C

Solved part 1 without a grid, looked at part 2, almost spit out my coffee. Didn't see that coming!

I used my visualisation mini-library to generate video with ffmpeg, stepped through it a bit, then thought better of it - this is a programming puzzle after all!

So I wrote a heuristic to find frames low on entropy (specifically: having many robots in the same line of column), where each record-breaking frame number was printed. That pointed right at the correct frame!

It was pretty slow though (.2 secs or such) because it required marking spots on a grid. I noticed the Christmas tree was neatly tucked into a corner, concluded that wasn't an accident, and rewrote the heuristic to check for a high concentration in a single quadrant. Reverted this because the tree-in-quadrant assumption proved incorrect for other inputs. Would've been cool though!

::: spoiler Code

#include "common.h"

#define SAMPLE 0
#define GW (SAMPLE ? 11 : 101)
#define GH (SAMPLE ?  7 : 103)
#define NR 501

int
main(int argc, char **argv)
{
	static char g[GH][GW];
	static int px[NR],py[NR], vx[NR],vy[NR];

	int p1=0, n=0, sec, i, x,y, q[4]={}, run;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));

	for (; scanf(" p=%d,%d v=%d,%d", px+n,py+n, vx+n,vy+n)==4; n++)
		assert(n+1 < NR);

	for (sec=1; !SAMPLE || sec <= 100; sec++) {
		memset(g, 0, sizeof(g));
		memset(q, 0, sizeof(q));

		for (i=0; i<n; i++) {
			px[i] = (px[i] + vx[i] + GW) % GW;
			py[i] = (py[i] + vy[i] + GH) % GH;

			g[py[i]][px[i]] = 1;

			if (sec == 100) {
				if (px[i] < GW/2) {
					if (py[i] < GH/2) q[0]++; else
					if (py[i] > GH/2) q[1]++;
				} else if (px[i] > GW/2) {
					if (py[i] < GH/2) q[2]++; else
					if (py[i] > GH/2) q[3]++;
				}
			}
		}

		if (sec == 100)
			p1 = q[0]*q[1]*q[2]*q[3];

		for (y=0; y<GH; y++)
		for (x=0, run=0; x<GW; x++)
			if (!g[y][x])
				run = 0;
			else if (++run >= 10)
				goto found_p2;
	}

found_p2:
	printf("14: %d %d\n", p1, sec);
	return 0;
}

:::

https://github.com/sjmulder/aoc/blob/master/2024/c/day14.c

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

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: ::: spoiler 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. ::: spoiler 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;
}

:::

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.

::: spoiler 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]
        ]

:::

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

Dart

Took far too long to work out a really stupidly simple method of finding the tree -- I ended up just checking the first height*width time slots to find when the most bots appear in any given row/column. The framing around the Christmas tree accidentally made this foolproof :-). Add a bit of Chinese Remainder Theorem and we're golden. (edit: forgot to mention that it's Dart code)

import 'dart:math';
import 'package:collection/collection.dart';
import 'package:more/more.dart';

List<List<Point<num>>> getBots(List<String> lines) {
  var bots = lines
      .map((e) => RegExp(r'(-?\d+)')
          .allMatches(e)
          .map((m) => int.parse(m.group(0)!))
          .toList())
      .map((p) => [Point<num>(p[0], p[1]), Point<num>(p[2], p[3])])
      .toList();
  return bots;
}

