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

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))

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

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.

deleted by creator

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)

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

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
}

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+)\\)"

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 m = match[4..^2].split(',')
      let mult = m[0].parseInt * m[1].parseInt
      result.part1 += mult
      if allow: result.part2 += mult

Codeberg repo

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))