Day 12: Garden Groups

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FAQ

  • Quant@programming.dev
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    2 months ago

    Uiua

    I spent a while thinking about how to best do a flood fill in Uiua when I saw that (partition) works beautifully with multidimensional markers: “Groups are formed from markers that are adjacent along any axis.”, meaning I just had to convert all letters into numbers and I’d get all indices belonging to a field into an array.
    For part 2, I cheated a bit by coming here and reading that you only need to count the edges. To my surprise, the second part is actually a bit faster than part 1. Takes less than 0.2 seconds each though :D

    Run with example input here

    $ RRRRIICCFF
    $ RRRRIICCCF
    $ VVRRRCCFFF
    $ VVRCCCJFFF
    $ VVVVCJJCFE
    $ VVIVCCJJEE
    $ VVIIICJJEE
    $ MIIIIIJJEE
    $ MIIISIJEEE
    $ MMMISSJEEE
    .
    N     ← +[0_¯1 0_1 ¯1_0 1_0]
    Areas ← ⊜□:⇡△.+1⍜♭⊛
    Peri  ← -/+≡(/+∊N¤)⟜¤⟜(×4⧻)
    Sides ← (
      ⊙(-¤)↯:▽⊙0×°⊟.+2⌵⊸-+1⊃⊣⊢⊸⍜⍉≡⍆
      ⧻⊚⊸∊1_3⧈(/+/+)2_2.⍜⊡=₀+1:
      +⊙(×2/+/+⧈(∊[[1_0 0_1][0_1 1_0]])2_2◌)
    )
    Cost! ← /+≡◇(×^0⟜⧻)
    
    PartOne ← (
      # &rs ∞ &fo "input-12.txt"
      ⊜∘≠@\n.
      Cost!Peri Areas
    )
    
    PartTwo ← (
      # &rs ∞ &fo "input-12.txt"
      ⊜∘≠@\n.
      Cost!Sides Areas
    )
    
    &p "Day 12:"
    &pf "Part 1: "
    &p PartOne
    &pf "Part 2: "
    &p PartTwo
    
  • RagingHungryPanda@lemm.ee
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    1 month ago

    I know I’m late, but it’s still fun and I’m sure no-one will see this.

    Part 2 took me way too long to get right. I was initially only returning the relative point to which a plot needed a fence. I ran into issues of knowing if it was a valid fence or not by my method of counting (later). I eventually went with returning a tuple of the plot and an enum flag of the sides that it has fences on.

    For counting I grouped the points by one axis then sorted on the other and counted the number of times the transition between two wasn’t contiguous.

    It could by done in parallel, but the original traversal would need de-duping, which I didn’t feel like doing. After that things are done on a region basis, which could be parallel.

    I also can’t help but notice mine is by far the longest ( > . < )

    F#

    Tap for spoiler
    type Plant = char
    type Plot = Plant * Point2I
    type Region = Plot list
    
    let area region = List.length region
    
    let perimeter (region:Region) (farm:Grid<Plant>) =
        (0, region)
        ||> List.fold (fun sum plot ->
            movements4 (snd plot) farm
            |> List.fold (fun acc po ->
                acc + match po with
                      | ValueNone -> 1
                      | ValueSome pother ->
                          get pother farm
                          |> fun x ->
                              match x with
                              | ValueSome v when v <> fst plot -> 1
                              | _ -> 0
                ) 0
            |> (fun x -> x + sum)
            )
    
    let movements (plot:Plot) g (visited:HashSet<Point2I>) =
        let plant, point = plot
        movements4 point g
        |> List.choosev (fun px ->
            let vx = get px g |> ifNoneFail "couldn't get point"
            struct(px,vx))
        |> List.filter (fun plotx ->
            let struct(px, vx) = plotx
            vx = plant && not (visited.Contains px))
    
