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

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)

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))

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

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

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 ;

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}");
}

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.

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

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

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)

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
}

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)
}

Go

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

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

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)
}

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 =: (<0 0) & {"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) & -) @: {:)
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

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+)\\)"

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

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.

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()}")

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 })
}

Rust

Regex made this one pretty straightforward. The second part additionally looks for do() and don't() in the same regex, then we do a case distinction on the match.

use regex::{Regex, Captures};

fn mul_cap(cap: Captures) -> i32 {
    let a = cap.get(1).unwrap().as_str().parse::<i32>().unwrap();
    let b = cap.get(2).unwrap().as_str().parse::<i32>().unwrap();
    a * b
}

fn part1(input: String) {
    let re = Regex::new(r"mul\((\d{1,3}),(\d{1,3})\)").unwrap();
    let res = re.captures_iter(&input).map(mul_cap).sum::<i32>();
    println!("{res}");
}

fn part2(input: String) {
    let re = Regex::new(r"do\(\)|don't\(\)|mul\((\d{1,3}),(\d{1,3})\)").unwrap();
    let mut enabled = true;
    let mut res = 0;
    for cap in re.captures_iter(&input) {
        match cap.get(0).unwrap().as_str() {
            "do()" => enabled = true,
            "don't()" => enabled = false,
            _ if enabled => res += mul_cap(cap),
            _ => {}
        }
    }
    println!("{res}");
}

util::aoc_main!();

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))

Nim

import ../aoc, re, sequtils, strutils, math

proc mulsum*(line:string):int=
  let matches = line.findAll(re"mul\([0-9]{1,3},[0-9]{1,3}\)")
  let pairs = matches.mapIt(it[4..^2].split(',').map(parseInt))
  pairs.mapIt(it[0]*it[1]).sum

proc filter*(line:string):int=
  var state = true;
  var i=0
  while i < line.len:
    if state:
      let off = line.find("don't()", i)
      if off == -1:
        break
      result += line[i..<off].mulsum
      i = off+6
      state = false
    else:
      let on = line.find("do()", i)
      if on == -1:
        break
      i = on+4
      state = true
      
  if state:
    result += line[i..^1].mulsum

proc solve*(input:string): array[2,int] =
  #part 1&2
  result = [input.mulsum, input.filter]

I had a nicer solution in mind for part 2, but for some reason nre didn't want to work for me, and re couldn't give me the start/end or all results, so I ended up doing this skip/toggle approach.

Also initially I was doing it line by line out of habit from other puzzles, but then ofc the don't()s didn't propagate to the next line.

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()
}

´´´

Uiua

Regex my beloved <3

Run with example input here

FindMul ← regex "mul\\((\\d+),(\\d+)\\)"

PartOne ← (
  &rs ∞ &fo "input-3.txt"
  FindMul
  /+≡(×°⊟⋕⊏1_2)
)

IdDont ← ⊗□"don't()"♭

PartTwo ← (
  &rs ∞ &fo "input-3.txt"
  regex "mul\\(\\d+,\\d+\\)|do\\(\\)|don't\\(\\)"
  ⍢(IdDont.
    ↘1⊃↘↙
    ⊗□"do()"♭.
    ⊂↘1↘
  | IdDont.
    ≠⧻,
  )
  ▽♭=0⌕□"do()".
  ≡(×°⊟⋕⊏1_2♭FindMul)♭
  /+
)

&p "Day 3:"
&pf "Part 1: "
&p PartOne
&pf "Part 2: "
&p PartTwo

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!

python

import re
import aoc

def setup():
    return (aoc.get_lines(3), 0)

def one():
    lines, acc = setup()
    for line in lines:
        ins = re.findall(r'mul\(\d+,\d+\)', line)
        for i in ins:
            p = [int(x) for x in re.findall(r'\d+', i)]
            acc += p[0] * p[1]
    print(acc)

def two():
    lines, acc = setup()
    on = 1
    for line in lines:
        ins = re.findall(r"do\(\)|don't\(\)|mul\(\d+,\d+\)", line)
        for i in ins:
            if i == "do()":
                on = 1
            elif i == "don't()":
                on = 0
            elif on:
                p = [int(x) for x in re.findall(r'\d+', i)]
                acc += p[0] * p[1]
    print(acc)

one()
two()

Julia

I did not try to make my solution concise and kept separate code for part 1 and part 2 with test cases for both to check if I broke anything. But after struggling with Day 2 I am quite pleased to have solved Day 3 with only a little bugfixing.

