Day 3: Mull It Over
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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, andre
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.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
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 ofMul !Int !Int
?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
Mul
s 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.
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.
Nim
From a first glance it was obviously a regex problem.
I’m using tinyre here instead of stdlibre
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
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!
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)
Of course it’s point-free
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}"); }
Sorry for the delay posting this one, Ategon seemed to have it covered, so I forgot :D I will do better.
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)
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.
Sub was my first instinct too, but I got a bad answer and saw that my input had unbalanced do/don’t.
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))
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!
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 }) }
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
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
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))
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)