In case anybody hasn't seen it, the relevant Oglaf (NSFW)
Most people would use "word", "half-word", "quarter-word" etc, but the Anglophiles insist on "tuppit", "ternary piece", "span" and "chunk" (that's 5 bits, or 12 old bits).
Maybe it was due to attempting the puzzles in real-time for the first time, but it felt like there was quite a spike in difficulty this year. Day 5 (If You Give A Seed A Fertilizer) in particular was pretty tough for an early puzzle.
Day 8 (Haunted Wasteland), Day 20 (Pulse Propagation) and Day 21 (Step Counter) were (I felt) a bit mean due to hidden properties of the input data.
I particularly liked Day 6 (Wait For It), Day 14 (Parabolic Reflector Dish) and Day 24 (Never Tell Me The Odds), although that one made my brain hurt.
Day 25 (Snowverload) had me reading research papers, although in the end I stumbled across Karger's algorithm. That's the first time I've used a probabilistic approach. This solution in particular was very clever.
I learned the Shoelace formula and Pick's theorem this year, which will be very helpful to remember.
Perhaps I'll try using Prolog or J next year :)
Oh, just like day 11! I hadn't thought of that. I was initially about to try something similar by separating into rectangular regions, as in ear-clipping triangulation. But that would require a lot of iterating, and something about "polygon" and "walking the edges" went ping in my memory...
Haskell
Wasn't able to start on time today, but this was a fun one! Got to apply the two theorems I learned from somebody else's solution to Day 10.
Solution
import Data.Char
import Data.List
readInput :: String -> (Char, Int, String)
readInput s =
let [d, n, c] = words s
in (head d, read n, drop 2 $ init c)
boundary :: [(Char, Int)] -> [(Int, Int)]
boundary = scanl' step (0, 0)
where
step (x, y) (d, n) =
let (dx, dy) = case d of
'U' -> (0, 1)
'D' -> (0, -1)
'L' -> (-1, 0)
'R' -> (1, 0)
in (x + n * dx, y + n * dy)
area :: [(Char, Int)] -> Int
area steps =
let a = -- shoelace formula
(abs . (`quot` 2) . sum)
. (zipWith (\(x, y) (x', y') -> x * y' - x' * y) <*> tail)
$ boundary steps
in a + 1 + sum (map snd steps) `quot` 2 -- Pick's theorem
part1, part2 :: [(Char, Int, String)] -> Int
part1 = area . map (\(d, n, _) -> (d, n))
part2 = area . map (\(_, _, c) -> decode c)
where
decode s = ("RDLU" !! digitToInt (last s), read $ "0x" ++ init s)
main = do
input <- map readInput . lines <$> readFile "input18"
print $ part1 input
print $ part2 input
Clever! And removing constraints doesn't increase the path cost, so it won't be an overestimate.
Some (not very insightful or helpful) observations:
- The shortest path is likely to be mostly monotonic (it's quite hard for the "long way round" to be cost-effective), so the Manhattan distance is probably a good metric.
- The center of the puzzle is expensive, so the straight-line distance is probably not a good metric
- I'm pretty sure that the shortest route (for part one at least) can't self-intersect. Implementing this constraint is probably not going to speed things up, and there might be some pathological case where it's not true.
Not an optimization, but I suspect that a heuristic-based "reasonably good" path such as a human would take will be fairly close to optimal.
Yeah, finding a good way to represent the "last three moves" constraint was a really interesting twist. You beat me to it, anyway!
Haskell
Wowee, I took some wrong turns solving today's puzzle! After fixing some really inefficient pruning I ended up with a Dijkstra search that runs in 2.971s (for a less-than-impressive 124.782 l-s).
Solution
import Control.Monad
import Data.Array.Unboxed (UArray)
import qualified Data.Array.Unboxed as Array
import Data.Char
import qualified Data.HashSet as Set
import qualified Data.PQueue.Prio.Min as PQ
readInput :: String -> UArray (Int, Int) Int
readInput s =
let rows = lines s
in Array.amap digitToInt
. Array.listArray ((1, 1), (length rows, length $ head rows))
$ concat rows
walk :: (Int, Int) -> UArray (Int, Int) Int -> Int
walk (minStraight, maxStraight) grid = go Set.empty initPaths
where
initPaths = PQ.fromList [(0, ((1, 1), (d, 0))) | d <- [(0, 1), (1, 0)]]
goal = snd $ Array.bounds grid
go done paths =
case PQ.minViewWithKey paths of
Nothing -> error "no route"
Just ((n, (p@(y, x), hist@((dy, dx), k))), rest)
| p == goal && k >= minStraight -> n
| (p, hist) `Set.member` done -> go done rest
| otherwise ->
let next = do
h'@((dy', dx'), _) <-
join
[ guard (k >= minStraight) >> [((dx, dy), 1), ((-dx, -dy), 1)],
guard (k < maxStraight) >> [((dy, dx), k + 1)]
]
let p' = (y + dy', x + dx')
guard $ Array.inRange (Array.bounds grid) p'
return (n + grid Array.! p', (p', h'))
in go (Set.insert (p, hist) done) $
(PQ.union rest . PQ.fromList) next
main = do
input <- readInput <$> readFile "input17"
print $ walk (0, 3) input
print $ walk (4, 10) input
(edited for readability)
Haskell
A pretty by-the-book "walk all paths" algorithm. This could be made a lot faster with some caching.
