Day 10: Hoof It

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FAQ

  • Quant@programming.dev
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    3 days ago

    Uiua

    After finally deciding to put aside Day 9 Part 2 for now, this was really easy actually. The longest was figuring out how many extra dimensions I had to give some arrays and where to remove those again (and how). Then part 2 came along and all I had to do was remove a single character (not removing duplicates when landing on the same field by going different ways from the same starting point). Basically, everything in the parentheses of the Trails! macro was my solution for part 1, just that the ^0 was (deduplicate). Once that was removed, the solution for part 2 was there as well.

    Run with example input here

    Note: in order to use the code here for the actual input, you have to replace =₈ with =₅₀ because I was too lazy to make it work with variable array sizes this time.

    $ 89010123
    $ 78121874
    $ 87430965
    $ 96549874
    $ 45678903
    $ 32019012
    $ 01329801
    $ 10456732
    .
    Adj ← ¤[0_¯1 0_1 ¯1_0 1_0]
    
    Trails! ← (
      ⊚=0.
      ⊙¤
      ≡(□¤)
      1
      ⍥(⊙(≡(□^0/⊂≡(+¤)⊙¤°□)⊙Adj
          ≡(□▽¬≡/++⊃=₋₁=₈.°□))
        +1⟜⊸⍚(▽=⊙(:⟜⊡))
      )9
      ⊙◌◌
      ⧻/◇⊂
    )
    
    PartOne ← (
      # &rs ∞ &fo "input-10.txt"
      ⊜∵⋕≠@\n.
      Trails!◴
    )
    
    PartTwo ← (
      # &rs ∞ &fo "input-10.txt"
      ⊜∵⋕≠@\n.
      Trails!∘
    )
    
    &p "Day 10:"
    &pf "Part 1: "
    &p PartOne
    &pf "Part 2: "
    &p PartTwo
    
  • sjmulder@lemmy.sdf.org
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    9 days ago

    C

    Tried a dynamic programming kind of thing first but recursion suited the problem much better.

    Part 2 seemed incompatible with my visited list representation. Then at the office I suddenly realised I just had to skip a single if(). Funny how that works when you let things brew in the back of your mind.

    Code
    #include "common.h"
    
    #define GZ 43
    
    /*
     * To avoid having to clear the 'seen' array after every search we mark
     * and check it with a per-search marker value ('id').
     */
    static char g[GZ][GZ];
    static int seen[GZ][GZ];
    
    static int
    score(int id, int x, int y, int p2)
    {
    	if (x<0 || x>=GZ ||
    	    y<0 || y>=GZ || (!p2 && seen[y][x] == id))
    		return 0;
    
    	seen[y][x] = id;
    
    	if (g[y][x] == '9')
    		return 1;
    
    	return
    	    (g[y-1][x] == g[y][x]+1 ? score(id, x, y-1, p2) : 0) +
    	    (g[y+1][x] == g[y][x]+1 ? score(id, x, y+1, p2) : 0) +
    	    (g[y][x-1] == g[y][x]+1 ? score(id, x-1, y, p2) : 0) +
    	    (g[y][x+1] == g[y][x]+1 ? score(id, x+1, y, p2) : 0);
    }
    
    int
    main(int argc, char **argv)
    {
    	int p1=0,p2=0, id=1, x,y;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    	for (y=0; y<GZ && fgets(g[y], GZ, stdin); y++)
    		;
    
    	for (y=0; y<GZ; y++)
    	for (x=0; x<GZ; x++)
    		if (g[y][x] == '0') {
    			p1 += score(id++, x, y, 0);
    			p2 += score(id++, x, y, 1);
    		}
    
    	printf("10: %d %d\n", p1, p2);
    	return 0;
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day10.c

  • lwhjp@lemmy.sdf.org
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    10 days ago

    Haskell

    A nice easy one today: didn’t even have to hit this with the optimization hammer.

