Day 11: Cosmic Expansion
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Man this one frustrated me because of a subtle difference in the wording of part 1 vs part 2. I had the correct logic from the start, but with an off-by-one error because of my interpretation of the wording. Part 1 says, “any rows or columns that contain no galaxies should all actually be twice as big” while part 2 says, “each empty column should be replaced with
1000000
empty columns”.I added 1 column/row in part
1
, and1_000_000
in part 2. But if you’re replacing an empty column with1_000_000
, you’re actually adding999_999
columns. It took me a good hour to discover where that off-by-one error was coming from.Yepp, this one got me as well! I found the discrepancy when testing against the sample through, which showed the result for a factor 100 (which needed to be 99). Knowing the correct outcome made debugging a lot easier.
I always make sure my solution passes all the samples before trying the full input.
Me too. I ran all the samples, and I was still banging my head. I can usually see the mistake if it’s an off-by-one error in a calculation, but this was a mistake in reading the problem description, so I couldn’t see it at first.
Yeah the descriptions contain a lot of story fluff, but also critical bits of information.
Uiua
As promised, just a little later than planned. I do like this solution as it’s actually using arrays rather than just imperative programming in fancy dress. Run it here
Grid ← =@# [ "...#......" ".......#.." "#........." ".........." "......#..." ".#........" ".........#" ".........." ".......#.." "#...#....." ] GetDist! ← ( # Build arrays of rows, cols of galaxies ⊙(⊃(◿)(⌊÷)⊃(⧻⊢)(⊚=1/⊂)). # check whether each row/col is just space # and so calculate its relative position ∩(\++1^1=0/+)⍉. # Map galaxy co-ords to these values ⊏:⊙(:⊏ :) # Map to [x, y] pairs, build cross product, # and sum all topright values. /+≡(/+↘⊗0.)⊠(/+⌵-).⍉⊟ ) GetDist!(×1) Grid GetDist!(×99) Grid
Rust
I was unsure in Part 1 whether to actually expand the grid or just count the number of empty lanes in each ranges. I ended up doing the latter which was obviously the right choice for part 2, but I think it could have gone either way.
I saw that coming, but decided to do it the naive way for part 1, then fixed that up for part 2. Thanks to AoC I can also recognise a Manhattan distance written in a complex manner.
Python
from __future__ import annotations import re import math import argparse import itertools def print_sky(sky:list): for r in sky: print("".join(r)) class Point: def __init__(self,x:int,y:int) -> None: self.x = x self.y = y def __repr__(self) -> str: return f"Point({self.x},{self.y})" def distance(self,point:Point): # Manhattan dist x = abs(self.x - point.x) y = abs(self.y - point.y) return x + y def expand_galaxies(galaxies:list,position:int,amount:int,index:str): for g in galaxies: if getattr(g,index) > position: c = getattr(g,index) setattr(g,index, c + amount) def main(line_list:list,part:int): ## list of lists is the plan for init idea expand_value = 2 -1 if part == 2: expand_value = 1e6 -1 if part > 2: expand_value = part -1 sky = list() for l in line_list: row_data = [*l] sky.append(row_data) print_sky(sky) # get galaxies gal_list = list() for r in range(0,len(sky)): for c in range(0,len(sky[r])): if sky[r][c] == '#': gal_list.append(Point(r,c)) print(gal_list) col_indexes = list(reversed(range(0,len(sky)))) # expand rows for i in col_indexes: if not '#' in sky[i]: expand_galaxies(gal_list,i,expand_value,'x') # check for expanding columns for i in reversed( range(0, len( sky[0] )) ): col = [sky[x][i] for x in col_indexes] if not '#' in col: expand_galaxies(gal_list,i,expand_value,'y') print(gal_list) # find all unique pair distance sum, part 1 sum = 0 for i in range(0,len(gal_list)): for j in range(i+1,len(gal_list)): sum += gal_list[i].distance(gal_list[j]) print(f"Sum distances: {sum}") if __name__ == "__main__": parser = argparse.ArgumentParser(description="template for aoc solver") parser.add_argument("-input",type=str) parser.add_argument("-part",type=int) args = parser.parse_args() filename = args.input if filename == None: parser.print_help() exit(1) part = args.part file = open(filename,'r') main([line.rstrip('\n') for line in file.readlines()],part) file.close()
Raku
Today I’m thankful that I have the combinations() method available. It’s not hard to implement combinations(), but it’s not particularly interesting. This code is a bit anachronistic because I solved part 1 by expanding the universe instead of contracting it, but this way makes the calculations for part 1 and part 2 symmetric. I was worried for a bit there that I’d have to do some “here are all the places where expansion happens, check this list when calculating distances” bookkeeping, and I was quite relieved when I realized that I could just use arithmetic.
