this post was submitted on 16 Dec 2024
8 points (83.3% liked)

Advent Of Code

920 readers
58 users here now

An unofficial home for the advent of code community on programming.dev!

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.

AoC 2024

Solution Threads

M T W T F S S
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 18 20 21 22
23 24 25

Rules/Guidelines

Relevant Communities

Relevant Links

Credits

Icon base by Lorc under CC BY 3.0 with modifications to add a gradient

console.log('Hello World')

founded 1 year ago
MODERATORS
 

Day 16: Reindeer Maze

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

top 24 comments
sorted by: hot top controversial new old
[–] [email protected] 1 points 1 day ago

Rust

Not sure if I should dump my full solution, its quite long. If its too long I'll delete it. Way over-engineered, and performs like it as well, quite slow.

Quite proud of my hack for pt2. I walk back along the path, which is nothing special. But because of the turn costs, whenever a turn joins a straight, it makes the straight discontinuous:

 ###### 11043 ######
 10041  10042 ######
 ###### 11041 ######

So I check the before and after cells, and make sure the previous is already marked as a short path, and check the after cell, to make sure its 2 steps apart, and ignore the middle. Dunno if anyone else has done the same thing, I've mostly managed to avoid spoilers today.

code

#[cfg(test)]
mod tests {
    use crate::day_16::tests::State::{CELL, END, SHORTPATH, START, WALL};
    use std::cmp::PartialEq;

    fn get_cell(board: &[Vec<MazeCell>], row: isize, col: isize) -> &MazeCell {
        &board[row as usize][col as usize]
    }

    fn set_cell(board: &mut [Vec<MazeCell>], value: &MazeStep) {
        let cell = &mut board[value.i as usize][value.j as usize];
        cell.dir = value.dir;
        cell.cost = value.cost;
        cell.state = value.state.clone();
    }

    fn find_cell(board: &mut [Vec<MazeCell>], state: State) -> (isize, isize) {
        for i in 0..board.len() {
            for j in 0..board[i].len() {
                if get_cell(board, i as isize, j as isize).state == state {
                    return (i as isize, j as isize);
                }
            }
        }
        unreachable!();
    }

    static DIRECTIONS: [(isize, isize); 4] = [(0, 1), (1, 0), (0, -1), (-1, 0)];

    #[derive(PartialEq, Debug, Clone)]
    enum State {
        CELL,
        WALL,
        START,
        END,
        SHORTPATH,
    }

    struct MazeCell {
        dir: i8,
        cost: isize,
        state: State,
    }

    struct MazeStep {
        i: isize,
        j: isize,
        dir: i8,
        cost: isize,
        state: State,
    }

    fn walk_maze(board: &mut [Vec<MazeCell>]) -> isize {
        let start = find_cell(board, START);
        let mut moves = vec![MazeStep {
            i: start.0,
            j: start.1,
            cost: 0,
            dir: 0,
            state: START,
        }];

        let mut best = isize::MAX;

        loop {
            if moves.is_empty() {
                break;
            }
            let cell = moves.pop().unwrap();
            let current_cost = get_cell(board, cell.i, cell.j);
            if current_cost.state == END {
                if cell.cost < best {
                    best = cell.cost;
                }
                continue;
            }
            if current_cost.state == WALL {
                continue;
            }
            if current_cost.cost < cell.cost {
                continue;
            }
            set_cell(board, &cell);

            for (i, dir) in DIRECTIONS.iter().enumerate() {
                let cost = match (i as i8) - cell.dir {
                    0 => cell.cost + 1,
                    -2 | 2 => continue,
                    _ => cell.cost + 1001,
                };
                moves.push(MazeStep {
                    i: cell.i + dir.0,
                    j: cell.j + dir.1,
                    dir: i as i8,
                    cost,
                    state: State::CELL,
                });
            }
        }

        best
    }

    fn unwalk_path(board: &mut [Vec<MazeCell>], total_cost: isize) -> usize {
        let end = find_cell(board, END);

        let mut cells = vec![MazeStep {
            i: end.0,
            j: end.1,
            dir: 0,
            cost: total_cost,
            state: State::END,
        }];

        set_cell(board, &cells[0]);

        while let Some(mut cell) = cells.pop() {
            for dir in DIRECTIONS {
                let next_cell = get_cell(board, cell.i + dir.0, cell.j + dir.1);
                if next_cell.cost == 0 {
                    continue;
                }
                if next_cell.state == WALL {
                    continue;
                }
                if next_cell.state == CELL
                    && (next_cell.cost == &cell.cost - 1001 || next_cell.cost == &cell.cost - 1)
                {
                    cells.push(MazeStep {
                        i: cell.i + dir.0,
                        j: cell.j + dir.1,
                        dir: 0,
                        cost: next_cell.cost,
                        state: CELL,
                    });
                } else {
                    let prev_cell = get_cell(board, cell.i - dir.0, cell.j - dir.1);
                    if prev_cell.state == SHORTPATH && prev_cell.cost - 2 == next_cell.cost {
                        cells.push(MazeStep {
                            i: cell.i + dir.0,
                            j: cell.j + dir.1,
                            dir: 0,
                            cost: next_cell.cost,
                            state: CELL,
                        });
                    }
                }
            }
            cell.state = SHORTPATH;
            set_cell(board, &cell);
        }
        let mut count = 0;
        for row in board {
            for cell in row {
                if cell.state == SHORTPATH {
                    count += 1;
                }
                if cell.state == END {
                    count += 1;
                }
                if cell.state == START {
                    count += 1;
                }
            }
        }
        count
    }

    #[test]
    fn day15_part2_test() {
        let input = std::fs::read_to_string("src/input/day_16.txt").unwrap();

        let mut board = input
            .split('\n')
            .map(|line| {
                line.chars()
                    .map(|c| match c {
                        '#' => MazeCell {
                            dir: 0,
                            cost: isize::MAX,
                            state: WALL,
                        },
                        'S' => MazeCell {
                            dir: 0,
                            cost: isize::MAX,
                            state: START,
                        },
                        'E' => MazeCell {
                            dir: 0,
                            cost: isize::MAX,
                            state: END,
                        },
                        '.' => MazeCell {
                            dir: 0,
                            cost: isize::MAX,
                            state: CELL,
                        },
                        _ => unreachable!(),
                    })
                    .collect::<Vec<MazeCell>>()
            })
            .collect::<Vec<Vec<MazeCell>>>();

        let cost = walk_maze(&mut board);

        let count = unwalk_path(&mut board, cost);

        println!("{count}");
    }
}

[–] [email protected] 4 points 1 day ago (1 children)

C

Yay more grids! Seemed like prime Dijkstra or A* material but I went with an iterative approach instead!

I keep an array cost[y][x][dir], which is seeded at 1 for the starting location and direction. Then I keep going over the array, seeing if any valid move (step or turn) would yield to a lower best-known-cost for this state. It ends when a pass does not yield changes.

This leaves us with the best-known-costs for every reachable state in the array, including the end cell (bit we have to take the min() of the four directions).

