this post was submitted on 04 Dec 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 2025

Solution Threads

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console.log('Hello World')

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Day 4: Printing Department

Megathread guidelines

  • 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

FAQ

top 27 comments
sorted by: hot top controversial new old
[–] VegOwOtenks@lemmy.world 5 points 2 weeks ago

Haskell

I tried rewriting part 2 to use a MutableArray, but it only made everything slower. So I left it at this. I saw somebody do a 1-second-challenge last year and I feel like that will be very hard unless I up my performance game.

Solution, Both Parts

{-# LANGUAGE OverloadedStrings #-}
{-# OPTIONS_GHC -Wall #-}
module Main (main) where
import qualified Data.Text as Text
import Data.Array.Unboxed (UArray)
import qualified Data.Array.IArray as Array
import qualified Data.List as List
import Control.Monad ((<$!>), guard)
import qualified Data.Text.IO as TextIO
import Data.Maybe (fromMaybe)
import Control.Arrow ((&&&))

parse :: Text.Text -> UArray (Int, Int) Bool
parse t = let
    gridLines = init $ Text.lines t
    lineSize = maybe 0 pred $ Text.findIndex (== '\n') t
    lineCount = Text.count "\n" t - 2
  in Array.listArray ((0, 0), (lineCount, lineSize)) $ List.concatMap (fmap (== '@') . Text.unpack) gridLines

neighbors8 :: (Int, Int) -> [(Int, Int)]
neighbors8 p@(x, y) = do
  x' <- [pred x .. succ x]
  y' <- [pred y .. succ y]
  let p' = (x', y')
  guard (p /= p')
  pure p'

main :: IO ()
main = do
  grid <- parse <$!> TextIO.getContents
  print $ part1 grid
  print $ part2 grid

part2 :: UArray (Int, Int) Bool -> Int
part2 grid = case accessiblePositions grid of
  [] -> 0
  xs -> List.length xs + part2 (grid Array.// fmap (id &&& const False) xs)

part1 :: UArray (Int, Int) Bool -> Int
part1 = List.length . accessiblePositions

accessiblePositions :: UArray (Int, Int) Bool -> [(Int, Int)]
accessiblePositions grid = let
     lookupPosition = fromMaybe False . (grid Array.!?)
     positions = Array.indices grid
     paperRollPositions = List.filter lookupPosition positions
     isPositionAccessible = (< 4) . List.length . List.filter lookupPosition . neighbors8
   in List.filter isPositionAccessible paperRollPositions

[–] mykl@lemmy.world 5 points 2 weeks ago* (last edited 2 weeks ago) (3 children)

Uiua

Suspiciously easy. I even included a free animation generator for your entertainment.

"..@@.@@@@.\n@@@.@.@.@@\n@@@@@.@.@@\n@.@@@@..@.\n@@.@@@@.@@\n.@@@@@@@.@\n.@.@.@.@@@\n@.@@@.@@@@\n.@@@@@@@@.\n@.@.@@@.@."
# You can run against your own input by dragging your file
# onto this pane and uncommenting the line below.
# &fras"day4.txt" # edit to match the filename.
⊜(=@@)βŠΈβ‰ @\n
Nβ‚ˆ ← βŠ‚Aβ‚‚Cβ‚‚
R  ← Γ—<4⊸(/+⬚0↻)Nβ‚ˆ
P₁ ← /+β™­R
Pβ‚‚ ← /+β™­-⊸β₯(-⊸R)∞
P₃ ← β₯(▽₃10)<1e6/Γ—βŠΈβ–³β₯⊸(-⊸R)∞
βŠƒ(P₁|Pβ‚‚|P₃)

[–] CameronDev@programming.dev 3 points 2 weeks ago (1 children)

Love a good visualisation <3

I was gonna do the same later when some free time, was wondering if it generated some kind of image.

[–] mykl@lemmy.world 3 points 2 weeks ago

If you click the link on that post, you'll see that the test data does resolve to a (very low res) elf!

[–] Deebster@programming.dev 3 points 2 weeks ago* (last edited 2 weeks ago)

Now you're just showing off!

Edit: ooh, this makes it obvious that my puzzle input takes more cycles to reach the done state.

[–] Quant@programming.dev 2 points 2 weeks ago (1 children)

That's a great addition :D

Running my own input I also noticed that your solution is a lot faster than mine (processing each roll individually). I'll keep that 2D-rotation in mind for the future.

[–] mykl@lemmy.world 1 points 2 weeks ago

Yeah, that's one thing the gurus keep hammering home: anything you can move out of loop constructs (inc rows, partition, etc as well as the obvious do, repeat) and handle pervasively is a big win.

[–] Pyro@programming.dev 4 points 2 weeks ago* (last edited 2 weeks ago)

Python

Simple brute-force is enough.

DIRECTIONS = [
    (0, 1), (1, 0), (0, -1), (-1, 0),   # cardinal
    (1, 1), (1, -1), (-1, 1), (-1, -1)  # diagonal
]
# yield all valid neighbor coordinates
def yield_neighbors(i, j, m, n):
    for di, dj in DIRECTIONS:
        ni, nj = i+di, j+dj
        if 0 <= ni < m and 0 <= nj < n:
            yield ni, nj

# build char grid from input data
def build_grid(data: str):
    return [list(row) for row in data.splitlines()]

def part1(data: str):
    grid = build_grid(data)
    m, n = len(grid), len(grid[0])

    # count accessible rolls
    accessible = 0
    for i in range(m):
        for j in range(n):
            if grid[i][j] != '@':
                continue

            neighbor_rolls = sum(grid[ni][nj] == '@' for ni, nj in yield_neighbors(i, j, m, n))
            if neighbor_rolls < 4:
                accessible += 1
    return accessible

def part2(data: str):
    grid = build_grid(data)
    m, n = len(grid), len(grid[0])

    total_accessible = 0    # total accessible rolls over all cycles
    accessible = None       # rolls accessible this cycle

    # repeat until no more rolls can be accessed
    while accessible != 0:
        accessible = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] != '@':
                    continue

                neighbor_rolls = sum(grid[ni][nj] == '@' for ni, nj in yield_neighbors(i, j, m, n))
                if neighbor_rolls < 4:
                    accessible += 1
                    # we can immediately remove this roll, no need to wait for next cycle
                    grid[i][j] = '.'
        
