diff --git a/2021/index.html b/2021/index.html index d9c4cebc8..60396164c 100644 --- a/2021/index.html +++ b/2021/index.html @@ -5,13 +5,13 @@ Scala Center Advent of Code | Scala Center Advent of Code - +
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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 8176f6b3b..e2e5e5fbb 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 6718faa58..32f75581c 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​

Solutions from the community​

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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 c5826964a..abd0ee160 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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Solutions from the community​

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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 d39c0c7b9..b92b6f592 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 7c5b26fe5..e07c4e43f 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 a25acd310..c70159f3b 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 f97dcd6ed..9392bad03 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)

Solutions from the community​

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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 4a1dc9774..cb9e299fb 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

Run it in the browser​

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/day08/index.html b/2022/puzzles/day08/index.html index 2ea4182ce..11a3ae4a3 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) }
}

Run it in the browser​

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/day09/index.html b/2022/puzzles/day09/index.html index 9aa3d6aa3..e523d4c82 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 bfeb13b03..54d8973e9 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 440eb962e..fd0bd7d9d 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 783bdc5f4..66b4f911c 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 4965dfc41..94dc275ab 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 2530f5135..52329fd0e 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 6e47f13b5..bcb5edd8b 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 b080172ff..0d2bf1ee9 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, @sjrd, and @bishabosha

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!

  val valvesLookup = IArray.from(valves)
  val valveCount = valvesLookup.size
  val _activeValveIndices = Array.fill[Boolean](valveCount + 1)(true) // add an extra valve for the initial state
  def valveIndexLeft(i: Int) = _activeValveIndices(i)
  def withoutValve(i: Int)(f: => Int) =
    _activeValveIndices(i) = false
    val result = f
    _activeValveIndices(i) = true
    result
  val roomsByIndices = IArray.tabulate(valveCount)(i => map.rooms(valvesLookup(i)))

  def recurse(hiddenValve: Int, current: Id, timeLeft: Int, totalValue: Int): Int = withoutValve(hiddenValve):
    // recursively consider all plausible options
    // we are finished when we no longer have time to reach another valve or all valves are open
    val routesOfCurrent = map.routes(current)
    var bestValue = totalValue
    for index <- 0 to valveCount do
      if valveIndexLeft(index) then
        val id = valvesLookup(index)
        val distance = routesOfCurrent(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
        if t > 0 then
          // 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 = roomsByIndices(index).flow * t
          val recValue = recurse(hiddenValve = index, id, t, totalValue + value)
          if recValue > bestValue then
            bestValue = recValue
        end if
      end if
    end for
    bestValue
  end recurse
  recurse(valveCount, start, 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

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Part 1​

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Part 2​

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

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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 8632a32a3..115321fcd 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 bc18a8374..e4c8438f1 100644 --- a/2022/puzzles/day18/index.html +++ b/2022/puzzles/day18/index.html @@ -5,13 +5,13 @@ Day 18: Boiling Boulders | Scala Center Advent of Code - +
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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.

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Part 1​

Part 2​

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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 bc2be5268..e0c0dca29 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 b48e20817..145e71f0b 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 d0975fc71..9abf6adf0 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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- + \ No newline at end of file diff --git a/2022/puzzles/day22/index.html b/2022/puzzles/day22/index.html index 425adfa72..9ac86c2ff 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 075561f68..1668d450a 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 e96419615..a32b920b4 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​

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- + \ No newline at end of file diff --git a/2022/puzzles/day25/index.html b/2022/puzzles/day25/index.html index 6d34707a9..89668dec4 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)

Run it in the browser​

Part 1 (Only 1 part today)​

Solutions from the community​

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- + \ No newline at end of file diff --git a/2023/index.html b/2023/index.html index 3886deb28..713586a04 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 1Day 2Day 3Day 4
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 a9fbfd2a6..7646cc23c 100644 --- a/2023/puzzles/day01/index.html +++ b/2023/puzzles/day01/index.html @@ -5,7 +5,7 @@ Day 1: Trebuchet?! | Scala Center Advent of Code - + @@ -22,7 +22,7 @@ Instead, we manually iterate over all the indices to see if a match starts there. This is equivalent to looking for prefix matches in all the suffixes of line. Conveniently, line.tails iterates over all such suffixes, and Regex.findPrefixOf will look only for prefixes.

Our fixed computation for matches is now:

val matchesIter =
  for
    lineTail <- line.tails
    oneMatch <- digitReprRegex.findPrefixOf(lineTail)
  yield
    oneMatch
val matches = matchesIter.toList

Final Code​

def part1(input: String): String =
  // Convert one line into the appropriate coordinates
  def lineToCoordinates(line: String): Int =
    val firstDigit = line.find(_.isDigit).get
    val lastDigit = line.findLast(_.isDigit).get
    s"$firstDigit$lastDigit".toInt

  // Convert each line to its coordinates and sum all the coordinates
  val result = input
    .linesIterator
    .map(lineToCoordinates(_))
    .sum
  result.toString()
end part1

/** The textual representation of digits. */
val stringDigitReprs = Map(
  "one" -> 1,
  "two" -> 2,
  "three" -> 3,
  "four" -> 4,
  "five" -> 5,
  "six" -> 6,
  "seven" -> 7,
  "eight" -> 8,
  "nine" -> 9,
)