    // visited is needed because I'm using similar logic to the trails, but no stopping point, so I
    // need to make sure that it doesn't retrace itself
    let rec traverse grid plot (visited:HashSet<Point2I>) =
        let plant, point = plot
        if visited.Contains(point) then []
        else
            visited.Add(point) |> ignore
            let path =
                movements plot grid visited
                |> List.filter (fun struct(newPoint, newPlant) -> newPlant = plant && visited.Contains(newPoint) |> not )
                |> List.map(fun struct(newPoint, newPlant) -> traverse grid (newPlant, newPoint) visited)
                |> List.concat
                |> List.distinct
            plot :: path
    
    /// Returns the list of plots walked and a new grid with the traversed plots set to '.'
    let walk (plot:Plot) (farm:Grid<Plant>) : Region * Grid<Plant> =
        let foundMovements = HashSet()
        let region:Region = traverse farm plot foundMovements
        let updatedGrid =
            let arr = Array2D.copy farm
            for _, p in region do
                set p arr '.' |> ignore
            arr
        (region, updatedGrid)
        
    let rec findRegions (farm:Grid<Plant>) : Region list=
        farm
        |> List.unfold (fun arr ->
            match (findFirst (fun struct(_,_,v) -> v <> '.' )) arr with
            | Some value ->
                let struct(x,y,v) = value
                let point = {X = x; Y = y}
                let plot : Plot = (v, point)
                let region, newFarm = walk plot arr
                Some (region, newFarm)
            | None -> None
        )
        
    let part1 input =
        (readGrid2D input (fun _ c -> c))
        |> fun grid ->
            findRegions grid
            |> List.sumBy (fun region ->
                let a = area region
                let p = perimeter region grid
                a * p)
    
    // Part 2 ---------------------------------------------
    [<Flags>]
    type Side = None = 0 | Top = 1 | Right = 2| Left = 4 | Bottom = 8
    type PlotWithFence = Plot * Side 
    
    type SideAndNextPoint = Side * Point2I
    
    // I couldn't use my original movement function because it filters out off-grid positions, so I modified it here
    let getSides plot grid =
        let toPlot (side:Side) (voption:Point2I voption) =
            match voption with | ValueNone -> None, side | ValueSome point -> (Some point, side)
        [
            l (snd plot) |> toPlot Side.Left
            r (snd plot) grid |> toPlot Side.Right
            u (snd plot) |> toPlot Side.Top
            d (snd plot) grid |> toPlot Side.Bottom
        ]
    
    
    let rec findEdges grid (plot:Plot) (visited:HashSet<Point2I>) =
        
        // this attempt that works is to attach the side information to each point
        // I didn't see a great way to use the existing movements function, so I copied it and added a mapping
        // so that I know in what direction the movement options are
        let sidesWithFence = getSides plot grid
    
        let plant, point = plot
        if visited.Add(point) |> not then Side.None
        else
            // use the movements to find edges
            let edges = 
                sidesWithFence
                |> List.fold (fun sides next ->
                    match next with
                    | None, side ->  sides ||| side
                    | Some point, side when (get point grid >>= fun v -> v <> plant) |> ifNone false -> sides ||| side
                    | _ -> sides
                ) Side.None
                
            edges
    
    let getSideCount2 (plotFences:(Point2I * Side) list) =
        // ok, now I have information on each SIDE that a fence is one... omg I suck at this.
        // I'll try to do the whole compare horizontally and vertically thing again, but this time
        // within each I'll have to check the top and bottom, or I'll iterate it twice
        
        let getX (pf:Point2I * Side) : int = (fst pf) |> _.X
        let getY (pf:Point2I * Side) : int = (fst pf) |> _.Y
        
