Day 17: Clumsy Crucible
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FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
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)
Rust: https://codeberg.org/Sekoia/adventofcode/src/branch/main/src/y2023/day17.rs
WOW that took me forever. Completely forgot how to dijkstra’s and then struggled with the actual puzzle too. 1h20 total, but I was still top 1500, which I guess means most people really struggled with this, huh.
Yeah, finding a good way to represent the “last three moves” constraint was a really interesting twist. You beat me to it, anyway!
Python
749 line-seconds
import collections import dataclasses import heapq import numpy as np from .solver import Solver @dataclasses.dataclass(order=True) class QueueEntry: price: int x: int y: int momentum_x: int momentum_y: int deleted: bool class Day17(Solver): lines: list[str] sx: int sy: int lower_bounds: np.ndarray def __init__(self): super().__init__(17) def presolve(self, input: str): self.lines = input.splitlines() self.sx = len(self.lines[0]) self.sy = len(self.lines) start = (self.sx - 1, self.sy - 1) self.lower_bounds = np.zeros((self.sx, self.sy)) + np.inf self.lower_bounds[start] = 0 queue: list[QueueEntry] = [QueueEntry(0, self.sx - 1, self.sy - 1, 0, 0, False)] queue_entries: dict[tuple[int, int], QueueEntry] = {start: queue[0]} while queue: cur_price, x, y, _, _, deleted = dataclasses.astuple(heapq.heappop(queue)) if deleted: continue del queue_entries[(x, y)] self.lower_bounds[x, y] = cur_price price = cur_price + int(self.lines[y][x]) for dx, dy in ((-1, 0), (1, 0), (0, -1), (0, 1)): nx, ny = x + dx, y + dy if not (0 <= nx < self.sx) or not (0 <= ny < self.sy): continue if price < self.lower_bounds[nx, ny]: self.lower_bounds[nx, ny] = price if (nx, ny) in queue_entries: queue_entries[(nx, ny)].deleted = True queue_entries[(nx, ny)] = QueueEntry(price, nx, ny, 0, 0, False) heapq.heappush(queue, queue_entries[(nx, ny)]) def _solve(self, maximum_run: int, minimum_run_to_turn: int): came_from: dict[tuple[int, int, int, int], tuple[int, int, int, int]] = {} start = (0, 0, 0, 0) queue: list[QueueEntry] = [QueueEntry(self.lower_bounds[0, 0], *start, False)] queue_entries: dict[tuple[int, int, int, int], QueueEntry] = {start: queue[0]} route: list[tuple[int, int]] = [] prices: dict[tuple[int, int, int, int], float] = collections.defaultdict(lambda: np.inf) prices[start] = 0 while queue: _, current_x, current_y, momentum_x, momentum_y, deleted = dataclasses.astuple(heapq.heappop(queue)) cur_price = prices[(current_x, current_y, momentum_x, momentum_y)] if deleted: continue if ((current_x, current_y) == (self.sx - 1, self.sy - 1) and (momentum_x >= minimum_run_to_turn or momentum_y >= minimum_run_to_turn)): previous = came_from.get((current_x, current_y, momentum_x, momentum_y)) route.append((current_x, current_y)) while previous: x, y, *_ = previous if x != 0 or y != 0: route.append((x, y)) previous = came_from.get(previous) break for dx, dy in ((-1, 0), (1, 0), (0, -1), (0, 1)): dot_product = dx * momentum_x + dy * momentum_y if dot_product < 0 or dot_product >= maximum_run: continue if ((momentum_x or momentum_y) and dot_product == 0 and abs(momentum_x) < minimum_run_to_turn and abs(momentum_y) < minimum_run_to_turn): continue new_x, new_y = current_x + dx, current_y + dy if not (0 <= new_x < self.sx) or not (0 <= new_y < self.sy): continue new_momentum_x, new_momentum_y = (dx, dy) if dot_product == 0 else (momentum_x + dx, momentum_y + dy) new_position = (new_x, new_y, new_momentum_x, new_momentum_y) potential_new_price = cur_price + int(self.lines[new_y][new_x]) if potential_new_price < prices[new_position]: queue_entry = queue_entries.get(new_position) if queue_entry: queue_entry.deleted = True queue_entries[new_position] = QueueEntry(potential_new_price + self.lower_bounds[new_x, new_y], *new_position, False) came_from[new_position] = (current_x, current_y, momentum_x, momentum_y) prices[new_position] = potential_new_price heapq.heappush(queue, queue_entries[new_position]) return sum(int(self.lines[y][x]) for x, y in route) def solve_first_star(self) -> int: return self._solve(3, 0) def solve_second_star(self) -> int: return self._solve(10, 4)
C
Very not pretty and not efficient. Using what I think is dynamic programming - essentially I just propagate cost total through a (x, y, heading, steps in heading) state space, so every cell in that N-dimensional array holds the minimum total cost known to get to that state which is updated iteratively from the neighbours until it settles down.
