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πŸŽ„ - 2023 DAY 8 SOLUTIONS -πŸŽ„

Day 8: Haunted Wasteland

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ , pastebin, or github (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)

FAQ

22 comments

  • Rust

    As others have shown, part 2 can be pretty simple if you allow one assumption: The distance from a start point to the nearest end point is always the same as cycling from that nearest end point back to itself. Under that assumption you can just take the lowest common multiple of these distances. And honestly, who can claim to understand ghost navigation and what you can and can't assume? Empirical evidence suggests that this is how ghosts travel.

  • 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)

  • Nim

    I like how if you have an error in your calculations, you end up wandering the haunted desert for eternity. Very flavorful.

    My solution for part 2 is pretty much the same as for part 1, just modified to work on a sequence of nodes instead of a single node. However, it doesn't find an answer in the time that I was willing to wait for it. I thought about trying to optimize it to run faster, but figured that if it was taking this long on Nim, then interpreted languages would have no chance, so that couldn't be the right approach.

    I suspected that maybe the ghosts arrived at the Z locations at predictable intervals. I added some code to output the step count each time each ghost reached a Z (see commented code), and my suspicion was correct. Just needed to calculate the least common multiple of the 6 cycle lengths.

  • APL

    I'm using this year's AoC to learn (Dyalog) APL, so this is most likely a pretty terrible solutions. I would've liked to use ⍣ instead of of my imitation of a while loop with a recursive function, but I couldn't figure out how to get to the number of iterations ⍣ performed to arrive at the destination. If someone here knows how to do that (or has other suggestions for improvement) I'm open for suggestions.

     apl
        
βŽ•IO ← 0       ⍝ use 0-based indexes
Iβ†βŒ·β¨βˆ˜βŠƒβ¨β€0 99  ⍝ "sane" indexing
βŽ•PP←17        ⍝ print integers with up to 17 significant digits without use of scientific notation

inputβ†βŠƒβŽ•NGET'inputs/day8.txt'1
instructionsβ†βŠƒinput
p←↑{0 2 4 I{6β΄βŽ•CSV⍠'Widths'(3 4 3 2 3 1)⊒⍡'S'1}⍡}Β¨ 2↓input
nodes←,1↑[1]p
L←(⍳⍴nodes) ∘.{nodes[⍺]≑⍡ 1 ⌷p} ⍳⍴nodes
R←(⍳⍴nodes) ∘.{nodes[⍺]≑⍡ 2 ⌷p} ⍳⍴nodes
S←(βŠƒ(+.Γ—)/{'R'≑⍡:R β‹„ ⍡≑'L':L}¨⌽instructions)
βŽ•β†(β‰’instructions) Γ— {⍺←0 β‹„ ⍡[⍸'ZZZ'βˆ˜β‰‘Β¨nodes]: ⍺ β‹„ ⍺+1 βˆ‡ S+.×⍡}'AAA'βˆ˜β‰‘Β¨nodes                             ⍝ part 1
βŽ•β†βˆ§/(β‰’instructions) Γ— {⍺←0 β‹„ 0≑+/⍡[⍸~('Z'∘=⊒/)Β¨nodes]: ⍺ β‹„ ⍺+1 βˆ‡ S+.×⍡}Β¨{(⍳≒nodes)=⍡}¨⍸('A'∘=⊒/)Β¨nodes  ⍝ part 2

  • Dart

    Part 1 was easy enough. For Part 2 I took a guess that the cycles were all well behaved and I could get away with just calculating the LCM, and it paid off.

      
        
(List, Map) parse(List lines) => (
      lines.first.split(''),
      {
        for (var l in lines.skip(2).map((e) => e.split(RegExp(r'[^A-Z0-9]'))))
          l[0]: [l[4], l[6]]
      }
    );
int movesTo(pos, steps, rules) {
  var stepNo = -1;
  while (!pos.endsWith('Z')) {
    pos = rules[pos]![steps[(stepNo += 1) % steps.length] == 'L' ? 0 : 1];
  }
  return stepNo + 1;
}

part1(List lines) {
  var (steps, rules) = parse(lines);
  return movesTo('AAA', steps, rules);
}

// All cycles are independent of state of `steps`, and have
// first instance == cycle length which makes this verrry simple.
part2(List lines) {
  var (steps, rules) = parse(lines);
  return [
    for (var start in rules.keys.where((e) => e.endsWith('A')))
      movesTo(start, steps, rules)
  ].reduce((s, t) => s.lcm(t));
}
}

