When doing functional programming, you can't really do loops (because of referential transparency, you can't update iterators or indices). However, recursion still works.
Agreed, i get annoyed when I can't actually solve the problem. I would be ok if the inputs are trivial special cases, as long as feasible (but harder) generalized solutions still existed.
Scala3
Learning about scala-graph yesterday seems to have paid off already. This explicitly constructs the entire graph of allowed moves, and then uses a naive dijkstra run. This works, and I don't have to write a lot of code, but it is fairly inefficient.
import day10._
import day10.Dir._
import day11.Grid
// standing on cell p, having entered from d
case class Node(p: Pos, d: Dir)
def connect(p: Pos, d: Dir, g: Grid[Int], dists: Range) =
val from = Seq(-1, 1).map(i => Dir.from(d.n + i)).map(Node(p, _))
val ends = List.iterate(p, dists.last + 1)(walk(_, d)).filter(g.inBounds)
val costs = ends.drop(1).scanLeft(0)(_ + g(_))
from.flatMap(f => ends.zip(costs).drop(dists.start).map((dest, c) => WDiEdge(f, Node(dest, d), c)))
def parseGrid(a: List[List[Char]], dists: Range) =
val g = Grid(a.map(_.map(_.getNumericValue)))
Graph() ++ g.indices.flatMap(p => Dir.all.flatMap(d => connect(p, d, g, dists)))
def compute(a: List[String], dists: Range): Long =
val g = parseGrid(a.map(_.toList), dists)
val source = Node(Pos(-1, -1), Right)
val sink = Node(Pos(-2, -2), Right)
val start = Seq(Down, Right).map(d => Node(Pos(0, 0), d)).map(WDiEdge(source, _, 0))
val end = Seq(Down, Right).map(d => Node(Pos(a(0).size - 1, a.size - 1), d)).map(WDiEdge(_, sink, 0))
val g2 = g ++ start ++ end
g2.get(source).shortestPathTo(g2.get(sink)).map(_.weight).getOrElse(-1.0).toLong
def task1(a: List[String]): Long = compute(a, 1 to 3)
def task2(a: List[String]): Long = compute(a, 4 to 10)
Scala3
def hash(s: String): Long = s.foldLeft(0)((h, c) => (h + c)*17 % 256)
extension [A] (a: List[A])
def mapAtIndex(idx: Long, f: A => A): List[A] =
a.zipWithIndex.map((e, i) => if i == idx then f(e) else e)
def runProcedure(steps: List[String]): Long =
@tailrec def go(boxes: List[List[(String, Int)]], steps: List[String]): List[List[(String, Int)]] =
steps match
case s"$label-" :: t =>
go(boxes.mapAtIndex(hash(label), _.filter(_._1 != label)), t)
case s"$label=$f" :: t =>
go(boxes.mapAtIndex(hash(label), b =>
val slot = b.map(_._1).indexOf(label)
if slot != -1 then b.mapAtIndex(slot, (l, _) => (l, f.toInt)) else (label, f.toInt) :: b
), t)
case _ => boxes
go(List.fill(256)(List()), steps).zipWithIndex.map((b, i) =>
b.zipWithIndex.map((lens, ilens) => (1 + i) * (b.size - ilens) * lens._2).sum
).sum
def task1(a: List[String]): Long = a.head.split(",").map(hash).sum
def task2(a: List[String]): Long = runProcedure(a.head.split(",").toList)
Scala3
type Grid = List[List[Char]]
def tiltUp(a: Grid): Grid =
@tailrec def go(c: List[Char], acc: List[Char]): List[Char] =
def shifted(c: List[Char]) =
val (h, t) = c.splitAt(c.count(_ == 'O'))
h.map(_ => 'O') ++ t.map(_ => '.') ++ acc
val d = c.indexOf('#')
if d == -1 then shifted(c) else go(c.slice(d + 1, c.size), '#'::shifted(c.slice(0, d)))
a.map(go(_, List()).reverse)
def weight(a: Grid): Long = a.map(d => d.zipWithIndex.filter((c, _) => c == 'O').map(1 + _._2).sum).sum
def rotateNeg90(a: Grid): Grid = a.reverse.transpose
def runCycle = Seq.fill(4)(tiltUp andThen rotateNeg90).reduceLeft(_ andThen _)
def stateAt(target: Long, a: Grid): Grid =
@tailrec def go(cycle: Int, state: Grid, seen: Map[Grid, Int]): Grid =
