Advent of Code: Day 10

Helpers

defmodule Helpers do
  @spec read_file_contents(String.t()) :: binary()
  def read_file_contents(filename) do
    file_path = "/Users/charlie/github.com/charlieroth/advent-of-code/2022/#{filename}"

    case File.read(file_path) do
      {:ok, contents} -> contents
      {:error, _} -> raise("Failed to read file contents")
    end
  end
end

Part 01: Cathode-Ray Tube

It seems to be some kind of cathode-ray tube screen and simple CPU that are both driven by a precise clock circuit. The clock circuit ticks at a constant rate; each tick is called a cycle.

Start by figuring out the signal being sent by the CPU. The CPU has a single register, X, which starts with the value 1. It supports only two instructions:

• addx V takes two cycles to complete. After two cycles, the X register is increased by the value V. (V can be negative.)
• noop takes one cycle to complete. It has no other effect.

The CPU uses these instructions in a program (your puzzle input) to, somehow, tell the screen what to draw.

Consider the following small program:

noop
addx 3
addx -5

Execution of this program proceeds as follows:

• At the start of the first cycle, the noop instruction begins execution. During the first cycle, X is 1. After the first cycle, the noop instruction finishes execution, doing nothing.
• At the start of the second cycle, the addx 3 instruction begins execution. During the second cycle, X is still 1.
• During the third cycle, X is still 1. After the third cycle, the addx 3 instruction finishes execution, setting X to 4.
• At the start of the fourth cycle, the addx -5 instruction begins execution. During the fourth cycle, X is still 4.
• During the fifth cycle, X is still 4. After the fifth cycle, the addx -5 instruction finishes execution, setting X to -1.

Maybe you can learn something by looking at the value of the X register throughout execution. For now, consider the signal strength (the cycle number multiplied by the value of the X register) during the 20th cycle and every 40 cycles after that (that is, during the 20th, 60th, 100th, 140th, 180th, and 220th cycles).

Example input:

addx 15
addx -11
addx 6
addx -3
addx 5
addx -1
addx -8
addx 13
addx 4
noop
addx -1
addx 5
addx -1
addx 5
addx -1
addx 5
addx -1
addx 5
addx -1
addx -35
addx 1
addx 24
addx -19
addx 1
addx 16
addx -11
noop
noop
addx 21
addx -15
noop
noop
addx -3
addx 9
addx 1
addx -3
addx 8
addx 1
addx 5
noop
noop
noop
noop
noop
addx -36
noop
addx 1
addx 7
noop
noop
noop
addx 2
addx 6
noop
noop
noop
noop
noop
addx 1
noop
noop
addx 7
addx 1
noop
addx -13
addx 13
addx 7
noop
addx 1
addx -33
noop
noop
noop
addx 2
noop
noop
noop
addx 8
noop
addx -1
addx 2
addx 1
noop
addx 17
addx -9
addx 1
addx 1
addx -3
addx 11
noop
noop
addx 1
noop
addx 1
noop
noop
addx -13
addx -19
addx 1
addx 3
addx 26
addx -30
addx 12
addx -1
addx 3
addx 1
noop
noop
noop
addx -9
addx 18
addx 1
addx 2
noop
noop
addx 9
noop
noop
noop
addx -1
addx 2
addx -37
addx 1
addx 3
noop
addx 15
addx -21
addx 22
addx -6
addx 1
noop
addx 2
addx 1
noop
addx -10
noop
noop
addx 20
addx 1
addx 2
addx 2
addx -6
addx -11
noop
noop
noop

The interesting signal strengths can be determined as follows:

• During the 20th cycle, register X has the value 21, so the signal strength is 20 * 21 = 420. (The 20th cycle occurs in the middle of the second addx -1, so the value of register X is the starting value, 1, plus all of the other addx values up to that point: 1 + 15 - 11 + 6 - 3 + 5 - 1 - 8 + 13 + 4 = 21.)
• During the 60th cycle, register X has the value 19, so the signal strength is 60 * 19 = 1140.
• During the 100th cycle, register X has the value 18, so the signal strength is 100 * 18 = 1800.
• During the 140th cycle, register X has the value 21, so the signal strength is 140 * 21 = 2940.
• During the 180th cycle, register X has the value 16, so the signal strength is 180 * 16 = 2880.
• During the 220th cycle, register X has the value 18, so the signal strength is 220 * 18 = 3960.

The sum of these signal strengths is 13140.

Find the signal strength during the 20th, 60th, 100th, 140th, 180th, and 220th cycles. What is the sum of these six signal strengths?

defmodule PartOne do
  def solution(input) do
    cycle_map =
      input
      |> parse()
      |> start()

    20..220//40
    |> Enum.to_list()
    |> Enum.reduce(0, fn cycle, signal_strength ->
      signal_strength + Map.get(cycle_map, cycle) * cycle
    end)
  end

  defp parse(input) do
    input
    |> String.split("\n", trim: true)
    |> Enum.map(&String.split(&1, " ", trim: true))
    |> Enum.map(fn line ->
      case length(line) do
        1 -> {:noop}
        2 -> {:addx, String.to_integer(Enum.at(line, 1))}
      end
    end)
  end

  defp start(instructions) do
    %{cycles: cycles} =
      instructions
      |> Enum.reduce(%{cycle: 1, reg: 1, cycles: []}, fn instruction,
                                                         %{cycle: cycle, reg: reg, cycles: cycles} =
                                                           state ->
        case instruction do
          {:noop} ->
            %{state | cycle: cycle + 1, cycles: [{cycle + 1, reg} | cycles]}

          {:addx, value} ->
            new_cycles = [{cycle + 1, reg} | cycles]
            new_cycles = [{cycle + 2, value + reg} | new_cycles]
            %{state | cycle: cycle + 2, reg: value + reg, cycles: new_cycles}
        end
      end)

    Map.new(cycles)
  end
end

Helpers.read_file_contents("inputs/Day10.txt")
|> PartOne.solution()