    import Data.Char
    import Data.List
    import Data.Map (Map)
    import Data.Map qualified as Map
    
    readInput :: String -> Map (Int, Int) Int
    readInput s =
      Map.fromList
        [ ((i, j), digitToInt c)
          | (i, l) <- zip [0 ..] (lines s),
            (j, c) <- zip [0 ..] l
        ]
    
    findTrails :: Map (Int, Int) Int -> [[[(Int, Int)]]]
    findTrails input =
      Map.elems . Map.map (filter ((== 10) . length)) $
        Map.restrictKeys accessible starts
      where
        starts = Map.keysSet . Map.filter (== 0) $ input
        accessible = Map.mapWithKey getAccessible input
        getAccessible (i, j) h
          | h == 9 = [[(i, j)]]
          | otherwise =
              [ (i, j) : path
                | (di, dj) <- [(-1, 0), (0, 1), (1, 0), (0, -1)],
                  let p = (i + di, j + dj),
                  input Map.!? p == Just (succ h),
                  path <- accessible Map.! p
              ]
    
    main = do
      trails <- findTrails . readInput <$> readFile "input10"
      mapM_
        (print . sum . (`map` trails))
        [length . nub . map last, length]
    
  • iAvicenna@lemmy.world
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    9 days ago

    Python

    Not surprisingly, trees

    import numpy as np
    from pathlib import Path
    
    cwd = Path(__file__).parent
    
    cross = np.array([[-1,0],[1,0],[0,-1],[0,1]])
    
    class Node():
      def __init__(self, coord, parent):
        self.coord = coord
        self.parent = parent
    
      def __repr__(self):
        return f"{self.coord}"
    
    def parse_input(file_path):
    
      with file_path.open("r") as fp:
        data = list(map(list, fp.read().splitlines()))
    
      return np.array(data, dtype=int)
    
    def find_neighbours(node_pos, grid):
    
      I = list(filter(lambda x: all([c>=0 and o-c>0 for c,o in zip(x,grid.shape)]),
                      list(cross + node_pos)))
    
      candidates = grid[tuple(np.array(I).T)]
      J = np.argwhere(candidates-grid[tuple(node_pos)]==1).flatten()
    
      return list(np.array(I).T[:, J].T)
    
    def construct_tree_paths(grid):
    
      roots = list(np.argwhere(grid==0))
      trees = []
    
      for root in roots:
    
        levels = [[Node(root, None)]]
        while len(levels[-1])>0 or len(levels)==1:
          levels.append([Node(node, root) for root in levels[-1] for node in
                         find_neighbours(root.coord, grid)])
        trees.append(levels)
    
      return trees
    
    def trace_back(trees, grid):
    
      paths = []
    
      for levels in trees:
        for node in levels[-2]:
    
          path = ""
          while node is not None:
            coord = ",".join(node.coord.astype(str))
            path += f"{coord} "
            node = node.parent
          paths.append(path)
    
      return paths
    
    def solve_problem(file_name):
    
      grid = parse_input(Path(cwd, file_name))
      trees = construct_tree_paths(grid)
      trails = trace_back(trees, grid)
      ntrails = len(set(trails))
      nreached = sum([len(set([tuple(x.coord) for x in levels[-2]])) for levels in trees])
    
      return nreached, ntrails
    
    • Acters@lemmy.world
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      8 days ago

      yay trees! my solution was really fast too! 😀

      edit: you can find it here, or look at my lemmy post

      should take only 1.5 milliseconds!

  • CameronDev@programming.devOPM
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    9 days ago

    Rust

    Definitely a nice and easy one, I accidentally solved part 2 first, because I skimmed the challenge and missed the unique part.

