Day 8: Playground

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FAQ

  • janAkali@lemmy.sdf.org
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    18 天前

    Nim

    view code
    type
      AOCSolution[T,U] = tuple[part1: T, part2: U]
      Vec3 = tuple[x,y,z: int]
      Node = ref object
        pos: Vec3
        cid: int
    
    proc dist(a,b: Vec3): float =
      sqrt(float((a.x-b.x)^2 + (a.y-b.y)^2 + (a.z-b.z)^2))
    
    proc solve(input: string, p1_limit: int): AOCSolution[int, int] =
      let boxes = input.splitLines().mapIt:
        let parts = it.split(',')
        let pos = Vec3 (parseInt parts[0], parseInt parts[1], parseInt parts[2])
        Node(pos: pos, cid: -1)
    
      var dists: seq[(float, (Node, Node))]
      for i in 0 .. boxes.high - 1:
        for j in i+1 .. boxes.high:
          dists.add (dist(boxes[i].pos, boxes[j].pos), (boxes[i], boxes[j]))
    
      var curcuits: Table[int, HashSet[Node]]
      var curcuitID = 0
    
      dists.sort(cmp = proc(a,b: (float, (Node, Node))): int = cmp(a[0], b[0]))
    
      for ind, (d, nodes) in dists:
        var (a, b) = nodes
        let (acid, bcid) = (a.cid, b.cid)
    
        if acid == -1 and bcid == -1: # new curcuit
          a.cid = curcuitID
          b.cid = curcuitID
          curcuits[curcuitId] = [a, b].toHashSet
          inc curcuitID
        elif bcid == -1: # add to a
          b.cid = acid
          curcuits[acid].incl b
        elif acid == -1: # add to b
          a.cid = bcid
          curcuits[bcid].incl a
        elif acid != bcid: # merge two curcuits
          for node in curcuits[bcid]:
            node.cid = acid
          curcuits[acid].incl curcuits[bcid]
          curcuits.del bcid
    
        if ind+1 == p1_limit:
          result.part1 = curcuits.values.toseq.map(len).sorted()[^3..^1].prod
    
        if not(acid == bcid and acid != -1): result.part2 = a.pos.x * b.pos.x
    

    Runtime: 364 ms

    Part 1:
    I compute all pairs of Euclidean distances between 3D points, sort them, then connect points into circuits, using, what I think is called a Union‑Find algorithm (circuits grow or merge). After exactly 1000 connections (including redundant ones), I take the three largest circuits and multiply their sizes.

    Part 2:
    While iterating through the sorted connections, I also calculate the product of each pair x‑coordinates. The last product is a result for part 2.

    Problems I encountered while doing this puzzle:

    • I’ve changed a dozen of data structures, before settled on curcuitIDs and ref objects stored in a HashTable (~ 40 min)
    • I did a silly mistake of mutating the fields of an object and then using new fields as keys for the HashTable (~ 20 min)
    • I am stil confused and don’t understand why do elves count already connected junction boxes (~ 40 min, had to look it up, otherwise it would be a lot more)

    Time to solve Part 1: 1 hour 56 minutes
    Time to solve Part 2: 4 minutes

    Full solution at Codeberg: solution.nim

    • gedhrel@lemmy.world
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      17 天前

      Upvote for mentioning union-find. It’s the little algorithm that could, and almost nobody has heard about it.