Day 3: Mull It Over

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FAQ

  • Amy@lemmy.sdf.orgdeleted by creator
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    2 years ago

    Haskell

    Oof, a parsing problem :/ This is some nasty-ass code. step is almost the State monad written out explicitly.

    Solution
    import Control.Monad
    import Data.Either
    import Data.List
    import Text.Parsec
    
    data Ins = Mul !Int !Int | Do | Dont
    
    readInput :: String -> [Ins]
    readInput = fromRight undefined . parse input ""
      where
        input = many ins <* many anyChar
        ins =
          choice . map try $
            [ Mul <$> (string "mul(" *> arg) <*> (char ',' *> arg) <* char ')',
              Do <$ string "do()",
              Dont <$ string "don't()",
              anyChar *> ins
            ]
        arg = do
          s <- many1 digit
          guard $ length s <= 3
          return $ read s
    
    run f = snd . foldl' step (True, 0)
      where
        step (e, a) i =
          case i of
            Mul x y -> (e, if f e then a + x * y else a)
            Do -> (True, a)
            Dont -> (False, a)
    
    main = do
      input <- readInput <$> readFile "input03"
      print $ run (const True) input
      print $ run id input
    
    • VegOwOtenks@lemmy.world
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      2 years ago

      Love to see you chewing through this parsing problem in Haskell, I didn’t dare use Parsec because I wasn’t confident enough.
      Why did you decide to have a strict definition of Mul !Int !Int?

      • kintrix@linux.community
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        2 years ago

        My guess is because a linter and/or HLS was suggesting it. I know HLS used to suggest making your fields strict in almost all cases. In this case I have a hunch that it slightly cuts down on memory usage because we use almost all Muls either way. So it does not need to keep the string it is parsed from in memory as part of the thunk.

        But it probably makes a small/negligible difference here.

        • Amy@lemmy.sdf.orgdeleted by creator
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          2 years ago

          Yep, HLS suggested it, and I figured since I’m definitely going to be using all of the values (in part one, at least), why not?

          Normally I ignore that kind of nitpicky suggestion though.

  • janAkali@lemmy.one
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    2 years ago

    Nim

    From a first glance it was obviously a regex problem.
    I’m using tinyre here instead of stdlib re library just because I’m more familiar with it.

    import pkg/tinyre
    
    proc solve(input: string): AOCSolution[int, int] =
      var allow = true
      for match in input.match(reG"mul\(\d+,\d+\)|do\(\)|don't\(\)"):
        if match == "do()": allow = true
        elif match == "don't()": allow = false
        else:
          let pair = match[4..^2].split(',')
          let mult = pair[0].parseInt * pair[1].parseInt
          result.part1 += mult
          if allow: result.part2 += mult
    

    Codeberg repo

    • sjmulder@lemmy.sdf.org
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      2 years ago

      I shy away from regexes for these parsing problems because part 2 likes to mess those up but here it worked beautifully. Nice and compact solution!

  • Ananace@lemmy.ananace.dev
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    2 years ago

    I started poking at doing a proper lexer/parser, but then I thought about how early in AoC it is and how low the chance is that the second part will require proper parsing.

    So therefore; hello regex my old friend, I’ve come to talk with you again.

    C#
    List<string> instructions = new List<string>();
    
    public void Input(IEnumerable<string> lines)
    {
      foreach (var line in lines)
        instructions.AddRange(Regex.Matches(line, @"mul\(\d+,\d+\)|do\(\)|don't\(\)").Select(m => m.Value));
    }
    
    public void Part1()
    {
      var sum = instructions.Select(mul => Regex.Match(mul, @"(\d+),(\d+)").Groups.Values.Skip(1).Select(g => int.Parse(g.Value))).Select(cc => cc.Aggregate(1, (acc, val) => acc * val)).Sum();
      Console.WriteLine($"Sum: {sum}");
    }
    public void Part2()
    {
      bool enabled = true;
      long sum = 0;
      foreach(var inst in instructions)
      {
        if (inst.StartsWith("don't"))
          enabled = false;
        else if (inst.StartsWith("do"))
          enabled = true;
        else if (enabled)
          sum += Regex.Match(inst, @"(\d+),(\d+)").Groups.Values.Skip(1).Select(g => int.Parse(g.Value)).Aggregate(1, (acc, val) => acc * val);
      }
      Console.WriteLine($"Sum: {sum}");
    }
    
  • LeixB@lemmy.world
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    2 years ago

