Int =
  solution(IndexedSeq.from(data.linesIterator), 'a')

And that's it!

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- + \ No newline at end of file diff --git a/2022/puzzles/day13/index.html b/2022/puzzles/day13/index.html index 0401cb04d..84f0c38f0 100644 --- a/2022/puzzles/day13/index.html +++ b/2022/puzzles/day13/index.html @@ -5,13 +5,13 @@ Day 13: Distress Signal | Scala Center Advent of Code - +
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Day 13: Distress Signal

by Jamie Thompson

Puzzle description​

https://adventofcode.com/2022/day/13

Final Code​

import scala.collection.immutable.Queue
import scala.math.Ordered.given
import Packet.*

def part1(input: String): Int =
  findOrderedIndices(input)

def part2(input: String): Int =
  findDividerIndices(input)

def findOrderedIndices(input: String): Int =
  val indices = (
    for
      case (Seq(a, b, _*), i) <- input.linesIterator.grouped(3).zipWithIndex
      if readPacket(a) <= readPacket(b)
    yield
      i + 1
  )
  indices.sum

def findDividerIndices(input: String): Int =
  val dividers = List("[[2]]", "[[6]]").map(readPacket)
  val lookup = dividers.toSet
  val packets = input
    .linesIterator
    .filter(_.nonEmpty)
    .map(readPacket)
  val indices = (dividers ++ packets)
    .sorted
    .iterator
    .zipWithIndex
    .collect { case (p, i) if lookup.contains(p) => i + 1 }
  indices.take(2).product

enum Packet:
  case Nested(packets: List[Packet])
  case Num(value: Int)

case class State(number: Int, values: Queue[Packet]):
  def nextWithDigit(digit: Int): State = // add digit to number
    copy(number = if number == -1 then digit else number * 10 + digit)

  def nextWithNumber: State =
    if number == -1 then this // no number to commit
    else
      // reset number, add accumulated number to values
      State.empty.copy(values = values :+ Num(number))

object State:
  val empty = State(-1, Queue.empty)

def readPacket(input: String): Packet =
  def loop(i: Int, state: State, stack: List[Queue[Packet]]): Packet =
    input(i) match // assume that list is well-formed.
      case '[' =>
        loop(i + 1, State.empty, state.values :: stack) // push old state to stack
      case ']' => // add trailing number, close packet
        val packet = Nested(state.nextWithNumber.values.toList)
        stack match
          case values1 :: rest => // restore old state
            loop(i + 1, State.empty.copy(values = values1 :+ packet), rest)
          case Nil => // terminating case
            packet
      case ',' => loop(i + 1, state.nextWithNumber, stack)
      case n => loop(i + 1, state.nextWithDigit(n.asDigit), stack)
  end loop
  if input.nonEmpty && input(0) == '[' then
    loop(i = 1, State.empty, stack = Nil)
  else
    throw IllegalArgumentException(s"Invalid input: `$input`")
end readPacket

given PacketOrdering: Ordering[Packet] with

  def nestedCompare(ls: List[Packet], rs: List[Packet]): Int = (ls, rs) match
    case (l :: ls1, r :: rs1) =>
      val res = compare(l, r)
      if res == 0 then nestedCompare(ls1, rs1) // equal, look at next element
      else res // less or greater

    case (_ :: _, Nil) => 1  // right ran out of elements first
    case (Nil, _ :: _) => -1 // left ran out of elements first
    case (Nil, Nil)    => 0  // equal size
  end nestedCompare

  def compare(left: Packet, right: Packet): Int = (left, right) match
    case (Num(l), Num(r))          => l compare r
    case (Nested(l), Nested(r))    => nestedCompare(l, r)
    case (num @ Num(_), Nested(r)) => nestedCompare(num :: Nil, r)
    case (Nested(l), num @ Num(_)) => nestedCompare(l, num :: Nil)
  end compare

end PacketOrdering

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- + \ No newline at end of file diff --git a/2022/puzzles/day14/index.html b/2022/puzzles/day14/index.html index 4bab809a7..faffb6115 100644 --- a/2022/puzzles/day14/index.html +++ b/2022/puzzles/day14/index.html @@ -5,13 +5,13 @@ Day 14: Regolith Reservoir | Scala Center Advent of Code - +
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Day 14: Regolith Reservoir

Puzzle description​

https://adventofcode.com/2022/day/14

Final Solution​

def part1(input: String): Int =
  val search = parseInput(input)
  search.states
    .takeWhile(_.fallingPath.head.y < search.lowestRock)
    .last
    .sand
    .size

def part2(input: String): Int =
  parseInput(input).states.last.sand.size

def parseInput(input: String): Scan =
  val paths = input.linesIterator
    .map { line =>
      line.split(" -> ").map { case s"$x,$y" => Point(x.toInt, y.toInt) }.toList
    }
  val rocks = paths.flatMap { path =>
    path.sliding(2).flatMap {
      case List(p1, p2) =>
        val dx = p2.x - p1.x
        val dy = p2.y - p1.y

        if dx == 0 then (p1.y to p2.y by dy.sign).map(Point(p1.x, _))
        else (p1.x to p2.x by dx.sign).map(Point(_, p1.y))
      case _ => None
    }
  }.toSet
  Scan(rocks)

case class Point(x: Int, y: Int)

case class Scan(rocks: Set[Point]):
  val lowestRock = rocks.map(_.y).max
  val floor = lowestRock + 2

  case class State(fallingPath: List[Point], sand: Set[Point]):
    def isFree(p: Point) = !sand(p) && !rocks(p)

    def next: Option[State] = fallingPath.headOption.map {
      case sandUnit @ Point(x, y) =>
        val down = Some(Point(x, y + 1)).filter(isFree)
        val downLeft = Some(Point(x - 1, y + 1)).filter(isFree)
        val downRight = Some(Point(x + 1, y + 1)).filter(isFree)

        down.orElse(downLeft).orElse(downRight).filter(_.y < floor) match
          case Some(fallingPos) =>
            State(fallingPos :: fallingPath, sand)
          case None =>
            State(fallingPath.tail, sand + sandUnit)
    }

  def states: LazyList[State] =
    val source = Point(500, 0)
    LazyList.unfold(State(List(source), Set.empty)) { _.next.map(s => s -> s) }
end Scan

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- + \ No newline at end of file diff --git a/2022/puzzles/day15/index.html b/2022/puzzles/day15/index.html index fe838e273..e70a28264 100644 --- a/2022/puzzles/day15/index.html +++ b/2022/puzzles/day15/index.html @@ -5,14 +5,14 @@ Day 15: Beacon Exclusion Zone | Scala Center Advent of Code - +
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Day 15: Beacon Exclusion Zone

Puzzle description​

https://adventofcode.com/2022/day/15

Explanation​

Part 1​

We first model and parse the input: 

case class Position(x: Int, y: Int)

def parse(input: String): List[(Position, Position)] =
  input.split("\n").toList.map{
    case s"Sensor at x=$sx, y=$sy: closest beacon is at x=$bx, y=$by" =>
      (Position(sx.toInt, sy.toInt), Position(bx.toInt, by.toInt))
  }

We then model the problem-specific knowledge:

def distance(p1: Position, p2: Position): Int =
  Math.abs(p1.x - p2.x) + Math.abs(p1.y - p2.y)

def distanceToLine(p: Position, y: Int): Int =
  Math.abs(p.y - y)

We use it to compute how much of a line is covered by one (by lineCoverage) and all (by coverOfLine) sensors:

def lineCoverage(sensor: Position, radius: Int, lineY: Int): Range =
  val radiusInLine = radius - distanceToLine(sensor, lineY)
  
  // if radiusInLine is smaller than 0, the range will be empty
  (sensor.x - radiusInLine) to (sensor.x + radiusInLine)

def coverOfLine(sensorsWithDistances: List[(Position, Int)], line: Int) =
  sensorsWithDistances.map( (sensor, radius) => lineCoverage(sensor, radius, line) ).filter(_.nonEmpty)

This is enought to solve part one:

def part1(input: String): Int =
  val parsed: List[(Position, Position)] = parse(input)
  val beacons: Set[Position] = parsed.map(_._2).toSet
  val sensorsWithDistances: List[(Position, Int)] =
    parsed.map( (sensor, beacon) => (sensor, distance(sensor, beacon)) )

  val line = 2000000
  val cover: List[Range] = coverOfLine(sensorsWithDistances, line)
  val beaconsOnLine: Set[Position] = beacons.filter(_.y == line)
  val count: Int = cover.map(_.size).sum - beaconsOnLine.size
  count

Part 2​

We wish to remove ranges from other ranges, sadly there is no built-in method to do so, instead rellying on a cast to a collection, which makes computation much much slower. Therefore we define our own difference method which returns zero, one or two ranges:

def smartDiff(r1: Range, r2: Range): List[Range] =
  val innit = r1.start to Math.min(r2.start - 1, r1.last)
  val tail = Math.max(r1.start, r2.last + 1) to r1.last
  val res = if innit == tail then
    List(innit)
  else
    List(innit, tail)
  res.filter(_.nonEmpty).toList

This allows us to subtract the cover from our target interval like so:

def remainingSpots(target: Range, cover: List[Range]): Set[Int] = 

  def rec(partialTarget: List[Range], remainingCover: List[Range]): List[Range] =
    if remainingCover.isEmpty then
      partialTarget
    else
      val (curr: Range) :: rest = remainingCover: @unchecked
      rec(
        partialTarget = partialTarget.flatMap( r => smartDiff(r, curr) ),
        remainingCover = rest
      )

  rec(List(target), cover).flatten.toSet

We can then iterate through all lines, and computing for each which positions are free. As per the problem statement, we know there will only be one inside the square of side 0 to 4_000_000. We then compute the solution's tuning frequency.

def part2(input: String): Any =

  val parsed: List[(Position, Position)] = parse(input)
  val beacons: Set[Position] = parsed.map(_._2).toSet
  val sensorsWithDistances: List[(Position, Int)] =
    parsed.map( (sensor, beacon) => (sensor, distance(sensor, beacon)) )

  val target: Range = 0 to 4_000_000
  val spots: Seq[Position] = target.flatMap{
    line => 
      val cover: List[Range] = coverOfLine(sensorsWithDistances, line)
      val beaconsOnLine: Set[Position] = beacons.filter(_.y == line)

      val remainingRanges: List[Range] = cover.foldLeft(List(target)){ 
        case (acc: List[Range], range: Range) => 
          acc.flatMap( r => smartDiff(r, range) )
      }
      val potential = remainingRanges.flatten.toSet

      val spotsOnLine = potential diff beaconsOnLine.map( b => b.x )
      spotsOnLine.map( x => Position(x, line) )
  }
  def tuningFrequency(p: Position): BigInt = BigInt(p.x) * 4_000_000 + p.y

  println(spots.mkString(", "))
  assert(spots.size == 1)
  tuningFrequency(spots.head)

Final Code​

case class Position(x: Int, y: Int)

def parse(input: String): List[(Position, Position)] =
  input.split("\n").toList.map{
    case s"Sensor at x=$sx, y=$sy: closest beacon is at x=$bx, y=$by" =>
      (Position(sx.toInt, sy.toInt), Position(bx.toInt, by.toInt))
  }

def distance(p1: Position, p2: Position): Int =
  Math.abs(p1.x - p2.x) + Math.abs(p1.y - p2.y)

def distanceToLine(p: Position, y: Int): Int =
  Math.abs(p.y - y)

def lineCoverage(sensor: Position, radius: Int, lineY: Int): Range =
  val radiusInLine = radius - distanceToLine(sensor, lineY)
  
  // if radiusInLine is smaller than 0, the range will be empty
  (sensor.x - radiusInLine) to (sensor.x + radiusInLine)

def coverOfLine(sensorsWithDistances: List[(Position, Int)], line: Int) =
  sensorsWithDistances.map( (sensor, radius) => lineCoverage(sensor, radius, line) ).filter(_.nonEmpty)

def smartDiff(r1: Range, r2: Range): List[Range] =
  val innit = r1.start to Math.min(r2.start - 1, r1.last)
  val tail = Math.max(r1.start, r2.last + 1) to r1.last
  val res = if innit == tail then
    List(innit)
  else
    List(innit, tail)
  res.filter(_.nonEmpty).toList

def remainingSpots(target: Range, cover: List[Range]): Set[Int] = 

  def rec(partialTarget: List[Range], remainingCover: List[Range]): List[Range] =
    if remainingCover.isEmpty then
      partialTarget
    else
      val (curr: Range) :: rest = remainingCover: @unchecked
      rec(
        partialTarget = partialTarget.flatMap( r => smartDiff(r, curr) ),
        remainingCover = rest
      )

  rec(List(target), cover).flatten.toSet

def part1(input: String): Int =
  val parsed: List[(Position, Position)] = parse(input)
  val beacons: Set[Position] = parsed.map(_._2).toSet
  val sensorsWithDistances: List[(Position, Int)] =
    parsed.map( (sensor, beacon) => (sensor, distance(sensor, beacon)) )

  val line = 2000000
  val cover: List[Range] = coverOfLine(sensorsWithDistances, line)
  val beaconsOnLine: Set[Position] = beacons.filter(_.y == line)
  val count: Int = cover.map(_.size).sum - beaconsOnLine.size
  count

def part2(input: String): Any =

  val parsed: List[(Position, Position)] = parse(input)
  val beacons: Set[Position] = parsed.map(_._2).toSet
  val sensorsWithDistances: List[(Position, Int)] =
    parsed.map( (sensor, beacon) => (sensor, distance(sensor, beacon)) )

  val target: Range = 0 to 4_000_000
  val spots: Seq[Position] = target.flatMap{
    line => 
      val cover: List[Range] = coverOfLine(sensorsWithDistances, line)
      val beaconsOnLine: Set[Position] = beacons.filter(_.y == line)

      val remainingRanges: List[Range] = cover.foldLeft(List(target)){ 
        case (acc: List[Range], range: Range) => 
          acc.flatMap( r => smartDiff(r, range) )
      }
      val potential = remainingRanges.flatten.toSet

      val spotsOnLine = potential diff beaconsOnLine.map( b => b.x )
      spotsOnLine.map( x => Position(x, line) )
  }
  def tuningFrequency(p: Position): BigInt = BigInt(p.x) * 4_000_000 + p.y

  println(spots.mkString(", "))
  assert(spots.size == 1)
  tuningFrequency(spots.head)

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- + \ No newline at end of file diff --git a/2022/puzzles/day16/index.html b/2022/puzzles/day16/index.html index a9700db69..f574b2529 100644 --- a/2022/puzzles/day16/index.html +++ b/2022/puzzles/day16/index.html @@ -5,13 +5,13 @@ Day 16: Proboscidea Volcanium | Scala Center Advent of Code - +
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Day 16: Proboscidea Volcanium

code by Tyler Coles (javadocmd.com) & Quentin Bernet

Puzzle description​

https://adventofcode.com/2022/day/16

Final Code​

type Id = String
case class Room(id: Id, flow: Int, tunnels: List[Id])

type Input = List[Room]
// $_ to avoid tunnel/tunnels distinction and so on
def parse(xs: String): Input = xs.split("\n").map{ case s"Valve $id has flow rate=$flow; tunnel$_ lead$_ to valve$_ $tunnelsStr" =>
  val tunnels = tunnelsStr.split(", ").toList
  Room(id, flow.toInt, tunnels)
}.toList

case class RoomsInfo(
  /** map of rooms by id */
  rooms: Map[Id, Room],
  /** map from starting room to a map containing the best distance to all other rooms */
  routes: Map[Id, Map[Id, Int]],
  /** rooms containing non-zero-flow valves */
  valves: Set[Id]
)

