advent-of-code-2022 VS advent-of-code

Slow dlang solution that simulates sand drops one grain at a time, one pixel at a time. Here's the main loop for part 2. The occupied[grain] = true assignment is due to the fact that the standard library doesn't have a hashset (to my knowledge), so instead I'm putting Points in an associative array while discarding the values.

Dlang solution, using JSON parsing from the standard library. At first I misunderstood the problem and kept getting failures in the second pair. I thought [2,3,4] vs [4] meant that the right list will run out of items after having successfully compared 2 and 4, but then I learned that this would only happen if 2 and 4 were equal (thanks to this thread). This made me realize that a binary comparison wasn't enough and pushed me to write the convoluted ternary comparison function below :

Dlang solution using BFS. Relevant part :

Dlang solution as usual, here's the class that handles monkey business :

Dlang solution. For part 2 I initially thought that the X register contains values exceeding 40, so I kept appending pixels to the CRT while comparing them to the sprite, and in the end I printed the string in chunks of 40. Only later did I realize my mistake

Dlang bruteforce solution. For some reason I thought the rope was 9 knots long and kept getting off by one results. Relevant part 2 portion :

Dlang bruteforce solution

Dlang solution, part 2. I'm embarrassed by this code but I'm sharing it because it took a lot of effort to write it. Previous solutions are available in this repo

[LANGUAGE: Python 3] 153/75 Raw solution

One could instead count |F7 (that's what I do in my refactored solution), but counting all the bends would miscount the vertical segments (FJ would end up canceling itself out).

That sounds like what I suggested here, actually. I don't have anything in my library with quite the right API yet, but I already have most of what you describe coded out. (It looks like I whipped it up for 2017 Day 13.)

[LANGUAGE: Python 3] Embarrassing/Embarrassing Ugly raw solution code

[LANGUAGE: Python 3] 66/101 Raw solution code

Python 3 21/12

It doesn't, but you can use a separate list, wrapper classes, and deque.index to find where the values live. I may be biased but I think that my solution (ultimately using deque) isn't as complex as a custom linked list.

Part 2 assumes you did part 1 properly. I did not! I'm pretty sure that the intended solution is to do a sort of reverse search (have a target number of geodes and work backwards to see if that's possible to achieve) but I was just not having success coming up with a way to do that. It's probably going to be blindingly obvious once I figure it out, but that might be an exercise for tomorrow.

Python 3 9/15!!!

Python 3 44/45