advent-of-code VS advent-2021

Python. Part a is whatever, part 2 is pretty clean. I wonder if there's a math-only solution to this?

Python.

Python. Brute force with a tiny bit of pruning, takes about 3 minutes. I've been thinking of switching from CL for days and having the chance to use matrix multiplications finally gave me a great reason to :)

Common Lisp Truly ugly day, took me about 4 hours for part 1 and submitted more than 24h past the puzzle for the first :(

Common Lisp. Very disappointing problem since you can just brute force over even unreasonable bounds quickly. Would have been more appropriate as a day 4 or 5 problem

Common Lisp. Christ this was a sad day. I probably had part 1 in 40 minutes (or I would have if emacs didn't start hanging which necessitated a huge apt-get update/apt-get upgrade 40 minutes in), but then I got hung up until ~3hrs by the "padding" stuff and trying to parse more than one root-level packet from the string. Uggghhh.

Simple Dijkstra's in Common Lisp, though I can think of a good heuristic to use with A*. Took me like 10 minute to find a priority queue lib I liked, and about another 15 to figure out I needed to initialize it with a predicate of #'< instead of #'= :/

Common Lisp. Pretty bad day, stumbled around for like an hour trying to actually construct the strings for part B and then finished in like 10 mins once I realized how to take the better approach :/

Common Lisp. Pretty ugly, I could have merged fold-left and fold-up and the printing logic leaves something to be desired

Common Lisp. Got bit really hard twice today, on part a by not knowing that equalp compares strings case-insensitively (wtf!?) and on part b by speedreading and thinking that all small caves could be revisited on a single path

Rust https://github.com/Crazytieguy/advent-2021/blob/master/src/bin/day22/main.rs Imagining the geometry of this was very tricky for me. For a while I contemplated if calculating the intersection of each pair of cuboids would give me enough information to know how many cubes are on at the end, and finally decided that I would also have to calculate the intersections of those and so on, so I gave up. Instead I decided that my state will be a vector of cuboids that are garuanteed to be non overlapping, and on each command I would subtract the current cuboid from each of these non overlapping cuboids (leaving 1 to 6 cuboids depending on intersection), and finally either add in the current cuboid (for on command) or leave it out (for off).