Advent of Code 2019 - Day 10

sb8244

Author of Real-Time Phoenix

Note: This topic is to talk about Day 10 of the Advent of Code 2019 .

There is a private leaderboard for elixirforum members. You can join it by following this link and entering the following code:

39276-eeb74f9a

16 comments

#2019 #advent-of-code

2 2006 16

2019-12-16 15:30:31 UTC

Showing Posts 1 to 10

Show Best Posts
Show All (oldest first)
Show All (newest first)

sb8244

Author of Real-Time Phoenix

My solution is here. I actually documented the approach in pt1 / pt2 at the top. I won’t paste it here to avoid spoilers, it’s still early on the day.

Post #1

aaronnamba

Part 1 and Part 2 (most days, I clone my Part 1 solution before starting on Part 2, to preserve it as it was).

I kinda like how this one turned out. Less brute-force-y than most of my other solutions, but boy did it take a long time to get there.

Post #2

xfix

gitlab.com/KonradBorowski/advent_of_code_2019

lib/day10.ex

master

defmodule AdventOfCode2019.Day10 do
  @moduledoc """
  Day 10 solutions
  """

  @doc ~S"""
  Solves the first riddle of day 10.

  ## Examples

      iex> AdventOfCode2019.Day10.part1(".#..#\n.....\n#####\n....#\n...##")
      8
      iex> AdventOfCode2019.Day10.part1("......#.#.\n#..#.#....\n..#######.\n.#.#.###..\n.#..#.....\n..#....#.#\n#..#....#.\n.##.#..###\n##...#..#.\n.#....####")
      33
      iex> AdventOfCode2019.Day10.part1("#.#...#.#.\n.###....#.\n.#....#...\n##.#.#.#.#\n....#.#.#.\n.##..###.#\n..#...##..\n..##....##\n......#...\n.####.###.")
      35
      iex> File.read!("inputs/day10.txt") |> AdventOfCode2019.Day10.part1
      214

  """

This file has been truncated. show original

I have no idea about trigonometry. Oh well. Didn’t implement the part with looping, because the task actually didn’t ask for it (look, I can tell that won’t be necessary when my first part solution was bigger than 200).

Post #3

ug0

My solution.

Post #4

jenarvaezg

Day 10 of elixir! My solution

It looks more messy than what I originally expected. Also my solution for part 2 only works on small inputs, I might refactor it a bit

Post #5

sasajuric

Author of Elixir In Action

My solution. Today’s challenge was finicky, but fun! Used a bit of trigonometry (atan2) and gb_trees for the second part.

Post #6

qhwa

Here’s my solution.

My hero functions today are Enum.group_by and :math.atan2.

There’s a trick in part 2. I was about to write an algorithm to find the result. But before start I inspected my data and found 200 is not enough to finish a single round. So the process is super easy.

I’m not sure if this works for others, because not everyone’s input is the same. I guess not because one time after I provided a wrong answer, it told me that my wrong answer is someone else’s right answer.

Post #7

bjorng

Erlang Core Team

Not my day today…

My solution, slightly cleaned up after I got both parts working.

I figured out how to solve part 1 without using any floating point arithmetic.

After some false starts trying to solve part 2, I searched some hints on the Internet, and found out how to use atan2(). I also saw that it didn’t actually seem necessary to go round more than once, since the 200th asteroid was reached before the first round ended, so I didn’t bother implementing going round more than once.

Post #8