Day 20: Race Condition
Megathread guidelines
- 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)
- You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL
FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- 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
Haskell
solution
import Control.Arrow import Data.Array.Unboxed import Data.Functor import Data.List import Data.Map qualified as M import Data.Set qualified as S type Pos = (Int, Int) type Board = Array Pos Char type Path = M.Map Pos Int parse board = listArray ((1, 1), (length l, length $ head l)) (concat l) where l = lines board moves :: Pos -> [Pos] moves p = [first succ p, first pred p, second succ p, second pred p] getOrigin :: Board -> Maybe Pos getOrigin = fmap fst . find ((== 'S') . snd) . assocs getPath :: Board -> Pos -> [Pos] getPath board p | not $ inRange (bounds board) p = [] | board ! p == 'E' = [p] | board ! p == '#' = [] | otherwise = p : (moves p >>= getPath (board // [(p, '#')])) taxiCab (xa, ya) (xb, yb) = abs (xa - xb) + abs (ya - yb) solve dist board = do path <- M.fromList . flip zip [1 ..] <$> (getOrigin board <&> getPath board) let positions = M.keys path jumps = [ (path M.! a) - (path M.! b) - d | a <- positions, b <- positions, d <- [taxiCab a b], d <= dist] return $ length $ filter (>=100) jumps main = getContents >>= print . (solve 2 &&& solve 20) . parse
C
Note to self: reread the puzzle after waking up properly! I initially wrote a great solution for the wrong question (pathfinding with a given number of allowed jumps).
For the actual question, a bit boring but flood fill plus some array iteration saved the day again. Find the cost for every open tile with flood fill, then for each position and offset combination, see if that jump yields a lower cost at the destination.
For arbitrary inputs this would require eliminating the non-optimal paths, but the one path covered all open tiles.
Code
#include "common.h" #define GB 2 /* safety border on X/Y plane */ #define GZ (GB+141+2+GB) static char G[GZ][GZ]; /* grid */ static int C[GZ][GZ]; /* costs */ static int sx,sy, ex,ey; /* start, end pos */ static void flood(int x, int y) { int lo = INT_MAX; if (x<1 || x>=GZ-1 || y<1 || y>=GZ-1 || G[y][x]!='.') return; if (C[y-1][x]) lo = MIN(lo, C[y-1][x]+1); if (C[y+1][x]) lo = MIN(lo, C[y+1][x]+1); if (C[y][x-1]) lo = MIN(lo, C[y][x-1]+1); if (C[y][x+1]) lo = MIN(lo, C[y][x+1]+1); if (lo != INT_MAX && (!C[y][x] || lo < C[y][x])) { C[y][x] = lo; flood(x, y-1); flood(x-1, y); flood(x, y+1); flood(x+1, y); } } static int count_shortcuts(int lim, int min) { int acc=0, x,y, dx,dy; for (y=GB; y<GZ-GB; y++) for (x=GB; x<GZ-GB; x++) for (dy = -lim; dy <= lim; dy++) for (dx = abs(dy)-lim; dx <= lim-abs(dy); dx++) acc += x+dx >= 0 && x+dx < GZ && y+dy >= 0 && y+dy < GZ && C[y][x] && C[y][x]+abs(dx)+abs(dy) <= C[y+dy][x+dx]-min; return acc; } int main(int argc, char **argv) { int x,y; if (argc > 1) DISCARD(freopen(argv[1], "r", stdin)); for (y=2; fgets(G[y]+GB, GZ-GB*2, stdin); y++) assert(y < GZ-3); for (y=GB; y<GZ-GB; y++) for (x=GB; x<GZ-GB; x++) if (G[y][x] == 'S') { G[y][x]='.'; sx=x; sy=y; } else if (G[y][x] == 'E') { G[y][x]='.'; ex=x; ey=y; } C[sy][sx] = 1; flood(sx, sy-1); flood(sx-1, sy); flood(sx, sy+1); flood(sx+1, sy); printf("20: %d %d\n", count_shortcuts(2, 100), count_shortcuts(20, 100)); return 0; }
https://codeberg.org/sjmulder/aoc/src/branch/master/2024/c/day20.c
Rust
I was stuck for a while, even after getting a few hints, until I read the problem more closely and realized: there is only one non-cheating path, and every free space is on it. This means that the target of any shortcut is guaranteed to be on the shortest path to the end.
