this post was submitted on 23 Dec 2023
9 points (100.0% liked)

Advent Of Code

763 readers
1 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 2023

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 19 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
9
submitted 10 months ago* (last edited 10 months ago) by CameronDev to c/advent_of_code
 

Day 22: A Long Walk

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 5 comments
sorted by: hot top controversial new old
[โ€“] cvttsd2si 2 points 10 months ago* (last edited 10 months ago)

Scala3

val allowed: Map[Char, List[Dir]] = Map('>'->List(Right), '<'->List(Left), '^'->List(Up), 'v'->List(Down), '.'->Dir.all)

def toGraph(g: Grid[Char], start: Pos, goal: Pos) =
    def nb(p: Pos) = allowed(g(p)).map(walk(p, _)).filter(g.inBounds).filter(g(_) != '#')

    @tailrec def go(q: List[Pos], seen: Set[Pos], acc: List[WDiEdge[Pos]]): List[WDiEdge[Pos]] =
        q match
            case h :: t =>
                @tailrec def findNext(prev: Pos, n: Pos, d: Int): Option[(Pos, Int)] =
                    val fwd = nb(n).filter(_ != prev)
                    if fwd.size == 1 then findNext(n, fwd.head, d + 1) else Option.when(fwd.size > 1 || n == goal)((n, d))

                val next = nb(h).flatMap(findNext(h, _, 1))
                go(next.map(_._1).filter(!seen.contains(_)) ++ t, seen ++ next.map(_._1), next.map((n, d) => WDiEdge(h, n, d)) ++ acc)
            case _ => acc
    
    Graph() ++ go(List(start), Set(start), List()) 

def parseGraph(a: List[String]) =
    val (start, goal) = (Pos(1, 0), Pos(a(0).size - 2, a.size - 1))
    (toGraph(Grid(a.map(_.toList)), start, goal), start, goal)

def task1(a: List[String]): Long = 
    val (g, start, goal) = parseGraph(a)
    val topo = g.topologicalSort.fold(failure => List(), order => order.toList.reverse)
    topo.tail.foldLeft(Map(topo.head -> 0.0))((m, n) => m + (n -> n.outgoing.map(e => e.weight + m(e.targets.head)).max))(g.get(start)).toLong

def task2(a: List[String]): Long = 
    val (g, start, goal) = parseGraph(a)

    // this problem is np hard (reduction from hamilton path)
    // on general graphs, and I can't see any special case
    // in the input.
    // => throw bruteforce at it
    def go(n: g.NodeT, seen: Set[g.NodeT], w: Double): Double =
        val m1 = n.outgoing.filter(e => !seen.contains(e.targets.head)).map(e => go(e.targets.head, seen + e.targets.head, w + e.weight)).maxOption
        val m2 = n.incoming.filter(e => !seen.contains(e.sources.head)).map(e => go(e.sources.head, seen + e.sources.head, w + e.weight)).maxOption
        List(m1, m2).flatMap(a => a).maxOption.getOrElse(if n.outer == goal then w else -1)
    
    val init = g.get(start)
    go(init, Set(init), 0).toLong
[โ€“] [email protected] 1 points 10 months ago

Rust Solution

For Part One I used a depth-first search which took too long for part two. Part Two I created an adjacency list of the junction points while keeping track of the distance to the adjacent nodes at the same time. Then depth-first search through the adjacency list.

[โ€“] [email protected] 1 points 10 months ago (1 children)

Nim

Part 1 was just a simple search. Part 2 looked like it just needed a trivial modification, but with the removal of the one-way tiles, the result I was getting was getting for the example was too large. I switched to a different method of determining the path length, but didn't yet figure out what what I had been doing wrong. Since the search space was now significantly larger, my part 2 code took almost an hour to come up with the answer.

I rewrote part 2 to simplify the maze into a graph with a node for each intersection and for the start and goal tiles, with edge costs equal to the path length between each. This resulted in significantly faster iteration (17 seconds instead of 52 minutes), but didn't actually reduce the search space. I'm assuming there's some clever optimization that can be done here, but I'm not sure what it is.

