AoC day 20: Jurassic Jigsaw


2020-12-20T21:28:27+01:00
advent of code aoc2020 haskell

Bitmap raster operations are the name of the game for today’s problem. Please ignore my literate Haskell imports so we can jump right in.

{-# LANGUAGE NamedFieldPuns #-}
{-# LANGUAGE RecordWildCards #-}
{-# LANGUAGE ViewPatterns #-}
import           Control.Applicative (liftA2)
import           Control.Lens        (view,_1,_2,(+~))
import           Control.Monad       (guard)
import           Data.Bool           (bool)
import           Data.Char           (isDigit)
import           Data.Function       ((&),on)
import           Data.Ix             (Ix,range)
import           Data.List           (delete,tails,transpose,groupBy)
import           Data.List.Split     (linesBy)
import           Data.Map.Strict     (Map,(!))
import qualified Data.Map.Strict as Map
import           Data.Set            (Set)
import qualified Data.Set as Set     
import           Linear              (V2(V2),zero)

My input file isn’t too long—144 tiles. I ought to be fine just using lists as a data structure.

type Image = [[Bool]]

The monochrome images come with an ID number. To avoid mixing them up, I’ll pack them together.

data Tile = Tile
  { tileId    :: Int
  , tileImage :: Image
  }
  deriving Eq

Now to actually parse the input. I’ll use the same trick as usual to split paragraphs.

parseInput :: String -> [Tile]
parseInput = map parseTile . linesBy null . lines

To parse the Tile structure itself, I’ll try something experimental1 to try and answer the great question: can view patterns be nested?

parseTile :: [String] -> Tile
parseTile ((break (== ' ') ->
              ("Tile",' ' : (span isDigit -> (read -> tileId,":"))))
           : ((map . map) (== '#') -> tileImage))
  = Tile{..}

…it seems like they can. I wouldn’t go as far as calling that a good idea, though. Ye gods this is horrendeous! Until someone can show me a visually-pleasing way to format that, I’ll classify this as “never again”.

Ok, so our image was split up in tiles, and the postman tripped on the way so now they’re a mess on the floor. And since they’re square and transparent, there’s no way to know which way to place them so all of them align.

Let’s formalize the various ways it could be messed up.

data Setup = Setup
  { sTranspose :: Bool
  , sOuterFlip :: Bool
  , sInnerFlip :: Bool
  }
  deriving (Eq,Ord,Ix,Bounded)

In real life I’d call the enumeration “4 right-angle rotations for 2 sides”. But we’re doing computer processing on lists, I’ll adjust my terminology.

For some reason GHC can’t derive Enum on this structure, so I’ll define my “universe” enumerator using Ix instead.

allSetups :: [Setup]
allSetups = range (minBound,maxBound)

Now I can transform my tile to bring it to and from all of those orientations.

reorient :: Tile -> Setup -> Tile
reorient tile Setup{..} = tile { tileImage = morph (tileImage tile) }
  where morph = bool id transpose     sTranspose .
                bool id reverse       sOuterFlip .
                bool id (map reverse) sInnerFlip

I chose to keep it in the same type, as there’s no information loss. I need the ID number to remain with it if I want to track it back when I’m done shifting bits around.

Now let’s move towards reassembly. The tiles are going to be arranged in a grid, so integral-V2-based cartesian coordinates are fine.

type Vec = V2 Int
type Pos = Vec

On the other hand, trying all permutations of the tiles in a square patterns seems a bit wasteful. I’ll likely get drastically better performance starting with an arbitrary tile and accreting the others on the fly.

But I can’t know in advance where in the final square the starting tile will end. So I’ll use a Map instead of a bidimensional array.

type Assembly = Map Pos Tile

Today’s neighborhood function only recognizes 4 neighbors per tile. Please indulge me while I try to go creative with linear’s V2 and lenses. I’m attempting to keep up the practice, here.2

neighbors :: Pos -> Set Pos
neighbors = Set.fromList . sequence (liftA2 (+~) [_1,_2] [-1,1])

No, it’s not shorter,3 more efficient or in any way better than a simple list comprehension-based implementation. But I welcomed the fun.

I’d better check it actually works.

λ> neighbors (V2 44 55)
fromList [V2 43 55,V2 44 54,V2 44 56,V2 45 55]

Looks good :-)

Let’s write a helper to determine whether two tiles can be placed next to each other.

compatible :: Tile -> Vec -> Tile -> Bool
compatible t1 delta t2 = edge1 (tileImage t1) == edge2 (tileImage t2)
  where
    (edge1,edge2) = case delta of
        V2 (-1) 0 -> (top,bottom)
        V2   1  0 -> (bottom,top)
        V2 0 (-1) -> (left,right)
        V2 0   1  -> (right,left)
    top    = head
    bottom = last
    left   = map head
    right  = map last

So to reassemble the big square from the tiles, I’ll place the tiles one by one, next to each other, starting with an arbitrary one—the first—, always ensuring the borders of the newly placed one are compatible with the previously placed others. Backtracking through the list monad.