    #[cfg(test)]
    mod tests {
    
        const DIR_ORDER: [(i8, i8); 4] = [(-1, 0), (0, 1), (1, 0), (0, -1)];
    
        fn walk_trail(board: &Vec<Vec<i8>>, level: i8, i: i8, j: i8) -> Vec<(i8, i8)> {
            let mut paths = vec![];
            if i < 0 || j < 0 {
                return paths;
            }
            let actual_level = match board.get(i as usize) {
                None => return paths,
                Some(line) => match line.get(j as usize) {
                    None => return paths,
                    Some(c) => c,
                },
            };
            if *actual_level != level {
                return paths;
            }
            if *actual_level == 9 {
                return vec![(i, j)];
            }
    
            for dir in DIR_ORDER.iter() {
                paths.extend(walk_trail(board, level + 1, i + dir.0, j + dir.1));
            }
            paths
        }
    
        fn count_unique(p0: &Vec<(i8, i8)>) -> u32 {
            let mut dedup = vec![];
            for p in p0.iter() {
                if !dedup.contains(p) {
                    dedup.push(*p);
                }
            }
            dedup.len() as u32
        }
    
        #[test]
        fn day10_part1_test() {
            let input = std::fs::read_to_string("src/input/day_10.txt").unwrap();
    
            let board = input
                .trim()
                .split('\n')
                .map(|line| {
                    line.chars()
                        .map(|c| {
                            if c == '.' {
                                -1
                            } else {
                                c.to_digit(10).unwrap() as i8
                            }
                        })
                        .collect::<Vec<i8>>()
                })
                .collect::<Vec<Vec<i8>>>();
    
            let mut total = 0;
    
            for (i, row) in board.iter().enumerate() {
                for (j, pos) in row.iter().enumerate() {
                    if *pos == 0 {
                        let all_trails = walk_trail(&board, 0, i as i8, j as i8);
                        total += count_unique(&all_trails);
                    }
                }
            }
    
            println!("{}", total);
        }
        #[test]
        fn day10_part2_test() {
            let input = std::fs::read_to_string("src/input/day_10.txt").unwrap();
    
            let board = input
                .trim()
                .split('\n')
                .map(|line| {
                    line.chars()
                        .map(|c| {
                            if c == '.' {
                                -1
                            } else {
                                c.to_digit(10).unwrap() as i8
                            }
                        })
                        .collect::<Vec<i8>>()
                })
                .collect::<Vec<Vec<i8>>>();
    
            let mut total = 0;
    
            for (i, row) in board.iter().enumerate() {
                for (j, pos) in row.iter().enumerate() {
                    if *pos == 0 {
                        total += walk_trail(&board, 0, i as i8, j as i8).len();
                    }
                }
            }
    
            println!("{}", total);
        }
    }
    
  • mykl@lemmy.world
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    4 days ago

    Uiua

    Run it here!

    How to read this

    Uiua has a very helpful path function built in which returns all valid paths that match your criteria (using diijkstra/a* depending on whether third function is provided), making a lot of path-finding stuff almost painfully simple, as you just need to provide a starting node and three functions: return next nodes, return confirmation if we’ve reached a suitable target node (here testing if it’s = 9), (optional) return heuristic cost to destination (here set to constant 1), .

    Data    ⊜≡⋕⊸≠@\n"89010123\n78121874\n87430965\n96549874\n45678903\n32019012\n01329801\n10456732"
    N₄      ≡+[0_1 1_0 0_¯1 ¯1_0]¤
    Ns      :(=1-:(0:Data))▽⊸≡(/×≥0)N₄. # Valid, in-bounds neighbours.
    Count!  /+≡(⧻^0⊙◌ path(Ns|(=9:Data)|1))⊚=0Data
    &p Count!(◴≡◇⊣)
    &p Count!
    
  • janAkali@lemmy.one
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    9 days ago

    Nim

    As many others today, I’ve solved part 2 first and then fixed a ‘bug’ to solve part 1. =)

    type Vec2 = tuple[x,y:int]
    const Adjacent = [(x:1,y:0),(-1,0),(0,1),(0,-1)]
    
    proc path(start: Vec2, grid: seq[string]): tuple[ends, trails: int] =
      var queue = @[@[start]]
      var endNodes: HashSet[Vec2]
      while queue.len > 0:
        let path = queue.pop()
        let head = path[^1]
        let c = grid[head.y][head.x]
    
        if c == '9':
          inc result.trails
          endNodes.incl head
          continue
    