Part 2 was interesting: I just happend to have written a dead end pruning function for part 1 and part 2 is, really, dead-end pruning for the cost map: remove any suboptimal step, keep doing so, and you end up with only the optimal steps. 'Suboptimal' here is a move that yields a higher total cost than the best-known-cost for that state.

It's fast enough too on my 2015 i5:

day16  0:00.05  1656 Kb  0+242 faults

Code

#include "common.h"

#define GZ 145

enum {NN, EE, SS, WW};

static const int dx[]={0,1,0,-1}, dy[]={-1,0,1,0};

static char g[GZ][GZ];		/* with 1 tile border */
static int cost[GZ][GZ][4];	/* per direction, starts at 1, 0=no info */

static int traversible(char c) { return c=='.' || c=='S' || c=='E'; }

static int
minat(int x, int y)
{
	int acc=0, d;

	for (d=0; d<4; d++)
		if (cost[y][x][d] && (!acc || cost[y][x][d] < acc))
			acc = cost[y][x][d];

	return acc;
}


static int
count_exits(int x, int y)
{
	int acc=0, i;

	assert(x>0); assert(x<GZ-1);
	assert(y>0); assert(y<GZ-1);

	for (i=0; i<4; i++)
		acc += traversible(g[y+dy[i]][x+dx[i]]);
	
	return acc;
}

/* remove all dead ends */
static void
prune_dead(void)
{
	int dirty=1, x,y;

	while (dirty) {
		dirty = 0;

		for (y=1; y<GZ-1; y++)
		for (x=1; x<GZ-1; x++)
			if (g[y][x]=='.' && count_exits(x,y) < 2)
				{ dirty = 1; g[y][x] = '#'; }
	}
}

/* remove all dead ends from cost[], leaves only optimal paths */
static void
prune_subopt(void)
{
	int dirty=1, x,y,d;

	while (dirty) {
		dirty = 0;

		for (y=1; y<GZ-1; y++)
		for (x=1; x<GZ-1; x++)
		for (d=0; d<4; d++) {
			if (!cost[y][x][d])
				continue;

			if (g[y][x]=='E') {
				if (cost[y][x][d] != minat(x,y))
					{ dirty = 1; cost[y][x][d] = 0; }
				continue;
			}

			if (cost[y][x][d]+1 > cost[y+dy[d]][x+dx[d]][d] &&
			    cost[y][x][d]+1000 > cost[y][x][(d+1)%4] &&
			    cost[y][x][d]+1000 > cost[y][x][(d+3)%4])
				{ dirty = 1; cost[y][x][d] = 0; }
		}
	}
}

static void
propagate_costs(void)
{
	int dirty=1, cost1, x,y,d;

	while (dirty) {
		dirty = 0;

		for (y=1; y<GZ-1; y++)
		for (x=1; x<GZ-1; x++)
		for (d=0; d<4; d++) {
			if (!traversible(g[y][x]))
				continue;

			/* from back */
			if ((cost1 = cost[y-dy[d]][x-dx[d]][d]) &&
			    (cost1+1 < cost[y][x][d] || !cost[y][x][d]))
				{ dirty = 1; cost[y][x][d] = cost1+1; }

			/* from right */
			if ((cost1 = cost[y][x][(d+1)%4]) &&
			    (cost1+1000 < cost[y][x][d] || !cost[y][x][d]))
				{ dirty = 1; cost[y][x][d] = cost1+1000; }

			/* from left */
			if ((cost1 = cost[y][x][(d+3)%4]) &&
			    (cost1+1000 < cost[y][x][d] || !cost[y][x][d]))
				{ dirty = 1; cost[y][x][d] = cost1+1000; }
		}
	}
}

int
main(int argc, char **argv)
{
	int p1=0,p2=0, sx=0,sy=0, ex=0,ey=0, x,y;
	char *p;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));

	for (y=1; fgets(g[y]+1, GZ-1, stdin); y++) {
		if ((p = strchr(g[y]+1, 'S'))) { sy=y; sx=p-g[y]; }
		if ((p = strchr(g[y]+1, 'E'))) { ey=y; ex=p-g[y]; }
		assert(y+1 < GZ-1);
	}

	cost[sy][sx][EE] = 1;

	prune_dead();
	propagate_costs();
	prune_subopt();

	p1 = minat(ex, ey) -1;	/* costs[] values start at 1! */

	for (y=1; y<GZ-1; y++)
	for (x=1; x<GZ-1; x++)
		p2 += minat(x,y) > 0;

	printf("16: %d %d\n", p1, p2);
	return 0;
}

[–] mykl 3 points 1 day ago

Very interesting approach. Pruning deadends by spawning additional walls is a very clever idea.

[–] [email protected] 3 points 1 day ago* (last edited 1 day ago)

C#

Ended up modifying part 1 to do part 2 and return both answers at once.

using System.Collections.Immutable;
using System.Diagnostics;
using Common;

namespace Day16;

static class Program
{
    static void Main()
    {
        var start = Stopwatch.GetTimestamp();

        var smallInput = Input.Parse("smallsample.txt");
        var sampleInput = Input.Parse("sample.txt");
        var programInput = Input.Parse("input.txt");

        Console.WriteLine($"Part 1 small: {Solve(smallInput)}");
        Console.WriteLine($"Part 1 sample: {Solve(sampleInput)}");
        Console.WriteLine($"Part 1 input: {Solve(programInput)}");

        Console.WriteLine($"That took about {Stopwatch.GetElapsedTime(start)}");
    }

    static (int part1, int part2) Solve(Input i)
    {
        State? endState = null;
        Dictionary<(Point, int), int> lowestScores = new();

        var queue = new Queue<State>();
        queue.Enqueue(new State(i.Start, 1, 0, ImmutableHashSet<Point>.Empty));
        while (queue.TryDequeue(out var state))
        {
            if (ElementAt(i.Map, state.Location) is '#')
            {
                continue;
            }

            if (lowestScores.TryGetValue((state.Location, state.DirectionIndex), out var lowestScoreSoFar))
            {
                if (state.Score > lowestScoreSoFar) continue;
            }

            lowestScores[(state.Location, state.DirectionIndex)] = state.Score;

            var nextStatePoints = state.Points.Add(state.Location);

            if (state.Location == i.End)
            {
                if ((endState is null) || (state.Score < endState.Score))
                    endState = state with { Points = nextStatePoints };
                else if (state.Score == endState.Score)
                    endState = state with { Points = nextStatePoints.Union(endState.Points) };
                continue;
            }

            // Walk forward
            queue.Enqueue(state with
            {
                Location = state.Location.Move(CardinalDirections[state.DirectionIndex]),
                Score = state.Score + 1,
                Points = nextStatePoints,
            });

            // Turn clockwise
            queue.Enqueue(state with
            {
                DirectionIndex = (state.DirectionIndex + 1) % CardinalDirections.Length,
                Score = state.Score + 1000,
                Points = nextStatePoints,
            });