I expect there to be only one solution up to orientations, and the input size is small, so I’m not throwing any optimization in. I pick, in order:

  1. an empty spot with at least one neighboring spot filled
  2. a loose tile
  3. an orientation for it

If it’s compatible with its neighbors, I recurse, else I backtrack.

reassemble :: [Tile] -> Assembly
reassemble (startingTile : tiles0) =
    head $ go (Map.singleton zero startingTile) tiles0
  where
    go placed [] = pure placed
    go placed tiles = do
      pos <- Map.keysSet placed                  &
             Set.map neighbors                   &
             Set.unions                          &
             Set.filter (`Map.notMember` placed) &
             Set.toList
      let nPoss = neighbors pos                    &
                  Set.filter (`Map.member` placed) &
                  Set.toList

      tile <- tiles
      setup <- allSetups
      let pl = reorient tile setup

      guard $ all (\nPos -> compatible pl (nPos - pos) (placed ! nPos)) nPoss
      go (Map.insert pos pl placed) (delete tile tiles)

If all goes well, this returns a “floating” map of oriented tiles. The following helper extracts the floating square’s corner ID numbers, for the part 1 answer field.

cornerIds :: Assembly -> [Int]
cornerIds arr = tileId . (arr !) <$>
    [ V2 top left, V2 top right, V2 bottom left, V2 bottom right ]
  where
    ixs = Map.keys arr
    top =    minimum (view _1 <$> ixs)
    left =   minimum (view _2 <$> ixs)
    bottom = maximum (view _1 <$> ixs)
    right =  maximum (view _2 <$> ixs)

You may have noticed I’m using the generic _1 and _2 lenses instead of linear’s V2-specialized _x and _y, here and for the neighborhood function. It’s the clash of worlds: in a linear algebra world, I’d normally consider X to be the first coordinate, horizontal and going right, and Y the second, vertical going up. In a raster world4, I rather see the first coordinate going down and the second right. I chose to go with the raster vision all the way, and evade that part of the confusion by using the numbered accessors.

λ> let assembly = reassemble (parseInput sample)
λ> cornerIds assembly
[2971,1171,1951,3079]

All good and a gold star!

For part 2 we merge the oriented tiles. I’m somewhat arbitrarily repacking it in a Tile so I can reuse the orientation routines later.

merge :: Assembly -> Tile
merge =
  Tile undefined                       . -- image to tile
  foldr1 (++)                          . -- fuse vertically
  map (foldr1 (zipWith (++)))          . -- fuse horizontally
  (map . map) snd                      . -- drop coordinates
  groupBy ((==) `on` (view (_1 . _1))) . -- rasterize
  Map.assocs                           . -- map to list
  fmap tileTrimmedImage                  -- trim

The tile trimmer is similar in style: composition of simpler functions.

tileTrimmedImage :: Tile -> Image
tileTrimmedImage = trimRight . trimLeft . trimBottom . trimTop . tileImage
  where
    trimBottom = init
    trimTop    = tail
    trimRight  = map init
    trimLeft   = map tail

Now to locate sea monsters. Let’s make sure we know what we’re looking for.

seaMonster :: Image
seaMonster = (map . map) (== '#')
  [ "                  # "
  , "#    ##    ##    ###"
  , " #  #  #  #  #  #   "
  ]

To count them, I’ll match by following both lists of lists in lockstep. There are two pitfalls here.

  1. I’m working on lists of lists of Bools. It’s all too natural to reach for zip-like functions, e.g. zipWith (==) or similar. But those constructs stop on the shortest sequence of those provided. When matching, it has a very different meaning to exhaust the pattern (success) or the input string (failure). I’ll alleviate with a dedicated helper.
  2. The comparison semantics are asymmetrical.
match :: Image -> Image -> Bool
match ref input = allMatch row ref input
  where
    allMatch p (a:as) (b:bs) = p a b && allMatch p as bs
    allMatch _   []     _    = True  -- out of pattern, success
    allMatch _   _      _    = False -- out of input, failure

    row = allMatch pixel

    pixel False = const True -- blank pattern, always match
    pixel True  = id         -- hash pattern, match on hash

This only matches if the pattern is found at (0, 0). Here’s a helper to attempt all shifts of the image .

lowerRights :: [[a]] -> [[[a]]]
lowerRights = concatMap tails . transpose . map tails

And I can now complete the monster counter.

countMonsters :: Tile -> Int
countMonsters = length . filter (match seaMonster). lowerRights . tileImage

A final helper before attempting part 2:

countHashes :: Image -> Int
countHashes = length . filter id . concat
λ> let image = merge assembly
λ> maximum $ map (countMonsters . reorient image) allSetups
2
λ> countHashes (tileImage image) -2*countHashes seaMonster
273

Am I playing my luck a bit here? This computation works for the sample, but theoretically fails if a hash can be a part of multiple sea monsters.

Does it happen in my puzzle input? There’s only one way to find out…5

In the meantime, here’s the rest of the code for completeness.

main :: IO ()
main = do
  tiles <- parseInput <$> readFile "day20.in"
  let assembly = reassemble tiles
  putStrLn $ "Assembly product: " ++ show (product (cornerIds assembly))
  let image = merge assembly
      monsters = maximum $ map (countMonsters . reorient image) allSetups
      roughness = countHashes (tileImage image)
        - monsters * countHashes seaMonster
  putStrLn $ "Water roughness: " ++ show roughness

Well, that turned out to be enough for the second gold star of the day. I’m not sure I can disprove sea monsters overlapping each other by reading the statement again, so I figure it’s just Eric being nice to us.

Or maybe only to me? O:-)

This concludes this day’s solution. Hope you enjoyed it; see you soon!


  1. Experimental for me. I expect them to be fully specified.↩︎

  2. I got back to lenses for day 14, but that part of the journey isn’t published yet. I’m on it!↩︎

  3. Well, ok, it is shorter, but the point would stand even if it weren’t.↩︎

  4. Matrix addressing, ironically, fits the raster vision.↩︎

  5. Of course there isn’t. I could go the rigorous route and mark the relevant pixels, perform their set union then difference with the rest of the hashes. But that would most likely take me more than a single minute. When the site grants us an attempt per minute, I’d be stupid not to try my luck before implementing the complex stuff.↩︎