        # accumulate accessible rolls this cycle
        total_accessible += accessible
    return total_accessible

sample = """<paste sample here>"""
assert part1(sample) == 13
assert part2(sample) == 43
[–] tranzystorek_io@beehaw.org 3 points 2 weeks ago
[–] lwhjp@piefed.blahaj.zone 3 points 2 weeks ago

Haskell

Very simple, this one.

import Data.List  
import Data.Set qualified as Set  

readInput s =  
  Set.fromDistinctAscList  
    [ (i, j) :: (Int, Int)  
      | (i, l) <- zip [0 ..] (lines s),  
        (j, c) <- zip [0 ..] l,  
        c == '@'  
    ]  

accessible ps = Set.filter ((< 4) . adjacent) ps  
  where  
    adjacent (i, j) =  
      length . filter (`Set.member` ps) $  
        [ (i + di, j + dj)  
          | di <- [-1 .. 1],  
            dj <- [-1 .. 1],  
            (di, dj) /= (0, 0)  
        ]  

main = do  
  input <- readInput <$> readFile "input04"  
  let removed =  
        (`unfoldr` input) $  
          \ps ->  
            case accessible ps of  
              d  
                | Set.null d -> Nothing  
                | otherwise -> Just (Set.size d, ps Set.\\ d)  
  print $ head removed  
  print $ sum removed  
[–] strlcpy@lemmy.sdf.org 3 points 2 weeks ago

C

For loops!

Code

#include <stdio.h>

#define GZ 144

static char g[GZ][GZ];

int
main()
{
	int p1=0,p2=0, nc=0, x,y;

	for (y=1; fgets(g[y]+1, GZ-2, stdin); y++)
		;

	for (y=1; y<GZ-1; y++)
	for (x=1; x<GZ-1; x++)
		p1 += g[y][x] == '@' &&
		      (g[y-1][x-1] == '@') +
		      (g[y-1][x  ] == '@') +
		      (g[y-1][x+1] == '@') +
		      (g[y  ][x-1] == '@') +
		      (g[y  ][x+1] == '@') +
		      (g[y+1][x-1] == '@') +
		      (g[y+1][x  ] == '@') +
		      (g[y+1][x+1] == '@') < 4;

	do {
		nc = 0;

		for (y=1; y<GZ-1; y++)
		for (x=1; x<GZ-1; x++)
			if (g[y][x] == '@' &&
			    (g[y-1][x-1] == '@') +
			    (g[y-1][x  ] == '@') +
			    (g[y-1][x+1] == '@') +
			    (g[y  ][x-1] == '@') +
			    (g[y  ][x+1] == '@') +
			    (g[y+1][x-1] == '@') +
			    (g[y+1][x  ] == '@') +
			    (g[y+1][x+1] == '@') < 4) {
				nc++;
				p2++;
				g[y][x] = '.';
			}
	} while (nc);

	printf("04: %d %d\n", p1, p2);
	return 0;
}

Repo

For my x86-16 version, the 20K input is pushing it over the 64K .COM limit, so I'll need to implement some better compression first.

[–] Avicenna@programming.dev 3 points 2 weeks ago

Turns out on part 2 you can remove on access rather than after a full sweep of the grid, which cuts down the number of iterations you need to do about 1/2 sometimes 1/3 (depending on input).

import itertools as it
from pathlib import Path

import numpy as np

cwd = Path(__file__).parent.resolve()


def parse_input(file_path):
  with file_path.open("r") as fp:
    grid = np.array(list(map(list, list(map(str.strip, fp.readlines())))),
                    dtype=str)
  return grid


def solve_problem(file_name, single_attempt=True):

  grid = parse_input(Path(cwd, file_name))
  nr, nc = grid.shape
  nacc_total = 0
  stop = False

  while not stop:

    nacc = 0

    for i,j in it.product(range(nr), range(nc)):

      if grid[i,j] != '@':
        continue

      if np.count_nonzero(grid[max(i-1, 0):min(i+2, nr),
                               max(j-1, 0):min(j+2, nc)] == '@')<5:
        nacc += 1

        if not single_attempt:
          grid[i,j] = '.'

    nacc_total += nacc

    if nacc==0 or single_attempt:
      stop = True

  return nacc_total
[–] Deebster@programming.dev 3 points 2 weeks ago* (last edited 2 weeks ago)

Rust

I pulled out some code from last year to make representing 2D grids as a vector easier, so this was quite straightforward. 2.5ms runtime (including reading/parsing the input twice cos of TDD).

impl Grid {
    fn neighbour(&self, i: usize, dir: Direction) -> Option<bool> {
        let width = self.width;
        let length = self.grid.len();

        use Direction::*;
        match dir {
            N if i >= width => Some(i - width),
            NE if i >= width && i % width != width - 1 => Some(i - width + 1),
            E if i % width != width - 1 => Some(i + 1),
            SE if i + width + 1 < length && i % width != width - 1 => Some(i + width + 1),
            S if i + width < length => Some(i + width),
            SW if i + width - 1 < length && !i.is_multiple_of(width) => Some(i + width - 1),
            W if !i.is_multiple_of(width) => Some(i - 1),
            NW if i >= width && !i.is_multiple_of(width) => Some(i - width - 1),
            _ => None,
        };
        .map(|i| self.grid[i])
    }

    #[rustfmt::skip]
    fn cell_accessible(&self, i: usize) -> bool {
        Direction::ALL
            .iter()
            .filter(|&&dir| self.neighbour(i, dir).unwrap_or(false))
            .count() < 4
    }

    fn num_accessible(&self) -> usize {
        self.grid
            .iter()
            .enumerate()
            .filter(|&(i, &is_occupied)| is_occupied && self.cell_accessible(i))
            .count()
    }

    fn remove_accessible(&mut self) -> Option<usize> {
        let removables = self
            .grid
            .iter()
            .enumerate()
            .filter_map(|(i, &is_occupied)| (is_occupied && self.cell_accessible(i)).then_some(i))
            .collect::<Vec<_>>();

        let count = removables.len();
        if count > 0 {
            for idx in removables {
                self.grid[idx] = false;
            }
            Some(count)
        } else {
            None
        }
    }

    fn remove_recursive(&mut self) -> usize {
        let mut total_removed = 0;
        while let Some(removed) = self.remove_accessible() {
            total_removed += removed;
        }
        total_removed
    }
}

