Day 3: Mull It Over

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FAQ

  • @[email protected]
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    35 months ago

    Rust

    Didn’t do anything crazy here – ended up using regex like a bunch of other folks.

    solution
    use regex::Regex;
    
    use crate::shared::util::read_lines;
    
    fn parse_mul(input: &[String]) -> (u32, u32) {
        // Lazy, but rejoin after having removed `\n`ewlines.
        let joined = input.concat();
        let re = Regex::new(r"mul\((\d+,\d+)\)|(do\(\))|(don't\(\))").expect("invalid regex");
    
        // part1
        let mut total1 = 0u32;
        // part2 -- adds `do()`s and `don't()`s
        let mut total2 = 0u32;
        let mut enabled = 1u32;
    
        re.captures_iter(&joined).for_each(|c| {
            let (_, [m]) = c.extract();
            match m {
                "do()" => enabled = 1,
                "don't()" => enabled = 0,
                _ => {
                    let product: u32 = m.split(",").map(|s| s.parse::<u32>().unwrap()).product();
                    total1 += product;
                    total2 += product * enabled;
                }
            }
        });
        (total1, total2)
    }
    
    pub fn solve() {
        let input = read_lines("inputs/day03.txt");
        let (part1_res, part2_res) = parse_mul(&input);
        println!("Part 1: {}", part1_res);
        println!("Part 2: {}", part2_res);
    }
    
    #[cfg(test)]
    mod test {
        use super::*;
    
        #[test]
        fn test_solution() {
            let test_input = vec![
                "xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))".to_string(),
            ];
            let (p1, p2) = parse_mul(&test_input);
            eprintln!("P1: {p1}, P2: {p2}");
            assert_eq!(161, p1);
            assert_eq!(48, p2);
        }
    }
    
    

    Solution on my github (Made it public now)

  • @[email protected]
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    5 months 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

  • @[email protected]
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    15 months ago

    Smalltalk

    I wrote matchesActual cause all of smalltalk’sstupid matchesDo: or whatever don’t give you the actual match with captures, only substrings (that wasted a good 40 minutes).

    Also smalltalk really needs an index operator

    day3p1: input
      | reg sum |
        reg := 'mul\((\d\d?\d?),(\d\d?\d?)\)' asRegex.
        sum := 0.
        
        reg matchesActual: input do: [ :m | " + sum at end cause operator precedence"
            sum := (m subexpression: 2) asInteger * (m subexpression: 3) asInteger + sum 
        ].
        
        ^ sum.
    
    day3p2: input
      | reg sum do val |
    
        reg := 'do(\:?n''t)?\(\)|mul\((\d{1,3}),(\d{1,3})\)' asRegex.
        sum := 0.
        do := true.
        reg matchesActual: input do: [ :m |
            val := m subexpression: 1.
            (val at: 1) = $d ifTrue: [ do := (val size < 5) ]
            ifFalse: [ 
                do ifTrue: [ 
                    sum := (m subexpression: 2) asInteger * (m subexpression: 3) asInteger + sum.
            ].  ].
        ].
        
        ^ sum.
    
  • @[email protected]
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    55 months ago

    I couldn’t figure it out in haskell, so I went with bash for the first part

    Shell

    cat example | grep -Eo "mul\([[:digit:]]{1,3},[[:digit:]]{1,3}\)" | cut -d "(" -f 2 | tr -d ")" | tr "," "*" | paste -sd+ | bc
    

    but this wouldn’t rock anymore in the second part, so I had to resort to python for it

    Python

    import sys
    
    f = "\n".join(sys.stdin.readlines())
    
    f = f.replace("don't()", "\ndon't()\n")
    f = f.replace("do()", "\ndo()\n")
    
    import re
    
    enabled = True
    muls = []
    for line in f.split("\n"):
        if line == "don't()":
            enabled = False
        if line == "do()":
            enabled = True
        if enabled:
            for match in re.finditer(r"mul\((\d{1,3}),(\d{1,3})\)", line):
                muls.append(int(match.group(1)) * int(match.group(2)))
            pass
        pass
    
    print(sum(muls))
    
    • @[email protected]
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      25 months ago

      Really cool trick. I did a bunch of regex matching that I’m sure I won’t remember how it works few weeks from now, this is so much readable

    • @[email protected]
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      15 months ago

      My first insinct was similar, add line breaks to the do and dont modifiers. But I got toa caught up thinking id have to keep track of the added characters, I wound up just abusing split()-

  • @[email protected]
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    5 months ago