Solution
import Control.Monad
import Data.Array.Unboxed (UArray)
import qualified Data.Array.Unboxed as A
import Data.Foldable
import Data.Set (Set)
import qualified Data.Set as Set
type Pos = (Int, Int)
readInput :: String -> UArray Pos Char
readInput s =
let rows = lines s
in A.listArray ((1, 1), (length rows, length $ head rows)) $ concat rows
energized :: (Pos, Pos) -> UArray Pos Char -> Set Pos
energized start grid = go Set.empty $ Set.singleton start
where
go seen beams
| Set.null beams = Set.map fst seen
| otherwise =
let seen' = seen `Set.union` beams
beams' = Set.fromList $ do
((y, x), (dy, dx)) <- toList beams
d'@(dy', dx') <- case grid A.! (y, x) of
'/' -> [(-dx, -dy)]
'\\' -> [(dx, dy)]
'|' | dx /= 0 -> [(-1, 0), (1, 0)]
'-' | dy /= 0 -> [(0, -1), (0, 1)]
_ -> [(dy, dx)]
let p' = (y + dy', x + dx')
beam' = (p', d')
guard $ A.inRange (A.bounds grid) p'
guard $ beam' `Set.notMember` seen'
return beam'
in go seen' beams'
part1 = Set.size . energized ((1, 1), (0, 1))
part2 input = maximum counts
where
(_, (h, w)) = A.bounds input
starts =
concat $
[[((y, 1), (0, 1)), ((y, w), (0, -1))] | y <- [1 .. h]]
++ [[((1, x), (1, 0)), ((h, x), (-1, 0))] | x <- [1 .. w]]
counts = map (\s -> Set.size $ energized s input) starts
main = do
input <- readInput <$> readFile "input16"
print $ part1 input
print $ part2 input
A whopping 130.050 line-seconds!
I'm not fluent in Rust, but is this something like the C++ placement new? Presumably just declaring a table of Vecs won't automatically call the default constructor? (Sorry for my total ignorance -- pointers to appropriate reading material appreciated)
Haskell
Took a while to figure out what part 2 was all about. Didn't have the energy to golf this one further today, so looking forward to seeing the other solutions!
Solution
0.3 line-seconds
import Data.Char
import Data.List
import Data.List.Split
import qualified Data.Vector as V
hash :: String -> Int
hash = foldl' (\a c -> ((a + ord c) * 17) `rem` 256) 0
hashmap :: [String] -> Int
hashmap = focus . V.toList . foldl' step (V.replicate 256 [])
where
focus = sum . zipWith focusBox [1 ..]
focusBox i = sum . zipWith (\j (_, z) -> i * j * z) [1 ..] . reverse
step boxes s =
let (label, op) = span isLetter s
i = hash label
in case op of
['-'] -> V.accum (flip filter) boxes [(i, (/= label) . fst)]
('=' : z) -> V.accum replace boxes [(i, (label, read z))]
replace ls (n, z) =
case findIndex ((== n) . fst) ls of
Just j ->
let (a, _ : b) = splitAt j ls
in a ++ (n, z) : b
Nothing -> (n, z) : ls
main = do
input <- splitOn "," . head . lines <$> readFile "input15"
print $ sum . map hash $ input
print $ hashmap input
Haskell
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
42.028 line-seconds
I probably should have made it clearer this is a somewhat tongue-in-cheek proposal :)
You're quite right - pretty much any program can be golfed into a single line.
Haskell
This was fun and (fairly) easy! Off-by-one errors are a likely source of bugs here.
import Control.Monad
import Data.List
import Data.List.Split
import Data.Maybe
score d pat = ((100 *) <$> search pat) `mplus` search (transpose pat)
where
search pat' = find ((d ==) . rdiff pat') [1 .. length pat' - 1]
rdiff pat' i =
let (a, b) = splitAt i pat'
in length $ filter (uncurry (/=)) $ zip (concat $ reverse a) (concat b)
main = do
input <- splitOn [""] . lines <$> readFile "input13"
let go d = print . sum . map (fromJust . score d) $ input
go 0
go 1
Line-seconds score: 0.102 π
Oh sure, it's only for fun - I was thinking of it more of a way to compare my own solutions to different problems.