seen.get(state) match
case Some(i) => if (target - cycle) % (cycle - i) == 0 then state else go(cycle + 1, runCycle(state), seen)
case None => go(cycle + 1, runCycle(state), seen + (state -> cycle))
go(0, a, Map())
def toColMajorGrid(a: List[String]): Grid = rotateNeg90(a.map(_.toList))
def task1(a: List[String]): Long = weight(tiltUp(toColMajorGrid(a)))
def task2(a: List[String]): Long = weight(stateAt(1_000_000_000, toColMajorGrid(a)))
Scala3
// i is like
// # # # # #
// 1 2 3 4
def smudgesAround(i: Int, s: List[List[Char]]): Long =
val toEdge = math.min(i, s.size - i)
(0 until toEdge).map(e => s(i - e - 1).lazyZip(s(i + e)).count(_ != _)).sum
def symmetries(g: List[List[Char]], smudges: Int) =
val rows = (1 until g.size).filter(smudgesAround(_, g) == smudges)
val g2 = g.transpose
val cols = (1 until g2.size).filter(smudgesAround(_, g2) == smudges)
100*rows.sum + cols.sum
def task1(a: List[String]): Long = a.chunk(_ == "").map(g => symmetries(g.map(_.toList), 0)).sum
def task2(a: List[String]): Long = a.chunk(_ == "").map(g => symmetries(g.map(_.toList), 1)).sum
Scala3
def diffs(a: Seq[Long]): List[Long] =
a.drop(1).zip(a).map(_ - _).toList
def predictNext(a: Seq[Long], combine: (Seq[Long], Long) => Long): Long =
if a.forall(_ == 0) then 0 else combine(a, predictNext(diffs(a), combine))
def predictAllNexts(a: List[String], combine: (Seq[Long], Long) => Long): Long =
a.map(l => predictNext(l.split(raw"\s+").map(_.toLong), combine)).sum
def task1(a: List[String]): Long = predictAllNexts(a, _.last + _)
def task2(a: List[String]): Long = predictAllNexts(a, _.head - _)
I can prove the opposite for you. The assumption that Gobbel2000 describes is wrong in general. For example, take
L
A -> B, X
B -> C, X
C -> Z, X
Z -> C, X
X -> X, X
the first Z is reached after 3 steps, but then every cycle only takes 2 steps.
The matches are still at consistent intervals, but you can easily find a counterexample for that as well:
L
A -> 1Z, X
1Z -> 2Z, X
2Z -> A, X
X -> X, X
now the intervals will be 2, 1, 2, 1, ...
However, it is easy to prove that there will be a loop of finite length, and that the intervals will behave somewhat nicely:
Identify a "position" by a node you are at, and your current index in the LRL instruction sequence. If you ever repeat a position P, you will repeat the exact path away from the position you took the last time, and the last time you later reached P, so you will keep reaching P again and again. There are finitely many positions, so you can't keep not repeating any forever, you will run out.
Walking in circles along this loop you eventually find yourself in, the intervals between Zs you see will definitely be a repeating sequence (as you will keep seeing not just same-length intervals, but in fact the exact same paths between Zs).
So in total, you will see some finite list of prefix-intervals, and then a repeating cycle of loop-intervals. I have no idea if this can be exploited to compute the answer efficiently, but see my solution-comment for something that only assumes that only one Z will be encountered each cycle.
Scala3
this is still not 100% general, as it assumes there is only one Z-node along the path for each start. But it doesn't assume anything else, I think. If you brute force combinations of entries of the zs-arrays, this assumption can be trivially removed, but if multiple paths see lots of Z-nodes, then the runtime will grow exponentially. I don't know if it's possible to do this any faster.
