Day 23: LAN Party
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Haskell
The solution for part two could now be used for part one as well but then I would have to rewrite part 1 .-.
import Control.Arrow import Data.Ord (comparing) import qualified Data.List as List import qualified Data.Map as Map import qualified Data.Set as Set parse = Map.fromListWith Set.union . List.map (second Set.singleton) . uncurry (++) . (id &&& List.map (uncurry (flip (,)))) . map (break (== '-') >>> second (drop 1)) . takeWhile (/= "") . lines depthSearch connections ps | length ps == 4 && head ps == last ps = [ps] | length ps == 4 = [] | otherwise = head >>> (connections Map.!) >>> Set.toList >>> List.map (:ps) >>> List.concatMap (depthSearch connections) $ ps interconnections (computer, connections) = depthSearch connections [computer] part1 = (Map.assocs &&& repeat) >>> first (List.map (uncurry Set.insert)) >>> first (Set.toList . Set.unions) >>> uncurry zip >>> List.concatMap interconnections >>> List.map (Set.fromList . take 3) >>> List.filter (Set.fold (List.head >>> (== 't') >>> (||)) False) >>> Set.fromList >>> Set.size getLANParty computer connections = (connections Map.!) >>> findLanPartyComponent connections [computer] $ computer filterCandidates connections participants candidates = List.map (connections Map.!) >>> List.foldl Set.intersection candidates >>> Set.filter ((connections Map.!) >>> \ s -> List.all (flip Set.member s) participants) $ participants findLanPartyComponent connections participants candidates | Set.null validParticipants = participants | otherwise = findLanPartyComponent connections (nextParticipant : participants) (Set.delete nextParticipant candidates) where nextParticipant = Set.findMin validParticipants validParticipants = filterCandidates connections participants candidates part2 = (Map.keys &&& repeat) >>> uncurry zip >>> List.map ((uncurry getLANParty) >>> List.sort) >>> List.nub >>> List.maximumBy (comparing List.length) >>> List.intercalate "," main = getContents >>= print . (part1 &&& part2) . parse
I initially thought that, but now I reconsider I’m not so sure. Isn’t it possible to have a 3-member clique overlapping two larger ones? In other words, there could be more than one way to partition the graph into completely connected components. Which means my solution to part 2 is technically incorrect. Bummer.
There probably are multiple ways to partition the graph. I haven’t applied any optimizations and my program checks members of already detected groups again, would that yield all possible partitions because I choose all the possible starting points for a k-clique?