My beginner’s solution, Day 14 part 1. Parabolic Reflector Dish

defmodule Day14 do
def part1(input), do:
  input
  |> String.split("\n")
  |> Enum.map(&String.graphemes/1)
  |> List.zip
  |> Enum.reduce(0, fn column, acc -> column
       |> Tuple.to_list
       |> Enum.chunk_by(&(&1=="#"))
       |> Enum.map(&Enum.sort(&1, :desc))
       |> List.flatten
       |> Enum.reverse
       |> Enum.with_index(1)
       |> then(&(for {"O", n} <- &1, reduce: 0 do acc -> acc + n end))
       |> Kernel.+(acc)
     end)
end

Not proud of this solution, although runs under 3 seconds for part two.
A few comments:

• This is probably the worst bit: I did not specifically look for when the repeated cycle begins, I just tried a few numbers for the length of the initial run. The point is that as long as it’s enough to get into the cycles, it will be good enough to find the cycle length. The upside is that I only need two grids at any given time.
• As others, just one direction of tilting (for me the easiest seemed “to the left”), and use grid transformations to get the other directions. This could definitely be optimised.

defmodule Main do
  def run() do
    get_input()
    |> Enum.map(&String.to_charlist/1)
    # |> solve1()
    |> solve2()
	end

  def get_input() do
    # "testinput14"
    "input14"
    |> File.read!()
    |> String.trim()
    |> String.split("\n")
  end

  def transpose(ls) do
    ls |> List.zip() |> Enum.map(&Tuple.to_list/1)
  end

  def flip_lr(ls) do
    ls |> Enum.map(&Enum.reverse/1)
  end

  def flip_ud(ls) do
    ls |> Enum.reverse()
  end

  def tilt_row_left(l) do
    (l ++ ~c"#")
    |> Enum.reduce({~c"", ~c"", ~c""}, fn c, {lsf, os, ds} ->
          # state: { line_so_far, accumulated_O_s, accumulated_dots }
          case c do
            ?. -> {lsf, os, ds ++ ~c"."}
            ?O -> {lsf, os ++ ~c"O", ds}
            ?# -> {lsf ++ os ++ ds ++ ~c"#", ~c"", ~c""}
          end
       end)
    |> elem(0)
    |> Enum.drop(-1)
  end

  def count_os(l) do
    l |> Enum.filter(fn c -> c == ?O end) |> Enum.count()
  end

  def value(ls) do
    ls
    |> Enum.map(&count_os/1)
    |> Enum.reverse()
    |> Enum.with_index(1)
    |> Enum.map(fn {n,i} -> n*i end)
    |> Enum.sum()
  end
  
  def solve1(ls) do
    ls
    # |> IO.inspect(width: 20)
    |> transpose()
    |> Enum.map(&tilt_row_left/1)
    |> transpose()
    # |> IO.inspect(width: 20)
    |> value()
  end

  def cycle(ls) do
    ls |> transpose() |> Enum.map(&tilt_row_left/1)
    |> transpose() |> Enum.map(&tilt_row_left/1) # after N,W, oriented orig
    |> transpose() |> flip_lr() |> Enum.map(&tilt_row_left/1)
    |> transpose() |> flip_lr() |> Enum.map(&tilt_row_left/1)
    |> flip_ud() |> flip_lr()
  end

  def run_cycles(ls, n) do
    1..n |> Enum.reduce(ls, fn _, gg -> cycle(gg) end)
  end

  def solve2(ls) do
    initial = 150
    ee = run_cycles(ls,initial)
    period = 1 .. 1_000
              |> Enum.reduce_while(ee, fn n, gg ->
                  if (ng = cycle(gg)) == ee do {:halt, n} else {:cont, ng} end
                end)
    run_cycles(ee, rem(1_000_000_000 - initial,period))
    |> value()
  end
  
end

:timer.tc(&Main.run/0)
|> IO.inspect()

My solution completes part 2 in less than a second but I have the same feeling that the tilt could be improved.

I lost so much time with wrong scores until I decided to print all scores in the loop and see that 64 was never coming. This was because my scoring function works with northbound rows but each cycle leaves the platform in eastbound rows.

So I had to add that final rotate() before scoring and it was fine:

    rows_loop_start
    |> apply_cycles(cycles_left)
    |> rotate()
    |> score()

But before I lost like 30 minutes to re-learn the concepts of division, multiplication, remainders and all… My 2nd grade teacher would be proud.

https://github.com/lud/adventofcode/blob/main/lib/solutions/2023/day14.ex