    Haskell

    module Main where
    
    import Control.Arrow hiding ((+++))
    import Data.Char
    import Data.Functor
    import Data.Maybe
    import Text.ParserCombinators.ReadP hiding (get)
    import Text.ParserCombinators.ReadP qualified as P
    
    data Op = Mul Int Int | Do | Dont deriving (Show)
    
    parser1 :: ReadP [(Int, Int)]
    parser1 = catMaybes <$> many ((Just <$> mul) <++ (P.get $> Nothing))
    
    parser2 :: ReadP [Op]
    parser2 = catMaybes <$> many ((Just <$> operation) <++ (P.get $> Nothing))
    
    mul :: ReadP (Int, Int)
    mul = (,) <$> (string "mul(" *> (read <$> munch1 isDigit <* char ',')) <*> (read <$> munch1 isDigit <* char ')')
    
    operation :: ReadP Op
    operation = (string "do()" $> Do) +++ (string "don't()" $> Dont) +++ (uncurry Mul <$> mul)
    
    foldOp :: (Bool, Int) -> Op -> (Bool, Int)
    foldOp (_, n) Do = (True, n)
    foldOp (_, n) Dont = (False, n)
    foldOp (True, n) (Mul a b) = (True, n + a * b)
    foldOp (False, n) _ = (False, n)
    
    part1 = sum . fmap (uncurry (*)) . fst . last . readP_to_S parser1
    part2 = snd . foldl foldOp (True, 0) . fst . last . readP_to_S parser2
    
    main = getContents >>= print . (part1 &&& part2)
    
  • mykl@lemmy.world
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    2 years ago

    Uiua

    Uses experimental feature of fold to track the running state of do/don’t.

    [edit] Slightly re-written to make it less painful :-) Try it online!

    # Experimental!
    DataP₁       ← $ xmul(2,4)%&mul[3,7]!@^do_not_mul(5,5)+mul(32,64]then(mul(11,8)mul(8,5))
    DataP₂       ← $ xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))
    GetMul       ← $ mul\((\d{1,3}),(\d{1,3})\)
    GetMulDoDont ← $ mul\(\d{1,3},\d{1,3}\)|do\(\)|don\'t\(\)
    
    &p/+≡(/×≡⋕↘1)regexGetMul DataP₁ # Part 1
    
    # Build an accumulator to track running state of do/don't
    Filter ← ↘1⊂:∧(⍣(0 °"don"|1 °"do("|.◌)) :1≡(↙3°□)
    ≡⊢ regex GetMulDoDont DataP₂
    ▽⊸≡◇(≍"mul"↙3)▽⊸Filter      # Apply Filter, remove the spare 'do's
    &p/+≡◇(/×≡◇⋕↘1⊢regexGetMul) # Get the digits and multiply, sum.
    
  • rwdf@lemmy.world
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    2 years ago

    Elixir

    First time writing Elixir. It’s probably janky af.

    I’ve had some help from AI to get some pointers along the way. I’m not competing in any way, just trying to learn and have fun.

    ~~Part 2 is currently not working, and I can’t figure out why. I’m trying to just remove everything from “don’t()” to “do()” and just pass the rest through the working solution for part 1. Should work, right?

    Any pointers?~~

    edit; working solution:

    defmodule Three do
      def get_input do
        File.read!("./input.txt")
      end
    
      def extract_operations(input) do
        Regex.scan(~r/mul\((\d{1,3}),(\d{1,3})\)/, input)
        |> Enum.map(fn [_op, num1, num2] ->
          num1 = String.to_integer(num1)
          num2 = String.to_integer(num2)
          [num1 * num2]
        end)
      end
    
      def sum_products(ops) do
        List.flatten(ops)
        |> Enum.filter(fn x -> is_integer(x) end)
        |> Enum.sum()
      end
    
      def part1 do
        extract_operations(get_input())
        |> sum_products()
      end
    
      def part2 do
        String.split(get_input(), ~r/don\'t\(\)[\s\S]*?do\(\)/)
        |> Enum.map(&extract_operations/1)
        |> sum_products()
      end
    end
    
    IO.puts("part 1: #{Three.part1()}")
    IO.puts("part 2: #{Three.part2()}")
    
    
    • vole@lemmy.world
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      2 years ago

      Part 2 is currently not working, and I can’t figure out why. I’m trying to just remove everything from “don’t()” to “do()” and just pass the rest through the working solution for part 1. Should work, right?

      I think I had the same issue. Consider what happens if there isn’t a do() after a don’t().

      • rwdf@lemmy.world
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        2 years ago

        Ah, yes, that’s it. The lazy solution would be to add a “do()” to the end of the input, right? Haha

        • rwdf@lemmy.world
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          2 years ago

          It was actually a line break that broke the regex. Changing from a “.” to “[\s\S]” fixed it.

