Advent of Code Problem 19b - Nondeterministic parsing with Yoctoparsec

Last year I took some time to solve the 2020 Advent of Code problems. Most of the problems were easy - the second part of each problem usually being slightly harder. I needed a couple of hints to get me through a couple of the problems when I felt I was missing a piece of the puzzle but I only did that twice - for problems 13b and 22b. However, for problem 19b I refused to cheat and pushed through until I got to the solution.

To recap - the challenge of Problem 19 is:

  • Read input formatted into a rules section and a set of candidate lines.
  • Output the number of candidate lines that match the rules.

An example input could be something like:

0: 1 | 3
1: 'a'
2: 'b'
3: 2 0 | 3 0


Therefore it’s your job to construct a matcher out of rule-0 and test it against each line, counting the ones that match.

I followed a similar approach to other dynamic-programming problems and constructed a data-recursive map of parsers for each rule:

    = show . length .  rights
    . (\(s,p) -> map (parse p "rules") s)
    . (last &&& build . map ((init . head &&& splitOn ["|"] . tail) . words) .  head)
    . splitOn [""] . lines

    build :: [(String,[[String]])] -> Parsec String () ()
    build r = a ! "0" *> eof
      a = M.fromList (map g r)
      f n@(x:xs)
        | isDigit x = a ! n
        | otherwise = () <$ string (init xs)
      g = second (choice . map (try . mapM_ f))
      m ! k = fromJust $ M.lookup k m

This was pretty trivial, however, part-2 allowed the rules to be recursive and my approach no longer worked. The reason why this confounded me for so long is that I really liked my solution for part-1 and saw no reason why I shouldn’t be able to use it for part-2.

It was actually an illuminating exercise to figure out why exactly a parsec-family solution wouldn’t be able to achieve what I wanted. It is a combination of the fact that EOF has to be encoded independently of the rules, and that backtracking can only occur on failure.

In particular:

choice . map (try . mapM_ f))

will recurse on referenced rules and backtrack if they fail, but that isn’t sufficient. They also need to backtrack if they succeed but the overall parser fails due to not consuming the entire line.

This took me on a search for a non-deterministic parser library and after asking around and googling furiously I found Yoctoparsec. A truly tiny (thus the pun in the name) variant of the Parsec family described as:

A monadic parsing library making use of the free monad transformer. All instances are provided by the FreeT monad.

(Pointed out on Reddit - Text.ParserCombinators.ReadP from base also provides a non-deterministic parser)

I need to stress just how tiny this library is and to do that, here is the entire source-code (sans comments):

module Control.Monad.Yoctoparsec where

import Data.List
import Control.Applicative
import Control.Monad
import Control.Monad.State
import Control.Monad.Trans.Free

type Parser b t a = FreeT ((->) t) b a

token :: Applicative b => Parser b t t
token = FreeT . pure . Free $ FreeT . pure . Pure

parseStream :: Monad b => (s -> b (t, s)) -> Parser b t a -> s -> b (a, s)
parseStream next = runStateT . iterTM (StateT next >>=)

parseString :: MonadPlus b => Parser b t a -> [t] -> b (a, [t])
parseString = parseStream (maybe empty pure . uncons)

This is honestly incredible. Not only that it allows for fully non-deterministic parsing but also that it facilitates other parsing philosophies too, including different failure and backtracking strategies simply by selecting the underlying monad.

This now enabled the exact solution I was looking for:

    = (\x -> map drawTree x ++ [show $ length x])
    . map (fst . head)
    . filter (not . null)
    . (\(s,p) -> map (parseString p) s)
    . (map (++"EOF") . last &&& build . map (sub . (init . head &&& splitOn ["|"] . tail) . words) .  head)
    . splitOn [""] . lines


    char c = mfilter (==c) token
    string = mapM char
    choice = foldl (<|>) empty

    sub t@("8", _) = t & _2 .~ [["42"],["42","8"]]
    sub t@("11",_) = t & _2 .~ [["42","31"],["42","11","31"]]
    sub t          = t

    m ! k = fromJust $ M.lookup k m

    build :: [(String,[[String]])] -> FreeT ((->) Char) [] (Tree String)
    build r = fmap (Node "0") $ a ! "0" <* string "EOF"
        a = M.fromList (map (second g) r)
        g = choice . map (traverse f)
        f n@(x:xs)
            | isDigit x  = Node n <$> a ! n
            | otherwise  = Node n [] <$ string (init xs)

A truly elegant and correct solution that exactly matches the approach from part-1 but extends the power via full non-determinism to allow for backtracking whenever the parser fails but other alternative paths still exist.

I was and am very happy with this and it allowed me to move on with the other puzzles. There are drawbacks with this approach, mainly performance and error reporting, however in this scenario and others like it, it’s very hard to beat the readability.

I think I’ll use Yoctoparsec again for sketching purposes where speed and error reporting aren’t as important as getting a parser up and running quickly. To help with that there are some libraries that export abstract parser combinators such as parser-combinators. This greatly extends the functionality of Yoctoparsec right out of the box.

You can see my solutions here on Github.