Debugging was annoying because terminals aren’t great at showing 4D grids. My mistakes were in the initial situation and missing the “4 steps to come to a stop at the end” aspect in part 2.
Haskell
Wowee, I took some wrong turns solving today’s puzzle! After fixing some really inefficient pruning I ended up with a Dijkstra search that runs in 2.971s (for a less-than-impressive 124.782 l-s).
Solution
import Control.Monad import Data.Array.Unboxed (UArray) import qualified Data.Array.Unboxed as Array import Data.Char import qualified Data.HashSet as Set import qualified Data.PQueue.Prio.Min as PQ readInput :: String -> UArray (Int, Int) Int readInput s = let rows = lines s in Array.amap digitToInt . Array.listArray ((1, 1), (length rows, length $ head rows)) $ concat rows walk :: (Int, Int) -> UArray (Int, Int) Int -> Int walk (minStraight, maxStraight) grid = go Set.empty initPaths where initPaths = PQ.fromList [(0, ((1, 1), (d, 0))) | d <- [(0, 1), (1, 0)]] goal = snd $ Array.bounds grid go done paths = case PQ.minViewWithKey paths of Nothing -> error "no route" Just ((n, (p@(y, x), hist@((dy, dx), k))), rest) | p == goal && k >= minStraight -> n | (p, hist) `Set.member` done -> go done rest | otherwise -> let next = do h'@((dy', dx'), _) <- join [ guard (k >= minStraight) >> [((dx, dy), 1), ((-dx, -dy), 1)], guard (k < maxStraight) >> [((dy, dx), k + 1)] ] let p' = (y + dy', x + dx') guard $ Array.inRange (Array.bounds grid) p' return (n + grid Array.! p', (p', h')) in go (Set.insert (p, hist) done) $ (PQ.union rest . PQ.fromList) next main = do input <- readInput <$> readFile "input17" print $ walk (0, 3) input print $ walk (4, 10) input
(edited for readability)
Haskell
import Data.Array.Unboxed import qualified Data.ByteString.Char8 as BS import Data.Char (digitToInt) import Data.Heap hiding (filter) import qualified Data.Heap as H import Relude type Pos = (Int, Int) type Grid = UArray Pos Int data Dir = U | D | L | R deriving (Eq, Ord, Show, Enum, Bounded, Ix) parse :: ByteString -> Maybe Grid parse input = do let l = fmap (fmap digitToInt . BS.unpack) . BS.lines $ input h = length l w <- fmap length . viaNonEmpty head $ l pure . listArray ((0, 0), (w - 1, h - 1)) . concat $ l move :: Dir -> Pos -> Pos move U = first pred move D = first succ move L = second pred move R = second succ nextDir :: Dir -> [Dir] nextDir U = [L, R] nextDir D = [L, R] nextDir L = [U, D] nextDir R = [U, D] -- position, previous direction, accumulated loss type S = (Int, Pos, Dir) doMove :: Grid -> Dir -> S -> Maybe S doMove g d (c, p, _) = do let p' = move d p guard $ inRange (bounds g) p' pure (c + g ! p', p', d) doMoveN :: Grid -> Dir -> Int -> S -> Maybe S doMoveN g d n = foldl' (>=>) pure . replicate n $ doMove g d doMoves :: Grid -> [Int] -> S -> Dir -> [S] doMoves g r s d = mapMaybe (flip (doMoveN g d) s) r allMoves :: Grid -> [Int] -> S -> [S] allMoves g r s@(_, _, prev) = nextDir prev >>= doMoves g r s solve' :: Grid -> [Int] -> UArray (Pos, Dir) Int -> Pos -> MinHeap S -> Maybe Int solve' g r distances target h = do ((acc, pos, dir), h') <- H.view h if pos == target then pure acc else do let moves = allMoves g r (acc, pos, dir) moves' = filter (\(acc, p, d) -> acc < distances ! (p, d)) moves distances' = distances // fmap (\(acc, p, d) -> ((p, d), acc)) moves' h'' = foldl' (flip H.insert) h' moves' solve' g r distances' target h'' solve :: Grid -> [Int] -> Maybe Int solve g r = solve' g r (emptyGrid ((lo, minBound), (hi, maxBound))) hi (H.singleton (0, (0, 0), U)) where (lo, hi) = bounds g emptyGrid = flip listArray (repeat maxBound) part1, part2 :: Grid -> Maybe Int part1 = (`solve` [1 .. 3]) part2 = (`solve` [4 .. 10])