Day 11: Cosmic Expansion

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  • @[email protected]
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    5 months ago

    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, and 1_000_000 in part 2. But if you’re replacing an empty column with 1_000_000, you’re actually adding 999_999 columns. It took me a good hour to discover where that off-by-one error was coming from.

    • Zarlin
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      45 months ago

      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.

      • @[email protected]
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        45 months ago

        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.

        • Zarlin
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          25 months ago

          Yeah the descriptions contain a lot of story fluff, but also critical bits of information.

  • @[email protected]
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    45 months ago

    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
    
  • @purplemonkeymad
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    5 months ago

    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()
    
  • @[email protected]
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    35 months ago

    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.

  • @[email protected]
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    5 months ago

    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 [Z].

    View the code on Github

    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";
    }
    
  • @[email protected]
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    25 months ago

    Python

    Also on Github.

    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)
    
  • cacheson
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    25 months ago

    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.

    • @[email protected]
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      25 months ago

      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.

  • @[email protected]
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    25 months ago

    Kotlin

    Github

    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
  • @[email protected]
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    25 months ago

    Crystal

    [email protected]

    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
    
  • @capitalpb
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    25 months ago

    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 an i64, 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()
        }
    }
    
  • janAkali
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    5 months ago

    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
    
  • @[email protected]
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    5 months ago

    Dart

    Nothing interesting here, just did it all explicitly. I might try something different in Uiua later.

    solve(List lines, {int age = 2}) {
      var grid = lines.map((e) => e.split('')).toList();
      var gals = [
        for (var r in grid.indices())
          for (var c in grid[r].indices().where((c) => grid[r][c] == '#')) (r, c)
      ];
      for (var row in grid.indices(step: -1)) {
        if (!grid[row].contains('#')) {
          gals = gals
              .map((e) => ((e.$1 > row) ? e.$1 + age - 1 : e.$1, e.$2))
              .toList();
        }
      }
      for (var col in grid.first.indices(step: -1)) {
        if (grid.every((r) => r[col] == '.')) {
          gals = gals
              .map((e) => (e.$1, (e.$2 > col) ? e.$2 + age - 1 : e.$2))
              .toList();
        }
      }
      var dists = [
        for (var ix1 in gals.indices())
          for (var ix2 in (ix1 + 1).to(gals.length))
            (gals[ix1].$1 - gals[ix2].$1).abs() +
                (gals[ix1].$2 - gals[ix2].$2).abs()
      ];
      return dists.sum;
    }
    
    part1(List lines) => solve(lines);
    part2(List lines) => solve(lines, age: 1000000);
    
  • Zarlin
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    15 months ago

    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 ^^

    Part 1 and 2 combined

  • @cvttsd2si
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    15 months ago

    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)