This made things relatively simple. I used Dijkstra to calculate the distance from the start to each space. I then looked at every pair of points: if they are a valid distance away from each other, check how much time I would save jumping from one to the next. If that amount of time is in the range we want, then this is a valid cheat.
https://gitlab.com/bricka/advent-of-code-2024-rust/-/blob/main/src/days/day20.rs?ref_type=heads
The Code
// Critical point to note: EVERY free space is on the shortest path. use itertools::Itertools; use crate::search::dijkstra; use crate::solver::DaySolver; use crate::grid::{Coordinate, Grid}; type MyGrid = Grid<MazeElement>; enum MazeElement { Wall, Free, Start, End, } impl MazeElement { fn is_free(&self) -> bool { !matches!(self, MazeElement::Wall) } } fn parse_input(input: String) -> (MyGrid, Coordinate) { let grid: MyGrid = input.lines() .map(|line| line.chars().map(|c| match c { '#' => MazeElement::Wall, '.' => MazeElement::Free, 'S' => MazeElement::Start, 'E' => MazeElement::End, _ => panic!("Invalid maze element: {}", c) }) .collect()) .collect::<Vec<Vec<MazeElement>>>() .into(); let start_pos = grid.iter().find(|(_, me)| matches!(me, MazeElement::Start)).unwrap().0; (grid, start_pos) } fn solve<R>(grid: &MyGrid, start_pos: Coordinate, min_save_time: usize, in_range: R) -> usize where R: Fn(Coordinate, Coordinate) -> bool { let (cost_to, _) = dijkstra( start_pos, |&c| grid.orthogonal_neighbors_iter(c) .filter(|&n| grid[n].is_free()) .map(|n| (n, 1)) .collect() ); cost_to.keys() .cartesian_product(cost_to.keys()) .map(|(&c1, &c2)| (c1, c2)) // We don't compare with ourself .filter(|&(c1, c2)| c1 != c2) // The two points need to be within range .filter(|&(c1, c2)| in_range(c1, c2)) // We need to save at least `min_save_time` .filter(|(c1, c2)| { // Because we are working with `usize`, the subtraction // could underflow. So we need to use `checked_sub` // instead, and check that a) no underflow happened, and // b) that the time saved is at least the minimum. cost_to.get(c2).copied() .and_then(|n| n.checked_sub(*cost_to.get(c1).unwrap())) .and_then(|n| n.checked_sub(c1.distance_to(c2))) .map(|n| n >= min_save_time) .unwrap_or(false) }) .count() } pub struct Day20Solver; impl DaySolver for Day20Solver { fn part1(&self, input: String) -> String { let (grid, start_pos) = parse_input(input); solve( &grid, start_pos, 100, |c1, c2| c1.distance_to(&c2) == 2, ).to_string() } fn part2(&self, input: String) -> String { let (grid, start_pos) = parse_input(input); solve( &grid, start_pos, 100, |c1, c2| c1.distance_to(&c2) <= 20, ).to_string() } } #[cfg(test)] mod tests { use super::*; #[test] fn test_part1() { let input = include_str!("../../inputs/test/20"); let (grid, start_pos) = parse_input(input.to_string()); let actual = solve(&grid, start_pos, 1, |c1, c2| c1.distance_to(&c2) == 2); assert_eq!(44, actual); } #[test] fn test_part2() { let input = include_str!("../../inputs/test/20"); let (grid, start_pos) = parse_input(input.to_string()); let actual = solve(&grid, start_pos, 50, |c1, c2| c1.distance_to(&c2) <= 20); assert_eq!(285, actual); } }
C++ / Boost
Ah, cunning - my favourite one so far this year, I think. Nothing too special compared to the other solutions - floods the map using Dijkstra, then checks “every pair” for how much of a time saver it is. 0.3s on my laptop; it iterates through every pair twice since it does part 1 and part 2 separately, which could easily be improved upon.