        let countContiguous side group sort =
            plotFences
            |> Seq.filter (fun pf -> (snd pf) &&& side = side)
            |> Seq.groupBy group 
            |> Seq.map(fun (key, g) -> g |> Seq.sortBy sort)
            // |> Array.ofSeq // debugging code
            |> Seq.sumBy(fun grouping ->
                let total =
                    grouping
                    |> Seq.splitByComparison (fun last current -> 
    // splitByComparison is an implementation I took from ChatGPT for a function that I commonly use in C# as part of the MoreLinq library that does the same thing. Here, I'm splitting whenever there is a difference. 
    // in hindsight, it probably could have been a window
                        let l = sort last 
                        let r = sort current
                        l <> (r - 1)
                    )
                    |> Seq.length
                total)
        
        let topCounts = countContiguous Side.Top getY getX
        let bottomCounts = countContiguous Side.Bottom getY getX
        let leftCounts = countContiguous Side.Left getX getY
        let rightCounts = countContiguous Side.Right getX getY
        
        topCounts + bottomCounts + leftCounts + rightCounts
    
    
    let part2 input =
        (readGrid2D input (fun _ c -> c))
        |> fun grid ->
            findRegions grid
            // |> List.take 1
            |> List.sumBy(fun region ->
                let (plotCount, edgecount) =
                    let visited = HashSet()
                    let regionCounts =
                        region
                        |> List.map(fun p -> (snd p), findEdges grid p visited)
                        |> getSideCount2
                    (visited.Count, regionCounts)
                
                let area = plotCount
                area * edgecount
                )
    
    
    • CameronDev@programming.devOPM
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      1 month ago

      I saw it :)

      If I understand your approach for pt2, you are getting all the fences and then grouping the connected ones? That definitely seems like a harder approach. I went with the counting corner method, which was also hard, but less iterating required.

      Keep the solutions coming, even as the sub wanes in activity, I still appreciate them :)

      • RagingHungryPanda@lemm.ee
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        1 month ago

        hey thanks!

        I didn’t check any other solutions before finishing (currently wondering way day 13 is too low), but I thought that trying to traverse fences would be a pain and since I have everything separated by regions and not traversing the array, counting corners never came to mind.

        But the thought that I had was that for each region, all points will be a straight line in the V or H orientations, so if I can go up and down and count when last != next - 1, then that’ll tell me that that is a contiguous piece of fence.

        The idea isn’t too hard, for tracking the XAxis it’s

        region.GroupBy(YAxis) // row
        .Select(group => 
            group.Sort(g => g.XAxis) // column
                .Window(a,b => a != b - 1 ? 1 : 0).Sum()
        .Sum()
        

        Except that I used a different splitting method and that came to me later.

  • mykl@lemmy.world
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    2 months ago

    Uiua

    Takes about 3 seconds to solve both parts for live data, caused primarily by my terrible fill function in FieldCoords which repeatedly refills and dedups already discovered cells. I promised myself when I wrote it that I would revisit it, but I really can’t be bothered right now. Sorry Kai.

    LATE EDIT: Thanks to Quant for the inspiration to revisit this. With his code snippet and the realisation that I should normalise all fields to remove wasted space, runtime is now down to 55ms.

    Data ← ⊜∘⊸≠@\n "AAAA\nBBCD\nBBCC\nEEEC"
    N₄     ← [¯1_0 1_0 0_¯1 0_1]               # Four orthogonal neighbours.
    Fences ← /+/+=0≡(⬚0⊡+N₄¤)⊙¤⊚.°⊚            # Fences for a field, by looking for edges.
    Cs     ← [0 1 1 0 1 0 2 1 1 2 0 1 0 1 1 0] # Number of corners keyed by bitarray of 2x2 grid.
    Sides  ← /+♭⬚0⧈(⊡:Cs°⋯♭)2_2°⊚              # Add border, look for corners in 2x2 windows.
    