// Solve system of congruences using the Chinese Remainder Theorem
int crt(int r1, int m1, int r2, int m2) {
  int inv = m1.modInverse(m2);
  int solution = (r1 + m1 * ((r2 - r1) % m2) * inv) % (m1 * m2);
  return (solution + (m1 * m2)) % (m1 * m2); // Ensure the result is positive
}

void moveBy(List<List<Point<num>>> bots, int t, int w, int h) {
  for (var b in bots) {
    b.first += b.last * t;
    b.first = Point(b.first.x % w, b.first.y % h);
  }
}

part1(List<String> lines, [width = 11, height = 7]) {
  var bots = getBots(lines);
  moveBy(bots, 100, width, height);
  var w = width ~/ 2, h = height ~/ 2;
  var quads = Multiset.fromIterable(
      bots.map((b) => (b.first.x.compareTo(w), b.first.y.compareTo(h))));
  return [(-1, -1), (-1, 1), (1, -1), (1, 1)]
      .map((k) => quads[k])
      .reduce((s, t) => s * t);
}

part2(List<String> lines, [width = 101, height = 103]) {
  var bots = getBots(lines);
  var t = 0;
  int rmax = 0, cmax = 0, rt = 0, ct = 0;
  while (t < width * height) {
    t += 1;
    moveBy(bots, 1, width, height);
    var r = Multiset.fromIterable(bots.map((e) => e.first.x)).counts.max;
    var c = Multiset.fromIterable(bots.map((e) => e.first.y)).counts.max;
    if (r > rmax) (rmax, rt) = (r, t);
    if (c > cmax) (cmax, ct) = (c, t);
  }
  t = crt(rt, width, ct, height);
  bots = getBots(lines);
  moveBy(bots, t, width, height);
  // printGrid(height, width, bots);
  return t;
}

Haskell

I solved part two interactively, I'm not very happy about it

::: spoiler 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

:::

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.

::: spoiler 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

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

Codeberg Repo

This one was wild. At first I started visualling it but wasn't seeing anything. I went up to almost 80 iterations before giving up. I then went with the "no overlapping points thing" but didn't think that was working for me either.

It was iteration 7,753. f's sake man.

I used the a modulo operation and an if check to skip ahead to the final state. It was also tricky b/c the grid is larger than I could get my console to work with.

F#

(just the important bits)

type Velocity = Point2I
[<Struct>]type Robot = {Position:Point2I; Velocity:Velocity}
type Boundary = {MaxX:int; MaxY: int}
let move ticks boundary robot =
    let newX = ((robot.Position.X + (robot.Velocity.X * ticks)) % boundary.MaxX) |> fun x -> if x < 0 then boundary.MaxX + x else x
    let newY = ((robot.Position.Y + (robot.Velocity.Y * ticks)) % boundary.MaxY) |> fun x -> if x < 0 then boundary.MaxY + x else x
    { robot with Position = {X = newX; Y = newY} }
    
let toQuadrantScore boundary robots =
    let removeX = ((boundary.MaxX |> float) / 2.0) |> int
    let removeY = ((boundary.MaxY |> float) / 2.0) |> int
    ((0,0,0,0), robots)
    ||> Seq.fold(fun (a,b,c,d) robot ->
        if robot.Position.X < removeX && robot.Position.Y < removeY then (a+1,b,c,d)
        else if robot.Position.X > removeX && robot.Position.Y < removeY then (a,b+1,c,d)
        else if robot.Position.X < removeX && robot.Position.Y > removeY then (a,b,c+1,d)
        else if (robot.Position.X > removeX && robot.Position.Y > removeY) then (a,b,c,d+1)
        else (a,b,c,d)
        )
    |> fun (a,b,c,d) -> a*b*c*d


let part1impl boundary ticks robots =
    robots
    |> Seq.map(move ticks boundary)
    |> toQuadrantScore boundary
    
let part1 input =
    (read input parse)
    |> part1impl {MaxX = 101; MaxY = 103 } 100


let part2 input =
    let robots = (read input parse) |> Array.ofSeq
    let boundary = {MaxX = 101; MaxY = 103 }
    
    // I'll steal the no overlapping robots approach
    // since I'll just be iterating through, I'll do batches of 100 numbers each in parallel and then draw it
    // up to 100 batches
    [0..100]
    |> List.find (fun batch ->
        try
            seq {0..100}
            |> Seq.pick (fun t ->
                // I could use PSeq here, but I'd have to remove the console stuff as the locking and fighting for the console in parallel really slows it
                let ticks = batch * 100 + t
                Console.Clear()
                Console.SetCursorPosition(0,0)
                Console.Write(ticks)
                
                let count =
                    robots
                    |> PSeq.map (move ticks boundary)
                    |> PSeq.distinctBy _.Position
                    |> PSeq.length
                    
                
                    
                if count = 500 then
                    ... write to file, console

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.

