Day 4: Printing Department
Megathread guidelines
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FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
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.
AnArray<Int>is an array of integer references, but anIntArrayis an array of primitive integers.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 } } }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 paperRollPositionsPython
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) == 43Rust
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
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: discardToday 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
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 removedUiua
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. ⊜(=@@)⊸≠@\n N₈ ← ⊂A₂C₂ R ← ×<4⊸(/+⬚0↻)N₈ P₁ ← /+♭R P₂ ← /+♭-⊸⍥(-⊸R)∞ P₃ ← ⍥(▽₃10)<1e6/×⊸△⍥⊸(-⊸R)∞ ⊃(P₁|P₂|P₃)
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.
If you click the link on that post, you’ll see that the test data does resolve to a (very low res) elf!
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.

Now you’re just showing off!

Edit: ooh, this makes it obvious that my puzzle input takes more cycles to reach the done state.
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; }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.
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).
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(()) }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
sedandvimto bring the input into a formfutharkcan 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 opaque 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 part1 (grid: [][]bool) = countIndexed2 (isAccessible grid) grid def part2 (grid: [][]bool): i32 = 0 def main (grid: [][]bool) = (part1 grid, part2 grid)The highlighting is a bit off because I used
ocamlas the language. There is no futhark highlighter (at least in Web UI) yet.Rust
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!();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) . parseThis 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))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))








