Sorry for the delay posting this one, Ategon seemed to have it covered, so I forgot :D I will do better.

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))

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.

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}");
}

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.

#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

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)

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

Haskell

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

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

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.

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 ;

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)

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)
}

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+)\\)"

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
}

deleted by creator

Rust

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

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)