this post was submitted on 09 Nov 2025
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Advent Of Code

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An unofficial home for the advent of code community on programming.dev! Other challenges are also welcome!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

Everybody Codes is another collection of programming puzzles with seasonal events.

EC 2025

AoC 2024

Solution Threads

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console.log('Hello World')

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Quest 4: Teeth of the Wind

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

Link to participate: https://everybody.codes/

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[โ€“] lwhjp@piefed.blahaj.zone 2 points 1 day ago

I liked this one!

import Control.Arrow  
import Control.Monad  
import Data.List  
import Data.Ratio  

simpleTrain = uncurry (%) . (head &&& last) . map read  

compoundTrain input =  
  let a = read $ head input  
      z = read $ last input  
      gs =  
        map  
          ( uncurry (%)  
              . (read *** read . tail)  
              . break (== '|')  
          )  
          $ (tail . init) input  
   in foldl' (/) (a % z) gs  

part1, part2, part3 :: [String] -> Integer  
part1 = floor . (2025 *) . simpleTrain  
part2 = ceiling . (10000000000000 /) . simpleTrain  
part3 = floor . (100 *) . compoundTrain  

main =  
  forM_  
    [ ("everybody_codes_e2025_q04_p1.txt", part1),  
      ("everybody_codes_e2025_q04_p2.txt", part2),  
      ("everybody_codes_e2025_q04_p3.txt", part3)  
    ]  
    $ \(input, solve) -> readFile input >>= print . solve . lines  
[โ€“] janAkali@lemmy.sdf.org 2 points 1 day ago

Nim

For part 3 I parse gears as tuples, with regular gears having same value on both ends e.g.

3|5 -> (3, 5)
3   -> (3, 3)
proc parseGears(input: string): seq[int] =
  for line in input.splitLines():
    result.add parseInt(line)

proc parseNestedGears(input: string): seq[(int, int)] =
  for line in input.splitLines():
    let nested = line.split('|').mapIt(it.parseInt)
    result.add:
      if nested.len == 1: (nested[0], nested[0])
      else: (nested[0], nested[1])

proc solve_part1*(input: string): Solution =
  let gears = parseGears(input)
  result := 2025 * gears[0] div gears[^1]

proc solve_part2*(input: string): Solution =
  let gears = parseGears(input)
  result := ceil(10000000000000'f64 / (gears[0] / gears[^1])).int

proc solve_part3*(input: string): Solution =
  let gears = parseNestedGears(input)
  let ratios = (0..gears.high-1).mapIt(gears[it][1] / gears[it+1][0])
  result := int(100 * ratios.prod)

Full solution at Codeberg: solution.nim

[โ€“] hades@programming.dev 2 points 1 day ago

Rust

use num::{BigInt, Integer};

pub fn solve_part_1(input: &str) -> String {
    let gears: Vec<i64> = input.trim().lines().map(|g| g.parse().unwrap()).collect();
    (2025 * gears[0] / gears.last().unwrap()).to_string()
}

pub fn solve_part_2(input: &str) -> String {
    let gears: Vec<i64> = input.trim().lines().map(|g| g.parse().unwrap()).collect();
    let res = (BigInt::parse_bytes(b"10000000000000", 10).unwrap() * gears.last().unwrap())
        .div_ceil(&(BigInt::ZERO + gears[0]));
    res.to_string()
}

pub fn solve_part_3(input: &str) -> String {
    let mut lines = input.trim().lines();
    let first_gear = BigInt::parse_bytes(lines.next().unwrap().as_bytes(), 10).unwrap();
    let mut nominator: BigInt = first_gear * 100;
    let mut denominator: BigInt = BigInt::ZERO + 1;
    for line in lines {
        let mut split = line.split("|");
        denominator *= BigInt::parse_bytes(split.next().unwrap().as_bytes(), 10).unwrap();
        match split.next() {
            Some(size) => {
                nominator *= BigInt::parse_bytes(size.as_bytes(), 10).unwrap();
            }
            None => {
                break;
            }
        }
    }
    (nominator / denominator).to_string()
}