        for d in Adjacent:
          let nd = (x:head.x + d.x, y:head.y + d.y)
          if nd.x < 0 or nd.y < 0 or nd.x > grid[0].high or nd.y > grid.high:
            continue
          if grid[nd.y][nd.x].ord - c.ord != 1: continue
          queue.add path & nd
      result.ends = endNodes.len
    
    proc solve(input: string): AOCSolution[int, int] =
      let grid = input.splitLines()
      var trailstarts: seq[Vec2]
    
      for y, line in grid:
        for x, c in line:
          if c == '0':
            trailstarts.add (x,y)
    
      for start in trailstarts:
        let (ends, trails) = start.path(grid)
        result.part1 += ends
        result.part2 += trails
    

    Codeberg Repo

  • ystael@beehaw.org
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    9 days ago

    J

    Who needs recursion or search algorithms? Over here in line noise array hell, we have built-in sparse matrices! :)

    data_file_name =: '10.data'
    grid =: "."0 ,. > cutopen fread data_file_name
    data =: , grid
    'rsize csize' =: $ grid
    inbounds =: monad : '(*/ y >: 0 0) * (*/ y &lt; rsize, csize)'
    coords =: ($ grid) &amp; #:
    uncoords =: ($ grid) &amp; #.
    NB. if n is the linear index of a point, neighbors n lists the linear indices
    NB. of its orthogonally adjacent points
    neighbors =: monad : 'uncoords (#~ inbounds"1) (coords y) +"1 (4 2 $ 1 0 0 1 _1 0 0 _1)'
    uphill1 =: dyad : '1 = (y { data) - (x { data)'
    uphill_neighbors =: monad : 'y ,. (#~ (y &amp; uphill1)) neighbors y'
    adjacency_of =: monad define
       edges =. ; (&lt; @: uphill_neighbors"0) i.#y
       NB. must explicitly specify fill of integer 0, default is float
       1 edges} 1 $. ((#y), #y); (0 1); 0
    )
    adjacency =: adjacency_of data
    NB. maximum path length is 9 so take 9th power of adjacency matrix
    leads_to_matrix =: adjacency (+/ . *)^:8 adjacency
    leads_to =: dyad : '({ &amp; leads_to_matrix) @: &lt; x, y'
    trailheads =: I. data = 0
    summits =: I. data = 9
    scores =: trailheads leads_to"0/ summits
    result1 =: +/, 0 &lt; scores
    result2 =: +/, scores
    
      • ystael@beehaw.org
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        9 days ago

        Yes. I don’t know whether this is a beehaw specific issue (that being my home instance) or a lemmy issue in general, but < and & are HTML escaped in all code blocks I see. Of course, this is substantially more painful for J code than many other languages.

  • Karmmah@lemmy.world
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    9 days ago

    Julia

    Quite happy that today went a lot smoother than yesterday even though I am not really familiar with recursion. Normally I never use recursion but I felt like today could be solved by it (or using trees, but I’m even less familiar with them). Surprisingly my solution actually worked and for part 2 only small modifications were needed to count peaks reached by each trail.

    Code
    function readInput(inputFile::String)
    	f = open(inputFile,"r")
    	lines::Vector{String} = readlines(f)
    	close(f)
    	topoMap = Matrix{Int}(undef,length(lines),length(lines[1]))
    	for (i,l) in enumerate(lines)
    		topoMap[i,:] = map(x->parse(Int,x),collect(l))
    	end
    	return topoMap
    end
    
    function getTrailheads(topoMap::Matrix{Int})
    	trailheads::Vector{Vector{Int}} = []
    	for (i,r) in enumerate(eachrow(topoMap))
    		for (j,c) in enumerate(r)
    			c==0 ? push!(trailheads,[i,j]) : nothing
    		end
    	end
    	return trailheads
    end
    