  • Crystal

    !crystal_lang@lemmy.ml

    part 1 was made way simpler after I made it a function for part 2
    also I accidentally saw someone elses code, so I can't take full credit for figuring out how to do part 2

     crystal
        
Input = File.read("input.txt").lines

Dirs = Input[0].chars.map {|c| c == 'L' ? 0 : 1}
Nodes = Hash(String, Tuple(String, String)).new
Input[2..].each {|line| Nodes[line[..2]] = {line[7..9], line[12..14]} }

# part 1
puts num_steps("AAA", &.== "ZZZ")

# part 2
currents = Nodes.keys.select(&.ends_with? 'A')
steps = currents.map {|cur| num_steps(cur, &.ends_with? 'Z')}
puts steps.reduce(1_i64) {|a, b| a.lcm b}

def num_steps(start)
    steps = 0
    Dirs.cycle do |dir|
        steps += 1
        start = Nodes[start][dir]
        break if yield start
    end
    steps
end

  • First part was simple enough. Second part was easy logically, but after running the brute force solution with a parallel iterator from rayon and maxing out all 12 cores of this CPU, it was still taking forever. I always get tripped up by these ones that need fancy math, because although I was always good at math, I've never been good at looking at these problems and figuring out what kind of formula would apply. So I cheated and looked at other people's comments for their solutions, and saw the least common multiple mentioned. This made sense to me, so I implemented it and got a result almost instantly. I hate having to look at other comments to solve these things, but I never would have came to that conclusion myself.

    The code still isn't the cleanest, and I'd love to tidy up the parsing, but it works and I'm happy.

    https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day08.rs

     rust
        
impl Solver for Day08 {
    fn star_one(&self, input: &str) -> String {
        let (directions, map) = input.split_once("\n\n").unwrap();

        let mut route_map = HashMap::new();

        for line in map.lines() {
            let line = line.replace(" ", "").replace("(", "").replace(")", "");
            let (position, destinations) = line.split_once('=').unwrap();
            let (left, right) = destinations.split_once(',').unwrap();
            route_map.insert(position.to_string(), (left.to_string(), right.to_string()));
        }

        let mut current_position = "AAA".to_string();
        for (step, direction) in directions.chars().cycle().enumerate() {
            current_position = match direction {
                'L' => route_map[Β€t_position].0.to_string(),
                'R' => route_map[Β€t_position].1.to_string(),
                _ => unreachable!(),
            };

            if current_position == "ZZZ" {
                return (step + 1).to_string();
            }
        }

        unreachable!()
    }

    fn star_two(&self, input: &str) -> String {
        let (directions, map) = input.split_once("\n\n").unwrap();

        let mut route_map = HashMap::new();

        for line in map.lines() {
            let line = line.replace(" ", "").replace("(", "").replace(")", "");
            let (position, destinations) = line.split_once('=').unwrap();
            let (left, right) = destinations.split_once(',').unwrap();
            route_map.insert(position.to_string(), (left.to_string(), right.to_string()));
        }

        let positions = route_map
            .keys()
            .filter(|pos| pos.ends_with('A'))
            .collect::>();

        let steps = positions
            .iter()
            .filter(|pos| pos.ends_with('A'))
            .map(|pos| {
                let mut current_position = pos.to_string();
                for (step, direction) in directions.chars().cycle().enumerate() {
                    current_position = match direction {
                        'L' => route_map[Β€t_position].0.to_string(),
                        'R' => route_map[Β€t_position].1.to_string(),
                        _ => unreachable!(),
                    };

                    if current_position.ends_with('Z') {
                        return step + 1;
                    }
                }
                unreachable!()
            })
            .collect::>();

        steps
            .into_iter()
            .reduce(|acc, steps| acc.lcm(&steps))
            .unwrap()
            .to_string()
    }
}

  • [language: Lean4]

    I did not make the assumption that the distance from the start to the goal is the same as the cycle length. This made the problem a bit harder, but I'm pretty sure my solution is actually generic. As always, I'll only post the direct solution. Helper functions (like, in this case, xgcd) are in other files, you can find them together with the full solution on github.

    Since my solution contains a ton of comments (that outline my thought process), I'll again split it in several posts here.

    • Actually, all my comments in Part 2 make it too big for a single post. The full solution consists of two functions (that could be combined with a bit more work...).

