this post was submitted on 08 Dec 2024
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Advent Of Code

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Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

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Day 8: Resonant Collinearity

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[โ€“] [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
[โ€“] [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!