    function getReachablePeaks(topoMap::Matrix{Int},trailheads::Vector{Vector{Int}})
    	reachablePeaks = Dict{Int,Vector{Vector{Int}}}()
    	function getPossibleMoves(topoMap::Matrix{Int},pos::Vector{Int})
    		possibleMoves::Vector{Vector{Int}} = []
    		pos[1]-1 in 1:size(topoMap)[1] && topoMap[pos[1]-1,pos[2]]==topoMap[pos[1],pos[2]]+1 ? push!(possibleMoves,[pos[1]-1,pos[2]]) : nothing #up?
    		pos[1]+1 in 1:size(topoMap)[1] && topoMap[pos[1]+1,pos[2]]==topoMap[pos[1],pos[2]]+1 ? push!(possibleMoves,[pos[1]+1,pos[2]]) : nothing #down?
    		pos[2]-1 in 1:size(topoMap)[2] && topoMap[pos[1],pos[2]-1]==topoMap[pos[1],pos[2]]+1 ? push!(possibleMoves,[pos[1],pos[2]-1]) : nothing #left?
    		pos[2]+1 in 1:size(topoMap)[2] && topoMap[pos[1],pos[2]+1]==topoMap[pos[1],pos[2]]+1 ? push!(possibleMoves,[pos[1],pos[2]+1]) : nothing #right?
    		return possibleMoves
    	end
    	function walkPossMoves(topoMap::Matrix{Int},pos::Vector{Int},reachedPeaks::Matrix{Bool},trailId::Int)
    		possMoves::Vector{Vector{Int}} = getPossibleMoves(topoMap,pos)
    		for m in possMoves
    			if topoMap[m[1],m[2]]==9
    				reachedPeaks[m[1],m[2]]=1
    				trailId += 1
    				continue
    			end
    			reachedPeaks,trailId = walkPossMoves(topoMap,m,reachedPeaks,trailId)
    		end
    		return reachedPeaks, trailId
    	end
    	peaksScore::Int = 0; trailsScore::Int = 0
    	trailId::Int = 0
    	for (i,t) in enumerate(trailheads)
    		if !haskey(reachablePeaks,i); reachablePeaks[i]=[]; end
    		reachedPeaks::Matrix{Bool} = zeros(size(topoMap))
    		trailId = 0
    		reachedPeaks,trailId = walkPossMoves(topoMap,t,reachedPeaks,trailId)
    		trailPeaksScore = sum(reachedPeaks)
    		peaksScore += trailPeaksScore
    		trailsScore += trailId
    	end
    	return peaksScore,trailsScore
    end #getReachablePeaks
    
    topoMap::Matrix{Int} = readInput("input/day10Input")
    trailheads::Vector{Vector{Int}} = getTrailheads(topoMap)
    @info "Part 1"
    reachablePeaks = getReachablePeaks(topoMap,trailheads)[1]
    println("reachable peaks: ",reachablePeaks)
    @info "Part 2"
    trailsScore::Int = getReachablePeaks(topoMap,trailheads)[2]
    println("trails score: $trailsScore")
    
  • LeixB@lemmy.world
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    9 days ago

    Haskell

    import Control.Arrow
    import Control.Monad.Reader
    import Data.Array.Unboxed
    import Data.List
    
    type Pos = (Int, Int)
    type Board = UArray Pos Char
    type Prob = Reader Board
    
    parse :: String -> Board
    parse s = listArray ((1, 1), (n, m)) $ concat l
      where
        l = lines s
        n = length l
        m = length $ head l
    
    origins :: Prob [Pos]
    origins =
        ask >>= \board ->
            return $ fmap fst . filter ((== '0') . snd) $ assocs board
    
    moves :: Pos -> Prob [Pos]
    moves pos =
        ask >>= \board ->
            let curr = board ! pos
             in return . filter ((== succ curr) . (board !)) . filter (inRange (bounds board)) $ fmap (.+. pos) deltas
      where
        deltas = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        (ax, ay) .+. (bx, by) = (ax + bx, ay + by)
    
    solve :: [Pos] -> Prob [Pos]
    solve p = do
        board <- ask
        nxt <- concat <$> mapM moves p
    
        let (nines, rest) = partition ((== '9') . (board !)) nxt
    
        fmap (++ nines) $ if null rest then return [] else solve rest
    
    scoreTrail = fmap (length . nub) . solve . pure
    scoreTrail' = fmap length . solve . pure
    
    part1 = sum . runReader (origins >>= mapM scoreTrail)
    part2 = sum . runReader (origins >>= mapM scoreTrail')
    
    main = getContents >>= print . (part1 &&& part2) . parse
    
  • VegOwOtenks@lemmy.world
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    9 days ago

