Day 5: Print Queue

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FAQ

  • RagingHungryPanda@lemm.ee
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    1 day ago

    I’ve got a “smart” solution and a really dumb one. I’ll start with the smart one (incomplete but you can infer). I did four different ways to try to get it faster, less memory, etc.

    // this is from a nuget package. My Mathy roommate told me this was a topological sort.
    // It's also my preferred, since it'd perform better on larger data sets.
    return lines
        .AsParallel()
        .Where(line => !IsInOrder(GetSoonestOccurrences(line), aggregateRules))
        .Sum(line => line.StableOrderTopologicallyBy(
                getDependencies: page =>
                    aggregateRules.TryGetValue(page, out var mustPreceed) ? mustPreceed.Intersect(line) : Enumerable.Empty<Page>())
            .Middle()
        );
    

    The dumb solution. These comparisons aren’t fully transitive. I can’t believe it works.

    public static SortedSet<Page> Sort3(Page[] line,
        Dictionary<Page, System.Collections.Generic.HashSet<Page>> rules)
    {
        // how the hell is this working?
        var sorted = new SortedSet<Page>(new Sort3Comparer(rules));
        foreach (var page in line)
            sorted.Add(page);
        return sorted;
    }
    
    public static Page[] OrderBy(Page[] line, Dictionary<Page, System.Collections.Generic.HashSet<Page>> rules)
    {
        return line.OrderBy(identity, new Sort3Comparer(rules)).ToArray();
    }
    
    sealed class Sort3Comparer : IComparer<Page>
    {
        private readonly Dictionary<Page, System.Collections.Generic.HashSet<Page>> _rules;
    
        public Sort3Comparer(Dictionary<Page, System.Collections.Generic.HashSet<Page>> rules) => _rules = rules;
    
        public int Compare(Page x, Page y)
        {
            if (_rules.TryGetValue(x, out var xrules))
            {
                if (xrules.Contains(y))
                    return -1;
            }
    
            if (_rules.TryGetValue(y, out var yrules))
            {
                if (yrules.Contains(x))
                    return 1;
            }
    
            return 0;
        }
    }
    
    Method Mean Error StdDev Gen0 Gen1 Allocated
    Part2_UsingList (literally just Insert) 660.3 us 12.87 us 23.20 us 187.5000 35.1563 1144.86 KB
    Part2_TrackLinkedList (wrong now) 1,559.7 us 6.91 us 6.46 us 128.9063 21.4844 795.03 KB
    Part2_TopologicalSort 732.3 us 13.97 us 16.09 us 285.1563 61.5234 1718.36 KB
    Part2_SortedSet 309.1 us 4.13 us 3.45 us 54.1992 10.2539 328.97 KB
    Part2_OrderBy 304.5 us 6.09 us 9.11 us 48.8281 7.8125 301.29 KB
  • Andy@programming.dev
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    7 days ago

    Factor

    : get-input ( -- rules updates )
      "vocab:aoc-2024/05/input.txt" utf8 file-lines
      { "" } split1
      "|" "," [ '[ [ _ split ] map ] ] bi@ bi* ;
    
    : relevant-rules ( rules update -- rules' )
      '[ [ _ in? ] all? ] filter ;
    
    : compliant? ( rules update -- ? )
      [ relevant-rules ] keep-under
      [ [ index* ] with map first2 < ] with all? ;
    
    : middle-number ( update -- n )
      dup length 2 /i nth-of string>number ;
    
    : part1 ( -- n )
      get-input
      [ compliant? ] with
      [ middle-number ] filter-map sum ;
    
    : compare-pages ( rules page1 page2 -- <=> )
      [ 2array relevant-rules ] keep-under
      [ drop +eq+ ] [ first index zero? +gt+ +lt+ ? ] if-empty ;
    
    : correct-update ( rules update -- update' )
      [ swapd compare-pages ] with sort-with ;
    
    : part2 ( -- n )
      get-input dupd
      [ compliant? ] with reject
      [ correct-update middle-number ] with map-sum ;
    

    on GitHub

  • Quant@programming.dev
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    5 days ago

    Uiua

    This is the first one that caused me some headache because I didn’t read the instructions carefully enough.
    I kept trying to create a sorted list for when all available pages were used, which got me stuck in an endless loop.

