Day 4: Restroom Redoubt
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FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
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Haskell, alternative approach
The x and y coordinates of robots are independent. 101 and 103 are prime. So, the pattern of x coordinates will repeat every 101 ticks, and the pattern of y coordinates every 103 ticks.
For the first 101 ticks, take the histogram of x-coordinates and test it to see if it’s roughly randomly scattered by performing a chi-squared test using a uniform distrobution as the basis. [That code’s not given below, but it’s a trivial transliteration of the formula on wikipedia, for instance.] In my case I found a massive peak at t=99.
Same for the first 103 ticks and y coordinates. Mine showed up at t=58.
You’re then just looking for solutions of t = 101m + 99, t = 103n + 58 [in this case]. I’ve a library function, maybeCombineDiophantine, which computes the intersection of these things if any exist; again, this is basic wikipedia stuff.
day14b ls = let rs = parse ls size = (101, 103) positions = map (\t -> process size t rs) [0..] -- analyse x coordinates. These should have period 101 xs = zip [0..(fst size)] $ map (\rs -> map (\(p,_) -> fst p) rs & C.count & chi_squared (fst size)) positions xMax = xs & sortOn snd & last & fst -- analyse y coordinates. These should have period 103 ys = zip [0..(snd size)] $ map (\rs -> map (\(p,_) -> snd p) rs & C.count & chi_squared (snd size)) positions yMax = ys & sortOn snd & last & fst -- Find intersections of: t = 101 m + xMax, t = 103 n + yMax ans = do (s,t) <- maybeCombineDiophantine (fromIntegral (fst size), fromIntegral xMax) (fromIntegral (snd size), fromIntegral yMax) pure $ minNonNegative s t in trace ("xs distributions: " ++ show (sortOn snd xs)) $ trace ("ys distributions: " ++ show (sortOn snd ys)) $ trace ("xMax = " ++ show xMax ++ ", yMax = " ++ show yMax) $ trace ("answer could be " ++ show ans) $ ans
I should add - it’s perfectly possible to draw pictures which won’t be spotted by this test, but in this case as it happens the distributions are exceedingly nonuniform at the critical point.
Python
I was very confused when I saw the second part. I was like “how the fuck am I supposed now how that shape will exactly look like?” I looked a couple of initial shapes all of which looked sufficiently random. So I constructed x and y marginal distributions of the robots to impose some non-symmetry conditions.
My initial attempt was to just require maximum of x marginal should be at the centre but the sneaky bastards apparently framed the tree and tree was shorter than I expected (realised this only later) so that did not return any hits. I played around a bit and settled for: most of the maximums of x marginal should be near the centre and y marginal should be asymmetric. I still had to play around with the thresholds for these a bit because initially there was a couple of shapes (some repeating every 100 cycles or so) that satisfied my requirements (I had a part which actually outputted found shapes to a text file but then removed that in the final code). So it wasn’t %100 free of manual labour but I can say mostly…