    # Use `classify` to find fields, then normalise to 0_0.
    Fields ← ≡⍚(-¤⊸/↧)⊜□:⇡△.+1⍜♭⊛Data # Thanks to Quant!
    
    /+×≡◇⊃⧻Fences Fields
    /+×≡◇⊃⧻Sides Fields
    
    • Quant@programming.dev
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      2 months ago

      I found multidimensional markers for partition to work really well for finding the fields: Areas ← ⊜□:⇡△.+1⍜♭⊛ It just groups the other array’s contents according to adjacent markers, horizontally and vertically. Took me quite a bit to figure out what’s actually happening in the example in the documentation ^^’

      • mykl@lemmy.world
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        2 months ago

        Ooh, interesting, I’ll have to give that a try. Thanks!

        (edit) Wow, that replaced my three lines of overly complex code without a hitch. classify is an operator I never really got the point of before. Beautiful.

        Data ← ⊜∘⊸≠@\n "AAAA\nBBCD\nBBCC\nEEEC"
        N₄     ← [¯1_0 1_0 0_¯1 0_1]               # Four orthogonal neighbours.
        Fences ← /+≡(/+=0⬚0⊡+N₄¤)⊙¤⊚.°⊚            # Fences for a field, by looking for edges.
        Cs     ← [0 1 1 0 1 0 2 1 1 2 0 1 0 1 1 0] # Number of corners keyed by bitarray of 2x2 grid.
        Sides  ← /+/+⧈(⊡:Cs°⋯♭)2_2⌝↘¯1_¯1⌝↘1_1°⊚   # Add border, look for corners in 2x2 windows.
        
        Fields ← ⊜□:⇡△.+1⍜♭⊛Data
        
        /+×≡◇⊃⧻Fences Fields
        /+×≡◇⊃⧻Sides Fields
        
          • mykl@lemmy.world
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            2 months ago

            1.8s now. 99% of that in Sides. I’ve just had an idea though… maybe too late for today though!

            edit: prepending ≡⍚(-¤⊸/↧) toFields spared me from manipulating hundreds of irrelevant 0’s, so time is very good now at 55ms.

            • Quant@programming.dev
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              2 months ago

              Damn that’s a lot time saved. I love how unassuming the addition looks for how great an effect it has

              • mykl@lemmy.world
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                2 months ago

                It was a real D’oh! moment when I visualised the data I was generating and saw all the zeros stretching across the page.

  • cabhan@discuss.tchncs.de
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    2 months ago

    Rust

    I essentially used flood fill to collect each region. Part 1 was then relatively easy: for each point, check how many neighbors are outside of the region.

    Part 2 took me forever, and I ended up looking for hints online, where I discovered that an easy way to count the number of sides is to instead count the number of corners. Doing this for “normal” corners (e.g. in a square) was relatively easy, but “reverse corners” took me a long time. Corners like here what we see in the NE corner of the first C in the third row here:

    ....
    ..C.
    ..CC
    ...C
    

    I’m more or less happy with my solution, but my brain is now totally fried.

    https://gitlab.com/bricka/advent-of-code-2024-rust/-/blob/main/src/days/day12.rs?ref_type=heads

    use std::collections::HashSet;
    
    use crate::grid::{Coordinate, Direction, Grid};
    use crate::solver::DaySolver;
    
    fn perimeter_score(c: Coordinate, grid: &MyGrid) -> usize {
        let plant_type = grid[c];
    
        Direction::orthogonal_iter()
            .map(|d| grid.neighbor_in_direction(c, d))
            .map(|c_opt| match c_opt {
                None => 1,
                Some(c) => if grid[c] == plant_type {
                    0
                } else {
                    1
                }
            })
            .sum()
    }
    
    type MyGrid = Grid<char>;
    
    struct Region {
        #[allow(dead_code)]
        plant_type: char,
        coordinates: HashSet<Coordinate>,
    }
    
    impl Region {
        fn new(plant_type: char, coordinates: HashSet<Coordinate>) -> Region {
            Region { plant_type, coordinates }
        }
    
        fn iter(&self) -> impl Iterator<Item = &Coordinate> {
            self.coordinates.iter()
        }
    