    Another fun part was figuring out to use memberof (∈) instead of find (⌕) in the last line of FindNext. So much time spent on debugging other areas of the code

    Run with example input here

    FindNext ← ⊙(
      ⊡1⍉,
      ⊃▽(▽¬)⊸∈
      ⊙⊙(⊡0⍉.)
      :⊙(⟜(▽¬∈))
    )
    
    # find the order of pages for a given set of rules
    FindOrder ← (
      ◴♭.
      []
      ⍢(⊂FindNext|⋅(>1⧻))
      ⊙◌⊂
    )
    
    PartOne ← (
      &rs ∞ &fo "input-5.txt"
      ∩°□°⊟⊜□¬⌕"\n\n".
      ⊙(⊜(□⊜⋕≠@,.)≠@\n.↘1)
      ⊜(⊜⋕≠@|.)≠@\n.
    
      ⊙.
      ¤
      ⊞(◡(°□:)
        ⟜:⊙(°⊟⍉)
        =2+∩∈
        ▽
        FindOrder
        ⊸≍°□:
        ⊙◌
      )
      ≡◇(⊡⌊÷2⧻.)▽♭
      /+
    )
    
    PartTwo ← (
      &rs ∞ &fo "input-5.txt"
      ∩°□°⊟⊜□¬⌕"\n\n".
      ⊙(⊜(□⊜⋕≠@,.)≠@\n.↘1)
      ⊜(⊜⋕≠@|.)≠@\n.
      ⊙.
      ⍜¤⊞(
        ◡(°□:)
        ⟜:⊙(°⊟⍉)
        =2+∩∈
        ▽
        FindOrder
        ⊸≍°□:
        ⊟∩□
      )
      ⊙◌
      ⊃(⊡0)(⊡1)⍉
      ≡◇(⊡⌊÷2⧻.)▽¬≡°□
      /+
    )
    
    &p "Day 5:"
    &pf "Part 1: "
    &p PartOne
    &pf "Part 2: "
    &p PartTwo
    
  • mykl@lemmy.world
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    6 days ago

    Uiua

    Well it’s still today here, and this is how I spent my evening. It’s not pretty or maybe even good, but it works on the test data…

    spoiler

    Uses Kahn’s algorithm with simplifying assumptions based on the helpful nature of the data.

    Try it here

    Data  ()⊸≠@\n "47|53\n97|13\n97|61\n97|47\n75|29\n61|13\n75|53\n29|13\n97|29\n53|29\n61|53\n97|53\n61|29\n47|13\n75|47\n97|75\n47|61\n75|61\n47|29\n75|13\n53|13\n\n75,47,61,53,29\n97,61,53,29,13\n75,29,13\n75,97,47,61,53\n61,13,29\n97,13,75,29,47"
    Rs    ≡◇(⊜⋕⊸≠@|)▽⊸≡◇(⧻⊚⌕@|)Data
    Ps    ≡⍚(⊜⋕⊸≠@,)▽⊸≡◇(¬⧻⊚⌕@|)Data
    
    NoPred   ⊢▽:((=0/+⌕)⊙¤)◴♭⟜≡⊣                # Find entry without predecessors.
    GetLead  (:((¬/+=))⊙¤)NoPred             # Remove that leading entry.
    Rules    ⇌⊂⊃(⇌⊢°□⊢|≡°□↘1)[□⍢(GetLead|≠1)] Rs # Repeatedly find rule without predecessors (Kaaaaaahn!).
    
    Sorted    ⊏⍏⊗,Rules
    IsSorted  /×>0≡/-◫2⊗°□: Rules
    MidVal    :(⌊÷ 2)
    
    ⇌⊕□⊸≡IsSorted Ps        # Group by whether the pages are in sort order.
    ≡◇(/+≡◇(MidVal Sorted)) # Find midpoints and sum.
    
    
    • mykl@lemmy.world
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      6 days ago

      Oh my. I just watched yernab’s video, and this becomes so much easier:

      # Order is totally specified, so sort by number of predecessors,
      # check to see which were already sorted, then group and sum each group.
      Data  (□⊜□⊸≠@\n)(¬⦷"\n\n")"47|53\n97|13\n97|61\n97|47\n75|29\n61|13\n75|53\n29|13\n97|29\n53|29\n61|53\n97|53\n61|29\n47|13\n75|47\n97|75\n47|61\n75|61\n47|29\n75|13\n53|13\n\n75,47,61,53,29\n97,61,53,29,13\n75,29,13\n75,97,47,61,53\n61,13,29\n97,13,75,29,47"
      Rs    ≡◇(⊜⋕⊸≠@|)°□⊢Data
      Ps    ≡⍚(⊜⋕⊸≠@,)°□⊣Data
      (/+≡◇(⊡⌊÷2⧻.))¬≡≍⟜:≡⍚(⊏⍏/+⊞(Rs)..).Ps
      
      • mykl@lemmy.world
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        6 days ago

        Ah, but the terseness of the code allows the beauty of the underlying algorithm to shine through :-)

  • Sparrow_1029@programming.dev
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    6 days ago

    Rust

    Real thinker. Messed around with a couple solutions before this one. The gist is to take all the pairwise comparisons given and record them for easy access in a ranking matrix.

