AoC has ended for the year! Congrats to everyone who participated

Here are some stats from the instance leaderboard

Top 10 Users

1. SekoiaTree
2. Leo Uino
3. Michon van Dooren
4. funnyboy-roks
5. PhiliPdB
6. jhuang97
7. Alex Utter
8. hades
9. Jindřich Ivánek
10. abclop99

Misc Stats

• 91 Users got at least 1 star
• 14 Users got 50 stars

Link to full results:

https://jeroenheijmans.github.io/advent-of-code-surveys/

Day 24: Odds

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Spoilers and explanation of solution:

Each vertex here is one intersection in our hike. We don't actually care about the parts in-between, because there's only one way to go. The above is a visualisation of the final path, the red edges are the edges taken. Our graph looks "like that" because it's a hiking trail, not a maze, so there's no dead ends. This took about 2 seconds to generate, due to all the cloning needed to keep track of paths. The two veeery long edges on the ends are pretty obvious choices, but one might notice that pretty much every vertex takes the two maximum paths it has, given the restrictions of the path. There's still some mildly surprising paths, such as (99, 29) -> (89, 37) with a weight of 38. I'm wondering if there's a way to dismiss more paths... This graph is actually pretty free in terms of movement.

My actual solution takes ~150 ms to run (and 8 microseconds for part one with barely any optimization, damnn)

Day 22: A Long Walk

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Day 22: Sand

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Day 21: Step

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Day 20: Pulse

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Day 19: Aplenty

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Day 18: Lavaduct Lagoon

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Anybody got some ideas to optimize today? I've got it down to 65ms (total) on my desktop, using A* with a visitation map. Each cell in the visitation map contains (in part 2) 16 entries; 4 per direction of movement, 1 for each level of straightaway. In part 2, I use a map with 11 entries per direction.

Optimizations I've implemented:

• use a 2D array instead of a hashset/map. No idea how much this saves, I did it in the first place.
• the minimum distance for a specific cell's direction + combo applies for higher combo levels as well for part 1. For part 2, if the current combo is greater than 4, we do the same*. Gains about 70(!!) ms
• A* heuristic weighting optimization, a weight of about 1% with a manhattan distance heuristic seems to gain about 15 ms (might be my input only tho)

*Correctness-wise: the reason we're splitting by direction is because there's a difference between being at a cell going up with a 3 combo but a really short path, and going right with a 0 combo but a long path. However, this is fine because a 3 combo in the same direction as a 0 combo is identical, just more restrictive.

Optimizations that could be done but I need to ensure correctness:

the same optimization for the combo, but for directions. If I'm on a specific combo+direction, does that imply something about the distance for another direction? Simply doing the same for every non-opposite direction isn't correct

Code: https://codeberg.org/Sekoia/adventofcode/src/branch/main/src/y2023/day17.rs

Warning: quite ugly, there's like 8 copy-pastes for adding to the queue

Day 17: Clumsy Crucible

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Day 16: The Floor Will Be Lava

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Day 15: Lens Library

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Thought I'd share mine because most of the visualisations out there show the rocks sliding one cell at a time, all together. That looks nice but for my solution that's not how it works - I walk every row/col with two cursors.

Code here: https://github.com/sjmulder/aoc/blob/master/2023/c/day14.c

The visualisation is emitted right from the solution using a small library which dumps frames to ffmpeg.

Day 14: Parabolic Reflector Dish

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Day 13: Point of Incidence

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cross-posted from: https://programming.dev/post/7188393

> This community still mostly has me running it and has 1 other mod but I typically like starting communities off with at least 2 mods. This community is a bit different in that theres the solution megathreads that need to be posted when a new day drops (and then unlocked when the global leaderboard fills up). When it has a dedicated mod team running it ill remove myself so its not admin run > > If youre interested feel free to say so below > > [email protected] > > Note: to be a mod its preferred if your account is within the instance since lemmy doesnt handle cross-instance modding that well and this makes it so you have access to features added to pangora that other instances may not have

We all know and love (!) the leaderboard, but how about a different method?

One can solve a problem with a simple, naive method resulting in a short program and long runtime, or put in lots of explicit optimizations for more code and shorter runtime. (Or if you're really good, a short, fast program!)

I propose the line-second.

Take the number of lines in your program (eg, 42 lines) and the runtime (eg 0.096 seconds). Multiply these together to get a score of 4.032 line-seconds.

A smaller score is a shorter, faster program.

Similarly, (for a particular solver), a larger score is a "harder" problem.

So I managed to get part 1 of the day, but it took 2 seconds to run on the real input, which is a bad sign.

I can't see any kind of optimisation that means I can skip checks and know how many combinations are in those skipped checks (aside from 0.) I can bail out of branches of combinations if the info so far won't fit, but that still leads me to visiting every valid combination which in one of the examples is 500k. (And probably way more in the input, since if I can't complete the example near instantly the input is not happening.)

Right now I take the string, then replace the first instance of a ? with the two possible options. Check it matches the check digits so far then use recursion on those two strings.

I can try to optimise the matching, but I don't think that solves the real problem of visiting every combination.