spoiler
#include <boost/log/trivial.hpp> #include <boost/unordered/unordered_flat_map.hpp> #include <boost/unordered/unordered_flat_set.hpp> #include <cstddef> #include <fstream> #include <limits> #include <queue> #include <stdexcept> namespace { using Loc = std::pair<int, int>; using Dir = std::pair<int, int>; template <class T> using Score = std::pair<size_t, T>; template <class T> using MinHeap = std::priority_queue<Score<T>, std::vector<Score<T>>, std::greater<Score<T>>>; using Map = boost::unordered_flat_set<Loc>; auto operator+(const Loc &l, const Dir &d) { return Loc{l.first + d.first, l.second + d.second}; } auto manhattan(const Loc &a, const Loc &b) { return std::abs(a.first - b.first) + std::abs(a.second - b.second); } auto dirs = std::vector<Dir>{ {0, -1}, {0, 1 }, {-1, 0 }, {1, 0 } }; struct Maze { Map map; Loc start; Loc end; }; auto parse() { auto rval = Maze{}; auto line = std::string{}; auto ih = std::ifstream{"input/20"}; auto row = 0; while (std::getline(ih, line)) { for (auto col = 0; col < int(line.size()); ++col) { auto t = line.at(col); switch (t) { case 'S': rval.start = Loc{col, row}; rval.map.insert(Loc{col, row}); break; case 'E': rval.end = Loc{col, row}; rval.map.insert(Loc{col, row}); break; case '.': rval.map.insert(Loc{col, row}); break; case '#': break; default: throw std::runtime_error{"oops"}; } } ++row; } return rval; } auto dijkstra(const Maze &m) { auto unvisited = MinHeap<Loc>{}; auto visited = boost::unordered_flat_map<Loc, size_t>{}; for (const auto &e : m.map) visited[e] = std::numeric_limits<size_t>::max(); visited[m.start] = 0; unvisited.push({0, {m.start}}); while (!unvisited.empty()) { auto next = unvisited.top(); unvisited.pop(); if (visited.at(next.second) < next.first) continue; for (const auto &dir : dirs) { auto prospective = Loc{next.second + dir}; if (!visited.contains(prospective)) continue; auto pscore = next.first + 1; if (visited.at(prospective) > pscore) { visited[prospective] = pscore; unvisited.push({pscore, prospective}); } } } return visited; } using Walk = decltype(dijkstra(Maze{})); constexpr auto GOOD_CHEAT = 100; auto evaluate_cheats(const Walk &walk, int skip) { auto rval = size_t{}; for (auto &start : walk) { for (auto &end : walk) { auto distance = manhattan(start.first, end.first); if (distance <= skip && end.second > start.second) { auto improvement = int(end.second) - int(start.second) - distance; if (improvement >= GOOD_CHEAT) ++rval; } } } return rval; } } // namespace auto main() -> int { auto p = parse(); auto walk = dijkstra(p); BOOST_LOG_TRIVIAL(info) << "01: " << evaluate_cheats(walk, 2); BOOST_LOG_TRIVIAL(info) << "02: " << evaluate_cheats(walk, 20); }
Rust
The important part here is to build two maps first that hold for every point on the path the distance to the start and the distance to the end respectively. Then calculating the path length for a cheat vector is a simple lookup. For part 2 I first generated all vectors with manhattan distance <= 20, or more specifically, exactly half of those vectors to avoid checking the same vector in both directions.
Part 2 takes 15ms. The code looks a bit unwieldy at parts and uses the pyramid of doom paradigm but works pretty well.
