Sorry for the delay posting this one, Ategon seemed to have it covered, so I forgot :D I will do better.
Advent Of Code
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AoC 2024
Solution Threads
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---|---|---|---|---|---|---|
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console.log('Hello World')
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))
Really cool trick. I did a bunch of regex matching that I'm sure I won't remember how it works few weeks from now, this is so much readable
Nice, sometimes a few extra linebreaks can do the trick...
My first insinct was similar, add line breaks to the do and dont modifiers. But I got toa caught up thinking id have to keep track of the added characters, I wound up just abusing split()-
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
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!
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.
Code
#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);
}
Got the code a little shorter:
Code
#include "common.h"
static int
parse_mul(const char **stringp, int *ap, int *bp)
{
const char *cur = *stringp, *end;
if (strncmp(cur, "mul(", 4)) { return 0; } cur += 4;
*ap = (int)strtol(cur, (char **)&end, 10);
if (end == cur) { return 0; } cur = end;
if (*cur != ',') { return 0; } cur += 1;
*bp = (int)strtol(cur, (char **)&end, 10);
if (end == cur) { return 0; } cur = end;
if (*cur != ')') { return 0; } cur += 1;
*stringp = cur;
return 1;
}
int
main(int argc, char **argv)
{
static char buf[32*1024];
const char *p;
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 (p = buf; *p; )
if (parse_mul(&p, &a, &b)) { p1 += a*b; p2 += a*b*!dont; }
else if (!strncmp(p, "do()", 4)) { dont = 0; p += 4; }
else if (!strncmp(p, "don't()", 7)) { dont = 1; p += 7; }
else p++;
printf("03: %d %d\n", p1, p2);
}
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}");
}
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
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)
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!
DataPβ β $ xmul(2,4)%&mul[3,7]!@^do_not_mul(5,5)+mul(32,64]then(mul(11,8)mul(8,5))
DataPβ β $ xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))
GetMul β $ mul\((\d{1,3}),(\d{1,3})\)
GetMulDoDont β $ mul\(\d{1,3},\d{1,3}\)|do\(\)|don\'t\(\)
&p/+β‘(/Γβ‘ββ1)regexGetMul DataPβ # Part 1
# Build an accumulator to track running state of do/don't
Filter β β1β:β§(β£(0 Β°"don"|1 Β°"do("|.β)) :1β‘(β3Β°β‘)
β‘β’ regex GetMulDoDont DataPβ
β½βΈβ‘β(β"mul"β3)β½βΈFilter # Apply Filter, remove the spare 'do's
&p/+β‘β(/Γβ‘βββ1β’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 ;
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 of Mul !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.
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)
Go
Part 1, just find the regex groups, parse to int, and done.
Part 1
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
Part2
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)
}
I also used Go - my solution for part 1 was essentially identical to yours. I went a different route for part 2 that I think ended up being simpler though.
I just prepended do()
and don't()
to the original regex with a |
, that way it captured all 3 in order and I just looped through all the matches once and toggled the isEnabled
flag accordingly.
Always interesting to see how other people tackle the same problem!
Part 2 Code
func part2() {
filePath := "input.txt"
file, _ := os.Open(filePath)
defer file.Close()
pattern := regexp.MustCompile(`do\(\)|don't\(\)|mul\((\d{1,3}),(\d{1,3})\)`)
productSum := 0
isEnabled := true
scanner := bufio.NewScanner(file)
for scanner.Scan() {
line := scanner.Text()
matches := pattern.FindAllStringSubmatch(line, -1)
for _, match := range matches {
if match[0] == "do()" {
isEnabled = true
} else if match[0] == "don't()" {
isEnabled = false
} else if isEnabled && len(match) == 3 {
n, _ := strconv.Atoi(match[1])
m, _ := strconv.Atoi(match[2])
productSum += n * m
}
}
}
fmt.Println("Total: ", productSum)
}
Honestly this is soo much better, I'm not proud of my code at all haha. Thanks for sharing, definitely adding that | to my bag of tricks
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"
/+β‘/Γβ‘ββ‘β1regex "mul\\((\\d+),(\\d+)\\)"
Part 2:
Filter β ββββ‘β
""βΈβ¦·Β°β‘
.&fras "day3/input.txt"
β§Filterβregex"don't\\(\\)?(.*?)(?:do\\(\\)|$)"
/+β‘/Γβ‘ββ‘β1regex "mul\\((\\d+),(\\d+)\\)"
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)
}
doit
I had to check if do
was reserved since I couldn't remember seeing it in the language, and it's in the reserved for future use section.
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
}
You can avoid having to escape the backslashes in regexps by using multiline strings:
val pattern = """mul\((\d{1,3}),(\d{1,3})\)""".toRegex()
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 })
}
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!