      The first function is a brute-force approach, that has a similar termination condition to the first part. It halts, once all paths are walking in circles. This function does not find a solution yet. Then there is a second function that looks for solutions that appear later, while all paths are already walking in circles. That function writes all goals as equations of the form x = start + n * period. The target condition is that all paths have the same value for x. This was then solved by looking at all permutations between two paths, and solving the Diophantine equations, creating a new, combined, path.

      In the end, the closest resulting valid goal was also the final result.

      • And last, but not least, here's the part that actually finds the result: :::spoiler Part 2, the part for after the brute force approach fails (don't be alarmed, it's mostly comments, very few actual code lines) ```Lean4 -- Okay, tried Brute Force, it did NOT work. Or rather, it might work, but I won't be able to prove -- termination for it. Not that it wouldn't be possible to prove, just I won't manage to. -- The problem is that the termination condition in part2_impl is too soon. -- You can actually see this in the example (for which part2_impl works, but by chance). -- While the goals in each part repeat, they repeat at different rates. -- Soo, we would need to continue even after each part has started walking in circles. -- However, just doing that runs for a very long time without finding a result. -- Sooo, let's be smarter.

        -- Every path consist of 2 segments. The part that leads up to a cycle, and the cycle. -- Both parts can contain goals, but once the cycle starts, the goals repeat with cycle-length. -- A cycle is at least one pass, but can (and does...) consist of multiple passes too.

        -- We can use part2_impl still - to verify that we do not reach the goals before all our paths start -- cycling. That allows us to only care about cycling paths in the second part of the solution, which -- we only reach if part2_impl does not yield a result (we know it doesn't, but that would be cheating).

        -- Soo, how can the second part look like?

        -- For simplicity let's not do this in parallel. Rather, let's find the goals for each start individually -- So, let's just consider a single path (like the one from part1) -- We need to write down the number of steps at which we reach a goal. -- Whenever we remove a remaining start from the possible starts list, we need to note it down, and -- how many steps it took us to get there. -- Once we detect a circle, we can look up -- how many steps we took in total till we startec cycling -- and how many steps it took us to reach the cycle start for the first time -- that's the period of each goal in the cycle. -- For each goal that was found between cycle-start and cycle-end, we can write down an equation: -- x = steps_from_start + cycle_length * n -- n is a Natural number here, not an integer. x is the number of steps at which we pass this goal

        -- Once we have that information for all goals of all starts, we can combine it: -- That's a set of Diophantine equations.

        -- Or, rather, several sets of Diophantine equations... -- For each combination of goals that are reached in the cycles of the participating paths, we need to -- solve the following system:

        -- We can write each goal for each run in the form x = g0 + n * cg -- Where x is the solution we are looking for, g0 is the number of steps from the start until -- we hit the goal for the first time in the cycle, and cg is the cycle length

        -- Once we have those equations, we can combine them pairwise: https://de.wikipedia.org/wiki/Lineare_diophantische_Gleichung -- This allows us to reduce all paths to a single one, which has multiple equations that -- describe when a goal is reached. -- For each of those equations we need to find the first solution that is larger than -- the point where all paths started to cycle. The smallest of those is the result.

        -- a recurring goal, that starts at "start" and afterwards appears at every "interval". private structure CyclingGoal where start : Nat interval : Nat deriving BEq

        instance : ToString CyclingGoal where toString g := s!"g = {g.start} + n * {g.interval}"

        private def CyclingGoal.nTh (goal : CyclingGoal) (n : Nat) : Nat := goal.start + n * goal.interval

        -- Combine two cycling goals into a new cycling goal. This might fail, if they never meet. -- This can for instance happen if they have the same frequency, but different starts. private def CyclingGoal.combine (a b : CyclingGoal) : Option CyclingGoal := -- a.start + n * a.interval = b.start + m * b.interval -- n * a.interval - m * b.interval = b.start - a.start -- we want to do as much as possible in Nat, such that we can easily reason about which numbers are -- positive. Soo let (a, b) := if a.start > b.start then (b,a) else (a,b) let (g, u, _) := Euclid.xgcd a.interval b.interval -- there is no solution if b.start - a.start is not divisible by g let c := (b.start - a.start) let s := c / g if s * g != c then none else let deltaN := b.interval / g let n0 := s * u -- we can use u directly - v would need its sign swapped, but we don't use v. -- a.start + (n0 + t * deltaN)a.interval -- a.start + n0a.interval + t * deltaN * a.interval -- we need the first value of t that yields a result >= max(a.start, b.start) -- because that's where our cycles start. let x := ((max a.start b.start : Int) - a.interval * n0 - a.start) let interval := a.interval * deltaN let t0 := x / interval let t0 := if t0 * interval == x || x < 0 then t0 else t0 + 1 -- int division rounds towards zero, so for x < 0 it's already ceil. let start := a.start + n0 * a.interval + t0 * deltaN * a.interval assert! (start β‰₯ max a.start b.start) let start := start.toNat some {start, interval }

