Day 11: Cosmic Expansion
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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
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.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() } }
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()
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)
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
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.