import numpy as np from pathlib import Path from collections import Counter cwd = Path(__file__).parent def parse_input(file_path): with file_path.open("r") as fp: robot_info = fp.readlines() _split = lambda x,p: [int(x.split(' ')[p].split(',')[0].split('=')[-1]), int(x.split(' ')[p].split(',')[-1])] robot_pos = np.array([_split(x, 0) for x in robot_info]) robot_vel = np.array([_split(x, 1) for x in robot_info]) return robot_pos, robot_vel def solve_problem1(file_name, nrows, ncols, nmoves): robot_pos, robot_vel = parse_input(Path(cwd, file_name)) final_pos = robot_pos + nmoves*robot_vel final_pos = [(x[0]%ncols, x[1]%nrows) for x in list(final_pos)] pos_counts = Counter(final_pos) coords = np.array(list(pos_counts.keys()))[:,::-1] #x is cols, y is rows coords = tuple(coords.T) grid = np.zeros((nrows, ncols), dtype=int) grid[coords] += list(pos_counts.values()) counts = [np.sum(grid[:nrows>>1, :ncols>>1]), np.sum(grid[:nrows>>1, -(ncols>>1):]), np.sum(grid[-(nrows>>1):, :ncols>>1]), np.sum(grid[-(nrows>>1):, -(ncols>>1):])] return int(np.prod(counts)) def solve_problem2(file_name, nrows, ncols): robot_pos, robot_vel = parse_input(Path(cwd, file_name)) grid = np.zeros((nrows, ncols), dtype=object) # update all positions in a vectorised manner final_positions = robot_pos[None, :, :] + np.arange(1,10000)[:,None,None]*robot_vel[None,:,:] final_positions[:,:,0] = final_positions[:,:,0]%ncols final_positions[:,:,1] = final_positions[:,:,1]%nrows for s in range(final_positions.shape[0]): grid[:,:] = 0 final_pos = map(tuple, tuple(final_positions[s,:,:])) pos_counts = Counter(final_pos) coords = np.array(list(pos_counts.keys()))[:,::-1] #x is cols, y is rows coords = tuple(coords.T) grid[coords] += list(pos_counts.values()) xmarg = np.sum(grid, axis=0) tops = set(np.argsort(xmarg)[::-1][:10]) p_near_center = len(tops.intersection(set(range((ncols>>1)-5, (ncols>>1) + 6))))/10 ymarg = np.sum(grid, axis=1) ysym = 1 - abs(ymarg[:nrows>>1].sum() - ymarg[nrows>>1:].sum())/ymarg.sum() if p_near_center>0.5 and ysym<0.8: return s+1
Haskell
Spent a lot of time trying to find symmetric quadrants. In the end made an interactive visualization and found that a weird pattern appeared on iterations (27 + 101k) and (75 + 103k’). Put those congruences in an online Chinese remainder theorem calculator and go to the answer:
x ≡ 8006 (mod 101*103)
import Data.Bifunctor import Data.Char import qualified Data.Set as S import Data.Functor import Data.List import Control.Monad import Text.ParserCombinators.ReadP import Data.IORef bounds = (101, 103) parseInt :: ReadP Int parseInt = (*) <$> option 1 (char '-' $> (-1)) <*> (read <$> munch1 isDigit) parseTuple = (,) <$> parseInt <*> (char ',' *> parseInt) parseRow = (,) <$> (string "p=" *> parseTuple) <*> (string " v=" *> parseTuple) parse = fst . last . readP_to_S (endBy parseRow (char '\n')) move t (x, y) (vx, vy) = bimap (mod (x + vx * t)) (mod (y + vy * t)) bounds getQuadrant :: (Int, Int) -> Int getQuadrant (x, y) | x == mx || y == my = 0 | otherwise = case (x > mx, y > my) of (True, True) -> 1 (True, False) -> 2 (False, True) -> 3 (False, False) -> 4 where (mx, my) = bimap (`div` 2) (`div` 2) bounds step (x, y) (vx, vy) = (,(vx, vy)) $ bimap (mod (x + vx)) (mod (y + vy)) bounds main = do p <- parse <$> readFile "input14" print . product . fmap length . group . sort . filter (/=0) . fmap (getQuadrant . uncurry (move 100)) $ p let l = iterate (fmap (uncurry step)) p current <- newIORef 0 actions <- lines <$> getContents forM_ actions $ \a -> do case a of "" -> modifyIORef current (+1) "+" -> modifyIORef current (+1) "-" -> modifyIORef current (subtract 1) n -> writeIORef current (read n) pos <- readIORef current putStr "\ESC[2J" -- clear screen print pos visualize $ fst <$> l !! pos visualize :: [(Int, Int)] -> IO () visualize pos = do let p = S.fromList pos forM_ [1..(snd bounds)] $ \y -> do forM_ [1..(fst bounds)] $ \x -> do putChar $ if S.member (x, y) p then '*' else '.' putChar '\n'
Haskell. For part 2 I just wrote 10000 text files and went through them by hand. I quickly noticed that every 103 seconds, an image started to form, so it didn’t take that long to find the tree.
