this post was submitted on 08 Dec 2024
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Advent Of Code

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Day 8: Resonant Collinearity

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  • 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)
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FAQ

top 31 comments
sorted by: hot top controversial new old
[–] [email protected] 8 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

Rust

For the first time, I can post my solution, because I actually solved it on the day :D Probably not the cleanest or optimal solution, but it does solve the problem.

Very long, looking forward to someone solving it in 5 lines of unicode :D

#[cfg(test)]
mod tests {

    fn get_frequences(input: &str) -> Vec<char> {
        let mut freq = vec![];
        for char in input.chars() {
            if char == '.' {
                continue;
            }
            if !freq.contains(&char) {
                freq.push(char);
            }
        }
        freq
    }

    fn find_antennas(board: &Vec<Vec<char>>, freq: char) -> Vec<(isize, isize)> {
        let mut antennas = vec![];
        for (i, line) in board.iter().enumerate() {
            for (j, char) in line.iter().enumerate() {
                if *char == freq {
                    antennas.push((i as isize, j as isize));
                }
            }
        }
        antennas
    }

    fn calc_antinodes(first: &(isize, isize), second: &(isize, isize)) -> Vec<(isize, isize)> {
        let deltax = second.0 - first.0;
        let deltay = second.1 - first.1;

        if deltax == 0 && deltay == 0 {
            return vec![];
        }

        vec![
            (first.0 - deltax, first.1 - deltay),
            (second.0 + deltax, second.1 + deltay),
        ]
    }

    #[test]
    fn test_calc_antinodes() {
        let expected = vec![(0, -1), (0, 2)];
        let actual = calc_antinodes(&(0, 0), &(0, 1));
        for i in &expected {
            assert!(actual.contains(i));
        }
        let actual = calc_antinodes(&(0, 1), &(0, 0));
        for i in &expected {
            assert!(actual.contains(i));
        }
    }

    fn calc_all_antinodes(board: &Vec<Vec<char>>, freq: char) -> Vec<(isize, isize)> {
        let antennas = find_antennas(&board, freq);

        let mut antinodes = vec![];

        for (i, first) in antennas.iter().enumerate() {
            for second in antennas[i..].iter() {
                antinodes.extend(calc_antinodes(first, second));
            }
        }

        antinodes
    }

    fn prune_nodes(
        nodes: &Vec<(isize, isize)>,
        height: isize,
        width: isize,
    ) -> Vec<(isize, isize)> {
        let mut pruned = vec![];
        for node in nodes {
            if pruned.contains(node) {
                continue;
            }
            if node.0 < 0 || node.0 >= height {
                continue;
            }
            if node.1 < 0 || node.1 >= width {
                continue;
            }
            pruned.push(node.clone());
        }
        pruned
    }

    fn print_board(board: &Vec<Vec<char>>, pruned: &Vec<(isize, isize)>) {
        for (i, line) in board.iter().enumerate() {
            for (j, char) in line.iter().enumerate() {
                if pruned.contains(&(i as isize, j as isize)) {
                    print!("#");
                } else {
                    print!("{char}");
                }
            }
            println!();
        }
    }

    #[test]
    fn day8_part1_test() {
        let input: String = std::fs::read_to_string("src/input/day_8.txt").unwrap();

        let frequencies = get_frequences(&input);

        let board = input
            .trim()
            .split('\n')
            .map(|line| line.chars().collect::<Vec<char>>())
            .collect::<Vec<Vec<char>>>();

        let mut all_nodes = vec![];
        for freq in frequencies {
            let nodes = calc_all_antinodes(&board, freq);
            all_nodes.extend(nodes);
        }

        let height = board.len() as isize;
        let width = board[0].len() as isize;

        let pruned = prune_nodes(&all_nodes, height, width);

        println!("{:?}", pruned);

        print_board(&board, &pruned);

        println!("{}", pruned.len());

        // 14 in test
    }

    fn calc_antinodes2(first: &(isize, isize), second: &(isize, isize)) -> Vec<(isize, isize)> {
        let deltax = second.0 - first.0;
        let deltay = second.1 - first.1;

        if deltax == 0 && deltay == 0 {
            return vec![];
        }
        let mut nodes = vec![];
        for n in 0..50 {
            nodes.push((first.0 - deltax * n, first.1 - deltay * n));
            nodes.push((second.0 + deltax * n, second.1 + deltay * n));
        }

        nodes
    }

    fn calc_all_antinodes2(board: &Vec<Vec<char>>, freq: char) -> Vec<(isize, isize)> {
        let antennas = find_antennas(&board, freq);

        let mut antinodes = vec![];

        for (i, first) in antennas.iter().enumerate() {
            for second in antennas[i..].iter() {
                antinodes.extend(calc_antinodes2(first, second));
            }
        }

        antinodes
    }

    #[test]
    fn day8_part2_test() {
        let input: String = std::fs::read_to_string("src/input/day_8.txt").unwrap();

        let frequencies = get_frequences(&input);

        let board = input
            .trim()
            .split('\n')
            .map(|line| line.chars().collect::<Vec<char>>())
            .collect::<Vec<Vec<char>>>();

        let mut all_nodes = vec![];
        for freq in frequencies {
            let nodes = calc_all_antinodes2(&board, freq);
            all_nodes.extend(nodes);
        }

        let height = board.len() as isize;
        let width = board[0].len() as isize;

        let pruned = prune_nodes(&all_nodes, height, width);

        println!("{:?}", pruned);

        print_board(&board, &pruned);

        println!("{}", pruned.len());
    }
}
[–] [email protected] 4 points 2 weeks ago

Any solution that solves the problem is a good solution :D

[–] [email protected] 5 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

Haskell

I overslept 26 minutes (AoC starts at 06:00 here) which upsets me more than it should.
I thought this one was going to be hard on performance or memory but it was surprisingly easy.

import Control.Arrow hiding (first, second)
import Data.Bifunctor

import Data.Array.Unboxed (UArray)

import qualified Data.List as List
import qualified Data.Set as Set
import qualified Data.Array.Unboxed as Array

parse :: String -> UArray (Int, Int) Char
parse s = Array.listArray ((1, 1), (n, m)) . filter (/= '\n') $ s :: UArray (Int, Int) Char

        where
                n = takeWhile   (/= '\n') >>> length $ s
                m = List.filter (== '\n') >>> length >>> pred $ s

groupSnd:: Eq b => (a, b) -> (a', b) -> Bool
groupSnd = curry (uncurry (==) <<< snd *** snd)

cartesianProduct xs ys = [(x, y) | x <- xs, y <- ys]

calculateAntitone ((y1, x1), (y2, x2)) = (y1 + dy, x1 + dx)
        where
                dy = y1 - y2
                dx = x1 - x2

antennaCombinations = Array.assocs
        >>> List.filter (snd >>> (/= '.'))
        >>> List.sortOn snd
        >>> List.groupBy groupSnd
        >>> map (map fst)
        >>> map (\ xs -> cartesianProduct xs xs)
        >>> map (filter (uncurry (/=)))

part1 a = antennaCombinations
        >>> List.concatMap (map calculateAntitone)
        >>> List.filter (Array.inRange (Array.bounds a))
        >>> Set.fromList
        >>> Set.size
        $ a

calculateAntitones ((y1, x1), (y2, x2)) = iterate (bimap (+dy) (+dx)) (y1, x1)
        where
                dy = y1 - y2
                dx = x1 - x2

part2 a = antennaCombinations
        >>> List.map (map calculateAntitones)
        >>> List.concatMap (List.concatMap (takeWhile (Array.inRange (Array.bounds a))))
        >>> Set.fromList
        >>> Set.size
        $ a

main = getContents
        >>= print
        . (part1 &&& part2)
        . parse
[–] [email protected] 2 points 2 weeks ago

D'oh. Computing antinodes in a single direction and permuting pairs is a much neater approach that what I did!