    For the sample input, this grid would look like this (I left out all the non-present integers, but it would be a 98 x 98 grid where all the empty spaces are filled with Ordering::Equal):

       13 29 47 53 61 75 97
    13  =  >  >  >  >  >  >
    29  <  =  >  >  >  >  >
    47  <  <  =  <  <  >  >
    53  <  <  >  =  >  >  >
    61  <  <  >  <  =  >  >
    75  <  <  <  <  <  =  >
    97  <  <  <  <  <  <  =
    

    I discovered this can’t be used for a total order on the actual puzzle input because there were cycles in the pairs given (see how rust changed sort implementations as of 1.81). I used usize for convenience (I did it with u8 for all the pair values originally, but kept having to cast over and over as usize). Didn’t notice a performance difference, but I’m sure uses a bit more memory.

    Also I Liked the simple_grid crate a little better than the grid one. Will have to refactor that out at some point.

    solution
    use std::{cmp::Ordering, fs::read_to_string};
    
    use simple_grid::Grid;
    
    type Idx = (usize, usize);
    type Matrix = Grid<Ordering>;
    type Page = Vec<usize>;
    
    fn parse_input(input: &str) -> (Vec<Idx>, Vec<Page>) {
        let split: Vec<&str> = input.split("\n\n").collect();
        let (pair_str, page_str) = (split[0], split[1]);
        let pairs = parse_pairs(pair_str);
        let pages = parse_pages(page_str);
        (pairs, pages)
    }
    
    fn parse_pairs(input: &str) -> Vec<Idx> {
        input
            .lines()
            .map(|l| {
                let (a, b) = l.split_once('|').unwrap();
                (a.parse().unwrap(), b.parse().unwrap())
            })
            .collect()
    }
    
    fn parse_pages(input: &str) -> Vec<Page> {
        input
            .lines()
            .map(|l| -> Page {
                l.split(",")
                    .map(|d| d.parse::<usize>().expect("invalid digit"))
                    .collect()
            })
            .collect()
    }
    
    fn create_matrix(pairs: &[Idx]) -> Matrix {
        let max = *pairs
            .iter()
            .flat_map(|(a, b)| [a, b])
            .max()
            .expect("iterator is non-empty")
            + 1;
        let mut matrix = Grid::new(max, max, vec![Ordering::Equal; max * max]);
        for (a, b) in pairs {
            matrix.replace_cell((*a, *b), Ordering::Less);
            matrix.replace_cell((*b, *a), Ordering::Greater);
        }
        matrix
    }
    
    fn valid_pages(pages: &[Page], matrix: &Matrix) -> usize {
        pages
            .iter()
            .filter_map(|p| {
                if check_order(p, matrix) {
                    Some(p[p.len() / 2])
                } else {
                    None
                }
            })
            .sum()
    }
    
    fn fix_invalid_pages(pages: &mut [Page], matrix: &Matrix) -> usize {
        pages
            .iter_mut()
            .filter(|p| !check_order(p, matrix))
            .map(|v| {
                v.sort_by(|a, b| *matrix.get((*a, *b)).unwrap());
                v[v.len() / 2]
            })
            .sum()
    }
    
    fn check_order(page: &[usize], matrix: &Matrix) -> bool {
        page.is_sorted_by(|a, b| *matrix.get((*a, *b)).unwrap() == Ordering::Less)
    }
    
    pub fn solve() {
        let input = read_to_string("inputs/day05.txt").expect("read file");
        let (pairs, mut pages) = parse_input(&input);
        let matrix = create_matrix(&pairs);
        println!("Part 1: {}", valid_pages(&pages, &matrix));
        println!("Part 2: {}", fix_invalid_pages(&mut pages, &matrix));
    }
    

    On github

    *Edit: I did try switching to just using std::collections::HashMap, but it was 0.1 ms slower on average than using the simple_grid::GridVec[idx] access is faster maybe?

  • mykl@lemmy.world
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    7 days ago

    Dart

    A bit easier than I first thought it was going to be.