Solution
use euclid::{default::*, vec2}; const DIRS: [Vector2D<i32>; 4] = [vec2(1, 0), vec2(0, 1), vec2(-1, 0), vec2(0, -1)]; const MIN_SAVE: u32 = 100; const MAX_DIST: i32 = 20; fn parse(input: &str) -> (Vec<Vec<bool>>, Point2D<i32>, Point2D<i32>) { let mut start = None; let mut end = None; let mut field = Vec::new(); for (y, line) in input.lines().enumerate() { let mut row = Vec::new(); for (x, b) in line.bytes().enumerate() { row.push(b == b'#'); if b == b'S' { start = Some(Point2D::new(x, y).to_i32()); } else if b == b'E' { end = Some(Point2D::new(x, y).to_i32()); } } field.push(row); } (field, start.unwrap(), end.unwrap()) } fn distances( field: &[Vec<bool>], start: Point2D<i32>, end: Point2D<i32>, ) -> (Vec<Vec<u32>>, Vec<Vec<u32>>) { let width = field[0].len(); let height = field.len(); let mut dist_to_start = vec![vec![u32::MAX; width]; height]; let bounds = Rect::new(Point2D::origin(), Size2D::new(width, height)).to_i32(); let mut cur = start; let mut dist = 0; dist_to_start[cur.y as usize][cur.x as usize] = dist; while cur != end { for dir in DIRS { let next = cur + dir; if bounds.contains(next) && !field[next.y as usize][next.x as usize] && dist_to_start[next.y as usize][next.x as usize] == u32::MAX { cur = next; break; } } dist += 1; dist_to_start[cur.y as usize][cur.x as usize] = dist; } let total_dist = dist_to_start[end.y as usize][end.x as usize]; let dist_to_end = dist_to_start .iter() .map(|row| { row.iter() .map(|&d| { if d == u32::MAX { u32::MAX } else { total_dist - d } }) .collect() }) .collect(); (dist_to_start, dist_to_end) } fn cheats( field: &[Vec<bool>], dist_to_start: &[Vec<u32>], dist_to_end: &[Vec<u32>], total_dist: u32, ) -> u32 { let width = field[0].len(); let height = field.len(); let bounds = Rect::new(Point2D::origin(), Size2D::new(width, height)).to_i32(); let mut count = 0; for (y, row) in field.iter().enumerate() { for (x, _w) in row.iter().enumerate().filter(|&(_i, w)| *w) { let pos = Point2D::new(x, y).to_i32(); for (d0, &dir0) in DIRS.iter().enumerate().skip(1) { for &dir1 in DIRS.iter().take(d0) { let p0 = pos + dir0; let p1 = pos + dir1; if bounds.contains(p0) && bounds.contains(p1) { let p0 = p0.to_usize(); let p1 = p1.to_usize(); if !field[p0.y][p0.x] && !field[p1.y][p1.x] { let dist = dist_to_start[p0.y][p0.x].min(dist_to_start[p1.y][p1.x]) + dist_to_end[p1.y][p1.x].min(dist_to_end[p0.y][p0.x]) + 2; // Add 2 for cutting across the wall if total_dist - dist >= MIN_SAVE { count += 1; } } } } } } } count } fn part1(input: String) { let (field, start, end) = parse(&input); let (dist_to_start, dist_to_end) = distances(&field, start, end); let total_dist = dist_to_start[end.y as usize][end.x as usize]; println!( "{}", cheats(&field, &dist_to_start, &dist_to_end, total_dist) ); } // Half of all vectors with manhattan distance <= MAX_DIST. // Only vectors with positive x or going straight down are considered to avoid using the same // vector twice in both directions. fn cheat_vectors() -> Vec<Vector2D<i32>> { let mut vectors = Vec::new(); for y in -MAX_DIST..=MAX_DIST { let start = if y > 0 { 0 } else { 1 }; for x in start..=(MAX_DIST - y.abs()) { assert!(x + y <= MAX_DIST); vectors.push(vec2(x, y)); } } vectors } fn cheats20( field: &[Vec<bool>], dist_to_start: &[Vec<u32>], dist_to_end: &[Vec<u32>], total_dist: u32, ) -> u32 { let vectors = cheat_vectors(); let width = field[0].len(); let height = field.len(); let bounds = Rect::new(Point2D::origin(), Size2D::new(width, height)).to_i32(); let mut count = 0; for (y, row) in field.iter().enumerate() { for (x, _w) in row.iter().enumerate().filter(|&(_i, w)| !