        private def findCyclingGoalsInPathPass (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (alreadyDoneSteps : Nat) (currentPosition : WaypointId) (instructions : List Instruction) (visitedGoals : List Nat) : Option (Nat Γ— WaypointId Γ— (List Nat)) := do let visitedGoals := if currentPosition.endsWith "Z" then alreadyDoneSteps :: visitedGoals else visitedGoals match instructions with | [] => some (alreadyDoneSteps, currentPosition, visitedGoals) | a :: as => let currentWaypoint := waypoints.find? currentPosition match currentWaypoint with | none => none -- should not happen | some currentWaypoint => findCyclingGoalsInPathPass waypoints (alreadyDoneSteps + 1) (currentWaypoint.get a) as visitedGoals

        private def findCyclingGoalsInPath_impl (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (instructions : List Instruction) (possibleStarts : List WaypointId) (visitedGoals : List Nat) (visitedStarts : List (WaypointId Γ— Nat)) (currentPosition : WaypointId) (currentSteps : Nat) : List CyclingGoal := let remainingStarts := possibleStarts.filter Ξ»s ↦ s != currentPosition if remainingStarts.length < possibleStarts.length then -- written this way to make termination_by easy let visitedStarts := (currentPosition, currentSteps) :: visitedStarts let passResult := findCyclingGoalsInPathPass waypoints currentSteps currentPosition instructions visitedGoals match passResult with | none => [] -- should not happen. Only possible if there's a dead end | some (currentSteps, currentPosition, visitedGoals) => findCyclingGoalsInPath_impl waypoints instructions remainingStarts visitedGoals visitedStarts currentPosition currentSteps else let beenHereWhen := visitedStarts.find? Ξ»s ↦ s.fst == currentPosition let beenHereWhen := beenHereWhen.get!.snd --cannot possibly fail let cycleLength := currentSteps - beenHereWhen visitedGoals.filterMap Ξ» n ↦ if n β‰₯ beenHereWhen then some {start := n, interval := cycleLength : CyclingGoal} else none -- goal was reached before we started to walk in cycles, ignore. termination_by findCyclingGoalsInPath_impl a b c d e f g => c.length

        private def findCyclingGoalsInPath (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (instructions : List Instruction) (possibleStarts : List WaypointId) (startPosition : WaypointId) : List CyclingGoal := findCyclingGoalsInPath_impl waypoints instructions possibleStarts [] [] startPosition 0

        -- returns the number of steps needed until the first commmon goal that cycles is found. private def findFirstCommonCyclingGoal (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (instructions : List Instruction) (possibleStarts : List WaypointId) (startPositions : List WaypointId) : Option Nat := let cyclingGoals := startPositions.map $ findCyclingGoalsInPath waypoints instructions possibleStarts let combinedGoals : List CyclingGoal := match cyclingGoals with | [] => [] | g :: gs => flip gs.foldl g Ξ»c n ↦ c.bind Ξ» cc ↦ n.filterMap Ξ» nn ↦ nn.combine cc let cyclingGoalStarts := combinedGoals.map CyclingGoal.start cyclingGoalStarts.minimum?

        open Lean in def part2 (input : List Instruction Γ— List Waypoint) : Option Nat := let possibleStarts := input.snd.map Waypoint.id let waypoints : HashMap WaypointId ConnectedWaypoints := HashMap.ofList $ input.snd.map Ξ»w ↦ (w.id, {left := w.left, right := w.right : ConnectedWaypoints}) let instructions := input.fst let positions : List WaypointId := (input.snd.filter Ξ»(w : Waypoint) ↦ w.id.endsWith "A").map Waypoint.id part2_impl waypoints instructions 0 (positions.map Ξ»_ ↦ possibleStarts) positions <|> -- if part2_impl fails (it does), we need to dig deeper. findFirstCommonCyclingGoal waypoints instructions possibleStarts positions

         
            
:::
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