Day 4: Scratchcards


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  • mykl@lemmy.world
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    1 year ago

    I had to give Uiua another go today. (run it here)

    {"Card 1: 41 48 83 86 17 | 83 86  6 31 17  9 48 53"
     "Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19"
     "Card 3:  1 21 53 59 44 | 69 82 63 72 16 21 14  1"
     "Card 4: 41 92 73 84 69 | 59 84 76 51 58  5 54 83"
     "Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36"
     "Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11"}
    
    LtoDec ← ∧(+ ×10:) :0
    StoDec ← LtoDec▽≥0. ▽≤9. -@0
    
    # Split on spaces, drop dross, parse ints
    ≡(⊜□≠0.⊐∵(StoDec)↘ 2⊜(□)≠@\s.⊔)
    
    # Find matches
    ≡(/+/+⊠(⌕)⊃(⊔⊢↙ 1)(⊔⊢↙¯1))
    
    # part 1
    /+ⁿ:2-1 ▽±..
    
    # part 2 - start with matches and initial counts
    =..:
    # len times: get 1st of each, rotate both, add new counts
    ⍥(⬚0+↯: ⊙⊙∩(↻1) ⊙:∩(⊢.))⧻.
    /+⊙;
    
  • Gobbel2000@feddit.de
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    1 year ago

    Rust

    This one wasn’t too bad. The example for part 2 even tells you how to process everything by visiting each card once in order. Another option could be to recursively look at all won copies, but that’s probably much less efficient.

      • cacheson@kbin.social
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        1 year ago

        I’m rather spoiled by python, so I feel like it could be more elegant. xD

        But yeah, I do like how this one turned out, and nim runs a whole lot faster than python does. I really like nim’s “method call syntax”. Instead of having methods associated with an individual type, you can just call any procedure as x.f(remaining_args) to call f with x as its first argument. Makes it easy to chain procedures. Since nim is strongly typed, it’ll know which procedure you mean to use by the signature.

        • Andy
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          1 year ago

          Aside from the general conciseness, the “universal function call syntax” is my favorite aspect of nim.

          If you want to take chaining procedures to the next level, try a concatenative language like Factor (I have a day 4 solution in this thread – with no assignment to variables).

          I also suggest having a look at Roc if you want a functional programming adventure, which offers great chaining syntax, a very friendly community, and is in an exciting development phase.

          • cacheson@kbin.social
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            1 year ago

            Thank you, I’ll keep those in mind. Functional programming seems interesting to me, but I don’t have any practical experience with it. At some point I want to learn one of the languages that are dedicated to it. Nim does have some features for enabling a functional style, but the overall flexibility of the language probably makes it harder to learn said style.

  • janAkali@lemmy.one
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    1 year ago

    LANGUAGE: Nim

    Welcome to the advent of parsing!
    Took me a lot more time than it should (Please, don’t check prior commits 😅).

    day_04.nim

  • Leo Uino@lemmy.sdf.org
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    1 year ago

    Haskell

    11:39 – I spent most of the time reading the scoring rules and (as usual) writing a parser…

    import Control.Monad
    import Data.Bifunctor
    import Data.List
    
    readCard :: String -> ([Int], [Int])
    readCard =
      join bimap (map read) . second tail . break (== "|") . words . tail . dropWhile (/= ':')
    
    countShared = length . uncurry intersect
    
    part1 = sum . map ((\n -> if n > 0 then 2 ^ (n - 1) else 0) . countShared)
    
    part2 = sum . foldr ((\n a -> 1 + sum (take n a) : a) . countShared) []
    
    main = do
      input <- map readCard . lines <$> readFile "input04"
      print $ part1 input
      print $ part2 input
    
  • mykl@lemmy.world
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    1 year ago

    Dart Solution

    Okay, that’s more like it. Simple parsing and a bit of recursion, and fits on one screen. Perfect for day 4 :-)

    int matchCount(String line) => line
        .split(RegExp('[:|]'))
        .skip(1)
        .map((ee) => ee.trim().split(RegExp(r'\s+')).map(int.parse))
        .map((e) => e.toSet())
        .reduce((s, t) => s.intersection(t))
        .length;
    
    late List matches;
    late List totals;
    
    int scoreFor(int ix) {
      if (totals[ix] != 0) return totals[ix];
      return totals[ix] =
          [for (var m in 0.to(matches[ix])) scoreFor(m + ix + 1) + 1].sum;
    }
    
    part1(List lines) =>
        lines.map((e) => pow(2, matchCount(e) - 1).toInt()).sum;
    
    part2(List lines) {
      matches = [for (var e in lines) matchCount(e)];
      totals = List.filled(matches.length, 0);
      return matches.length + 0.to(matches.length).map(scoreFor).sum;
    }
    
  • corristo
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    1 year ago

    APL

    I’m using this years’ AoC to learn (Dyalog) APL, so this is probably terrible code. I’m happy to receive pointers for improvement, particularly if there is a way to write the same logic with tacit functions or inner/outer products that I missed.

