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author | Julien Dessaux | 2024-12-23 20:59:26 +0100 |
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committer | Julien Dessaux | 2024-12-23 20:59:26 +0100 |
commit | b900a37be0154db733981fc6e1409fbb94c44f99 (patch) | |
tree | 7dde9f9fdabe221ef0ea5a6ba911427a89daca68 /content/blog/haskell | |
parent | fixed the yaml and tofu mangling in the /dev/shm on kubernetes blog article (diff) | |
download | www-b900a37be0154db733981fc6e1409fbb94c44f99.tar.gz www-b900a37be0154db733981fc6e1409fbb94c44f99.tar.bz2 www-b900a37be0154db733981fc6e1409fbb94c44f99.zip |
add advent of code 2023 in haskell blog article
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-rw-r--r-- | content/blog/haskell/advent-of-code-2023-in-haskell.md | 242 |
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diff --git a/content/blog/haskell/advent-of-code-2023-in-haskell.md b/content/blog/haskell/advent-of-code-2023-in-haskell.md new file mode 100644 index 0000000..cd8340a --- /dev/null +++ b/content/blog/haskell/advent-of-code-2023-in-haskell.md @@ -0,0 +1,242 @@ +--- +title: Advent of code 2023 in haskell +description: I improved in haskell this year and still love parsing +date: 2024-11-22 +tags: +- haskell +--- + +## Introduction + +I did the [advent of code 2023](https://adventofcode.com/2023) in haskell, it was a fun experience as always! Why writing about this now? Because I just finished the last puzzle as a warm up for the upcoming year's puzzles! + +I did the first 11 puzzles on time last December but the "one puzzle a day" schedule is a bit much when life happens around you. I therefore took a break and did a few more puzzles in mid January. Upon reaching [the 17th puzzle](https://adventofcode.com/2023/day/17) (the shortest paths with weird constraints puzzle) I took another break until June were I pushed through until [Day 24th](https://adventofcode.com/2023/day/24) (the hailstorm that forces you to do math). I took another break only to pick it up this week. I just finished days 24 and 25, completing the set! + +This article explains some patterns I used for solving the puzzles. I always use megaparsec to parse the input, even when it is overkill... just because I find it so fun to work with. + +## Haskell for puzzles + +### Parsing permutations + +Relying on megaparsec payed off from day 2 where you need to parse this beauty: + +``` +Game 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green +Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue +Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red +Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red +Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green +``` + +You got an ID, then some draws separated by `;`. A draw is a set of colors given out of order, which I see as a clear cut case of running permutations: + +```haskell +data Draw = Draw Int Int Int deriving (Eq, Show) +data Game = Game Int [Draw] deriving Show +type Input = [Game] + +type Parser = Parsec Void String + +parseColor :: String -> Parser Int +parseColor color = read <$> try (some digitChar <* hspace <* string color <* optional (string ", ")) + +parseDraw :: Parser Draw +parseDraw = do + (blue, green, red) <- runPermutation $ + (,,) <$> toPermutationWithDefault 0 (parseColor "blue") + <*> toPermutationWithDefault 0 (parseColor "green") + <*> toPermutationWithDefault 0 (parseColor "red") + void . optional $ string "; " + return $ Draw blue green red + +parseGame :: Parser Game +parseGame = do + id <- read <$> (string "Game " *> some digitChar <* optional (string ": ")) + Game id <$> someTill parseDraw eol + +parseInput' :: Parser Input +parseInput' = some parseGame <* eof +``` + +### Functors and applicatives + +I also got better at understanding functors and applicatives, using them to simplify mapping things to types. For example on day 12 you got a map that looks like: + +``` +???.### 1,1,3 +.??..??...?##. 1,1,3 +?#?#?#?#?#?#?#? 1,3,1,6 +????.#...#... 4,1,1 +????.######..#####. 1,6,5 +?###???????? 