The rewrite was still getting the wrong answer, though. I eventually figured out that it was including paths that didn't actually reach the goal, as long as they didn't revisit any nodes. I changed my recursive search function to return a large negative result at dead ends, which fixed the issue.

[โ€“] [email protected] 1 points 10 months ago

Python

from .solver import Solver


def _directions_from(char: str) -> list[tuple[int, int]]:
  if char == '.':
    return [(0, 1), (0, -1), (1, 0), (-1, 0)]
  if char == 'v':
    return [(1, 0)]
  if char == '^':
    return [(-1, 0)]
  if char == '<':
    return [(0, -1)]
  if char == '>':
    return [(0, 1)]
  raise ValueError(f'unknown char: {char}')

class Day23(Solver):
  lines: list[str]

  def __init__(self):
    super().__init__(23)

  def presolve(self, input: str):
    self.lines = input.splitlines()

  def _find_longest(self, current: tuple[int, int], visited: set[tuple[int, int]]) -> int|None:
    i, j = current
    if i == len(self.lines) - 1:
      return 0
    visited.add(current)
    options = []
    for di, dj in _directions_from(self.lines[i][j]):
      ni, nj = i + di, j + dj
      if ni < 0 or ni >= len(self.lines) or nj < 0 or nj >= len(self.lines[0]):
        continue
      if self.lines[ni][nj] == '#':
        continue
      if (ni, nj) in visited:
        continue
      result = self._find_longest((ni, nj), visited)
      if result is not None:
        options.append(result)
    visited.remove(current)
    if options:
      return max(options) + 1
    return None

  def solve_first_star(self) -> int:
    start_i = 0
    start_j = self.lines[0].find('.')
    result = self._find_longest((start_i, start_j), set())
    assert result
    return result

  def _find_longest_2(self, current: tuple[int, int],
                      connections: dict[tuple[int, int], list[tuple[int, int, int]]],
                      visited: set[tuple[int, int]]) -> int|None:
    i, j = current
    if i == len(self.lines) - 1:
      return 0
    visited.add(current)
    options = []
    for ni, nj, length in connections[(i, j)]:
      if (ni, nj) in visited:
        continue
      result = self._find_longest_2((ni, nj), connections, visited)
      if result is not None:
        options.append(result + length)
    visited.remove(current)
    if options:
      return max(options)
    return None

  def solve_second_star(self) -> int:
    start_i = 0
    start_j = self.lines[0].find('.')

    stack = [(start_i, start_j)]
    connections = {}
    visited = set()
    while stack:
      edge_i, edge_j = stack.pop()
      i, j = edge_i, edge_j
      path_length = 0
      options = []
      connections[(edge_i, edge_j)] = []
      while True:
        options = []
        path_length += 1
        visited.add((i, j))
        for di, dj in ((0, 1), (0, -1), (1, 0), (-1, 0)):
          ni, nj = i + di, j + dj
          if ni < 0 or ni >= len(self.lines) or nj < 0 or nj >= len(self.lines[0]):
            continue
          if self.lines[ni][nj] == '#':
            continue
          if (ni, nj) in visited:
            continue
          options.append((ni, nj))
        if len(options) == 1:
          i, j = options[0]
        else:
          connections[(edge_i, edge_j)].append((i, j, path_length - 1))
          break
      connections[(i, j)] = [(ni, nj, 1) for ni, nj in options]
      stack.extend(options)

    connections_pairs = list(connections.items())
    for (i, j), connected_nodes in connections_pairs:
      for (ni, nj, d) in connected_nodes:
        if (ni, nj) not in connections:
          connections[(ni, nj)] = [(i, j, d)]
        elif (i, j, d) not in connections[(ni, nj)]:
          connections[(ni, nj)].append((i, j, d))

    result = self._find_longest_2((start_i, start_j), connections, set())
    assert result
    return result