    Haskell

    Cool task, nothing to optimize

    import Control.Arrow
    
    import Data.Array.Unboxed (UArray)
    import Data.Set (Set)
    
    import qualified Data.Char as Char
    import qualified Data.List as List
    import qualified Data.Set as Set
    import qualified Data.Array.Unboxed as UArray
    
    parse :: String -> UArray (Int, Int) Int
    parse s = UArray.listArray ((1, 1), (n, m)) . map Char.digitToInt . filter (/= '\n') $ s
            where
                    n = takeWhile (/= '\n') >>> length $ s
                    m = filter (== '\n') >>> length >>> pred $ s
    
    reachableNeighbors :: (Int, Int) -> UArray (Int, Int) Int -> [(Int, Int)]
    reachableNeighbors p@(py, px) a = List.filter (UArray.inRange (UArray.bounds a))
            >>> List.filter ((a UArray.!) >>> pred >>> (== (a UArray.! p)))
            $ [(py-1, px), (py+1, px), (py, px-1), (py, px+1)]
    
    distinctTrails :: (Int, Int) -> UArray (Int, Int) Int -> Int
    distinctTrails p a
            | a UArray.! p == 9 = 1
            | otherwise = flip reachableNeighbors a
                    >>> List.map (flip distinctTrails a)
                    >>> sum
                    $ p
    
    reachableNines :: (Int, Int) -> UArray (Int, Int) Int -> Set (Int, Int)
    reachableNines p a
            | a UArray.! p == 9 = Set.singleton p
            | otherwise = flip reachableNeighbors a
                    >>> List.map (flip reachableNines a)
                    >>> Set.unions
                    $ p
    
    findZeros = UArray.assocs
            >>> filter (snd >>> (== 0))
            >>> map fst
    
    part1 a = findZeros
            >>> map (flip reachableNines a)
            >>> map Set.size
            >>> sum
            $ a
    part2 a = findZeros
            >>> map (flip distinctTrails a)
            >>> sum
            $ a
    
    main = getContents
            >>= print
            . (part1 &&& part2)
            . parse
    
  • Gobbel2000@programming.dev
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    9 days ago

    Rust

    This was a nice one. Basically 9 rounds of Breadth-First-Search, which could be neatly expressed using fold. The only difference between part 1 and part 2 turned out to be the datastructure for the search frontier: The HashSet in part 1 unifies paths as they join back to the same node, the Vec in part 2 keeps all paths separate.

    Solution
    use std::collections::HashSet;
    
    fn parse(input: &str) -> Vec<&[u8]> {
        input.lines().map(|l| l.as_bytes()).collect()
    }
    
    fn adj(grid: &[&[u8]], (x, y): (usize, usize)) -> Vec<(usize, usize)> {
        let n = grid[y][x];
        let mut adj = Vec::with_capacity(4);
        if x > 0 && grid[y][x - 1] == n + 1 {
            adj.push((x - 1, y))
        }
        if y > 0 && grid[y - 1][x] == n + 1 {
            adj.push((x, y - 1))
        }
        if x + 1 < grid[0].len() && grid[y][x + 1] == n + 1 {
            adj.push((x + 1, y))
        }
        if y + 1 < grid.len() && grid[y + 1][x] == n + 1 {
            adj.push((x, y + 1))
        }
        adj
    }
    