::: spoiler Full code


use std::{fs, str::FromStr};

use color_eyre::eyre::{Report, Result};

#[derive(Debug, Copy, Clone)]
enum Direction {
    N, NE, E, SE, S, SW, W, NW,
}

impl Direction {
    const ALL: [Direction; 8] = [
        Direction::N,
        Direction::NE,
        Direction::E,
        Direction::SE,
        Direction::S,
        Direction::SW,
        Direction::W,
        Direction::NW,
    ];
}

#[derive(Debug, PartialEq, Eq, Clone)]
struct Grid {
    grid: Vec<bool>,
    width: usize,
    height: usize,
}

impl FromStr for Grid {
    type Err = Report;

    fn from_str(s: &str) -> Result<Self, Self::Err> {
        let grid: Vec<_> = s
            .chars()
            .filter_map(|ch| match ch {
                '.' => Some(false),
                '@' => Some(true),
                '\n' => None,
                _ => panic!("Invalid input"),
            })
            .collect();
        let width = s
            .chars()
            .position(|ch| ch == '\n')
            .ok_or_else(|| Report::msg("grid width cannot be zero, or one line"))?;
        let height = grid.len() / width;
        Ok(Self {
            grid,
            width,
            height,
        })
    }
}

impl Grid {
    fn neighbour(&self, i: usize, dir: Direction) -> Option<bool> {
        let width = self.width;
        let length = self.grid.len();

        use Direction::*;
        match dir {
            N if i >= width => Some(i - width),
            NE if i >= width && i % width != width - 1 => Some(i - width + 1),
            E if i % width != width - 1 => Some(i + 1),
            SE if i + width + 1 < length && i % width != width - 1 => Some(i + width + 1),
            S if i + width < length => Some(i + width),
            SW if i + width - 1 < length && !i.is_multiple_of(width) => Some(i + width - 1),
            W if !i.is_multiple_of(width) => Some(i - 1),
            NW if i >= width && !i.is_multiple_of(width) => Some(i - width - 1),
            _ => None,
        };
        .map(|i| self.grid[i])
    }

    #[rustfmt::skip]
    fn cell_accessible(&self, i: usize) -> bool {
        Direction::ALL
            .iter()
            .filter(|&&dir| self.neighbour(i, dir).unwrap_or(false))
            .count() < 4
    }

    fn num_accessible(&self) -> usize {
        self.grid
            .iter()
            .enumerate()
            .filter(|&(i, &is_occupied)| is_occupied && self.cell_accessible(i))
            .count()
    }

    fn remove_accessible(&mut self) -> Option<usize> {
        let removables = self
            .grid
            .iter()
            .enumerate()
            .filter_map(|(i, &is_occupied)| (is_occupied && self.cell_accessible(i)).then_some(i))
            .collect::<Vec<_>>();

        let count = removables.len();
        if count > 0 {
            for idx in removables {
                self.grid[idx] = false;
            }
            Some(count)
        } else {
            None
        }
    }

    fn remove_recursive(&mut self) -> usize {
        let mut total_removed = 0;
        while let Some(removed) = self.remove_accessible() {
            total_removed += removed;
        }
        total_removed
    }
}

fn part1(filepath: &str) -> Result<usize> {
    let input = fs::read_to_string(filepath)?;
    let grid = Grid::from_str(&input)?;
    Ok(grid.num_accessible())
}

fn part2(filepath: &str) -> Result<usize> {
    let input = fs::read_to_string(filepath)?;
    let mut grid = Grid::from_str(&input)?;
    Ok(grid.remove_recursive())
}

fn main() -> Result<()> {
    color_eyre::install()?;

    println!("Part 1: {}", part1("d04/input.txt")?);
    println!("Part 2: {}", part2("d04/input.txt")?);
    Ok(())
}
[–] Zikeji@programming.dev 3 points 2 weeks ago

Javascript

After smashing out a functional version in 20 minutes, I converted it into a OOP approach for a more appealing solution.

Solution

const input = require('fs').readFileSync('input-day4.txt', 'utf-8');

class PrintingDepartmentMap {
    /** @param {string} input */
    constructor(input) {
        /** @type {number[][]} */
        this.map = input.split("\n").map(r => r.split('').map(v => v === '@' ? 1 : 0));
    }

    /**
     * @param {number} x
     * @param {number} y
     * @returns {number} the value of that tile, or -1 if invalid
     */
    getCoord(x, y) {
        if (x < 0 || y < 0 || x >= this.map[0].length || y >= this.map.length) {
            return -1;
        }
        return this.map[y][x];
    }

    /**
     * @param {number} x
     * @param {number} y
     * @returns {number} the number of adjacent tiles that are >= 1
     */
    countAdjacent(x, y) {
        return [
            // top-left
            this.getCoord(x - 1, y - 1) >= 1,
            // top
            this.getCoord(x, y - 1) >= 1,
            // top-right
            this.getCoord(x + 1, y - 1) >= 1,
            // right
            this.getCoord(x + 1, y) >= 1,
            // bottom-right
            this.getCoord(x + 1, y + 1) >= 1,
            // bottom
            this.getCoord(x, y + 1) >= 1,
            // bottom-left
            this.getCoord(x - 1, y + 1) >= 1,
            // left
            this.getCoord(x - 1, y) >= 1
        ].reduce((acc, v) => !!v ? acc + 1 : acc, 0);
    }