/** All the string representation of digits, including the digits themselves. */
val digitReprs = stringDigitReprs ++ (1 to 9).map(i => i.toString() -> i)

def part2(input: String): String =
  // A regex that matches any of the keys of `digitReprs`
  val digitReprRegex = digitReprs.keysIterator.mkString("|").r

  def lineToCoordinates(line: String): Int =
    // Find all the digit representations in the line
    val matchesIter =
      for
        lineTail <- line.tails
        oneMatch <- digitReprRegex.findPrefixOf(lineTail)
      yield
        oneMatch
    val matches = matchesIter.toList

    // Convert the string representations into actual digits and form the result
    val firstDigit = digitReprs(matches.head)
    val lastDigit = digitReprs(matches.last)
    s"$firstDigit$lastDigit".toInt
  end lineToCoordinates

  // Process lines as in part1
  val result = input
    .linesIterator
    .map(lineToCoordinates(_))
    .sum
  result.toString()
end part2

Run it in the browser​

Part 1​

Part 2​

Solutions from the community​

Share your solution to the Scala community by editing this page.

- + \ No newline at end of file diff --git a/2023/puzzles/day02/index.html b/2023/puzzles/day02/index.html index 1637df1ed..b8cce5b4f 100644 --- a/2023/puzzles/day02/index.html +++ b/2023/puzzles/day02/index.html @@ -5,7 +5,7 @@ Day 2: Cube Conundrum | Scala Center Advent of Code - + @@ -14,7 +14,7 @@ In a single pass over the puzzle input it will:

case class Colors(color: String, count: Int)
case class Game(id: Int, hands: List[List[Colors]])
type Summary = Game => Int

def solution(input: String, summarise: Summary): Int =
  input.linesIterator.map(parse andThen summarise).sum

def parse(line: String): Game = ???

part1 and part2 will use this framework, plugging in the appropriate summarise function.

Parsing​

Let's fill in the parse function as follows:

def parseColors(pair: String): Colors =
  val (s"$value $name") = pair: @unchecked
  Colors(color = name, count = value.toInt)

def parse(line: String): Game =
  val (s"Game $id: $hands0") = line: @unchecked
  val hands1 = hands0.split("; ").toList
  val hands2 = hands1.map(_.split(", ").toList.map(parseColors))
  Game(id = id.toInt, hands = hands2)

Summary​

As described above, to summarise each game, we evaluate it as a possibleGame, where if it is a validGame summarise as the game's id, otherwise 0.

A game is valid if for all hands in the game, all the colors in each hand has a count that is less-than or equal-to the count of same color from the possibleCubes configuration.

val possibleCubes = Map(
  "red" -> 12,
  "green" -> 13,
  "blue" -> 14,
)

def validGame(game: Game): Boolean =
  game.hands.forall: hand =>
    hand.forall:
      case Colors(color, count) =>
        count <= possibleCubes.getOrElse(color, 0)

val possibleGame: Summary =
  case game if validGame(game) => game.id
  case _ => 0

def part1(input: String): Int = solution(input, possibleGame)

Part 2​

Summary​

In part 2, the summary of a game requires us to find the minimumCubes necessary to make a possible game. What this means is for any given game, across all hands calculating the maximum cubes drawn for each color.

In Scala we can accumulate the maximum counts for each cube in a Map from color to count. Take the initial maximums as all zero:

val initial = Seq("red", "green", "blue").map(_ -> 0).toMap

Then for each game we can compute the maximum cubes drawn in each game as follows

def minimumCubes(game: Game): Int =
  var maximums = initial
  for
    hand <- game.hands
    Colors(color, count) <- hand
  do
    maximums += (color -> (maximums(color) `max` count))
  maximums.values.product

Finally we can complete the solution by using minimumCubes to summarise each game:

def part2(input: String): Int = solution(input, minimumCubes)

Final Code​

case class Colors(color: String, count: Int)
case class Game(id: Int, hands: List[List[Colors]])
type Summary = Game => Int

def parseColors(pair: String): Colors =
  val (s"$value $name") = pair: @unchecked
  Colors(color = name, count = value.toInt)

def parse(line: String): Game =
  val (s"Game $id: $hands0") = line: @unchecked
  val hands1 = hands0.split("; ").toList
  val hands2 = hands1.map(_.split(", ").toList.map(parseColors))
  Game(id = id.toInt, hands = hands2)

def solution(input: String, summarise: Summary): Int =
  input.linesIterator.map(parse andThen summarise).sum

val possibleCubes = Map(
  "red" -> 12,
  "green" -> 13,
  "blue" -> 14,
)

def validGame(game: Game): Boolean =
  game.hands.forall: hand =>
    hand.forall:
      case Colors(color, count) =>
        count <= possibleCubes.getOrElse(color, 0)

val possibleGame: Summary =
  case game if validGame(game) => game.id
  case _ => 0

def part1(input: String): Int = solution(input, possibleGame)

val initial = Seq("red", "green", "blue").map(_ -> 0).toMap

def minimumCubes(game: Game): Int =
  var maximums = initial
  for
    hand <- game.hands
    Colors(color, count) <- hand
  do
    maximums += (color -> (maximums(color) `max` count))
  maximums.values.product

def part2(input: String): Int = solution(input, minimumCubes)

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/2023/puzzles/day03/index.html b/2023/puzzles/day03/index.html index 0c5d05069..857d1a554 100644 --- a/2023/puzzles/day03/index.html +++ b/2023/puzzles/day03/index.html @@ -5,13 +5,13 @@ Day 3: Gear Ratios | Scala Center Advent of Code - +
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Day 3: Gear Ratios

by @bishabosha and @iusildra

Puzzle description​

https://adventofcode.com/2023/day/3

Solution summary​

The solution models the input as a grid of numbers and symbols.

