Day 12: Hot Springs
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Haskell
Phew! I struggled with this one. A lot of the code here is from my original approach, which cuts down the search space to plausible positions for each group. Unfortunately, that was still way too slow…
It took an embarrassingly long time to try memoizing the search (which made precomputing valid points far less important). Anyway, here it is!
Solution
{-# LANGUAGE LambdaCase #-} import Control.Monad import Control.Monad.State import Data.List import Data.List.Split import Data.Map (Map) import qualified Data.Map as Map import Data.Maybe readInput :: String -> ([Maybe Bool], [Int]) readInput s = let [a, b] = words s in ( map (\case '#' -> Just True; '.' -> Just False; '?' -> Nothing) a, map read $ splitOn "," b ) arrangements :: ([Maybe Bool], [Int]) -> Int arrangements (pat, gs) = evalState (searchMemo 0 groups) Map.empty where len = length pat groups = zipWith startPoints gs $ zip minStarts maxStarts where minStarts = scanl (\a g -> a + g + 1) 0 $ init gs maxStarts = map (len -) $ scanr1 (\g a -> a + g + 1) gs startPoints g (a, b) = let ps = do (i, pat') <- zip [a .. b] $ tails $ drop a pat guard $ all (\(p, x) -> maybe True (== x) p) $ zip pat' $ replicate g True ++ [False] return i in (g, ps) clearableFrom i = fmap snd $ listToMaybe $ takeWhile ((<= i) . fst) $ dropWhile ((< i) . snd) clearableRegions where clearableRegions = let go i [] = [] go i pat = let (a, a') = span (/= Just True) pat (b, c) = span (== Just True) a' in (i, i + length a - 1) : go (i + length a + length b) c in go 0 pat searchMemo :: Int -> [(Int, [Int])] -> State (Map (Int, Int) Int) Int searchMemo i gs = do let k = (i, length gs) cached <- gets (Map.!? k) case cached of Just x -> return x Nothing -> do x <- search i gs modify (Map.insert k x) return x search i gs | i >= len = return $ if null gs then 1 else 0 search i [] = return $ case clearableFrom i of Just b | b == len - 1 -> 1 _ -> 0 search i ((g, ps) : gs) = do let maxP = maybe i (1 +) $ clearableFrom i ps' = takeWhile (<= maxP) $ dropWhile (< i) ps sum <$> mapM (\p -> let i' = p + g + 1 in searchMemo i' gs) ps' expand (pat, gs) = (intercalate [Nothing] $ replicate 5 pat, concat $ replicate 5 gs) main = do input <- map readInput . lines <$> readFile "input12" print $ sum $ map arrangements input print $ sum $ map (arrangements . expand) input
Rust
Took me way too long, but I’m happy with my solution now. I spent probably half an hour looking at my naive backtracking program churning away unsuccessfully before I thought of dynamic programming, meaning caching all intermediate results in a hashtable under their current state. The state is just the index into the spring array and the index into the range array, meaning there really can’t be too many different entries. Doing so worked very well, solving part 2 in 4ms.
Adding the caching required me to switch from a loop to a recursive function, which turned out way easier. Why did no one tell me to just go recursive from the start?
Nim
I don’t want to talk about it. -_-
And finally:
Haskell
Abused
ParserCombinators
for the first part. For the second, I took quite a while to figure out dynamic programming in Haskell.Solution
module Day12 where import Data.Array import Data.Char (isDigit) import Data.List ((!!)) import Relude hiding (get, many) import Relude.Unsafe (read) import Text.ParserCombinators.ReadP type Spring = (String, [Int]) type Problem = [Spring] parseStatus :: ReadP Char parseStatus = choice $ char <$> ".#?" parseSpring :: ReadP Spring parseSpring = do status <- many1 parseStatus <* char ' ' listFailed <- (read <$> munch1 isDigit) `sepBy` char ',' return (status, listFailed) parseProblem :: ReadP Problem parseProblem = parseSpring `sepBy` char '\n' parse :: ByteString -> Maybe Problem parse = fmap fst . viaNonEmpty last . readP_to_S parseProblem . decodeUtf8 good :: ReadP () good = choice [char '.', char '?'] $> () bad :: ReadP () bad = choice [char '#', char '?'] $> () buildParser :: [Int] -> ReadP () buildParser l = do _ <- many good sequenceA_ $ intersperse (many1 good) [count x bad | x <- l] _ <- many good <* eof return () combinations :: Spring -> Int combinations (s, l) = length $ readP_to_S (buildParser l) s part1, part2 :: Problem -> Int part1 = sum . fmap combinations part2 = sum . fmap (combinations' . toSpring' . bimap (join . intersperse "?" . replicate 5) (join . replicate 5)) run1, run2 :: FilePath -> IO Int run1 f = readFileBS f >>= maybe (fail "parse error") (return . part1) . parse run2 f = readFileBS f >>= maybe (fail "parse error") (return . part2) . parse data Status = Good | Bad | Unknown deriving (Eq, Show) type Spring' = ([Status], [Int]) type Problem' = [Spring'] toSpring' :: Spring -> Spring' toSpring' (s, l) = (fmap toStatus s, l) where toStatus :: Char -> Status toStatus '.' = Good toStatus '#' = Bad toStatus '?' = Unknown toStatus _ = error "impossible" isGood, isBad :: Status -> Bool isGood Bad = False isGood _ = True isBad Good = False isBad _ = True combinations' :: Spring' -> Int combinations' (s, l) = t ! (0, 0) where n = length s m = length l t = listArray ((0, 0), (n, m)) [f i j | i <- [0 .. n], j <- [0 .. m]] f :: Int -> Int -> Int f n' m' | n' >= n = if m' >= m then 1 else 0 | v == Unknown = tGood + tBad | v == Good = tGood | v == Bad = tBad | otherwise = error "impossible" where v = s !! n' x = l !! m' ss = drop n' s (bads, rest) = splitAt x ss badsDelimited = maybe True isGood (viaNonEmpty head rest) off = if null rest then 0 else 1 tGood = t ! (n' + 1, m') tBad = if m' + 1 <= m && length bads == x && all isBad bads && badsDelimited then t ! (n' + x + off, m' + 1) else 0
Scala3
def countDyn(a: List[Char], b: List[Int]): Long = // Simple dynamic programming approach // We fill a table T, where // T[ ai, bi ] -> number of ways to place b[bi..] in a[ai..] // T[ ai, bi ] = 0 if an-ai >= b[bi..].sum + bn-bi // T[ ai, bi ] = 1 if bi == b.size - 1 && ai == a.size - b[bi] - 1 // T[ ai, bi ] = // (place) T [ ai + b[bi], bi + 1] if ? or # // (skip) T [ ai + 1, bi ] if ? or . // def t(ai: Int, bi: Int, tbl: Map[(Int, Int), Long]): Long = if ai >= a.size then if bi >= b.size then 1L else 0L else val place = Option.when( bi < b.size && // need to have piece left ai + b(bi) <= a.size && // piece needs to fit a.slice(ai, ai + b(bi)).forall(_ != '.') && // must be able to put piece there (ai + b(bi) == a.size || a(ai + b(bi)) != '#') // piece needs to actually end )((ai + b(bi) + 1, bi + 1)).flatMap(tbl.get).getOrElse(0L) val skip = Option.when(a(ai) != '#')((ai + 1, bi)).flatMap(tbl.get).getOrElse(0L) place + skip @tailrec def go(ai: Int, tbl: Map[(Int, Int), Long]): Long = if ai == 0 then t(ai, 0, tbl) else go(ai - 1, tbl ++ b.indices.inclusive.map(bi => (ai, bi) -> t(ai, bi, tbl)).toMap) go(a.indices.inclusive.last + 1, Map()) def countLinePossibilities(repeat: Int)(a: String): Long = a match case s"$pattern $counts" => val p2 = List.fill(repeat)(pattern).mkString("?") val c2 = List.fill(repeat)(counts).mkString(",") countDyn(p2.toList, c2.split(",").map(_.toInt).toList) case _ => 0L def task1(a: List[String]): Long = a.map(countLinePossibilities(1)).sum def task2(a: List[String]): Long = a.map(countLinePossibilities(5)).sum
(Edit: fixed mangling of &<)
I’m struggling to fully understand your solution. Could you tell me, why do you return
1
when at the end ofa
andb
? And why do you start fromsize + 1
?T counts the number of ways to place the blocks with lengths specified in b in the remaining a.size - ai slots. If there are no more slots left, there are two cases: Either there are also no more blocks left, then everything is fine, and the current situation is 1 way to place the blocks in the slots. Otherwise, there are still blocks left, and no more space to place them in. This means the current sitution is incorrect, so we contribute 0 ways to place the blocks. This is what the
if bi >= b.size then 1L else 0L
does.The start at
size + 1
is necessary, as we need to compute every table entry before it may get looked up. When placing the last block, we may check the entry(ai + b(bi) + 1, bi + 1)
, whereai + b(bi)
may already equala.size
(in the case where the block ends exactly at the end ofa
). The+ 1
in the entry is necessary, as we need to skip a slot after every block: If we looked at(ai + b(bi), bi + 1)
, we could start ata.size
, but then, for e.g.b = [2, 3]
, we would consider...#####.
a valid placement.Let me know if there are still things unclear :)
Thanks for the detailed explanation. It helped a lot, especially what the
tbl
actually holds.I’ve read your code again and I get how it works, but it still feels kinda strange that we are considering values outside of range of
a
andb
, and that we are marking them as correct. Like in first row of the example???.### 1,1,3
, there is no spring at8
and no group at3
but we are marking(8,3)
and(7,3)
as correct. In my mind, first position that should be marked as correct is4,2
, because that’s where group of 3 can fit.If you make the recurrent case a little more complicated, you can sidestep the weird base cases, but I like reducing the endpoints down to things like this that are easily implementable, even if they sound a little weird at first.
You are probably right. Just my rumblings. Thanks for the help.
Python
Let me know if you have any questions or feedback!
import dataclasses import functools from .solver import Solver class MatchState: pass @dataclasses.dataclass class NotMatching(MatchState): pass @dataclasses.dataclass class Matching(MatchState): current_length: int desired_length: int @functools.cache def _match_one_template(template: str, groups: tuple[int, ...]) -> int: if not groups: if '#' in template: return 0 else: return 1 state: MatchState = NotMatching() remaining_groups: list[int] = list(groups) options_in_other_branches: int = 0 for i in range(len(template)): match (state, template[i]): case (NotMatching(), '.'): pass case (NotMatching(), '?'): options_in_other_branches += _match_one_template(template[i+1:], tuple(remaining_groups)) if not remaining_groups: return options_in_other_branches group, *remaining_groups = remaining_groups state = Matching(1, group) case (NotMatching(), '#'): if not remaining_groups: return options_in_other_branches group, *remaining_groups = remaining_groups state = Matching(1, group) case (Matching(current_length, desired_length), '.') if current_length == desired_length: state = NotMatching() case (Matching(current_length, desired_length), '.') if current_length < desired_length: return options_in_other_branches case (Matching(current_length, desired_length), '?') if current_length == desired_length: state = NotMatching() case (Matching(current_length, desired_length), '?') if current_length < desired_length: state = Matching(current_length + 1, desired_length) case (Matching(current_length, desired_length), '#') if current_length < desired_length: state = Matching(current_length + 1, desired_length) case (Matching(current_length, desired_length), '#') if current_length == desired_length: return options_in_other_branches case _: raise RuntimeError(f'unexpected {state=} with {template=} position {i} and {remaining_groups=}') match state, remaining_groups: case NotMatching(), []: return options_in_other_branches + 1 case Matching(current, desired), [] if current == desired: return options_in_other_branches + 1 case (NotMatching(), _) | (Matching(_, _), _): return options_in_other_branches raise RuntimeError(f'unexpected {state=} with {template=} at end of template and {remaining_groups=}') def _unfold(template: str, groups: tuple[int, ...]) -> tuple[str, tuple[int, ...]]: return '?'.join([template] * 5), groups * 5 class Day12(Solver): def __init__(self): super().__init__(12) self.input: list[tuple[str, tuple[int]]] = [] def presolve(self, input: str): lines = input.rstrip().split('\n') for line in lines: template, groups = line.split(' ') self.input.append((template, tuple(int(group) for group in groups.split(',')))) def solve_first_star(self) -> int: return sum(_match_one_template(template, groups) for template, groups in self.input) def solve_second_star(self) -> int: return sum(_match_one_template(*_unfold(template, groups)) for template, groups in self.input)
C
That was something! I quickly settled on the main approach for part 1 but it took some unit testing to get it all right. Then part 2 had me stumped for a bit. It was clear some kind of pruning was necessary, possibly with memoization.
Hashmaps are possible but annoying with C so I was happy to realise that, for my implementation,
(num chars, num runs)
is a suitable key within the context of a single recursive search. That space is small enough to index with an array 😁Dart
Terrible monkey-coding approach of banging strings together and counting the resulting shards. Just kept to a reasonable 300ms runtime by a bit of memoisation of results. I’m sure this can all be replaced by a single line of clever combinatorial wizardry.
var __countMatches = {}; int _countMatches(String pattern, List counts) => __countMatches.putIfAbsent( pattern + counts.toString(), () => countMatches(pattern, counts)); int countMatches(String pattern, List counts) { if (!pattern.contains('#') && counts.isEmpty) return 1; if (pattern.startsWith('..')) return _countMatches(pattern.skip(1), counts); if (pattern == '.' || counts.isEmpty) return 0; var thisShape = counts.first; var minSpaceForRest = counts.length == 1 ? 0 : counts.skip(1).sum + counts.skip(1).length + 1; var lastStart = pattern.length - minSpaceForRest - thisShape; if (lastStart < 1) return 0; var total = 0; for (var start in 1.to(lastStart + 1)) { // Skipped a required spring. Bad, and will be for all successors. if (pattern.take(start).contains('#')) break; // Includes a required separator. Also bad. if (pattern.skip(start).take(thisShape).contains('.')) continue; var rest = pattern.skip(start + thisShape); if (rest.startsWith('#')) continue; // force '.' or '?' to be '.' and count matches. total += _countMatches('.${rest.skip(1)}', counts.skip(1).toList()); } return total; } solve(List lines, {stretch = 1}) { var ret = []; for (var line in lines) { var ps = line.split(' '); var pattern = List.filled(stretch, ps.first).join('?'); var counts = List.filled(stretch, ps.last) .join(',') .split(',') .map(int.parse) .toList(); ret.add(countMatches('.$pattern.', counts)); // top and tail. } return ret.sum; } part1(List lines) => solve(lines); part2(List lines) => solve(lines, stretch: 5);