I don't think (or hope) it's just bad code but this is my code so far (python.)

edit:

Day 12: Hot Springs

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I wanted to show how the maze follows from individual letters but it was way too large

Day 11: Cosmic Expansion

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Day 10: Pipe Maze

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https://www.reddit.com/r/adventofcode/comments/18dry49/2023_day_8_part_2glsl_brute_forced_in_under_a/

Day 9: Mirage Maintenance

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This map was really opaque until yEd helped explain the problem...

I had 6 parallel paths, and for all of them, the goal node was found at the point where the directions string repeated. So I actually only had to search each of the paths for the point where I was on the goal position and the first direction. This worked on the input data, but not the example. See the assert in the code, and note that the statement println!("Found non-repeat at {node} step {step} (num dirs {num_directions})"); is never executed.

At the end there's also the first attempt of brute-forcing the solution which seems like it would have taken nearly forever.

```rust use std::collections::HashMap;

use regex::Regex;

use crate::input::Source;

pub fn execute(source: Source) -> impl std::fmt::Debug { let lines = source.lines(); let ([directions], [empty, nodes @ ..]) = lines.split_at(1) else { panic!("Wrong format"); }; assert!(empty.is_empty());

let node_re = Regex::new(r"(...) = \((...), (...)\)").unwrap(); let nodes: HashMap = nodes .iter() .map(|line| { let caps = node_re.captures(line).unwrap(); let name = caps[1].to_string(); let left = caps[2].to_string(); let right = caps[3].to_string(); (name, (left, right)) }) .collect();

let start_nodes: Vec = nodes.keys().filter(|n| n.ends_with('A')).cloned().collect(); let repeat_lengths: Vec = start_nodes .iter() .map(|node| { let (repeat_length, goal_steps) = find_repeats(node, directions, &nodes); assert!(goal_steps.is_empty()); // not a given, but that's what the input looks like dbg!(node, repeat_length, goal_steps); repeat_length }) .collect();

repeat_lengths.iter().cloned().reduce(num::integer::lcm) }

/// (repeat length, steps to goal node within repeat) fn find_repeats( starting_node: &str, directions: &str, nodes: &HashMap, ) -> (usize, Vec) { let mut node = starting_node.to_string(); let num_directions = directions.chars().count(); let mut goals = Vec::new(); for (step, dir) in directions.chars().cycle().enumerate() { if node.ends_with('Z') { if step % num_directions == 0 { println!("Found repeat at {node} step {step} (num dirs {num_directions})"); return (step, goals); } println!("Found non-repeat at {node} step {step} (num dirs {num_directions})"); goals.push(step); } let (left, right) = &nodes[&node]; node = match dir { 'L' => left.clone(), 'R' => right.clone(), other => panic!("Unknown dir {other}"), } } unreachable!() }

// too slow #[allow(dead_code)] fn find_parallel( start_nodes: &[String], directions: &str, nodes: &HashMap, ) -> usize { let mut current_nodes = start_nodes.to_vec(); for (step, dir) in directions.chars().cycle().enumerate() { if current_nodes.iter().all(|n| n.ends_with('Z')) { return step; } for node in current_nodes.iter_mut() { let (left, right) = nodes[node].clone(); *node = match dir { 'L' => left, 'R' => right, other => panic!("Unknown dir {other}"), } } } unreachable!() } ```

Day 8: Haunted Wasteland

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The instructions for part 2 are missing info about how to handle the Four Of A Kind and a joker case. Wasted quite some time because I assumed that the remaining joker would remain unused, but turns out it turns your deck into Five Of A Kind.

Did everybody else just expect this?

Day 7: Camel Cards

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Survey by u/jeroenheijmans Closes Dec 22

Credits to u/Felix_Tholomyes over on r/adventofcode

I messed around in desmos for like 20 mins trying to find an algebraic solution to day 6 and made this. I feel as if this is the best way to solve the problem by graphing f(x) = (x-max_time/2)^2. Finding its y-intercept, finding its vertex (max_time/2) and then finding the minimum x needed to get the max distance by solving for max_distance = f(x).

```rust fn main() { let mut raw_input = include_str!("input.txt") .lines() .map(|f| {

f.split_ascii_whitespace() .skip(1) .collect::<String>() .parse::<usize>() .unwrap()

//for part 1, first parse then collect to a Vec<&str>

}) ;

let max_time = raw_input.next().unwrap(); let max_distance = raw_input.next().unwrap();

print!("{}", determine_range(max_time, max_distance));

//let max_times = raw_input.next().unwrap(); //let max_distances = raw_input.next().unwrap();

// let races = (0..max_times.len()).map(|i| { // determine_range(max_times[i], max_distances[i]) // });

// let total = races.reduce(|acc,x| x*acc).unwrap(); }

fn determine_range(max_time: usize, max_dist: usize) -> f64 { let vertex = max_time as f64/2.0; let min_y = vertex * vertex - max_dist as f64; let min_x = vertex - min_y.sqrt()+ 0.001;

let mut res = 2.0 * ( (vertex-0.001).floor() - min_x.ceil() + 1.0); if vertex.fract() == 0.0 { res+=1.0; } res } ```