function calcLineResult(line::String)
	lineResult::Int = 0
	enabled::Bool = true
	for i=1 : length(line)
		line[i]!='m' ? continue : (i<length(line) ? i+=1 : continue)
		line[i]!='u' ? continue : (i<length(line) ? i+=1 : continue)
		line[i]!='l' ? continue : (i<length(line) ? i+=1 : continue)
		line[i]!='(' ? continue : (i<length(line) ? i+=1 : continue)
		num1Str::String = ""
		while line[i] in ['0','1','2','3','4','5','6','7','8','9'] #should check for length of digits < 3, but works without
			num1Str = num1Str*line[i]; (i<length(line) ? i+=1 : continue)
		end
		line[i]!=',' ? continue : (i<length(line) ? i+=1 : continue)
		num2Str::String = ""
		while line[i] in ['0','1','2','3','4','5','6','7','8','9'] #should check for length of digits < 3, but works without
			num2Str = num2Str*line[i]; (i<length(line) ? i+=1 : continue)
		end
		line[i]==')' ? lineResult+=parse(Int,num1Str)*parse(Int,num2Str) : continue
	end
	return lineResult
end

function calcLineResultWithEnabling(line::String,enabled::Bool)
	lineResult::Int = 0
	for i=1 : length(line)
		if enabled && line[i] == 'm'
			i<length(line) ? i += 1 : continue
			line[i]!='u' ? continue : (i<length(line) ? i+=1 : continue)
			line[i]!='l' ? continue : (i<length(line) ? i+=1 : continue)
			line[i]!='(' ? continue : (i<length(line) ? i+=1 : continue)
			num1Str::String = ""
			while line[i] in ['0','1','2','3','4','5','6','7','8','9']
				num1Str = num1Str*line[i]; (i<length(line) ? i+=1 : continue)
			end
			line[i]!=',' ? continue : (i<length(line) ? i+=1 : continue)
			num2Str::String = ""
			while line[i] in ['0','1','2','3','4','5','6','7','8','9']
				num2Str = num2Str*line[i]; (i<length(line) ? i+=1 : continue)
			end
			line[i]==')' ? lineResult+=parse(Int,num1Str)*parse(Int,num2Str) : continue
		elseif line[i] == 'd'
			i<length(line) ? i += 1 : continue
			line[i]!='o' ? continue : (i<length(line) ? i+=1 : continue)
			if line[i] == '('
				i<length(line) ? i += 1 : continue
				line[i]==')' ? enabled=true : continue
				#@info i,line[i-3:i]
			elseif line[i] == 'n'
				i<length(line) ? i += 1 : continue
				line[i]!=''' ? continue : (i<length(line) ? i+=1 : continue)
				line[i]!='t' ? continue : (i<length(line) ? i+=1 : continue)
				line[i]!='(' ? continue : (i<length(line) ? i+=1 : continue)
				line[i]==')' ? enabled=false : continue
				#@info i,line[i-6:i]
			else
				nothing
			end
		end
	end
	return lineResult,enabled
end

function calcMemoryResult(inputFile::String,useEnabling::Bool)
	memoryRes::Int = 0
	f = open(inputFile,"r")
	lines = readlines(f)
	close(f)
	enabled::Bool = true
	for line in lines
		if useEnabling
			lineRes::Int,enabled = calcLineResultWithEnabling(line,enabled)
			memoryRes += lineRes
		else
			memoryRes += calcLineResult(line)
		end
	end
	return memoryRes
end

if abspath(PROGRAM_FILE) == @__FILE__
	@info "Part 1"
	@debug "checking test input"
	inputFile::String = "day03InputTest"
	memoryRes::Int = calcMemoryResult(inputFile,false)
	try
		@assert memoryRes==161
	catch e
		throw(ErrorException("$e memoryRes=$memoryRes"))
	end
	@debug "test input ok"
	@debug "running real input"
	inputFile::String = "day03Input"
	memoryRes::Int = calcMemoryResult(inputFile,false)
	try
		@assert memoryRes==153469856
	catch e
		throw(ErrorException("$e memoryRes=$memoryRes"))
	end
	println("memory result: $memoryRes")
	@debug "real input ok"

	@info "Part 2"
	@debug "checking test input"
	inputFile::String = "day03InputTest"
	memoryRes::Int = calcMemoryResult(inputFile,true)
	try
		@assert memoryRes==48
	catch e
		throw(ErrorException("$e memoryRes=$memoryRes"))
	end
	@debug "test input ok"
	@debug "running real input"
	inputFile::String = "day03Input"
	memoryRes::Int = calcMemoryResult(inputFile,true)
	try
		@assert memoryRes==77055967
	catch e
		throw(ErrorException("$e memoryRes=$memoryRes"))
	end
	println("memory result: $memoryRes")
	@debug "real input ok"

end

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(())
}

I did part 2 live with the python interactive shell. I deleted all the stuff where I was just exploring ideas.

part 1:

import re

def multiply_and_add(data: "str") -> int:
    digit_matches = re.findall(r"mul\(\d{0,3},\d{0,3}\)", data)
    result = 0
    for _ in digit_matches:
        first = _.split("(")[1].split(")")[0].split(",")[0]
        second = _.split("(")[1].split(")")[0].split(",")[1]
        result += int(first) * int(second)

    return result

with open("input") as file:
    data = file.read()


answer = multiply_and_add(data)
print(answer)

part 2:

Python 3.11.2 (main, Aug 26 2024, 07:20:54) [GCC 12.2.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import solution2
<re.Match object; span=(647, 651), match='do()'>
>>> from solution2 import *
>>> split_on_dont = data.split("don't()")
>>> valid = []
>>> valid.append(split_on_dont[0])
>>> for substring in split_on_dont[1:]:
...     subsubstrings = substring.split("do()", maxsplit=1)
...     for subsubstring in subsubstrings[1:]:
...             valid.append(subsubstring)
...
>>> answer = 0
>>> for _ in valid:
...     answer += multiply_and_add(_)
...
>>> answer
103811193