    /**
     * transform in place the map replacing any rolls (1) with (2) if they are accessible
     * @returns {PrintingDepartmentMap}
     */
    markAccessable() {
        for (let y = 0; y < this.map.length; y += 1) {
            for (let x = 0; x < this.map[0].length; x += 1) {
                if (this.map[y][x] < 1) continue;
                if (this.countAdjacent(x, y) < 4) {
                    this.map[y][x] = 2;
                }

            }
        }
        return this;
    }

    /** @returns {number} the number of removed rolls */
    removeAccessable() {
        let removed = 0;
        for (let y = 0; y < this.map.length; y += 1) {
            for (let x = 0; x < this.map[0].length; x += 1) {
                if (this.map[y][x] === 2) {
                    removed += 1;
                    this.map[y][x] = 0;
                }
            }
        }
        return removed;
    }
}

const map = new PrintingDepartmentMap(input);

let removed = 0;
while (true) {
    const justRemoved = map.markAccessable().removeAccessable();
    if (removed === 0) {
        console.log(`Part 1 Answer: ${justRemoved}`);
    }
    if (justRemoved === 0) break;
    removed += justRemoved;
}

console.log(`Part 2 Answer: ${removed}`);

[–] janAkali@lemmy.sdf.org 3 points 2 weeks ago

Nim

type
  AOCSolution[T,U] = tuple[part1: T, part2: U]
  Vec2 = tuple[x,y: int]

proc removePaper(rolls: var seq[string]): int =
  var toRemove: seq[Vec2]
  for y, line in rolls:
    for x, c in line:
      if c != '@': continue
      var adjacent = 0
      for (dx, dy) in [(-1,-1),(0,-1),(1,-1),
                       (-1, 0),       (1, 0),
                       (-1, 1),(0, 1),(1, 1)]:
        let pos: Vec2 = (x+dx, y+dy)
        if pos.x < 0 or pos.x >= rolls[0].len or
           pos.y < 0 or pos.y >= rolls.len: continue
        if rolls[pos.y][pos.x] == '@': inc adjacent

      if adjacent < 4:
        inc result
        toRemove.add (x, y)

  for (x, y) in toRemove: rolls[y][x] = '.'

proc solve(input: string): AOCSolution[int, int] =
  var rolls = input.splitLines()
  result.part1 = rolls.removePaper()
  result.part2 = result.part1
  while (let cnt = rolls.removePaper(); result.part2 += cnt; cnt) > 0:
    discard

Today was so easy, that I decided to solve it twice, just for fun. First is a 2D traversal (see above). And then I did a node graph solution in a few minutes (in repo below). Both run in ~27 ms.

It's a bit concerning, because a simple puzzle can only mean that tomorrow will be a nightmare. Good Luck everyone, we will need it.

Full solution is at Codeberg: solution.nim

[–] eco_game@discuss.tchncs.de 3 points 2 weeks ago (2 children)

Kotlin

Pretty simple solution, just plain count / remove the rolls until none can be removed anymore. I would've liked to try using imaginary numbers this year (due to this article), but sadly Kotlin doesn't natively support them and I was too lazy to use a library.

Solution

class Day04 : Puzzle {

    val grid = mutableListOf<MutableList<Char>>()

    override fun readFile() {
        val input = readInputFromFile("src/main/resources/a2025/day04.txt")
        for ((r, line) in input.lines().filter(String::isNotBlank).withIndex()) {
            grid.add(mutableListOf())
            for (char in line) {
                grid[r].add(char)
            }
        }
    }

    override fun solvePartOne(): String {
        return countRolls(grid).toString()
    }

    override fun solvePartTwo(): String {
        var sum = 0
        var removed = -1
        while (removed != 0) {
            // main grid is modified here, not the best but doesn't really matter
            // also no idea how to clone 2D list in Kotlin
            removed = countRolls(grid, true)
            sum += removed
        }
        return sum.toString()
    }

    private fun countRolls(grid: MutableList<MutableList<Char>>, removeRolls: Boolean = false): Int {
        val dr = listOf(-1, -1, -1, 0, 0, 1, 1, 1)
        val dc = listOf(-1, 0, 1, -1, 1, -1, 0, 1)

        var sum = 0
        for (r in grid.indices) {
            for (c in grid[r].indices) {
                if (grid[r][c] != '@') continue

                var neighborCount = 0
                for (i in dr.indices) {
                    if (gridGet(grid, r + dr[i], c + dc[i]) == '@') neighborCount++
                }
                if (neighborCount < 4) {
                    sum++
                    if (removeRolls) grid[r][c] = '.'
                }
            }
        }
        return sum
    }

    private fun gridGet(grid: List<List<Char>>, r: Int, c: Int, default: Char = '.'): Char {
        return if (r in grid.indices && c in grid[r].indices) {
            grid[r][c]
        } else {
            default
        }
    }
}

full code on Codeberg

[–] Deebster@programming.dev 3 points 2 weeks ago

Ha, I've got that article half-read in a tab somewhere. Same problem here though - they're not in the standard library for anything I plan to use for AoC.

[–] chunkystyles@sopuli.xyz 2 points 2 weeks ago* (last edited 2 weeks ago)

Edit: looking at your code, I had forgotten about .indices. That would have made this a little easier to write.

I completely forgot to do the puzzle yesterday somehow. I struggled a bit on this one for a while because I'd used a <= 4 where I should have used a < 4. Just a complete brainfart of thinking, "It needs to be 4 or less". I wasted more time on that than I'd like to admit.