I didn't notice there was a challenges community! That's awesome. (Maybe a more casual honor-based version where anybody can submit puzzles would be easier? Creating puzzles sounds like fun!)
Rating problems and solutions
We all know and love (!) the leaderboard, but how about a different method?
One can solve a problem with a simple, naive method resulting in a short program and long runtime, or put in lots of explicit optimizations for more code and shorter runtime. (Or if you're really good, a short, fast program!)
I propose the line-second.
Take the number of lines in your program (eg, 42 lines) and the runtime (eg 0.096 seconds). Multiply these together to get a score of 4.032 line-seconds.
A smaller score is a shorter, faster program.
Similarly, (for a particular solver), a larger score is a "harder" problem.
Haskell
Phew! I struggled with this one. A lot of the code here is from my original approach, which cuts down the search space to plausible positions for each group. Unfortunately, that was still way too slow...
It took an embarrassingly long time to try memoizing the search (which made precomputing valid points far less important). Anyway, here it is!
Solution
{-# LANGUAGE LambdaCase #-}
import Control.Monad
import Control.Monad.State
import Data.List
import Data.List.Split
import Data.Map (Map)
import qualified Data.Map as Map
import Data.Maybe
readInput :: String -> ([Maybe Bool], [Int])
readInput s =
let [a, b] = words s
in ( map (\case '#' -> Just True; '.' -> Just False; '?' -> Nothing) a,
map read $ splitOn "," b
)
arrangements :: ([Maybe Bool], [Int]) -> Int
arrangements (pat, gs) = evalState (searchMemo 0 groups) Map.empty
where
len = length pat
groups = zipWith startPoints gs $ zip minStarts maxStarts
where
minStarts = scanl (\a g -> a + g + 1) 0 $ init gs
maxStarts = map (len -) $ scanr1 (\g a -> a + g + 1) gs
startPoints g (a, b) =
let ps = do
(i, pat') <- zip [a .. b] $ tails $ drop a pat
guard $
all (\(p, x) -> maybe True (== x) p) $
zip pat' $
replicate g True ++ [False]
return i
in (g, ps)
clearableFrom i =
fmap snd $
listToMaybe $
takeWhile ((<= i) . fst) $
dropWhile ((< i) . snd) clearableRegions
where
clearableRegions =
let go i [] = []
go i pat =
let (a, a') = span (/= Just True) pat
(b, c) = span (== Just True) a'
in (i, i + length a - 1) : go (i + length a + length b) c
in go 0 pat
searchMemo :: Int -> [(Int, [Int])] -> State (Map (Int, Int) Int) Int
searchMemo i gs = do
let k = (i, length gs)
cached <- gets (Map.!? k)
case cached of
Just x -> return x
Nothing -> do
x <- search i gs
modify (Map.insert k x)
return x
search i gs | i >= len = return $ if null gs then 1 else 0
search i [] = return $
case clearableFrom i of
Just b | b == len - 1 -> 1
_ -> 0
search i ((g, ps) : gs) = do
let maxP = maybe i (1 +) $ clearableFrom i
ps' = takeWhile (<= maxP) $ dropWhile (< i) ps
sum <$> mapM (\p -> let i' = p + g + 1 in searchMemo i' gs) ps'
expand (pat, gs) =
(intercalate [Nothing] $ replicate 5 pat, concat $ replicate 5 gs)
main = do
input <- map readInput . lines <$> readFile "input12"
print $ sum $ map arrangements input
print $ sum $ map (arrangements . expand) input
Haskell
This problem has a nice closed form solution, but brute force also works.
(My keyboard broke during part two. Yet another day off the bottom of the leaderboard...)
import Control.Monad
import Data.Bifunctor
import Data.List
readInput :: String -> [(Int, Int)]
readInput = map (\[t, d] -> (read t, read d)) . tail . transpose . map words . lines
-- Quadratic formula
wins :: (Int, Int) -> Int
wins (t, d) =
let c = fromIntegral t / 2 :: Double
h = sqrt (fromIntegral $ t * t - 4 * d) / 2
in ceiling (c + h) - floor (c - h) - 1
main = do
input <- readInput <$> readFile "input06"
print $ product . map wins $ input
print $ wins . join bimap (read . concatMap show) . unzip $ input
Puzzle #6 - "Green"
Tried a little too hard to go with a theme on this one, and some of the clues are a bit contrived. Feel free to suggest alternatives!
Puzzle #5
Here's an old puzzle of mine to get started. One of the clues (at least!) is a little unfair, but the puzzle has been solved by others so it should be possible. Comments much appreciated, and more to come...