edit: Also, first time using the Slip class in Raku, which is mind bending, but very useful for expanding/contracting the universe and generating the lists of galaxy coordinates. And I learned a neat way to transpose 2D arrays using
[
. ]Code
use v6; sub MAIN($input) { my $file = open $input; my @map = $file.lines».comb».Array; my @galaxies-original = @map».grep("#", :k).grep(*.elems > 0, :kv).rotor(2).map({($_[0] X $_[1]).Slip}); my $distances-original = @galaxies-original.List.combinations(2).map({($_[0] Z- $_[1])».abs.sum}).sum; # contract the universe @map = @map.map: {$_.all eq '.' ?? slip() !! $_}; @map = [Z] @map; @map = @map.map: {$_.all eq '.' ?? slip() !! $_}; @map = [Z] @map; my @galaxies = @map».grep("#", :k).grep(*.elems > 0, :kv).rotor(2).map({($_[0] X $_[1]).Slip}); my $distances-contracted = @galaxies.List.combinations(2).map({($_[0] Z- $_[1])».abs.sum}).sum; my $distances-twice-expanded = ($distances-original - $distances-contracted) * 2 + $distances-contracted; say "part 1: $distances-twice-expanded"; my $distances-many-expanded = ($distances-original - $distances-contracted) * 1000000 + $distances-contracted; say "part 2: $distances-many-expanded"; }
That was a fun one. Especially after yesterday. As soon as I saw that star 1 was expanding each gap by 1, I just had a feeling that star 2 would be doing the same calculation with a larger expansion, so I wrote my code in a way that would make that quite simple to modify. When I saw the factor of 1,000,000 I was scared that it was going to be one of those processor-destroying AoC challenges where you either wait for 2 hours to get an answer, or have to come up with a fancy mathematical way of solving things, but after changing my
i32
distance to ani64
, it calculated just fine and instantly. I guess only storing the locations of galaxies and not dealing with the entire grid was good enough to keep the performance down.https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day11.rs
use crate::Solver; use itertools::Itertools; use num::abs; #[derive(Debug)] struct Point { x: usize, y: usize, } struct GalaxyMap { locations: Vec, } impl GalaxyMap { fn from(input: &str) -> GalaxyMap { let locations = input .lines() .rev() .enumerate() .map(|(x, row)| { row.chars() .enumerate() .filter_map(|(y, digit)| { if digit == '#' { Some(Point { x, y }) } else { None } }) .collect::>() }) .flatten() .collect::>(); GalaxyMap { locations } } fn empty_rows(&self) -> Vec { let occupied_rows = self .locations .iter() .map(|point| point.y) .unique() .collect::>(); let max_y = *occupied_rows.iter().max().unwrap(); (0..max_y) .filter(move |y| !occupied_rows.contains(&y)) .collect() } fn empty_cols(&self) -> Vec { let occupied_cols = self .locations .iter() .map(|point| point.x) .unique() .collect::>(); let max_x = *occupied_cols.iter().max().unwrap(); (0..max_x) .filter(move |x| !occupied_cols.contains(&x)) .collect() } fn expand(&mut self, factor: usize) { let delta = factor - 1; for y in self.empty_rows().iter().rev() { for galaxy in &mut self.locations { if galaxy.y > *y { galaxy.y += delta; } } } for x in self.empty_cols().iter().rev() { for galaxy in &mut self.locations { if galaxy.x > *x { galaxy.x += delta; } } } } fn galactic_distance(&self) -> i64 { self.locations .iter() .combinations(2) .map(|pair| { abs(pair[0].x as i64 - pair[1].x as i64) + abs(pair[0].y as i64 - pair[1].y as i64) }) .sum::() } } pub struct Day11; impl Solver for Day11 { fn star_one(&self, input: &str) -> String { let mut galaxy = GalaxyMap::from(input); galaxy.expand(2); galaxy.galactic_distance().to_string() } fn star_two(&self, input: &str) -> String { let mut galaxy = GalaxyMap::from(input); galaxy.expand(1_000_000); galaxy.galactic_distance().to_string() } }
Nim
Part 1 and 2: I solved today’s puzzle without expanding the universe. Path in expanded universe is just a path in the original grid + expansion rate times the number of crossed completely-empty lines (both horizontal and vertical). For example, if a single tile after expansion become 5 tiles (rate = +4), original path was 12 and it crosses 7 lines, new path will be:
12 + 4 * 7 = 40
.