My first stab at this set all of the adjacency counts to 0, and that lead to a few rolls that had no rolls adjacent to them staying on the map by accident.

const val toiletPaperRoll = '@'
const val adjacentLimit = 4
var height = 0
var width = 0

fun main() {
    val input = getInput(4)
    val map = parseInput1(input)
    height = map.size
    width = map[0].size
    val adjacencyMap = createAdjacencyMap(map)
    var anyRemoved = true
    var count = 0
    while (anyRemoved) {
        anyRemoved = false
        for (y in 0..<height) {
            for (x in 0..<width) {
                if (adjacencyMap[y][x] in 0..<adjacentLimit) {
                    count++
                    anyRemoved = true
                    removeToiletPaperRoll(adjacencyMap, x, y)
                }
            }
        }
    }
    println(count)
}

fun parseInput1(input: String): List<List<Char>> = input
    .lines()
    .filter { it.isNotBlank() }
    .map { it.toCharArray().toList() }

fun createAdjacencyMap(map: List<List<Char>>): List<MutableList<Int>> {
    val adjacencyMap = List(height) { MutableList(width) { -1 } }
    for (y in 0..<height) {
        for (x in 0..<width) {
            if (map[y][x] != toiletPaperRoll) {
                continue
            }
            adjacencyMap[y][x]++
            for (y2 in y - 1..y + 1) {
                for (x2 in x - 1..x + 1) {
                    if (y == y2 && x == x2 || (y2 < 0 || x2 < 0 || y2 >= height || x2 >= width)) {
                        continue
                    }
                    if (map[y2][x2] == toiletPaperRoll) {
                        adjacencyMap[y2][x2]++
                    }
                }
            }
        }
    }
    return adjacencyMap
}

fun removeToiletPaperRoll(adjacencyMap: List<MutableList<Int>>, x: Int, y: Int) {
    adjacencyMap[y][x] = -1
    for (y2 in y - 1..y + 1) {
        for (x2 in x - 1..x + 1) {
            if (y == y2 && x == x2 || (y2 < 0 || x2 < 0 || y2 >= height || x2 >= width)) {
                continue
            }
            if (adjacencyMap[y2][x2] >= adjacentLimit) {
                adjacencyMap[y2][x2]--
            }
        }
    }
}
[–] ystael@beehaw.org 3 points 2 weeks ago

This is the first day I've wished I were working in J, like last year, instead of Lisp. Common Lisp arrays show the age of the language design. They're basically C arrays with less convenient syntax; if you want higher-order array iteration you have to write it yourself or use a package that provides it. This solution is also less efficient than it could be for part 2, because I recompute the full neighbors array on each iteration rather than updating it incrementally.

(defun read-inputs (filename)
  (let* ((input-lines (uiop:read-file-lines filename))
         (rows (length input-lines))
         (cols (length (car input-lines)))
         (result (make-array (list rows cols) :initial-element 0)))
    (loop for line in input-lines
          for y from 0 to (1- rows)
          do (loop for x from 0 to (1- cols)
                   if (eql (char line x) #\@)
                     do (setf (aref result y x) 1)))
    result))

(defun neighbors (grid)
  (let* ((dimensions (array-dimensions grid))
         (rows (car dimensions))
         (cols (cadr dimensions))
         (result (make-array dimensions :initial-element 0)))
    (flet ((neighbors-at (y x)
             (loop for dy from -1 to 1
                   sum (loop for dx from -1 to 1
                             sum (let ((yy (+ y dy))
                                       (xx (+ x dx)))
                                   (if (and (>= yy 0)
                                            (< yy rows)
                                            (>= xx 0)
                                            (< xx cols)
                                            (not (and (= dx 0) (= dy 0)))
                                            (> (aref grid yy xx) 0))
                                       1
                                       0))))))
      (loop for y from 0 to (1- rows)
            do (loop for x from 0 to (1- cols)
                     do (setf (aref result y x) (neighbors-at y x))))
      result)))

(defun main-1 (filename)
  (let* ((grid (read-inputs filename))
         (dimensions (array-dimensions grid))
         (neighbors-grid (neighbors grid)))
    (loop for y from 0 to (1- (car dimensions))
          sum (loop for x from 0 to (1- (cadr dimensions))
                    sum (if (and (> (aref grid y x) 0)
                                 (< (aref neighbors-grid y x) 4))
                            1
                            0)))))

(defun remove-accessible (grid)
  (let* ((dimensions (array-dimensions grid))
         (neighbors-grid (neighbors grid))
         (removed 0))
    (loop for y from 0 to (1- (car dimensions))
          do (loop for x from 0 to (1- (cadr dimensions))
                   do (if (and (> (aref grid y x) 0)
                               (< (aref neighbors-grid y x) 4))
                          (progn
                            (setf (aref grid y x) 0)
                            (incf removed)))))
    removed))

(defun main-2 (filename)
  (let ((grid (read-inputs filename))
        (removed 0))
    (loop for this-removed = (remove-accessible grid)
          until (zerop this-removed)
          do (incf removed this-removed))
    removed))
[–] LeixB@lemmy.world 2 points 2 weeks ago

Haskell

import Data.Array.Unboxed
import Control.Arrow
import Data.Foldable

type Coord = (Int, Int)
type Diagram = UArray Coord Char

moves :: Coord -> [Coord]
moves pos = (.+. pos) <$> deltas
  where
    deltas = [(x, y) | x <- [-1, 0, 1], y <- [-1, 0, 1], not (x == 0 && y == 0)]
    (ax, ay) .+. (bx, by) = (ax + bx, ay + by)

parse :: String -> Diagram
parse s = listArray ((1, 1), (n, m)) $ concat l
  where
    l = lines s
    n = length l
    m = length $ head l

isRoll = (== '@')
numRolls = length . filter isRoll

neighbors d p = (d !) <$> filter (inRange (bounds d)) (moves p)

removable d = filter ((<4) . numRolls . neighbors d . fst) . filter (isRoll . snd) $ assocs d

part1 :: Diagram -> Int
part1 = length . removable

part2 d = fmap ((initial -) . fst) . find (uncurry (==)) $ zip stages (tail stages)
  where
    initial = numRolls $ elems d
    stages = numRolls . elems <$> iterate (\x -> x // toX (removable x)) d
    toX = fmap (second (const 'x'))

main = getContents >>= print . (part1 &&& part2) . parse
[–] Gobbel2000@programming.dev 2 points 2 weeks ago