    I had a look at the Uiua discussion, and this one looks to be beyond my pay grade, so this will be it for today.

    import 'package:collection/collection.dart';
    import 'package:more/more.dart';
    
    (int, int) solve(List<String> lines) {
      var parts = lines.splitAfter((e) => e == '');
      var pred = SetMultimap.fromEntries(parts.first.skipLast(1).map((e) {
        var ps = e.split('|').map(int.parse);
        return MapEntry(ps.last, ps.first);
      }));
      ordering(a, b) => pred[a].contains(b) ? 1 : 0;
    
      var pageSets = parts.last.map((e) => e.split(',').map(int.parse).toList());
      var partn = pageSets.partition((ps) => ps.isSorted(ordering));
      return (
        partn.truthy.map((e) => e[e.length ~/ 2]).sum,
        partn.falsey.map((e) => (e..sort(ordering))[e.length ~/ 2]).sum
      );
    }
    
    part1(List<String> lines) => solve(lines).$1;
    part2(List<String> lines) => solve(lines).$2;
    
  • ystael@beehaw.org
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    7 days ago

    J

    This is a problem where J’s biases lead one to a very different solution from most of the others. The natural representation of a directed graph in J is an adjacency matrix, and sorting is specified in terms of a permutation to apply rather than in terms of a comparator: x /: y (respectively x \: y) determines the permutation that would put y in ascending (descending) order, then applies that permutation to x.

    data_file_name =: '5.data'
    lines =: cutopen fread data_file_name
    NB. manuals start with the first line where the index of a comma is &lt; 5
    start_of_manuals =: 1 i.~ 5 > ',' i.~"1 > lines
    NB. ". can't parse the | so replace it with a space
    edges =: ". (' ' &amp; (2}))"1 > start_of_manuals {. lines
    NB. don't unbox and parse yet because they aren't all the same length
    manuals =: start_of_manuals }. lines
    max_page =: >./ , edges
    NB. adjacency matrix of the page partial ordering; e.i. makes identity matrix
    adjacency =: 1 (&lt; edges)} e. i. >: max_page
    NB. ordered line is true if line is ordered according to the adjacency matrix
    ordered =: monad define
       pages =. ". > y
       NB. index pairs 0 &lt;: i &lt; j &lt; n; box and raze to avoid array fill
       page_pairs =. ; (&lt; @: (,~"0 i.)"0) i. # pages
       */ adjacency {~ &lt;"1 pages {~ page_pairs
    )
    midpoint =: ({~ (&lt;. @: -: @: #)) @: ". @: >
    result1 =: +/ (ordered"0 * midpoint"0) manuals
    
    NB. toposort line yields the pages of line topologically sorted by adjacency
    NB. this is *not* a general topological sort but works for our restricted case:
    NB. we know that each individual manual will be totally ordered
    toposort =: monad define
       pages =. ". > y
       NB. for each page, count the pages which come after it, then sort descending
       pages \: +/"1 adjacency {~ &lt;"1 pages ,"0/ pages
    )
    NB. midpoint2 doesn't parse, but does remove trailing zeroes
    midpoint2 =: ({~ (&lt;. @: -: @: #)) @: ({.~ (i. &amp; 0))
    result2 =: +/ (1 - ordered"0 manuals) * midpoint2"1 toposort"0 manuals
    
  • wer2@lemm.ee
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    6 days ago

    Lisp

    Part 1 and 2
    
    (defun p1-process-rules (line)
      (mapcar #'parse-integer (uiop:split-string line :separator "|")))
    
    (defun p1-process-pages (line)
      (mapcar #'parse-integer (uiop:split-string line :separator ",")))
    
    (defun middle (pages)
      (nth (floor (length pages) 2) pages))
    
    (defun check-rule-p (rule pages)
      (let ((p1 (position (car rule) pages))
            (p2 (position (cadr rule) pages)))
        (or (not p1) (not p2) (< p1 p2))))
    
    (defun ordered-p (pages rules)
      (loop for r in rules
            unless (check-rule-p r pages)
              return nil
            finally
               (return t)))
    
    (defun run-p1 (rules-file pages-file) 
      (let ((rules (read-file rules-file #'p1-process-rules))
            (pages (read-file pages-file #'p1-process-pages)))
        (loop for p in pages
              when (ordered-p p rules)
                sum (middle p)
              )))
    
    (defun fix-pages (rules pages)
      (sort pages (lambda (p1 p2) (ordered-p (list p1 p2) rules)) ))
    
    (defun run-p2 (rules-file pages-file) 
      (let ((rules (read-file rules-file #'p1-process-rules))
            (pages (read-file pages-file #'p1-process-pages)))
        (loop for p in pages
              unless (ordered-p p rules)
                sum (middle (fix-pages rules p))
              )))
    