  1. Define some models to represent the input:
    • case class Coord(x: Int, y: Int) to represent one coordinate on the grid
    • case class Symbol(sym: String, pos: Coord) to represent one symbol and its location
    • case class PartNumber(value: Int, start: Coord, end: Coord) to represent one number and its starting/ending location
  2. Parse the input to create a dense collection of symbols and numbers
  3. Separate the symbols from the numbers
  4. Then summarise the whole grid as follows:
    • in part1, find all numbers adjacent to a symbol, and sum the total of the resulting number values,
    • in part2,
      1. Find all numbers adjacent to a symbol whose char value is *
      2. Filter out the * symbol with less/more than 2 adjacent numbers
      3. For each * symbol remaining, take the product of its two number values
      4. Sum the resulting products
    • a symbol is adjacent to a number (and vice-versa) if that symbol is inside the number's bounding box on the grid at 1 unit away (see manhattan distance)

Global​

We want a convenient way to represent a coordinate to be able to compute whether one element is within the bounding box of another.

case class Coord(x: Int, y: Int):
  def within(start: Coord, end: Coord) =
    if y < start.y || y > end.y then false
    else if x < start.x || x > end.x then false
    else true

We also want to easily distinguish a Symbol from a Number, and to know wether a Symbol is adjacent to a Number:

case class PartNumber(value: Int, start: Coord, end: Coord)
case class Symbol(sym: String, pos: Coord):
  def neighborOf(number: PartNumber) = pos.within(
    Coord(number.start.x - 1, number.start.y - 1),
    Coord(number.end.x + 1, number.end.y + 1)
  )

Then we need to parse the input to get every Symbol and Number:

import scala.util.matching.Regex.Match

object IsInt:
  def unapply(in: Match): Option[Int] = in.matched.toIntOption

def findPartsAndSymbols(source: String) =
  val extractor = """(\d+)|[^.\d]""".r
  source.split("\n").zipWithIndex.flatMap: (line, i) =>
    extractor
      .findAllMatchIn(line)
      .map:
        case m @ IsInt(nb) =>
          PartNumber(nb, Coord(m.start, i), Coord(m.end - 1, i))
        case s => Symbol(s.matched, Coord(s.start, i))

The object IsInt with the .unapply method is called an extractor. It allows to define patterns to match on. Here it will give me a number if it can parse it from a string

The findPartsAndSymbols does the parsing and returns a collection of PartNumber and Symbol. What we want to match on is either a number or a symbol (which is anything except the . and a digit). The regex match gives us some information (such as starting / ending position of the matched string) which we use to create the PartNumber and Symbol instances.

The m @ IsInt(nb) is a pattern match that will match on the IsInt extractor and binds the parsed integer to nb and the value being matched to m. A similar way to achieve this is:

.map: m =>
  m match
    case IsInt(nb) => PartNumber(nb, Coord(m.start, i), Coord(m.end - 1, i))
    case s => Symbol(s.matched, Coord(s.start, i))

Part 1​

Compute part1 as described above:

  1. Find all numbers and symbols in the grid
  2. Filter out the symbols in a separate collection
  3. For each number element of the grid and if it has a least one symbol neighbor, return its value
  4. Sum the resulting values
def part1(input: String) =
  val all = findPartsAndSymbols(input)
  val symbols = all.collect { case s: Symbol => s }
  all
    .collect:
      case n: PartNumber if symbols.exists(_.neighborOf(n)) =>
        n.value
    .sum

Part 2​

We might want to represent a Gear to facilitate the computation of the gear ratios:

case class Gear(part: PartNumber, symbol: Symbol)

(Note: a case class is not necessary here, a tuple would do the job)

Compute part2 as described above:

  1. Find all numbers and symbols in the grid
  2. Filter out the symbols in a separate collection
  3. For each number element of the grid and if it has one * neighbor, return a Gear with the number and the * symbol. For any other cases, return None
    • The .flatMap method will filter out the None values when flattening, so we get a collection of Gear only
  4. Group them by symbol and map the values to the number values
    • So we obtain a Map[Symbol, List[Int]] instead of a Map[Symbol, List[Gear]]
  5. Filter out the symbols with less/more than 2 adjacent numbers
  6. For each entry remaining, take the product of its two number values and sum the resulting products
def part2(input: String) =
  val all = findPartsAndSymbols(input)
  val symbols = all.collect { case s: Symbol => s }
  all
    .flatMap:
      case n: PartNumber =>
        symbols
          .find(_.neighborOf(n))
          .filter(_.sym == "*")
          .map(Gear(n, _))
      case _ => None
    .groupMap(_.symbol)(_.part.value)
    .filter(_._2.length == 2)
    .foldLeft(0) { _ + _._2.product }

