Advent Of Code 2024

ideapad-320

Day 12
I have done part 1, just a basic flood fill. Part 2 was pretty straightforward for me.

Last edited by ideapad-320 (Dec. 13, 2024 20:14:04)

gilbert_given_189

Day 12 - Discussion
I had to confess that I can't be so hasty with doing the days right now, since I have a final test until next week. I'll still try to do them, though.
Initially I tried to be smart [by combining the flood fill to work with multiple plants], but that doesn't work so I had to [track the unvisited plots instead.]

Also, I like to procrastinate on this day… Part 2 is just too much for me. I'm sure I can do it, but I had to leave that tomorrow.

davidtheplatform

Day 12 - Discussion
For part 1 I did a [flood fill to find each region. Finding the edges for part 1 is as simple as iterating over every tile, then checking its adjacent neighbors. If they're different, there is an edge there.]
Part 2: [I realized that you only have to find the start of every side. So the code is: for every tile: for every edge that tile has: check if there is an edge to the left/top of that edge. If there isn't, increment the number of sides. There is an edge case if two corners touch each other, like in the 368 price example. I just added another if statement to check if this happens and increment the sides by 2.]
I had a lot of bugs in this one. The edge case in part 2 also took a while to figure out. This was probably the most annoying problem so far, although part of that is because my code isn't very good.

Looking at adventofcode.com/2024/stats/, this day had a significant drop in # of solvers and ratio of people who solved both parts. I definitely felt like it was harder than previous days, especially part 2.

gilbert_given_189

Day 12 - Solution
Ugh, what a day. [Once again, I'm trying to be smart by combining tasks of searching sides of all regions, while really I should've done them in isolation. And then there's the edge case, or rather the point case, but that's not a problem.] This day really brings me back the experience of programming my bitmap vectorizer I made when trying to make the BiY font.

I'm not proud of my solution, but here it is:

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mybearworld

Day 12 - Discussion

That was frustrating. I couldn't get it to work yesterday, on the actual day 12, at all - and then I woke up today, tried something else and got it pretty much immediately. Oh well. Hopefully day 13 will be nicer on me?

gilbert_given_189

Day 13 - Discussion
Ooh, it's linear programming! I'm sure SymPy can handle this.
…oh, it's a new feature. Guess that means I had to do it locally then.

gilbert_given_189

Day 13 - Solution
Well, that was tremendously easy, after the nightmare of Day 12. If only I started earlier…

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gilbert_given_189

General Discussion
So I did some analysis of the player counts (from 4 hours ago). The thing I'm trying to figure out is the difficulty across days and between parts.
Here's what I got:

The point of interest is the Odds row, in J:J. Higher numbers means there's more people that've done 2 parts than a single one. Closer to 1 means they're around the same, and lower than 1 means there's more people that only done 1 part instead of both.

Spoilers?

As I've suspected, Day 12 has lower odds than usual, about 5:2, since day 12's part 2 is really hard. Day 10 has the highest odds of 48:1, perhaps because part 2's solution is the “faulty” case of part 1. Day 6's odds is relatively low with ~3:1, maybe because of the looping test.

Another point of interest is the Expected (E) row in H:H and the Difference row alongside it (ignore the Expected (L) row, that's using linear regression which is wildly inaccurate). The Expected row represents the ideal player count, with constantly increasing difficulty, such that the rate of which players quit the next day is constant. In this case, that rate (the ratio) is ~0.84, which means each day only around 84% will do the next day. We can also predict the people that will do day 25, which is around 3500. Although I'll probably pick around 4000, since day 13's data is incomplete.

Here's the graph of the values:

Last edited by gilbert_given_189 (Dec. 13, 2024 18:06:12)

mybearworld

Day 13 - Discussion

Yeah, I think these have reached the point where I'll not be able to solve them. I can't figure out a way to do part 2 today.

gilbert_given_189

mybearworld wrote:
Day 13 - Discussion

Yeah, I think these have reached the point where I'll not be able to solve them. I can't figure out a way to do part 2 today.

Did you actually try to brute force a solution? That, or maybe using an external library is unfair or something.

Well, I'm not going to spoil my solution (that's encrypted on my post), but here's something I used on my solution:

ideapad-320

gilbert_given_189 wrote:
General Discussion
So I did some analysis of the player counts (from 4 hours ago). The thing I'm trying to figure out is the difficulty across days and between parts.
Here's what I got:
<removed image>
The point of interest is the Odds row, in J:J. Higher numbers means there's more people that've done 2 parts than a single one. Closer to 1 means they're around the same, and lower than 1 means there's more people that only done 1 part instead of both.

Spoilers?

As I've suspected, Day 12 has lower odds than usual, about 5:2, since day 12's part 2 is really hard. Day 10 has the highest odds of 48:1, perhaps because part 2's solution is the “faulty” case of part 1. Day 6's odds is relatively low with ~3:1, maybe because of the looping test.

Another point of interest is the Expected (E) row in H:H and the Difference row alongside it (ignore the Expected (L) row, that's using linear regression which is wildly inaccurate). The Expected row represents the ideal player count, with constantly increasing difficulty, such that the rate of which players quit the next day is constant. In this case, that rate (the ratio) is ~0.84, which means each day only around 84% will do the next day. We can also predict the people that will do day 25, which is around 3500. Although I'll probably pick around 4000, since day 13's data is incomplete.