*w) { let p0 = Point2D::new(x, y); for &v in &vectors { let pi1 = p0.to_i32() + v; if bounds.contains(pi1) { let p1 = pi1.to_usize(); if !field[p1.y][p1.x] { let dist = dist_to_start[p0.y][p0.x].min(dist_to_start[p1.y][p1.x]) + dist_to_end[p1.y][p1.x].min(dist_to_end[p0.y][p0.x]) + v.x.unsigned_abs() // Manhattan distance of vector + v.y.unsigned_abs(); if total_dist - dist >= MIN_SAVE { count += 1; } } } } } } count } fn part2(input: String) { let (field, start, end) = parse(&input); let (dist_to_start, dist_to_end) = distances(&field, start, end); let total_dist = dist_to_start[end.y as usize][end.x as usize]; println!( "{}", cheats20(&field, &dist_to_start, &dist_to_end, total_dist) ); } util::aoc_main!();
Also on github
Dart
Simple path-finding + brute force. O(n*sqrt-n) I think but only takes ~200 milliseconds so that’ll do for today. Time to walk the dog!
After being so pleased to have written my own path-finding algorithm the other day, I discovered that my regular
more
library already includes one. D’oh.(edit: shaved some milliseconds off)
import 'dart:math'; import 'package:more/more.dart'; List<Point<num>> d4 = [Point(1, 0), Point(-1, 0), Point(0, 1), Point(0, -1)]; List<Point<num>> findPath(List<String> lines) { var maze = { for (var r in lines.indexed()) for (var c in r.value.split('').indexed().where((e) => e.value != '#')) Point<num>(c.index, r.index): c.value }.withDefault('#'); var path = AStarSearchIterable<Point>( startVertices: [maze.entries.firstWhere((e) => e.value == 'S').key], successorsOf: (p) => d4.map((e) => (e + p)).where((n) => maze[n] != '#'), costEstimate: (p) => 1, targetPredicate: (p) => maze[p] == 'E').first.vertices; return path; } num dist(Point p1, Point p2) => (p1.x - p2.x).abs() + (p1.y - p2.y).abs(); solve(List<String> lines, int cheat, int target) { List<Point<num>> path = findPath(lines); var cheats = 0; for (int s = 0; s < path.length - cheat; s++) { for (int e = s + cheat; e < path.length; e++) { var d = dist(path[e], path[s]); var diff = e - s - d; if (d <= cheat && diff >= target) cheats += 1; } } return cheats; }
Haskell
I should probably do something about the n2 loop inSomewhat better (0.575s). Can’t shake the feeling that I’m missing an obvious closed-form solution, though.findCheats
, but it’s fast enough for now. Besides, my brain has melted.import Control.Monad import Data.List import Data.Map (Map) import Data.Map qualified as Map import Data.Maybe import Data.Set qualified as Set type Pos = (Int, Int) readInput :: String -> Map Pos Char readInput s = Map.fromList [((i, j), c) | (i, l) <- zip [0 ..] (lines s), (j, c) <- zip [0 ..] l] solveMaze :: Map Pos Char -> Maybe [Pos] solveMaze maze = listToMaybe $ go [] start where walls = Map.keysSet $ Map.filter (== '#') maze Just [start, end] = traverse (\c -> fst <$> find ((== c) . snd) (Map.assocs maze)) ['S', 'E'] go h p@(i, j) | p == end = return [end] | otherwise = do p' <- [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)] guard $ p' `notElem` h guard $ p' `Set.notMember` walls (p :) <$> go [p] p' dist (i1, j1) (i2, j2) = abs (i2 - i1) + abs (j2 - j1) findCheats :: [Pos] -> Int -> Int -> [((Pos, Pos), Int)] findCheats path minScore maxLen = do (t2, end) <- zip [0 ..] path (t1, start) <- zip [0 .. t2 - minScore] path let len = dist start end score = t2 - t1 - len guard $ len <= maxLen guard $ score >= minScore return ((start, end), score) main = do Just path <- solveMaze . readInput <$> readFile "input20" mapM_ (print . length . findCheats path 100) [2, 20]
Ah, the number of potential start points is much smaller than the length of the path. I guess a map from position to offset would do it, but I’m not sure it’s really worth it.
