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

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.

Python

I’m surprised I don’t see more people taking advantage of eval I thought it was pretty slick.

import operator 
import re

with open('input.txt', 'r') as file:
    memory = file.read()

matches = re.findall("mul\(\d{1,3},\d{1,3}\)|don't\(\)|do\(\)", memory)

enabled = 1
filtered_matches = []
for instruction in matches:
    if instruction == "don't()":
        enabled = 0
        continue
    elif instruction == "do()":
        enabled = 1
        continue
    elif enabled: 
        filtered_matches.append(instruction)
multipled = [eval(f"operator.{x}") for x in filtered_matches]
print(sum(multiples))

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()
}

´´´

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)

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 })
}

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

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.

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)

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+)\\)"

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

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))

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()

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

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