Here's the graph of the values:
<removed image>

I have some issues with your analysis:
1. People probably drop out on the first few days just because they are not interested, so people drop out faster at the start. At some point, almost everyone solves everything they can.
2. Older puzzles have more solves and more people have done both parts, so that data might be skewed. Especially the newest ones, not everyone has gotten around to solving them.
3. The difficulty is not meant to increase slowly. Weekends are harder, and the difficulty is sometimes intentionally varied (like putting tricky puzzles at start and easy one towards the end). Source from creator of AoC

ideapad-320

Day 13
Because I used math to find an exact solution in 1 go, part 2 was easy.
Hint: (select to see) There is only one solution for each claw machine.

Hint 2: (select to see) There are 2 linear equations, one for x and one for y. Make a general solution. Pay attention to the no solution at all case, and the non-integer solution case.

Hint 3: (select to see) The equations are: (btn a uses)*(btn a x change)+(btn b uses)*(btn b x change)=(target x), and (btn a uses)*(btn a y change)+(btn b uses)*(btn b y change)=(target y).

Part 2 is trivial if you do it that way.

Last edited by ideapad-320 (Dec. 14, 2024 00:28:37)

gilbert_given_189

Day 14 - Discussion
Finally, something I can do immediately for once.
[…Christmas tree? What even is that? How do I know? Is this a reference to an earlier calendar?]

gilbert_given_189

Day 14 - Solution

Spoilers!

This is the first time I had to look at the AoC “discussion room” for hints (which is fascinating by itself, since I'm not normally allowed to do it). Apparently somebody represented the locations of the robots into an image file so they can literally hunt for the Christmas tree. That's how I managed to do it. By cheating, essentially.

Theoretically you can find it by testing for the biggest clusters of robots, either by flood-fill or some other cluster analysis. I could've thought of this, alas I was too keen on the specifics. Besides, I'm not sure if I want to do cluster analysis 10000 times…

If you want to know how the Christmas tree looks, it's stored in the Assets with this hash: ce43e936d9c002bca282c59139df7a21.png

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gilbert_given_189

Day 15 - Discussion
It feels like I'm the only one standing in this topic. Oh well.
It feels weird when your solution works the first time. Not so on part 2 though. I had to rework most of my movement code to make it work.
[There's a funny bug where the boxes being pushed got pushed anyway even though the pushing box cannot because of the walls, but I managed to fix that. It's hilarious when I found it out.]

Now I know how it feels like to be a sokoban programmer.
Day 15 - Solution

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ideapad-320

Day 14
Part 2 is wayyyyyy to vauge. Floodfill came to the rescue.

Last edited by ideapad-320 (Dec. 17, 2024 00:38:32)

ideapad-320

Day 16
This looks like a graph theory problem.

gilbert_given_189

Day 16 - Discussion
Initially I thought of using A*, but I'm concerned that the target heuristic would mess with the score, so Dijkstra it is.
[For part 2, I tried to use recursive DFS, but it's slow. Too slow. Then I thought of using the path from the Dijkstra algorithm. I didn't pursue on that idea and instead used a prioritized BFS with reconsidering instead (with score as priority). It has some merit, but is also still slow. (also, it's filling up my RAM, though I'm using Windows and from experience it has better resiliency with programs doing them)]

ideapad-320

gilbert_given_189 wrote:
Day 16 - Discussion
Initially I thought of using A*, but I'm concerned that the target heuristic would mess with the score, so Dijkstra it is.
[For part 2, I tried to use recursive DFS, but it's slow. Too slow. Then I thought of using the path from the Dijkstra algorithm. I didn't pursue on that idea and instead used a prioritized BFS with reconsidering instead (with score as priority). It has some merit, but is also still slow. (also, it's filling up my RAM, though I'm using Windows and from experience it has better resiliency with programs doing them)]

I am trying to implement Dijkstra, but I haven't had much time yet.

gilbert_given_189

Day 16 - Solution
TIL: You never know, when your program already has the correct solution while it's still crunching numbers…

Spoilers!

Seeing that there's a large disparity between reconsidering explored points than not, I thought instead of putting explored points into the queue, I store these points in a stack I call the reconsider stack. When the queue is emptied, I can pop a value from the reconsider stack, reset the explored set, and worked on that point instead.
That allows my program to finally find multiple paths given my input, and both the queue and stack's size never got into a thousand. But it's still very slow.
At this point, I've added a progress printout (not quite a progress bar), so I know my program is doing something and have not froze without my knowing (especially since I'm about to sleep when I added that). I thought of adding the places count there (what the problem asks for) just for the heck of it. When the program stopped printing more found paths, I put that as my answer, and lo and behold that got the part solved. Welp.

This one is very big, because I decided to include all the variations I've tried.

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Note that the script I used for part 2 is day16-2_reconsider. I highly recommend not opening day16-2_cursed.

Last edited by gilbert_given_189 (Dec. 17, 2024 04:17:07)

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