The shortest path is easy to calculate in O(1) time:abs(start.x - finish.x) + abs(start.y - finish.y)
.
And to count crossed lines I just check if line is between the start and finish indexes.Total runtime: 2.5 ms
Puzzle rating: 7/10 Code: day_11/solution.nim
Snippet:proc solve(lines: seq[string]): AOCSolution[int] = let galaxies = lines.getGalaxies() emptyLines = lines.emptyLines() emptyColumns = lines.emptyColumns() for gi, g1 in galaxies: for g2 in galaxies[gi+1..^1]: let path = shortestPathLength(g1, g2) let crossedLines = countCrossedLines(g1, g2, emptyColumns, emptyLines) block p1: result.part1 += path + crossedLines * 1 block p2: result.part2 += path + crossedLines * 999_999
Nim
Approached part 1 in the expected way, by expanding the grid. For part 2, I left the grid alone and just adjusted the galaxy location vectors based on how many empty rows and columns there were above and to the left of them. I divided my final totals by 2 instead of bothering with any fancy combinatoric iterators.
I divided my final totals by 2 instead of bothering with any fancy combinatoric iterators.
same lmao, it’s only double the calculations. If x2 is mucking your code up, it’s too slow anyway.
Kotlin
Was lazy at the beginning, so I started to do actual expansion in memory, then decided to do it properly. And it paid off in the second part.
- find galaxy coordinates + hold few bit masks to find unused rows and columns → convert to indexes
- transform original galaxies coordinates considering number of previous empty lines (time for simple binary search)
- use the Manhattan distance
Crystal
wording in part 2 threw me off too
I could have done this in 2 loops, but this method is way easier to do
And it’s gorgeous code imo
(except for the fact that lemmy’s huge tab sizes make it look weird)code
E = ARGV[0].to_i input = File.read("input.txt") sky = input.lines.map &.chars # find galaxies galaxies = Array(Tuple(Int32, Int32)).new sky.size.times do |y| sky[0].size.times do |x| if sky[y][x] == '#' galaxies << {y, x} end end end # puts galaxies # vertical expansion locations expandsy = Array(Int32).new sky.size.times do |i| unless galaxies.any? {|gal| gal[0] == i} expandsy << i end end # horizontal expansion locations expandsx = Array(Int32).new sky[0].size.times do |i| unless galaxies.any? {|gal| gal[1] == i} expandsx << i end end # calculate expansion for each galaxy adds = Array.new(galaxies.size) { [0, 0] } expandsy.each do |y| galaxies.each_with_index do |gal, i| if gal[0] > y adds[i][0] += 1 end end end # calculate expansion for each galaxy expandsx.each do |x| galaxies.each_with_index do |gal, i| if gal[1] > x adds[i][1] += 1 end end end # expaaaaaaaand galaxies.map_with_index! {|gal, i| {gal[0] + adds[i][0]*E, gal[1] + adds[i][1]*E} } # distances sum = 0_u64 galaxies.each do |gal| galaxies.each do |gal2| if gal2 != gal sum += (gal2[0] - gal[0]).abs + (gal2[1] - gal[1]).abs end end end puts sum/2
Python
from .solver import Solver class Day11(Solver): def __init__(self): super().__init__(11) self.galaxies: list = [] self.blank_x: set[int] = set() self.blank_y: set[int] = set() def presolve(self, input: str): lines = input.rstrip().split('\n') self.galaxies = [] max_x = 0 max_y = 0 for y, line in enumerate(lines): for x, c in enumerate(line): if c == '#': self.galaxies.append((x, y)) max_x = max(max_x, x) max_y = max(max_y, y) self.blank_x = set(range(max_x + 1)) - {x for x, _ in self.galaxies} self.blank_y = set(range(max_y + 1)) - {y for _, y in self.galaxies} def solve(self, expansion_factor: int) -> int: galaxies = list(self.galaxies) total = 0 for i in range(len(galaxies)): for j in range(i + 1, len(galaxies)): sx, sy = galaxies[i] dx, dy = galaxies[j] if sx > dx: sx, dx = dx, sx if sy > dy: sy, dy = dy, sy dist = sum((dx - sx, dy - sy, max(0, expansion_factor - 1) * len([x for x in self.blank_x if sx < x < dx]), max(0, expansion_factor - 1) * len([y for y in self.blank_y if sy < y < dy]))) total += dist return total def solve_first_star(self): return self.solve(2) def solve_second_star(self): return self.solve(1000000)