        fn part1_score(&self, grid: &MyGrid) -> usize {
            self.coordinates.len() * self.coordinates.iter().map(|c| perimeter_score(*c, grid)).sum::<usize>()
        }
    
        fn part2_score(&self, grid: &MyGrid) -> usize {
            let area = self.coordinates.len();
            let sides = self.number_of_corners(grid);
    
            area * sides
        }
    
        fn number_of_corners(&self, grid: &MyGrid) -> usize {
            self.coordinates.iter().cloned()
                .map(|coordinate| {
                    // How many corners do we have from here?
                    // println!("Checking {}", border_coordinate);
    
                    let corner_count = Direction::diagonal_iter()
                        .filter(|corner_direction| {
                            // Either:
                            // 1) Both neighbor directions are not 100% in the region
                            // 2) Both neighbors are in the region, but the corner itself isn't
    
                            let corner_in_region = match grid.neighbor_in_direction(coordinate, *corner_direction) {
                                None => false,
                                Some(c) => self.coordinates.contains(&c),
                            };
    
                            let both_neighbors_not_in_region = corner_direction.neighbor_directions().iter()
                                .all(|direction| match grid.neighbor_in_direction(coordinate, *direction) {
                                    None => true,
                                    Some(c) => !self.coordinates.contains(&c),
                                });
    
                            let both_neighbors_in_region = corner_direction.neighbor_directions().iter()
                                .all(|direction| match grid.neighbor_in_direction(coordinate, *direction) {
                                    None => false,
                                    Some(c) => self.coordinates.contains(&c),
                                });
    
                            both_neighbors_not_in_region || (both_neighbors_in_region && !corner_in_region)
                        })
                        .count();
                    // println!("Corner count = {}", corner_count);
                    corner_count
                })
                .sum()
        }
    }
    
    fn parse_input(input: String) -> MyGrid {
        input.lines()
            .map(|line| line.chars().collect())
            .collect::<Vec<Vec<char>>>()
            .into()
    }
    
    fn find_region_at(grid: &MyGrid, start: Coordinate) -> Region {
        let plant_type = grid[start];
        let mut coordinates = HashSet::new();
        let mut frontier = vec![start];
    
        while let Some(coordinate) = frontier.pop() {
            if grid[coordinate] == plant_type  && !coordinates.contains(&coordinate) {
                coordinates.insert(coordinate);
                frontier.extend(grid.orthogonal_neighbors_iter(coordinate));
            }
        }
    
        Region::new(plant_type, coordinates)
    }
    
    fn find_regions(grid: &MyGrid) -> Vec<Region> {
        let mut visited_coordinates: HashSet<Coordinate> = HashSet::new();
        let mut regions = vec![];
    
        for coordinate in grid.coordinates_iter() {
            if !visited_coordinates.contains(&coordinate) {
                let region = find_region_at(grid, coordinate);
                visited_coordinates.extend(region.iter().cloned());
                regions.push(region);
            }
        }
    
        regions
    }
    
    pub struct Day12Solver;
    
    impl DaySolver for Day12Solver {
        fn part1(&self, input: String) -> usize {
            let grid = parse_input(input);
            let regions = find_regions(&grid);
    
            regions.into_iter()
                .map(|region| region.part1_score(&grid))
                .sum()
        }
    
        fn part2(&self, input: String) -> usize {
            let grid = parse_input(input);
            let regions = find_regions(&grid);
    
            regions.into_iter()
                .map(|region| region.part2_score(&grid))
                .sum()
        }
    }
    
    • Quant@programming.dev
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      2 months ago

      Counting the number of corners was a very useful hint for part 2. I had the most trouble with detecting the double corners, i.e. like in the example where the two B fields touch diagonally:

      AAAAAA
      AAABBA
      AAABBA
      ABBAAA
      ABBAAA
      AAAAAA
      

      Still, I would’ve taken a lot longer and probably made really-bad-performance-code without reading this :D