Day 11: Plutonian Pebbles

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FAQ

  • lwhjp
    3•2 days ago

    Haskell

    Yay, mutation! Went down the route of caching the expanded lists of stones at first. Oops.

    import Data.IORef
import Data.Map.Strict (Map)
import Data.Map.Strict qualified as Map

blink :: Int -> [Int]
blink 0 = [1]
blink n
  | s <- show n,
    l <- length s,
    even l =
      let (a, b) = splitAt (l `div` 2) s in map read [a, b]
  | otherwise = [n * 2024]

countExpanded :: IORef (Map (Int, Int) Int) -> Int -> [Int] -> IO Int
countExpanded _ 0 = return . length
countExpanded cacheRef steps = fmap sum . mapM go
  where
    go n =
      let key = (n, steps)
          computed = do
            result <- countExpanded cacheRef (steps - 1) $ blink n
            modifyIORef' cacheRef (Map.insert key result)
            return result
       in readIORef cacheRef >>= maybe computed return . (Map.!? key)

main = do
  input <- map read . words <$> readFile "input11"
  cache <- newIORef Map.empty
  mapM_ (\steps -> countExpanded cache steps input >>= print) [25, 75]
    • @VegOwOtenks@lemmy.world
      2•1 day ago

      Does the IORef go upwards the recursion tree? If you modify the IORef at some depth of 15, does the calling function also receive the update, is there also a Non-IO-Ref?

      • lwhjp
        2•1 day ago

        The IORef is like a mutable box you can stick things in, so readIORef returns whatever was last put in it (in this case using modifyIORef'). “last” makes sense here because operations are sequenced thanks to the IO monad, so yes: values get carried back up the tree to the caller. There’s also STRef for the ST monad, or I could have used the State monad which (kind of) encapsulates a single ref.