def parse(a: List[String]): (List[Char], Map[String, Map[Char, String]]) =
def parseNodes(n: List[String]) =
n.flatMap{case s"$from = ($l, $r)" => Some(from -> Map('L'->l, 'R'->r)); case _ => None}.toMap
a match{case instr::""::nodes => ( instr.toList, parseNodes(nodes) ); case _ => (List(), Map())}
def task1(a: List[String]): Long =
val (instr, nodes) = parse(a)
def go(i: Stream[Char], pos: String, n: Long): Long =
if pos != "ZZZ" then go(i.tail, nodes(pos)(i.head), n+1) else n
go(instr.cycle, "AAA", 0)
// ok so i originally thought this was going to be difficult, so
// we parse a lot of info we won't use
case class PathInfo(zs: List[Long], loopFrom: Long, loopTo: Long):
def stride: Long = loopTo - loopFrom
type Cycle = Long
def getInfo(instr: List[Char], isEnd: String => Boolean, map: Map[String, Map[Char, String]], start: String): PathInfo =
def go(i: Cycle, pos: String, is: List[Char], seen: Map[(Long, String), Cycle], acc: List[Long]): PathInfo =
val current: (Long, String) = (is.size % instr.size, pos)
val nis = if is.isEmpty then instr else is
val nacc = if isEnd(pos) then i::acc else acc
seen.get(current) match
case Some(l) => PathInfo(acc, l, i)
case _ => go(i + 1, map(pos)(nis.head), nis.tail, seen + (current -> i), nacc)
go(0, start, instr, Map(), List())
// turns out we don't need to check all the different positions
// in each cycle where we are on a Z position, as a) every start
// walks into unique cycles with b) exactly one Z position,
// and c) the length of the cycle is equivalent to the steps to first
// encounter a Z (so a->b->c->Z->c->Z->... is already more complicated,
// as the cycle length is 2 but the first encounter is after 3 steps)
// anyway let's do some math
// this is stolen code
def gcd(a: Long, b: Long): Long =
if(b ==0) a else gcd(b, a%b)
def primePowers(x: Long): Map[Long, Long] =
// for each prime p for which p^k divides x: p->p^k
def go(r: Long, t: Long, acc: Map[Long, Long]): Map[Long, Long] =
if r == 1 then acc else
if r % t == 0 then
go(r/t, t, acc + (t -> acc.getOrElse(t, 1L)*t))
else
go(r, t+1, acc)
go(x, 2, Map())
// this algorithm is stolen, but I scalafied the impl
def vanillaCRT(congruences: Map[Long, Long]): Long =
val N = congruences.keys.product
val x = congruences.map((n, y) => y*(N/n)*((N/n) % n)).sum
if x <= 0 then x + N else x
def generalizedHardcoreCRTThatCouldHaveBeenAnLCMBecauseTheInputIsVeryConvenientlyTrivialButWeWantToDoThisRight(ys: List[Long], xs: List[Long]): Option[Long] =
// finds the smallest k s.t. k === y_i (mod x_i) for each i
// even when stuff is not nice
// pre-check if everything is sufficiently coprime
// https://math.stackexchange.com/questions/1644677/what-to-do-if-the-modulus-is-not-coprime-in-the-chinese-remainder-theorem
val vars = for
((y1, n1), i1) <- ys.zip(xs).zipWithIndex
((y2, n2), i2) <- ys.zip(xs).zipWithIndex
if i2 > i1
yield
val g = gcd(n1, n2)
y1 % g == y2 % g
if !vars.forall(a => a) then
None
else
// we need to split k === y (mod mn) into k === y (mod m) and k === y (mod n) for m, n coprime
val congruences = for
(x, y) <- xs.zip(ys)
(p, pl) <- primePowers(x)
yield
p -> (y % pl -> pl)
// now we eliminate redundant congruences
// since our input is trivial, this is essentially
// the step in which we solve the task by
// accidentaly computing an lcm
val r = congruences.groupMap(_._1)(_._2).mapValues(l => l.map(_._2).max -> l.head._1).values.toMap
// ok now we meet the preconditions
// for doing this
Some(vanillaCRT(r))
def task2(a: List[String]): Long =
val (instr, nodes) = parse(a)
val infos = nodes.keySet.filter(_.endsWith("A")).map(s => getInfo(instr, _.endsWith("Z"), nodes, s))