Final code​

case class Coord(x: Int, y: Int):
  def within(start: Coord, end: Coord) =
    if y < start.y || y > end.y then false
    else if x < start.x || x > end.x then false
    else true
case class PartNumber(value: Int, start: Coord, end: Coord)
case class Symbol(sym: String, pos: Coord):
  def neighborOf(number: PartNumber) = pos.within(
    Coord(number.start.x - 1, number.start.y - 1),
    Coord(number.end.x + 1, number.end.y + 1)
  )

object IsInt:
  def unapply(in: Match): Option[Int] = in.matched.toIntOption

def findPartsAndSymbols(source: String) =
  val extractor = """(\d+)|[^.\d]""".r
  source.split("\n").zipWithIndex.flatMap: (line, i) =>
    extractor
      .findAllMatchIn(line)
      .map:
        case m @ IsInt(nb) =>
          PartNumber(nb, Coord(m.start, i), Coord(m.end - 1, i))
        case s => Symbol(s.matched, Coord(s.start, i))

def part1(input: String) =
  val all = findPartsAndSymbols(input)
  val symbols = all.collect { case s: Symbol => s }
  all
    .collect:
      case n: PartNumber if symbols.exists(_.neighborOf(n)) =>
        n.value
    .sum

case class Gear(part: PartNumber, symbol: Symbol)

def part2(input: String) =
  val all = findPartsAndSymbols(input)
  val symbols = all.collect { case s: Symbol => s }
  all
    .flatMap:
      case n: PartNumber =>
        symbols
          .find(_.neighborOf(n))
          .filter(_.sym == "*")
          .map(Gear(n, _))
      case _ => None
    .groupMap(_.symbol)(_.part.value)
    .filter(_._2.length == 2)
    .foldLeft(0) { _ + _._2.product }

Run it in the browser​

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/2023/puzzles/day04/index.html b/2023/puzzles/day04/index.html index 605f64de5..88aa2d06a 100644 --- a/2023/puzzles/day04/index.html +++ b/2023/puzzles/day04/index.html @@ -5,7 +5,7 @@ Day 4: Scratchcards | Scala Center Advent of Code - + @@ -33,8 +33,8 @@ the current card.

        .zipWithIndex
        .map((mult, idx) => if idx < winning then mult + thisMult else mult)
      (numCards + thisMult, restMult)
    }
    ._1.toString()
end part2

Throughout the iteration, the vector satisfies the following invariants:

Why track by relative index instead of absolute?

Final code​

def countWinning(card: String): Int =
  val numbers = card
    .substring(card.indexOf(":") + 1)   // discard "Card X:"
    .split(" ")
    .filterNot(_.isEmpty())
  val (winningNumberStrs, givenNumberStrs) = numbers.span(_ != "|")
  val winningNumbers = winningNumberStrs.map(_.toInt).toSet
  // drop the initial "|"
  val givenNumbers = givenNumberStrs.drop(1).map(_.toInt).toSet
  winningNumbers.intersect(givenNumbers).size
end countWinning

def winningCounts(input: String): Iterator[Int] =
  input.linesIterator.map(countWinning)
end winningCounts

def part1(input: String): String =
  winningCounts(input)
    .map(winning => if winning > 0 then Math.pow(2, winning - 1).toInt else 0)
    .sum.toString()
end part1

def part2(input: String): String =
  winningCounts(input)
    // we only track the multiplicities of the next few cards as needed, not all of them;
    // and the first element always exists, and corresponds to the current card;
    // and the elements are always positive (because there is at least 1 original copy of each card)
    .foldLeft((0, Vector(1))){ case ((numCards, multiplicities), winning) =>
      val thisMult = multiplicities(0)
      val restMult = multiplicities
        .drop(1)
        // these are the original copies of the next few cards
        .padTo(Math.max(1, winning), 1)
        .zipWithIndex
        // these are the extra copies we just won
        .map((mult, idx) => if idx < winning then mult + thisMult else mult)
      (numCards + thisMult, restMult)
    }
    ._1.toString()
end part2

Solutions from the community​

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

- +possible number of winning numbers.
  • (Personal opinion) It makes for a nicer, purely functional solution!

    • Final code​

    def countWinning(card: String): Int =
      val numbers = card
        .substring(card.indexOf(":") + 1)   // discard "Card X:"
        .split(" ")
        .filterNot(_.isEmpty())
      val (winningNumberStrs, givenNumberStrs) = numbers.span(_ != "|")
      val winningNumbers = winningNumberStrs.map(_.toInt).toSet
      // drop the initial "|"
      val givenNumbers = givenNumberStrs.drop(1).map(_.toInt).toSet
      winningNumbers.intersect(givenNumbers).size
    end countWinning

    def winningCounts(input: String): Iterator[Int] =
      input.linesIterator.map(countWinning)
    end winningCounts

    def part1(input: String): String =
      winningCounts(input)
        .map(winning => if winning > 0 then Math.pow(2, winning - 1).toInt else 0)
        .sum.toString()
    end part1

    def part2(input: String): String =
      winningCounts(input)
        // we only track the multiplicities of the next few cards as needed, not all of them;
        // and the first element always exists, and corresponds to the current card;
        // and the elements are always positive (because there is at least 1 original copy of each card)
        .foldLeft((0, Vector(1))){ case ((numCards, multiplicities), winning) =>
          val thisMult = multiplicities(0)
          val restMult = multiplicities
            .drop(1)
            // these are the original copies of the next few cards
            .padTo(Math.max(1, winning), 1)
            .zipWithIndex
            // these are the extra copies we just won
            .map((mult, idx) => if idx < winning then mult + thisMult else mult)
          (numCards + thisMult, restMult)
        }
        ._1.toString()
    end part2