[–] [email protected] 4 points 2 weeks ago (1 children)

Haskell

Not a very pretty solution today, I'm afraid.

import Control.Arrow
import Control.Monad
import Data.Biapplicative
import Data.Ix
import Data.Map (Map)
import Data.Map qualified as Map
import Data.Set qualified as Set

type Coords = (Int, Int)

readInput :: String -> Map Coords Char
readInput s =
  Map.fromAscList
    [ ((i, j), c)
      | (i, l) <- zip [0 ..] (lines s),
        (j, c) <- zip [0 ..] l
    ]

(.+.), (.-.) :: Coords -> Coords -> Coords
(.+.) = join biliftA2 (+)
(.-.) = join biliftA2 (-)

part1, part2 :: (Coords -> Bool) -> (Coords, Coords) -> [Coords]
part1 valid (p1, p2) =
  let s = p2 .-. p1
   in filter valid [p1 .-. s, p2 .+. s]
part2 valid (p1, p2) =
  let (si, sj) = p2 .-. p1
      d = gcd si sj
      s = (si `div` d, sj `div` d)
   in takeWhile valid (iterate (.+. s) p1)
        ++ takeWhile valid (drop 1 $ iterate (.-. s) p2)

pairs (x : xs) = map (x,) xs ++ pairs xs
pairs _ = []

main = do
  input <- readInput <$> readFile "input08"
  let antennas = Map.filter (/= '.') input
      antennaGroups =
        Map.foldrWithKey
          (\p c m -> Map.insertWith (++) c [p] m)
          Map.empty
          antennas
      valid =
        inRange
          . (Set.findMin &&& Set.findMax)
          $ Map.keysSet input
      antiNodes model =
        Set.fromList
          . concatMap (concatMap (model valid) . pairs)
          $ antennaGroups
  print . Set.size $ antiNodes part1
  print . Set.size $ antiNodes part2
[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

Whaaat? It is possible to declare mutliple signatures on one line? 🀯
Does that function (.+.) add tuples/coordinates?

[–] [email protected] 2 points 2 weeks ago (1 children)

Yup, that's right! The function monad is a bit of a mind-bender, but (join f) x == f x x is a useful thing to remember.

[–] [email protected] 2 points 2 weeks ago (1 children)

This is so cool, it's going to replace the lambda in my function pipeline for calculating pairs.

[–] [email protected] 1 points 2 weeks ago

BTW, for more in-depth vector stuff I usually use the Linear package.

[–] [email protected] 4 points 2 weeks ago

C#

public class Day08 : Solver
{
  private ImmutableArray<string> data;
  private int width, height;

  public void Presolve(string input) {
    data = input.Trim().Split("\n").ToImmutableArray();
    width = data[0].Length;
    height = data.Length;
  }

  public string SolveFirst() {
    Dictionary<char, List<(int, int)>> antennae = [];
    HashSet<(int, int)> antinodes = [];
    for (int i = 0; i < width; i++) {
      for (int j = 0; j < height; j++) {
        if ('.' == data[j][i]) continue;
        antennae.TryAdd(data[j][i], []);
        foreach (var (oi, oj) in antennae[data[j][i]]) {
          int di = i - oi;
          int dj = j - oj;
          int ai = i + di;
          int aj = j + dj;
          if (ai >= 0 && aj >= 0 && ai < width && aj < height) {
            antinodes.Add((ai, aj));
          }
          ai = oi - di;
          aj = oj - dj;
          if (ai >= 0 && aj >= 0 && ai < width && aj < height) {
            antinodes.Add((ai, aj));
          }
        }
        antennae[data[j][i]].Add((i, j));
      }
    }
    return antinodes.Count.ToString();
  }

  public string SolveSecond() {
    Dictionary<char, List<(int, int)>> antennae = [];
    HashSet<(int, int)> antinodes = [];
    for (int i = 0; i < width; i++) {
      for (int j = 0; j < height; j++) {
        if ('.' == data[j][i]) continue;
        antennae.TryAdd(data[j][i], []);
        foreach (var (oi, oj) in antennae[data[j][i]]) {
          int di = i - oi;
          int dj = j - oj;
          for (int ai = i, aj = j;
               ai >= 0 && aj >= 0 && ai < width && aj < height; 
               ai += di, aj +=dj) {
            antinodes.Add((ai, aj));
          }
          for (int ai = oi, aj = oj;
               ai >= 0 && aj >= 0 && ai < width && aj < height; 
               ai -= di, aj -=dj) {
            antinodes.Add((ai, aj));
          }
        }
        antennae[data[j][i]].Add((i, j));
      }
    }
    return antinodes.Count.ToString();
  }
}
[–] [email protected] 4 points 2 weeks ago

Uiua

Adapting the part one solution for part two took me longer than part one did today, but I didn't want to change much anymore.

I even got scolded by the interpreter to split the evaluating line onto multiple ones because it got too long.
Can't say it's pretty but it does it's job ^^'

Run with example input here

PartOne ← (
  &rs ∞ &fo "input-8.txt"
  ⟜(β–½Β¬βˆˆ".\n".β—΄)
  βŠœβˆ˜β‰ @\n.
  :€⟜(:Β€-1β–³)
  ≑(β–‘βŠšβŒ•)
  β—΄/β—‡βŠ‚βš(≑(-:⟜-°⊟)β§…β‰ 2)
  ⧻▽¬:βŠ™(/+⍉+)βŸœβŠ“><,0
)

PartTwo ← (
  &rs ∞ &fo "input-8.txt"
  ⟜(β–½Β¬βˆˆ".\n".β—΄βŸœΒ€
    β–½:βŸœβ‰‘(>1β§»βŠšβŒ•)
  )
  βŠœβˆ˜β‰ @\n.
  :€⟜(:Β€-1β–³)
  ≑(β–‘βŠšβŒ•)
  ⊸⍚(
    β§…β‰ 2βŠ™Β€
    ≑(:€⟜-°⊟
      ⍒(βŠ™βŠ‚βŸœ-βŠ™βŠΈβŠ’
      | β‹…(=0/++βŠ“><,0⊒))
      β–‘βŠ™β—Œβ—Œ
    )
  )
  β—΄/β—‡βŠ‚/β—‡βŠ‚
  ⧻▽¬:βŠ™(/+⍉+)βŸœβŠ“><,0
)