Rust
use std::fs; use std::path::PathBuf; use clap::Parser; use itertools::Itertools; use std::ops::Range; #[derive(Parser)] #[command(author, version, about, long_about = None)] struct Cli { input_file: PathBuf, } fn main() { // Parse CLI arguments let cli = Cli::parse(); // Read file let input_text = fs::read_to_string(&cli.input_file) .expect(format!("File \"{}\" not found", cli.input_file.display()).as_str()); let map: Vec> = input_text.lines().map(|r| r.chars().collect()).collect(); let mut expanding_rows = vec![]; // Expand rows map.iter().enumerate().for_each(|(row_num, r)| { if r.iter().filter(|c| **c == '.').count() == r.len() { expanding_rows.push(row_num); } }); let mut expanding_cols = vec![]; // Find expanding cols for col in 0..map[0].len() { // Determine if all '.' let expand = map.iter().filter(|row| row[col] == '.').count() == map.len(); if expand { expanding_cols.push(col); } } // Find galaxies let mut galaxies: Vec<(usize, usize)> = vec![]; for (row_num, row) in map.iter().enumerate() { for (col_num, char) in row.iter().enumerate() { if *char == '#' { galaxies.push((row_num.try_into().unwrap(), col_num.try_into().unwrap())); } } } println!("Num Galaxies: ({})", galaxies.len()); println!( "Expanding rows: {:?}\tExpanding cols: {:?}", expanding_rows, expanding_cols ); // Distance between each pair let expansion_1 = 2; let expansion_2 = 1_000_000; let mut sum_distances_1 = 0; let mut sum_distances_2 = 0; println!( "Total combinations: {}", galaxies.iter().tuple_combinations::<(_, _)>().count() ); for (g1, g2) in galaxies.iter().tuple_combinations() { let vert_range: Range = if g1.0 < g2.0 { (g1.0)..(g2.0) } else { (g2.0)..(g1.0) }; let horr_range: Range = if g1.1 < g2.1 { (g1.1)..(g2.1) } else { (g2.1)..(g1.1) }; let vert_expansions = expanding_rows .iter() .filter(|r| vert_range.contains(*r)) .count(); let horr_expansions = expanding_cols .iter() .filter(|r| horr_range.contains(*r)) .count(); let expansions = vert_expansions + horr_expansions; let distance = vert_range.len() + horr_range.len(); sum_distances_1 += distance + (expansions * (expansion_1 - 1)); sum_distances_2 += distance + (expansions * (expansion_2 - 1)); } println!("Total distance: {}", sum_distances_1); println!("Part 2 distance: {}", sum_distances_2); }
Scala3
def compute(a: List[String], growth: Long): Long = val gaps = Seq(a.map(_.toList), a.transpose).map(_.zipWithIndex.filter((d, i) => d.forall(_ == '.')).map(_._2).toSet) val stars = for y <- a.indices; x <- a(y).indices if a(y)(x) == '#' yield List(x, y) def dist(gaps: Set[Int], a: Int, b: Int): Long = val i = math.min(a, b) until math.max(a, b) i.size.toLong + (growth - 1)*i.toSet.intersect(gaps).size.toLong (for Seq(p, q) <- stars.combinations(2); m <- gaps.lazyZip(p).lazyZip(q).map(dist) yield m).sum def task1(a: List[String]): Long = compute(a, 2) def task2(a: List[String]): Long = compute(a, 1_000_000)
Nim
Happy I decided to not actually expand anything. Manhattan distance and counting the number of empty rows and columns was plenty. Also made part 2 an added oneliner :) It’s still pretty inefficient iterating over the grid multiple times to gather the galaxies and empty rows, runtime is about 17ms
I could also extract and re-use my 2D Coord and Grid classes from day 10, and learned more about Nim in the process ^^
Multiplied the manhattan distance by the number of empty space lines crossed.