    Solutions from the community​

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

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    - + \ No newline at end of file diff --git a/assets/js/3c0fac00.701f4e27.js b/assets/js/3c0fac00.71f35976.js similarity index 97% rename from assets/js/3c0fac00.701f4e27.js rename to assets/js/3c0fac00.71f35976.js index ba7c7908a..e77d9251d 100644 --- a/assets/js/3c0fac00.701f4e27.js +++ b/assets/js/3c0fac00.71f35976.js @@ -1 +1 @@ -"use strict";(self.webpackChunkwebsite=self.webpackChunkwebsite||[]).push([[5692],{7693:(e,t,n)=>{n.r(t),n.d(t,{assets:()=>o,contentTitle:()=>l,default:()=>p,frontMatter:()=>r,metadata:()=>s,toc:()=>u});var a=n(7462),i=(n(7294),n(3905));n(6340);const r={},l="Day 4: Scratchcards",s={unversionedId:"2023/puzzles/day04",id:"2023/puzzles/day04",title:"Day 4: Scratchcards",description:"by @shardulc",source:"@site/target/mdoc/2023/puzzles/day04.md",sourceDirName:"2023/puzzles",slug:"/2023/puzzles/day04",permalink:"/scala-advent-of-code/2023/puzzles/day04",draft:!1,editUrl:"https://github.com/scalacenter/scala-advent-of-code/edit/website/docs/2023/puzzles/day04.md",tags:[],version:"current",frontMatter:{},sidebar:"adventOfCodeSidebar",previous:{title:"Day 3: Gear Ratios",permalink:"/scala-advent-of-code/2023/puzzles/day03"},next:{title:"Day 1: Calorie Counting",permalink:"/scala-advent-of-code/2022/puzzles/day01"}},o={},u=[{value:"Puzzle description",id:"puzzle-description",level:2},{value:"Solution summary",id:"solution-summary",level:2},{value:"Count number of winning numbers",id:"count-number-of-winning-numbers",level:3},{value:"Part 1",id:"part-1",level:3},{value:"Part 2",id:"part-2",level:3},{value:"Final code",id:"final-code",level:3},{value:"Solutions from the community",id:"solutions-from-the-community",level:2}],c={toc:u};function p(e){let{components:t,...n}=e;return(0,i.kt)("wrapper",(0,a.Z)({},c,n,{components:t,mdxType:"MDXLayout"}),(0,i.kt)("h1",{id:"day-4-scratchcards"},"Day 4: Scratchcards"),(0,i.kt)("p",null,"by ",(0,i.kt)("a",{parentName:"p",href:"https://github.com/shardulc"},"@shardulc")),(0,i.kt)("h2",{id:"puzzle-description"},"Puzzle description"),(0,i.kt)("p",null,(0,i.kt)("a",{parentName:"p",href:"https://adventofcode.com/2023/day/4"},"https://adventofcode.com/2023/day/4")),(0,i.kt)("h2",{id:"solution-summary"},"Solution summary"),(0,i.kt)("p",null,"First, we note that both parts rely on counting how many winning numbers there\nare on a card, so we write the helper function ",(0,i.kt)("inlineCode",{parentName:"p"},"countWinning"),". Then, part\xa01\nconverts each card's winning numbers count to a number of points and adds them\nup. Part\xa02 cannot be expressed as a map over the winning counts because the\nvalue for a card possibly depends on the cards before it. Thus, we write it as a\nfold, where the accumulator tracks: (1)\xa0the number of cards won up to the\ncurrent card; and (2)\xa0the number of copies of cards to be processed, i.e.\xa0their\n",(0,i.kt)("em",{parentName:"p"},"multiplicities"),", for as many of the following cards as needed."),(0,i.kt)("h3",{id:"count-number-of-winning-numbers"},"Count number of winning numbers"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def countWinning(card: String): Int =\n")),(0,i.kt)("p",null,"Most of this function is string manipulation:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},' val numbers = card\n .substring(card.indexOf(":") + 1) // discard "Card X:"\n .split(" ")\n .filterNot(_.isEmpty())\n val (winningNumberStrs, givenNumberStrs) = numbers.span(_ != "|")\n val winningNumbers = winningNumberStrs.map(_.toInt).toSet\n // drop the initial "|"\n val givenNumbers = givenNumberStrs.drop(1).map(_.toInt).toSet\n')),(0,i.kt)("p",null,"Although this is not specified in the puzzle description, it seems like a number\ncan appear at most once in the list of winning numbers as well as in the given\nnumbers, so it is valid to perform ",(0,i.kt)("inlineCode",{parentName:"p"},"toSet")," and the final counting step, which is\nwhat we return:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," winningNumbers.intersect(givenNumbers).size\nend countWinning\n")),(0,i.kt)("p",null,"Lastly, the standard library conveniently gives us an iterator over lines."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def winningCounts(input: String): Iterator[Int] =\n input.linesIterator.map(countWinning)\n")),(0,i.kt)("h3",{id:"part-1"},"Part 1"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def part1(input: String): String =\n winningCounts(input)\n .map(winning => if winning > 0 then Math.pow(2, winning - 1).toInt else 0)\n .sum.toString()\nend part1\n")),(0,i.kt)("h3",{id:"part-2"},"Part 2"),(0,i.kt)("p",null,"(This solution is presented in literate programming style with explanations\ninterspersed with lines of code.)"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def part2(input: String): String =\n winningCounts(input)\n")),(0,i.kt)("p",null,"The key insight here is that when we process one card, the cards we win (if any)\nare always at the immediately following indices. So instead of keeping track of\n",(0,i.kt)("em",{parentName:"p"},"absolute"),' indices (e.g.\xa0"3 copies of card 5"), we only keep track of how many\ncards we\'ve won ',(0,i.kt)("em",{parentName:"p"},"relative to the current index"),' (e.g.