    
  • Zarlin@lemmy.world
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    6 days ago

    Nim

    import ../aoc, strutils, sequtils, tables
    
    type
      Rules = ref Table[int, seq[int]]
    
    #check if an update sequence is valid
    proc valid(update:seq[int], rules:Rules):bool =
      for pi, p in update:
        for r in rules.getOrDefault(p):
          let ri = update.find(r)
          if ri != -1 and ri < pi:
            return false
      return true
    
    proc backtrack(p:int, index:int, update:seq[int], rules: Rules, sorted: var seq[int]):bool =
      if index == 0:
        sorted[index] = p
        return true
      
      for r in rules.getOrDefault(p):
        if r in update and r.backtrack(index-1, update, rules, sorted):
          sorted[index] = p
          return true
      
      return false
    
    #fix an invalid sequence
    proc fix(update:seq[int], rules: Rules):seq[int] =
      echo "fixing", update
      var sorted = newSeqWith(update.len, 0);
      for p in update:
        if p.backtrack(update.len-1, update, rules, sorted):
          return sorted
      return @[]
    
    proc solve*(input:string): array[2,int] =
      let parts = input.split("\r\n\r\n");
      
      let rulePairs = parts[0].splitLines.mapIt(it.strip.split('|').map(parseInt))
      let updates = parts[1].splitLines.mapIt(it.split(',').map(parseInt))
      
      # fill rules table
      var rules = new Rules
      for rp in rulePairs:
        if rules.hasKey(rp[0]):
          rules[rp[0]].add rp[1];
        else:
          rules[rp[0]] = @[rp[1]]
          
      # fill reverse rules table
      var backRules = new Rules
      for rp in rulePairs:
        if backRules.hasKey(rp[1]):
          backRules[rp[1]].add rp[0];
        else:
          backRules[rp[1]] = @[rp[0]]
      
      for u in updates:
        if u.valid(rules):
          result[0] += u[u.len div 2]
        else:
          let uf = u.fix(backRules)
          result[1] += uf[uf.len div 2]
    

    I thought of doing a sort at first, but dismissed it for some reason, so I came up with this slow and bulky recursive backtracking thing which traverses the rules as a graph until it reaches a depth equal to the given sequence. Not my finest work, but it does solve the puzzle :)

  • hosaka@programming.dev
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    6 days ago

    Zig

    const std = @import("std");
    const List = std.ArrayList;
    const Map = std.AutoHashMap;
    
    const tokenizeScalar = std.mem.tokenizeScalar;
    const splitScalar = std.mem.splitScalar;
    const parseInt = std.fmt.parseInt;
    const print = std.debug.print;
    const contains = std.mem.containsAtLeast;
    const eql = std.mem.eql;
    
    var gpa = std.heap.GeneralPurposeAllocator(.{}){};
    const alloc = gpa.allocator();
    
    const Answer = struct {
        middle_sum: i32,
        reordered_sum: i32,
    };
    
    pub fn solve(input: []const u8) !Answer {
        var rows = splitScalar(u8, input, '\n');
    
        // key is a page number and value is a
        // list of pages to be printed before it
        var rules = Map(i32, List(i32)).init(alloc);
        var pages = List([]i32).init(alloc);
        defer {
            var iter = rules.iterator();
            while (iter.next()) |rule| {
                rule.value_ptr.deinit();
            }
            rules.deinit();
            pages.deinit();
        }
    
        var parse_rules = true;
        while (rows.next()) |row| {
            if (eql(u8, row, "")) {
                parse_rules = false;
                continue;
            }
    
            if (parse_rules) {
                var rule_pair = tokenizeScalar(u8, row, '|');
                const rule = try rules.getOrPut(try parseInt(i32, rule_pair.next().?, 10));
                if (!rule.found_existing) {
                    rule.value_ptr.* = List(i32).init(alloc);
                }
                try rule.value_ptr.*.append(try parseInt(i32, rule_pair.next().?, 10));
            } else {
                var page = List(i32).init(alloc);
                var page_list = tokenizeScalar(u8, row, ',');
                while (page_list.next()) |list| {
                    try page.append(try parseInt(i32, list, 10));
                }
                try pages.append(try page.toOwnedSlice());
            }
        }
    
        var middle_sum: i32 = 0;
        var reordered_sum: i32 = 0;
    
        var wrong_order = false;
        for (pages.items) |page| {
            var index: usize = page.len - 1;
            while (index > 0) : (index -= 1) {
                var page_rule = rules.get(page[index]) orelse continue;
    