&p "Day 8:"
&pf "Part 1: "
&p PartOne
&pf "Part 2: "
&p PartTwo
[–] [email protected] 3 points 2 weeks ago (1 children)

Dart

This really does feel like a weekend break this year, maybe Eric and co have begun to realise that family time is more precious than work time :-)

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

solve(List<String> lines, int min, int max) {
  var map = ListMultimap<String, Point<int>>();
  for (var r in lines.indices()) {
    for (var ci in lines[r].split('').indexed()) {
      if (ci.value != '.') map[ci.value].add(Point(ci.index, r));
    }
  }
  var anti = <Point<int>>{};
  for (var k in map.keys) {
    for (var p in map[k].combinations(2, repetitions: false)) {
      var diff = p.last - p.first;
      for (var m in min.to(max)) {
        anti.addAll([p.first - diff * m, p.last + diff * m]);
      }
    }
  }

  return anti.count((e) =>
      e.x.between(0, lines.first.length - 1) &&
      e.y.between(0, lines.length - 1));
}

part1(List<String> lines) => solve(lines, 1, 2);

part2(List<String> lines) => solve(lines, 0, 50);
[–] [email protected] 4 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

maybe Eric and co have begun to realise that family time is more precious than work time

Last year the difficulty was fluctuating from 0 to 100 each day.
This year all problems so far are suspiciously easy. Maybe the second half of the month will be extra hard?

[–] [email protected] 4 points 2 weeks ago

Maybe we've been good all year :-)

[–] [email protected] 2 points 2 weeks ago

Julia

I was surprised when my solution worked for part 2 since I thought you also had to include fractions of antenna distances, but apparently not.

Code

function readInput(inputFile::String)::Matrix{Char}
	f = open(inputFile,"r")
	lines::Vector{String} = readlines(f)
	close(f)
	cityMap = Matrix{Char}(undef,length(lines),length(lines[1]))
	for (i,l) in enumerate(lines)
		cityMap[i,:] = collect(l)
	end
	return cityMap
end

function getAntennaLocations(cityMap::Matrix{Char})::Dict
	antennaLocations = Dict{Char,Vector{Vector{Int}}}()
	for l=1 : size(cityMap)[1]
		for c=1 : size(cityMap)[2]
			cityMap[l,c]=='.' ? continue : nothing
			if !haskey(antennaLocations,cityMap[l,c])
				antennaLocations[cityMap[l,c]] = []
			end
			push!(antennaLocations[cityMap[l,c]],[l,c])
		end
	end
	return antennaLocations
end

function countAntinodes(cityMap::Matrix{Char},antLoc::Dict{Char,Vector{Vector{Int}}},withHarmonics::Bool)::Int #antLoc: antenna locations
	lBounds = 1:size(cityMap)[1]; cBounds = 1:size(cityMap)[2]
	anodeLocs::Matrix{Bool} = zeros(size(cityMap))
	for key in keys(antLoc)
		for i=1 : length(antLoc[key])
			withHarmonics&&length(antLoc[key])>1 ? anodeLocs[antLoc[key][i][1],antLoc[key][i][2]]=1 : nothing #add antenna locations as antinodes
			#should also add fractions of antenna distances, but works without
			for j=i+1 : length(antLoc[key])
				harmonic::Int = 1
				while true
					n1l = antLoc[key][i][1]+harmonic*(antLoc[key][i][1]-antLoc[key][j][1])
					n1c = antLoc[key][i][2]+harmonic*(antLoc[key][i][2]-antLoc[key][j][2])
					n2l = antLoc[key][j][1]+harmonic*(antLoc[key][j][1]-antLoc[key][i][1])
					n2c = antLoc[key][j][2]+harmonic*(antLoc[key][j][2]-antLoc[key][i][2])
					if n1l in lBounds && n1c in cBounds
						anodeLocs[n1l,n1c] = 1
					end
					if n2l in lBounds && n2c in cBounds
						anodeLocs[n2l,n2c] = 1
					end
					withHarmonics ? nothing : break
					!(n1l in lBounds) && !(n1c in cBounds) && !(n2l in lBounds) && !(n2c in cBounds) ? break : harmonic+=1
				end
			end
		end
	end
	return sum(anodeLocs)
end

@info "Part 1"
println("antinode count $(countAntinodes(getAntennaLocations(readInput("day08Input"))),false)")
@info "Part 2"
println("antinode count $(countAntinodes(getAntennaLocations(readInput("day08Input"))),faltrue)")

[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago)

Lisp

Could probably just write points right to the results instead of to an intermediate list, but it runs instantly, so my motivation to do so was low.

Code

(defun p1-process-line (line)
   (to-symbols line 'advt2024-d8))
  
(defun count-results (results)
  (loop for i from 0 below (array-total-size results)
        count (row-major-aref results i)))

(defun place-annode (pos results)
  (let ((x (first pos)) (y (second pos)))
    (when (in-map results x y) 
      (setf (aref results y x) t))))

(defun create-annodes-p1 (x1 y1 x2 y2)
  (let ((delta-x (- x2 x1)) (delta-y (- y2 y1)))
    (list (list (- x1 delta-x) (- y1 delta-y)) (list (+ x2 delta-x) (+ y2 delta-y)))))

(defun place-annodes (positions results create-annodes)
  (when positions
     (loop with a = (car positions)
           with x1 = (first a)
           with y1 = (second a)
           for b in (cdr positions)
           for ans = (funcall create-annodes x1 y1 (first b) (second b))
           do (dolist (a ans) (place-annode a results)))
     (place-annodes (cdr positions) results create-annodes)))

(defun place-all-annodes (xmits map &optional (create-annodes #'create-annodes-p1))
  (let ((results (make-array (array-dimensions map) :element-type 'boolean :initial-element nil)))
    (loop for k being the hash-key of xmits
          do (place-annodes (gethash k xmits) results create-annodes))
    results))

(defun find-transmitters (map)
  "look throught the map and record where the transmitters are in a hash map"
  (let ((h (make-hash-table)))
    (destructuring-bind (rows cols) (array-dimensions map)
      (loop for j from 0 below rows
            do (loop for i from 0 below cols
                     for v = (aref map j i)
                     unless (eql v '|.|)
                       do (push (list i j) (gethash v h))
                     )))
    h))

(defun run-p1 (file) 
  (let* ((map (list-to-2d-array (read-file file #'p1-process-line))))
    (count-results (place-all-annodes (find-transmitters map) map))
    ))

(defun create-annodes-2 (x1 y1 x2 y2 map)
  (destructuring-bind (rows cols) (array-dimensions map)
    (let* ((m (/ (- y2 y1) (- x2 x1) ))
           (b (- y2 (* m x2))))
      (loop for x from 0 below cols
            for y = (+ b (* x m))
            for r = (nth-value 1 (floor y))
            when (and (= r 0) (>= y 0) (< y rows))
              collect (list x y)))))

(defun run-p2 (file) 
  (let* ((map (list-to-2d-array (read-file file #'p1-process-line))))
    (count-results (place-all-annodes (find-transmitters map) map
                                      (lambda (x1 y1 x2 y2)
                                        (create-annodes-2 x1 y1 x2 y2 map))))))

[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago)

Uiua

Getting closer to an elegant solution, but still a way to go...