    input←⊃⎕NGET'inputs/day4.txt'1
    num_matches←'Card [ \d]+: ([ 0-9]+) \| ([ 0-9]+)'⎕S{≢↑∩/0~⍨¨{,⎕CSV⍠'Separator' ' '⊢⍵'S'3}¨⍵.(1↓Lengths↑¨Offsets↓¨⊂Block)} input
    ⎕←+/2*1-⍨0~⍨num_matches ⍝ part 1
    ⎕←+/{⍺←0 ⋄ ⍺=≢⍵:⍵ ⋄ (⍺+1)∇⍵ + (≢⍵)↑∊((⍺+1)⍴0)(num_matches[⍺]⍴⍵[⍺])((≢⍵)⍴0)}(≢num_matches)⍴1 ⍝ part 2
    
    • mykl@lemmy.world
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      1 year ago

      I just posted a solution in Uiua, which is also probably equally terrible, but if you squint you can see some similarities in our approaches.

      • corristo
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        1 year ago

        I haven’t heard of Uiua before, but I can read some things :D I like the idea of rotating the vector instead of manually padding it with the required number of leading zeroes!

        • mykl@lemmy.world
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          1 year ago

          I think it’s only a few months old. I’ve enjoyed playing with it because it allows me to use stack manipulation as an alternative to combinators and every symbol has a fixed arity both of which make it feel a lot more accessible to me.

          I was very pleased with myself when I thought of that rotation trick :-)

    • Nighed@sffa.community
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      1 year ago

      Edit: Sorry, should have read your code first, you made it work too. if it works it works, Recursive solutions just click for me over other solutions.

      I made the recursion work, went to a depth of 24 for my input set.

      Recursive C#
      internal class Day4Task2 : IRunnable
          {
              private Regex _regex = new Regex("Card\\s*\\d*: ([\\d\\s]{2} )*\\|( [\\d\\s]{2})*");
              private Dictionary _matchCountCache = new Dictionary();
              private int _maxDepth = 0;
      
              public void Run()
              {
                  var inputLines = File.ReadAllLines("Days/Four/Day4Input.txt");
                  int sumScore = 0;
      
                  for (int i = 0; i < inputLines.Length; i++)
                  {
                      sumScore += ScoreCard(i, inputLines, 0);
                      Console.WriteLine("!!!" + i + "!!!");
                  }
      
                  Console.WriteLine("Sum:"+sumScore.ToString());
                  Console.WriteLine("Max Recursion Depth:"+ _maxDepth.ToString());
              }
      
              private int ScoreCard(int lineId, string[] inputLines, int depth)
              {
                  if( depth > _maxDepth )
                  {
                      _maxDepth = depth;
                  }
      
                  if(lineId >= inputLines.Length)
                  {
                      return 0;
                  }
      
                  int matchCount = 0;
      
                  if (!_matchCountCache.ContainsKey(lineId)) {
      
                      var winningSet = new HashSet();
                      var matches = _regex.Match(inputLines[lineId]);
                      foreach (Capture capture in matches.Groups[1].Captures)
                      {
                          winningSet.Add(capture.Value.Trim());
                      }
      
                      foreach (Capture capture in matches.Groups[2].Captures)
                      {
                          if (winningSet.Contains(capture.Value.Trim()))
                          {
                              matchCount++;
                          }
                      }
      
                      _matchCountCache[lineId] = matchCount;
                  }
      
                  matchCount = _matchCountCache[lineId];
      
                  int totalCards = 1;
                  while(matchCount > 0)
                  {
                      totalCards += ScoreCard(lineId+matchCount, inputLines, depth+1);
                      matchCount--;
                  }
                  //Console.WriteLine("Finished processing id: " + lineId + " Sum is: " + totalCards);
                  return totalCards;
              }
          }
      
  • capitalpb
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    1 year ago

    I enjoyed this one. It was a nice simple break after Days 1 and 3; the type of basic puzzle I expect from the first few days of Advent of Code. Pretty simple logic in this one, I don’t think I would change too much. I’m sure I’ll find a way to clean up how it’s written a bit, but I’m happy with this one today.

    https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day04.rs

    struct Scratchcard {
        winning_numbers: HashSet,
        player_numbers: HashSet,
    }
    
    impl Scratchcard {
        fn from(input: &str) -> Scratchcard {
            let (_, numbers) = input.split_once(':').unwrap();
            let (winning_numbers, player_numbers) = numbers.split_once('|').unwrap();
            let winning_numbers = winning_numbers
                .split_ascii_whitespace()
                .filter_map(|number| number.parse::().ok())
                .collect::>();
            let player_numbers = player_numbers
                .split_ascii_whitespace()
                .filter_map(|number| number.parse::().ok())
                .collect::>();
    
