Nim

Solution: sort numbers using custom rules and compare if sorted == original. Part 2 is trivial.
Runtime for both parts: 1.05 ms

proc parseRules(input: string): Table[int, seq[int]] =
  for line in input.splitLines():
    let pair = line.split('|')
    let (a, b) = (pair[0].parseInt, pair[1].parseInt)
    discard result.hasKeyOrPut(a, newSeq[int]())
    result[a].add b

proc solve(input: string): AOCSolution[int, int] =
  let chunks = input.split("\n\n")
  let later = parseRules(chunks[0])
  for line in chunks[1].splitLines():
    let numbers = line.split(',').map(parseInt)
    let sorted = numbers.sorted(cmp =
      proc(a,b: int): int =
        if a in later and b in later[a]: -1
        elif b in later and a in later[b]: 1
        else: 0
    )
    if numbers == sorted:
      result.part1 += numbers[numbers.len div 2]
    else:
      result.part2 += sorted[sorted.len div 2]

Codeberg repo

Kotlin

Took me a while to figure out how to sort according to the rules. 🤯

fun part1(input: String): Int {
    val (rules, listOfNumbers) = parse(input)
    return listOfNumbers
        .filter { numbers -> numbers == sort(numbers, rules) }
        .sumOf { numbers -> numbers[numbers.size / 2] }
}

fun part2(input: String): Int {
    val (rules, listOfNumbers) = parse(input)
    return listOfNumbers
        .filterNot { numbers -> numbers == sort(numbers, rules) }
        .map { numbers -> sort(numbers, rules) }
        .sumOf { numbers -> numbers[numbers.size / 2] }
}

private fun sort(numbers: List<Int>, rules: List<Pair<Int, Int>>): List<Int> {
    return numbers.sortedWith { a, b -> if (rules.contains(a to b)) -1 else 1 }
}

private fun parse(input: String): Pair<List<Pair<Int, Int>>, List<List<Int>>> {
    val (rulesSection, numbersSection) = input.split("\n\n")
    val rules = rulesSection.lines()
        .mapNotNull { line -> """(\d{2})\|(\d{2})""".toRegex().matchEntire(line) }
        .map { match -> match.groups[1]?.value?.toInt()!! to match.groups[2]?.value?.toInt()!! }
    val numbers = numbersSection.lines().map { line -> line.split(',').map { it.toInt() } }
    return rules to numbers
}

C#

using QuickGraph;
using QuickGraph.Algorithms.TopologicalSort;
public class Day05 : Solver
{
  private List<int[]> updates;
  private List<int[]> updates_ordered;

  public void Presolve(string input) {
    var blocks = input.Trim().Split("\n\n");
    List<(int, int)> rules = new();
    foreach (var line in blocks[0].Split("\n")) {
      var pair = line.Split('|');
      rules.Add((int.Parse(pair[0]), int.Parse(pair[1])));
    }
    updates = new();
    updates_ordered = new();
    foreach (var line in input.Trim().Split("\n\n")[1].Split("\n")) {
      var update = line.Split(',').Select(int.Parse).ToArray();
      updates.Add(update);

      var graph = new AdjacencyGraph<int, Edge<int>>();
      graph.AddVertexRange(update);
      graph.AddEdgeRange(rules
        .Where(rule => update.Contains(rule.Item1) && update.Contains(rule.Item2))
        .Select(rule => new Edge<int>(rule.Item1, rule.Item2)));
      List<int> ordered_update = [];
      new TopologicalSortAlgorithm<int, Edge<int>>(graph).Compute(ordered_update);
      updates_ordered.Add(ordered_update.ToArray());
    }
  }

  public string SolveFirst() => updates.Zip(updates_ordered)
    .Where(unordered_ordered => unordered_ordered.First.SequenceEqual(unordered_ordered.Second))
    .Select(unordered_ordered => unordered_ordered.First)
    .Select(update => update[update.Length / 2])
    .Sum().ToString();

  public string SolveSecond() => updates.Zip(updates_ordered)
    .Where(unordered_ordered => !unordered_ordered.First.SequenceEqual(unordered_ordered.Second))
    .Select(unordered_ordered => unordered_ordered.Second)
    .Select(update => update[update.Length / 2])
    .Sum().ToString();
}

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

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…

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.