b. Pretty much just changed the input parsing and ran the same algorithm. Thought that it might be too slow, but I put in some stopwatch ticks and found out that it should take about 5 minutes to run my a) algorithm to get the answer.

The actual time elapsed was 5m50s.

I miiiiight try make a faster version but my head is too fucked up to care right now, hence me letting the slow algorithm run.

a) I used a regex to do some parsing because I haven't looked at dart regex much and wanted to dip my toes a little.

I considered doing this "properly" with OO classes and subclasses for the different rules. I felt that it would be too difficult and just wrote something janky instead. In hindsight, this was probably the wrong choice, especially since grappling with all the nullable types I had in my single rule class became a little too complex for my melting brain (it is HOT in Australia right now; also my conjunctivae are infected from my sinus infection. So my current IQ is like down 40 IQ points from its normal value of probably -12)

b) There may have been a trick here to simplify the programming (not the processing). Again, I felt that directly implementing the specified algorithm was the only real way forward. In brief:

1. Start with an "open" set containing a part with 4 ranges from [1, 4001) and an "accepted" set that is empty.
2. Start at the workflow "in"
3. For each rule in the current workflow:

• If the rule accepts part of the ranges in the open set, remember those ranges in a closed set and remove them from the open set.
• Remove anything the rule rejects from the open set.
• If the rule redirects to a different workflow W, split off the applicable ranges and recurse at 3 with the current workflow as W.
• Keep in the open set anything the rule doesn't consider.

Because this is AOC, I assumed that the input would be nice and wouldn't have anything problematic like overlapping ranges, and I was right. I had a very stupid off by one error that took me a while to find as well.

I read the problem as follows: for N galaxies, find N/2 pairings such that the sum of distances is minimal.

At this point, I was like, wow, the difficulty has ramped up. A DP? That will probably run out of memory with most approaches, requiring a BitSet. I dove in, eager to implement a janky data structure to solve this humdinger.

I wrote the whole damn thing. The bitset, the DP, everything. I ran the code, and WOAH, that number was much smaller than the sample answer. I reread the prompt and realised I had it all wrong.

It wasn't all for naught, though. A lot of the convenience stuff I'd written was fine. Also, I implemented a sparse parser, which helped for b.

b: I was hoping they were asking for what I had accidentally implemented for a. My hopes were squandered.

Anyway, this was pretty trivial with a sparse representation of the galaxies.

part b: Before diving into the discussion, I timed how long 1000 cycles takes to compute, and apparently, it would take 1643175 seconds or just over 19 days to compute 1 billion cycles naively. How fun!

So, how do you cut down that number? First, that number includes a sparse map representation, so you can tick that box off.

Second is intuiting that the result of performing a cycle is cyclical after a certain point. You can confirm this after you've debugged whatever implementation you have for performing cycles- run it a few times on the sample input and you'll find that the result has a cycle length of 7 after the second interaction.

Once you've got that figured out, it's a matter of implementing some kind of cycle detection and modular arithmetic to get the right answer without having to run 1000000000 cycles. For the record, mine took about 400 cycles to find the loop.

b was interesting- luckily, I remembered how to test if a point lives inside a curve. The line-crossing count algorithm was a little tricky to write from scratch, but I finally got there. Also, I screwed up for like an hour when I didn't realise you were supposed to count junk pipes as part of the territory. Code wise it got messy, and I've thrown something up on my github, but might try clean it up a little later.

So in my first year in university, in my intro DSA class, we learned A*. Have I learned any other ways to search since? Not really. So I used A* here.

It took way longer than it should have to solve, which I blame on my ongoing illness. The main sticking point was that I implemented a class to represent the search state, and since I was going to use it as a key to a map, I implemented a hashcode for it. The rest of the A* code freely flowed from my brain (really the wikipedia pseudocode).

Cue like 40 mins plus of wondering how the test input was searching over millions of states, wondering if I’d fucked up the A* implementation, wondering if the problem was too big for A*, and wondering if it was finally time to take a days break from all the aoc nonsense.

Anyway at some point I realised I forgot to implement the corresponding equals method to my hashcode. Once I had that my code ran in seconds and everything was fine. This sickness is the worst!!!

a. was a fun coding exercise. Not much more to say.

b. I lucked out by making a recursive traversal function for a), which let me specify the entry and direction of where my traversal would start. Besides that, similar to a., this was a fun coding exercise. I was surprised that my code (which just ran the function from a) on every edge tile) It only took 2s to run; I thought I might need to memoize some of the results.

This problem was mainly testing:

• Are you a bad enough dude to formulate this DP correctly?
• Are you a bad enough dude to choose a good DP state storage schema?
• Is your computer a bad enough dude to store all the DP states if you choose a bad storage schema?

...which is true of all DP problems, honestly. So given you know and understand how to approach DP problems, this would be more an engineering issue than anything.

b. It's not much of a difference to solve b. The trick here is that if you did the bit stuff I suggested above, you'd quickly realise that a smudge interpreted as a binary number is a power of two. Finding a smudge is equivalent to if the bitwise XOR of two numbers is a power of 2, which can be done with some bitwise magic.