Day 3: Mull It Over

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FAQ

  • sjmulder@lemmy.sdf.org
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    24 days ago

    C

    Yay parsers! I’ve gotten quite comfortable writing these with C. Using out pointers arguments for the cursor that are only updated if the match is successful makes for easy bookkeeping.

    Code
    #include "common.h"
    
    static int
    parse_exact(const char **stringp, const char *expect)
    {
    	const char *s = *stringp;
    	int i;
    
    	for (i=0; s[i] && expect[i] && s[i] == expect[i]; i++)
    		;
    	if (expect[i])
    		return 0;
    
    	*stringp  = &s[i];
    	return 1;
    }
    
    static int
    parse_int(const char **stringp, int *outp)
    {
    	char *end;
    	int val;
    
    	val = (int)strtol(*stringp, &end, 10);
    	if (end == *stringp)
    		return 0;
    
    	*stringp = end;
    	if (outp) *outp = val;
    	return 1;
    }
    
    static int
    parse_mul(const char **stringp, int *ap, int *bp)
    {
    	const char *cur = *stringp;
    	int a,b;
    
    	if (!parse_exact(&cur, "mul(") ||
    	    !parse_int(&cur, &a) ||
    	    !parse_exact(&cur, ",") ||
    	    !parse_int(&cur, &b) ||
    	    !parse_exact(&cur, ")"))
    		return 0;
    
    	*stringp = cur;
    	if (ap) *ap = a;
    	if (bp) *bp = b;
    	return 1;
    }
    
    int
    main(int argc, char **argv)
    {
    	static char buf[32*1024];
    	const char *cur;
    	size_t nr;
    	int p1=0,p2=0, a,b, dont=0;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	nr = fread(buf, 1, sizeof(buf), stdin);
    	assert(!ferror(stdin));
    	assert(nr != sizeof(buf));
    	buf[nr] = '\0';
    
    	for (cur = buf; *cur; )
    		if (parse_exact(&cur, "do()"))
    			dont = 0;
    		else if (parse_exact(&cur, "don't()"))
    			dont = 1;
    		else if (parse_mul(&cur, &a, &b)) {
    			p1 += a * b;
    			if (!dont) p2 += a * b;
    		} else
    			cur++;
    
    	printf("03: %d %d\n", p1, p2);
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day03.c

    • sjmulder@lemmy.sdf.org
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      24 days ago

      Got the code a little shorter:

      Code
      #include "common.h"
      
      static int
      parse_mul(const char **stringp, int *ap, int *bp)
      {
      	const char *cur = *stringp, *end;
      
      	if (strncmp(cur, "mul(", 4)) { return 0; } cur += 4;
      	*ap = (int)strtol(cur, (char **)&end, 10);
      	if (end == cur)  { return 0; } cur = end;
      	if (*cur != ',') { return 0; } cur += 1;
      	*bp = (int)strtol(cur, (char **)&end, 10);
      	if (end == cur)  { return 0; } cur = end;
      	if (*cur != ')') { return 0; } cur += 1;
      
      	*stringp = cur;
      	return 1;
      }
      
      int
      main(int argc, char **argv)
      {
      	static char buf[32*1024];
      	const char *p;
      	size_t nr;
      	int p1=0,p2=0, a,b, dont=0;
      
      	if (argc > 1)
      		DISCARD(freopen(argv[1], "r", stdin));
      
      	nr = fread(buf, 1, sizeof(buf), stdin);
      	assert(!ferror(stdin));
      	assert(nr != sizeof(buf));
      	buf[nr] = '\0';
      
      	for (p = buf; *p; )
      		if (parse_mul(&p, &a, &b)) { p1 += a*b; p2 += a*b*!dont; }
      		else if (!strncmp(p, "do()", 4))    { dont = 0; p += 4; }
      		else if (!strncmp(p, "don't()", 7)) { dont = 1; p += 7; }
      		else p++;
      
      	printf("03: %d %d\n", p1, p2);
      }
      
  • janAkali@lemmy.one
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    24 days 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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      24 days 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!