How to read this

Try it live!

Grid      ← βŠœβˆ˜βŠΈβ‰ @\n "............\n........0...\n.....0......\n.......0....\n....0.......\n......A.....\n............\n............\n........A...\n.........A..\n............\n............"
Pairs     ← ∧(βŠ‚β§…β‰ 2βŠšβŒ•)βŠ™Β€βŠΈ(β—΄β–½βŠΈβ‰ @.β™­)Grid[] # get pairs of nodes in grid (both ways round).
InGrid    ← ▽≑/Γ—Γ—<⧻Grid:β‰₯0..            # (posns) Keeps those in range.
AntiNodes ← β†―βˆž_2≑(+Β€βŠ™(Γ—Β€)βŠƒβŠ£/-):Β€        # (range, pairs) Find antinodes by taking offset, mul by range, add to last node, check it's in grid.
&p ⧻InGridβ—΄ AntiNodes β†˜1⇑2 Pairs
&p ⧻InGridβ—΄ AntiNodes β†˜0⇑50 Pairs
[–] [email protected] 2 points 2 weeks ago

C

Not hard but a little fiddly.

Code

#include "common.h"

#define GZ 52

static char g[GZ][GZ];
#define ANTI_P1 1
#define ANTI_P2 2
static uint8_t anti[GZ][GZ];
static int w,h;

int
main(int argc, char **argv)
{
	int p1=0,p2=0, x,y, x1,y1, ax,ay, i;
	char *lf;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));
	for (h=0; h<GZ && fgets(g[h], GZ, stdin); h++)
		;

	assert(feof(stdin));
	lf = strchr(g[0], '\n');
	assert(lf);
	w = lf - g[0];

	/*
	 * Find antenna pairs, then project backwards from the first,
	 * forwards from the second. Don't like the repetition but it
	 * makes for easy code.
	 */
	for (y=0; y<h; y++)
	for (x=0; x<w; x++) {
		if (!isalnum(g[y][x]))
			continue;

		for (y1=y; y1<h; y1++)
		for (x1=(y==y1?x+1:0); x1<w; x1++) {
			if (g[y][x] != g[y1][x1])
				continue;

			for (i=0; ; i++) {
				if ((ax = x-(x1-x)*i) <0 || ax>w ||
				    (ay = y-(y1-y)*i) <0 || ay>h)
					break;
				anti[ay][ax] |= ANTI_P1 * i==1;
				anti[ay][ax] |= ANTI_P2;
			}

			for (i=0; ; i++) {
				if ((ax = x1+(x1-x)*i) <0 || ax>w ||
				    (ay = y1+(y1-y)*i) <0 || ay>h)
					break;
				anti[ay][ax] |= ANTI_P1 * i==1;
				anti[ay][ax] |= ANTI_P2;
			}
		}
	}

	for (y=0; y<h; y++)
	for (x=0; x<w; x++) {
		p1 += !!(anti[y][x] & ANTI_P1);
		p2 += !!(anti[y][x] & ANTI_P2);
	}

	printf("08: %d %d\n", p1, p2);
	return 0;
}

https://github.com/sjmulder/aoc/blob/master/2024/c/day08.c

[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago)

python

solution

import aoc

def setup():
    lines = aoc.get_lines(8, stripped=True)
    ll = len(lines)
    fm = {f: [(x, y) for y, r in enumerate(lines)
              for x, z in enumerate(r) if z == f]
          for f in {z for r in lines for z in r if z != '.'}}
    return ll, fm

def fa(fm, ll, rh=False):
    ans = set()
    for cd in fm.values():
        l = len(cd)
        for i in range(l):
            x1, y1 = cd[i]
            for j in range(i + 1, l):
                x2, y2 = cd[j]
                dx, dy = x2 - x1, y2 - y1
                if rh:
                    for k in range(-ll, ll):
                        x, y = x1 + k * dx, y1 + k * dy
                        if 0 <= x < ll and 0 <= y < ll:
                            ans.add((x, y))
                else:
                    x3, y3, x4, y4 = x1 - dx, y1 - dy, x2 + dx, y2 + dy
                    if 0 <= x3 < ll and 0 <= y3 < ll:
                        ans.add((x3, y3))
                    if 0 <= x4 < ll and 0 <= y4 < ll:
                        ans.add((x4, y4))
    return len(ans)

def one():
    ll, fm = setup()
    print(fa(fm, ll))

def two():
    ll, fm = setup()
    print(fa(fm, ll, rh=True))

one()
two()

[–] [email protected] 1 points 2 weeks ago

J

J really doesn't have hashes! Or anything like hashes! And it's really annoying after a while!

What it does have is automatic internal optimization via hashing of the "index of" operation m i. n where m is a fixed list (the object being searched) and n is the query, which can vary. But as soon as you update m the hash table is thrown away. And you still have to choose some kind of numeric key, or store a list of boxed pairs where the first coordinate is the key -- effectively this is an old-style Lisp association list, but with extra steps because you have to use boxing to defeat J's automatic array concatenation and reshaping. If you want non-cubical shapes (J calls these "ragged arrays"), or heterogeneous lists, you end up writing u &amp;.> a lot -- this means "unbox, apply u then rebox". J arrays are required to be rectangular and homogeneous, but a boxed anything is a single atom just like a number is.

It's just a really bad choice of language if you want data structures other than essentially-cubical arrays. On the other hand, once you beat the list manipulation primitives into producing your 1970s Lisp data structure of choice, the rest of the program is as nice as it usually is.