\xa0"3 copies of the\nnext-to-next card"). This is the ',(0,i.kt)("inlineCode",{parentName:"p"},"Vector")," in the accumulator of our fold."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," .foldLeft((0, Vector(1))){ case ((numCards, multiplicities), winning) =>\n")),(0,i.kt)("p",null,"In each iteration, we remove its first element, i.e.\xa0the multiplicity of the\ncurrent card..."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," val thisMult = multiplicities(0)\n")),(0,i.kt)("p",null,"... and carry forward the rest:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," val restMult = multiplicities\n .drop(1)\n")),(0,i.kt)("p",null,"If we just won no new cards, then we extend the vector by a single ",(0,i.kt)("inlineCode",{parentName:"p"},"1")," for the 1\noriginal copy of the next card to be processed in the next iteration. Else, we\nextend the vector by as many elements as required to keep track of the cards we\njust won."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," .padTo(Math.max(1, winning), 1)\n")),(0,i.kt)("p",null,"Remember that we win a copy of a later card for every copy we'd already won of\nthe current card."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," .zipWithIndex\n .map((mult, idx) => if idx < winning then mult + thisMult else mult)\n (numCards + thisMult, restMult)\n }\n ._1.toString()\nend part2\n")),(0,i.kt)("p",null,"Throughout the iteration, the vector satisfies the following invariants:"),(0,i.kt)("ul",null,(0,i.kt)("li",{parentName:"ul"},"It has at least one element."),(0,i.kt)("li",{parentName:"ul"},"All its elements are positive."),(0,i.kt)("li",{parentName:"ul"},"When processing a card, the first element is the final multiplicity of that\ncard. 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1).toInt else 0)\n .sum.toString()\nend part1\n\ndef part2(input: String): String =\n winningCounts(input)\n // we only track the multiplicities of the next few cards as needed, not all of them;\n // and the first element always exists, and corresponds to the current card;\n // and the elements are always positive (because there is at least 1 original copy of each card)\n .foldLeft((0, Vector(1))){ case ((numCards, multiplicities), winning) =>\n val thisMult = multiplicities(0)\n val restMult = multiplicities\n .drop(1)\n // these are the original copies of the next few cards\n .padTo(Math.max(1, winning), 1)\n .zipWithIndex\n // these are the extra copies we just won\n .map((mult, idx) => if idx < winning then mult + thisMult else mult)\n (numCards + thisMult, restMult)\n }\n ._1.toString()\nend part2\n')),(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/spamegg1/advent-of-code-2023-scala/blob/solutions/04.worksheet.sc#L116"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/spamegg1"},"Spamegg")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/pkarthick/AdventOfCode/blob/master/2023/scala/src/main/scala/day04.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/pkarthick"},"Karthick Pachiappan")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/YannMoisan/advent-of-code/blob/master/2023/src/main/scala/Day4.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/YannMoisan"},"Yann Moisan")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/prinsniels/AdventOfCode2023/blob/main/src/main/scala/solutions/day04.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/prinsniels"},"Niels Prins")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/Philippus/adventofcode/blob/main/src/main/scala/adventofcode2023/day4/Day4.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/philippus"},"Philippus Baalman")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/jnclt/adventofcode2023/blob/main/day04/scratchcards.sc"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/jnclt"},"jnclt")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/guycastle/advent_of_code_2023/blob/main/src/main/scala/days/day04/DayFour.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/guycastle"},"Guillaume Vandecasteele")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/mpilquist/aoc/blob/main/2023/day4.sc"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/mpilquist"},"Michael Pilquist")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://git.dtth.ch/nki/aoc2023/src/branch/master/Day4.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/natsukagami"},"natsukagami")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/kbielefe/advent-of-code/blob/master/2023/src/main/scala/4.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/kbielefe"},"Karl Bielefeldt")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/GrigoriiBerezin/advent_code_2023/tree/master/task04/src/main/scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/GrigoriiBerezin"},"g.berezin")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/alexandru/advent-of-code/blob/main/scala3/2023/src/main/scala/day4.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/alexandru/"},"Alexandru Nedelcu")),(0,i.kt)("li",{parentName:"ul"},(0,i.kt)("a",{parentName:"li",href:"https://github.com/bishabosha/advent-of-code-2023/blob/main/2023-day04.scala"},"Solution")," by ",(0,i.kt)("a",{parentName:"li",href:"https://github.com/bishabosha"},"Jamie Thompson"))),(0,i.kt)("p",null,"Share your solution to the Scala community by editing this page. (You can even write the whole article!)"))