AoC24 - Some stone blinking

February 9, 2025 - 2 minutes read - 342 words

Who said AoC25-11 should be complicated?

The Challenge

The ancient civilization on Pluto created stones that change every time you blink. Each stone follows specific transformation rules:
- A stone marked 0 becomes 1.
- A stone with an even number of digits splits into two stones, each half of the original number.
- Any other stone is replaced by a new one, with its number multiplied by 2024.
The stones remain in order, and their transformations continue with each blink. For example, the sequence [0, 1, 10, 99, 999] would change to [1, 2024, 1, 0, 9, 9, 2021976] after one blink.
Part 1: blink 25 times. Part 2: blink 75 times.

Full description available here.

Simple wouldn’t you say? While the first challenge required only 25 iterations, the answer was already of the order 10⁵ processing an input of only 8 stones.

Reaching 75 iterations for the part 2 of the challenge easily tricked me into increasing the number of iterations in my original code.

text

After waiting over 10 minutes, I had barely reached 29 iterations—seeing others on Reddit who faced the same struggle was somewhat reassuring!

Being relatively new to Python while tackling this challenge, I opted for a simple yet effective approach. While many shared more intricate and likely faster solutions, using a straightforward Counter got the job done in just 14 ms.

import numpy as np
from functools import cache
from collections import Counter


def get_input():
    with open('example.txt', 'r') as file:
        stones = []
        for line in file:
            for char in line.split(' '):
                stones.append(int(char))
        return stones

def blink_at_stones(blinks, stones):
    stones = Counter(stones)
    for i in range(blinks):
        print(f'Blink: {i}')
        output = Counter()
        for stone, occurences in stones.items():
            if stone == 0:
                #print('Found 0')
                output[1] += occurences
                #print('New array is' + str(stones))
                continue
            elif len(str(stone)) % 2 == 0:
                #print(f'Found even nb {len(str(stone))}')
                strSt = str(stone)
                size = len(strSt)
                output[int(strSt[:size//2])] += occurences
                output[int(strSt[size//2:])] += occurences
                continue
            else:
                #print('Found nothing!')
                output[stone*2024] += occurences
                #print('New array is' + str(stones))
                continue
        stones = output
    return sum(stones.values())

if __name__ == '__main__':
    print(blink_at_stones(75, get_input()))

Algos
Python