// precalculate useful things like pathfinding
def constructInfo(input: Input): RoomsInfo =
  val rooms: Map[Id, Room]          = Map.from(for r <- input yield r.id -> r)
  val valves: Set[Id]               = Set.from(for r <- input if r.flow > 0 yield r.id)
  val tunnels: Map[Id, List[Id]]    = rooms.mapValues(_.tunnels).toMap
  val routes: Map[Id, Map[Id, Int]] = (valves + "AA").iterator.map{ id => id -> computeRoutes(id, tunnels) }.toMap
  RoomsInfo(rooms, routes, valves)

// a modified A-star to calculate the best distance to all rooms rather then the best path to a single room
def computeRoutes(start: Id, neighbors: Id => List[Id]): Map[Id, Int] =

  case class State(frontier: List[(Id, Int)], scores: Map[Id, Int]):

    private def getScore(id: Id): Int = scores.getOrElse(id, Int.MaxValue)
    private def setScore(id: Id, s: Int) = State((id, s + 1) :: frontier, scores + (id -> s))

    def dequeued: (Id, State) =
      val sorted = frontier.sortBy(_._2)
      (sorted.head._1, copy(frontier = sorted.tail))

    def considerEdge(from: Id, to: Id): State =
      val toScore = getScore(from) + 1
      if toScore >= getScore(to) then this
      else setScore(to, toScore)
  end State

  object State:
    def initial(start: Id) = State(List((start, 0)), Map(start -> 0))

  def recurse(state: State): State =
    if state.frontier.isEmpty then
      state
    else
      val (curr, currState) = state.dequeued
      val newState = neighbors(curr)
        .foldLeft(currState) { (s, n) =>
          s.considerEdge(curr, n)
        }
      recurse(newState)

  recurse(State.initial(start)).scores

end computeRoutes


// find the best path (the order of valves to open) and the total pressure released by taking it
def bestPath(map: RoomsInfo, start: Id, valves: Set[Id], timeAllowed: Int): Int =
  // each step involves moving to a room with a useful valve and opening it
  // we don't need to track each (empty) room in between
  // we limit our options by only considering the still-closed valves
  // and `valves` has already culled any room with a flow value of 0 -- no point in considering these rooms!

  def recurse(path: List[Id], valvesLeft: Set[Id], timeLeft: Int, totalValue: Int): Int =
    // recursively consider all plausible options
    // we are finished when we no longer have time to reach another valve or all valves are open
    valvesLeft
      .flatMap{ id =>
        val current  = path.head
        val distance = map.routes(current)(id)
        // how much time is left after we traverse there and open the valve?
        val t = timeLeft - distance - 1
        // if `t` is zero or less this option can be skipped
        Option.when(t > 0) {
          // the value of choosing a particular valve (over the life of our simulation)
          // is its flow rate multiplied by the time remaining after opening it
          val value = map.rooms(id).flow * t
          recurse(id :: path, valvesLeft - id, t, totalValue + value)
        }
      }
      .maxOption
      .getOrElse { totalValue }
  end recurse
  recurse(start :: Nil, valves, timeAllowed, 0)

def part1(input: String) =
  val time   = 30
  val map    = constructInfo(parse(input))
  bestPath(map, "AA", map.valves, time)
end part1

def part2(input: String) =
  val time = 26
  val map  = constructInfo(parse(input))

  // in the optimal solution, the elephant and I will have divided responsibility for switching the valves
  // 15 (useful valves) choose 7 (half) yields only 6435 possible divisions which is a reasonable search space!
  val valvesA = map.valves.toList
    .combinations(map.valves.size / 2)
    .map(_.toSet)

  // NOTE: I assumed an even ditribution of valves would be optimal, and that turned out to be true.
  // However I suppose it's possible an uneven distribution could have been optimal for some graphs.
  // To be safe, you could re-run this using all reasonable values of `n` for `combinations` (1 to 7) and
  // taking the best of those.

  // we can now calculate the efforts separately and sum their values to find the best
  val allPaths =
    for va <- valvesA yield
      val vb              = map.valves -- va
      val scoreA = bestPath(map, "AA", va, time)
      val scoreB = bestPath(map, "AA", vb, time)
      scoreA + scoreB

  allPaths.max
end part2

Run it in the browser​

Part 1​

Warning: This is pretty slow and may cause the UI to freeze (close tab if problematic)

Part 2​

Warning: This is pretty slow and may cause the UI to freeze (close tab if problematic)

Solutions from the community​

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- + \ No newline at end of file diff --git a/2022/puzzles/day17/index.html b/2022/puzzles/day17/index.html index 8531112bd..21e148384 100644 --- a/2022/puzzles/day17/index.html +++ b/2022/puzzles/day17/index.html @@ -5,13 +5,13 @@ Day 17: Pyroclastic Flow | Scala Center Advent of Code - +
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- + \ No newline at end of file diff --git a/2022/puzzles/day18/index.html b/2022/puzzles/day18/index.html index 4114459cb..350e2396d 100644 --- a/2022/puzzles/day18/index.html +++ b/2022/puzzles/day18/index.html @@ -3,15 +3,15 @@ -Day 18: Boiling Boulders | Scala Center Advent of Code +Day 18: Boiling Boulders | Scala Center Advent of Code - +
-
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Day 18: Boiling Boulders

Puzzle description​

https://adventofcode.com/2022/day/18

Solution​

Part 1​

To solve the first part, we can first count the total number of cubes and multiply this by six (as a cube has six sides), and then subtract the number of sides which are connected.

As this requires checking if two cubes are adjacent, let's first define a function which we can use to determine cubes adjacent to a given cube:

def adjacent(x: Int, y: Int, z: Int): Set[(Int, Int, Int)] = {
  Set(
    (x + 1, y, z),
    (x - 1, y, z),
    (x, y + 1, z),
    (x, y - 1, z),
    (x, y, z + 1),
    (x, y, z - 1)
  )
}
info

Note that since cubes are given to be 1⨉1⨉1, they can be represented as a single integral (x, y, z) coordinate which makes up the input for the adjacent function. Then two cubes are adjacent (one of each of their sides touch) if and only if exactly one of their (x, y, z) components differ by one, and the rest by zero.

Now given our cubes, we can implement our strategy with a fold:

def sides(cubes: Set[(Int, Int, Int)]): Int = {
  cubes.foldLeft(0) { case (total, (x, y, z)) =>
    val adj = adjacent(x, y, z)
    val numAdjacent = adj.filter(cubes).size
    total + 6 - numAdjacent
  }
}

We use a Set for fast fast membership lookups which we need to determine which adjacent spaces for a given cube contain other cubes.

Part 2​

The second part is a bit more tricky. Lets introduce some nomenclature: we'll say a 1⨉1⨉1 empty space is on the interior if it lies in an air pocket, else we'll say the space is on the exterior.

A useful observation is that if we consider empty spaces which have a taxicab distance of at most two from any cube, and join these spaces into connected components, then the connected components we are left with form distinct air pockets in addition to one component containing empty spaces on the exterior.

This component can always be identified since the space with the largest x component will always lie in it. So we can determine empty spaces in the interior adjacent to cubes like so:

def interior(cubes: Set[(Int, Int, Int)]): Set[(Int, Int, Int)] = {
  val allAdj = cubes.flatMap((x, y, z) => adjacent(x, y, z).filterNot(cubes))
  val sts = allAdj.map { case adj @ (x, y, z) =>
    adjacent(x, y, z).filterNot(cubes) + adj
  }
  def cc(sts: List[Set[(Int, Int, Int)]]): List[Set[(Int, Int, Int)]] = {
    sts match {
      case Nil => Nil
      case set :: rst =>
        val (matching, other) = rst.partition(s => s.intersect(set).nonEmpty)
        val joined = matching.foldLeft(set)(_ ++ _)
        if (matching.nonEmpty) cc(joined :: other) else joined :: cc(other)
    }
  }
  val conn = cc(sts.toList)
  val exterior = conn.maxBy(_.maxBy(_(0)))
  conn.filterNot(_ == exterior).foldLeft(Set())(_ ++ _)
}

Where the nested function cc is used to generate a list of connected components. We can now slightly modify our sides function to complete part two:

def sidesNoPockets(cubes: Set[(Int, Int, Int)]): Int = {
  val int = interior(cubes)
  val allAdj = cubes.flatMap(adjacent)
  allAdj.foldLeft(sides(cubes)) { case (total, (x, y, z)) =>
    val adj = adjacent(x, y, z)
    if (int((x, y, z))) total - adj.filter(cubes).size else total
  }
}

Let's put this all together:

def part1(input: String): Int = sides(cubes(input))
def part2(input: String): Int = sidesNoPockets(cubes(input))

def cubes(input: String): Set[(Int, Int, Int)] =
  val cubesIt = input.linesIterator.collect {
    case s"$x,$y,$z" => (x.toInt, y.toInt, z.toInt)
  }
  cubesIt.toSet

Which gives use our desired results.

Run it in the browser​

Part 1​

Part 2​

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- +
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Day 18: Boiling Boulders

by LaurenceWarne

Puzzle description​

https://adventofcode.com/2022/day/18

Solution​

Part 1​

To solve the first part, we can first count the total number of cubes and multiply this by six (as a cube has six sides), and then subtract the number of sides which are connected.

As this requires checking if two cubes are adjacent, let's first define a function which we can use to determine cubes adjacent to a given cube:

def adjacent(x: Int, y: Int, z: Int): Set[(Int, Int, Int)] = {
  Set(
    (x + 1, y, z),
    (x - 1, y, z),
    (x, y + 1, z),
    (x, y - 1, z),
    (x, y, z + 1),
    (x, y, z - 1)
  )
}
info

Note that since cubes are given to be 1⨉1⨉1, they can be represented as a single integral (x, y, z) coordinate which makes up the input for the adjacent function. Then two cubes are adjacent (one of each of their sides touch) if and only if exactly one of their (x, y, z) components differ by one, and the rest by zero.

Now given our cubes, we can implement our strategy with a fold:

def sides(cubes: Set[(Int, Int, Int)]): Int = {
  cubes.foldLeft(0) { case (total, (x, y, z)) =>
    val adj = adjacent(x, y, z)
    val numAdjacent = adj.filter(cubes).size
    total + 6 - numAdjacent
  }
}

We use a Set for fast fast membership lookups which we need to determine which adjacent spaces for a given cube contain other cubes.

Part 2​

The second part is a bit more tricky. Lets introduce some nomenclature: we'll say a 1⨉1⨉1 empty space is on the interior if it lies in an air pocket, else we'll say the space is on the exterior.

A useful observation is that if we consider empty spaces which have a taxicab distance of at most two from any cube, and join these spaces into connected components, then the connected components we are left with form distinct air pockets in addition to one component containing empty spaces on the exterior.

This component can always be identified since the space with the largest x component will always lie in it. So we can determine empty spaces in the interior adjacent to cubes like so:

def interior(cubes: Set[(Int, Int, Int)]): Set[(Int, Int, Int)] = {
  val allAdj = cubes.flatMap((x, y, z) => adjacent(x, y, z).filterNot(cubes))
  val sts = allAdj.map { case adj @ (x, y, z) =>
    adjacent(x, y, z).filterNot(cubes) + adj
  }
  def cc(sts: List[Set[(Int, Int, Int)]]): List[Set[(Int, Int, Int)]] = {
    sts match {
      case Nil => Nil
      case set :: rst =>
        val (matching, other) = rst.partition(s => s.intersect(set).nonEmpty)
        val joined = matching.foldLeft(set)(_ ++ _)
        if (matching.nonEmpty) cc(joined :: other) else joined :: cc(other)
    }
  }
  val conn = cc(sts.toList)
  val exterior = conn.maxBy(_.maxBy(_(0)))
  conn.filterNot(_ == exterior).foldLeft(Set())(_ ++ _)
}

Where the nested function cc is used to generate a list of connected components. We can now slightly modify our sides function to complete part two:

def sidesNoPockets(cubes: Set[(Int, Int, Int)]): Int = {
  val int = interior(cubes)
  val allAdj = cubes.flatMap(adjacent)
  allAdj.foldLeft(sides(cubes)) { case (total, (x, y, z)) =>
    val adj = adjacent(x, y, z)
    if (int((x, y, z))) total - adj.filter(cubes).size else total
  }
}

Let's put this all together:

def part1(input: String): Int = sides(cubes(input))
def part2(input: String): Int = sidesNoPockets(cubes(input))

def cubes(input: String): Set[(Int, Int, Int)] =
  val cubesIt = input.linesIterator.collect {
    case s"$x,$y,$z" => (x.toInt, y.toInt, z.toInt)
  }
  cubesIt.toSet

Which gives use our desired results.

Run it in the browser​

Part 1​

Part 2​

Solutions from the community​

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+ \ No newline at end of file diff --git a/2022/puzzles/day19/index.html b/2022/puzzles/day19/index.html index c4e2eaf98..008b1a791 100644 --- a/2022/puzzles/day19/index.html +++ b/2022/puzzles/day19/index.html @@ -5,13 +5,13 @@ Day 19: Not Enough Minerals | Scala Center Advent of Code - +
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- + \ No newline at end of file diff --git a/2022/puzzles/day20/index.html b/2022/puzzles/day20/index.html index 3556d08fc..11e3fe28c 100644 --- a/2022/puzzles/day20/index.html +++ b/2022/puzzles/day20/index.html @@ -5,13 +5,13 @@ Day 20: Grove Positioning System | Scala Center Advent of Code - +
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- + \ No newline at end of file diff --git a/2022/puzzles/day21/index.html b/2022/puzzles/day21/index.html index 4fda73253..2172a5789 100644 --- a/2022/puzzles/day21/index.html +++ b/2022/puzzles/day21/index.html @@ -5,13 +5,13 @@ Day 21: Monkey Math | Scala Center Advent of Code - +
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Day 21: Monkey Math

Puzzle description​

https://adventofcode.com/2022/day/21

Final Code​

import annotation.tailrec
import Operation.*

def part1(input: String): Long =
  resolveRoot(input)

def part2(input: String): Long =
  whichValue(input)

enum Operator(val eval: BinOp, val invRight: BinOp, val invLeft: BinOp):
  case `+` extends Operator(_ + _, _ - _, _ - _)
  case `-` extends Operator(_ - _, _ + _, (x, y) => y - x)
  case `*` extends Operator(_ * _, _ / _, _ / _)
  case `/` extends Operator(_ / _, _ * _, (x, y) => y / x)

enum Operation:
  case Binary(op: Operator, depA: String, depB: String)
  case Constant(value: Long)

type BinOp = (Long, Long) => Long
type Resolved = Map[String, Long]
type Source = Map[String, Operation]
type Substitutions = List[(String, PartialFunction[Operation, Operation])]

def readAll(input: String): Map[String, Operation] =
  Map.from(
    for case s"$name: $action" <- input.linesIterator yield
      name -> action.match
        case s"$x $binop $y" =>
          Binary(Operator.valueOf(binop), x, y)
        case n =>
          Constant(n.toLong)
  )

@tailrec
def reachable(names: List[String], source: Source, resolved: Resolved): Resolved = names match
  case name :: rest =>
    source.get(name) match
      case None => resolved // return as name is not reachable
      case Some(operation) => operation match
        case Binary(op, x, y) =>
          (resolved.get(x), resolved.get(y)) match
            case (Some(a), Some(b)) =>
              reachable(rest, source, resolved + (name -> op.eval(a, b)))
            case _ =>
              reachable(x :: y :: name :: rest, source, resolved)
        case Constant(value) =>
          reachable(rest, source, resolved + (name -> value))
  case Nil =>
    resolved
end reachable

def resolveRoot(input: String): Long =
  val values = reachable("root" :: Nil, readAll(input), Map.empty)
  values("root")

def whichValue(input: String): Long =
  val source = readAll(input) - "humn"

  @tailrec
  def binarySearch(name: String, goal: Option[Long], resolved: Resolved): Long =

    def resolve(name: String) =
      val values = reachable(name :: Nil, source, resolved)
      values.get(name).map(_ -> values)

    def nextGoal(inv: BinOp, value: Long): Long = goal match
      case Some(prev) => inv(prev, value)
      case None => value

    (source.get(name): @unchecked) match
      case Some(Operation.Binary(op, x, y)) =>
        ((resolve(x), resolve(y)): @unchecked) match
          case (Some(xValue -> resolvedX), _) => // x is known, y has a hole
            binarySearch(y, Some(nextGoal(op.invLeft, xValue)), resolvedX)
          case (_, Some(yValue -> resolvedY)) => // y is known, x has a hole
            binarySearch(x, Some(nextGoal(op.invRight, yValue)), resolvedY)
      case None =>
        goal.get // hole found
  end binarySearch

  binarySearch(goal = None, name = "root", resolved = Map.empty)
end whichValue

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Part 1​

Part 2​

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/2022/puzzles/day22/index.html b/2022/puzzles/day22/index.html index df762b371..f37334c3f 100644 --- a/2022/puzzles/day22/index.html +++ b/2022/puzzles/day22/index.html @@ -5,13 +5,13 @@ Day 22: Monkey Map | Scala Center Advent of Code - +
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- + \ No newline at end of file diff --git a/2022/puzzles/day23/index.html b/2022/puzzles/day23/index.html index ae75ede45..cd518d03d 100644 --- a/2022/puzzles/day23/index.html +++ b/2022/puzzles/day23/index.html @@ -5,13 +5,13 @@ Day 23: Unstable Diffusion | Scala Center Advent of Code - +
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- + \ No newline at end of file diff --git a/2022/puzzles/day24/index.html b/2022/puzzles/day24/index.html index 52d058abc..fa1c21303 100644 --- a/2022/puzzles/day24/index.html +++ b/2022/puzzles/day24/index.html @@ -5,13 +5,13 @@ Day 24: Blizzard Basin | Scala Center Advent of Code - +
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Day 24: Blizzard Basin

Puzzle description​

https://adventofcode.com/2022/day/24

Solution​

Today's problem is similar to Day 12, where we need to find our way through a maze. It's made more challenging by impassable blizzards moving through the maze. We can use a similar approach to that of Day 12 still, but we'll improve a little bit further by using A* search instead of a standard breadth first search.