            // Turn counter clockwise
            queue.Enqueue(state with
            {
                DirectionIndex = (state.DirectionIndex + CardinalDirections.Length - 1) % CardinalDirections.Length,
                Score = state.Score + 1000,
                Points = nextStatePoints, 
            });
        }

        if (endState is null) throw new Exception("No end state found!");
        return (endState.Score, endState.Points.Count);
    }

    public static void DumpMap(Input i, ISet<Point>? points, Point current)
    {
        for (int row = 0; row < i.Bounds.Row; row++)
        {
            for (int col = 0; col < i.Bounds.Col; col++)
            {
                var p = new Point(row, col);
                Console.Write(
                    (p == current) ? 'X' :
                    (points?.Contains(p) ?? false) ? 'O' :
                    ElementAt(i.Map, p));
            }

            Console.WriteLine();
        }

        Console.WriteLine();
    }

    public static char ElementAt(string[] map, Point location) => map[location.Row][location.Col];

    public record State(Point Location, int DirectionIndex, int Score, ImmutableHashSet<Point> Points);

    public static readonly Direction[] CardinalDirections =
        [Direction.Up, Direction.Right, Direction.Down, Direction.Left];
}

public class Input
{
    public string[] Map { get; init; } = [];
    public Point Start { get; init; } = new(-1, -1);
    public Point End { get; init; } = new(-1, -1);
    public Point Bounds => new(this.Map.Length, this.Map[0].Length);

    public static Input Parse(string file)
    {
        var map = File.ReadAllLines(file);
        Point start = new(-1, -1), end = new(-1, -1);
        foreach (var p in map
            .SelectMany((line, i) => new []
            {
                 new Point(i, line.IndexOf('S')),
                 new Point(i, line.IndexOf('E')),
            })
            .Where(p => p.Col >= 0)
            .Take(2))
        {
            if (map[p.Row][p.Col] is 'S') start = p;
            else end = p;
        }

        return new Input()
        {
            Map = map,
            Start = start,
            End = end,
        };
    }
}
[–] LeixB 3 points 1 day ago

Haskell

::: spoiler code

import Control.Arrow
import Control.Monad
import Control.Monad.RWS
import Control.Monad.Trans.Maybe
import Data.Array.Unboxed
import Data.List
import Data.Map qualified as M
import Data.Maybe
import Data.Set qualified as S

data Dir = N | S | W | E deriving (Show, Eq, Ord)
type Maze = UArray Pos Char
type Pos = (Int, Int)
type Node = (Pos, Dir)
type CostNode = (Int, Node)
type Problem = RWS Maze [(Node, [Node])] (M.Map Node Int, S.Set (CostNode, Maybe Node))

parse = toMaze . lines

toMaze :: [String] -> Maze
toMaze b = listArray ((0, 0), (n - 1, m - 1)) $ concat b
  where
    n = length b
    m = length $ head b

next :: Int -> (Pos, Dir) -> Problem [CostNode]
next c (p, d) = do
    m <- ask

    let straigth = fmap ((1,) . (,d)) . filter ((/= '#') . (m !)) . return $ move d p
        turn = (1000,) . (p,) <$> rot d

    return $ first (+ c) <$> straigth ++ turn

move N = first (subtract 1)
move S = first (+ 1)
move W = second (subtract 1)
move E = second (+ 1)

rot d
    | d `elem` [N, S] = [E, W]
    | otherwise = [N, S]

dijkstra :: MaybeT Problem ()
dijkstra = do
    m <- ask
    visited <- gets fst
    Just (((cost, vertex@(p, _)), father), queue) <- gets (S.minView . snd)

    let (prevCost, visited') = M.insertLookupWithKey (\_ a _ -> a) vertex cost visited

    case prevCost of
        Nothing -> do
            queue' <- lift $ foldr S.insert queue <$> (fmap (,Just vertex) <$> next cost vertex)
            put (visited', queue')
            tell [(vertex, maybeToList father)]
        Just c -> do
            if c == cost
                then tell [(vertex, maybeToList father)]
                else guard $ m ! p /= 'E'
            put (visited, queue)
    dijkstra

solve b = do
    start <- getStart b
    end <- getEnd b
    let ((m, _), w) = execRWS (runMaybeT dijkstra) b (M.empty, S.singleton (start, Nothing))
        parents = M.fromListWith (++) w
        endDirs = (end,) <$> [N, S, E, W]
        min = minimum $ mapMaybe (`M.lookup` m) endDirs
        ends = filter ((== Just min) . (`M.lookup` m)) endDirs
        part2 =
            S.size . S.fromList . fmap fst . concat . takeWhile (not . null) $
                iterate (>>= flip (M.findWithDefault []) parents) ends
    return (min, part2)

getStart :: Maze -> Maybe CostNode
getStart = fmap ((0,) . (,E) . fst) . find ((== 'S') . snd) . assocs

getEnd :: Maze -> Maybe Pos
getEnd = fmap fst . find ((== 'E') . snd) . assocs

main = getContents >>= print . solve . parse
[–] [email protected] 1 points 1 day ago* (last edited 1 day ago)

Javascript

So my friend tells me my solution is close to Dijkstra but honestly I just tried what made sense until it worked. I originally wanted to just bruteforce it and get every single possible path explored but uh... Yeah that wasn't gonna work, I terminated that one after 1B operations.

I created a class to store the state of the current path being explored, and basically just clone it, sending it in each direction (forward, 90 degrees, -90 degrees), then queue it up if it didn't fail. Using a priority queue (array based) to store them, I inverted it for the second answer to reduce the memory footprint (though ultimately once I fixed the issue with the algorithm, which turned out to just be a less than or equal to that should have been a less than, I didn't really need this).

Part two "only" took 45 seconds to run on my Thinkpad P14 Gen1.

My code was too powerful for Lemmy (or verbose): https://blocks.programming.dev/Zikeji/ae06ca1ca88649c99581eefce97a708e

[–] VegOwOtenks 3 points 1 day ago* (last edited 1 day ago)

Haskell

This one was surprisingly slow to run

Big codeblock

import Control.Arrow

import Data.Map (Map)
import Data.Set (Set)
import Data.Array.ST (STArray)
import Data.Array (Array)
import Control.Monad.ST (ST, runST)

import qualified Data.Char as Char
import qualified Data.List as List
import qualified Data.Map as Map
import qualified Data.Set as Set
import qualified Data.Array.ST as MutableArray
import qualified Data.Array as Array
import qualified Data.Maybe as Maybe

data Direction = East | West | South | North
        deriving (Show, Eq, Ord)
data MazeTile = Start | End | Wall | Unknown | Explored (Map Direction ExplorationScore)
        deriving Eq

--      instance Show MazeTile where
--              show Wall = "#"
--              show Start = "S"
--              show End = "E"
--              show Unknown = "."
--              show (Explored (East, _))  = ">"
--              show (Explored (South, _)) = "v"
--              show (Explored (West, _))  = "<"
--              show (Explored (North, _)) = "^"

type Position = (Int, Int)
type ExplorationScore = Int

translate '#' = Wall
translate '.' = Unknown
translate 'S' = Start
translate 'E' = End

parse :: String -> Array (Int, Int) MazeTile
parse s = Array.listArray ((1, 1), (height - 1, width)) . map translate . filter (/= '\n') $ s
        where
                width = length . takeWhile (/= '\n') $ s
                height = length . filter (== '\n') $ s

