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

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console.log('Hello World')

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Day 4: Printing Department

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[โ€“] 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