                // check the rest of the pages
                var remaining: usize = 0;
                while (remaining < page[0..index].len) {
                    if (contains(i32, page_rule.items, 1, &[_]i32{page[remaining]})) {
                        // re-order the wrong page
                        const element = page[remaining];
                        page[remaining] = page[index];
                        page[index] = element;
                        wrong_order = true;
    
                        if (rules.get(element)) |next_rule| {
                            page_rule = next_rule;
                        }
    
                        continue;
                    }
                    remaining += 1;
                }
            }
            if (wrong_order) {
                reordered_sum += page[(page.len - 1) / 2];
                wrong_order = false;
            } else {
                // middle page number
                middle_sum += page[(page.len - 1) / 2];
            }
        }
        return Answer{ .middle_sum = middle_sum, .reordered_sum = reordered_sum };
    }
    
    pub fn main() !void {
        const answer = try solve(@embedFile("input.txt"));
        print("Part 1: {d}\n", .{answer.middle_sum});
        print("Part 2: {d}\n", .{answer.reordered_sum});
    }
    
    test "test input" {
        const answer = try solve(@embedFile("test.txt"));
        try std.testing.expectEqual(143, answer.middle_sum);
        try std.testing.expectEqual(123, answer.reordered_sum);
    }
    
    
  • Gobbel2000@programming.dev
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    7 days ago

    Rust

    While part 1 was pretty quick, part 2 took me a while to figure something out. I figured that the relation would probably be a total ordering, and obtained the actual order using topological sorting. But it turns out the relation has cycles, so the topological sort must be limited to the elements that actually occur in the lists.

    Solution
    use std::collections::{HashSet, HashMap, VecDeque};
    
    fn parse_lists(input: &str) -> Vec<Vec<u32>> {
        input.lines()
            .map(|l| l.split(',').map(|e| e.parse().unwrap()).collect())
            .collect()
    }
    
    fn parse_relation(input: String) -> (HashSet<(u32, u32)>, Vec<Vec<u32>>) {
        let (ordering, lists) = input.split_once("\n\n").unwrap();
        let relation = ordering.lines()
            .map(|l| {
                let (a, b) = l.split_once('|').unwrap();
                (a.parse().unwrap(), b.parse().unwrap())
            })
            .collect();
        (relation, parse_lists(lists))
    }
    
    fn parse_graph(input: String) -> (Vec<Vec<u32>>, Vec<Vec<u32>>) {
        let (ordering, lists) = input.split_once("\n\n").unwrap();
        let mut graph = Vec::new();
        for l in ordering.lines() {
            let (a, b) = l.split_once('|').unwrap();
            let v: u32 = a.parse().unwrap();
            let w: u32 = b.parse().unwrap();
            let new_len = v.max(w) as usize + 1;
            if new_len > graph.len() {
                graph.resize(new_len, Vec::new())
            }
            graph[v as usize].push(w);
        }
        (graph, parse_lists(lists))
    }
    
    
    fn part1(input: String) {
        let (relation, lists) = parse_relation(input); 
        let mut sum = 0;
        for l in lists {
            let mut valid = true;
            for i in 0..l.len() {
                for j in 0..i {
                    if relation.contains(&(l[i], l[j])) {
                        valid = false;
                        break
                    }
                }
                if !valid { break }
            }
            if valid {
                sum += l[l.len() / 2];
            }
        }
        println!("{sum}");
    }
    
    
    // Topological order of graph, but limited to nodes in the set `subgraph`.
    // Otherwise the graph is not acyclic.
    fn topological_sort(graph: &[Vec<u32>], subgraph: &HashSet<u32>) -> Vec<u32> {
        let mut order = VecDeque::with_capacity(subgraph.len());
        let mut marked = vec![false; graph.len()];
        for &v in subgraph {
            if !marked[v as usize] {
                dfs(graph, subgraph, v as usize, &mut marked, &mut order)
            }
        }
        order.into()
    }
    
    fn dfs(graph: &[Vec<u32>], subgraph: &HashSet<u32>, v: usize, marked: &mut [bool], order: &mut VecDeque<u32>) {
        marked[v] = true;
        for &w in graph[v].iter().filter(|v| subgraph.contains(v)) {
            if !marked[w as usize] {
                dfs(graph, subgraph, w as usize, marked, order);
            }
        }
        order.push_front(v as u32);
    }
    
    fn rank(order: &[u32]) -> HashMap<u32, u32> {
        order.iter().enumerate().map(|(i, x)| (*x, i as u32)).collect()
    }
    