    fn solve(input: String, trailhead: fn(&[&[u8]], (usize, usize)) -> u32) -> u32 {
        let grid = parse(&input);
        let mut sum = 0;
        for (y, row) in grid.iter().enumerate() {
            for (x, p) in row.iter().enumerate() {
                if *p == b'0' {
                    sum += trailhead(&grid, (x, y));
                }
            }
        }
        sum
    }
    
    fn part1(input: String) {
        fn score(grid: &[&[u8]], start: (usize, usize)) -> u32 {
            (1..=9)
                .fold(HashSet::from([start]), |frontier, _| {
                    frontier.iter().flat_map(|p| adj(grid, *p)).collect()
                })
                .len() as u32
        }
        println!("{}", solve(input, score))
    }
    
    fn part2(input: String) {
        fn rating(grid: &[&[u8]], start: (usize, usize)) -> u32 {
            (1..=9)
                .fold(vec![start], |frontier, _| {
                    frontier.iter().flat_map(|p| adj(grid, *p)).collect()
                })
                .len() as u32
        }
        println!("{}", solve(input, rating))
    }
    
    util::aoc_main!();
    

    Also on github

  • vole@lemmy.world
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    9 days ago

    Raku

    Pretty straight-forward problem today.

    sub MAIN($input) {
        my $file = open $input;
        my @map = $file.slurp.trim.lines>>.comb>>.Int;
    
        my @pos-tracking = [] xx 10;
        for 0..^@map.elems X 0..^@map[0].elems -> ($row, $col) {
            @pos-tracking[@map[$row][$col]].push(($row, $col).List);
        }
    
        my %on-possible-trail is default([]);
        my %trail-score-part2 is default(0);
        for 0..^@pos-tracking.elems -> $height {
            for @pos-tracking[$height].List -> ($row, $col) {
                if $height == 0 {
                    %on-possible-trail{"$row;$col"} = set ("$row;$col",);
                    %trail-score-part2{"$row;$col"} = 1;
                } else {
                    for ((1,0), (-1, 0), (0, 1), (0, -1)) -> @neighbor-direction {
                        my @neighbor-position = ($row, $col) Z+ @neighbor-direction;
                        next if @neighbor-position.any < 0 or (@neighbor-position Z>= (@map.elems, @map[0].elems)).any;
                        next if @map[@neighbor-position[0]][@neighbor-position[1]] != $height - 1;
                        %on-possible-trail{"$row;$col"} ∪= %on-possible-trail{"{@neighbor-position[0]};{@neighbor-position[1]}"};
                        %trail-score-part2{"$row;$col"} += %trail-score-part2{"{@neighbor-position[0]};{@neighbor-position[1]}"};
                    }
                }
            }
        }
    
        my $part1-solution = @pos-tracking[9].map({%on-possible-trail{"{$_[0]};{$_[1]}"}.elems}).sum;
        say "part 1: $part1-solution";
    
        my $part2-solution = @pos-tracking[9].map({%trail-score-part2{"{$_[0]};{$_[1]}"}}).sum;
        say "part 2: $part2-solution";
    }
    
  • Ananace@lemmy.ananace.dev
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    9 days ago

    Nice to have a really simple one for a change, both my day 1 and 2 solutions worked on their very first attempts.
    I rewrote the code to combine the two though, since the implementations were almost identical for both solutions, and also to replace the recursion with a search list instead.

    C#
    int[] heights = new int[0];
    (int, int) size = (0, 0);
    
    public void Input(IEnumerable<string> lines)
    {
      size = (lines.First().Length, lines.Count());
      heights = string.Concat(lines).Select(c => int.Parse(c.ToString())).ToArray();
    }
    
    int trails = 0, trailheads = 0;
    public void PreCalc()
    {
      for (int y = 0; y < size.Item2; ++y)
        for (int x = 0; x < size.Item1; ++x)
          if (heights[y * size.Item1 + x] == 0)
          {
            var unique = new HashSet<(int, int)>();
            trails += CountTrails((x, y), unique);
            trailheads += unique.Count;
          }
    }
    
    public void Part1()
    {
      Console.WriteLine($"Trailheads: {trailheads}");
    }
    public void Part2()
    {
      Console.WriteLine($"Trails: {trails}");
    }
    
    int CountTrails((int, int) from, HashSet<(int,int)> unique)
    {
      int found = 0;
    
