I tried to use combinatorial to solve today’s puzzles but failed (my brain burned out 
). In the end I just used brute force with memoization.
Task.async_stream turned out to be very helpful. It both let me handle each line of input concurrently, and allows me to abuse process dictionaries 
defmodule AoC2023.Day12 do
# `input` for both parts are things like
#
# [
# {"???.###", [1,1,3]},
# {".??..??...###.", [1,1,3]},
# ...
# ]
@spec part1([{String.t(), [pos_integer()]}]) :: non_neg_integer()
def part1(input) do
input
|> Task.async_stream(fn {springs, counts} ->
aux(springs, ".", counts)
end, ordered: false)
|> Stream.map(&elem(&1, 1))
|> Enum.sum()
end
@spec part2([{String.t(), [pos_integer()]}]) :: non_neg_integer()
def part2(input) do
input
|> Enum.map(fn {springs, counts} ->
{
List.duplicate(springs, 5) |> Enum.join("?"),
List.duplicate(counts, 5) |> List.flatten()
}
end)
|> part1()
end
@spec aux(
springs :: String.t(),
previous_spring :: String.t(),
counts :: [pos_integer()]
) :: non_neg_integer()
defp aux("", _, []), do: 1
defp aux("", _, [0]), do: 1
defp aux("", _, _), do: 0
defp aux("#" <> _, _, []), do: 0
defp aux("#" <> _, _, [0 | _]), do: 0
defp aux("#" <> rest, _, [h | t]), do: aux(rest, "#", [h - 1 | t])
defp aux("." <> rest, _, []), do: aux(rest, ".", [])
defp aux("." <> rest, "#", [0 | t]), do: aux(rest, ".", t)
defp aux("." <> _, "#", [_ | _]), do: 0
defp aux("." <> rest, ".", counts), do: aux(rest, ".", counts)
defp aux("?" <> rest, "#", []), do: aux(rest, ".", [])
defp aux("?" <> rest, "#", [0 | t]), do: aux(rest, ".", t)
defp aux("?" <> rest, "#", [h | t]), do: aux(rest, "#", [h - 1 | t])
defp aux("?" <> rest, ".", []), do: aux(rest, ".", [])
defp aux("?" <> rest, ".", [0 | t]), do: aux(rest, ".", t)
defp aux("?" <> rest, ".", [h | t]) do
memoized({rest, [h | t]}, fn ->
aux(rest, "#", [h - 1 | t]) + aux(rest, ".", [h | t])
end)
end
defp memoized(key, fun) do
with nil <- Process.get(key) do
fun.() |> tap(&Process.put(key, &1))
end
end
end
