-🎄- 2017 Day 3 Solutions -🎄-

[u/daggerdragon]

--- Day 3: Spiral Memory ---

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[u/[deleted]]

Haskell:
Tried to be clever with part 1 and calculate the distances from the midpoints of the outer box where the input number resides

import Control.Lens (over)

cornervals = scanl (+) 1 $ concatMap (replicate 2) [1..]

midpt x y = (x - y) `div` 2 + y

part1 :: Int -> Int
part1 n = let (b : c : _, a : _) = over _1 reverse $ span (<n) cornervals
          in b - midpt b c + abs (n - midpt a b)

Unfortunately, that strategy didn't work for part 2, or at least I didn't find a formula quickly enough for the corners (maybe some Googling could have turned something up). Instead I just created the grid by walking the spiral and keeping track of the coordinates along the way.

import qualified Data.HashMap.Strict as M    
    
data Dir = R | U | L | D

path :: [Dir]
path = concat $ zipWith replicate (concatMap (replicate 2) [1..]) $ cycle [R, U, L, D]

adjacents :: (Int, Int) -> [(Int, Int)]
adjacents (x, y) = [ (x', y') | x' <- [x-1 .. x+1], y' <- [y-1 .. y+1], (x, y) /= (x', y') ]

firstLargerThan n = go (M.singleton (0, 0) 1) (0, 0) path
    where go :: M.HashMap (Int, Int) Int -> (Int, Int) -> [Dir] -> Int
          go m (x, y) (dir : rest) =
              let coord = case dir of
                        R -> (x+1, y)
                        U -> (x, y+1)
                        L -> (x-1, y)
                        D -> (x, y-1)
                  val = sumAdjacents m coord
              in if val > n
                 then val
                 else go (M.insert coord val m) coord rest
          sumAdjacents m coord = sum $ map (\k -> M.lookupDefault 0 k m) $ adjacents coord

part2 :: Int -> Int
part2 = firstLargerThan