We'll need some kind of point and a few functions that are useful on the 2d grid. A simple tuple (Int, Int) will suffice, and we'll add the functions as extension methods. We'll use Manhattan distance as the A* heuristic function, and we'll need the neighbours in cardinal directions.

type Coord = (Int, Int)
extension (coord: Coord)
  def x = coord._1
  def y = coord._2
  def up = (coord.x, coord.y - 1)
  def down = (coord.x, coord.y + 1)
  def left = (coord.x - 1, coord.y)
  def right = (coord.x + 1, coord.y)
  def cardinals = Seq(coord.up, coord.down, coord.left, coord.right)
  def manhattan(rhs: Coord) = (coord.x - rhs.x).abs + (coord.y - rhs.y).abs
  def +(rhs: Coord) = (coord.x + rhs.x, coord.y + rhs.y)

Before we get to the search, let's deal with the input.

case class Blizzard(at: Coord, direction: Coord)

def parseMaze(in: Seq[String]) =
  val start = (in.head.indexOf('.'), 0) // start in the empty spot in the top row
  val end = (in.last.indexOf('.'), in.size - 1) // end in the empty spot in the bottom row
  val xDomain = 1 to in.head.size - 2 // where blizzards are allowed to go
  val yDomain = 1 to in.size - 2
  val initialBlizzards =
    for
      y <- in.indices
      x <- in(y).indices
      if in(y)(x) != '.' // these aren't blizzards!
      if in(y)(x) != '#'
    yield in(y)(x) match
      case '>' => Blizzard(at = (x, y), direction = (1, 0))
      case '<' => Blizzard(at = (x, y), direction = (-1, 0))
      case '^' => Blizzard(at = (x, y), direction = (0, -1))
      case 'v' => Blizzard(at = (x, y), direction = (0, 1))

  ??? // ...to be implemented

Ok, let's deal with the blizzards. The blizzards move toroidally, which is to say they loop around back to the start once they fall off an edge. This means that, eventually, the positions and directions of all blizzards must loop at some point. Naively, after xDomain.size * yDomain.size minutes, every blizzard must have returned to it's original starting location. Let's model that movement and calculate the locations of all the blizzards up until that time. With it, we'll have a way to tell us where the blizzards are at a given time t, for any t.

def move(blizzard: Blizzard, xDomain: Range, yDomain: Range) =
  blizzard.copy(at = cycle(blizzard.at + blizzard.direction, xDomain, yDomain))

def cycle(coord: Coord, xDomain: Range, yDomain: Range): Coord = (cycle(coord.x, xDomain), cycle(coord.y, yDomain))

def cycle(n: Int, bounds: Range): Int =
  if n > bounds.max then bounds.min // we've fallen off the end, go to start
  else if n < bounds.min then bounds.max // we've fallen off the start, go to the end
  else n // we're chillin' in bounds still

We can replace the ??? in parseMaze now. And we'll need a return type for the function. We can cram everything into a Maze case class. For the blizzards, we actually only need to care about where they are after this point, as they'll prevent us from moving to those locations. We'll throw away the directions and just keep the set of Coords the blizzards are at.

case class Maze(xDomain: Range, yDomain: Range, blizzards: Seq[Set[Coord]], start: Coord, end: Coord)

def parseMaze(in: Seq[String]): Maze =
  /* ...omitted for brevity... */
  def tick(blizzards: Seq[Blizzard]) = blizzards.map(move(_, xDomain, yDomain))
  val allBlizzardLocations = Iterator.iterate(initialBlizzards)(tick)
      .take(xDomain.size * yDomain.size)
      .map(_.map(_.at).toSet)
      .toIndexedSeq

  Maze(xDomain, yDomain, allBlizzardLocations, start, end)

But! We can do a little better for the blizzards. The blizzards actually cycle for any common multiple of xDomain.size and yDomain.size. Using the least common multiple would be sensible to do the least amount of computation.

def gcd(a: Int, b: Int): Int = if b == 0 then a else gcd(b, a % b)
def lcm(a: Int, b: Int): Int = a * b / gcd(a, b)
def tick(blizzards: Seq[Blizzard]) = blizzards.map(move(_, xDomain, yDomain))
val allBlizzardLocations = Iterator.iterate(initialBlizzards)(tick)
    .take(lcm(xDomain.size, yDomain.size))
    .map(_.map(_.at).toSet)
    .toIndexedSeq

Great! Let's solve the maze.

import scala.collection.mutable
case class Step(at: Coord, time: Int)

def solve(maze: Maze): Step =
  // order A* options by how far we've taken + an estimated distance to the end
  given Ordering[Step] = Ordering[Int].on((step: Step) => step.at.manhattan(maze.end) + step.time).reverse
  val queue = mutable.PriorityQueue[Step]()
  val visited = mutable.Set.empty[Step]
  
  def inBounds(coord: Coord) = coord match
    case c if c == maze.start || c == maze.end => true
    case c => maze.xDomain.contains(c.x) && maze.yDomain.contains(c.y)

  queue += Step(at = maze.start, time = 0)
  while queue.head.at != maze.end do
    val step = queue.dequeue
    val time = step.time + 1
    // where are the blizzards for our next step? we can't go there
    val blizzards = maze.blizzards(time % maze.blizzards.size)
    // we can move in any cardinal direction, or chose to stay put; but it needs to be in the maze
    val options = (step.at.cardinals :+ step.at).filter(inBounds).map(Step(_, time))
    // queue up the options if they are possible; and if we have not already queued them
    queue ++= options
      .filterNot(o => blizzards(o.at)) // the option must not be in a blizzard
      .filterNot(visited) // avoid duplicate work
      .tapEach(visited.add) // keep track of what we've enqueued

  queue.dequeue

That's pretty much it! Part 1 is then:

def part1(in: Seq[String]) = solve(parseMaze(in)).time

Part 2 requires solving the maze 3 times. Make it to the end (so, solve part 1 again), go back to the start, then go back to the end. We can use the same solve function, but we need to generalize a bit so we can start the solver at an arbitrary time. This will allow us to keep the state of the blizzards for subsequent runs. We actually only need to change one line!

def solve(maze: Maze, startingTime: Int = 0): Step =
  /* the only line we need to change is... */
  queue += Step(at = maze.start, time = startingTime)

Then part 2 requires calling solve 3 times. We need to be a little careful with the start/end locations and starting times.

def part2(in: Seq[String]) =
  val maze = parseMaze(in)
  val first = solve(maze)
  val second = solve(maze.copy(start = maze.end, end = maze.start), first.time)
  solve(maze, second.time).time

That's Day 24. Huzzah!

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/2022/puzzles/day25/index.html b/2022/puzzles/day25/index.html index 088f87361..43e0e3bb0 100644 --- a/2022/puzzles/day25/index.html +++ b/2022/puzzles/day25/index.html @@ -5,13 +5,13 @@ Day 25: Full of Hot Air | Scala Center Advent of Code - +
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Day 25: Full of Hot Air

Puzzle description​

https://adventofcode.com/2022/day/25

Final Code​

def part1(input: String): String =
  totalSnafu(input)

val digitToInt = Map(
  '0' -> 0,
  '1' -> 1,
  '2' -> 2,
  '-' -> -1,
  '=' -> -2,
)
val intToDigit = digitToInt.map(_.swap)

def showSnafu(value: Long): String =
  val reverseDigits = Iterator.unfold(value)(v =>
    Option.when(v != 0) {
      val mod = math.floorMod(v, 5).toInt
      val digit = if mod > 2 then mod - 5 else mod
      intToDigit(digit) -> (v - digit) / 5
    }
  )
  if reverseDigits.isEmpty then "0"
  else reverseDigits.mkString.reverse

def readSnafu(line: String): Long =
  line.foldLeft(0L)((acc, digit) =>
    acc * 5 + digitToInt(digit)
  )

def totalSnafu(input: String): String =
  showSnafu(value = input.linesIterator.map(readSnafu).sum)

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Part 1 (Only 1 part today)​

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/2023/index.html b/2023/index.html index 091ae950e..d04b55b0d 100644 --- a/2023/index.html +++ b/2023/index.html @@ -5,13 +5,13 @@ Scala Center Advent of Code | Scala Center Advent of Code - +
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Scala Advent of Code 2023 byScala Center

Day 1
Credit to https://github.com/OlegIlyenko/scala-icon

Learn Scala 3

A simpler, safer and more concise version of Scala, the famous object-oriented and functional programming language.

Solve Advent of Code puzzles

Challenge your programming skills by solving Advent of Code puzzles.

Share with the community

Get or give support to the community. Share your solutions with the community.

- + \ No newline at end of file diff --git a/2023/puzzles/day01/index.html b/2023/puzzles/day01/index.html index adeb9a9cd..55249b0ea 100644 --- a/2023/puzzles/day01/index.html +++ b/2023/puzzles/day01/index.html @@ -5,13 +5,13 @@ Day 1 | Scala Center Advent of Code - +
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Day 1

Puzzle description​

Come back on December 1st 2023 for the challenge description. While you wait, read more at https://adventofcode.com/2023.

Solutions from the community​

When the challenge is opened, feel free to edit this page to add a link to your solution here.

- + \ No newline at end of file diff --git a/404.html b/404.html index 8fb73d855..64e1afcf7 100644 --- a/404.html +++ b/404.html @@ -5,13 +5,13 @@ Page Not Found | Scala Center Advent of Code - +
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Page Not Found

We could not find what you were looking for.

Please contact the owner of the site that linked you to the original URL and let them know their link is broken.