    Kotlin

    fun part1(input: String): Int {
        val pattern = "mul\\((\\d{1,3}),(\\d{1,3})\\)".toRegex()
        var sum = 0
        pattern.findAll(input).forEach { match ->
            val first = match.groups[1]?.value?.toInt()!!
            val second = match.groups[2]?.value?.toInt()!!
            sum += first * second
    
        }
        return sum
    }
    
    fun part2(input: String): Int {
        val pattern = "mul\\((\\d{1,3}),(\\d{1,3})\\)|don't\\(\\)|do\\(\\)".toRegex()
        var sum = 0
        var enabled = true
        pattern.findAll(input).forEach { match ->
            if (match.value == "do()") enabled = true
            else if (match.value == "don't()") enabled = false
            else if (enabled) {
                val first = match.groups[1]?.value?.toInt()!!
                val second = match.groups[2]?.value?.toInt()!!
                sum += first * second
            }
        }
        return sum
    }
    
    • @[email protected]
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      35 months ago

      You can avoid having to escape the backslashes in regexps by using multiline strings:

      val pattern = """mul\((\d{1,3}),(\d{1,3})\)""".toRegex()
      
  • Ananace
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    55 months 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}");
    }
    
  • lwhjp
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    45 months 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
    
    • @[email protected]
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      25 months 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?

      • @[email protected]
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        35 months 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.

        • lwhjp
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          35 months 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.

  • @[email protected]
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    5 months ago

    Python

    After a bunch of fiddling yesterday and today I finally managed to arrive at a regex-only solution for part 2. That re.DOTALL is crucial here.

    import re
    from pathlib import Path
    
    
    def parse_input_one(input: str) -> list[tuple[int]]:
        p = re.compile(r"mul\((\d{1,3}),(\d{1,3})\)")
        return [(int(m[0]), int(m[1])) for m in p.findall(input)]
    
    
    def parse_input_two(input: str) -> list[tuple[int]]:
        p = re.compile(r"don't\(\).*?do\(\)|mul\((\d{1,3}),(\d{1,3})\)", re.DOTALL)
        return [(int(m[0]), int(m[1])) for m in p.findall(input) if m[0] and m[1]]
    
    
    def part_one(input: str) -> int:
        pairs = parse_input_one(input)
        return sum(map(lambda v: v[0] * v[1], pairs))
    
    
    def part_two(input: str) -> int:
        pairs = parse_input_two(input)
        return sum(map(lambda v: v[0] * v[1], pairs))
    
    
    if __name__ == "__main__":
        input = Path("input").read_text("utf-8")
        print(part_one(input))
        print(part_two(input))
    
  • bugsmith
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    25 months 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 })
    }
    
  • janAkali
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    5 months 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

    • @[email protected]
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      45 months 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!

  • @[email protected]
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    55 months 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)
    
  • @[email protected]
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    15 months ago

    TypeScript

    Solution
    import { AdventOfCodeSolutionFunction } from "./solutions";
    
    export const solution_3: AdventOfCodeSolutionFunction = (input) => {
        const mul_regex = /mul\((\d+),(\d+)\)/g; // mul()
        const do_regex = /do\(\)/g;              // do()
        const do_not_regex = /don\'t\(\)/g;      // don't()
    
        const doLength = "do()".length;
        const doNotLength = "don't()".length;
    
        let input_copy = "" + input;
        let part_1 = 0;
        let part_2 = 0;
        let isEnabled = true;
        while (true) {
            const nextMul = input_copy.search(mul_regex);
            const nextDo = input_copy.search(do_regex);
            const nextDoNot = input_copy.search(do_not_regex);
            let pointer = Number.POSITIVE_INFINITY;
    
            // find the smallest while ignoring items that are not found
            if (nextMul != -1)
                pointer = Math.min(pointer, nextMul);
    
            if (nextDo != -1)
                pointer = Math.min(pointer, nextDo);
    
            if (nextDoNot != -1)
                pointer = Math.min(pointer, nextDoNot);
    
            // no matches
            if (pointer == Number.POSITIVE_INFINITY)
                break
    
            // handle found command
            switch (pointer) {
                case nextDo: {
                    pointer += doLength;
                    isEnabled = true;
                    break;
                }
    
                case nextDoNot: {
                    pointer += doNotLength;
                    isEnabled = false;
                    break;
                }
    
                case nextMul: {
                    const res = input_copy.matchAll(mul_regex).next().value;
                    if (!res) {
                        // this should never happen but here's an escape hatch
                        throw Error("regex result is undefined or null");
                    }
    