(a1, b1) .+. (a2, b2) = (a1+a2, b1+b2)
(a1, b1) .-. (a2, b2) = (a1-a2, b1-b2)

directions = [East, West, South, North]
directionVector East  = (0,  1)
directionVector West  = (0, -1)
directionVector North = (-1, 0)
directionVector South = ( 1, 0)

turnRight East  = South
turnRight South = West
turnRight West  = North
turnRight North = East

walkableNeighbors a p = do
        let neighbors = List.map ((.+. p) . directionVector) directions
        tiles <- mapM (MutableArray.readArray a) neighbors
        let neighborPosition = List.map fst . List.filter ((/= Wall). snd) . zip neighbors $ tiles
        return $ neighborPosition


findDeadEnds a = Array.assocs
        >>> List.filter (snd >>> (== Unknown))
        >>> List.map (fst)
        >>> List.filter (isDeadEnd a)
        $ a
isDeadEnd a p = List.map directionVector
        >>> List.map (.+. p)
        >>> List.map (a Array.!)
        >>> List.filter (/= Wall)
        >>> List.length
        >>> (== 1)
        $ directions

fillDeadEnds :: Array (Int, Int) MazeTile -> ST s (Array (Int, Int) MazeTile)
fillDeadEnds a = do
        ma <- MutableArray.thaw a
        let deadEnds = findDeadEnds a
        mapM_ (fillDeadEnd ma) deadEnds
        MutableArray.freeze ma

fillDeadEnd :: STArray s (Int, Int) MazeTile -> Position -> ST s ()
fillDeadEnd a p = do
        MutableArray.writeArray a p Wall
        p' <- walkableNeighbors a p >>= return . head
        t <- MutableArray.readArray a p'
        n <- walkableNeighbors a p' >>= return . List.length
        if n == 1 && t == Unknown then fillDeadEnd a p' else return ()

thawArray :: Array (Int, Int) MazeTile -> ST s (STArray s (Int, Int) MazeTile)
thawArray a = do
        a' <- MutableArray.thaw a
        return a'

solveMaze a = do
        a' <- fillDeadEnds a
        a'' <- thawArray a'
        let s = Array.assocs
                >>> List.filter ((== Start) . snd)
                >>> Maybe.listToMaybe
                >>> Maybe.maybe (error "Start not in map") fst
                $ a
        let e = Array.assocs
                >>> List.filter ((== End) . snd)
                >>> Maybe.listToMaybe
                >>> Maybe.maybe (error "End not in map") fst
                $ a
        MutableArray.writeArray a'' s $ Explored (Map.singleton East 0)
        MutableArray.writeArray a'' e $ Unknown
        solveMaze' (s, East) a''
        fa <- MutableArray.freeze a''
        t <- MutableArray.readArray a'' e
        case t of
                Wall  -> error "Unreachable code"
                Start -> error "Unreachable code"
                End   -> error "Unreachable code"
                Unknown -> error "End was not explored yet"
                Explored m -> return (List.minimum . List.map snd . Map.toList $ m, countTiles fa s e)

countTiles a s p = Set.size . countTiles' a s p $ South

countTiles' :: Array (Int, Int) MazeTile -> Position -> Position -> Direction -> Set Position
countTiles' a s p d
        | p == s    = Set.singleton p
        | otherwise = Set.unions 
                . List.map (Set.insert p) 
                . List.map (uncurry (countTiles' a s)) 
                $ (zip minCostNeighbors minCostDirections)
        where
                minCostNeighbors   = List.map ((p .-.) . directionVector) minCostDirections
                minCostDirections  = List.map fst . List.filter ((== minCost) . snd) . Map.toList $ visits
                visits = case a Array.! p of
                        Explored m -> Map.adjust (+ (-1000)) d m
                minCost = List.minimum . List.map snd . Map.toList $ visits

maybeExplore c p d a = do
        t <- MutableArray.readArray a p
        case t of
                Wall     -> return ()
                Start    -> error "Unreachable code"
                End      -> error "Unreachable code"
                Unknown  -> do
                        MutableArray.writeArray a p $ Explored (Map.singleton d c)
                        solveMaze' (p, d) a
                Explored m -> do
                        let c' = Maybe.maybe c id (m Map.!? d)
                        if c <= c' then do
                                let m' = Map.insert d c m
                                MutableArray.writeArray a p (Explored m')
                                solveMaze' (p, d) a
                        else
                                return ()

solveMaze' :: (Position, Direction) -> STArray s (Int, Int) MazeTile -> ST s ()
solveMaze' s@(p, d) a = do
        t <- MutableArray.readArray a p
        case t of
                Wall -> return ()
                Start -> error "Unreachable code"
                End -> error "Unreachable code"
                Unknown -> error "Starting on unexplored field"
                Explored m -> do
                        let c = m Map.! d
                        maybeExplore (c+1)    (p .+. directionVector d)  d a
                        let d' = turnRight d
                        maybeExplore (c+1001) (p .+. directionVector d') d' a
                        let d'' = turnRight d'
                        maybeExplore (c+1001) (p .+. directionVector d'') d'' a
                        let d''' = turnRight d''
                        maybeExplore (c+1001) (p .+. directionVector d''') d''' a

part1 a = runST (solveMaze a)

main = getContents
        >>= print
        . part1
        . parse

[–] mykl 2 points 1 day ago* (last edited 1 day ago)

Dart

I liked the flexibility of the path operator in the Uiua solution so much that I built a similar search function in Dart. Not quite as compact, but still an interesting piece of code that I will keep on hand for other path-finding puzzles.

About 80 lines of code, about half of which is the super-flexible search function.

import 'dart:math';
import 'package:collection/collection.dart';
import 'package:more/more.dart';

List<Point<num>> d4 = [Point(1, 0), Point(-1, 0), Point(0, 1), Point(0, -1)];

/// Returns cost to destination, plus list of routes to destination.
/// Does Dijkstra/A* search depending on whether heuristic returns 1 or
/// something better.
(num, List<List<T>>) aStarSearch<T>(T start, Map<T, num> Function(T) fNext,
    int Function(T) fHeur, bool Function(T) fAtEnd) {
  var cameFrom = SetMultimap<T, T>.fromEntries([MapEntry(start, start)]);

  var ends = <T>{};
  var front = PriorityQueue<T>((a, b) => fHeur(a).compareTo(fHeur(b)))
    ..add(start);
  var cost = <T, num>{start: 0};
  while (front.isNotEmpty) {
    var here = front.removeFirst();
    if (fAtEnd(here)) {
      ends.add(here);
      continue;
    }
    var ns = fNext(here);
    for (var n in ns.keys) {
      var nCost = cost[here]! + ns[n]!;
      if (!cost.containsKey(n) || nCost < cost[n]!) {
        cost[n] = nCost;
        front.add(n);
        cameFrom.removeAll(n);
      }
      if (cost[n] == nCost) cameFrom[n].add(here);
    }
  }

