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)
Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ , pastebin, or github (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)
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
π Thread is locked until there's at least 100 2 star entries on the global leaderboard
Getting caught up slowly after spending way too long on day 12. I'll be busy this weekend though, so I'll probably fall further behind.
Part 2 looked daunting at first, as I knew brute-forcing 1 billion iterations wouldn't be practical. I did some premature optimization anyway, pre-calculating north/south and east/west runs in which the round rocks would be able to travel.
At first I figured maybe the rocks would eventually reach a stable configuration, so I added a check to detect if the current iteration matches the previous one. It never triggered, so I dumped some of the grid states and it became obvious that there was a cycle occurring. I probably should have guessed this in advance. The spin cycle is effectively a pseudorandom number generator, and all PRNGs eventually cycle. Good PRNGs have a very long cycle length, but this one isn't very good.
I added a hash table, mapping the state of each iteration to the next one. Once a value is added that already exists in the table as a key, there's a complete cycle. At that point it's just a matter of walking the cycle to determine it's length, and calculating from there.
A little slow (1.106s on my machine), but list operations made this really easy to write. I expect somebody more familiar with Haskell than me will be able to come up with a more elegant solution.
Nevertheless, 59th on the global leaderboard today! Woo!
Solution
import Data.List
import qualified Data.Map.Strict as Map
import Data.Semigroup
rotateL, rotateR, tiltW :: Endo [[Char]]
rotateL = Endo $ reverse . transpose
rotateR = Endo $ map reverse . transpose
tiltW = Endo $ map tiltRow
where
tiltRow xs =
let (a, b) = break (== '#') xs
(os, ds) = partition (== 'O') a
rest = case b of
('#' : b') -> '#' : tiltRow b'
[] -> []
in os ++ ds ++ rest
load rows = sum $ map rowLoad rows
where
rowLoad = sum . map (length rows -) . elemIndices 'O'
lookupCycle xs i =
let (o, p) = findCycle 0 Map.empty xs
in xs !! if i < o then i else (i - o) `rem` p + o
where
findCycle i seen (x : xs) =
case seen Map.!? x of
Just j -> (j, i - j)
Nothing -> findCycle (i + 1) (Map.insert x i seen) xs
main = do
input <- lines <$> readFile "input14"
print . load . appEndo (tiltW <> rotateL) $ input
print $
load $
lookupCycle
(iterate (appEndo $ stimes 4 (rotateR <> tiltW)) $ appEndo rotateL input)
1000000000
import numpy as np
from .solver import Solver
def _tilt(row: list[int], reverse: bool = False) -> list[int]:
res = row[::-1] if reverse else row[:]
rock_x = 0
for x, item in enumerate(res):
if item == 1:
rock_x = x + 1
if item == 2:
if rock_x < x:
res[rock_x] = 2
res[x] = 0
rock_x += 1
return res[::-1] if reverse else res
class Day14(Solver):
data: np.ndarray
def __init__(self):
super().__init__(14)
def presolve(self, input: str):
lines = input.splitlines()
self.data = np.zeros((len(lines), len(lines[0])), dtype=np.int8)
for x, line in enumerate(lines):
for y, char in enumerate(line):
if char == '#':
self.data[x, y] = 1
elif char == 'O':
self.data[x, y] = 2
def solve_first_star(self) -> int:
for y in range(self.data.shape[1]):
self.data[:, y] = _tilt(self.data[:, y].tolist())
return sum((self.data.shape[0] - x) * (self.data[x] == 2).sum() for x in range(self.data.shape[0]))
def solve_second_star(self) -> int:
seen = {}
order = []
for i in range(1_000_000_000):
order += [self.data.copy()]
s = self.data.tobytes()
if s in seen:
loop_size = i - seen[s]
remainder = (1_000_000_000 - i) % loop_size
self.data = order[seen[s] + remainder]
break
seen[s] = i
for y in range(self.data.shape[1]):
self.data[:, y] = _tilt(self.data[:, y].tolist())
for x in range(self.data.shape[0]):
self.data[x, :] = _tilt(self.data[x, :].tolist())
for y in range(self.data.shape[1]):
self.data[:, y] = _tilt(self.data[:, y].tolist(), reverse=True)
for x in range(self.data.shape[0]):
self.data[x, :] = _tilt(self.data[x, :].tolist(), reverse=True)
return sum((self.data.shape[0] - x) * (self.data[x] == 2).sum() for x in range(self.data.shape[0]))
33.938 line-seconds (ranks 3rd hardest after days 8 and 12 so far).