generalizedHardcoreCRTThatCouldHaveBeenAnLCMBecauseTheInputIsVeryConvenientlyTrivialButWeWantToDoThisRight(
infos.map(_.zs.head).toList, infos.map(_.stride).toList
).getOrElse(0)
@main def main: Unit =
val data = readlines("/home/tim/test/aoc23/aoc23/inputs/day8/task1.txt")
for f <- List(
task1,
task2,
).map(_ andThen println)
do f(data)
Scala3
val tiers = List(List(1, 1, 1, 1, 1), List(1, 1, 1, 2), List(1, 2, 2), List(1, 1, 3), List(2, 3), List(1, 4), List(5))
val cards = List('2', '3', '4', '5', '6', '7', '8', '9', 'T', 'J', 'Q', 'K', 'A')
def cardValue(base: Long, a: List[Char], cards: List[Char]): Long =
a.foldLeft(base)(cards.size * _ + cards.indexOf(_))
def hand(a: List[Char]): List[Int] =
a.groupMapReduce(s => s)(_ => 1)(_ + _).values.toList.sorted
def power(a: List[Char]): Long =
cardValue(tiers.indexOf(hand(a)), a, cards)
def power3(a: List[Char]): Long =
val x = hand(a.filter(_ != 'J'))
val t = tiers.lastIndexWhere(x.zipAll(_, 0, 0).forall(_ <= _))
cardValue(t, a, 'J'::cards)
def win(a: List[String], pow: List[Char] => Long) =
a.flatMap{case s"$hand $bid" => Some((pow(hand.toList), bid.toLong)); case _ => None}
.sorted.map(_._2).zipWithIndex.map(_ * _.+(1)).sum
def task1(a: List[String]): Long = win(a, power)
def task2(a: List[String]): Long = win(a, power3)
Scala3
kind of convoluted, but purely functional
import scala.collection.immutable.NumericRange.Exclusive
import math.max
import math.min
extension [A] (l: List[A])
def chunk(pred: A => Boolean): List[List[A]] =
def go(l: List[A], partial_acc: List[A], acc: List[List[A]]): List[List[A]] =
l match
case (h :: t) if pred(h) => go(t, List(), partial_acc.reverse :: acc)
case (h :: t) => go(t, h :: partial_acc, acc)
case _ => partial_acc.reverse :: acc
go(l, List(), List()).reverse
type R = Exclusive[Long]
def intersectTranslate(r: R, c: R, t: Long): R =
(t + max(r.start, c.start) - c.start) until (t + min(r.end, c.end) - c.start)
case class MappingEntry(from: R, to: Long)
case class Mapping(entries: List[MappingEntry], produces: String):
def resolveRange(in: R): List[R] =
entries.map(e => intersectTranslate(in, e.from, e.to)).filter(!_.isEmpty)
def completeEntries(a: List[MappingEntry]): List[MappingEntry] =
a ++ ((0L until 0L) +: a.map(_.from).sorted :+ (Long.MaxValue until Long.MaxValue)).sliding(2).flatMap{ case List(a, b) => Some(MappingEntry(a.end until b.start, a.end)); case _ => None}.toList
def parse(a: List[String], init: List[Long] => List[R]): (List[R], Map[String, Mapping]) =
def parseEntry(s: String): MappingEntry =
s match
case s"$end $start $range" => MappingEntry(start.toLong until start.toLong + range.toLong, end.toLong)
a.chunk(_ == "") match
case List(s"seeds: $seeds") :: maps =>
init(seeds.split(raw"\s+").map(_.toLong).toList) -> (maps.flatMap{ case s"$from-to-$to map:" :: entries => Some(from -> Mapping(completeEntries(entries.map(parseEntry)), to)); case _ => None }).toMap
case _ => (List(), Map()).ensuring(false)
def singletons(a: List[Long]): List[R] = a.map(s => s until s + 1)
def paired(a: List[Long]): List[R] = a.grouped(2).flatMap{ case List(x, y) => Some(x until x+y); case _ => None }.toList
def chase(d: (List[R], Map[String, Mapping]), initial: String, target: String) =
val (init, m) = d
def go(a: List[R], s: String): List[R] =
if trace(s) == target then a else
val x = m(s)
go(a.flatMap(x.resolveRange), x.produces)
go(trace(init), initial)
def task1(a: List[String]): Long =
chase(parse(a, singletons), "seed", "location").min.start
def task2(a: List[String]): Long =
chase(parse(a, paired), "seed", "location").min.start
Scala3
all done!