  Iterable<List<T>> routes(T h) sync* {
    if (h == start) {
      yield [h];
      return;
    }
    for (var p in cameFrom[h]) {
      yield* routes(p).map((e) => e + [h]);
    }
  }

  var minCost = ends.map((e) => cost[e]!).min;
  ends = ends.where((e) => cost[e]! == minCost).toSet();
  return (minCost, ends.fold([], (s, t) => s..addAll(routes(t).toList())));
}

typedef PP = (Point, Point);

(num, List<List<PP>>) solve(List<String> lines) {
  var grid = {
    for (var r in lines.indexed())
      for (var c in r.value.split('').indexed().where((e) => e.value != '#'))
        Point<num>(c.index, r.index): c.value
  };
  var start = grid.entries.firstWhere((e) => e.value == 'S').key;
  var end = grid.entries.firstWhere((e) => e.value == 'E').key;
  var dir = Point<num>(1, 0);

  fHeur(PP pd) => 1; // faster than euclidean distance.
  fNextAndCost(PP pd) => <PP, int>{
        for (var n in d4
            .where((n) => n != pd.last * -1 && grid.containsKey(pd.first + n)))
          (pd.first + n, n): ((n == pd.last) ? 1 : 1001) // (Point, Dir) : Cost
      };
  fAtEnd(PP pd) => pd.first == end;

  return aStarSearch<PP>((start, dir), fNextAndCost, fHeur, fAtEnd);
}

part1(List<String> lines) => solve(lines).first;

part2(List<String> lines) => solve(lines)
    .last
    .map((l) => l.map((e) => e.first).toSet())
    .flattenedToSet
    .length;
[–] mykl 3 points 2 days ago* (last edited 1 day ago) (1 children)

Uiua

Uiua's new builtin path operator makes this a breeze. Given a function that returns valid neighbours for a point and their relative costs, and another function to test whether you have reached a valid goal, it gives the minimal cost, and all relevant paths. We just need to keep track of the current direction as we work through the maze.

(edit: forgot the Try It Live! link)

Data  ← ≑°░°/$"_\n_" "#################\n#...#...#...#..E#\n#.#.#.#.#.#.#.#^#\n#.#.#.#...#...#^#\n#.#.#.#.###.#.#^#\n#>>v#.#.#.....#^#\n#^#v#.#.#.#####^#\n#^#v..#.#.#>>>>^#\n#^#v#####.#^###.#\n#^#v#..>>>>^#...#\n#^#v###^#####.###\n#^#v#>>^#.....#.#\n#^#v#^#####.###.#\n#^#v#^........#.#\n#^#v#^#########.#\n#S#>>^..........#\n#################"
Dβ‚„    ← [1_0 Β―1_0 0_1 0_Β―1]
End   ← ⊒⊚=@EData
Costs ← :βˆ©β–½βŸœ:≑(β‰ @#⊑:Data⊒).β‰‘βŠŸβŠ™βŸœ(+1Γ—1000¬≑/Γ—=)+⟜:Dβ‚„βˆ©Β€Β°βŠŸ
path(Costs|≍EndβŠ™β—ŒΒ°βŠŸ)⊟:1_0⊒⊚=@SData
&p β§»β—΄β‰‘βŠ’/β—‡βŠ‚ &p :

[–] Acters 2 points 1 day ago

uiua is insane, such small code footprint to fit a maze solver like algo

[–] [email protected] 2 points 1 day ago

Rust

Dijkstra's algorithm. While the actual shortest path was not needed in part 1, only the distance, in part 2 the path is saved in the parent hashmap, and crucially, if we encounter two paths with the same distance, both parent nodes are saved. This ensures we end up with all shortest paths in the end.

Solution

use std::cmp::{Ordering, Reverse};

use euclid::{default::*, vec2};
use priority_queue::PriorityQueue;
use rustc_hash::{FxHashMap, FxHashSet};

const DIRS: [Vector2D<i32>; 4] = [vec2(1, 0), vec2(0, 1), vec2(-1, 0), vec2(0, -1)];

type Node = (Point2D<i32>, u8);

fn parse(input: &str) -> (Vec<Vec<bool>>, Point2D<i32>, Point2D<i32>) {
    let mut start = None;
    let mut end = None;
    let mut field = Vec::new();
    for (y, l) in input.lines().enumerate() {
        let mut row = Vec::new();
        for (x, b) in l.bytes().enumerate() {
            if b == b'S' {
                start = Some(Point2D::new(x, y).to_i32());
            } else if b == b'E' {
                end = Some(Point2D::new(x, y).to_i32());
            }
            row.push(b == b'#');
        }
        field.push(row);
    }
    (field, start.unwrap(), end.unwrap())
}

fn adj(field: &[Vec<bool>], (v, dir): Node) -> Vec<(Node, u32)> {
    let mut adj = Vec::with_capacity(3);
    let next = v + DIRS[dir as usize];
    if !field[next.y as usize][next.x as usize] {
        adj.push(((next, dir), 1));
    }
    adj.push(((v, (dir + 1) % 4), 1000));
    adj.push(((v, (dir + 3) % 4), 1000));
    adj
}

fn shortest_path_length(field: &[Vec<bool>], start: Node, end: Point2D<i32>) -> u32 {
    let mut dist: FxHashMap<Node, u32> = FxHashMap::default();
    dist.insert(start, 0);
    let mut pq: PriorityQueue<Node, Reverse<u32>> = PriorityQueue::new();
    pq.push(start, Reverse(0));
    while let Some((v, _)) = pq.pop() {
        for (w, weight) in adj(field, v) {
            let dist_w = dist.get(&w).copied().unwrap_or(u32::MAX);
            let new_dist = dist[&v] + weight;
            if dist_w > new_dist {
                dist.insert(w, new_dist);
                pq.push_increase(w, Reverse(new_dist));
            }
        }
    }
    // Shortest distance to end, regardless of final direction
    (0..4).map(|dir| dist[&(end, dir)]).min().unwrap()
}

fn part1(input: String) {
    let (field, start, end) = parse(&input);
    let distance = shortest_path_length(&field, (start, 0), end);
    println!("{distance}");
}

fn shortest_path_tiles(field: &[Vec<bool>], start: Node, end: Point2D<i32>) -> u32 {
    let mut parents: FxHashMap<Node, Vec<Node>> = FxHashMap::default();
    let mut dist: FxHashMap<Node, u32> = FxHashMap::default();
    dist.insert(start, 0);
    let mut pq: PriorityQueue<Node, Reverse<u32>> = PriorityQueue::new();
    pq.push(start, Reverse(0));
    while let Some((v, _)) = pq.pop() {
        for (w, weight) in adj(field, v) {
            let dist_w = dist.get(&w).copied().unwrap_or(u32::MAX);
            let new_dist = dist[&v] + weight;
            match dist_w.cmp(&new_dist) {
                Ordering::Greater => {
                    parents.insert(w, vec![v]);
                    dist.insert(w, new_dist);
                    pq.push_increase(w, Reverse(new_dist));
                }
                // Remember both parents if distance is equal
                Ordering::Equal => parents.get_mut(&w).unwrap().push(v),
                Ordering::Less => {}
            }
        }
    }
    let mut path_tiles: FxHashSet<Point2D<i32>> = FxHashSet::default();
    path_tiles.insert(end);