  • urquell@lemm.ee
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    2 years ago

    Python

    Part1:

    matches = re.findall(r"(mul\((\d+),(\d+)\))", input)
    muls = [int(m[1]) * int(m[2]) for m in matches]
    print(sum(muls))
    

    Part2:

    instructions = list(re.findall(r"(do\(\)|don't\(\)|(mul\((\d+),(\d+)\)))", input)
    mul_enabled = True
    muls = 0
    
    for inst in instructions:
        if inst[0] == "don't()":
            mul_enabled = False
        elif inst[0] == "do()":
            mul_enabled = True
        elif mul_enabled:
            muls += int(inst[2]) * int(inst[3])
    
    print(muls)
    
  • reboot6675@sopuli.xyz
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    2 years ago

    Go

    Part 1, just find the regex groups, parse to int, and done.

    Part 1
    func part1() {
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    	scanner := bufio.NewScanner(file)
    
    	re := regexp.MustCompile(`mul\(([0-9]{1,3}),([0-9]{1,3})\)`)
    	product := 0
    
    	for scanner.Scan() {
    		line := scanner.Text()
    		submatches := re.FindAllStringSubmatch(line, -1)
    
    		for _, s := range submatches {
    			a, _ := strconv.Atoi(s[1])
    			b, _ := strconv.Atoi(s[2])
    			product += (a * b)
    		}
    	}
    
    	fmt.Println(product)
    }
    

    Part 2, not so simple. Ended up doing some weird hack with a map to check if the multiplication was enabled or not. Also instead of finding regex groups I had to find the indices, and then interpret what those mean… Not very readable code I’m afraid

    Part2
    func part2() {
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    	scanner := bufio.NewScanner(file)
    
    	mulRE := regexp.MustCompile(`mul\(([0-9]{1,3}),([0-9]{1,3})\)`)
    	doRE := regexp.MustCompile(`do\(\)`)
    	dontRE := regexp.MustCompile(`don't\(\)`)
    	product := 0
    	enabled := true
    
    	for scanner.Scan() {
    		line := scanner.Text()
    		doIndices := doRE.FindAllStringIndex(line, -1)
    		dontIndices := dontRE.FindAllStringIndex(line, -1)
    		mulSubIndices := mulRE.FindAllStringSubmatchIndex(line, -1)
    
    		mapIndices := make(map[int]string)
    		for _, do := range doIndices {
    			mapIndices[do[0]] = "do"
    		}
    		for _, dont := range dontIndices {
    			mapIndices[dont[0]] = "dont"
    		}
    		for _, mul := range mulSubIndices {
    			mapIndices[mul[0]] = "mul"
    		}
    
    		nextMatch := 0
    
    		for i := 0; i < len(line); i++ {
    			val, ok := mapIndices[i]
    			if ok && val == "do" {
    				enabled = true
    			} else if ok && val == "dont" {
    				enabled = false
    			} else if ok && val == "mul" {
    				if enabled {
    					match := mulSubIndices[nextMatch]
    					a, _ := strconv.Atoi(string(line[match[2]:match[3]]))
    					b, _ := strconv.Atoi(string(line[match[4]:match[5]]))
    					product += (a * b)
    				}
    				nextMatch++
    			}
    		}
    	}
    
    	fmt.Println(product)
    }
    
    • gothink@lemmy.ca
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      2 years ago

      I also used Go - my solution for part 1 was essentially identical to yours. I went a different route for part 2 that I think ended up being simpler though.

      I just prepended do() and don't() to the original regex with a |, that way it captured all 3 in order and I just looped through all the matches once and toggled the isEnabled flag accordingly.

      Always interesting to see how other people tackle the same problem!

      Part 2 Code
      func part2() {
      	filePath := "input.txt"
      	file, _ := os.Open(filePath)
      	defer file.Close()
      
      	pattern := regexp.MustCompile(`do\(\)|don't\(\)|mul\((\d{1,3}),(\d{1,3})\)`)
      	productSum := 0
      	isEnabled := true
      
      	scanner := bufio.NewScanner(file)
      	for scanner.Scan() {
      		line := scanner.Text()
      		matches := pattern.FindAllStringSubmatch(line, -1)
      
      		for _, match := range matches {
      			if match[0] == "do()" {
      				isEnabled = true
      			} else if match[0] == "don't()" {
      				isEnabled = false
      			} else if isEnabled && len(match) == 3 {
      				n, _ := strconv.Atoi(match[1])
      				m, _ := strconv.Atoi(match[2])
      				productSum += n * m
      			}
      		}
      	}
      
      	fmt.Println("Total: ", productSum)
      }
      
  • vole@lemmy.world
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    2 years ago