3,2,1 +``` + +Here is how I parsed it: + +```haskell +data Tile = Broken | Operational | Unknown deriving Eq +instance Show Tile where + show Broken = "#" + show Operational = "." + show Unknown = "?" +data Row = Row [Tile] [Int] deriving Show +type Input = [Row] + +type Parser = Parsec Void String + +parseNumber :: Parser Int +parseNumber = read <$> some digitChar <* optional (char ',') + +parseTile :: Parser Tile +parseTile = char '#' $> Broken + <|> char '.' $> Operational + <|> char '?' $> Unknown + +parseRow :: Parser Row +parseRow = Row <$> some parseTile <* space + <*> some parseNumber <* eol + +parseInput' :: Parser Input +parseInput' = some parseRow <* eof +``` + +The functor usage is very useful for parts where you want to parse one thing but return another thing like: + +```haskell +char '#' $> Broken +``` + +I also used it to parse the integers from the digit characters without any intermediate step, which I find really clean and powerful: + +```haskell +parseNumber = read <$> some digitChar <* optional (char ',') +``` + +The applicative (which is an extension of functors but for types instead of functions) allows this clever structure: + +```haskell +parseRow :: Parser Row +parseRow = Row <$> some parseTile <* space + <*> some parseNumber <* eol +``` + +### Playing poker + +Parsing also did all the heavy lifting on day 7 where you need to rank poker like hands. Your input is a list of hands of five cards and a bid: + +``` +32T3K 765 +T55J5 684 +KK677 28 +KTJJT 220 +QQQJA 483 +``` + +Here is the data structure I settled on: +```haskell +data Card = Two | Three | Four | Five | Six | Seven | Eight | Nine | T | J | Q | K | A deriving (Eq, Ord) + +data Rank = HighCard + | Pair + | Pairs + | Brelan + | FullHouse + | Quartet + | Quintet + deriving (Eq, Ord, Show) + +data Hand = Hand Rank [Card] Int deriving (Eq, Show) +compareCards :: [Card] -> [Card] -> Ordering +compareCards (x:xs) (y:ys) | x == y = compareCards xs ys + | otherwise = x `compare` y +instance Ord Hand where + (Hand a x _) `compare` (Hand b y _) | a == b = compareCards x y + | otherwise = a `compare` b + +type Input = [Hand] +``` + +The hard part of the puzzle is to rank hands, which I decided to compute while parsing because why not! +```haskell +parseCard :: Parser Card +parseCard = char '2' $> Two + <|> char '3' $> Three + <|> char '4' $> Four + <|> char '5' $> Five + <|> char '6' $> Six + <|> char '7' $> Seven + <|> char '8' $> Eight + <|> char '9' $> Nine + <|> char 'T' $> T + <|> char 'J' $> J + <|> char 'Q' $> Q + <|> char 'K' $> K + <|> char 'A' $> A + +evalRank :: [Card] -> Rank +evalRank n@(a:b:c:d:e:_) | not (a<=b && b<=c && c<=d && d<=e) = evalRank $ L.sort n + | a==b && b==c && c==d && d==e = Quintet + | (a==b && b==c && c==d) || (b==c && c==d && d==e) = Quartet + | a==b && (b==c || c==d) && d==e = FullHouse + | (a==b && b==c) || (b==c && c==d) || (c==d && d==e) = Brelan + | (a==b && (c==d || d==e)) || (b==c && d==e) = Pairs + | a==b || b==c || c==d || d==e = Pair + | otherwise = HighCard + +parseHand :: Parser Hand +parseHand = do + cards <- some parseCard <* char ' ' + bid <- read <$> (some digitChar <* eol) + return $ Hand (evalRank cards) cards bid + +parseInput' :: Parser Input +parseInput' = some parseHand <* eof +``` + +With all the heavy lifting already done, computing the solution for part1 of the puzzle is simply: +```haskell +compute :: Input -> Int +compute = sum . zipWith (*) [1..] . map (\(Hand _ _ bid) -> bid) . L.sort +``` + +This was particularly interesting for part 2 where there is a twist: `J` cards are now jokers, so you need to handle this as a wildcard when ranking hands! After raking my brain for a while, I decided to make the type system bear the complexity by adjusting the data structure to this: + +```haskell +data Card = J | Two | Three | Four | Five | Six | Seven | Eight | Nine | T | Q | K | A + +instance Eq Card where + J == _ = True + _ == J = True + a == b = show a == show b + +instance Ord Card where + a `compare` b = show a `compare` show b + a <= b = show a <= show b +``` + +With this change, I could now rank the hands with: +```haskell +evalRank :: [Card] -> Rank +evalRank [J, J, J, J, _] = Quintet +evalRank [J, J, J, d, e] | d==e = Quintet + | otherwise = Quartet +evalRank [J, J, c, d, e] | c==d && d==e = Quintet + | c==d || d==e = Quartet + | otherwise = Brelan +evalRank [J, b, c, d, e] | b==c && c==d && d==e = Quintet + | (b==c || d==e) && c==d = Quartet + | b==c && d==e = FullHouse + | b==c || c==d || d==e = Brelan + | otherwise = Pair +evalRank [a, b, c, d, e] | a==b && a==c && a==d && a==e = Quintet + | (a==b && a==c && a==d) || (b==c && b==d && b==e) = Quartet + | a==b && (b==c || c==d) && d==e = FullHouse + | (a==b && b==c) || (b==c && c==d) || (c==d && d==e) = Brelan + | (a==b && (c==d || d==e)) || (b==c && d==e) = Pairs + | a==b || b==c || c==d || d==e = Pair + | otherwise = HighCard +``` + +## Conclusion + +I love haskell, I wish I could use it daily and not just for seasonal puzzles. |