    // Shortest distance to end, regardless of final direction
    let shortest_dist = (0..4).map(|dir| dist[&(end, dir)]).min().unwrap();
    for dir in 0..4 {
        if dist[&(end, dir)] == shortest_dist {
            collect_tiles(&parents, &mut path_tiles, (end, dir));
        }
    }
    path_tiles.len() as u32
}

fn collect_tiles(
    parents: &FxHashMap<Node, Vec<Node>>,
    tiles: &mut FxHashSet<Point2D<i32>>,
    cur: Node,
) {
    if let Some(pars) = parents.get(&cur) {
        for p in pars {
            tiles.insert(p.0);
            collect_tiles(parents, tiles, *p);
        }
    }
}

fn part2(input: String) {
    let (field, start, end) = parse(&input);
    let tiles = shortest_path_tiles(&field, (start, 0), end);
    println!("{tiles}");
}

util::aoc_main!();

Also on github

[–] [email protected] 3 points 2 days ago

Haskell

Rather busy today so late and somewhat messy! (Probably the same tomorrow...)

import Data.List
import Data.Map (Map)
import Data.Map qualified as Map
import Data.Maybe
import Data.Set (Set)
import Data.Set qualified as Set

readInput :: String -> Map (Int, Int) Char
readInput s = Map.fromList [((i, j), c) | (i, l) <- zip [0 ..] (lines s), (j, c) <- zip [0 ..] l]

bestPath :: Map (Int, Int) Char -> (Int, Set (Int, Int))
bestPath maze = go (Map.singleton start (0, Set.singleton startPos)) (Set.singleton start)
  where
    start = (startPos, (0, 1))
    walls = Map.keysSet $ Map.filter (== '#') maze
    [Just startPos, Just endPos] = map (\c -> fst <$> find ((== c) . snd) (Map.assocs maze)) ['S', 'E']
    go best edge
      | Set.null edge = Map.mapKeysWith mergePaths fst best Map.! endPos
      | otherwise =
          let nodes' =
                filter (\(x, (c, _)) -> maybe True ((c <=) . fst) $ best Map.!? x) $
                  concatMap (step . (\x -> (x, best Map.! x))) (Set.elems edge)
              best' = foldl' (flip $ uncurry $ Map.insertWith mergePaths) best nodes'
           in go best' $ Set.fromList (map fst nodes')
    step ((p@(i, j), d@(di, dj)), (cost, path)) =
      let rots = [((p, d'), (cost + 1000, path)) | d' <- [(-dj, di), (dj, -di)]]
          moves =
            [ ((p', d), (cost + 1, Set.insert p' path))
              | let p' = (i + di, j + dj),
                p `Set.notMember` walls
            ]
       in moves ++ rots
    mergePaths a@(c1, p1) b@(c2, p2) =
      case compare c1 c2 of
        LT -> a
        GT -> b
        EQ -> (c1, Set.union p1 p2)

main = do
  (score, visited) <- bestPath . readInput <$> readFile "input16"
  print score
  print (Set.size visited)
[–] [email protected] 2 points 1 day ago* (last edited 1 day ago)

C#

using QuickGraph;
using QuickGraph.Algorithms.ShortestPath;

namespace aoc24;

[ForDay(16)]
public class Day16 : Solver {
  private string[] data;
  private int width, height;
  private int start_x, start_y;
  private int end_x, end_y;

  private readonly (int, int)[] directions = [(1, 0), (0, 1), (-1, 0), (0, -1)];
  private record class Edge((int, int, int) Source, (int, int, int) Target) : IEdge<(int, int, int)>;

  private DelegateVertexAndEdgeListGraph<(int, int, int), Edge> graph;
  private AStarShortestPathAlgorithm<(int, int, int), Edge> search;

  private long min_distance;
  private List<(int, int, int)> min_distance_targets;

  public void Presolve(string input) {
    data = input.Trim().Split("\n");
    width = data[0].Length;
    height = data.Length;
    for (int i = 0; i < width; i++) {
      for (int j = 0; j < height; j++) {
        if (data[j][i] == 'S') {
          start_x = i;
          start_y = j;
        } else if (data[j][i] == 'E') {
          end_x = i;
          end_y = j;
        }
      }
    }
    graph = MakeGraph();
    var start = (start_x, start_y, 0);
    search = new AStarShortestPathAlgorithm<(int, int, int), Edge>(
      graph,
      edge => edge.Source.Item3 == edge.Target.Item3 ? 1 : 1000,
      vertex => Math.Abs(vertex.Item1 - start_x) + Math.Abs(vertex.Item2 - start_y) + 1000 *
          Math.Min(vertex.Item3, 4 - vertex.Item3)
      );
    Dictionary<(int, int, int), long> distances = [];
    search.SetRootVertex(start);
    search.ExamineVertex += vertex => {
      if (vertex.Item1 == end_x && vertex.Item2 == end_y) {
        distances[vertex] = (long)search.Distances[vertex];
      }
    };
    search.Compute();
    min_distance = distances.Values.Min();
    min_distance_targets = distances.Keys.Where(v => distances[v] == min_distance).ToList();
  }

  private DelegateVertexAndEdgeListGraph<(int, int, int), Edge> MakeGraph() => new(GetAllVertices(), GetOutEdges);

  private bool GetOutEdges((int, int, int) arg, out IEnumerable<Edge> result_enumerable) {
    List<Edge> result = [];
    var (x, y, dir) = arg;
    result.Add(new Edge(arg, (x, y, (dir + 1) % 4)));
    result.Add(new Edge(arg, (x, y, (dir + 3) % 4)));
    var (tx, ty) = (x + directions[dir].Item1, y + directions[dir].Item2);
    if (data[ty][tx] != '#') result.Add(new Edge(arg, (tx, ty, dir)));
    result_enumerable = result;
    return true;
  }

  private IEnumerable<(int, int, int)> GetAllVertices() {
    for (int i = 0; i < width; i++) {
      for (int j = 0; j < height; j++) {
        if (data[j][i] == '#') continue;
        yield return (i, j, 0);
        yield return (i, j, 1);
        yield return (i, j, 2);
        yield return (i, j, 3);
      }
    }
  }

  private HashSet<(int, int, int)> GetMinimumPathNodesTo((int, int, int) vertex) {
    var (x, y, dir) = vertex;
    if (x == start_x && y == start_y && dir == 0) return [vertex];
    if (!search.Distances.TryGetValue(vertex, out var distance_to_me)) return [];
    List<(int, int, int)> candidates = [
          (x, y, (dir + 1) % 4),
          (x, y, (dir + 3) % 4),
          (x - directions[dir].Item1, y - directions[dir].Item2, dir),
      ];
    HashSet<(int, int, int)> result = [vertex];
    foreach (var (cx, cy, cdir) in candidates) {
      if (!search.Distances.TryGetValue((cx, cy, cdir), out var distance_to_candidate)) continue;
      if (distance_to_candidate > distance_to_me - (dir == cdir ? 1 : 1000)) continue;
      result = result.Union(GetMinimumPathNodesTo((cx, cy, cdir))).ToHashSet();
    }
    return result;
  }

  public string SolveFirst() => min_distance.ToString();

  public string SolveSecond() => min_distance_targets
    .SelectMany(v => GetMinimumPathNodesTo(v))
    .Select(vertex => (vertex.Item1, vertex.Item2))
    .ToHashSet()
    .Count
    .ToString();
}
[–] [email protected] 2 points 2 days ago* (last edited 2 days ago) (2 children)

Python

Part 1: Run Dijkstra's algorithm to find shortest path.

