this post was submitted on 02 Dec 2024
14 points (100.0% liked)

NotAwfulTech

385 readers
3 users here now

a community for posting cool tech news you don’t want to sneer at

non-awfulness of tech is not required or else we wouldn’t have any posts

founded 1 year ago
MODERATORS
[email protected]
14
Advent of Code 2024 - Historian goes looking for history in all the wrong places (adventofcode.com)
submitted 3 weeks ago by [email protected] to c/[email protected]
54 comments fedilink hide all child comments
 

copy pasting the rules from last year's thread:

Rules: no spoilers.

The other rules are made up aswe go along.

Share code by link to a forge, home page, pastebin (Eric Wastl has one here) or code section in a comment.

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 1 points 2 weeks ago (2 children)

days 5 and 6.

5:

p1, p2:Initially, I was thrown for a loop. It wasn't apparent to me what data structure to use or the problem's properties. My first (and correct) instinct was to interpret the data as a directed graph, but then what? Try to find some total ordering, if such a thing was possible?

As it turns out, that instinct was also correct. By drawing the sample data (or counting or printing it out), I noticed that every page number had a defined relation with every other page number. This meant that a total ordering (rather than a lattice) existed, meaning it was possible to construct a comparison function.

So, the algorithm for part 1 was to check if a list was sorted, and part 2 was to sort the list. There's probably a 1-3 line solution for both parts a and b, but that's for Mr. The Reader.

6:

p1, p2as discussed in a different part of the thread, I consider the input size for square inputs to be N, the "side length" of the square.

Context: I participated (and completed!) in AoC last year and pragmatically wrote my code as a set of utility modules for solving these pathological problems. So, I had about 80% of the boilerplate for this problem written, waiting for me to read and relearn.

Anyway, the analysis: P1. was pretty straightforward. Just walk along the map, turn right if you hit an obstacle, and stop when you leave the map. I guessed that there may be a case where one needs to turn in place more than once to escape an obstacle, but I never checked if that was true. Either way, I got the answer out. This is a worst-case O(n^2^) solution, which was fast for n = 130.

P2. I chose to brute force this, and it was fine. I iterated through the grid, placing a wall if possible, and checked if this produced a loop using an explored set. This is worst case O(n^4^), which, for n=130, takes a few seconds to spit out the answer. It's parallelisable, though, so there's that. If a faster solution existed, I'd love to know.

[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago)

day 6

part 2

I also brute-forced this. I figured there's a bit optimization to be done if you "draw a line" to the next obstacle instead of going step by step, and also maybe exclude some areas, but in the end I just set an exit value to break if the number of steps exceeded a certain value and say that was a loop. Took almost 20m but a star is a star.

update I only added obstructions in the original path, which cut the time down to 5 minutes or so.

[–] [email protected] 1 points 2 weeks ago* (last edited 2 weeks ago)

code for day 5, written in dart to be short:

spoiler

void d5(bool isPart2) {
  Map<int, Set<int>> ordering = {};
  int comp(int a, int b) => ordering[a]?.contains(b) ?? false ? -1 : 1;

  List<String> lines = getLines();
  int rulesEnd = lines.indexWhere((line) => line.isEmpty);

  lines
      .sublist(0, rulesEnd)
      .map((line) => line.split('|').map((e) => int.parse(e)).toList())
      .forEach((rule) => ordering.putIfAbsent(rule[0], Set.new).add(rule[1]));

  print(lines
      .sublist(rulesEnd + 1)
      .map((line) => line.split(",").map((s) => int.parse(s)).toList())
      .fold<int>(
          0,
          (p, e) =>
              p +
              (e.isSorted(comp) != isPart2
                  ? e.sorted(comp)[e.length >> 1]
                  : 0)));
}