- + \ No newline at end of file diff --git a/assets/js/935f2afb.0795cfa4.js b/assets/js/935f2afb.3f9e49c0.js similarity index 66% rename from assets/js/935f2afb.0795cfa4.js rename to assets/js/935f2afb.3f9e49c0.js index 1c38970b8..f8aa8c945 100644 --- a/assets/js/935f2afb.0795cfa4.js +++ b/assets/js/935f2afb.3f9e49c0.js @@ -1 +1 @@ -"use strict";(self.webpackChunkwebsite=self.webpackChunkwebsite||[]).push([[53],{1109:e=>{e.exports=JSON.parse('{"pluginId":"default","version":"current","label":"Next","banner":null,"badge":false,"noIndex":false,"className":"docs-version-current","isLast":true,"docsSidebars":{"adventOfCodeSidebar":[{"type":"link","label":"Introduction","href":"/scala-advent-of-code/introduction","docId":"introduction"},{"type":"link","label":"Setup","href":"/scala-advent-of-code/setup","docId":"setup"},{"type":"category","label":"2023 Puzzles","items":[{"type":"link","label":"Day 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Reservoir","href":"/scala-advent-of-code/2022/puzzles/day14","docId":"2022/puzzles/day14"},{"type":"link","label":"Day 15: Beacon Exclusion Zone","href":"/scala-advent-of-code/2022/puzzles/day15","docId":"2022/puzzles/day15"},{"type":"link","label":"Day 16: Proboscidea Volcanium","href":"/scala-advent-of-code/2022/puzzles/day16","docId":"2022/puzzles/day16"},{"type":"link","label":"Day 17: Pyroclastic Flow","href":"/scala-advent-of-code/2022/puzzles/day17","docId":"2022/puzzles/day17"},{"type":"link","label":"Day 18: Boiling Boulders","href":"/scala-advent-of-code/2022/puzzles/day18","docId":"2022/puzzles/day18"},{"type":"link","label":"Day 19: Not Enough Minerals","href":"/scala-advent-of-code/2022/puzzles/day19","docId":"2022/puzzles/day19"},{"type":"link","label":"Day 20: Grove Positioning System","href":"/scala-advent-of-code/2022/puzzles/day20","docId":"2022/puzzles/day20"},{"type":"link","label":"Day 21: Monkey Math","href":"/scala-advent-of-code/2022/puzzles/day21","docId":"2022/puzzles/day21"},{"type":"link","label":"Day 22: Monkey Map","href":"/scala-advent-of-code/2022/puzzles/day22","docId":"2022/puzzles/day22"},{"type":"link","label":"Day 23: Unstable Diffusion","href":"/scala-advent-of-code/2022/puzzles/day23","docId":"2022/puzzles/day23"},{"type":"link","label":"Day 24: Blizzard Basin","href":"/scala-advent-of-code/2022/puzzles/day24","docId":"2022/puzzles/day24"},{"type":"link","label":"Day 25: Full of Hot Air","href":"/scala-advent-of-code/2022/puzzles/day25","docId":"2022/puzzles/day25"}],"collapsed":true,"collapsible":true},{"type":"category","label":"2021 Puzzles","items":[{"type":"link","label":"Day 1: Sonar Sweep","href":"/scala-advent-of-code/puzzles/day1","docId":"puzzles/day1"},{"type":"link","label":"Day 2: Dive!","href":"/scala-advent-of-code/puzzles/day2","docId":"puzzles/day2"},{"type":"link","label":"Day 3: Binary 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solve the first part, we can first count the total number of cubes and multiply this by six (as a cube has six sides), and then subtract the number of sides which are connected."),(0,i.kt)("p",null,"As this requires checking if two cubes are adjacent, let's first define a function which we can use to determine cubes adjacent to a given cube:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def adjacent(x: Int, y: Int, z: Int): Set[(Int, Int, Int)] = {\n Set(\n (x + 1, y, z),\n (x - 1, y, z),\n (x, y + 1, z),\n (x, y - 1, z),\n (x, y, z + 1),\n (x, y, z - 1)\n )\n}\n")),(0,i.kt)("admonition",{type:"info"},(0,i.kt)("p",{parentName:"admonition"},"Note that since cubes are given to be 1\u2a091\u2a091, they can be represented as a single integral ",(0,i.kt)("inlineCode",{parentName:"p"},"(x, y, z)")," coordinate which makes up the input for the ",(0,i.kt)("inlineCode",{parentName:"p"},"adjacent")," function. Then two cubes are adjacent (one of each of their sides touch) if and only if exactly one of their ",(0,i.kt)("inlineCode",{parentName:"p"},"(x, y, z)")," components differ by one, and the rest by zero.")),(0,i.kt)("p",null,"Now given our cubes, we can implement our strategy with a ",(0,i.kt)("inlineCode",{parentName:"p"},"fold"),":"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def sides(cubes: Set[(Int, Int, Int)]): Int = {\n cubes.foldLeft(0) { case (total, (x, y, z)) =>\n val adj = adjacent(x, y, z)\n val numAdjacent = adj.filter(cubes).size\n total + 6 - numAdjacent\n }\n}\n")),(0,i.kt)("p",null,"We use a ",(0,i.kt)("inlineCode",{parentName:"p"},"Set")," for fast ",(0,i.kt)("a",{parentName:"p",href:"https://docs.scala-lang.org/overviews/collections-2.13/performance-characteristics.html"},"fast")," membership lookups which we need to determine which adjacent spaces for a given cube contain other cubes."),(0,i.kt)("h3",{id:"part-2"},"Part 2"),(0,i.kt)("p",null,"The second part is a bit more tricky. Lets introduce some nomenclature: we'll say a 1\u2a091\u2a091 empty space is on the ",(0,i.kt)("em",{parentName:"p"},"interior")," if it lies in an air pocket, else we'll say the space is on the ",(0,i.kt)("em",{parentName:"p"},"exterior"),"."),(0,i.kt)("p",null,"A useful observation is that if we consider empty spaces which have a ",(0,i.kt)("a",{parentName:"p",href:"https://adventofcode.com/2022/day/18"},"taxicab")," distance of at most two from any cube, and join these spaces into ",(0,i.kt)("a",{parentName:"p",href:"https://en.wikipedia.org/wiki/Component_(graph_theory)"},"connected components"),", then the connected components we are left with form distinct air pockets in addition to one component containing empty spaces on the exterior."),(0,i.kt)("p",null,"This component can always be identified since the space with the largest ",(0,i.kt)("inlineCode",{parentName:"p"},"x")," component will always lie in it. So we can determine empty spaces in the interior adjacent to cubes like so:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def interior(cubes: Set[(Int, Int, Int)]): Set[(Int, Int, Int)] = {\n val allAdj = cubes.flatMap((x, y, z) => adjacent(x, y, z).filterNot(cubes))\n val sts = allAdj.map { case adj @ (x, y, z) =>\n adjacent(x, y, z).filterNot(cubes) + adj\n }\n def cc(sts: List[Set[(Int, Int, Int)]]): List[Set[(Int, Int, Int)]] = {\n sts match {\n case Nil => Nil\n case set :: rst =>\n val (matching, other) = rst.partition(s => s.intersect(set).nonEmpty)\n val joined = matching.foldLeft(set)(_ ++ _)\n if (matching.nonEmpty) cc(joined :: other) else joined :: cc(other)\n }\n }\n val conn = cc(sts.toList)\n val exterior = conn.maxBy(_.maxBy(_(0)))\n conn.filterNot(_ == exterior).foldLeft(Set())(_ ++ _)\n}\n")),(0,i.kt)("p",null,"Where the nested function ",(0,i.kt)("inlineCode",{parentName:"p"},"cc")," is used to generate a list of connected components. We can now slightly modify our ",(0,i.kt)("inlineCode",{parentName:"p"},"sides")," function to complete part two:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def sidesNoPockets(cubes: Set[(Int, Int, Int)]): Int = {\n val int = interior(cubes)\n val allAdj = cubes.flatMap(adjacent)\n allAdj.foldLeft(sides(cubes)) { case (total, (x, y, z)) =>\n val adj = adjacent(x, y, z)\n if (int((x, y, z))) total - adj.filter(cubes).size else total\n }\n}\n")),(0,i.kt)("p",null,"Let's put this all together:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},'def part1(input: String): Int = sides(cubes(input))\ndef part2(input: String): Int = sidesNoPockets(cubes(input))\n\ndef cubes(input: String): Set[(Int, Int, Int)] =\n val cubesIt = input.linesIterator.collect {\n case s"$x,$y,$z" => (x.toInt, y.toInt, z.toInt)\n }\n cubesIt.toSet\n')),(0,i.kt)("p",null,"Which gives use our desired results."),(0,i.kt)("h3",{id:"run-it-in-the-browser"},"Run it in the browser"),(0,i.kt)("h4",{id:"part-1-1"},"Part 1"),(0,i.kt)(o.Z,{puzzle:"day18-part1",year:"2022",mdxType:"Solver"}),(0,i.kt)("h4",{id:"part-2-1"},"Part 2"),(0,i.kt)(o.Z,{puzzle:"day18-part2",year:"2022",mdxType:"Solver"}),(0,i.kt)("h2",{id:"solutions-from-the-community"},"Solutions from the community"),(0,i.kt)("ul",null,(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/erikvanoosten/advent-of-code/blob/main/src/main/scala/nl/grons/advent/y2022/Day18.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/erikvanoosten"},"Erik van Oosten")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/cosminci/advent-of-code/blob/master/src/main/scala/com/github/cosminci/aoc/_2022/Day18.scala"},"Solution")," by Cosmin Ciobanu"),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/AvaPL/Advent-of-Code-2022/tree/main/src/main/scala/day18"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/AvaPL"},"Pawe\u0142 Cembaluk")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/w-r-z-k/aoc2022/blob/main/src/main/scala/Day18.scala"},"Solution")," by Richard W")),(0,i.kt)("p",null,"Share your solution to the Scala community by editing this page. (You can even write the whole article!)"))}u.isMDXComponent=!0}}]); \ No newline at end of file diff --git a/assets/js/e7ad7ee9.b6bf4f78.js b/assets/js/e7ad7ee9.b6bf4f78.js deleted file mode 100644 index 13a53b404..000000000 --- a/assets/js/e7ad7ee9.b6bf4f78.js +++ /dev/null @@ -1 +0,0 @@ -"use strict";(self.webpackChunkwebsite=self.webpackChunkwebsite||[]).push([[3360],{6640:(e,t,n)=>{n.r(t),n.d(t,{assets:()=>c,contentTitle:()=>l,default:()=>u,frontMatter:()=>s,metadata:()=>r,toc:()=>d});var a=n(7462),i=(n(7294),n(3905)),o=n(6340);const s={},l="Day 18: Boiling Boulders",r={unversionedId:"2022/puzzles/day18",id:"2022/puzzles/day18",title:"Day 18: Boiling Boulders",description:"Puzzle description",source:"@site/target/mdoc/2022/puzzles/day18.md",sourceDirName:"2022/puzzles",slug:"/2022/puzzles/day18",permalink:"/scala-advent-of-code/2022/puzzles/day18",draft:!1,editUrl:"https://github.com/scalacenter/scala-advent-of-code/edit/website/docs/2022/puzzles/day18.md",tags:[],version:"current",frontMatter:{},sidebar:"adventOfCodeSidebar",previous:{title:"Day 17: Pyroclastic Flow",permalink:"/scala-advent-of-code/2022/puzzles/day17"},next:{title:"Day 19: Not Enough Minerals",permalink:"/scala-advent-of-code/2022/puzzles/day19"}},c={},d=[{value:"Puzzle description",id:"puzzle-description",level:2},{value:"Solution",id:"solution",level:2},{value:"Part 1",id:"part-1",level:3},{value:"Part 2",id:"part-2",level:3},{value:"Run it in the browser",id:"run-it-in-the-browser",level:3},{value:"Part 1",id:"part-1-1",level:4},{value:"Part 2",id:"part-2-1",level:4},{value:"Solutions from the community",id:"solutions-from-the-community",level:2}],p={toc:d};function u(e){let{components:t,...n}=e;return(0,i.kt)("wrapper",(0,a.Z)({},p,n,{components:t,mdxType:"MDXLayout"}),(0,i.kt)("h1",{id:"day-18-boiling-boulders"},"Day 18: Boiling Boulders"),(0,i.kt)("h2",{id:"puzzle-description"},"Puzzle description"),(0,i.kt)("p",null,(0,i.kt)("a",{parentName:"p",href:"https://adventofcode.com/2022/day/18"},"https://adventofcode.com/2022/day/18")),(0,i.kt)("h2",{id:"solution"},"Solution"),(0,i.kt)("h3",{id:"part-1"},"Part 1"),(0,i.kt)("p",null,"To solve the first part, we can first count the total number of cubes and multiply this by six (as a cube has six sides), and then subtract the number of sides which are connected."),(0,i.kt)("p",null,"As this requires checking if two cubes are adjacent, let's first define a function which we can use to determine cubes adjacent to a given cube:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def adjacent(x: Int, y: Int, z: Int): Set[(Int, Int, Int)] = {\n Set(\n (x + 1, y, z),\n (x - 1, y, z),\n (x, y + 1, z),\n (x, y - 1, z),\n (x, y, z + 1),\n (x, y, z - 1)\n )\n}\n")),(0,i.kt)("admonition",{type:"info"},(0,i.kt)("p",{parentName:"admonition"},"Note that since cubes are given to be 1\u2a091\u2a091, they can be represented as a single integral ",(0,i.kt)("inlineCode",{parentName:"p"},"(x, y, z)")," coordinate which makes up the input for the ",(0,i.kt)("inlineCode",{parentName:"p"},"adjacent")," function. Then two cubes are adjacent (one of each of their sides touch) if and only if exactly one of their ",(0,i.kt)("inlineCode",{parentName:"p"},"(x, y, z)")," components differ by one, and the rest by zero.")),(0,i.kt)("p",null,"Now given our cubes, we can implement our strategy with a ",(0,i.kt)("inlineCode",{parentName:"p"},"fold"),":"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def sides(cubes: Set[(Int, Int, Int)]): Int = {\n cubes.foldLeft(0) { case (total, (x, y, z)) =>\n val adj = adjacent(x, y, z)\n val numAdjacent = adj.filter(cubes).size\n total + 6 - numAdjacent\n }\n}\n")),(0,i.kt)("p",null,"We use a ",(0,i.kt)("inlineCode",{parentName:"p"},"Set")," for fast ",(0,i.kt)("a",{parentName:"p",href:"https://docs.scala-lang.org/overviews/collections-2.13/performance-characteristics.html"},"fast")," membership lookups which we need to determine which adjacent spaces for a given cube contain other cubes."),(0,i.kt)("h3",{id:"part-2"},"Part 2"),(0,i.kt)("p",null,"The second part is a bit more tricky. Lets introduce some nomenclature: we'll say a 1\u2a091\u2a091 empty space is on the ",(0,i.kt)("em",{parentName:"p"},"interior")," if it lies in an air pocket, else we'll say the space is on the ",(0,i.kt)("em",{parentName:"p"},"exterior"),"."),(0,i.kt)("p",null,"A useful observation is that if we consider empty spaces which have a ",(0,i.kt)("a",{parentName:"p",href:"https://adventofcode.com/2022/day/18"},"taxicab")," distance of at most two from any cube, and join these spaces into ",(0,i.kt)("a",{parentName:"p",href:"https://en.wikipedia.org/wiki/Component_(graph_theory)"},"connected components"),", then the connected components we are left with form distinct air pockets in addition to one component containing empty spaces on the exterior."),(0,i.kt)("p",null,"This component can always be identified since the space with the largest ",(0,i.kt)("inlineCode",{parentName:"p"},"x")," component will always lie in it. So we can determine empty spaces in the interior adjacent to cubes like so:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def interior(cubes: Set[(Int, Int, Int)]): Set[(Int, Int, Int)] = {\n val allAdj = cubes.flatMap((x, y, z) => adjacent(x, y, z).filterNot(cubes))\n val sts = allAdj.map { case adj @ (x, y, z) =>\n adjacent(x, y, z).filterNot(cubes) + adj\n }\n def cc(sts: List[Set[(Int, Int, Int)]]): List[Set[(Int, Int, Int)]] = {\n sts match {\n case Nil => Nil\n case set :: rst =>\n val (matching, other) = rst.partition(s => s.intersect(set).nonEmpty)\n val joined = matching.foldLeft(set)(_ ++ _)\n if (matching.nonEmpty) cc(joined :: other) else joined :: cc(other)\n }\n }\n val conn = cc(sts.toList)\n val exterior = conn.maxBy(_.maxBy(_(0)))\n conn.filterNot(_ == exterior).foldLeft(Set())(_ ++ _)\n}\n")),(0,i.kt)("p",null,"Where the nested function ",(0,i.kt)("inlineCode",{parentName:"p"},"cc")," is used to generate a list of connected components. We can now slightly modify our ",(0,i.kt)("inlineCode",{parentName:"p"},"sides")," function to complete part two:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def sidesNoPockets(cubes: Set[(Int, Int, Int)]): Int = {\n val int = interior(cubes)\n val allAdj = cubes.flatMap(adjacent)\n allAdj.foldLeft(sides(cubes)) { case (total, (x, y, z)) =>\n val adj = adjacent(x, y, z)\n if (int((x, y, z))) total - adj.filter(cubes).size else total\n }\n}\n")),(0,i.kt)("p",null,"Let's put this all together:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},'def part1(input: String): Int = sides(cubes(input))\ndef part2(input: String): Int = sidesNoPockets(cubes(input))\n\ndef cubes(input: String): Set[(Int, Int, Int)] =\n val cubesIt = input.linesIterator.collect {\n case s"$x,$y,$z" => (x.toInt, y.toInt, z.toInt)\n }\n cubesIt.toSet\n')),(0,i.kt)("p",null,"Which gives use our desired results."),(0,i.kt)("h3",{id:"run-it-in-the-browser"},"Run it in the browser"),(0,i.kt)("h4",{id:"part-1-1"},"Part 1"),(0,i.kt)(o.Z,{puzzle:"day18-part1",year:"2022",mdxType:"Solver"}),(0,i.kt)("h4",{id:"part-2-1"},"Part 2"),(0,i.kt)(o.Z,{puzzle:"day18-part2",year:"2022",mdxType:"Solver"}),(0,i.kt)("h2",{id:"solutions-from-the-community"},"Solutions from the community"),(0,i.kt)("ul",null,(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/erikvanoosten/advent-of-code/blob/main/src/main/scala/nl/grons/advent/y2022/Day18.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/erikvanoosten"},"Erik van Oosten")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/cosminci/advent-of-code/blob/master/src/main/scala/com/github/cosminci/aoc/_2022/Day18.scala"},"Solution")," by Cosmin Ciobanu"),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/AvaPL/Advent-of-Code-2022/tree/main/src/main/scala/day18"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/AvaPL"},"Pawe\u0142 Cembaluk")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/w-r-z-k/aoc2022/blob/main/src/main/scala/Day18.scala"},"Solution")," by Richard W")),(0,i.kt)("p",null,"Share your solution to the Scala community by editing this page. 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Scala Advent of Code byScala Center

202120222023
Credit to https://github.com/OlegIlyenko/scala-icon

Learn Scala 3

A simpler, safer and more concise version of Scala, the famous object-oriented and functional programming language.

Solve Advent of Code puzzles

Challenge your programming skills by solving Advent of Code puzzles.

Share with the community

Get or give support to the community. Share your solutions with the community.

- + \ No newline at end of file diff --git a/introduction/index.html b/introduction/index.html index 552080751..03a750181 100644 --- a/introduction/index.html +++ b/introduction/index.html @@ -5,7 +5,7 @@ Introduction | Scala Center Advent of Code - + @@ -18,7 +18,7 @@ We will strive to only use the Scala standard library to solve the puzzles, so that no specific knowledge of external libraries is necessary to follow along.

