Sorry for the necropost: I have completed all the problems! One of them completely stumped me and I had to cheat. Not going to do a writeup unless requested :)
this post was submitted on 01 Dec 2023
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congrats! I have officially checked out of the competition for the time being. Maybe if I get some spare energy later.
What problem had you stumped?
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Day 16: The Floor Will Be Lava
[Language: Perl]
https://github.com/gustafe/aoc2023/blob/main/d16-The-Floor-Will-Be-Lava.pl
Day 20: Pulse Propagation
It feels weird to kick one of these threads off, but hey, here we go.
Code as always: https://github.com/Fluxward/aoc2023/blob/main/20.dart
a,b
A
So following from yesterday where I was punished by not going full OO, I decided, hey, this is a problem in which I can do some OOP, so I did. This took very long to do but I regret nothing. If you look at my code, feel free to click your tongue and shake your head at every opportunity I missed to use a design pattern.
Anyway, after a slight snafu with misunderstanding the FF logic and not spotting that some modules can be dummy modules, it all just worked, and I got my answer.
B
This was a bit of a headscratcher, but the answer was surprisingly simple.
First, the answer. Here's how to do it:
- Look for the "rx" module in your input.
- If the module that outputs to rx is a conjunction, keep track of how many button presses it takes for each input of the conjunction to change. The answer is the LCM of all those numbers.
- If the module is a FF, you can also work backwards to get it, but this was not my input so I didn't need to try this.
Getting here was a bit weird. I hoped that I could just run the code from A and spit out the answer when rx went low, but as of time of writing I've been running it now on a separate machine for about an hour and still no result.
My next instinct was to simply work it out from pen and paper. I thought it might be possible (it probably is) but decided to at least experimentally see if the states of the modules connected to rx were cyclic first. I did, and that was enough for me to get to the answer.
My answer was about 230 trillion BPs, which, extrapolating on how long execution is taking on my other machine, might take just under 137 years to calculate naively. Fun!
Completed when waiting for the second leg of my Christmas holidays flight. (It was a long wait, can I blame jet-lag?).
Have a more compact implementation of LCM/GCD, something tells me it will come in handy In future editions. (I’ve also progressively been doing past years)
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21 Step Counter
Starting this thread having only solved a.
A
Pretty straightforward. Could probably be done in a few lines with the right syntactic sugar.
B
This is some game of life thing that I've never implemented or seen an implementation of, so I am altogether lost.
My current code (https://github.com/Fluxward/aoc2023/blob/main/21.dart) has a memoisation based approach but my current ailments are preventing me from really thinking about this properly so I am bowing out until I have the wherewithal.
This is the hardest problem of the year so far, based on leaderboard completion times. I'm busy wrapping up work for this year, and looking for a new job, so this will have to be put on the TODO pile
At this point I have officially given up and started looking at other people’s code. I’ll work on it after Imm fully better, it’s too much for me right now.
Only solved by receving heavy hints from other's solution, and it still took me forever. By far the hardest this year.
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Happy Holidays everyone. I’ve decided I am going to take a break from aoc to properly rest and recover from my mystery illness. Perhaps I will attempt solves again in the new year.
Happy holidays to you too! I decided this morning that I'm not gonna work myself up missing days, so they are on hold until after xmas for me!
Get well soon!
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Happy holidays!
Day 5: If You Give A Seed A Fertilizer
https://adventofcode.com/2023/day/5
Leaderboard completion time: 26m37s, so it's the toughest problem this year so far.
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Starting a new comment thread for my solutions to 10-19. Double digits, baby! Code here: https://github.com/Fluxward/aoc2023/
a,b, not much to say
The hardest part has finding the right dart ascii library to use (by default dart treats everything as UTF-16, which is horrible for this sort of thing) and the right data structure (linked hash map, which is a map that remembers insertion order.)
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- After this problem, I will create a new reply in the OP if it is not there already, and will discuss under that thread.
a,b
So, like many other problems from this year, this is one of those direct solution problems where there isn't much of a neat trick to getting the answer. You just have to implement the algorithm they specify and hope you can do it correctly.
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:
- Start with an "open" set containing a part with 4 ranges from [1, 4001) and an "accepted" set that is empty.
- Start at the workflow "in"
- 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.
The code I have up as of this comment is pretty long and boring, I might try clean it up later.
update: have cleaned up the code.
Replying in OP: Yeah, Lemmy punishes old threads/posts a bit too much for my taste ^^.
Good note for next year!
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Day 9: Mirage Maintenance
My solution: https://github.com/gustafe/aoc2023/blob/main/d09-Mirage-Maintenance.pl
discussion
What can I say. Shockingly simple.
I just literally followed the instructions, and got a solution in 20ms. This despite literally creating each intermediate array yet only using the ends. I'm sure I used way to much memory but you know? I'm using a $5/mo VPS for everything and unless I'm barking totally up the wrong tree I've never exceeded its memory limits.
On the subreddit I see people discussing recursion and "dynamic programming" (which is an empty signifier imho) but I really don't see the need, unless you wanna be "elegant"
spoiler
DP to me is when you use memoisation and sometimes recursion and you want to feel smarter about what you did.
I also struggle to think of the need for DP, even in a more “elegant” approach. Maybe if you wanted to do an O(n) memory solution instead of n^2, or something. Not saying this out of derision. I do like looking at elegant code, sometimes you learn something.
I feel like there’s an unreadable Perl one line solution to this problem, wanna give that a go, @gerikson?
spoiler
Part 2 only, but Part 1 is very similar.
#!/usr/bin/env jq -n -R -f
[
# For each line, get numbers eg: [ [1,2,3] ]
inputs / " " | map(tonumber) | [ . ] |
# Until latest row is all zeroes
until (.[-1] | [ .[] == 0] | all;
. += [
# Add next row, where for element(i) = prev(i+1) - prev(i)
[ .[-1][1:] , .[-1][0:-1] ] | transpose | map(.[0] - .[1])
]
)
# Get extrapolated previous element for first row
| [ .[][0] ] | reverse | reduce .[] as $i (0; $i - . )
]
# Output sum of extapolations for all lines
| add
I'm pretty sure you could make this one line and unreadable ^^.
Now this is content
Here's where I landed in dart
no comments
d9(bool s) {
print(getLines().fold(0, (p, e) {
int pre(List h, bool s) {
return h.every((e) => e == 0)
? 0
: (pre(List.generate(h.length - 1, (i) => h[i + 1] - h[i]), s)) *
(s ? -1 : 1) +
(s ? h.first : h.last);
}
return p + pre(stois(e), s);
}));
}
Day 6: Wait For It
https://adventofcode.com/2023/day/6
Alternate spoiler name - for part 2
~~Do you remember highschool algebra?~~ Can you (or your compiler) remember highschool algebra fast enough to beat out a naïve implementation?
nice cleanser after yesterday
spoiler
it would have taken me longer to figure out the algebra than to just mush the inputs together and get the solution that way (16s runtime)
I have come up with a more horrifying way to solve Day 1 Part 1 involving C++ implementation defined behavior, but it requires launching two subprocesses for every test case so I'm not sure I have the motivation right now.
Proof of concept: https://www.animeprincess.net/blog/?p=60
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Day 7: Camel Cards
https://adventofcode.com/2023/day/7
So far, my favorite puzzle. Challenging but fair. Also plays to Perl's strengths.
Leaderboard completion time: 16 minutes flat, so not a pushover.
Day 3: Gear Ratios
https://adventofcode.com/2023/day/3
Writeup of my solution: https://github.com/gustafe/aoc2023#day-3-gear-ratios