this post was submitted on 11 Dec 2023
10 points (77.8% liked)

Advent Of Code

761 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
 

Day 11: Cosmic Expansion

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)
  • Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ , pastebin, or github (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)

FAQ


🔒 Thread is locked until there's at least 100 2 star entries on the global leaderboard

🔓 Unlocked after 9 mins

you are viewing a single comment's thread
view the rest of the comments
[–] janAkali@lemmy.one 2 points 11 months ago* (last edited 11 months ago)

Nim

Part 1 and 2: I solved today's puzzle without expanding the universe. Path in expanded universe is just a path in the original grid + expansion rate times the number of crossed completely-empty lines (both horizontal and vertical). For example, if a single tile after expansion become 5 tiles (rate = +4), original path was 12 and it crosses 7 lines, new path will be: 12 + 4 * 7 = 40.
The shortest path is easy to calculate in O(1) time: abs(start.x - finish.x) + abs(start.y - finish.y).
And to count crossed lines I just check if line is between the start and finish indexes.

Total runtime: 2.5 ms
Puzzle rating: 7/10 Code: day_11/solution.nim
Snippet:

proc solve(lines: seq[string]): AOCSolution[int] =
  let
    galaxies = lines.getGalaxies()
    emptyLines = lines.emptyLines()
    emptyColumns = lines.emptyColumns()

  for gi, g1 in galaxies:
    for g2 in galaxies[gi+1..^1]:
      let path = shortestPathLength(g1, g2)
      let crossedLines = countCrossedLines(g1, g2, emptyColumns, emptyLines)
      block p1:
        result.part1 += path + crossedLines * 1
      block p2:
        result.part2 += path + crossedLines * 999_999