                    // add the length of the whole capture to the pointer
                    pointer += res[0].length;
                    
                    // multiply capture groups
                    const comp = Number(res[1]) * Number(res[2]);
    
                    // part 1 sum
                    part_1 += comp;
    
                    // part 2 sum
                    if(isEnabled)
                        part_2 += comp;
    
                    break;
                }
            }
    
            // shorten the start of the string
            input_copy = input_copy.slice(pointer);
        }
    
        return {
            part_1,
            part_2,
        };
    }
    

    This one was harder but still. I feel like I can improve it for sure :)

  • @[email protected]
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    5 months 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()}")
    
    
    • @[email protected]
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      25 months 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().

      • @[email protected]
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        15 months ago

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

        • @[email protected]
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          15 months ago

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

  • @[email protected]
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    35 months ago

    J

    We can take advantage of the manageable size of the input to avoid explicit looping and mutable state; instead, construct vectors which give, for each character position in the input, the position of the most recent do() and most recent don't(); for part 2 a multiplication is enabled if the position of the most recent do() (counting start of input as 0) is greater than that of the most recent don't() (counting start of input as minus infinity).

    load 'regex'
    
    raw =: fread '3.data'
    mul_matches =: 'mul\(([[:digit:]]{1,3}),([[:digit:]]{1,3})\)' rxmatches raw
    
    NB. a b sublist y gives elements [a..b) of y
    sublist =: ({~(+i.)/)~"1 _
    
    NB. ". is number parsing
    mul_results =: */"1 ". (}."2 mul_matches) sublist raw
    result1 =: +/ mul_results
    
    do_matches =: 'do\(\)' rxmatches raw
    dont_matches =: 'don''t\(\)' rxmatches raw
    match_indices =: (&lt;0 0) &amp; {"2
    do_indices =: 0 , match_indices do_matches  NB. start in do mode
    dont_indices =: match_indices dont_matches
    NB. take successive diffs, then append length from last index to end of string
    run_lengths =: (}. - }:) , (((#raw) &amp; -) @: {:)
    do_map =: (run_lengths do_indices) # do_indices
    dont_map =: (({. , run_lengths) dont_indices) # __ , dont_indices
    enabled =: do_map > dont_map
    result2 =: +/ ((match_indices mul_matches) { enabled) * mul_results
    
  • Deebster
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    5 months ago

    Rust feat. pest

    No Zalgo here! I wasted a huge amount of time by not noticing that the second part’s example input was different - my code worked fine but my test failed 🤦‍♂️

    pest.rs is lovely, although part two made my PEG a bit ugly.

    part1    =  { SOI ~ (mul_expr | junk)+ ~ EOI }
    part2    =  { (enabled | disabled)+ ~ EOI }
    mul_expr =  { "mul(" ~ number ~ "," ~ number ~ ")" }
    number   =  { ASCII_DIGIT{1,3} }
    junk     = _{ ASCII }
    on       = _{ "do()" }
    off      = _{ "don't()" }
    enabled  = _{ (SOI | on) ~ (!(off) ~ (mul_expr | junk))+ }
    disabled = _{ off ~ (!(on) ~ junk)+ }
    
    use std::fs;
    
    use color_eyre::eyre;
    use pest::Parser;
    use pest_derive::Parser;
    
    #[derive(Parser)]
    #[grammar = "memory.pest"]
    pub struct MemoryParser;
    
    fn parse(input: &str, rule: Rule) -> eyre::Result<usize> {
        let sum = MemoryParser::parse(rule, input)?
            .next()
            .expect("input must be ASCII")
            .into_inner()
            .filter(|pair| pair.as_rule() == Rule::mul_expr)
            .map(|pair| {
                pair.into_inner()
                    .map(|num| num.as_str().parse::<usize>().unwrap())
                    .product::<usize>()
            })
            .sum();
        Ok(sum)
    }
    
    fn part1(filepath: &str) -> eyre::Result<usize> {
        let input = fs::read_to_string(filepath)?;
        parse(&input, Rule::part1)
    }
    
    fn part2(filepath: &str) -> eyre::Result<usize> {
        let input = fs::read_to_string(filepath)?;
        parse(&input, Rule::part2)
    }
    
    fn main() -> eyre::Result<()> {
        color_eyre::install()?;
    
        let part1 = part1("d03/input.txt")?;
        let part2 = part2("d03/input.txt")?;
        println!("Part 1: {part1}\nPart 2: {part2}");
        Ok(())
    }