Day 20: Race Condition

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FAQ

  • VegOwOtenks@lemmy.world
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    19 hours ago

    Haskell

    First parse and floodfill from start, each position then holds the distance from the start

    For part 1, I check all neighbor tiles of neighbor tiles that are walls and calculate the distance that would’ve been in-between.

    In part 2 I check all tiles within a manhattan distance <= 20 and calculate the distance in-between on the path. Then filter out all cheats <100 and count

    Takes 1.4s sadly, I believe there is still potential for optimization.

    Edit: coding style

    import Control.Arrow
    
    import qualified Data.List as List
    import qualified Data.Set as Set
    import qualified Data.Map as Map
    import qualified Data.Maybe as Maybe
    
    parse s = Map.fromList [ ((y, x), c) | (l, y) <- zip ls [0..], (c, x) <- zip l [0..]]
            where
            ls = lines s
    
    floodFill m = floodFill' m startPosition (Map.singleton startPosition 0)
            where
                    startPosition = Map.assocs
                            >>> filter ((== 'S') . snd)
                            >>> head
                            >>> fst
                            $ m
    
    neighbors (p1, p2) = [(p1-1, p2), (p1, p2-1), (p1, p2+1), (p1+1, p2)]
    
    floodFill' m p f
            | m Map.! p == 'E' = f
            | otherwise = floodFill' m n f'
            where
                    seconds = f Map.! p
                    ns = neighbors p
                    n = List.filter ((m Map.!) >>> (`Set.member` (Set.fromList ".E")))
                            >>> List.filter ((f Map.!?) >>> Maybe.isNothing)
                            >>> head
                            $ ns
                    f' = Map.insert n (succ seconds) f
    
    taxiCabDistance (a1, a2) (b1, b2) = abs (a1 - b1) + abs (a2 - b2)
    
    calculateCheatAdvantage f (p1, p2) = c2 - c1 - taxiCabDistance p1 p2
            where
                    c1 = f Map.! p1
                    c2 = f Map.! p2
    
    cheatDeltas :: Int -> Int -> [(Int, Int)]
    cheatDeltas l h = [(y, x) | x <- [-h..h], y <- [-h..h], let d = abs x + abs y, d <= h, d >= l]
    
    (a1, a2) .+. (b1, b2) = (a1 + b1, a2 + b2)
    
    solve l h (f, ps) = Set.toList
            >>> List.map ( repeat
                    >>> zip (cheatDeltas l h)
                    >>> List.map (snd &&& uncurry (.+.))
                    >>> List.filter (snd >>> (`Set.member` ps))
                    >>> List.map (calculateCheatAdvantage f)
                    >>> List.filter (>= 100)
                    >>> List.length
                    )
            >>> List.sum
            $ ps
    part1 = solve 2 2
    part2 = solve 1 20
    
    main = getContents
            >>= print
            . (part1 &&& part2)
            . (id &&& Map.keysSet)
            . floodFill
            . parse
    
    • gedhrel@lemmy.world
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      14 hours ago

      Hey - I’ve a question about this. Why is it correct? (Or is it?)

      If you have two maps for positions in the maze that give (distance to end) and (distance from start), then you can select for points p1, p2 such that

      d(p1, p2) + distance-to-end(p1) + distance-to-start(p2) <= best - 100

      however, your version seems to assume that distance-to-end(p) = best - distance-to-start(p) - surely this isn’t always the case?

      • Gobbel2000@programming.dev
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        13 hours ago

        There is exactly one path without cheating, so yes, the distance to one end is always the total distance minus the distance to the other end.

        • gedhrel@lemmy.world
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          13 hours ago

          Gotcha, thanks. I just re-read the problem statement and looked at the input and my input has the strongest possible version of that constraint: the path is unbranching and has start and end at the extremes. Thank-you!

          • Deebster@programming.dev
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            12 hours ago

            I missed that line too:

            Because there is only a single path from the start to the end

            So I also did my pathfinding for every variation in the first part, but realised something must be wrong with my approach when I saw part 2.

      • gedhrel@lemmy.world
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        14 hours ago

        (I ask because everyone’s solution seems to make the same assumption - that is, that you’re finding a shortcut onto the same path, as opposed to onto a different path.)

        • VegOwOtenks@lemmy.world
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          12 hours ago

          Some others have answered already, but yes, there was a well-hidden line in the problem description about the map having only a single path from start to end…