# Advent of Code 2020 — day 24 remade

I wanted to get back to yesterday’s puzzle, and I have a little time before I need to run off…

Yes, I was fed up with manual parsing of strings so decided to do it using the re module for once. :p

— Yassine Alouini (@YassineAlouini) December 24, 2020

I will have a look at your cleaned code for sure. Enjoy the day. 🥳

The use of `findall()`

I refer to in the tweet is this. You can split the input using a regular expression, and then map each input code to a direction. If you combine that with a `reduce()`

you get a very succinct parser:

```
f = open('/Users/mailund/Projects/adventofcode/2020/24/input.txt')
tiles = f.read().strip().split()
DIRECTIONS = {
'ne': 1 + 1j,
'e': 2,
'se': 1 - 1j,
'sw': -1 - 1j,
'w': -2,
'nw': -1 + 1j
}
import re
from functools import reduce
def tile_location(tile):
return reduce(
lambda x, y: x + DIRECTIONS[y],
re.findall(r'(ne|e|se|sw|w|nw)', tile),
0
)
```

That piece I am shamelessly stealing. It is much nicer than my original code.

To continue solving Puzzle #1 (again), to compute the black tiles, we need the tiles that have been flipped an odd number of times. The way I used a set earlier is smarter than the code below—it is a little easier to read and a lot faster—but this was a fun way to do it:

```
from collections import Counter
def flip_tiles(tiles):
counts = Counter(tiles)
return { tile for tile in counts if counts[tile] % 2 }
```

And that is all it takes to solve Puzzle #1:

```
tiles = flip_tiles(map(tile_location, tiles))
print(f"Puzzle #1: {len(tiles)}")
```

For Puzzle #2, I had a class for tracking the floor, but since I was only keeping track of the black tiles anyway, I might as well solve everything with set operations. With a few helper functions, we can navigate the floor from the black tiles:

```
def neighbours(tile):
return { tile + n for n in DIRECTIONS.values() }
def black_neighbours(tile, black):
return len(black & neighbours(tile))
def black_tiles(tiles):
return tiles
def white_tiles(tiles):
return set.union( *({ x for x in neighbours(tile) if x not in tiles }
for tile in tiles) )
```

When updating the floor, there are black tiles we need to remove and white tiles we need to add (by flipping them to black), and it is a simple case of mapping through the black and white tiles to compute a set of tiles to remove and a set of tiles to add:

```
def update(tiles):
to_remove = {
b for b in black_tiles(tiles)
if not (1 <= black_neighbours(b, tiles) <= 2)
}
to_add = {
w for w in white_tiles(tiles)
if black_neighbours(w, tiles) == 2
}
tiles.difference_update(to_remove)
tiles.update(to_add)
def evolve(tiles, n):
for _ in range(n):
update(tiles)
return tiles
```

And that solves Puzzle #2.

`print(f"Puzzle #2: {len(evolve(tiles, 100))}")`