Day 3: Mull It Over

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FAQ

  • Ananace ( ace@lemmy.ananace.dev ) 
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    2 years ago

    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 instructions = new List();
    
    public void Input(IEnumerable 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}");
    }
    
  • Amy ( lwhjp@lemmy.sdf.org ) 
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    2 years ago

    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
    
  • Andy ( Andy@programming.dev ) 
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    2 years ago

    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 ;
    
  • janAkali ( janAkali@lemmy.one ) 
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    2 years ago

    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

  • sjmulder ( sjmulder@lemmy.sdf.org ) 
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    2 years ago

    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);
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day03.c

    • 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);
      }
      
  • madmo ( madmo@programming.dev ) 
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    2 years ago

    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, Token)), nom::Err>>;
    
    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)
    }
    
  • 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::().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
    }
    
    • Faresh ( Faresh@lemmy.ml ) 
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      2 years ago

      You can avoid having to escape the backslashes in regexps by using multiline strings:

      val pattern = """mul\((\d{1,3}),(\d{1,3})\)""".toRegex()
      
  • ystael ( ystael@beehaw.org ) 
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    2 years ago

    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
    
  • 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)
    }
    
    • gothink ( gothink@lemmy.ca ) 
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      2 years ago

      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)
      }
      
  • 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::().unwrap();
        let b = cap.get(2).unwrap().as_str().parse::().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::();
        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 dont() anddo() `, 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 :)

  • 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 })
    }
    
  • urquell ( urquell@lemm.ee ) 
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    2 years ago

    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)