Code
import Data.Maybe import Text.ParserCombinators.ReadP import qualified Data.Map.Strict as M type Coord = (Int, Int) type Robot = (Coord, Coord) int :: ReadP Int int = fmap read $ many1 $ choice $ map char $ '-' : ['0' .. '9'] coord :: ReadP Coord coord = (,) <$> int <*> (char ',' *> int) robot :: ReadP Robot robot = (,) <$> (string "p=" *> coord) <*> (string " v=" *> coord) robots :: ReadP [Robot] robots = sepBy robot (char '\n') simulate :: Coord -> Int -> Robot -> Coord simulate (x0, y0) t ((x, y), (vx, vy)) = ((x + t * vx) `mod` x0, (y + t * vy) `mod` y0) quadrant :: Coord -> Coord -> Maybe Int quadrant (x0, y0) (x, y) = case (compare (2*x + 1) x0, compare (2*y + 1) y0) of (LT, LT) -> Just 0 (LT, GT) -> Just 1 (GT, LT) -> Just 2 (GT, GT) -> Just 3 _ -> Nothing freqs :: (Foldable t, Ord a) => t a -> M.Map a Int freqs = foldr (\x -> M.insertWith (+) x 1) M.empty solve :: Coord -> Int -> [Robot] -> Int solve grid t = product . freqs . catMaybes . map (quadrant grid . simulate grid t) showGrid :: Coord -> [Coord] -> String showGrid (x0, y0) cs = unlines [ [if (x, y) `M.member` m then '#' else ' ' | x <- [0 .. x0]] | let m = M.fromList [(c, ()) | c <- cs] , y <- [0 .. y0] ] main :: IO () main = do rs <- fst . last . readP_to_S robots <$> getContents let g = (101, 103) print $ solve g 100 rs sequence_ [ writeFile ("tree_" ++ show t) $ showGrid g $ map (simulate g t) rs | t <- [0 .. 10000] ]
Rust
Part 2 was very surprising in that it had a very vague requirement: “Find christmas tree!”. But my idea of finding the first round where no robots overlap turned out to just work when printing the map, so that was nice. I’m glad I did not instead start looking for symmetric patterns, because the christmas tree map is not symmetric at all.
Solution
use euclid::default::*; use regex::Regex; fn parse(input: &str) -> Vec<(Point2D<i32>, Vector2D<i32>)> { let re = Regex::new(r"p=(\d+),(\d+) v=(-?\d+),(-?\d+)").unwrap(); re.captures_iter(input) .map(|cap| { let (_, [p0, p1, v0, v1]) = cap.extract(); ( Point2D::new(p0.parse().unwrap(), p1.parse().unwrap()), Vector2D::new(v0.parse().unwrap(), v1.parse().unwrap()), ) }) .collect() } const ROOM: Size2D<i32> = Size2D::new(101, 103); const TIME: i32 = 100; fn part1(input: String) { let robots = parse(&input); let new_pos: Vec<Point2D<i32>> = robots.iter() .map(|&(p, v)| (p + v * TIME).rem_euclid(&ROOM)) .collect(); assert_eq!(ROOM.width % 2, 1); assert_eq!(ROOM.height % 2, 1); let mid_x = ROOM.width / 2; let mid_y = ROOM.height / 2; let mut q = [0u32; 4]; for p in new_pos { use std::cmp::Ordering::*; match (p.x.cmp(&mid_x), p.y.cmp(&mid_y)) { (Less, Less) => q[0] += 1, (Greater, Less) => q[1] += 1, (Less, Greater) => q[2] += 1, (Greater, Greater) => q[3] += 1, _ => {} }; } let prod = q[0] * q[1] * q[2] * q[3]; println!("{prod}"); } fn print_map(map: &[Vec<bool>]) { for row in map { for p in row { if *p { print!("#") } else { print!(".") } } println!(); } println!(); } fn part2(input: String) { let mut robots = parse(&input); let mut map = vec![vec![false; ROOM.width as usize]; ROOM.height as usize]; for i in 1.. { let mut overlap = false; for (p, v) in &mut robots { *p = (*p + *v).rem_euclid(&ROOM); if map[p.y as usize][p.x as usize] { // Found two robots on the same spot, // which is used as a heuristic for detecting the easter egg. overlap = true; } else { map[p.y as usize][p.x as usize] = true; } } if !overlap { print_map(&map); println!("Round: {i}"); break; } for row in &mut map { row.fill(false); } } } util::aoc_main!();
Also on github
Haskell
I solved part two interactively, I’m not very happy about it
Reveal Code