data_file_name =: '8.data'
grid =: ,. > cutopen fread data_file_name
'rsize csize' =: $ grid
inbounds =: monad : '(*/ y >: 0 0) * (*/ y &lt; rsize, csize)'
antenna_types =: (#~ (~: &amp; '.')) ~. , grid
NB. list_antennas gives a list of boxed matrices of shape 2 n_k in cell k, where
NB. n_k is the number of antennas of type k and the rows are coordinates of that type
list_antennas =: monad define
   antenna_locs =. (# antenna_types) $ a:
   for_r. i. rsize do.
      for_c. i. csize do.
         cell =. y {~ &lt;(r, c)
         if. '.' ~: cell do.
            at =. antenna_types i. cell
            antenna_locs =. ((&lt;(r, c)) ,&amp;.> at { antenna_locs) at} antenna_locs
         end.
      end.
   end.
   NB. _2 ]\ l reshapes l into length 2 rows without finding its length ahead of time
   (_2 &amp; (]\))&amp;.> antenna_locs
)
NB. a1 pair_antinodes a2 gives the two antinodes from that pair
pair_antinodes =: dyad : '(#~ inbounds"1) ((2 * x) - y) ,: (2 * y) - x'
NB. if u is a symmetric dyad expecting rank 1 arguments, u on_pairs is a monad
NB. expecting a list of rank 1 arguments, and yields the concatenation of x u y
NB. where (x, y) is drawn from the (unordered) pairs of elements of the argument
NB. see page_pairs in 5.ijs for a non-point-free version of pair enumeration
on_pairs =: adverb define
   ; @: (&lt; @: u/"2) @: ({~ (; @: (&lt; @: (,~"0 i.)"0) @: i. @: #))
)
NB. antinodes antennas gives a list (may contain duplicates) of all the antinodes from
NB. that set of antennas
antinodes =: pair_antinodes on_pairs
NB. on_antennas concatenates and uniquifies result lists from all antennas
on_antennas =: adverb define
   ~. @: ; @: (u &amp;.>) @: list_antennas
)
result1 =: # antinodes on_antennas grid

NB. a1 res_antinodes a2 gives the list of antinodes from that pair with resonance
res_antinodes =: dyad define
   step =. (% +./) x - y
   NB. lazy: max_steps doesn't take location of x into account
   max_steps =. &lt;. (rsize % 1 >. | 0 { step) &lt;. (csize % 1 >. 1 { step)
   (#~ inbounds"1) x +"1 step *"1 0 i: max_steps
)
result2 =: # res_antinodes on_pairs on_antennas grid
[–] [email protected] 1 points 2 weeks ago

Haskell

import Control.Arrow
import Control.Monad
import Data.List
import Data.Map qualified as M

type Pos = [Int]

parse :: String -> (Pos, [(Char, Pos)])
parse s = ([n, m], [(c, [i, j]) | i <- [0 .. n], j <- [0 .. m], c <- [l !! i !! j], c /= '.'])
  where
    l = lines s
    n = pred $ length $ head l
    m = pred $ length l

buildMap :: [(Char, Pos)] -> M.Map Char [Pos]
buildMap = M.fromListWith (++) . fmap (second pure)

allPairs :: [Pos] -> [(Pos, Pos)]
allPairs l = [(x, y) | (x : xs) <- tails l, y <- xs]

add = zipWith (+)
sub = zipWith (-)

antinodes :: Pos -> Pos -> [Pos]
antinodes a b = [a `sub` ab, b `add` ab]
  where
    ab = b `sub` a

inBounds [x', y'] [x, y] = x >= 0 && y >= 0 && x <= x' && y <= y'

antinodes' :: Pos -> Pos -> Pos -> [Pos]
antinodes' l a b = al ++ bl
  where
    ab = b `sub` a
    al = takeWhile (inBounds l) $ iterate (`sub` ab) a
    bl = takeWhile (inBounds l) $ iterate (`add` ab) b

part1 l = length . nub . filter (inBounds l) . concat . M.elems . fmap (allPairs >=> uncurry antinodes)
part2 l = length . nub . concat . M.elems . fmap (allPairs >=> uncurry (antinodes' l))

main = getContents >>= print . (uncurry part1 &&& uncurry part2) . second buildMap . parse
[–] [email protected] 1 points 2 weeks ago

Rust

Pretty happy with my solution today. I took my time today as it was a bit of a slow day and did it in Rust instead of python. Having proper Vec2 types is very nice.

Tap for spoiler

use std::{collections::HashMap, error::Error, io::Read};

use glam::{IVec2, Vec2};

fn permutations_of_size_two(antennas: &[Vec2]) -> Vec<[&Vec2; 2]> {
    let mut permutations = vec![];
    for (i, antenna) in antennas.iter().enumerate() {
        for j in 0..antennas.len() {
            if i == j {
                continue;
            }
            permutations.push([antenna, &antennas[j]])
        }
    }
    permutations
}

fn main() -> Result<(), Box<dyn Error>> {
    let mut input = String::new();
    std::io::stdin().read_to_string(&mut input)?;

    let height = input.lines().count() as i32;
    let width = input.lines().next().unwrap().len() as i32;

    let antenna_positions = input
        .lines()
        .enumerate()
        .flat_map(|(y, l)| 
            l.chars().enumerate().map(move |(x, c)| (Vec2::new(x as f32, y as f32), c))
        )
        .filter(|(_v, c)| *c != '.')
        .fold(HashMap::new(), |mut acc: HashMap<char, Vec<_>> , current| {
            acc.entry(current.1).or_default().push(current.0);
            acc
        });

    let mut antinodes = vec![];
    for (_c, antennas) in antenna_positions {
        let perms = permutations_of_size_two(&antennas);
        for [first, second] in perms {
            let mut i = 1.;
            loop {
                let antinode = (first + (second-first) * i).round();
                if (0..height).contains(&(antinode.x as i32)) &&
                    (0..width).contains(&(antinode.y as i32)) {
                        antinodes.push(antinode);
                } else {
                    break;
                }
                i += 1.;
            }
        }
    }

    let mut antinode_count = 0;
    let map = input
        .lines()
        .enumerate()
        .map(|(y, l)| 
            l.chars().enumerate().map(|(x, c)| {
                if antinodes.contains(&Vec2::new(x as f32, y as f32)) {
                    println!("({x},{y})");
                    antinode_count += 1;
                    return '#';
                }
                c
            }).collect::<String>()
        )
        .collect::<Vec<_>>()
        .join("\n");

    println!("{map}");
    println!("{antinode_count}");

    Ok(())
}

[–] [email protected] 1 points 2 weeks ago

TypeScript

I was a little confuzzled with this one, but I managed to get it. :) Happy to know that I managed to reuse more of my code from previous days. I should write something to handle Vectors. It was sad to write my own basic, non-reusable thing.

Solution

import { AdventOfCodeSolutionFunction } from "./solutions";

// imported code:
export const check_coords = (grid: Array<Array<any>>, x: number, y: number) => {
    return y >= grid.length ||
        y < 0 ||
        x >= grid[y].length ||
        x < 0
}

export const makeGridFromMultilineString =
    (input: string) => input.split("\n").map(st => st.trim()).map(v => v.split(""));

export const MakeEmptyGenericArray = <T>(length: number, fn: (index: number) => T) => {
    const newArray = [];
    let i = 0;
    while (i++ < length)
        newArray.push(fn(i));

    return newArray;
}

export const MakeEmptyArray = (length: number) => MakeEmptyGenericArray(length, () => 0);
export const MakeEmpty2DArray = (x: number, y: number) => MakeEmptyArray(y).map(() => MakeEmptyArray(x));