    // Part 1 with topological sorting, which is slower
    fn _part1(input: String) {
        let (graph, lists) = parse_graph(input);
        let mut sum = 0;
        for l in lists {
            let subgraph = HashSet::from_iter(l.iter().copied());
            let rank = rank(&topological_sort(&graph, &subgraph));
            if l.is_sorted_by_key(|x| rank[x]) {
                sum += l[l.len() / 2];
            }
        }
        println!("{sum}");
    }
    
    fn part2(input: String) {
        let (graph, lists) = parse_graph(input);
        let mut sum = 0;
        for mut l in lists {
            let subgraph = HashSet::from_iter(l.iter().copied());
            let rank = rank(&topological_sort(&graph, &subgraph));
            if !l.is_sorted_by_key(|x| rank[x]) {
                l.sort_unstable_by_key(|x| rank[x]);            
                sum += l[l.len() / 2];
            }
        }
        println!("{sum}");
    }
    
    util::aoc_main!();
    

    also on github

  • LeixB@lemmy.world
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    7 days ago

    Haskell

    I should probably have used sortBy instead of this ad-hoc selection sort.

    import Control.Arrow
    import Control.Monad
    import Data.Char
    import Data.List qualified as L
    import Data.Map
    import Data.Set
    import Data.Set qualified as S
    import Text.ParserCombinators.ReadP
    
    parse = (,) <$> (fromListWith S.union <$> parseOrder) <*> (eol *> parseUpdate)
    parseOrder = endBy (flip (,) <$> (S.singleton <$> parseInt <* char '|') <*> parseInt) eol
    parseUpdate = endBy (sepBy parseInt (char ',')) eol
    parseInt = read <$> munch1 isDigit
    eol = char '\n'
    
    verify :: Map Int (Set Int) -> [Int] -> Bool
    verify m = and . (zipWith fn <*> scanl (flip S.insert) S.empty)
      where
        fn a = flip S.isSubsetOf (findWithDefault S.empty a m)
    
    getMiddle = ap (!!) ((`div` 2) . length)
    
    part1 m = sum . fmap getMiddle
    
    getOrigin :: Map Int (Set Int) -> Set Int -> Int
    getOrigin m l = head $ L.filter (S.disjoint l . preds) (S.toList l)
      where
        preds = flip (findWithDefault S.empty) m
    
    order :: Map Int (Set Int) -> Set Int -> [Int]
    order m s
      | S.null s = []
      | otherwise = h : order m (S.delete h s)
        where
          h = getOrigin m s
    
    part2 m = sum . fmap (getMiddle . order m . S.fromList)
    
    main = getContents >>= print . uncurry runParts . fst . last . readP_to_S parse
    runParts m = L.partition (verify m) >>> (part1 m *** part2 m)
    
  • VegOwOtenks@lemmy.world
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    7 days ago

    I was very much unhappy because my previous implementation took 1 second to execute and trashed through 2GB RAM in the process of doing so, I sat down again with some inspiration about the sorting approach.
    I am very much happy now, the profiler tells me that most of time is spend in the parsing functions now.

    I am also grateful for everyone else doing haskell, this way I learned about Arrays, Bifunctors and Arrows which (I think) improved my code a lot.

    Haskell

    import Control.Arrow hiding (first, second)
    
    import Data.Map (Map)
    import Data.Set (Set)
    import Data.Bifunctor
    
    import qualified Data.Maybe as Maybe
    import qualified Data.List as List
    import qualified Data.Map as Map
    import qualified Data.Set as Set
    import qualified Data.Ord as Ord
    
    
    parseRule :: String -> (Int, Int)
    parseRule s = (read . take 2 &&& read . drop 3) s
    
    replace t r c = if t == c then r else c
    
    parse :: String -> (Map Int (Set Int), [[Int]])
    parse s = (map parseRule >>> buildRuleMap $ rules, map (map read . words) updates)
            where
                    rules = takeWhile (/= "") . lines $ s
                    updates = init . map (map (replace ',' ' ')) . drop 1 . dropWhile (/= "") . lines $ s
    
    middleElement :: [a] -> a
    middleElement us = (us !!) $ (length us `div` 2)
    
    ruleGroup :: Eq a => (a, b) -> (a, b') -> Bool
    ruleGroup = curry (uncurry (==) <<< fst *** fst)
    
    buildRuleMap :: [(Int, Int)] -> Map Int (Set Int)
    buildRuleMap rs = List.sortOn fst
            >>> List.groupBy ruleGroup 
            >>> map ((fst . head) &&& map snd) 
            >>> map (second Set.fromList) 
            >>> Map.fromList 
            $ rs
    