import Control.Arrow import Data.Bifunctor hiding (first, second) import Control.Monad import qualified Data.List as List import qualified Data.Set as Set import qualified Data.Map as Map import qualified Data.Maybe as Maybe parse :: String -> [((Int, Int), (Int, Int))] parse = map (break (== ' ') >>> second (drop 1) >>> join bimap (drop 2) >>> join bimap (break (== ',')) >>> join bimap (second (drop 1)) >>> join bimap (join bimap read)) . filter (/= "") . lines moveRobot ((px, py), (vx, vy)) t = (px + t * vx, py + t * vy) constrainCoordinates (mx, my) (px, py) = (px `mod` mx, py `mod` my) coordinateConstraints = (101, 103) robotQuadrant (mx, my) (px, py) | px > middleX && py < middleY = Just 1 -- upper right | px > middleX && py > middleY = Just 2 -- lower right | px < middleX && py > middleY = Just 3 -- lower left | px < middleX && py < middleY = Just 4 -- upper left | otherwise = Nothing where middleX = (mx `div` 2) middleY = (my `div` 2) countQuadrants (q1, q2, q3, q4) 1 = (succ q1, q2, q3, q4) countQuadrants (q1, q2, q3, q4) 2 = (q1, succ q2, q3, q4) countQuadrants (q1, q2, q3, q4) 3 = (q1, q2, succ q3, q4) countQuadrants (q1, q2, q3, q4) 4 = (q1, q2, q3, succ q4) part1 = map (flip moveRobot 100 >>> constrainCoordinates coordinateConstraints) >>> map (robotQuadrant coordinateConstraints) >>> Maybe.catMaybes >>> foldl (countQuadrants) (0, 0, 0, 0) >>> \ (a, b, c, d) -> a * b * c * d showMaybe (Just i) = head . show $ i showMaybe Nothing = ' ' buildRobotString robotMap = [ [ showMaybe (robotMap Map.!? (x, y)) | x <- [0..fst coordinateConstraints] ] | y <- [0..snd coordinateConstraints]] part2 rs t = map (flip moveRobot t >>> constrainCoordinates coordinateConstraints) >>> flip zip (repeat 1) >>> Map.fromListWith (+) >>> buildRobotString $ rs showConstellation (i, s) = do putStrLn (replicate 49 '#' ++ show i ++ replicate 49 '#') putStrLn $ s main = do f <- getContents let i = parse f print $ part1 i let constellations = map (id &&& (part2 i >>> List.intercalate "\n")) . filter ((== 86) . (`mod` 103)) $ [0..1000000] mapM_ showConstellation constellations print 7502
Haskell
Part 2 could be improved significantly now that I know what to look for, but this is the (very inefficient) heuristic I eventually found the answer with.
Solution
import Control.Arrow import Data.Char import Data.List import Data.Map qualified as Map import Data.Maybe import Text.Parsec (w, h) = (101, 103) readInput :: String -> [((Int, Int), (Int, Int))] readInput = either (error . show) id . parse (robot `endBy` newline) "" where robot = (,) <$> (string "p=" >> coords) <*> (string " v=" >> coords) coords = (,) <$> num <* char ',' <*> num num = read <$> ((++) <$> option "" (string "-") <*> many1 digit) runBots :: [((Int, Int), (Int, Int))] -> [[(Int, Int)]] runBots = transpose . map botPath where botPath (p, (vx, vy)) = iterate (incWrap w vx *** incWrap h vy) p incWrap s d = (`mod` s) . (+ d) safetyFactor :: [(Int, Int)] -> Int safetyFactor = product . Map.fromListWith (+) . map (,1) . mapMaybe quadrant where cx = w `div` 2 cy = h `div` 2 quadrant (x, y) | x == cx || y == cy = Nothing | otherwise = Just (x `div` (cx + 1), y `div` (cy + 1)) render :: [(Int, Int)] -> [String] render bots = let counts = Map.fromListWith (+) $ map (,1) bots in flip map [0 .. h - 1] $ \y -> flip map [0 .. w - 1] $ \x -> maybe '.' intToDigit $ counts Map.!? (x, y) isImage :: [String] -> Bool isImage = (> 4) . length . filter hasRun where hasRun = any ((> 3) . length) . filter head . group . map (/= '.') main = do positions <- runBots . readInput <$> readFile "input14" print . safetyFactor $ positions !! 100 let (Just (t, image)) = find (isImage . snd) $ zip [0 ..] $ map render positions print t mapM_ putStrLn image
TypeScript
Part 2 was a major curveball for sure. I was expecting something like the grid size and number of seconds multiplying by a large amount to make iterative solutions unfeasible.
First I was baffled how we’re supposed to know what shape we’re looking for exactly. I just started printing out cases where many robots were next to each other and checked them by hand and eventually found it. For my input the correct picture looked like this:
The Christmas tree
Later it turned out that a much simpler way is to just check for the first time none of the robots are overlapping each other. I cannot say for sure if this works for every input, but I suspect the inputs are generated in such a way that this approach always works.