  • LeixB@lemmy.world
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    24 days 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)
    
  • Ananace@lemmy.ananace.dev
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    24 days 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}");
    }
    
  • VegOwOtenks@lemmy.world
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    24 days 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))
    
    • reboot6675@sopuli.xyz
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      24 days 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

    • Hammerheart@programming.dev
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      23 days 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()-

  • lwhjp@lemmy.sdf.org
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    24 days 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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      24 days 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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        24 days 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@lemmy.sdf.org
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          24 days 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.

  • urquell@lemm.ee
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    24 days 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)
    
  • Andy@programming.dev
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    23 days ago

    Factor

    : get-input ( -- corrupted-input )
      "vocab:aoc-2024/03/input.txt" utf8 file-contents ;
    
    : get-muls ( corrupted-input -- instructions )
      R/ mul\(\d+,\d+\)/ all-matching-subseqs ;
    
    : process-mul ( instruction -- n )
      R/ \d+/ all-matching-subseqs
      [ string>number ] map-product ;
    
    : solve ( corrupted-input -- n )
      get-muls [ process-mul ] map-sum ;
    
    : part1 ( -- n )
      get-input solve ;
    
    : part2 ( -- n )
      get-input
      R/ don't\(\)(.|\n)*?do\(\)/ split concat
      R/ don't\(\)(.|\n)*/ "" re-replace
      solve ;
    
  • mykl@lemmy.world
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    24 days 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.
    
  • madmo@programming.dev
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    24 days ago

    Rust with nom parser

    Decided to give it a go with the nom parser (first time using this crate). Turned out quite nicely. Had some issues with the alt combinator: All alternatives have to return the same type, using a enum to wrap all options did the trick.

    use memmap2::Mmap;
    use nom::{
        branch::alt, bytes::complete::*, character::complete::*, combinator::map, multi::many_till,
        sequence::tuple, AsBytes, IResult,
    };
    
    #[derive(Debug)]
    enum Token {
        Do,
        Dont,
        Mul(u64, u64),
    }
    
    fn main() -> anyhow::Result<()> {
        let file = std::fs::File::open("input.txt")?;
        let mmap = unsafe { Mmap::map(&file)? };
    
        let mut sum_part1 = 0;
        let mut sum_part2 = 0;
        let mut enabled = true;
    
        let mut cursor = mmap.as_bytes();
        while let Ok(token) = parse(cursor) {
            match token.1 .1 {
                Token::Do => enabled = true,
                Token::Dont => enabled = false,
                Token::Mul(left, right) => {
                    let prod = left * right;
                    sum_part1 += prod;
                    if enabled {
                        sum_part2 += prod;
                    }
                }
            }
    
            cursor = token.0;
        }
    
        println!("part1: {} part2: {}", sum_part1, sum_part2);
    
        Ok(())
    }
    
    type ParseResult<'a> =
        Result<(&'a [u8], (Vec<char>, Token)), nom::Err<nom::error::Error<&'a [u8]>>>;
    
    fn parse(input: &[u8]) -> ParseResult {
        many_till(
            anychar,
            alt((
                map(doit, |_| Token::Do),
                map(dont, |_| Token::Dont),
                map(mul, |el| Token::Mul(el.2, el.4)),
            )),
        )(input)
    }
    
    fn doit(input: &[u8]) -> IResult<&[u8], &[u8]> {
        tag("do()")(input)
    }
    
    fn dont(input: &[u8]) -> IResult<&[u8], &[u8]> {
        tag("don't()")(input)
    }
    
    type ParsedMulResult<'a> = (&'a [u8], &'a [u8], u64, &'a [u8], u64, &'a [u8]);
    
    fn mul(input: &[u8]) -> IResult<&[u8], ParsedMulResult> {
        tuple((tag("mul"), tag("("), u64, tag(","), u64, tag(")")))(input)
    }
    
  • Sparrow_1029@programming.dev
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    24 days 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)

  • RollingRagdoll@lemmy.world
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    24 days ago

    Uiua

    Part 1:

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

    Part 2:

    Filter  ⍜⊜∘≡⋅""⊸⦷°□
    .&fras "day3/input.txt"
    Filterregex"don't\\(\\)?(.*?)(?:do\\(\\)|$)"
    /+≡/×≡⋕≡↘1regex "mul\\((\\d+),(\\d+)\\)"
    
  • proved_unglue@programming.dev
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    24 days 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
    }
    
    • Faresh@lemmy.ml
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      24 days 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()
      
  • ystael@beehaw.org
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    24 days 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