}p.isMDXComponent=!0}}]); \ No newline at end of file +"use strict";(self.webpackChunkwebsite=self.webpackChunkwebsite||[]).push([[5692],{7693:(e,t,n)=>{n.r(t),n.d(t,{assets:()=>o,contentTitle:()=>l,default:()=>p,frontMatter:()=>r,metadata:()=>s,toc:()=>u});var a=n(7462),i=(n(7294),n(3905));n(6340);const r={},l="Day 4: Scratchcards",s={unversionedId:"2023/puzzles/day04",id:"2023/puzzles/day04",title:"Day 4: Scratchcards",description:"by @shardulc",source:"@site/target/mdoc/2023/puzzles/day04.md",sourceDirName:"2023/puzzles",slug:"/2023/puzzles/day04",permalink:"/scala-advent-of-code/2023/puzzles/day04",draft:!1,editUrl:"https://github.com/scalacenter/scala-advent-of-code/edit/website/docs/2023/puzzles/day04.md",tags:[],version:"current",frontMatter:{},sidebar:"adventOfCodeSidebar",previous:{title:"Day 3: Gear Ratios",permalink:"/scala-advent-of-code/2023/puzzles/day03"},next:{title:"Day 1: Calorie Counting",permalink:"/scala-advent-of-code/2022/puzzles/day01"}},o={},u=[{value:"Puzzle description",id:"puzzle-description",level:2},{value:"Solution summary",id:"solution-summary",level:2},{value:"Count number of winning numbers",id:"count-number-of-winning-numbers",level:3},{value:"Part 1",id:"part-1",level:3},{value:"Part 2",id:"part-2",level:3},{value:"Final code",id:"final-code",level:3},{value:"Solutions from the community",id:"solutions-from-the-community",level:2}],c={toc:u};function p(e){let{components:t,...n}=e;return(0,i.kt)("wrapper",(0,a.Z)({},c,n,{components:t,mdxType:"MDXLayout"}),(0,i.kt)("h1",{id:"day-4-scratchcards"},"Day 4: Scratchcards"),(0,i.kt)("p",null,"by ",(0,i.kt)("a",{parentName:"p",href:"https://github.com/shardulc"},"@shardulc")),(0,i.kt)("h2",{id:"puzzle-description"},"Puzzle description"),(0,i.kt)("p",null,(0,i.kt)("a",{parentName:"p",href:"https://adventofcode.com/2023/day/4"},"https://adventofcode.com/2023/day/4")),(0,i.kt)("h2",{id:"solution-summary"},"Solution summary"),(0,i.kt)("p",null,"First, we note that both parts rely on counting how many winning numbers there\nare on a card, so we write the helper function ",(0,i.kt)("inlineCode",{parentName:"p"},"countWinning"),". Then, part\xa01\nconverts each card's winning numbers count to a number of points and adds them\nup. Part\xa02 cannot be expressed as a map over the winning counts because the\nvalue for a card possibly depends on the cards before it. Thus, we write it as a\nfold, where the accumulator tracks: (1)\xa0the number of cards won up to the\ncurrent card; and (2)\xa0the number of copies of cards to be processed, i.e.\xa0their\n",(0,i.kt)("em",{parentName:"p"},"multiplicities"),", for as many of the following cards as needed."),(0,i.kt)("h3",{id:"count-number-of-winning-numbers"},"Count number of winning numbers"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def countWinning(card: String): Int =\n")),(0,i.kt)("p",null,"Most of this function is string manipulation:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},' val numbers = card\n .substring(card.indexOf(":") + 1) // discard "Card X:"\n .split(" ")\n .filterNot(_.isEmpty())\n val (winningNumberStrs, givenNumberStrs) = numbers.span(_ != "|")\n val winningNumbers = winningNumberStrs.map(_.toInt).toSet\n // drop the initial "|"\n val givenNumbers = givenNumberStrs.drop(1).map(_.toInt).toSet\n')),(0,i.kt)("p",null,"Although this is not specified in the puzzle description, it seems like a number\ncan appear at most once in the list of winning numbers as well as in the given\nnumbers, so it is valid to perform ",(0,i.kt)("inlineCode",{parentName:"p"},"toSet")," and the final counting step, which is\nwhat we return:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," winningNumbers.intersect(givenNumbers).size\nend countWinning\n")),(0,i.kt)("p",null,"Lastly, the standard library conveniently gives us an iterator over lines."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def winningCounts(input: String): Iterator[Int] =\n input.linesIterator.map(countWinning)\n")),(0,i.kt)("h3",{id:"part-1"},"Part 1"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def part1(input: String): String =\n winningCounts(input)\n .map(winning => if winning > 0 then Math.pow(2, winning - 1).toInt else 0)\n .sum.toString()\nend part1\n")),(0,i.kt)("h3",{id:"part-2"},"Part 2"),(0,i.kt)("p",null,"(This solution is presented in literate programming style with explanations\ninterspersed with lines of code.)"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},"def part2(input: String): String =\n winningCounts(input)\n")),(0,i.kt)("p",null,"The key insight here is that when we process one card, the cards we win (if any)\nare always at the immediately following indices. So instead of keeping track of\n",(0,i.kt)("em",{parentName:"p"},"absolute"),' indices (e.g.\xa0"3 copies of card 5"), we only keep track of how many\ncards we\'ve won ',(0,i.kt)("em",{parentName:"p"},"relative to the current index"),' (e.g.\xa0"3 copies of the\nnext-to-next card"). This is the ',(0,i.kt)("inlineCode",{parentName:"p"},"Vector")," in the accumulator of our fold."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," .foldLeft((0, Vector(1))){ case ((numCards, multiplicities), winning) =>\n")),(0,i.kt)("p",null,"In each iteration, we remove its first element, i.e.\xa0the multiplicity of the\ncurrent card..."