Rust

View on github

fn parse_input(input: &str) -> Vec<Vec<bool>> {
    input
        .lines()
        .map(|l| l.chars().map(|c| c == '@').collect())
        .collect()
}

fn count_adj(grid: &[Vec<bool>], (x, y): (usize, usize)) -> usize {
    let width = grid[0].len();
    let height = grid.len();
    grid.iter()
        .take((y + 2).min(height))
        .skip(y.saturating_sub(1))
        .map(|r| {
            r.iter()
                .take((x + 2).min(width))
                .skip(x.saturating_sub(1))
                .take(3)
                .filter(|e| **e)
                .count()
        })
        .sum::<usize>()
}

fn part1(input: String) {
    let grid = parse_input(&input);
    let mut count = 0u32;
    for (y, row) in grid.iter().enumerate() {
        for (x, _) in row.iter().enumerate().filter(|(_, r)| **r) {
            let n_adj = count_adj(&grid, (x, y));
            // Center roll is counted too
            if n_adj < 5 {
                count += 1;
            }
        }
    }
    println!("{count}");
}

fn part2(input: String) {
    let mut grid = parse_input(&input);
    let mut removed = 0u32;
    loop {
        let mut next_grid = grid.clone();
        let prev_removed = removed;
        for (y, row) in grid.iter().enumerate() {
            for (x, _) in row.iter().enumerate().filter(|(_, r)| **r) {
                let n_adj = count_adj(&grid, (x, y));
                // Center roll is counted too
                if n_adj < 5 {
                    next_grid[y][x] = false;
                    removed += 1;
                }
            }
        }
        if removed == prev_removed {
            break;
        }
        grid = next_grid;
    }
    println!("{}", removed);
}

util::aoc_main!();
[–] VegOwOtenks@lemmy.world 2 points 2 weeks ago* (last edited 2 weeks ago)

Futhark

Only part 1 so far, I want to do part 2 later too.

This is my first ever futhark program. I have not yet figured out whether string parsing is possible or intended with this language. I used a combination of sed and vim to bring the input into a form futhark can read.

def neighbors (x: i32, y: i32): [8](i32, i32) = [(x+1, y+1), (x+1, y), (x+1, y-1), (x, y+1), (x, y-1), (x-1, y+1), (x-1, y), (x-1, y-1)]

def count 't (p: t -> bool) (xs: []t) : i32 = reduce (+) 0 (map (\ x -> i32.bool (p x)) xs)
def count2 't (p: t -> bool) (xs: [][]t) : i32 = reduce (+) 0 (map (count p) xs)

def zipIndices [n] 't (xs: [n]t): [n](i32, t) = zip (map i32.i64 (indices xs)) xs
def zipIndices2 [n][m] 't (xs: [m][n]t): [m][n]((i32, i32), t) = 
  let innerIndices = map zipIndices xs in
  let innerAndOuterIndices = zipIndices innerIndices in
  map (\ (r, a) -> map (\ (c, x) -> ((r, c), x)) a) innerAndOuterIndices

def countIndexed2 't (p: (i32, i32) -> t -> bool) (xs: [][]t): i32 = 
  let withIndices = zipIndices2 xs in
  count2 (\ (i, x) -> p i x) withIndices

type option 't
  = #single t
  | #empty

def safeIndex 't (xs: []t) (i: i32): option t = if i32.i64 (length xs) > i && i >= 0
  then #single xs[i]
  else #empty

def safeIndex2 't (xs: [][]t) ((r, c): (i32, i32)): option t = 
  match safeIndex xs r
    case #single a -> safeIndex a c
    case #empty -> #empty

def orElse 't (o: option t) (d: t): t =
  match o
    case #single x -> x
    case #empty    -> d

def isAccessible (grid: [][]bool) (p: (i32, i32)) (x:bool): bool =
  let neighborsOptions = map (safeIndex2 grid) (neighbors p) in
  let neighborsFilled = map (`orElse` false) neighborsOptions in
  x && count id neighborsFilled < 4

def mapIndexed2 'a 'b (f: (i32, i32) -> a -> b) (xs: [][]a): [][]b =
  let withIndices = zipIndices2 xs in
  map (map (\ (i, x) -> f i x)) withIndices

def removeAccessibles (grid: [][]bool): [][]bool = mapIndexed2 (\ p x -> x && not (isAccessible grid p x)) grid

def part1 (grid: [][]bool): i32 = countIndexed2 (isAccessible grid) grid
def part2 (grid: [][]bool): i32 =
  let (reducedGrid, _) = 
    loop (current, last) = (removeAccessibles grid, grid)
    while current != last
    do 
      let current' = removeAccessibles current in
      let last'    = copy current in
      (current', last')
  in count2 id grid - count2 id reducedGrid

def main (grid: [][]bool) = (part1 grid, part2 grid)

The highlighting is a bit off because I used ocaml as the language. There is no futhark highlighter (at least in Web UI) yet.
Edit: Part2

Also, it runs blazingly fast πŸš€ :O, even in sequential C mode

[–] GiantTree@feddit.org 2 points 1 week ago* (last edited 1 week ago)

Kotlin

I'm catching up on this year's AOC.
This one was rather easy. I already have a pretty versatile grid class that I have just iterated as often as needed.

Doing this one also lead me into the rabbit hole that is source code generation in Gradle. I used this to generate all the implementations for the primitive types of the grid class as primitive arrays are not generic in the JVM.
An Array<Int> is an array of integer references, but an IntArray is an array of primitive integers.

Code on GitHub

Code

class Day04 : AOCSolution {
    override val year = 2025
    override val day = 4

    override fun part1(inputFile: String): String {
        val grid = readResourceLines(inputFile).map(CharSequence::toList).toCharGrid()

        var accessiblePaperRolls = 0

        // Quickly iterate the grid in top-left to bottom-right order
        for (y in 0 until grid.height) {
            for (x in 0 until grid.width) {
                // Count the neighbours of each paper roll.
                if (grid[x, y] == PAPER_ROLL &&
                    grid.countNeighbours(x, y, 1) { it == PAPER_ROLL } < 4
                ) {
                    accessiblePaperRolls++
                }
            }
        }
        return accessiblePaperRolls.toString()
    }

    override fun part2(inputFile: String): String {
        val grid = readResourceLines(inputFile).map(CharSequence::toList).toCharGrid()

        var count = 0
        while (true) {
            var iterationCount = 0

            // Quickly iterate the grid in top-left to bottom-right order
            for (y in 0 until grid.height) {
                for (x in 0 until grid.width) {
                    if (grid[x, y] == PAPER_ROLL &&
                        grid.countNeighbours(x, y, 1) { it == PAPER_ROLL } < 4
                    ) {
                        // Remove the paper roll for the next iteration
                        grid[x, y] = REMOVED_PAPER_ROLL
                        iterationCount++
                    }
                }
            }
            count += iterationCount