    Raku

    sub MAIN($input) {
        grammar Muls {
            token TOP { .*? <mul>+%.*? .* }
            token mul { "mul(" <number> "," <number> ")" }
            token number { \d+ }
        }
    
        my $parsedMuls = Muls.parsefile($input);
        my @muls = $parsedMuls<mul>.map({.<number>».Int});
        my $part-one-solution = @muls.map({[*] $_.List}).sum;
        say "part 1: $part-one-solution";
    
        grammar EnabledMuls {
            token TOP { .*? [<.disabled> || <mul>]+%.*? .* }
            token mul { "mul(" <number> "," <number> ")" }
            token number { \d+ }
            token disabled { "don't()" .*? ["do()" || $] }
        }
    
        my $parsedEnabledMuls = EnabledMuls.parsefile($input);
        my @enabledMuls = $parsedEnabledMuls<mul>.map({.<number>».Int});
        my $part-two-solution = @enabledMuls.map({[*] $_.List}).sum;
        say "part 2: $part-two-solution";
    }
    

    github

  • thirteen37@lemmy.world
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    2 years ago

    Python

    def process(input, part2=False):
        if part2:
            input = re.sub(r'don\'t\(\).+?do\(\)', '', input) # remove everything between don't() and do()
        total = [ int(i[0]) * int(i[1]) for i in re.findall(r'mul\((\d+),(\d+)\)', input) ]
        return sum(total)
    

    Given the structure of the input file, we just have to ignore everything between don’t() and do(), so remove those from the instructions before processing.

  • bugsmith@programming.dev
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    2 years ago

    Gleam

    Struggled with the second part as I am still very new to this very cool language, but got there after scrolling for some inspiration.

    import gleam/int
    import gleam/io
    import gleam/list
    import gleam/regex
    import gleam/result
    import gleam/string
    import simplifile
    
    pub fn main() {
      let assert Ok(data) = simplifile.read("input.in")
      part_one(data) |> io.debug
      part_two(data) |> io.debug
    }
    
    fn part_one(data) {
      let assert Ok(multiplication_pattern) =
        regex.from_string("mul\\(\\d{1,3},\\d{1,3}\\)")
      let assert Ok(digit_pattern) = regex.from_string("\\d{1,3},\\d{1,3}")
      let multiplications =
        regex.scan(multiplication_pattern, data)
        |> list.flat_map(fn(reg) {
          regex.scan(digit_pattern, reg.content)
          |> list.map(fn(digits) {
            digits.content
            |> string.split(",")
            |> list.map(fn(x) { x |> int.parse |> result.unwrap(0) })
            |> list.reduce(fn(a, b) { a * b })
            |> result.unwrap(0)
          })
        })
        |> list.reduce(fn(a, b) { a + b })
        |> result.unwrap(0)
    }
    
    fn part_two(data) {
      let data = "do()" <> string.replace(data, "\n", "") <> "don't()"
      let assert Ok(pattern) = regex.from_string("do\\(\\).*?don't\\(\\)")
      regex.scan(pattern, data)
      |> list.map(fn(input) { input.content |> part_one })
      |> list.reduce(fn(a, b) { a + b })
    }
    
  • Sleepless One@lemmy.ml
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    2 years ago

    Rust

    use crate::utils::read_lines;
    
    pub fn solution1() {
        let lines = read_lines("src/day3/input.txt");
        let sum = lines
            .map(|line| {
                let mut sum = 0;
                let mut command_bytes = Vec::new();
                for byte in line.bytes() {
                    match (byte, command_bytes.as_slice()) {
                        (b')', [.., b'0'..=b'9']) => {
                            handle_mul(&mut command_bytes, &mut sum);
                        }
                        _ if matches_mul(byte, &command_bytes) => {
                            command_bytes.push(byte);
                        }
                        _ => {
                            command_bytes.clear();
                        }
                    }
                }
    
                sum
            })
            .sum::<usize>();
    
        println!("Sum of multiplication results = {sum}");
    }
    
    pub fn solution2() {
        let lines = read_lines("src/day3/input.txt");
    