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," val thisMult = multiplicities(0)\n")),(0,i.kt)("p",null,"... and carry forward the rest:"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," val restMult = multiplicities\n .drop(1)\n")),(0,i.kt)("p",null,"If we just won no new cards, then we extend the vector by a single ",(0,i.kt)("inlineCode",{parentName:"p"},"1")," for the 1\noriginal copy of the next card to be processed in the next iteration. Else, we\nextend the vector by as many elements as required to keep track of the cards we\njust won."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," .padTo(Math.max(1, winning), 1)\n")),(0,i.kt)("p",null,"Remember that we win a copy of a later card for every copy we'd already won of\nthe current card."),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"}," .zipWithIndex\n .map((mult, idx) => if idx < winning then mult + thisMult else mult)\n (numCards + thisMult, restMult)\n }\n ._1.toString()\nend part2\n")),(0,i.kt)("p",null,"Throughout the iteration, the vector satisfies the following invariants:"),(0,i.kt)("ul",null,(0,i.kt)("li",{parentName:"ul"},"It has at least one element."),(0,i.kt)("li",{parentName:"ul"},"All its elements are positive."),(0,i.kt)("li",{parentName:"ul"},"When processing a card, the first element is the final multiplicity of that\ncard. (It is summed up in ",(0,i.kt)("inlineCode",{parentName:"li"},"numCards")," in the accumulator.)")),(0,i.kt)("p",null,"Why track by relative index instead of absolute?"),(0,i.kt)("ul",null,(0,i.kt)("li",{parentName:"ul"},"We don't have to parse or store the card numbers."),(0,i.kt)("li",{parentName:"ul"},"We can discard information as soon as it is no longer needed, and keep only\nlimited information about the future, in this case bounded by the maximum\npossible number of winning numbers."),(0,i.kt)("li",{parentName:"ul"},"(Personal opinion) It makes for a nicer, purely functional solution!")),(0,i.kt)("h3",{id:"final-code"},"Final code"),(0,i.kt)("pre",null,(0,i.kt)("code",{parentName:"pre",className:"language-scala"},'def countWinning(card: String): Int =\n val numbers = card\n .substring(card.indexOf(":") + 1) // discard "Card X:"\n .split(" ")\n .filterNot(_.isEmpty())\n val (winningNumberStrs, givenNumberStrs) = numbers.span(_ != "|")\n val winningNumbers = winningNumberStrs.map(_.toInt).toSet\n // drop the initial "|"\n val givenNumbers = givenNumberStrs.drop(1).map(_.toInt).toSet\n winningNumbers.intersect(givenNumbers).size\nend countWinning\n\ndef winningCounts(input: String): Iterator[Int] =\n input.linesIterator.map(countWinning)\nend winningCounts\n\ndef part1(input: String): String =\n winningCounts(input)\n .map(winning => if winning > 0 then Math.pow(2, winning - 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e={1303:0,532:0};r.f.j=(c,a)=>{var f=r.o(e,c)?e[c]:void 0;if(0!==f)if(f)a.push(f[2]);else if(/^(1303|532)$/.test(c))e[c]=0;else{var d=new Promise(((a,d)=>f=e[c]=[a,d]));a.push(f[2]=d);var t=r.p+r.u(c),b=new Error;r.l(t,(a=>{if(r.o(e,c)&&(0!==(f=e[c])&&(e[c]=void 0),f)){var d=a&&("load"===a.type?"missing":a.type),t=a&&a.target&&a.target.src;b.message="Loading chunk "+c+" failed.\n("+d+": "+t+")",b.name="ChunkLoadError",b.type=d,b.request=t,f[1](b)}}),"chunk-"+c,c)}},r.O.j=c=>0===e[c];var c=(c,a)=>{var f,d,t=a[0],b=a[1],o=a[2],n=0;if(t.some((c=>0!==e[c]))){for(f in b)r.o(b,f)&&(r.m[f]=b[f]);if(o)var i=o(r)}for(c&&c(a);n Scala Center Advent of Code | Scala Center Advent of Code - +

    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 c72ddba4b..038bb3c25 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 2996d98f0..dd7586476 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 c5f8cc5b8..087edbb20 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 9b1602dce..bca2fb733 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 0ab5d9e12..da51078c0 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 54d70204c..f46448370 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 060215029..a33122e0d 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 b7764c1d3..4865d6595 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 13c6bea69..e5a40bb8f 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 024a9e164..61aa27b33 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 5e564e408..60a7bc1e4 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 bfe393004..4e6517b74 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 4ffbd2441..250d46c20 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 6a6ecdca0..bba202fd5 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 b2ab1f699..eddf11730 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 5eb69189a..4bb36e3bd 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 92c45ef68..73cbf1224 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 1dc879335..5dcedb392 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 13ac26859..1aec7c02c 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 a14682fe0..caabb246f 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 4c7cb8174..4252a7ba5 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 3ccb88d6f..47dfaae61 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 e585d724a..326799781 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 f8bf6b229..16ecefdcc 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 3a29edb28..9bb1e0fdd 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 485b80d5d..f59ac2f06 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 19f06e873..3224c22c5 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