            // Repeat the count until no paper rolls are accessible.
            if (iterationCount == 0) {
                break
            }
        }

        return count.toString()
    }

    private companion object {
        const val PAPER_ROLL = '@'
        const val REMOVED_PAPER_ROLL = 'x'

        /**
         * Count the neighbours of the given cell in the given [radius] of cells that satisfy the given predicate.
         *
         * @param startX the horizontal position of the center of the count in the grid
         * @param startY the vertical position of the center of the count in the grid
         * @param radius the radius counted in Manhattan distance to the center
         * @param predicate the test that needs to pass for a neighbour to count.
         */
        private fun CharGrid.countNeighbours(
            startX: Int,
            startY: Int,
            radius: Int,
            predicate: Predicate<Char>,
        ): Int {
            var count = 0
            for (y in maxOf(startY - radius, 0)..minOf(startY + radius, height - 1)) {
                for (x in maxOf(startX - radius, 0)..minOf(startX + radius, width - 1)) {
                    if (y == startY && x == startX) {
                        continue
                    }
                    if (predicate.test(this[x, y])) {
                        count++
                    }
                }
            }
            return count
        }
    }
}

[–] Quant@programming.dev 2 points 2 weeks ago

Uiua

Quite simple this one. Part 2 still takes a few seconds because I'm essentially checking off each roll individually.

Run with example input

Code

$ ..@@.@@@@.
$ @@@.@.@.@@
$ @@@@@.@.@@
$ @.@@@@..@.
$ @@.@@@@.@@
$ .@@@@@@@.@
$ .@.@.@.@@@
$ @.@@@.@@@@
$ .@@@@@@@@.
$ @.@.@@@.@.
βŠœβˆ˜βŠΈβ‰ @\n
=@@

Rolls ← ⍣(β§»βŠšΛ™β€)∞⊸⊑1_1
Removable ← (
  ⬚0⧈Rolls[3_3 1_1 1_1]
  βŠšβ‰€4
)

Remove ← ⍜⊑(Λœβ†―0β§»)

PartOne ← β§»Removable

PartTwo ← (
  βŠ™0
  β₯(
    βŠ™(βŠ™+⟜⧻)⟜Removable
    ˜Remove
  )∞
  β—Œ
)
&pf "Part One: "
&p ⊸PartOne

&pf "Part Two: "
&p PartTwo

Old Part 2Before seeing mykl's solution this was my solution for part 2

PartTwoOld ← (
  0βŠ™0
  ⍒(βŠ™βŠ™β—Œ
    βŠ™βŠΈRemovable
    +βŠ™βŠΈβ§»
    βŠ™βŠΈRemove
  | Β¬β‰β—Œ)
  βŠ™β‹…β—Œ
)

It's basically the same, just that I used a while-do-loop, making the check for the ending condition myself (which took me a bit to get right because I still find loops in Uiua a bit confusing).
Using the repeat-loop as above also gets rid of the dip's (βŠ™). I could've removed them here as well but I was already deep in the trouble of getting the loop to work correctly and I liked the little face at the beginning 0βŠ™0

[–] CameronDev@programming.dev 2 points 2 weeks ago

Rust

   fn count_sides(grid: &[Vec<char>], x: usize, y: usize) -> usize {
        let mut count = 0;
        for i in y.saturating_sub(1)..=y + 1 {
            for j in x.saturating_sub(1)..=x + 1 {
                if i == y && j == x {
                    continue;
                }

                if let Some(row) = grid.get(i) {
                    if let Some(col) = row.get(j) {
                        if *col == '@' {
                            count += 1;
                        }
                    }
                }
            }
        }
        count
    }

    #[test]
    fn test_y2025_day4_part1() {
        let input = std::fs::read_to_string("input/2025/day_4.txt").unwrap();
        let grid = input
            .lines()
            .map(|l| l.chars().collect::<Vec<_>>())
            .collect::<Vec<_>>();
        let mut total = 0;
        let width = grid[0].len();
        let height = grid.len();
        for y in 0..height {
            for x in 0..width {
                if grid[y][x] != '@' {
                    continue;
                }

                if count_sides(&grid, x, y) < 4 {
                    total += 1
                }
            }
        }
        println!("Total = {total}")
    }

    #[test]
    fn test_y2025_day4_part2() {
        let input = std::fs::read_to_string("input/2025/day_4.txt").unwrap();
        let grid = input
            .lines()
            .map(|l| l.chars().collect::<Vec<_>>())
            .collect::<Vec<_>>();
        let mut grid = input
            .lines()
            .map(|l| l.chars().collect::<Vec<_>>())
            .collect::<Vec<_>>();
        let mut total = 0;
        let width = grid[0].len();
        let height = grid.len();
        loop {
            let mut iter_total = 0;
            for y in 0..height {
                for x in 0..width {
                    if grid[y][x] != '@' {
                        continue;
                    }

                    if count_sides(&grid, x, y) < 4 {
                        grid[y][x] = '*';
                        iter_total += 1
                    }
                }
            }
            println!("Moved = {iter_total}");
            if iter_total == 0 {
                break;
            }
            total += iter_total;
        }
        println!("Total = {total}");
    }

Nothing really exciting here, was pretty straightforward

[–] Chais@sh.itjust.works 2 points 1 week ago

Python

Send in the object orientation!
Honestly though, it was just a convenient way to keep things contained.