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()
}
´´´
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!();
Our part1's are basically identical, but for part 2 I just regex deleted everything between don
t() and
do() `, after sanitizing out the newlines.
I guess it would have failed if my test data had a mul after the last dont, but it worked out, so good enough :)
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()}")
Part 2 is currently not working, and I canβt figure out why. Iβm trying to just remove everything from βdonβt()β to βdo()β and just pass the rest through the working solution for part 1. Should work, right?
I think I had the same issue. Consider what happens if there isn't a do() after a don't().
Ah, yes, that's it. The lazy solution would be to add a "do()" to the end of the input, right? Haha
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>Β».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>Β».Int});
my $part-two-solution = @enabledMuls.map({[*] $_.List}).sum;
say "part 2: $part-two-solution";
}
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.
I did wonder if that might be the case, I must have been lucky with my input.
Uiua
Regex my beloved <3
Run with example input here
FindMul β regex "mul\\((\\d+),(\\d+)\\)"
PartOne β (
&rs β &fo "input-3.txt"
FindMul
/+β‘(ΓΒ°βββ1_2)
)
IdDont β ββ‘"don't()"β
PartTwo β (
&rs β &fo "input-3.txt"
regex "mul\\(\\d+,\\d+\\)|do\\(\\)|don't\\(\\)"
β’(IdDont.
β1βββ
ββ‘"do()"β.
ββ1β
| IdDont.
β ⧻,
)
β½β=0ββ‘"do()".
β‘(ΓΒ°βββ1_2βFindMul)β
/+
)
&p "Day 3:"
&pf "Part 1: "
&p PartOne
&pf "Part 2: "
&p PartTwo
C#
public partial class Day03 : Solver
{
[GeneratedRegex(@"mul[(](\d{1,3}),(\d{1,3})[)]")]
private partial Regex mulRegex();
[GeneratedRegex(@"(do)[(][)]|(don't)[(][)]|(mul)[(](\d{1,3}),(\d{1,3})[)]")]
private partial Regex fullRegex();
private string input;
public void Presolve(string input)
{
this.input = input.Trim();
}
public string SolveFirst() => mulRegex().Matches(input)
.Select(match => int.Parse(match.Groups[1].Value) * int.Parse(match.Groups[2].Value))
.Sum().ToString();
public string SolveSecond()
{
bool enabled = true;
int sum = 0;
foreach (Match match in fullRegex().Matches(input)) {
if (match.Groups[1].Length > 0) {
enabled = true;
} else if (match.Groups[2].Length > 0) {
enabled = false;
} else if (enabled) {
sum += int.Parse(match.Groups[4].Value) * int.Parse(match.Groups[5].Value);
}
}
return sum.ToString();
}
}
python
solution
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()
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 π€¦ββοΈ
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(())
}
TypeScript
Solution
import { AdventOfCodeSolutionFunction } from "./solutions";
export const solution_3: AdventOfCodeSolutionFunction = (input) => {
const mul_regex = /mul\((\d+),(\d+)\)/g; // mul()
const do_regex = /do\(\)/g; // do()
const do_not_regex = /don\'t\(\)/g; // don't()
const doLength = "do()".length;
const doNotLength = "don't()".length;
let input_copy = "" + input;
let part_1 = 0;
let part_2 = 0;
let isEnabled = true;
while (true) {
const nextMul = input_copy.search(mul_regex);
const nextDo = input_copy.search(do_regex);
const nextDoNot = input_copy.search(do_not_regex);
let pointer = Number.POSITIVE_INFINITY;
// find the smallest while ignoring items that are not found
if (nextMul != -1)
pointer = Math.min(pointer, nextMul);
if (nextDo != -1)
pointer = Math.min(pointer, nextDo);
if (nextDoNot != -1)
pointer = Math.min(pointer, nextDoNot);
// no matches
if (pointer == Number.POSITIVE_INFINITY)
break
// handle found command
switch (pointer) {
case nextDo: {
pointer += doLength;
isEnabled = true;
break;
}
case nextDoNot: {
pointer += doNotLength;
isEnabled = false;
break;
}
case nextMul: {
const res = input_copy.matchAll(mul_regex).next().value;
if (!res) {
// this should never happen but here's an escape hatch
throw Error("regex result is undefined or null");
}
// add the length of the whole capture to the pointer
pointer += res[0].length;
// multiply capture groups
const comp = Number(res[1]) * Number(res[2]);
// part 1 sum
part_1 += comp;
// part 2 sum
if(isEnabled)
part_2 += comp;
break;
}
}
// shorten the start of the string
input_copy = input_copy.slice(pointer);
}
return {
part_1,
part_2,
};
}
This one was harder but still. I feel like I can improve it for sure :)
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