Haskell
First parse and floodfill from start, each position then holds the distance from the start
For part 1, I check all neighbor tiles of neighbor tiles that are walls and calculate the distance that would’ve been in-between.
In part 2 I check all tiles within a manhattan distance
<= 20
and calculate the distance in-between on the path. Then filter out all cheats<100
and countTakes 1.4s sadly, I believe there is still potential for optimization.
Edit: coding style
import Control.Arrow 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 s = Map.fromList [ ((y, x), c) | (l, y) <- zip ls [0..], (c, x) <- zip l [0..]] where ls = lines s floodFill m = floodFill' m startPosition (Map.singleton startPosition 0) where startPosition = Map.assocs >>> filter ((== 'S') . snd) >>> head >>> fst $ m neighbors (p1, p2) = [(p1-1, p2), (p1, p2-1), (p1, p2+1), (p1+1, p2)] floodFill' m p f | m Map.! p == 'E' = f | otherwise = floodFill' m n f' where seconds = f Map.! p ns = neighbors p n = List.filter ((m Map.!) >>> (`Set.member` (Set.fromList ".E"))) >>> List.filter ((f Map.!?) >>> Maybe.isNothing) >>> head $ ns f' = Map.insert n (succ seconds) f taxiCabDistance (a1, a2) (b1, b2) = abs (a1 - b1) + abs (a2 - b2) calculateCheatAdvantage f (p1, p2) = c2 - c1 - taxiCabDistance p1 p2 where c1 = f Map.! p1 c2 = f Map.! p2 cheatDeltas :: Int -> Int -> [(Int, Int)] cheatDeltas l h = [(y, x) | x <- [-h..h], y <- [-h..h], let d = abs x + abs y, d <= h, d >= l] (a1, a2) .+. (b1, b2) = (a1 + b1, a2 + b2) solve l h (f, ps) = Set.toList >>> List.map ( repeat >>> zip (cheatDeltas l h) >>> List.map (snd &&& uncurry (.+.)) >>> List.filter (snd >>> (`Set.member` ps)) >>> List.map (calculateCheatAdvantage f) >>> List.filter (>= 100) >>> List.length ) >>> List.sum $ ps part1 = solve 2 2 part2 = solve 1 20 main = getContents >>= print . (part1 &&& part2) . (id &&& Map.keysSet) . floodFill . parse
Hey - I’ve a question about this. Why is it correct? (Or is it?)
If you have two maps for positions in the maze that give (distance to end) and (distance from start), then you can select for points p1, p2 such that
d(p1, p2) + distance-to-end(p1) + distance-to-start(p2) <= best - 100
however, your version seems to assume that distance-to-end(p) = best - distance-to-start(p) - surely this isn’t always the case?
There is exactly one path without cheating, so yes, the distance to one end is always the total distance minus the distance to the other end.
Gotcha, thanks. I just re-read the problem statement and looked at the input and my input has the strongest possible version of that constraint: the path is unbranching and has start and end at the extremes. Thank-you!
I missed that line too:
Because there is only a single path from the start to the end
So I also did my pathfinding for every variation in the first part, but realised something must be wrong with my approach when I saw part 2.
(I ask because everyone’s solution seems to make the same assumption - that is, that you’re finding a shortcut onto the same path, as opposed to onto a different path.)
Some others have answered already, but yes, there was a well-hidden line in the problem description about the map having only a single path from start to end…