// solution code
type v2 = [x: number, y: number];

const Sub = (x1: number, y1: number, x2: number, y2: number): v2 => {
    return [x1 - x2, y1 - y2];
}

const Add = (x1: number, y1: number, x2: number, y2: number): v2 => {
    return [x1 + x2, y1 + y2];
}

export const solution_8: AdventOfCodeSolutionFunction = (input) => {
    const grid = makeGridFromMultilineString(input);
    const nodes = new Map<string, Array<v2>>();
    const nodeKinds: Array<string> = [];
    const singleAntinodeLocations = MakeEmpty2DArray(grid.length, grid[0].length);
    const resonantAntinodeLocations = MakeEmpty2DArray(grid.length, grid[0].length);

    // find all the nodes
    grid.forEach((row, y) => row.forEach((item, x) => {
        if (item == ".")
            return;

        if (nodes.has(item))
            nodes.get(item)!.push([x, y]);

        else {
            nodes.set(item, [[x, y]]);
            nodeKinds.push(item);
        }
    }));

    nodeKinds.forEach((nodeKind) => {
        const nodesOfKind = nodes.get(nodeKind)!;
        for (let bunn = 0; bunn < nodesOfKind.length; bunn++) {
            const first = nodesOfKind[bunn];
            for (let tort = bunn + 1; tort < nodesOfKind.length; tort++) {
                // find antinode
                const second = nodesOfKind[tort];
                const diff = Sub(...first, ...second);
                const [x1, y1] = Add(...first, ...diff);
                const [x2, y2] = Sub(...second, ...diff);

                if(!check_coords(singleAntinodeLocations, x1, y1)) singleAntinodeLocations[y1][x1]++;
                if(!check_coords(singleAntinodeLocations, x2, y2)) singleAntinodeLocations[y2][x2]++;

                // find all resonances
                // starting
                resonantAntinodeLocations[first[1]][first[0]]++;
                resonantAntinodeLocations[second[1]][second[0]]++;

                // go forward
                let newFirst = [x1, y1] as v2;
                while(!check_coords(resonantAntinodeLocations, ...newFirst)) {
                    let [x, y] = newFirst;
                    resonantAntinodeLocations[y][x]++;
                    newFirst = Add(...newFirst, ...diff);
                }

                // go back
                newFirst = [x2, y2] as v2;
                while(!check_coords(resonantAntinodeLocations, ...newFirst)) {
                    let [x, y] = newFirst;
                    resonantAntinodeLocations[y][x]++;
                    newFirst = Sub(...newFirst, ...diff);
                }
            }
        }
    });

    const antinodeCount = (prev: number, curr: Array<number>) => prev + curr.reduce((prev, curr) => prev + (curr > 0 ? 1 : 0), 0);
    const part_1 = singleAntinodeLocations.reduce<number>(antinodeCount, 0);
    const part_2 = resonantAntinodeLocations.reduce<number>(antinodeCount, 0);

    return {
        part_1, //390
        part_2, //1246
    }
}

Loops on loops on loops on loops...

[–] [email protected] 1 points 2 weeks ago (1 children)

Rust

Proper Point and Vector types made this pretty simple, part 2 was just a tiny change (basically while instead of if), but left with a lot of copy-pasted code.

Solution

use euclid::default::*;

const N_ANTENNAS: usize = (b'z' - b'0') as usize + 1;
// For each frequency (from b'0' to b'z') the list of antenna positions
type Antennas = Box<[Vec<Point2D<i32>>]>;

fn parse(input: String) -> (Antennas, Rect<i32>) {
    let mut antennas = vec![Vec::new(); N_ANTENNAS].into_boxed_slice();
    let mut width = 0;
    let mut height = 0;
    for (y, l) in input.lines().enumerate() {
        height = y + 1;
        if width == 0 {
            width = l.len()
        } else {
            assert!(width == l.len())
        }
        for (x, b) in l.bytes().enumerate().filter(|(_, b)| *b != b'.') {
            antennas[(b - b'0') as usize].push(Point2D::new(x, y).to_i32())
        }
    }
    let bounds = Rect::new(Point2D::origin(), Size2D::new(width, height).to_i32());
    (antennas, bounds)
}

fn part1(input: String) {
    let (antennas, bounds) = parse(input);
    let mut antinodes = vec![vec![false; bounds.width() as usize]; bounds.height() as usize];
    for list in antennas.iter().filter(|l| !l.is_empty()) {
        for (i, &a) in list.iter().enumerate().skip(1) {
            for &b in list.iter().take(i) {
                let diff = b - a;
                let ax = a - diff;
                if bounds.contains(ax) {
                    antinodes[ax.y as usize][ax.x as usize] = true;
                }
                let bx = b + diff;
                if bounds.contains(bx) {
                    antinodes[bx.y as usize][bx.x as usize] = true;
                }
            }
        }
    }
    let sum = antinodes
        .iter()
        .map(|row| row.iter().map(|b| u32::from(*b)).sum::<u32>())
        .sum::<u32>();
    println!("{sum}");
}

fn part2(input: String) {
    let (antennas, bounds) = parse(input);
    let mut antinodes = vec![vec![false; bounds.width() as usize]; bounds.height() as usize];
    for list in antennas.iter().filter(|l| !l.is_empty()) {
        for (i, &a) in list.iter().enumerate().skip(1) {
            for &b in list.iter().take(i) {
                let diff = b - a;
                // Start at antenna a, keep going until hitting bounds
                let mut ax = a;
                while bounds.contains(ax) {
                    antinodes[ax.y as usize][ax.x as usize] = true;
                    ax -= diff;
                }
                let mut bx = b;
                while bounds.contains(bx) {
                    antinodes[bx.y as usize][bx.x as usize] = true;
                    bx += diff;
                }
            }
        }
    }
    let sum = antinodes
        .iter()
        .map(|row| row.iter().map(|b| u32::from(*b)).sum::<u32>())
        .sum::<u32>();
    println!("{sum}");
}

util::aoc_main!();

also on github

[–] [email protected] 2 points 2 weeks ago (1 children)

I also did it in Rust, though mine is quite a bit more verbose than yours :). https://gitlab.com/bricka/advent-of-code-2024-rust/-/blob/main/src/days/day08.rs?ref_type=heads

Have you considered using a HashSet<(usize, usize)> to store the visited locations instead of a 2d vector?

[–] [email protected] 1 points 2 weeks ago

I try to use Vecs instead of HashSets and maps whenever the key domain is reasonably small (just the grid in this case), simply because in the end direct memory access is a lot faster than always hashing values.

But looking at this case again, it is certainly a lot easier to have just antinodes.len() at the end instead of counting all true values. This datastructure is also not really performance-critical, so a HashSet is probably the cleaner choice here.