    elementSort :: Map Int (Set Int) -> Int -> Int -> Ordering 
    elementSort rs a b
            | Maybe.maybe False (Set.member b) (rs Map.!? a) = LT
            | Maybe.maybe False (Set.member a) (rs Map.!? b) = GT
            | otherwise = EQ
    
    isOrdered rs u = (List.sortBy (elementSort rs) u) == u
    
    part1 (rs, us) = filter (isOrdered rs)
            >>> map middleElement
            >>> sum
            $ us
    part2 (rs, us) = filter (isOrdered rs >>> not)
            >>> map (List.sortBy (elementSort rs))
            >>> map middleElement
            >>> sum
            $ us
    
    main = getContents >>= print . (part1 &&& part2) . parse
    
  • Chais@sh.itjust.works
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    edit-2
    7 days ago

    Python

    Also took advantage of cmp_to_key.

    from functools import cmp_to_key
    from pathlib import Path
    
    
    def parse_input(input: str) -> tuple[dict[int, list[int]], list[list[int]]]:
        rules, updates = tuple(input.strip().split("\n\n")[:2])
        order = {}
        for entry in rules.splitlines():
            values = entry.split("|")
            order.setdefault(int(values[0]), []).append(int(values[1]))
        updates = [[int(v) for v in u.split(",")] for u in updates.splitlines()]
        return (order, updates)
    
    
    def is_ordered(update: list[int], order: dict[int, list[int]]) -> bool:
        return update == sorted(
            update, key=cmp_to_key(lambda a, b: 1 if a in order.setdefault(b, []) else -1)
        )
    
    
    def part_one(input: str) -> int:
        order, updates = parse_input(input)
        return sum([u[len(u) // 2] for u in (u for u in updates if is_ordered(u, order))])
    
    
    def part_two(input: str) -> int:
        order, updates = parse_input(input)
        return sum(
            [
                sorted(u, key=cmp_to_key(lambda a, b: 1 if a in order[b] else -1))[
                    len(u) // 2
                ]
                for u in (u for u in updates if not is_ordered(u, order))
            ]
        )
    
    
    if __name__ == "__main__":
        input = Path("input").read_text("utf-8")
        print(part_one(input))
        print(part_two(input))
    
  • sjmulder@lemmy.sdf.org
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    edit-2
    7 days ago

    C

    I got the question so wrong - I thought a|b and b|c would imply a|c so I went and used dynamic programming to propagate indirect relations through a table.

    It worked beautifully but not for the input, which doesn’t describe an absolute global ordering at all. It may well give a|c and b|c AND c|a. Nothing can be deduced then, and nothing needs to, because all required relations are directly specified.

    The table works great though, the sort comparator is a simple 2D array index, so O(1).

    Code
    #include "common.h"
    
    #define TSZ 100
    #define ASZ 32
    
    /* tab[a][b] is -1 if a<b and 1 if a>b */
    static int8_t tab[TSZ][TSZ];
    
    static int
    cmp(const void *a, const void *b)
    {
    	return tab[*(const int *)a][*(const int *)b];
    }
    
    int
    main(int argc, char **argv)
    {
    	char buf[128], *rest, *tok;
    	int p1=0,p2=0, arr[ASZ],srt[ASZ], n,i, a,b;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    	
    	while (fgets(buf, sizeof(buf), stdin)) {
    		if (sscanf(buf, "%d|%d", &a, &b) != 2)
    			break;
    		assert(a>=0); assert(a<TSZ);
    		assert(b>=0); assert(b<TSZ);
    		tab[a][b] = -(tab[b][a] = 1);
    	}
    
    	while ((rest = fgets(buf, sizeof(buf), stdin))) {
    		for (n=0; (tok = strsep(&rest, ",")); n++) {
    			assert(n < (int)LEN(arr));
    			sscanf(tok, "%d", &arr[n]);
    		}
    
    		memcpy(srt, arr, n*sizeof(*srt));
    		qsort(srt, n, sizeof(*srt), cmp);
    		*(memcmp(srt, arr, n*sizeof(*srt)) ? &p1 : &p2) += srt[n/2];
    	}
    
    	printf("05: %d %d\n", p1, p2);
    	return 0;
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day05.c

    • reboot6675@sopuli.xyz
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      7 days ago

      Same, I initially also thought a|b and a|c implies a|c. However when I drew the graph of the example on paper, I suspected that all relations will be given, and coded it with that assumption, that turned out to be correct