The code
import fs from "fs"; type Coord = {x: number, y: number}; type Robot = {start: Coord, velocity: Coord}; const SIZE: Coord = {x: 101, y: 103}; const input: Robot[] = fs.readFileSync("./14/input.txt", "utf-8") .split(/[\r\n]+/) .map(row => /p=(-?\d+),(-?\d+)\sv=(-?\d+),(-?\d+)/.exec(row)) .filter(matcher => matcher != null) .map(matcher => { return { start: {x: parseInt(matcher[1]), y: parseInt(matcher[2])}, velocity: {x: parseInt(matcher[3]), y: parseInt(matcher[4])} }; }); console.info("Part 1: " + safetyFactor(input.map(robot => calculatePosition(robot, SIZE, 100)), SIZE)); // Part 2 // Turns out the Christmas tree is arranged the first time none of the robots are overlapping for (let i = 101; true; i++) { const positions = input.map(robot => calculatePosition(robot, SIZE, i)); if (positions.every((position, index, arr) => arr.findIndex(pos => pos.x === position.x && pos.y === position.y) === index)) { console.info("Part 2: " + i); break; } } function calculatePosition(robot: Robot, size: Coord, seconds: number): Coord { return { x: ((robot.start.x + robot.velocity.x * seconds) % size.x + size.x) % size.x, y: ((robot.start.y + robot.velocity.y * seconds) % size.y + size.y) % size.y }; } function safetyFactor(positions: Coord[], size: Coord): number { const midX = Math.floor(size.x / 2); const midY = Math.floor(size.y / 2); let quadrant0 = 0; // Top-left let quadrant1 = 0; // Top-right let quadrant2 = 0; // Bottom-left let quadrant3 = 0; // Bottom-right for (const {x,y} of positions) { if (x === midX || y === midY) { continue; } if (x < midX && y < midY) { quadrant0++; } else if (x > midX && y < midY) { quadrant1++; } else if (x < midX && y > midY) { quadrant2++; } else if (x > midX && y > midY) { quadrant3++; } } return quadrant0 * quadrant1 * quadrant2 * quadrant3; }
Checking for no overlaps is an interesting one. Intuitively I’d expect that to happen more often due to the low density, but as you say perhaps it’s deliberate.
Nim
Part 1: there’s no need to simulate each step, final position for each robot is
(position + velocity * iterations) modulo grid
Part 2: I solved it interactively: Maybe I just got lucky, but my input has certain pattern: after 99th iteration every 101st iteration looking very different from other. I printed first couple hundred iterations, noticed a pattern and started looking only at “interesting” grids. It took 7371 iterations (I only had to check 72 manually) to reach an easter egg.type Vec2 = tuple[x,y: int] Robot = object pos, vel: Vec2 var GridRows = 101 GridCols = 103 proc examine(robots: seq[Robot]) = for y in 0..<GridCols: for x in 0..<GridRows: let c = robots.countIt(it.pos == (x, y)) stdout.write if c == 0: '.' else: char('0'.ord + c) stdout.write '\n' stdout.flushFile() proc solve(input: string): AOCSolution[int, int] = var robots: seq[Robot] for line in input.splitLines(): let parts = line.split({' ',',','='}) robots.add Robot(pos: (parts[1].parseInt,parts[2].parseInt), vel: (parts[4].parseInt,parts[5].parseInt)) block p1: var quads: array[4, int] for robot in robots: let newX = (robot.pos.x + robot.vel.x * 100).euclmod GridRows newY = (robot.pos.y + robot.vel.y * 100).euclmod GridCols relRow = cmp(newX, GridRows div 2) relCol = cmp(newY, GridCols div 2) if relRow == 0 or relCol == 0: continue inc quads[int(relCol>0)*2 + int(relRow>0)] result.part1 = quads.foldl(a*b) block p2: if GridRows != 101: break p2 var interesting = 99 var interval = 101 var i = 0 while true: for robot in robots.mitems: robot.pos.x = (robot.pos.x + robot.vel.x).euclmod GridRows robot.pos.y = (robot.pos.y + robot.vel.y).euclmod GridCols inc i if i == interesting: robots.examine() echo "Iteration #", i, "; Do you see an xmas tree?[N/y]" if stdin.readLine().normalize() == "y": result.part2 = i break interesting += interval