Participate​

In addition, if you have your own solution for a puzzle and would like to share it, you can add a link to a gist in the solution page of the puzzles.

If you would like to discuss the puzzles with other developers, or discuss our solutions on the following day, drop by the the Scala Discord server, in the #advent-of-code channel. There you can also find a pinned message with the invite code for a private leaderboard including those from the Scala community for some friendly competition.

Do you want to get your hands dirty and solve the Advent of Code puzzles in Scala? Read ahead to the Setup page!

- + \ No newline at end of file diff --git a/puzzles/day1/index.html b/puzzles/day1/index.html index 1b0ba95a2..06325711e 100644 --- a/puzzles/day1/index.html +++ b/puzzles/day1/index.html @@ -5,7 +5,7 @@ Day 1: Sonar Sweep | Scala Center Advent of Code - + @@ -16,7 +16,7 @@ For example, the sliding window of size 3 of Seq(10, 20, 30, 40, 50) is:

Seq(Seq(10, 20, 30), Seq(20, 30, 40), Seq(30, 40, 50)).

We can generalize this procedure in a method that compute a sliding window of some size n on any sequence of elements. Such a method already exists in the Scala standard library under the name sliding. It returns an iterator of arrays.

$ Seq(10, 20, 30, 40, 50).sliding(3).toSeq
Seq(Array(10, 20, 30), Array(20, 30, 40), Array(30, 40, 50))

We can use the sliding method to make our code shorter and faster.

Final solution​

def part1(input: String): Int =
  val depths = input.linesIterator.map(_.toInt)
  val pairs = depths.sliding(2).map(arr => (arr(0), arr(1)))
  pairs.count((prev, next) => prev < next)

def part2(input: String): Int =
  val depths = input.linesIterator.map(_.toInt)
  val sums = depths.sliding(3).map(_.sum)
  val pairs = sums.sliding(2).map(arr => (arr(0), arr(1)))
  pairs.count((prev, next) => prev < next)

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

The you can run it with scala-cli:

$ scala-cli 2021 -M day1.part1
The answer is 1559

$ scala-cli 2021 -M template1.part2
The answer is 1600

You can replace the content of the input/day1 file with your own input from adventofcode.com to get your own solution.

Run it in the browser​

Part 1​

Part 2​

Bonus​

There is a trick to make the solution of part 2 even smaller.

Indeed comparing depths(i) + depths(i + 1) + depths(i + 2) with depths(i + 1) + depths(i + 2) + depths(i + 3) is the same as comparing depths(i) with depths(i + 3). So the second part of the puzzle is almost the same as the first part of the puzzle.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day10/index.html b/puzzles/day10/index.html index 5ba09100c..b5d73374c 100644 --- a/puzzles/day10/index.html +++ b/puzzles/day10/index.html @@ -5,7 +5,7 @@ Day 10: Syntax Scoring | Scala Center Advent of Code - + @@ -47,7 +47,7 @@ the element in the middle:

def part2(input: String): BigInt =
  val rows: LazyList[List[Symbol]] =
    input.linesIterator
      .to(LazyList)
      .map(parseRow)

  val scores =
    rows.map(checkChunks)
      .collect { case incomplete: CheckResult.Incomplete => incomplete.score }
      .toVector
      .sorted

  scores(scores.length / 2)

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day10.part1
The solution is 367059
$ scala-cli 2021 -M day10.part2
The solution is 1952146692

You can replace the content of the input/day10 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day11/index.html b/puzzles/day11/index.html index 4285cfd5d..2f88807c2 100644 --- a/puzzles/day11/index.html +++ b/puzzles/day11/index.html @@ -5,13 +5,13 @@ Day 11: Dumbo Octopus | Scala Center Advent of Code - +

Day 11: Dumbo Octopus

by @tgodzik

Puzzle description​

https://adventofcode.com/2021/day/11

Final Solution​

trait Step:
  def increment: Step
  def addFlashes(f: Int): Step
  def shouldStop: Boolean
  def currentFlashes: Int
  def stepNumber: Int

case class MaxIterStep(currentFlashes: Int, stepNumber: Int, max: Int) extends Step:
  def increment = this.copy(stepNumber = stepNumber + 1)
  def addFlashes(f: Int) = this.copy(currentFlashes = currentFlashes + f)
  def shouldStop = stepNumber == max

case class SynchronizationStep(
    currentFlashes: Int,
    stepNumber: Int,
    maxChange: Int,
    lastFlashes: Int
) extends Step:
  def increment = this.copy(stepNumber = stepNumber + 1)
  def addFlashes(f: Int) =
    this.copy(currentFlashes = currentFlashes + f, lastFlashes = currentFlashes)
  def shouldStop = currentFlashes - lastFlashes == maxChange

case class Point(x: Int, y: Int)
case class Octopei(inputMap: Map[Point, Int]):

  @tailrec
  private def propagate(
      toVisit: Queue[Point],
      alreadyFlashed: Set[Point],
      currentSituation: Map[Point, Int]
  ): Map[Point, Int] =
    toVisit.dequeueOption match
      case None => currentSituation
      case Some((point, dequeued)) =>
        currentSituation.get(point) match
          case Some(value) if value > 9 && !alreadyFlashed(point) =>
            val propagated =
              Seq(
                point.copy(x = point.x + 1),
                point.copy(x = point.x - 1),
                point.copy(y = point.y + 1),
                point.copy(y = point.y - 1),
                point.copy(x = point.x + 1, y = point.y + 1),
                point.copy(x = point.x + 1, y = point.y - 1),
                point.copy(x = point.x - 1, y = point.y + 1),
                point.copy(x = point.x - 1, y = point.y - 1)
              )
            val newSituation = propagated.foldLeft(currentSituation) {
              case (map, point) =>
                map.get(point) match
                  case Some(value) => map.updated(point, value + 1)
                  case _       => map
            }
            propagate(
              dequeued.appendedAll(propagated),
              alreadyFlashed + point,
              newSituation
            )
          case _ =>
            propagate(dequeued, alreadyFlashed, currentSituation)
  end propagate

  def simulate(step: Step) = simulateIter(step, inputMap)

  @tailrec
  private def simulateIter(
      step: Step,
      currentSituation: Map[Point, Int]
  ): Step =
    if step.shouldStop then step
    else
      val incremented = currentSituation.map { case (point, value) =>
        (point, value + 1)
      }
      val flashes = incremented.collect {
        case (point, value) if value > 9 => point
      }.toList
      val propagated = propagate(Queue(flashes*), Set.empty, incremented)
      val newFlashes = propagated.collect {
        case (point, value) if value > 9 => 1
      }.sum
      val zeroed = propagated.map {
        case (point, value) if value > 9 => (point, 0)
        case same            => same
      }
      simulateIter(step.increment.addFlashes(newFlashes), zeroed)
  end simulateIter

end Octopei

def part1(input: String) =
  val octopei = parse(input)
  val step = MaxIterStep(0, 0, 100)
  octopei.simulate(step).currentFlashes

def part2(input: String) =
  val octopei = parse(input)
  val step = SynchronizationStep(0, 0, octopei.inputMap.size, 0)
  octopei.simulate(step).stepNumber

Run it in the browser​

Part 1​

Part 2​

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day11.part1
The answer is: 1673

$ scala-cli 2021 -M day11.part2
The answer is: 279

You can replace the content of the input/day11 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day12/index.html b/puzzles/day12/index.html index 7ca5b0feb..3eb5e7a5e 100644 --- a/puzzles/day12/index.html +++ b/puzzles/day12/index.html @@ -5,13 +5,13 @@ Day 12: Passage Pathing | Scala Center Advent of Code - +
- + \ No newline at end of file diff --git a/puzzles/day13/index.html b/puzzles/day13/index.html index bdb7fb10e..86b8f97c6 100644 --- a/puzzles/day13/index.html +++ b/puzzles/day13/index.html @@ -5,7 +5,7 @@ Day 13: Transparent Origami | Scala Center Advent of Code - + @@ -22,7 +22,7 @@ Finally we convert this double array to a String with .map(_.mkString).mkString('\n').

def part2(input: String): String =
  val (dots, folds) = parseInstructions(input)
  val foldedDots = folds.foldLeft(dots)((dots, fold) => dots.map(fold.apply))

  val (width, height) = (foldedDots.map(_.x).max + 1, foldedDots.map(_.y).max + 1)
  val paper = Array.fill(height, width)('.')
  for dot <- foldedDots do paper(dot.y)(dot.x) = '#'

  paper.map(_.mkString).mkString("\n")

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day13.part1
The answer is: 788

$ scala-cli 2021 -M day10.part2
The answer is:
#..#...##.###..#..#.####.#..#.###...##.
#.#.....#.#..#.#.#..#....#..#.#..#.#..#
##......#.###..##...###..#..#.###..#...
#.#.....#.#..#.#.#..#....#..#.#..#.#.##
#.#..#..#.#..#.#.#..#....#..#.#..#.#..#
#..#..##..###..#..#.####..##..###...###

You can replace the content of the input/day13 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day14/index.html b/puzzles/day14/index.html index d76469abc..2341132a8 100644 --- a/puzzles/day14/index.html +++ b/puzzles/day14/index.html @@ -5,7 +5,7 @@ Day 14: Extended Polymerization | Scala Center Advent of Code - + @@ -25,7 +25,7 @@ We can use the sum of multisets, noted with ++ and Σ\Sigma, which accumulates counts.

With the definitions above, we can proceed with solving our problem.

For any string longer than 2 chars, we have the following property:

S(x1x2x3⋯xp,n)=Σi=1p−1(S(xixi+1,n))S(x_1 x_2 x_3 \cdots x_p, n) = \Sigma_{i=1}^{p-1} (S(x_i x_{i+1}, n))

In other words, SS for a string is the sum (a multiset sum) of SS for all the adjacent pairs in the string, with the same number of iterations nn. That is because each initial pair of letters (such as NN, NC and CB) expands independently of the others. Each initial char is counted exactly once in the final frequencies because it is counted as part of the expansion of the pair on its left, and not the expansion of the pair on its right (we always exclude the first char).

As a particular case of the above, for a string of 3 chars xzyxzy, we have

S(xzy,n)=S(xz,n)+S(zy,n)S(xzy, n) = S(xz, n) + S(zy, n)

For strings of length 2, we have two cases: n=0n = 0 and n>0n > 0.

Base case: a pair xyxy, and n=0n = 0

S(xy,0)={y→1} for all x,y (by definition)S(xy, 0) = \{ y \rightarrow 1 \} \text{ for all } x, y \text{ (by definition)}

Inductive case: a pair xyxy, and n>0n > 0

S(xy,n)=S(xzy,n−1) where z is the insertion char for the pair xy (by definition)=S(xz,n−1)+S(zy,n−1) – the particular case of 3 chars above\begin{aligned} S(xy, n) & = S(xzy, n-1) \text{ where $z$ is the insertion char for the pair $xy$ (by definition)} \\ & = S(xz, n-1) + S(zy, n-1) \text{ -- the particular case of 3 chars above} \end{aligned}

This means that the frequencies of letters after xyxy has produced its final polymer are equal to the sum of frequencies that xzyxzy produces after 1 fewer steps.

Now that we have an inductive definition of S(xy,n)S(xy, n) for all pairs, we can write an iterative algorithm that computes that for all possible pairs xyxy, for n∈[0,40]n \in \lbrack 0, 40 \rbrack :

// S : (charPair, n) -> frequencies
val S = mutable.Map.empty[(CharPair, Int), Frequencies]

// Base case: S(xy, 0) = {y -> 1} for all x, y
for (pair @ (first, second), insert) <- insertionRules do
  S((pair, 0)) = Map(second -> 1L)

// Recursive case S(xy, n) = S(xz, n - 1) + S(zy, n - 1) with z = insertionRules(xy)
for n <- 1 to 40 do
  for (pair, insert) <- insertionRules do
    val (x, y) = pair
    val z = insertionRules(pair)
    S((pair, n)) = addFrequencies(S((x, z), n - 1), S((z, y), n - 1))

where addFrequencies implements the multiset sum of two bags of frequencies:

def addFrequencies(a: Frequencies, b: Frequencies): Frequencies =
  b.foldLeft(a) { case (prev, (char, frequency)) =>
    prev + (char -> (prev.getOrElse(char, 0L) + frequency))
  }

Using the initial property of SS for strings longer than 2 chars, we can compute S(initialPolymer,40)S(\text{initialPolymer}, 40) from the compute S(xy,40)S(xy, 40):

// S(polymer, 40) = ∑(S(pair, 40))
val pairsInPolymer = initialPolymer.zip(initialPolymer.tail)
val polymerS = (for pair <- pairsInPolymer yield S(pair, 40)).reduce(addFrequencies)

And finally, we add the very first character, once, to compute the full frequencies:

// We have to add the very first char to get all the frequencies
val frequencies = addFrequencies(polymerS, Map(initialPolymer.head -> 1L))

After which we have all the pieces for part 2.

Final code for part 2​

def part2(input: String): Long =
  val (initialPolymer, insertionRules) = parseInput(input)

  // S : (charPair, n) -> frequencies of everything but the first char after n iterations from charPair
  val S = mutable.Map.empty[(CharPair, Int), Frequencies]

  // Base case: S(xy, 0) = {y -> 1} for all x, y
  for (pair @ (first, second), insert) <- insertionRules do
    S((pair, 0)) = Map(second -> 1L)

  // Recursive case S(xy, n) = S(xz, n - 1) + S(zy, n - 1) with z = insertionRules(xy)
  for n <- 1 to 40 do
    for (pair, insert) <- insertionRules do
      val (x, y) = pair
      val z = insertionRules(pair)
      S((pair, n)) = addFrequencies(S((x, z), n - 1), S((z, y), n - 1))

  // S(polymer, 40) = ∑(S(pair, 40))
  val pairsInPolymer = initialPolymer.zip(initialPolymer.tail)
  val polymerS = (for pair <- pairsInPolymer yield S(pair, 40)).reduce(addFrequencies)

  // We have to add the very first char to get all the frequencies
  val frequencies = addFrequencies(polymerS, Map(initialPolymer.head -> 1L))

  // Finally, we can finish the computation as in part 1
  val max = frequencies.values.max
  val min = frequencies.values.min
  max - min

def addFrequencies(a: Frequencies, b: Frequencies): Frequencies =
  b.foldLeft(a) { case (prev, (char, frequency)) =>
    prev + (char -> (prev.getOrElse(char, 0L) + frequency))
  }

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day14.part1
The answer is: 3306

$ scala-cli 2021 -M day14.part2
The answer is: 3760312702877

You can replace the content of the input/day14 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day15/index.html b/puzzles/day15/index.html index d8323fc2f..adadb239b 100644 --- a/puzzles/day15/index.html +++ b/puzzles/day15/index.html @@ -5,13 +5,13 @@ Day 15: Chiton | Scala Center Advent of Code - +

Day 15: Chiton

By @anatoliykmetyuk

Puzzle description​

https://adventofcode.com/2021/day/15

Problem​

The problem in its essence is that of finding the least-costly path through a graph. This problem is solved by Dijkstra's algorithm, nicely explained in this Computerphile video.

Domain Model​

The two domain entities we are working with are the game map and an individual cell of that map. In presence of the game map, a cell is fully described by a pair of its coordinates.

type Coord = (Int, Int)

The game map contains all the cells from the challenge input. It also defines the neighbours of a given cell, which we need to know for Dijkstra's algorithm. Finally, it defines a function to get the cost of entering a given cell.

class GameMap(cells: IndexedSeq[IndexedSeq[Int]]):
  val maxRow = cells.length - 1
  val maxCol = cells.head.length - 1

  def neighboursOf(c: Coord): List[Coord] =
    val (row, col) = c
    val lb = mutable.ListBuffer.empty[Coord]
    if row < maxRow then lb.append((row+1, col))
    if row > 0      then lb.append((row-1, col))
    if col < maxCol then lb.append((row, col+1))
    if col > 0      then lb.append((row, col-1))
    lb.toList

  def costOf(c: Coord): Int = c match
    case (row, col) => cells(row)(col)
end GameMap

IndexedSeq in the cells type is important for this algorithm since we are doing a lot of index-based accesses, so we need to use a data structure optimized for that.

