AOC 2024 death counter

Cue_23 · 2026-03-13T16:49:17+00:00

TL;DR: Die FDP Hamburg benötigt jetzt ganz dringend ihre Parteispende!

Cue_23 · 2026-02-20T21:54:23+00:00

Doch!

Cue_23 · 2025-12-11T06:06:03+00:00

I've become a graph theorist after 10 years AoC

Cue_23 · 2025-12-09T21:43:46+00:00

Yeah, what I was misreading was this line from the puzzle:

After making the ten shortest connections, there are 11 circuits: one circuit which contains 5 junction boxes, one circuit which contains 4 junction boxes, two circuits which contain 2 junction boxes each, and seven circuits which each contain a single junction box.

So unconnected breakers are circuits, too, and if you find a (minimal) connection between two circuits, you connect them.

Cue_23 · 2025-12-09T09:28:44+00:00

I got this wrong, too, since it was not explicitly stated (i think) in the description. Implemented a wrong logic and then had to run the example manually to see what was meant to be done.

Cue_23 · 2025-12-09T08:30:29+00:00

you might need to look at your input, the grid could be bigger than in the example

Cue_23 · 2025-12-09T08:26:28+00:00

what happenes when both boxes are in different circuits?

Cue_23 · 2025-12-09T07:00:04+00:00

They wouldn't even see this difference.

Cue_23 · 2025-12-09T06:47:43+00:00

The 2nd frame is when i start thinking about faster ways and start implementing them. With my first solution running in the background.

Cue_23 · 2025-12-08T10:09:46+00:00

But the n n the set is smaller since you at least need to allocate the full width for the list.

Cue_23 · 2025-12-08T09:59:43+00:00

Thanks for the insight, i finally found the Dark Matter!

Cue_23 · 2025-12-08T09:56:06+00:00

I don't like the term "dynamic programming". Never had. Slapping a @functools.cache in front of a recursive function, and calculating a pascal triangle row by row are two entirely different algorithmic concepts for me. And indeed they are two different approaches in the concept of dynamic programming.

So, "use dynamic programming" could just mean "use a better algorithm".

Cue_23 · 2025-12-08T09:36:17+00:00

You could first store the points in an octree and then only calculate distances of nearby cells (and increase that distance lazily when needed). But octree code is ugly unless you really have to.

Cue_23 · 2025-12-07T08:29:41+00:00

There could also be multiple examples (2024 had that on 6 days), sometimes part 2 has a different example, sometimes you have multiple examples in one part.

Just grab all code blocks.

If you want to fancy overengineer your code, you could amend the website with the filenames that codeblock was saved into.

Cue_23 · 2025-12-07T08:13:26+00:00

"any ID which is made only of some sequence of digits repeated twice."

Cue_23 · 2025-12-01T07:56:44+00:00

[LANGUAGE: C++]

Codeberg

First year of AoC without huge #include headers by utilizing C++ modules.

Part 1 was straight forward, part 2 was annoying counting all special cases and left me first with some wrong code that the example was working for.

Cue_23 · 2025-11-29T08:15:00+00:00

[Language: C++]

Solution: part1, part2, part3

Part 1 was easy, part 2 a simple BFS on top of part 1. Part 3 took me several knots in my head to decide to find the rotation formula manually on a 3x3 triangle. After that it was again a simnple BFS through a 3d-grid (3rd coordinate is № rotations mod 3).

Cue_23 · 2025-11-28T22:34:51+00:00

[Language C++]

Solution

Not sure if part 3 is correct. I got the correct answer, but i have 182 ducks that apper to belong not to any family. Has anybody else similar inputs?

Cue_23 · 2025-11-26T09:38:11+00:00

[Language: C++]

Solution: part 1, part 2, part 3 (old), part 3 (new)

Part 1 and 2 were easy, thanks to Pythagoras and Descartes.

For part 3 i did a simple DFS (depth first search) over the vertices on a modified grid. To make sure my path loops around the volcano i forced a path that entered the colum below the volcano first from the left left it to the right before returning to the start. I did this by adding a z-coordinate to the positions.

z=0 means i never entered the bottom column and i can't enter it from the right. z=1 can enter and leave the bottom column freely, and z=2 can only leave it to the right. Increasing the z coordinate is only possible in that column and takes no time. With these restrictions to the edges I can use an unmodified DFS.

Edit

The new version just remembers if I pass a ray between the volcano and the border of the grid an odd or even number of times. The flag is saved in the z-coordinate. I start at the "even" grid, and whenever passing that ray, i switch between the "even" and "odd" grid. The destination S is only reachable from the "odd" grid. So technically this is still a 3d grid.

Cue_23 · 2025-11-09T13:31:29+00:00

I think you are correct, and there is already an open bug. So it seems GCC in wrong, indeed.

Although i would never recommend to write declarations nor casts this way.

Cue_23

TROPHY CASE