      List<(int,int)> toSearch = new List<(int, int)>();
      toSearch.Add(from);
    
      while (toSearch.Any())
      {
        var cur = toSearch.First();
        toSearch.RemoveAt(0);
    
        int height = heights[cur.Item2 * size.Item1 + cur.Item1];
        for (int y = -1; y <= 1; ++y)
          for (int x = -1; x <= 1; ++x)
          {
            if ((y != 0 && x != 0) || (y == 0 && x == 0))
              continue;
    
            var newAt = (cur.Item1 + x, cur.Item2 + y);
            if (newAt.Item1 < 0 || newAt.Item1 >= size.Item1 || newAt.Item2 < 0 || newAt.Item2 >= size.Item2)
              continue;
    
            int newHeight = heights[newAt.Item2 * size.Item1 + newAt.Item1];
            if (newHeight - height != 1)
              continue;
    
            if (newHeight == 9)
            {
              unique.Add(newAt);
              found++;
              continue;
            }
    
            toSearch.Add(newAt);
          }
      }
    
      return found;
    }
    
  • SteveDinn@lemmy.ca
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    edit-2
    9 days ago

    C#

    using System.Diagnostics;
    using Common;
    
    namespace Day10;
    
    static class Program
    {
        static void Main()
        {
            var start = Stopwatch.GetTimestamp();
    
            var sampleInput = Input.ParseInput("sample.txt");
            var programInput = Input.ParseInput("input.txt");
    
            Console.WriteLine($"Part 1 sample: {Part1(sampleInput)}");
            Console.WriteLine($"Part 1 input: {Part1(programInput)}");
    
            Console.WriteLine($"Part 2 sample: {Part2(sampleInput)}");
            Console.WriteLine($"Part 2 input: {Part2(programInput)}");
    
            Console.WriteLine($"That took about {Stopwatch.GetElapsedTime(start)}");
        }
    
        static object Part1(Input i) => GetTrailheads(i)
            .Sum(th => CountTheNines(th, i, new HashSet<Point>(), false));
    
        static object Part2(Input i) => GetTrailheads(i)
            .Sum(th => CountTheNines(th, i, new HashSet<Point>(), true));
    
        static int CountTheNines(Point loc, Input i, ISet<Point> visited, bool allPaths)
        {
            if (!visited.Add(loc)) return 0;
            
            var result =
                (ElevationAt(loc, i) == 9) ? 1 :
                loc.GetCardinalMoves()
                    .Where(move => move.IsInBounds(i.Bounds.Row, i.Bounds.Col))
                    .Where(move => (ElevationAt(move, i) - ElevationAt(loc, i)) == 1)
                    .Where(move => !visited.Contains(move))
                    .Sum(move => CountTheNines(move, i, visited, allPaths));
            
            if(allPaths) visited.Remove(loc);
            
            return result;
        }
    
        static IEnumerable<Point> GetTrailheads(Input i) => Grid.EnumerateAllPoints(i.Bounds)
            .Where(loc => ElevationAt(loc, i) == 0);
    
        static int ElevationAt(Point p, Input i) => i.Map[p.Row][p.Col];
    }
    
    public class Input
    {
        public required Point Bounds { get; init; }
        public required int[][] Map { get; init; }
        
        public static Input ParseInput(string file)
        {
            using var reader = new StreamReader(file);
            var map = reader.EnumerateLines()
                .Select(l => l.Select(c => (int)(c - '0')).ToArray())
                .ToArray();
            var bounds = new Point(map.Length, map.Max(l => l.Length));
            return new Input()
            {
                Map = map,
                Bounds = bounds,
            };
        }
    }
    
    • SteveDinn@lemmy.ca
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      9 days ago

      Straightforward depth first search. I found that the only difference for part 2 was to remove the current location from the HashSet of visited locations when the recurive call finished so that it could be visited again in other unique paths.