Algorithm – Part 1​

We start the solution by defining three data structures for the algorithm:

val visited = mutable.Set.empty[Coord]
val dist = mutable.Map[Coord, Int]((0, 0) -> 0)
val queue = java.util.PriorityQueue[Coord](Ordering.by(dist))
queue.add((0, 0))

The first one is a Set of all visited nodes – the ones the algorithm will not look at again. The second one is a Map of distances containing the smallest currently known distance from the top-left corner of the map to the given cell. Finally, the third one is a java.util.PriorityQueue that defines in which order to examine cells. We are using Java's PriorityQueue, not the Scala's one since the Java PriorityQueue implementation defines the remove operation on the queue which is necessary for efficient implementation and which the Scala queue lacks.

We also initialize the queue with the first node we are going to examine – the top-left corner of the map.

Once we have the data structures, there's a loop which runs Dijkstra's algorithm on those structures:

while queue.peek() != null do
  val c = queue.poll()
  visited += c
  val newNodes: List[Coord] = gameMap.neighboursOf(c).filterNot(visited)
  val cDist = dist(c)
  for n <- newNodes do
    val newDist = cDist + gameMap.costOf(n)
    if !dist.contains(n) || dist(n) > newDist then
      dist(n) = newDist
      queue.remove(n)
      queue.add(n)
dist((gameMap.maxRow, gameMap.maxCol))

We use queue.remove(n) followed by queue.add(n) here – this is to recompute the position of n in the queue following the change in the ordering of the queue (that is, the mutation of dist). Ideally, you would need a decreaseKey operation on the priority queue for the best performance – but that would require writing a dedicated data structure, which is out of scope for this solution.

Part 2​

Part 2 is like Part 1 but 25 times larger. The Part 1 algorithm is capable of dealing with scale, and so the only challenge is to construct the game map for part 2.

We generate the Part 2 game map from the Part 1 map using three nested loops:

val seedTile = readInput()
val gameMap = GameMap(
  (0 until 5).flatMap { tileIdVertical =>
    for row <- seedTile yield
      for
        tileIdHorizontal <- 0 until 5
        cell <- row
      yield (cell + tileIdHorizontal + tileIdVertical - 1) % 9 + 1
  }
)

The innermost loop generates individual cells according to the challenge spec. The second-level loop pads the 100x100 tiles of the map horizontally, starting from the seedTile (the one used in Part 1). Finally, the outermost loop pads the tiles vertically.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day16/index.html b/puzzles/day16/index.html index 17e817352..2701583d2 100644 --- a/puzzles/day16/index.html +++ b/puzzles/day16/index.html @@ -5,7 +5,7 @@ Day 16: Packet Decoder | Scala Center Advent of Code - + @@ -54,7 +54,7 @@ that will calculate the equation. We can do it similarly to the versionsSum function in the previous part:

  def value: Long =
    this match
      case Sum(version, exprs)            => exprs.map(_.value).sum
      case Product(version, exprs)        => exprs.map(_.value).reduce(_ * _)
      case Minimum(version, exprs)        => exprs.map(_.value).min
      case Maximum(version, exprs)        => exprs.map(_.value).max
      case Literal(version, value)        => value
      case GreaterThan(version, lhs, rhs) => if lhs.value > rhs.value then 1 else 0
      case LesserThan(version, lhs, rhs)  => if lhs.value < rhs.value then 1 else 0
      case Equals(version, lhs, rhs)      => if lhs.value == rhs.value then 1 else 0

Full solution​

package day16

import scala.util.Using
import scala.io.Source
import scala.annotation.tailrec

@main def part1(): Unit =
  println(s"The solution is ${part1(readInput())}")

@main def part2(): Unit =
  println(s"The solution is ${part2(readInput())}")

def readInput(): String =
  Using.resource(Source.fromFile("input/day16"))(_.mkString)

val hexadecimalMapping =
  Map(
    '0' -> "0000",
    '1' -> "0001",
    '2' -> "0010",
    '3' -> "0011",
    '4' -> "0100",
    '5' -> "0101",
    '6' -> "0110",
    '7' -> "0111",
    '8' -> "1000",
    '9' -> "1001",
    'A' -> "1010",
    'B' -> "1011",
    'C' -> "1100",
    'D' -> "1101",
    'E' -> "1110",
    'F' -> "1111"
  )

/*
 * Structures for all possible operators
 */
enum Packet(version: Int, typeId: Int):
  case Sum(version: Int, exprs: List[Packet]) extends Packet(version, 0)
  case Product(version: Int, exprs: List[Packet]) extends Packet(version, 1)
  case Minimum(version: Int, exprs: List[Packet]) extends Packet(version, 2)
  case Maximum(version: Int, exprs: List[Packet]) extends Packet(version, 3)
  case Literal(version: Int, literalValue: Long) extends Packet(version, 4)
  case GreaterThan(version: Int, lhs: Packet, rhs: Packet) extends Packet(version, 5)
  case LesserThan(version: Int, lhs: Packet, rhs: Packet) extends Packet(version, 6)
  case Equals(version: Int, lhs: Packet, rhs: Packet) extends Packet(version, 7)

  def versionSum: Int =
    this match
      case Sum(version, exprs)            => version + exprs.map(_.versionSum).sum
      case Product(version, exprs)        => version + exprs.map(_.versionSum).sum
      case Minimum(version, exprs)        => version + exprs.map(_.versionSum).sum
      case Maximum(version, exprs)        => version + exprs.map(_.versionSum).sum
      case Literal(version, value)        => version
      case GreaterThan(version, lhs, rhs) => version + lhs.versionSum + rhs.versionSum
      case LesserThan(version, lhs, rhs)  => version + lhs.versionSum + rhs.versionSum
      case Equals(version, lhs, rhs)      => version + lhs.versionSum + rhs.versionSum

  def value: Long =
    this match
      case Sum(version, exprs)            => exprs.map(_.value).sum
      case Product(version, exprs)        => exprs.map(_.value).reduce(_ * _)
      case Minimum(version, exprs)        => exprs.map(_.value).min
      case Maximum(version, exprs)        => exprs.map(_.value).max
      case Literal(version, value)        => value
      case GreaterThan(version, lhs, rhs) => if lhs.value > rhs.value then 1 else 0
      case LesserThan(version, lhs, rhs)  => if lhs.value < rhs.value then 1 else 0
      case Equals(version, lhs, rhs)      => if lhs.value == rhs.value then 1 else 0
end Packet

type BinaryData = List[Char]

inline def toInt(chars: BinaryData): Int =
  Integer.parseInt(chars.mkString, 2)

inline def toLong(chars: BinaryData): Long =
  java.lang.Long.parseLong(chars.mkString, 2)

@tailrec
def readLiteralBody(tail: BinaryData, numAcc: BinaryData): (Long, BinaryData) =
  val (num, rest) = tail.splitAt(5)
  if num(0) == '1' then readLiteralBody(rest, numAcc.appendedAll(num.drop(1)))
  else
    val bits = numAcc.appendedAll(num.drop(1))
    (toLong(bits), rest)
end readLiteralBody

def readOperatorBody(current: BinaryData): (List[Packet], BinaryData) =
  val (lenId, rest) = current.splitAt(1)

  @tailrec
  def readMaxBits(
      current: BinaryData,
      remaining: Int,
      acc: List[Packet]
  ): (List[Packet], BinaryData) =
    if remaining == 0 then (acc, current)
    else
      val (newExpr, rest) = decodePacket(current)
      readMaxBits(rest, remaining - (current.size - rest.size), acc :+ newExpr)

  @tailrec
  def readMaxPackets(
      current: BinaryData,
      remaining: Int,
      acc: List[Packet]
  ): (List[Packet], BinaryData) =
    if remaining == 0 then (acc, current)
    else
      val (newExpr, rest) = decodePacket(current)
      readMaxPackets(rest, remaining - 1, acc :+ newExpr)

  lenId match
    // read based on length
    case List('0') =>
      val (size, packets) = rest.splitAt(15)
      readMaxBits(packets, toInt(size), Nil)

    // read based on number of packages
    case _ =>
      val (size, packets) = rest.splitAt(11)
      readMaxPackets(packets, toInt(size), Nil)
  end match
end readOperatorBody

def decodePacket(packet: BinaryData): (Packet, BinaryData) =
  val (versionBits, rest) = packet.splitAt(3)
  val version = toInt(versionBits)
  val (typeBits, body) = rest.splitAt(3)
  val tpe = toInt(typeBits)

  tpe match
    case 4 =>
      val (value, remaining) = readLiteralBody(body, Nil)
      (Packet.Literal(version, value), remaining)
    case otherTpe =>
      val (values, remaining) = readOperatorBody(body)
      otherTpe match
        case 0 => (Packet.Sum(version, values), remaining)
        case 1 => (Packet.Product(version, values), remaining)
        case 2 => (Packet.Minimum(version, values), remaining)
        case 3 => (Packet.Maximum(version, values), remaining)
        case 5 => (Packet.GreaterThan(version, values(0), values(1)), remaining)
        case 6 => (Packet.LesserThan(version, values(0), values(1)), remaining)
        case 7 => (Packet.Equals(version, values(0), values(1)), remaining)
  end match
end decodePacket

def parse(input: String) =
  val number = input.toList.flatMap(hex => hexadecimalMapping(hex).toCharArray)
  val (operator, _) = decodePacket(number)
  operator

def part1(input: String) =
  val packet = parse(input)
  packet.versionSum

def part2(input: String) =
  val packet = parse(input)
  packet.value
end part2

You might have noticed that we had to slightly modify the versionsSum function to work with our new structure.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day17/index.html b/puzzles/day17/index.html index f56b0faed..da6290c46 100644 --- a/puzzles/day17/index.html +++ b/puzzles/day17/index.html @@ -5,7 +5,7 @@ Day 17: Trick Shot | Scala Center Advent of Code - + @@ -102,7 +102,7 @@ edge of the target (to travel to the furthest edge in one step).

We adapt allMaxHeights with this new rule:

def allMaxHeights(target: Target)(positiveOnly: Boolean): Seq[Int] =
  val Target(xs, ys) = target
  val upperBoundX = xs.max
  val upperBoundY = -ys.min -1
  val lowerBoundY = if positiveOnly then 0 else ys.min
  for
    vx <- 0 to upperBoundX
    vy <- lowerBoundY to upperBoundY
    maxy <- simulate(Probe(initial, Velocity(vx, vy)), target)
  yield
    maxy

Computing the Solution​

As our input has not changed, we can update part 1 and give the code for part 2 as follows:

def part1(input: String) =
  allMaxHeights(Input(input.trim))(positiveOnly = true).max

def part2(input: String) =
  allMaxHeights(Input(input.trim))(positiveOnly = false).size

Notice that in part 2 we only need the number of possible max heights, rather than find the highest.

Final Code​

case class Target(xs: Range, ys: Range)

case class Velocity(x: Int, y: Int)

case class Position(x: Int, y: Int)

val initial = Position(x = 0, y = 0)

case class Probe(position: Position, velocity: Velocity)

def step(probe: Probe): Probe =
  val Probe(Position(px, py), Velocity(vx, vy)) = probe
  Probe(Position(px + vx, py + vy), Velocity(vx - vx.sign, vy - 1))

def collides(probe: Probe, target: Target): Boolean =
  val Probe(Position(px, py), _) = probe
  val Target(xs, ys) = target
  xs.contains(px) && ys.contains(py)

def beyond(probe: Probe, target: Target): Boolean =
  val Probe(Position(px, py), Velocity(vx, vy)) = probe
  val Target(xs, ys) = target
  val beyondX = (vx == 0 && px < xs.min) || px > xs.max
  val beyondY = vy < 0 && py < ys.min
  beyondX || beyondY

def simulate(probe: Probe, target: Target): Option[Int] =
  LazyList
    .iterate((probe, 0))((probe, maxY) => (step(probe), maxY `max` probe.position.y))
    .dropWhile((probe, _) => !collides(probe, target) && !beyond(probe, target))
    .headOption
    .collect { case (probe, maxY) if collides(probe, target) => maxY }

def allMaxHeights(target: Target)(positiveOnly: Boolean): Seq[Int] =
  val upperBoundX = target.xs.max
  val upperBoundY = target.ys.min.abs
  val lowerBoundY = if positiveOnly then 0 else -upperBoundY
  for
    vx <- 0 to upperBoundX
    vy <- lowerBoundY to upperBoundY
    maxy <- simulate(Probe(initial, Velocity(vx, vy)), target)
  yield
    maxy

type Parser[A] = PartialFunction[String, A]

val IntOf: Parser[Int] =
  case s if s.matches(raw"-?\d+") => s.toInt

val RangeOf: Parser[Range] =
  case s"${IntOf(begin)}..${IntOf(end)}" => begin to end

val Input: Parser[Target] =
  case s"target area: x=${RangeOf(xs)}, y=${RangeOf(ys)}" => Target(xs, ys)

def part1(input: String) =
  allMaxHeights(Input(input.trim))(positiveOnly = true).max

def part2(input: String) =
  allMaxHeights(Input(input.trim))(positiveOnly = false).size

Run it in the browser​

Part 1​

Part 2​

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day17.part1
The answer is: 4851

$ scala-cli 2021 -M day17.part2
The answer is: 1739

You can replace the content of the input/day14 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day18/index.html b/puzzles/day18/index.html index 90c9d0eeb..deabfe8ec 100644 --- a/puzzles/day18/index.html +++ b/puzzles/day18/index.html @@ -5,13 +5,13 @@ Day 18: Snailfish | Scala Center Advent of Code - +
- + \ No newline at end of file diff --git a/puzzles/day19/index.html b/puzzles/day19/index.html index 1c5117789..d73dc5967 100644 --- a/puzzles/day19/index.html +++ b/puzzles/day19/index.html @@ -5,13 +5,13 @@ Day 19: Beacon Scanner | Scala Center Advent of Code - +
- + \ No newline at end of file diff --git a/puzzles/day2/index.html b/puzzles/day2/index.html index 7d9f4b768..837a1d8e2 100644 --- a/puzzles/day2/index.html +++ b/puzzles/day2/index.html @@ -5,7 +5,7 @@ Day 2: Dive! | Scala Center Advent of Code - + @@ -19,7 +19,7 @@ and then add a method move that will translate the puzzle's rules to a position.

case class Position(horizontal: Int, depth: Int):
  def move(p: Command): Position =
    p match
      case Command.Forward(x) => Position(horizontal + x, depth)
      case Command.Down(x)    => Position(horizontal, depth + x)
      case Command.Up(x)      => Position(horizontal, depth - x)

To apply all the commands from the input file, we use foldLeft

val firstPosition = Position(0, 0)
val lastPosition = entries.foldLeft(firstPosition)((position, command) => position.move(command))

foldLeft is a method from the standard library on iterable collections: Seq, List, Iterator...

It's a super convenient method that allows to iterate from left to right on a list.

