Day 11: Plutonian Pebbles
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FAQ
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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]
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?
The IORef is like a mutable box you can stick things in, so
readIORef
returns whatever was last put in it (in this case usingmodifyIORef'
). “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 alsoSTRef
for the ST monad, or I could have used the State monad which (kind of) encapsulates a single ref.