I chose to represent nodes using the location (i, j) as well as the direction dir faced by the reindeer.
Initially I tried creating the complete adjacency graph but that lead to max recursion so I ended up populating graph for only the nodes I was currently exploring.

Part 2: Track paths while performing Dijkstra's algorithm.

First, I modified the algorithm to look through neighbors with equal cost along with the ones with lesser cost, so that it would go through all shortest paths.
Then, I keep track of the list of previous nodes for every node explored.
Finally, I use those lists to run through the paths backwards, taking note of all unique locations.

Code:
import os

# paths
here = os.path.dirname(os.path.abspath(__file__))
filepath = os.path.join(here, "input.txt")

# read input
with open(filepath, mode="r", encoding="utf8") as f:
    data = f.read()

from collections import defaultdict
from dataclasses import dataclass
import heapq as hq
import math

# up, right, down left
DIRECTIONS = [(-1, 0), (0, 1), (1, 0), (0, -1)]


# Represent a node using its location and the direction
@dataclass(frozen=True)
class Node:
    i: int
    j: int
    dir: int


maze = data.splitlines()
m, n = len(maze), len(maze[0])

# we always start from bottom-left corner (facing east)
start_node = Node(m - 2, 1, 1)
# we always end in top-right corner (direction doesn't matter)
end_node = Node(1, n - 2, -1)

# the graph will be updated lazily because it is too much processing
#   to completely populate it beforehand
graph = defaultdict(list)
# track nodes whose all edges have been explored
visited = set()
# heap to choose next node to explore
# need to add id as middle tuple element so that nodes dont get compared
min_heap = [(0, id(start_node), start_node)]
# min distance from start_node to node so far
# missing values are treated as math.inf
min_dist = {}
min_dist[start_node] = 0
# keep track of all previous nodes for making path
prev_nodes = defaultdict(list)


# utility method for debugging (prints the map)
def print_map(current_node, prev_nodes):
    pns = set((n.i, n.j) for n in prev_nodes)
    for i in range(m):
        for j in range(n):
            if i == current_node.i and j == current_node.j:
                print("X", end="")
            elif (i, j) in pns:
                print("O", end="")
            else:
                print(maze[i][j], end="")
        print()


# Run Dijkstra's algo
while min_heap:
    cost_to_node, _, node = hq.heappop(min_heap)
    if node in visited:
        continue
    visited.add(node)

    # early exit in the case we have explored all paths to the finish
    if node.i == end_node.i and node.j == end_node.j:
        # assign end so that we know which direction end was reached by
        end_node = node
        break

    # update adjacency graph from current node
    di, dj = DIRECTIONS[node.dir]
    if maze[node.i + di][node.j + dj] != "#":
        moved_node = Node(node.i + di, node.j + dj, node.dir)
        graph[node].append((moved_node, 1))
    for x in range(3):
        rotated_node = Node(node.i, node.j, (node.dir + x + 1) % 4)
        graph[node].append((rotated_node, 1000))

    # explore edges
    for neighbor, cost in graph[node]:
        cost_to_neighbor = cost_to_node + cost
        # The following condition was changed from > to >= because we also want to explore
        #   paths with the same cost, not just better cost
        if min_dist.get(neighbor, math.inf) >= cost_to_neighbor:
            min_dist[neighbor] = cost_to_neighbor
            prev_nodes[neighbor].append(node)
            # need to add id as middle tuple element so that nodes dont get compared
            hq.heappush(min_heap, (cost_to_neighbor, id(neighbor), neighbor))

print(f"Part 1: {min_dist[end_node]}")

# PART II: Run through the path backwards, making note of all coords

visited = set([start_node])
path_locs = set([(start_node.i, start_node.j)])  # all unique locations in path
stack = [end_node]

while stack:
    node = stack.pop()
    if node in visited:
        continue
    visited.add(node)

    path_locs.add((node.i, node.j))

    for prev_node in prev_nodes[node]:
        stack.append(prev_node)

print(f"Part 2: {len(path_locs)}")

[–] Acters 2 points 1 day ago* (last edited 1 day ago) (1 children)

only improvement I can think of is to implement a dead end finder to block for the search algorithm to skip all dead ends that do not have the end tile ("E"). by block I mean artificially add a wall to the entrance of the dead end. this should help make it so that it doesn't go down dead ends. It would be improbable but there might be an input with a ridiculously long dead end.

[–] [email protected] 1 points 1 day ago (2 children)

Interesting, how would one write such a finder? I can only think of backtracking DFS, but that seems like it would outweigh the savings.

[–] Acters 2 points 15 hours ago* (last edited 14 hours ago) (1 children)

took some time out of my day to implement a solution that beats only running your solution by like 90 ms. This is because the algorithm for filling in all dead ends takes like 9-10 milliseconds and reduces the time it takes your algorithm to solve this by like 95-105 ms!

decent improvement for so many lines of code, but it is what it is. using .index and .rindex on strings is just way too fast. there might be a faster way to replace with # or just switch to complete binary bit manipulation for everything, but that is like incredibly difficult to think of rn.

but here is the monster script that seemingly does it in ~90 milliseconds faster than your current script version. because it helps eliminated time waste in your Dijkstra’s algorithm and fills all dead ends with minimal impact on performance. Could there be corner cases that I didn't think of? maybe, but saving time on your algo is better than just trying to be extra sure to eliminate all dead ends, and I am skipping loops because your algorithm will handle that better than trying to do a flood fill type algorithm. (remember first run of a modified script will run a little slow.)

as of rn, the slowest parts of the script is your Dijkstra’s algorithm. I could try to implement my own solver that isn't piggy-backing off your Dijkstra’s algorithm. however, I think that is just more than I care to do rn. I also was not going to bother with reducing LOC for the giant match case. its fast and serves it purpose good enough.

[ BigFastScript Paste ]

[–] [email protected] 1 points 2 hours ago (3 children)

Those are some really great optimizations, thank you! I understand what you're doing generally, but I'll have to step through the code myself to get completely familiar with it.

It's interesting that string operations win out here over graph algorithms even though this is technically a graph problem. Honestly your write-up and optimizations deserve its own post.