        let mut can_mul = true;
        let sum = lines
            .map(|line| {
                let mut sum = 0;
                let mut command_bytes = Vec::new();
                for byte in line.bytes() {
                    match (byte, command_bytes.as_slice()) {
                        (b')', [.., b'0'..=b'9']) if can_mul => {
                            handle_mul(&mut command_bytes, &mut sum);
                        }
                        (b')', [b'd', b'o', b'(']) => {
                            can_mul = true;
                            command_bytes.clear();
                        }
                        (b')', [.., b't', b'(']) => {
                            can_mul = false;
                            command_bytes.clear();
                        }
                        _ if matches_do_or_dont(byte, &command_bytes)
                            || matches_mul(byte, &command_bytes) =>
                        {
                            command_bytes.push(byte);
                        }
                        _ => {
                            command_bytes.clear();
                        }
                    }
                }
    
                sum
            })
            .sum::<usize>();
    
        println!("Sum of enabled multiplication results = {sum}");
    }
    
    fn matches_mul(byte: u8, command_bytes: &[u8]) -> bool {
        matches!(
            (byte, command_bytes),
            (b'm', [])
                | (b'u', [.., b'm'])
                | (b'l', [.., b'u'])
                | (b'(', [.., b'l'])
                | (b'0'..=b'9', [.., b'(' | b'0'..=b'9' | b','])
                | (b',', [.., b'0'..=b'9'])
        )
    }
    
    fn matches_do_or_dont(byte: u8, command_bytes: &[u8]) -> bool {
        matches!(
            (byte, command_bytes),
            (b'd', [])
                | (b'o', [.., b'd'])
                | (b'n', [.., b'o'])
                | (b'\'', [.., b'n'])
                | (b'(', [.., b'o' | b't'])
                | (b't', [.., b'\''])
        )
    }
    
    fn handle_mul(command_bytes: &mut Vec<u8>, sum: &mut usize) {
        let first_num_index = command_bytes
            .iter()
            .position(u8::is_ascii_digit)
            .expect("Guarunteed to be there");
        let comma_index = command_bytes
            .iter()
            .position(|&c| c == b',')
            .expect("Guarunteed to be there.");
    
        let num1 = bytes_to_num(&command_bytes[first_num_index..comma_index]);
        let num2 = bytes_to_num(&command_bytes[comma_index + 1..]);
    
        *sum += num1 * num2;
        command_bytes.clear();
    }
    
    fn bytes_to_num(bytes: &[u8]) -> usize {
        bytes
            .iter()
            .rev()
            .enumerate()
            .map(|(i, digit)| (*digit - b'0') as usize * 10usize.pow(i as u32))
            .sum::<usize>()
    }
    

    Definitely not my prettiest code ever. It would probably look nicer if I used regex or some parsing library, but I took on the self-imposed challenge of not using third party libraries. Also, this is already further than I made it last year!

  • RollingRagdoll@lemmy.world
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    2 years ago

    Uiua

    Part 1:

    &fras "day3/input.txt"
    /+≡/×≡⋕≡↘1regex "mul\\((\\d+),(\\d+)\\)"
    

    Part 2:

    Filter ← ⍜⊜∘≡⋅""⊸⦷°□
    .&fras "day3/input.txt"
    ∧Filter♭regex"don't\\(\\)?(.*?)(?:do\\(\\)|$)"
    /+≡/×≡⋕≡↘1regex "mul\\((\\d+),(\\d+)\\)"
    
  • the_beber@lemm.ee
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    2 years ago

    Kotlin

    Just the standard Regex stuff. I found this website to be very helpful to write the patterns. (Very useful in general)

    fun main() {
        fun part1(input: List<String>): Int =
            Regex("""mul\(\d+,\d+\)""").findAll(input.joinToString()).sumOf {
                with(Regex("""\d+""").findAll(it.value)) { this.first().value.toInt() * this.last().value.toInt() }
            }
    
        fun part2(input: List<String>): Int {
            var isMultiplyInstructionEnabled = true  // by default
            return Regex("""mul\(\d+,\d+\)|do\(\)|don't\(\)""").findAll(input.joinToString()).fold(0) { acc, instruction ->
                when (instruction.value) {
                    "do()" -> acc.also { isMultiplyInstructionEnabled = true }
                    "don't()" -> acc.also { isMultiplyInstructionEnabled = false }
                    else -> {
                        if (isMultiplyInstructionEnabled) {
                            acc + with(Regex("""\d+""").findAll(instruction.value)) { this.first().value.toInt() * this.last().value.toInt() }
                        } else acc
                    }
                }
            }
        }
    
        val testInputPart1 = readInput("Day03_test_part1")
        val testInputPart2 = readInput("Day03_test_part2")
        check(part1(testInputPart1) == 161)
        check(part2(testInputPart2) == 48)
    
        val input = readInput("Day03")
        part1(input).println()
        part2(input).println()
    }
    
    ´´´