The signature of foldLeft is:

def foldLeft[B](initialElement: B)(op: (B, A) => B): B

Let's see an example:

// Implementing a sum on a List
List(1, 3, 2, 4).foldLeft(0)((accumulator, current) => accumulator + current) // 10

It is the same as:

(((0 + 1) + 3) + 2) + 4

Final code for part 1​

def part1(input: String): Int =
  val entries = input.linesIterator.map(Command.from)
  val firstPosition = Position(0, 0)
  // we iterate on each entry and move it following the received command
  val lastPosition = entries.foldLeft(firstPosition)((position, command) => position.move(command))
  lastPosition.result

case class Position(horizontal: Int, depth: Int):
  def move(p: Command): Position =
    p match
      case Command.Forward(x) => Position(horizontal + x, depth)
      case Command.Down(x)    => Position(horizontal, depth + x)
      case Command.Up(x)      => Position(horizontal, depth - x)

  def result = horizontal * depth

enum Command:
  case Forward(x: Int)
  case Down(x: Int)
  case Up(x: Int)

object Command:
  def from(s: String): Command =
    s match
      case s"forward $x" if x.toIntOption.isDefined => Forward(x.toInt)
      case s"up $x"      if x.toIntOption.isDefined => Up(x.toInt)
      case s"down $x"    if x.toIntOption.isDefined => Down(x.toInt)
      case _ => throw new Exception(s"value $s is not valid command")

Solution of Part 2​

The part 2 introduces new rules to move the sonar. So we need a new position that takes into account the aim and a new method move with the new rules. The remaining code remains the same.

Moving the sonar part 2​

case class PositionWithAim(horizontal: Int, depth: Int, aim: Int):
  def move(p: Command): PositionWithAim =
    p match
      case Command.Forward(x) => PositionWithAim(horizontal + x, depth + x * aim, aim)
      case Command.Down(x)    => PositionWithAim(horizontal, depth, aim + x)
      case Command.Up(x)      => PositionWithAim(horizontal, depth, aim - x)

Final code for part 2​

case class PositionWithAim(horizontal: Int, depth: Int, aim: Int):
  def move(p: Command): PositionWithAim =
    p match
      case Command.Forward(x) => PositionWithAim(horizontal + x, depth + x * aim, aim)
      case Command.Down(x)    => PositionWithAim(horizontal, depth, aim + x)
      case Command.Up(x)      => PositionWithAim(horizontal, depth, aim - x)

  def result = horizontal * depth

enum Command:
  case Forward(x: Int)
  case Down(x: Int)
  case Up(x: Int)

object Command:
  def from(s: String): Command =
    s match
      case s"forward $x" if x.toIntOption.isDefined => Forward(x.toInt)
      case s"up $x"      if x.toIntOption.isDefined => Up(x.toInt)
      case s"down $x"    if x.toIntOption.isDefined => Down(x.toInt)
      case _ => throw new Exception(s"value $s is not valid command")

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

The you can run it with scala-cli:

$ scala-cli 2021 -M day2.part1
The answer is 2070300

$ scala-cli 2021 -M day2.part2
The answer is 2078985210

You can replace the content of the input/day2 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day20/index.html b/puzzles/day20/index.html index b264b9ff7..bd3e45dce 100644 --- a/puzzles/day20/index.html +++ b/puzzles/day20/index.html @@ -5,7 +5,7 @@ Day 20: Trench Map | Scala Center Advent of Code - + @@ -52,7 +52,7 @@ element n. Then, we compute its 50th element by calling .apply(50). As a consequence, only the first 50 elements will be computed at all.

Finally, we call countLitPixels() on the output image to count its number of lit pixels.

Run it in the browser​

Part 1​

Part 2​

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day20.part1
The solution is: 5301
$ scala-cli 2021 -M day20.part2
The solution is: 19492

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day21/index.html b/puzzles/day21/index.html index 83839b6f5..beda48b7c 100644 --- a/puzzles/day21/index.html +++ b/puzzles/day21/index.html @@ -5,7 +5,7 @@ Day 21: Dirac Dice | Scala Center Advent of Code - + @@ -23,7 +23,7 @@ There are only 7 different outcomes to the roll of three dice, with most of them occurring several times. The rest of the game is not affected by anything but the sum, although it will happen in several universes, which we need to count. We can implement that by remembering in how many universes the current state of the game gets played, and add that amount to the number of times player 1 or 2 wins.

We first compute how many times each outcome happens:

/** For each 3-die throw, how many of each total sum do we have? */
val dieCombinations: List[(Int, Long)] =
  val possibleRolls: List[Int] =
    for
      die1 <- List(1, 2, 3)
      die2 <- List(1, 2, 3)
      die3 <- List(1, 2, 3)
    yield
      die1 + die2 + die3
  possibleRolls.groupMapReduce(identity)(_ => 1L)(_ + _).toList

Then, we add a parameter inHowManyUniverses to playWithDiracDie, and multiply it in the recursive calls by the number of times that each outcome happens:

def playWithDiracDie(players: Players, player1Turn: Boolean, wins: Wins, inHowManyUniverses: Long): Unit =
  for (diesValue, count) <- dieCombinations do
    val newInHowManyUniverses = inHowManyUniverses * count
    val player = players(0)
    val newCell = (player.cell + diesValue) % 10
    val newScore = player.score + (newCell + 1)
    if newScore >= 21 then
      if player1Turn then
        wins.player1Wins += newInHowManyUniverses
      else
        wins.player2Wins += newInHowManyUniverses
    else
      val newPlayer = Player(newCell, newScore)
      playWithDiracDie((players(1), newPlayer), !player1Turn, wins, newInHowManyUniverses)
  end for

We start with 1 universe, so the initial call to playWithDiracDie is:

playWithDiracDie(players, player1Turn = true, wins, inHowManyUniverses = 1L)

The reduction of the branching factor from 27 to 7 is enough to simulate all the possible universes in seconds, whereas I stopped waiting for the naive solution after a few minutes.

Solution for part 2​

Here is the full code for part 2:

final class Wins(var player1Wins: Long, var player2Wins: Long)

def part2(input: String): Long =
  val players = parseInput(input)
  val wins = new Wins(0L, 0L)
  playWithDiracDie(players, player1Turn = true, wins, inHowManyUniverses = 1L)
  Math.max(wins.player1Wins, wins.player2Wins)

/** For each 3-die throw, how many of each total sum do we have? */
val dieCombinations: List[(Int, Long)] =
  val possibleRolls: List[Int] =
    for
      die1 <- List(1, 2, 3)
      die2 <- List(1, 2, 3)
      die3 <- List(1, 2, 3)
    yield
      die1 + die2 + die3
  possibleRolls.groupMapReduce(identity)(_ => 1L)(_ + _).toList

def playWithDiracDie(players: Players, player1Turn: Boolean, wins: Wins, inHowManyUniverses: Long): Unit =
  for (diesValue, count) <- dieCombinations do
    val newInHowManyUniverses = inHowManyUniverses * count
    val player = players(0)
    val newCell = (player.cell + diesValue) % 10
    val newScore = player.score + (newCell + 1)
    if newScore >= 21 then
      if player1Turn then
        wins.player1Wins += newInHowManyUniverses
      else
        wins.player2Wins += newInHowManyUniverses
    else
      val newPlayer = Player(newCell, newScore)
      playWithDiracDie((players(1), newPlayer), !player1Turn, wins, newInHowManyUniverses)
  end for

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day21.part1
The answer is: 855624

$ scala-cli 2021 -M day21.part2
The answer is: 187451244607486

You can replace the content of the input/day21 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day22/index.html b/puzzles/day22/index.html index 0a2e4ced5..40ac23280 100644 --- a/puzzles/day22/index.html +++ b/puzzles/day22/index.html @@ -5,7 +5,7 @@ Day 22: Reactor Reboot | Scala Center Advent of Code - + @@ -68,7 +68,7 @@ only while they fit the initialisation sequence, and then summarise the set of cuboids:

def part1(input: String): BigInt =
  val steps = input.linesIterator.map(StepOf)
  summary(run(steps.takeWhile(s => isInit(s.cuboid))))

Solution of Part 2​

Part 2 is identical to part 1, except that we run all steps, not just the initialisation sequence:

def part2(input: String): BigInt =
  summary(run(input.linesIterator.map(StepOf)))

Run it in the browser​

Part 1​

Part 2​

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day22.part1
The answer is: 647062

$ scala-cli 2021 -M day22.part2
The answer is: 1319618626668022

You can replace the content of the input/day22 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day23/index.html b/puzzles/day23/index.html index e62f12187..fad36dba9 100644 --- a/puzzles/day23/index.html +++ b/puzzles/day23/index.html @@ -5,7 +5,7 @@ Day 23: Amphipod | Scala Center Advent of Code - + @@ -14,7 +14,7 @@ Our intuition here is that the puzzle can be modeled as a graph and solved using Dijkstra's algorithm.

A graph of situations​

We can think of the puzzle as a graph of situations, where a node is an instance of Situation and an edge is an amphipod's move whose weight is the energy cost of the move.

In such a graph, two situations are connected if there is an amphipod move that transform the first situation into the second.

Implementing the Dijkstra's solver​

We want to find the minimal energy cost to go from the initial situation to the final situation, where all amphipods are located in their destination room. This is the energy cost of the shortest path between the two situations in the graph described above. We can use Dijkstra's algorithm to find it.

Here is our implementation:

class DijkstraSolver(initialSituation: Situation):
  private val bestSituations = mutable.Map(initialSituation -> 0)
  private val situationsToExplore =
    mutable.PriorityQueue((initialSituation, 0))(Ordering.by((_, energy) => -energy))

  @tailrec
  final def solve(): Energy =
    val (situation, energy) = situationsToExplore.dequeue
    if situation.isFinal then energy
    else if bestSituations(situation) < energy then solve()
    else
      for
        (nextSituation, consumedEnergy) <- situation.moveAllAmphipodsOnce
        nextEnergy = energy + consumedEnergy
        knownEnergy = bestSituations.getOrElse(nextSituation, Int.MaxValue)
        if nextEnergy < knownEnergy
      do
        bestSituations.update(nextSituation, nextEnergy)
        situationsToExplore.enqueue((nextSituation, nextEnergy))
      solve()

At the beginning we only know the cost of the initial situation which is 0.

The solve method is recursive:

  1. First we dequeue the best known situation in the situationToExplore queue.
  2. If it is the final situation, we return the associated energy cost.
  3. If it is not:
  • We compute all the situations connected to it by moving all amphipods once.
  • For each of these new situations, we check if the energy cost is better than before and if so we add it into the queue.
  • We recurse by calling solve again.

Final solution​

// using scala 3.1.0

package day23

import scala.util.Using
import scala.io.Source
import scala.annotation.tailrec
import scala.collection.mutable


@main def part1(): Unit =
  val answer = part1(readInput())
  println(s"The answer is: $answer")

@main def part2(): Unit =
  val answer = part2(readInput())
  println(s"The answer is: $answer")

def readInput(): String =
  Using.resource(Source.fromFile("input/day23"))(_.mkString)

case class Position(x: Int, y: Int)

enum Room(val x: Int):
  case A extends Room(3)
  case B extends Room(5)
  case C extends Room(7)
  case D extends Room(9)

type Energy = Int

enum Amphipod(val energy: Energy, val destination: Room):
  case A extends Amphipod(1, Room.A)
  case B extends Amphipod(10, Room.B)
  case C extends Amphipod(100, Room.C)
  case D extends Amphipod(1000, Room.D)

object Amphipod:
  def tryParse(input: Char): Option[Amphipod] =
    input match
      case 'A' => Some(Amphipod.A)
      case 'B' => Some(Amphipod.B)
      case 'C' => Some(Amphipod.C)
      case 'D' => Some(Amphipod.D)
      case _ => None

val hallwayStops: Seq[Position] = Seq(
  Position(1, 1),
  Position(2, 1),
  Position(4, 1),
  Position(6, 1),
  Position(8, 1),
  Position(10, 1),
  Position(11, 1)
)

case class Situation(positions: Map[Position, Amphipod], roomSize: Int):
  def moveAllAmphipodsOnce: Seq[(Situation, Energy)] =
    for
      (start, amphipod) <- positions.toSeq
      stop <- nextStops(amphipod, start)
      path = getPath(start, stop)
      if path.forall(isEmpty)
    yield
      val newPositions = positions - start + (stop -> amphipod)
      val energy = path.size * amphipod.energy
      (copy(positions = newPositions), energy)

  def isFinal =
    positions.forall((position, amphipod) => position.x == amphipod.destination.x)

  /**
   * Return a list of positions to which an amphipod at position `from` can go:
   * - If the amphipod is in its destination room and the room is free it must not go anywhere.
   * - If the amphipod is in its destination room and the room is not free it can go to the hallway.
   * - If the amphipod is in the hallway it can only go to its destination.
   * - Otherwise it can go to the hallway.
   */
  private def nextStops(amphipod: Amphipod, from: Position): Seq[Position] =
    from match
      case Position(x, y) if x == amphipod.destination.x =>
        if isDestinationFree(amphipod) then Seq.empty
        else hallwayStops
      case Position(_, 1) =>
        if isDestinationFree(amphipod) then
          (roomSize + 1).to(2, step = -1)
            .map(y => Position(amphipod.destination.x, y))
            .find(isEmpty)
            .toSeq
        else Seq.empty
      case _ => hallwayStops


  private def isDestinationFree(amphipod: Amphipod): Boolean =
    2.to(roomSize + 1)
      .flatMap(y => positions.get(Position(amphipod.destination.x, y)))
      .forall(_ == amphipod)

  // Build the path to go from `start` to `stop`
  private def getPath(start: Position, stop: Position): Seq[Position] =
    val hallway =
      if start.x < stop.x
      then (start.x + 1).to(stop.x).map(Position(_, 1))
      else (start.x - 1).to(stop.x, step = -1).map(Position(_, 1))
    val startRoom = (start.y - 1).to(1, step = -1).map(Position(start.x, _))
    val stopRoom = 2.to(stop.y).map(Position(stop.x, _))
    startRoom ++ hallway ++ stopRoom

  private def isEmpty(position: Position) =
    !positions.contains(position)

object Situation:
  def parse(input: String, roomSize: Int): Situation =
    val positions =
      for
        (line, y) <- input.linesIterator.zipWithIndex
        (char, x) <- line.zipWithIndex
        amphipod <- Amphipod.tryParse(char)
      yield Position(x, y) -> amphipod
    Situation(positions.toMap, roomSize)

class DijkstraSolver(initialSituation: Situation):
  private val bestSituations = mutable.Map(initialSituation -> 0)
  private val situationsToExplore =
    mutable.PriorityQueue((initialSituation, 0))(Ordering.by((_, energy) => -energy))

  @tailrec
  final def solve(): Energy =
    val (situation, energy) = situationsToExplore.dequeue
    if situation.isFinal then energy
    else if bestSituations(situation) < energy then solve()
    else
      for
        (nextSituation, consumedEnergy) <- situation.moveAllAmphipodsOnce
        nextEnergy = energy + consumedEnergy
        knownEnergy = bestSituations.getOrElse(nextSituation, Int.MaxValue)
        if nextEnergy < knownEnergy
      do
        bestSituations.update(nextSituation, nextEnergy)
        situationsToExplore.enqueue((nextSituation, nextEnergy))
      solve()

def part1(input: String): Energy =
  val initialSituation = Situation.parse(input, roomSize = 2)
  DijkstraSolver(initialSituation).solve()

def part2(input: String): Energy =
  val lines = input.linesIterator
  val unfoldedInput = (lines.take(3) ++ Seq("  #D#C#B#A#", "  #D#B#A#C#") ++ lines.take(2)).mkString("\n")
  val initialSituation = Situation.parse(unfoldedInput, roomSize = 4)
  DijkstraSolver(initialSituation).solve()

Run it in the browser​

Part 1​

Part 2​

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day21.part1
The answer is: 855624

$ scala-cli 2021 -M day21.part2
The answer is: 187451244607486

You can replace the content of the input/day21 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day24/index.html b/puzzles/day24/index.html index dc86eecd3..957d771d4 100644 --- a/puzzles/day24/index.html +++ b/puzzles/day24/index.html @@ -5,13 +5,13 @@ Day 24: Arithmetic Logic Unit | Scala Center Advent of Code - +
- + \ No newline at end of file diff --git a/puzzles/day25/index.html b/puzzles/day25/index.html index c17c30bc0..3f0c5bb3a 100644 --- a/puzzles/day25/index.html +++ b/puzzles/day25/index.html @@ -5,13 +5,13 @@ Day 25: Sea Cucumber | Scala Center Advent of Code - +