[–] [email protected] 1 points 2 weeks ago

Rust

use std::collections::{HashMap, HashSet};

use crate::solver::DaySolver;
use crate::grid::{Coordinate, Grid};

fn add_distance(coordinate: Coordinate, distance: (i64, i64)) -> Option<Coordinate> {
    coordinate.try_add(distance)
}

fn sub_distance(coordinate: Coordinate, distance: (i64, i64)) -> Option<Coordinate> {
    coordinate.try_sub(distance)
}

fn part2_possible_antinodes<F>(
    grid: &Grid<Option<char>>,
    coordinate: Coordinate,
    distance: (i64, i64),
    op: F,
    mut accumulator: Vec<Coordinate>
) -> Vec<Coordinate>
where F: Fn(Coordinate, (i64, i64)) -> Option<Coordinate> {
    match op(coordinate, distance).filter(|c| grid.get(*c).is_some()) {
        None => accumulator,
        Some(next_coord) => {
            accumulator.push(next_coord);
            part2_possible_antinodes(grid, next_coord, distance, op, accumulator)
        }
    }
}

trait Pairable<T> {
    fn pairs(&self) -> Vec<(&T, &T)>;
}

impl<T> Pairable<T> for HashSet<T> {
    fn pairs(&self) -> Vec<(&T, &T)> {
        let v: Vec<&T> = self.iter().collect();

        let mut p = vec![];

        for i in 0..v.len() {
            let thing1 = v[i];

            for thing2 in &v[i+1..] {
                p.push((thing1, *thing2));
            }
        }

        p
    }
}

fn parse_input(input: String) -> (Grid<Option<char>>, HashMap<char, HashSet<Coordinate>>) {
    let g: Grid<Option<char>> =
        input.lines()
        .map(|line| line.chars()
             .map(|c| if c == '.' {
                 None
             } else {
                 Some(c)
             }).collect::<Vec<Option<char>>>()
        )
        .collect::<Vec<Vec<Option<char>>>>()
        .into();

    let mut freq_to_coords: HashMap<char, HashSet<Coordinate>> = HashMap::new();

    for (coord, freq_opt) in g.iter() {
        match freq_opt {
            None => (),
            Some(freq) => {
                freq_to_coords.entry(*freq)
                    .and_modify(|coords| {
                        coords.insert(coord);
                    })
                    .or_insert(HashSet::from([coord]));
            }
        }
    }

    (g, freq_to_coords)
}

pub struct Day08Solver;

impl DaySolver for Day08Solver {
    fn part1(&self, input: String) -> usize {
        let (g, freq_to_coords) = parse_input(input);

        let mut antinodes: HashSet<Coordinate> = HashSet::new();

        for (_, coords) in freq_to_coords {
            // println!("Freq = {}", freq);
            for (c1, c2) in coords.pairs() {
                let distance = c1.xy_distance_to(c2);
                let possible_antinodes: Vec<Coordinate> = [c1.try_sub(distance), c2.try_add(distance)].into_iter()
                    .flat_map(|co| co.filter(|c| g.get(*c).is_some()))
                    .collect();

                // println!("Pair = ({},{}), antinodes = {:?}", c1, c2, possible_antinodes);

                for antinode in possible_antinodes {
                    antinodes.insert(antinode);
                }
            }
        }

        antinodes.len()
    }

    fn part2(&self, input: String) -> usize {
        let (g, freq_to_coords) = parse_input(input);

        let mut antinodes: HashSet<Coordinate> = HashSet::new();

        for (freq, coords) in freq_to_coords {
            println!("Freq = {}", freq);
            for (c1, c2) in coords.pairs() {
                let distance = c1.xy_distance_to(c2);

                let possible_antinodes: Vec<Coordinate> = [
                    part2_possible_antinodes(&g, *c1, distance, add_distance, vec![*c1]),
                    part2_possible_antinodes(&g, *c1, distance, sub_distance, vec![*c1]),
                    part2_possible_antinodes(&g, *c2, distance, add_distance, vec![*c2]),
                    part2_possible_antinodes(&g, *c2, distance, sub_distance, vec![*c2]),
                ].into_iter().flatten().collect();

                println!("Pair = ({},{}), antinodes = {:?}", c1, c2, possible_antinodes);

                for antinode in possible_antinodes {
                    antinodes.insert(antinode);
                }
            }
        }

        antinodes.len()
    }
}

https://gitlab.com/bricka/advent-of-code-2024-rust/-/blob/main/src/days/day08.rs?ref_type=heads

[–] [email protected] 1 points 2 weeks ago* (last edited 2 weeks ago)

Nim

Overall really simple puzzle, but description is so confusing, that I mostly solved it based on example diagrams.
Edit: much shorter and faster one-pass solution. Runtime: 132 us

type Vec2 = tuple[x,y: int]
func delta(a, b: Vec2): Vec2 = (a.x-b.x, a.y-b.y)
func outOfBounds[T: openarray | string](pos: Vec2, grid: seq[T]): bool =
  pos.x < 0 or pos.y < 0 or pos.x > grid[0].high or pos.y > grid.high

proc solve(input: string): AOCSolution[int, int] =
  var grid = input.splitLines()
  var antennas: Table[char, seq[Vec2]]

  for y, line in grid:
    for x, c in line:
      if c != '.':
        discard antennas.hasKeyOrPut(c, newSeq[Vec2]())
        antennas[c].add (x, y)

  var antinodesP1: HashSet[Vec2]
  var antinodesP2: HashSet[Vec2]

  for _, list in antennas:
    for ind, ant1 in list:
      antinodesP2.incl ant1 # each antenna is antinode
      for ant2 in list.toOpenArray(ind+1, list.high):
        let d = delta(ant1, ant2)
        for dir in [-1, 1]:
          var i = dir
          while true:
            let antinode = (x: ant1.x+d.x*i, y: ant1.y+d.y*i)
            if antinode.outOfBounds(grid): break
            if i in [1, -2]: antinodesP1.incl antinode
            antinodesP2.incl antinode
            i += dir
  result.part1 = antinodesP1.len
  result.part2 = antinodesP2.len


Codeberg repo

[–] [email protected] 1 points 2 weeks ago

Kotlin

A bit late to the party, but here's my solution. I don't know, if you even need to search for the smallest integer vector in the same direction in part 2, but I did it anyway.

Code:

import kotlin.math.abs
import kotlin.math.pow

fun main() {
    fun part1(input: List<String>): Int {
        val inputMap = Day08Map(input)
        return inputMap.isoFrequencyNodeVectorsByLocations
            .flatMap { (location, vectors) ->
                vectors.map { (2.0 scaleVec it) + location }
            }
            .toSet()
            .count { inputMap.isInGrid(it) }
    }

    fun part2(input: List<String>): Int {
        val inputMap = Day08Map(input)
        return buildSet {
            inputMap.isoFrequencyNodeVectorsByLocations.forEach { (location, vectors) ->
                vectors.forEach { vector ->
                    var i = 0.0
                    val scaledDownVector = smallestIntegerVectorInSameDirection2D(vector)
                    while (inputMap.isInGrid(location + (i scaleVec scaledDownVector))) {
                        add(location + (i scaleVec scaledDownVector))
                        i++
                    }
                }
            }
        }.count()
    }

    val testInput = readInput("Day08_test")
    check(part1(testInput) == 14)
    check(part2(testInput) == 34)

    val input = readInput("Day08")
    part1(input).println()
    part2(input).println()
}

tailrec fun gcdEuclid(a: Int, b: Int): Int =
    if (b == 0) a
    else if (a == 0) b
    else if (a > b) gcdEuclid(a - b, b)
    else gcdEuclid(a, b - a)

fun smallestIntegerVectorInSameDirection2D(vec: VecNReal): VecNReal {
    assert(vec.dimension == 2)  // Only works in two dimensions.
    assert(vec == vec.roundComponents())  // Only works on integer vectors.