from pathlib import Path


class Grid:
    def __init__(self, input: str) -> None:
        self.grid = list(map(lambda l: list(map(lambda c: 1 if c == "@" else 0, l)), input.splitlines()))
        self.dims = (len(self.grid[0]), len(self.grid))

    def get(self, x: int, y: int) -> int:
        if x < 0 or x >= self.dims[0] or y < 0 or y >= self.dims[1]:
            return 0
        return self.grid[x][y]

    def set(self, x: int, y: int, value: int):
        self.grid[x][y] = value

    def is_accessible(self, x: int, y: int) -> bool:
        if self.get(x, y):
            return sum(map(lambda t: self.get(*t), [(ix, iy) for ix in range(x - 1, x + 2) for iy in range(y - 1, y + 2)])) - 1 < 4
        return False


def part_one(input: str) -> int:
    grid = Grid(input)
    return len(list(filter(None, map(lambda t: grid.is_accessible(*t), [(x, y) for x in range(grid.dims[0]) for y in range(grid.dims[1])]))))


def part_two(input: str) -> int:
    grid = Grid(input)
    total = 0
    while True:
        remove = list(filter(None, map(lambda t: t if grid.is_accessible(*t) else None, [(x, y) for x in range(grid.dims[0]) for y in range(grid.dims[1])])))
        total += len(remove)
        if remove:
            [grid.set(*t, 0) for t in remove]
        else:
            break
    return total


if __name__ == "__main__":
    input = Path("_2025/_4/input").read_text("utf-8")
    print(part_one(input))
    print(part_two(input))
[–] Jayjader@jlai.lu 1 points 1 week ago

(Browser-based) Javascript

This was a good opportunity to refresh my grasp on the math involved in losslessly stuffing a tuple into a single number. JS-in-the-browser has Sets and Maps but no Tuples, and Arrays are indexed on their id / memory handle instead of their value contents, so if you want to put coordinates into a set or map and have the collection behave as expected you need to serialize the coordinates into a primitive type. Stuff it into a string if you don't want to think too hard. For this specific problem we don't even need to be able to compute the original coordinates (just count the unique removed points) but implementing that computation was a handy way to verify the "serializer" was working correctly.

Seeing as the record tuple proposal was withdrawn in February of this year this is still a technique worth knowing when working with coords in JS.

Code

function part1(inputText) {
  const gridWidth = inputText.indexOf('\n');
  const lines = inputText.trim().split('\n');
  const gridHeight = lines.length;
  let accessibleRolls = 0;
  for (let y = 0; y < gridHeight; y++) {
    for (let x = 0; x < gridWidth; x++) {
      if (lines[y][x] === '@') {
        let occupiedNeighbors = 0;
        for (const [neighborX, neighborY] of [
            [x - 1, y],
            [x + 1, y],
            [x, y - 1],
            [x, y + 1],
            [x - 1, y - 1],
            [x - 1, y + 1],
            [x + 1, y - 1],
            [x + 1, y + 1],
          ]) {
          if (neighborX < 0 || neighborX >= gridWidth || neighborY < 0 || neighborY >= gridHeight) {
            continue;
          }
          if (lines[neighborY][neighborX] === '@') {
            occupiedNeighbors++;
          }
        }
        if (occupiedNeighbors < 4) {
          accessibleRolls++;
        }
      }
    }
  }
  return accessibleRolls;
}
{
  const start = performance.now()
  const result = part1(document.body.textContent)
  const end = performance.now()
  console.info({day: 4, part: 1, time: end - start, result})
}

function serializeCoords(x, y, gridWidth) {
  const leftShiftAmount = Math.ceil(Math.log10(gridWidth));
  return x * (10 ** leftShiftAmount) + y;
}
/*
{
  const x = 3;
  const y = 4;
  const gridWidth = 13;
  const serialized = serializeCoords(x, y, gridWidth);
  console.debug({ x, y, gridWidth, serialized });
}
*/

function deserializeCoords(serialized, gridWidth) {
  const leftShiftAmount = Math.ceil(Math.log10(gridWidth));
  const x = Math.floor(serialized / (10 ** leftShiftAmount));
  const y = serialized - x * 10 ** leftShiftAmount;
  return [x, y];
}
/*
{
  const serialized = 304;
  const gridWidth = 13;
  const [x, y] = deserializeCoords(serialized, gridWidth);
  console.debug({ serialized, gridWidth, x, y });
}
*/
function part2(inputText) {
  const gridWidth = inputText.indexOf('\n');
  const lines = inputText.trim().split('\n');
  const gridHeight = lines.length;
  let removed = new Set();
  while (true) {
    const toRemove = new Set();
    for (let y = 0; y < gridHeight; y++) {
      for (let x = 0; x < gridWidth; x++) {
        const serialized = serializeCoords(x, y, gridWidth);
        if (lines[y][x] === '@' && !removed.has(serialized)) {
          let occupiedNeighbors = 0;
          for (const [neighborX, neighborY] of [
              [x - 1, y],
              [x + 1, y],
              [x, y - 1],
              [x, y + 1],
              [x - 1, y - 1],
              [x - 1, y + 1],
              [x + 1, y - 1],
              [x + 1, y + 1],
            ]) {
            if (neighborX < 0 || neighborX >= gridWidth || neighborY < 0 || neighborY >= gridHeight) {
              continue;
            }
            const serializedNeighbor = serializeCoords(neighborX, neighborY, gridWidth);
            if (lines[neighborY][neighborX] === '@' && !removed.has(serializedNeighbor)) {
              occupiedNeighbors++;
            }
          }
          if (occupiedNeighbors < 4) {
            toRemove.add(serialized);
          }
        }
      }
    }
    if (toRemove.size === 0) {
      break;
    }
    removed = removed.union(toRemove);
  }
  return removed.size;
}
/*
{
  const exampleText = `..@@.@@@@.
@@@.@.@.@@
@@@@@.@.@@
@.@@@@..@.
@@.@@@@.@@
.@@@@@@@.@
.@.@.@.@@@
@.@@@.@@@@
.@@@@@@@@.
@.@.@@@.@.
`;
  const start = performance.now();
  const result = part2(exampleText);
  const end = performance.now();
  console.info({ day: 4, part: 2, time: end - start, result });
}
*/

{
  const start = performance.now()
  const result = part2(document.body.textContent)
  const end = performance.now()
  console.info({ day: 4, part: 2, time: end - start, result });
}