Day 25: Sea Cucumber

by @Sporarum, student at EPFL, and @adpi2

Puzzle description​

https://adventofcode.com/2021/day/25

Solution of Part 1​

enum SeaCucumber:
  case Empty, East, South

object SeaCucumber:
  def fromChar(c: Char) = c match
    case '.' => Empty
    case '>' => East
    case 'v' => South

type Board = Seq[Seq[SeaCucumber]]

def part1(input: String): Int =
  val board: Board = input.linesIterator.map(_.map(SeaCucumber.fromChar(_))).toSeq
  fixedPoint(board)

def fixedPoint(board: Board, step: Int = 1): Int =
  val next = move(board)
  if board == next then step else fixedPoint(next, step + 1)

def move(board: Board) = moveSouth(moveEast(board))
def moveEast(board: Board) = moveImpl(board, SeaCucumber.East)
def moveSouth(board: Board) = moveImpl(board.transpose, SeaCucumber.South).transpose

def moveImpl(board: Board, cucumber: SeaCucumber): Board =
  board.map { l =>
    zip3(l.last +: l.init, l, (l.tail :+ l.head)).map{
      case (`cucumber`, SeaCucumber.Empty, _) => `cucumber`
      case (_, `cucumber`, SeaCucumber.Empty) => SeaCucumber.Empty
      case (_, curr, _)  => curr
    }
  }

def zip3[A,B,C](l1: Seq[A], l2: Seq[B], l3: Seq[C]): Seq[(A,B,C)] =
  l1.zip(l2).zip(l3).map { case ((a, b), c) => (a,b,c) }

Run it in the browser​

Part 1​

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day25.part1
The answer is: 435

You can replace the content of the input/day25 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day3/index.html b/puzzles/day3/index.html index cc5716e80..d3c354cd1 100644 --- a/puzzles/day3/index.html +++ b/puzzles/day3/index.html @@ -5,7 +5,7 @@ Day 3: Binary Diagnostic | Scala Center Advent of Code - + @@ -34,7 +34,7 @@ Here is an example of partition that separates odd numbers from even numbers:

val numbers = List(4, 6, 5, 12, 75, 3, 10)
val (oddNumbers, evenNumbers) = numbers.partition(x => x % 2 != 0)
// oddNumbers = List(5, 75, 3)
// evenNumbers = List(4, 6, 12, 10)

We use it as follows to separate our lines in two lists:

val (bitLinesWithOne, bitLinesWithZero) =
  bitLines.partition(line => line(bitPosition) == 1)

We can determine whether there are more 1s than 0s (or a tie) by comparing the size of the two lists. Comparing the sizes of two collections is best done with sizeCompare:

val onesAreMostCommon = bitLinesWithOne.sizeCompare(bitLinesWithZero) >= 0

Finally, we decide which list we keep to go further:

val bitLinesToKeep =
  if onesAreMostCommon then
    if keepMostCommon then bitLinesWithOne else bitLinesWithZero
  else
    if keepMostCommon then bitLinesWithZero else bitLinesWithOne
recursiveFilter(bitLinesToKeep, bitPosition + 1, keepMostCommon)

(The two tests could be combined as if onesAreMostCommon == keepMostCommon, but I found that less readable.)

Final code for part 2​

def part2(input: String): Int =
  val bitLines: List[BitLine] = input.linesIterator.map(parseBitLine).toList

  val oxygenGeneratorRatingLine: BitLine =
    recursiveFilter(bitLines, 0, keepMostCommon = true)
  val oxygenGeneratorRating = bitLineToInt(oxygenGeneratorRatingLine)

  val co2ScrubberRatingLine: BitLine =
    recursiveFilter(bitLines, 0, keepMostCommon = false)
  val co2ScrubberRating = bitLineToInt(co2ScrubberRatingLine)

  oxygenGeneratorRating * co2ScrubberRating

@scala.annotation.tailrec
def recursiveFilter(bitLines: List[BitLine], bitPosition: Int,
    keepMostCommon: Boolean): BitLine =
  bitLines match
    case Nil =>
      throw new AssertionError("this shouldn't have happened")
    case lastRemainingLine :: Nil =>
      lastRemainingLine
    case _ =>
      val (bitLinesWithOne, bitLinesWithZero) =
        bitLines.partition(line => line(bitPosition) == 1)
      val onesAreMostCommon = bitLinesWithOne.sizeCompare(bitLinesWithZero) >= 0
      val bitLinesToKeep =
        if onesAreMostCommon then
          if keepMostCommon then bitLinesWithOne else bitLinesWithZero
        else
          if keepMostCommon then bitLinesWithZero else bitLinesWithOne
      recursiveFilter(bitLinesToKeep, bitPosition + 1, keepMostCommon)

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli. Since today's solution is written in Scala.js, you will need a local setup of Node.js to run it.

$ scala-cli 2021 -M day3.part1 --js-module-kind commonjs
The answer is 1025636

$ scala-cli 2021 -M day3.part2 --js-module-kind commonjs
The answer is 793873

You can replace the content of the input/day3 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day4/index.html b/puzzles/day4/index.html index 7d2fc38c0..e5ea67919 100644 --- a/puzzles/day4/index.html +++ b/puzzles/day4/index.html @@ -5,7 +5,7 @@ Day 4: Giant Squid | Scala Center Advent of Code - + @@ -24,7 +24,7 @@ We filter them with lines.filter(_ > turn).

However, only taking the sum would be wrong, as we are using the turns, and not the original numbers! We thus need to map them to their original values:

val sumNumsNotDrawn = board.lines.map{ line =>
  line.filter(_ > turn).map(turnToNumber(_)).sum
}.sum

The score is then:

turnToNumber(turn) * sumUnmarkedNums

Solution of Part 1​

In part one, we have to compute the score of the first board to win. This is the board whith the smallest winning turn.

val (winnerBoard, winnerTurn) = winningTurns.minBy((_, turn) => turn)

And so the score is:

val winnerScore = score(winnerBoard, winnerTurn)

Solution of Part 2​

In part two, we have to find the score of the last board to win, so we swap the minBy by a maxBy to get our result:

val (loserBoard, loserTurn) = winningTurns.maxBy((_, turn) => turn)
val loserScore = score(loserBoard, loserTurn)

Run it in the browser​

Part 1​

Part 2​

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day4.run
The answer of part 1 is 14093.
The answer of part 2 is 17388.

You can replace the content of the input/day4 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day5/index.html b/puzzles/day5/index.html index 945712e06..85b25215f 100644 --- a/puzzles/day5/index.html +++ b/puzzles/day5/index.html @@ -5,7 +5,7 @@ Day 5: Hydrothermal Venture | Scala Center Advent of Code - + @@ -24,7 +24,7 @@ both x and y positions increment by 1 at each step of the range. So we can add additional condition to our solution:

else for (px, py) <- rangex.zip(rangey) do update(Point(px, py))

We can just use the 2 previously defined ranges for this. So the full method will look like this:

def findDangerousPoints(vents: Seq[Vent]) =
  val map = mutable.Map[Point, Int]().withDefaultValue(0)
  def update(p: Point) =
    val current = map(p)
    map.update(p, current + 1)

  for vent <- vents do
    def rangex =
      val stepx = if vent.end.x > vent.start.x then 1 else -1
      vent.start.x.to(vent.end.x, stepx)
    def rangey =
      val stepy = if vent.end.y > vent.start.y then 1 else -1
      vent.start.y.to(vent.end.y, stepy)
    // vent is horizontal
    if vent.start.x == vent.end.x then
      for py <- rangey do update(Point(vent.start.x, py))
    // vent is vertical
    else if vent.start.y == vent.end.y then
      for px <- rangex do update(Point(px, vent.start.y))
    // vent is diagonal
    else for (px, py) <- rangex.zip(rangey) do update(Point(px, py))
  end for

  map.count { case (_, v) => v > 1 }
end findDangerousPoints

Run solution in the browser​

Part 1​

Part 2​

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day5.part1
The answer is: 7674

$ scala-cli 2021 -M day5.part2
The answer is: 20898

You can replace the content of the input/day5 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day6/index.html b/puzzles/day6/index.html index 42a01449e..9ae718c8d 100644 --- a/puzzles/day6/index.html +++ b/puzzles/day6/index.html @@ -5,7 +5,7 @@ Day 6: Lanternfish | Scala Center Advent of Code - + @@ -68,7 +68,7 @@ achieves this by summing the groups of fish: the method values returns a collection of groups of fish (each containing the number of fish in that group), finally the method sum sums up the groups.

Final code for part 2​

// "How many lanternfish would there be after 256 days?"
def part2(input: String): BigInt =
  simulate(
    days = 256,
    Fish.parseSeveral(input).groupMapReduce(_.timer)(_ => BigInt(1))(_ + _)
  )

def simulate(days: Int, initialPopulation: Map[Int, BigInt]): BigInt =
  (1 to days)
    .foldLeft(initialPopulation)((population, _) => tick(population))
    .values
    .sum

def tick(population: Map[Int, BigInt]): Map[Int, BigInt] =
  def countPopulation(daysLeft: Int): BigInt = population.getOrElse(daysLeft, BigInt(0))
  Map(
    0 -> countPopulation(1),
    1 -> countPopulation(2),
    2 -> countPopulation(3),
    3 -> countPopulation(4),
    4 -> countPopulation(5),
    5 -> countPopulation(6),
    6 -> (countPopulation(7) + countPopulation(0)),
    7 -> countPopulation(8),
    8 -> countPopulation(0)
  )

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day6.part1
The solution is 345793

$ scala-cli 2021 -M day6.part2
The solution is 1572643095893

You can replace the content of the input/day6 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day7/index.html b/puzzles/day7/index.html index d597242ec..f0f24d4be 100644 --- a/puzzles/day7/index.html +++ b/puzzles/day7/index.html @@ -5,7 +5,7 @@ Day 7: The Treachery of Whales | Scala Center Advent of Code - + @@ -40,7 +40,7 @@ solution.

Solutions from the community​

There are most likely some other solutions that we could have used. In particular some advent coders had luck with using median and average for determining the final horizontal positions of the crabmarines.

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day8/index.html b/puzzles/day8/index.html index bfce63366..4fb3b5fb6 100644 --- a/puzzles/day8/index.html +++ b/puzzles/day8/index.html @@ -5,7 +5,7 @@ Day 8: Seven Segment Search | Scala Center Advent of Code - + @@ -82,7 +82,7 @@ Each display has 4 digits, so after decoding the digits we will have a sequence of 4 Digit.

To convert a sequence of Digit to an integer value, we can convert each digit to its corresponding integer representation by calling .ordinal, and then we can accumulate a sum by (from the left), multiplying the current total by 10 for each new digit, and then adding the current digit:

def digitsToInt(digits: Seq[Digit]): Int =
  digits.foldLeft(0)((acc, d) => acc * 10 + d.ordinal)

Final Result​

Finally, we use our digitsToInt function to convert each solution to an integer value, and sum the result:

solutions.map(digitsToInt).sum

Final Code​

The final code for part 2 can be appended to the code of part 1:

import Digit.*

def part2(input: String): Int =

  def parseSegmentsSeq(segments: String): Seq[Segments] =
    segments.trim.split(" ").toSeq.map(Segment.parseSegments)

  def splitParts(line: String): (Seq[Segments], Seq[Segments]) =
    val Array(cipher, plaintext) = line.split('|').map(parseSegmentsSeq)
    (cipher, plaintext)

  def digitsToInt(digits: Seq[Digit]): Int =
    digits.foldLeft(0)((acc, d) => acc * 10 + d.ordinal)

  val problems = input.linesIterator.map(splitParts)

  val solutions = problems.map((cipher, plaintext) =>
    plaintext.map(substitutions(cipher))
  )

  solutions.map(digitsToInt).sum

end part2

def substitutions(cipher: Seq[Segments]): Map[Segments, Digit] =

  def lookup(section: Seq[Segments], withSegments: Segments): (Segments, Seq[Segments]) =
    val (Seq(uniqueMatch), remaining) = section.partition(withSegments.subsetOf)
    (uniqueMatch, remaining)

  val uniques: Map[Digit, Segments] =
    Map.from(
      for
        segments <- cipher
        digit <- Digit.lookupUnique(segments)
      yield
        digit -> segments
    )

  val ofSizeFive = cipher.filter(_.sizeIs == 5)
  val ofSizeSix = cipher.filter(_.sizeIs == 6)

  val one = uniques(One)
  val four = uniques(Four)
  val seven = uniques(Seven)
  val eight = uniques(Eight)
  val (three, remainingFives) = lookup(ofSizeFive, withSegments = one)
  val (nine, remainingSixes) = lookup(ofSizeSix, withSegments = three)
  val (zero, Seq(six)) = lookup(remainingSixes, withSegments = seven)
  val (five, Seq(two)) = lookup(remainingFives, withSegments = four &~ one)

  val decode: Map[Segments, Digit] =
    Seq(zero, one, two, three, four, five, six, seven, eight, nine)
      .zip(Digit.index)
      .toMap

  decode
end substitutions

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day8.part1
The solution is 521

$ scala-cli 2021 -M day8.part2
The solution is 1016804

You can replace the content of the input/day8 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/puzzles/day9/index.html b/puzzles/day9/index.html index a7ddba145..3942686e3 100644 --- a/puzzles/day9/index.html +++ b/puzzles/day9/index.html @@ -5,7 +5,7 @@ Day 9: Smoke Basin | Scala Center Advent of Code - + @@ -37,7 +37,7 @@ retrieve neighbors of neighbors, I add the cells that still need to be processed in the queue. The algorithm stops when there are no more cells to visit:

def basin(lowPoint: Position, heightMap: Heightmap): Set[Position] =
  @scala.annotation.tailrec
  def iter(visited: Set[Position], toVisit: Queue[Position], basinAcc: Set[Position]): Set[Position] =
    // No cells to visit, we are done
    if toVisit.isEmpty then basinAcc
    else
      // Select next cell to visit
      val (currentPos, remaining) = toVisit.dequeue
      // Collect the neighboring cells that should be part of the basin
      val newNodes = heightMap.neighborsOf(currentPos).toList.collect {
        case (pos, height) if !visited(currentPos) && height != 9 => pos
      }
      // Continue to next neighbor
      iter(visited + currentPos, remaining ++ newNodes, basinAcc ++ newNodes)

  iter(Set.empty, Queue(lowPoint), Set(lowPoint))

Run it locally​

You can get this solution locally by cloning the scalacenter/scala-advent-of-code repository.

$ git clone https://github.com/scalacenter/scala-advent-of-code
$ cd scala-advent-of-code

You can run it with scala-cli.

$ scala-cli 2021 -M day9.part1
The solution is 448
$ scala-cli 2021 -M day9.part2
The solution is 1417248

You can replace the content of the input/day9 file with your own input from adventofcode.com to get your own solution.

Solutions from the community​

Share your solution to the Scala community by editing this page. (You can even write the whole article!)

- + \ No newline at end of file diff --git a/setup/index.html b/setup/index.html index 8596d4f51..d58df2f66 100644 --- a/setup/index.html +++ b/setup/index.html @@ -5,7 +5,7 @@ Setup | Scala Center Advent of Code - + @@ -18,7 +18,7 @@ It supports an incredible number of languages through its extension system.

Its more popular extension for Scala is called Metals. We will use VS Code and Metals to write and navigate Scala code.

VS Code​

Download the right VS Code for your operating system on the download page of VS Code and then install it.

Install Metals​

1. Open VS Code and Click the extensions icon in the left bar​

Open Extensions

2. Search metals and click the Scala (Metals) extension and click the Install button​

Install Metals

- + \ No newline at end of file