    return (gcdEuclid(abs(vec[0].toInt()), abs(vec[1].toInt())).toDouble().pow(-1) scaleVec vec).roundComponents()
}

class Day08Map(input: List<String>): Grid2D<Char>(input.reversed().map { it.toList() }) {
    init {
        transpose()
    }

    val isoFrequencyNodesLocations = asIterable().toSet().filter { it != '.' }.map { frequency -> asIterable().indicesWhere { frequency == it } }
    val isoFrequencyNodeVectorsByLocations = buildMap {
        isoFrequencyNodesLocations.forEach { isoFrequencyLocationList ->
            isoFrequencyLocationList.mapIndexed { index, nodeLocation ->
                this[VecNReal(nodeLocation)] = isoFrequencyLocationList
                    .slice((0 until index) + ((index + 1)..isoFrequencyLocationList.lastIndex))
                    .map { VecNReal(it) - VecNReal(nodeLocation) }
            }
        }
    }
}

[–] [email protected] 1 points 2 weeks ago

Raku

Solution

sub MAIN($input) {
    my $file = open $input;
    my @map = $file.slurp.lines>>.comb>>.List.List;
    my %freqs;
    for 0..^@map.elems -> $row {
        for 0..^@map[0].elems -> $col {
            if @map[$row; $col] ne "." {
                my $freq = @map[$row; $col];
                %freqs{$freq} = [] if %freqs{$freq}:!exists;
                %freqs{$freq}.push(($row, $col));
            }
        }
    }
    my %antinodes is SetHash;
    for %freqs.kv -> $freq, @locations {
        for (0..^@locations.elems) X (0..^@locations.elems) -> ($loc1, $loc2) {
            next if $loc1 == $loc2;
            my @base = @locations[$loc1].List;
            my @vector = @locations[$loc2].List Z- @base;
            my @antinode1 = @base Z+ @vector.map(* * 2);
            %antinodes{@antinode1.List.raku}++ if point-is-in-map(@map, @antinode1);
            my @antinode2 = @base Z+ @vector.map(* * -1);
            %antinodes{@antinode2.List.raku}++ if point-is-in-map(@map, @antinode2);
        }
    }
    my $part1-solution = %antinodes.elems;
    say "part 1: $part1-solution";


    my %antinodes2 is SetHash;
    for %freqs.kv -> $freq, @locations {
        for (0..^@locations.elems) X (0..^@locations.elems) -> ($loc1, $loc2) {
            next if $loc1 == $loc2;
            my @base = @locations[$loc1].List;
            my @vector = @locations[$loc2].List Z- @base;
            # make integer unit-ish vector
            for 2..@vector[0] -> $divisor {
                if @vector[0] %% $divisor and @vector[1] %% $divisor {
                    @vector[0] = @vector[0] div $divisor;
                    @vector[1] = @vector[1] div $divisor;
                }
            }
            for 0..max(@map.elems, @map[0].elems) -> $length {
                my @antinode = @base Z+ @vector.map(* * $length);
                if point-is-in-map(@map, @antinode) {
                    %antinodes2{@antinode.List.raku}++ 
                } else {
                    last
                }
            }
            for 1..max(@map.elems, @map[0].elems) -> $length {
                my @antinode = @base Z+ @vector.map(* * -$length);
                if point-is-in-map(@map, @antinode) {
                    %antinodes2{@antinode.List.raku}++ 
                } else {
                    last
                }
            }
        }
    }
    my $part2-solution = %antinodes2.elems;
    say "part 2: $part2-solution";
}

sub point-is-in-map(@map, @point) {
    return False if !(0 <= @point[0] < @map.elems);
    return False if !(0 <= @point[1] < @map[0].elems);
    return True;
}

[–] [email protected] 1 points 2 weeks ago

And I of course misread and wasted a bunch of time debugging the second part, entirely missed the fact that antinodes occurred on top of the emanating antennae as well...

C#

public static class LINQExt
{
  public static IEnumerable<(T,T)> PermutatePairs<T>(this IEnumerable<T> source) {
    return source.SelectMany(k => source.Where(v => !v?.Equals(k) ?? false).Select(v => (k, v)));
  }
}

struct Antenna
{
  public int X, Y;
  public char Frequency;
}

List<Antenna> antennae = new List<Antenna>();
int width, height;

public void Input(IEnumerable<string> lines)
{
  char[] map = string.Join("", lines).ToCharArray();
  width = lines.First().Length;
  height = lines.Count();

  for (int y = 0; y < height; ++y)
    for (int x = 0; x < width; ++x)
    {
      char at = map[y * width + x];
      if (at == '.')
        continue;

      antennae.Add(new Antenna{ X = x, Y = y, Frequency = at });
    }
}

public void Part1()
{
  HashSet<(int, int)> antinodes = new HashSet<(int, int)>();

  foreach (var antinode in antennae.GroupBy(k => k.Frequency).SelectMany(g => g.PermutatePairs()).SelectMany(v => GetOpposing(v.Item1, v.Item2)).Where(InRange))
    antinodes.Add(antinode);

  Console.WriteLine($"Unique antinodes: {antinodes.Count}");
}
public void Part2()
{
  HashSet<(int, int)> antinodes = new HashSet<(int, int)>();

  foreach (var antennaePair in antennae.GroupBy(k => k.Frequency).SelectMany(g => g.PermutatePairs()))
  {
    // Iterate separately, to make the handling of bound exit easier
    foreach (var antinode in GetAllOpposing(antennaePair.Item1, antennaePair.Item2).TakeWhile(InRange))
      antinodes.Add(antinode);
    foreach (var antinode in GetAllOpposing(antennaePair.Item2, antennaePair.Item1).TakeWhile(InRange))
      antinodes.Add(antinode);
  }
  Console.WriteLine($"Unique antinodes: {antinodes.Count}");
}

bool InRange((int, int) point) {
  return point.Item1 >= 0 && point.Item1 < width && point.Item2 >= 0 && point.Item2 < height;
}
(int, int)[] GetOpposing(Antenna a, Antenna b) {
  return new[] { (a.X + (a.X - b.X), a.Y + (a.Y - b.Y)), (b.X + (b.X - a.X), b.Y + (b.Y - a.Y)) };
}
IEnumerable<(int, int)> GetAllOpposing(Antenna a, Antenna b) {
  (int, int) diff = (a.X - b.X, a.Y - b.Y);
  (int, int) at = (a.X, a.Y);
  yield return at;

  while (true)
  {
    at.Item1 += diff.Item1;
    at.Item2 += diff.Item2;

    yield return at;
  }
}