            Scratchcard {
                winning_numbers,
                player_numbers,
            }
        }
    
        fn matches(&self) -> u32 {
            self.winning_numbers
                .intersection(&self.player_numbers)
                .count() as u32
        }
    }
    
    pub struct Day04;
    
    impl Solver for Day04 {
        fn star_one(&self, input: &str) -> String {
            input
                .lines()
                .map(Scratchcard::from)
                .map(|card| {
                    let matches = card.matches();
                    if matches == 0 {
                        0
                    } else {
                        2u32.pow(matches - 1)
                    }
                })
                .sum::()
                .to_string()
        }
    
        fn star_two(&self, input: &str) -> String {
            let cards: Vec = input.lines().map(Scratchcard::from).collect();
            let mut card_counts = vec![1usize; cards.len()];
    
            for card_number in 0..cards.len() {
                let matches = cards[card_number].matches();
    
                if matches == 0 {
                    continue;
                }
    
                for i in 1..=matches {
                    card_counts[card_number + i as usize] += card_counts[card_number];
                }
            }
    
            card_counts.iter().sum::().to_string()
        }
    }
    
  • __init__
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    1 year ago

    (python) Much easier than day 3.

    code
    import pathlib
    
    base_dir = pathlib.Path(__file__).parent
    filename = base_dir / "day4_input.txt"
    
    with open(base_dir / filename) as f:
        lines = f.read().splitlines()
    
    score = 0
    
    extra_cards = [0 for _ in lines]
    n_cards = [1 for _ in lines]
    
    for i, line in enumerate(lines):
        _, numbers = line.split(":")
        winning, have = numbers.split(" | ")
    
        winning_numbers = {int(n) for n in winning.split()}
        have_numbers = {int(n) for n in have.split()}
    
        have_winning_numbers = winning_numbers & have_numbers
        n_matches = len(have_winning_numbers)
    
        if n_matches:
            score += 2 ** (n_matches - 1)
    
        j = i + 1
        for _ in range(n_matches):
            if j >= len(lines):
                break
            n_cards[j] += n_cards[i]
            j += 1
    
    answer_p1 = score
    print(f"{answer_p1=}")
    
    answer_p2 = sum(n_cards)
    print(f"{answer_p2=}")
    
  • Andy
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    11 months ago

    Factor on github (with comments and imports):

    : line>cards ( line -- winning-nums player-nums )
      ":|" split rest
      [
        [ CHAR: space = ] trim
        split-words harvest [ string>number ] map
      ] map first2
    ;
    
    : points ( winning-nums player-nums -- n )
      intersect length
      dup 0 > [ 1 - 2^ ] when
    ;
    
    : part1 ( -- )
      "vocab:aoc-2023/day04/input.txt" utf8 file-lines
      [ line>cards points ] map-sum .
    ;
    
    : follow-card ( i commons -- n )
      [ 1 ] 2dip
      2dup nth swapd
      over + (a..b]
      [ over follow-card ] map-sum
      nip +
    ;
    
    : part2 ( -- )
      "vocab:aoc-2023/day04/input.txt" utf8 file-lines
      [ line>cards intersect length ] map
      dup length <iota> swap '[ _ follow-card ]
      map-sum .
    ;
    
  • hades@lemm.ee
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    1 year ago

    Python

    Questions and feedback welcome!

    import collections
    import re
    
    from .solver import Solver
    
    class Day04(Solver):
      def __init__(self):
        super().__init__(4)
        self.cards = []
    
      def presolve(self, input: str):
        lines = input.rstrip().split('\n')
        self.cards = []
        for line in lines:
          left, right = re.split(r' +\| +', re.split(': +', line)[1])
          left, right = map(int, re.split(' +', left)), map(int, re.split(' +', right))
          self.cards.append((list(left), list(right)))
    
      def solve_first_star(self):
        points = 0
        for winning, having in self.cards:
          matches = len(set(winning) &amp; set(having))
          if not matches:
            continue
          points += 1 &lt;&lt; (matches - 1)
        return points
    
      def solve_second_star(self):
        factors = collections.defaultdict(lambda: 1)
        count = 0
        for i, (winning, having) in enumerate(self.cards):
          count += factors[i]
          matches = len(set(winning) &amp; set(having))
          if not matches:
            continue
          for j in range(i + 1, i + 1 + matches):
            factors[j] = factors[j] + factors[i]
        return count
    
  • sjmulder@lemmy.sdf.org
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    1 year ago

    Language: C

    Another day of parsing, another day of strsep() to the rescue. Today was one of those satisfying days where the puzzle text is complicated but the solution is simple once well understood.

    GitHub link

    Code (29 lines)
    int main()
    {
    	char line[128], *rest, *tok;
    	int nextra[200]={0}, nums[10], nnums;
    	int p1=0,p2=0, id,val,nmatch, i;
    
    	for (id=0; (rest = fgets(line, sizeof(line), stdin)); id++) {
    		nnums = nmatch = 0;
    
    		while ((tok = strsep(&amp;rest, " ")) &amp;&amp; !strchr(tok, ':'))
    			;
    		while ((tok = strsep(&amp;rest, " ")) &amp;&amp; !strchr(tok, '|'))
    			if ((val = atoi(tok)))
    				nums[nnums++] = val;
    		while ((tok = strsep(&amp;rest, " ")))
    			if ((val = atoi(tok)))
    				for (i=0; i
        • mykl@lemmy.world
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          1 year ago

          Ohh that’s interesting, I’ve seen a few comments about mishandling of special chars in code blocks and assumed it was a server issue, maybe it’s fixed in newer releases or perhaps it’s client side.