[–] Acters 1 points 1 hour ago* (last edited 33 minutes ago)

ah yes, I was right. simply string slicing away lines that were checked does make the regex faster. while the code is smaller, it is still slower than the more verbose option. which is only because of the iterative approach of checking each node in the While(True) loop, instead of building 2 lists of lines and manipulating them with .index() and .rindex() [ Paste ]

However, if you notice, even the regex is slower than my iterative approach with index by 3-5 milliseconds. While having one line for the regex is nice, I do think it is harder to read and could prove to be slightly more cumbersome as it could be useless in other types of challenges, while the iterative approach is nice and easily manipulable for most circumstances that may have some quirks.

Also, it shows that the more verbose option is still faster by 7 ms because of the fact that checking each node in the While(True) loop is rather slow approach. So really, there is nothing to it overall, and the main slow down is in you solver that I didn't touch at all, because I wanted to only show the dead end filling part.

[–] Acters 1 points 1 hour ago* (last edited 1 hour ago)

I tried to compartmentalize it. the search is on its own function, and while that fill_in_dead_ends function is extremely large, it is a lot of replicated code. match k case statement could just be removed. A lot of the code is extremely verbose and has an air of being "unrolled" for purposes of me just tweaking each part of the process individually to see what could be done. The entire large af match case all seemingly ended up being very similar code. I could condense it down a lot. however, I know doing so would impact performance unless plenty of time is spent on tweaking it. So unrolled copy pasta was good.

The real shining star is the find_next_dead_end function because the regex before took 99% of the time of about ~300 ms seconds. Even with this fast iterative function, the find_next_dead_end still takes about 75% of the time for the entire thing to finish filling in dead ends. This is because as the search ran deeper into the string, it would start slowing down because it was like O(n*m) time complexity, where n in line width and m is line count being searched through until next match. My approach was to store the relative position for each search which conveniently was the curr_row,curr_col. Another aspect to reduce cost/time complexity on the logic that would make sure it filled in newly created dead-ends was to simply check if the current search for the next dead end was from the start after it finished checking the final line. Looking at the line by line profiler from iPython, the entire function spends most of the time at the while('.' in r[:first_loc]): and first_loc = r[:first_loc].rindex('.') which is funny because that is still fast at over 11k+ hits on the same line with only a 5-5.5 microsecond impact for each time it ran the lines.

though I could likely remove that strange logic by moving it into the find_next_dead_end instead of having that strange if elif else statement in the fill_in_dead_ends logic.

there is so much possible to make it improved, but it was quick and dirty.

Now that I am thinking about it, there would be a way to make the regex faster by simply string slicing lines off the search, so that the regex doesn't spend time looking at the same start of string.

[–] Acters 1 points 7 minutes ago* (last edited 5 minutes ago)

If you are wondering how my string operations is able to be fast, it is because of the simple fact that python's index and rindex are pactically O(1) time. Also, the more verbose option is simply tricks in batch processing. why bother checking each node individually, when we already know that a dead end is simply straight lines?

If there was an exceedingly large maze was just a simple two spirals design, where one is a dead end and another has the "end flag" then my batch processing would simply outpace the slower per node iterator. in this scenario, there is a 50/50 chance you pick the right spiral, while it is just easier to look for which one is a dead end and just backtrack to chose the other option. technically it is slower than just guessing correctly first try, but that feels awfully similar to how a bogosort works. you just randomly choose paths(removing previously checked paths) or deterministically enumerate all paths. while a dead end is extremely easy to find and culls all those paths as extremely low priority, or in this spiral scenario, it is the more safe option than accidentally choosing the wrong path.

[–] Acters 2 points 1 day ago* (last edited 1 day ago)

ah well, my idea is at high level view. Here is a naive approach that should accomplish this. Not sure how else I would accomplish this without more thought put in to make it faster:

[ Paste ]

edit: whoops, sorry had broke the regex string and had to check for E and S is not deleted lol

This is how the first example would look like:

###############
#...#####....E#
#.#.#####.###.#
#.....###...#.#
#.###.#####.#.#
#.###.......#.#
#.#######.###.#
#...........#.#
###.#.#####.#.#
#...#.....#.#.#
#.#.#.###.#.#.#
#.....#...#.#.#
#.###.#.#.#.#.#
#S###.....#...#
###############

This is how the second example would look like:

#################
#...#...#...#..E#
#.#.#.#.#.#.#.#.#
#.#.#.#...#...#.#
#.#.#.#####.#.#.#
#...#.###.....#.#
#.#.#.###.#####.#
#.#...###.#.....#
#.#.#####.#.###.#
#.#.###.....#...#
#.#.###.#####.###
#.#.#...###...###
#.#.#.#####.#####
#.#.#.......#####
#.#.#.###########
#S#...###########
#################

for this challenge, it will only have a more noticeable improvement on larger maps, and especially effective if there are no loops! (i.e. one path) because it would just remove all paths that will lead to a dead end.

For smaller maps, there is no improvement or worse performance as there is not enough dead ends for any search algorithm to waste enough time on. So for more completeness sake, you would make a check to test various sizes with various amount of dead ends and find the optimal map size for where it would make sense to try to fill in all dead ends with walls. Also, when you know a maze would only have one path, then this is more optimal than any path finding algorithm, that is if the map is big enough. That is because you can just find the path fast enough that filling in dead ends is not needed and can just path find it.

for our input, I think this would not help as the map should NOT be large enough. This is naive approach is too costly. It would probably be better if there is a faster approach than this naive approach.

actually, testing this naive approach on the smaller examples, it does have a slight edge over not filling in dead ends. This means that the regex is likely slowing down as the map get larger. so something that can find dead ends faster would be a better choice than the one line regex we have right now.

I guess location of both S and E for the input does matter, because the maze map could end up with S and E being close enough that most, if not all, dead ends are never wasting the time of the Dijkstra's algorithm. however, my input had S and E being on opposite corners. So the regex is likely the culprit in why the larger map makes filling in dead ends slower.

if you notice from the profiler output, on the smaller examples, the naive approach makes a negligible loss in time and improves the time by a few tenths of a millisecond for your algorithm to do both part1 and part 2. however, on the larger input, the naive approach starts to take a huge hit and loses about 350 ms to 400 ms on filling in dead ends, while only improving the time of your algorithm by 90 ms. while filling in dead ends does improve performance for your algorithm, it just has too much overhead. That means that with a less naive approach, there would be a significant way to improve time on the solving algorithm.

[–] [email protected] 3 points 1 day ago (1 children)

prev_nodes[neighbor].append(node)

I think you're adding too many neighbours to the prev_nodes here potentially. At the time you explore the edge, you're not yet sure if the path to the edge's target via the current node will be the cheapest.

[–] [email protected] 3 points 1 day ago* (last edited 1 day ago)

Good catch! IIRC, only when a node is selected from the min heap can we guarantee that the cost to that node will not go any lower. This definitely seems like a bug, but I still got the correct answer for the samples and my input somehow Β―\_(ツ)_/Β―