::: spoiler 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

:::

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'

Uiua

Ok, so part one wasn't too hard, and since uiua also takes negative values for accessing arrays, I didn't even have to care about converting my modulus results (though I did later for part two).
I'm a bit conflicted about the way I detected the quadrants the robots are in, or rather the way the creation of the mask-array happens. I basically made a 11x7 field of 0's, then picked out each quadrant and added 1-4 respectively. Uiua's group (⊕) function then takes care of putting all the robots in separate arrays for each quadrant. Simple.

For part two, I didn't even think long before I came here to see other's approaches. The idea to look for the first occurrence where no robots' positions overlapped was my starting point for what follows.

::: spoiler Example input stuff Run with example input here

$ p=0,4 v=3,-3
$ p=6,3 v=-1,-3
$ p=10,3 v=-1,2
$ p=2,0 v=2,-1
$ p=0,0 v=1,3
$ p=3,0 v=-2,-2
$ p=7,6 v=-1,-3
$ p=3,0 v=-1,-2
$ p=9,3 v=2,3
$ p=7,3 v=-1,2
$ p=2,4 v=2,-3
$ p=9,5 v=-3,-3
.
PartOne ← (
  # &rs ∞ &fo "input-14.txt"
  ⊜(↯2_2⋕regex"-?\\d+")≠@\n.
  ≡(⍜⌵(◿11_7)+°⊟⍜⊡₁×₁₀₀)
  ↯⟜(▽×°⊟)7_11 0
  ⍜↙₃(⍜≡↙₅+₁⍜≡↘₆+₂)
  ⍜↘₄(⍜≡↙₅+₃⍜≡↘₆+₄)
  /×≡◇⧻⊕□-₁⊸(⊡:)⍉
)

PartTwo ← (
  # &rs ∞ &fo "input-14.txt"
  ⊜(↯2_2⋕regex"-?\\d+")≠@\n.
  0 # number of seconds to start at
  0_0
  ⍢(◡(≡(⍜⌵(◿11_7)+°⊟⍜⊡₁×):)◌
    ◿[11_7]≡+[11_7]
    ⊙+₁
  | ≠⊙(⧻◴)⧻.)
  ⊙◌◌
  -₁
)

&p "Day 14:"
&pf "Part 1: "
&p PartOne
&pf "Part 2: "
&p PartTwo

:::

---

Now on to the more layered approach of how I got my solution.

In my case, there's two occasions of non-overlapping positions before the christmas tree appears.
I had some fun trying to get those frames and kept messing up with going back and forth between 7x11 vs 103x101 fields, often forgetting to adjust the modulus and other parts, so that was great.

In the end, I uploaded my input to the online uiua pad to make visualizing possible frames easier since uiua is able to output media if the arrays match a defined format.

::: spoiler Try it out yourself with your input

1. Open the uiua pad with code here
2. Replace the 0 in the first line with your solution for part two
3. If necessary, change the name of the file containing your input
4. Drag a file containing your input onto the pad to upload it and run the code
5. An image should be displayed :::

I used this code to find the occurrence of non-overlapping positions (running this locally):

&rs ∞ &fo "input-14.txt"
⊜(↯2_2⋕regex"-?\\d+")≠@\n.
0 # number of seconds to start at
0_0
⍢(◡(≡(⍜⌵(◿101_103)+°⊟⍜⊡₁×):)◌
  ◿[101_103]≡+[101_103]
  ⊙+₁
| ≠⊙(⧻◴)⧻.)
⊙◌◌
-₁

Whenever a new case was found, I put the result into the code in the online pad to check the generated image, and finally got this at the third try: