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Ecography

@ecography.bsky.social

A test of the abundant-center hypothesis for stream fishes vist.ly/3zde8 #StreamFish #DenistyEstimation #RangeSize

July 28, 2025 at 9:25 AM

oliver

@eikopf.com

very happy with my solution for today

$data Db = Db [(Int, Int)] [Int] deriving (Show) parse :: String -> Db parse = uncurry Db . bimap (map readRange) (map read . drop 1) . break null . lines where readRange :: String -> (Int, Int) readRange = bimap read (read . drop 1) . break (== '-') solvePart1 :: Db -> Int solvePart1 (Db rs is) = length $ filter (flip any rs . flip inRange) is solvePart2 :: Db -> Int solvePart2 (Db rs _) = sum . map rangeSize $ fix (\rec rs -> let rs' = foldl' upsertRange [] rs in if rs == rs' then rs else rec rs') rs where upsertRange :: [(Int, Int)] -> (Int, Int) -> [(Int, Int)] upsertRange [] r' = [r'] upsertRange (r : rs) r' = case unionMaybe r r' of Nothing -> r : upsertRange rs r' Just r'' -> r'' : rs unionMaybe :: (Int, Int) -> (Int, Int) -> Maybe (Int, Int) unionMaybe lhs@(l1, l2) rhs@(r1, r2) = if inRange lhs r1 || inRange rhs l1 then Just (min l1 r1, max l2 r2) else Nothing main :: IO () main = interact $ \s -> let input = parse s part1 = solvePart1 input part2 = solvePart2 input in "part 1: " ++ show part1 ++ "\npart 2: " ++ show part2 ++ "\n"$

December 5, 2025 at 10:35 AM