In the previous post, we had briefly discuss about the solution of Part 1. It is fairly straightforward. I thought Part 2 is going to be easy too. But man, I was wrong. It is a bit tricky.

My initial solution takes around 17 seconds to compute the answer. I made it 34x faster by changing the data structure. In the end, the improved solution solves the puzzle in just 0.5 seconds.

Code is available at GitHub. For those who just want to know how the solution improved by 34x faster, see here.

Okay, that’s the end of some irrelavant stuff. Let’s jump straight to the puzzle of Part 2.

Part 2

Part 2 still uses the same inputs as Part 1.

-9
+7 +5
-13
+6
...
-23
-46
-27
-11
-75223

This time, the question and scenario are abit different, which make it tricky.

The problem description states that:

You notice that the device repeats the same frequency change list over and over. To calibrate the device, you need to find the first frequency it reaches twice.

For example:

Here are other examples:

    +1, -1 first reaches 0 twice.
    +3, +3, +4, -2, -4 first reaches 10 twice.
    -6, +3, +8, +5, -6 first reaches 5 twice.
    +7, +7, -2, -7, -4 first reaches 14 twice.

Note that your device might need to repeat its list of frequency changes many times before a duplicate frequency is found, and that duplicates might be found while in the middle of processing the list.

And the question is:

What is the first frequency your device reaches twice?

Problem Breakdown

So the question is asking for the first frequency that reaches twice. But, do note that the frequency list can be repeated many times before an asnwer is found.

If after the first interation of the frequency list, there is no duplicate found, the frequency list will go through the second interation again. This only stops until a duplicate frequency found.

To summarize:

  • Find first duplicate of frequency.
  • Repeat the frequency list if no duplicate frequency found. 1
  • The initial frequency is 0.
  • Upon second iteration, the current frequency will be resumed from the previosu iteration. 2

Initial Solution

defmodule Finder do
  def is_duplicate?(frequency, appeared_frequencies) do
    # Check if the frequency is duplicate in appeared_frequencies
    if Enum.member?(appeared_frequencies, frequency) do

      # Return :halt to halt the reduce_while loop
      # and the duplicate
      {:halt, frequency}
    else
      # Update the appeared_frequencies to include the new frequency

      # Prepend [ element_to_add | rest_of_list ] is used since it
      # has a constant time of adding element to list.
      appeared_frequencies = [ frequency | appeared_frequencies ]

      # Return :cont to continue the reduce_while loop
      # with the frequency and updated appeared_frequencies
      {:cont, {frequency, appeared_frequencies}}
    end
  end

  def accumulate_and_find(frequencies, initial, past_frequencies) do
    frequencies
     # The accumulator for reduce_while is:

     #   initial/prev_frequency -> Initial/Previous Frequency,
     #   past_frequencies/appeared_frequencies -> Past Appeared Frequencies /
     #                                            Appeared Frequencies
    |> Enum.reduce_while({initial, past_frequencies},
      fn (x, {prev_frequency, appeared_frequencies})  ->

        # Calculate the new frequency.
        new_frequency = x + prev_frequency

        # Check if the new frequency is a duplicate

        # Halt if it is.
        # Continue the reduce_while if it is not.
        is_duplicate?(new_frequency, appeared_frequencies)
      end)
  end

  def find_first_duplicate_frequency(frequencies, initial, past_frequencies) do
    case accumulate_and_find(frequencies, initial, past_frequencies) do
      # No duplicates found
      {result, appeared_frequencies} ->
        # Recursively call this method
        # by passing in the frequencies again.

        # However, the next iteration `initial` frequency will be the previous
        # iteration frequency, which is the value of `result`.

        # The `appeared_frequencies in the previous iteration will be pass
        # forward to the next iteration too.

        find_first_duplicate_frequency(frequencies, result, appeared_frequencies)

      # Duplicate found
      result -> result # Return duplicate and end recursive call.
    end
  end
end

case File.read("input.txt") do
  {:ok, content} ->
    frequencies = content
                  |> String.split()
                  |> Enum.map(&String.to_integer/1)

    Finder.find_first_duplicate_frequency(frequencies, 0, [0])
    |> IO.inspect
  {:error, _} -> IO.puts "Error opening files"
end

If the comments in the code is not sufficient, here are the more detailed explanation to help you understand the code (hopefully):

  1. First, same as Part 1, we read the content from the file, split it and conver every element into integer first.
  2. Then, we pass it into Finder.find_first_duplicate_frequency/3 method, with the frequencies, initial frequency, and appeared frequencies.
  3. In ind_first_duplicate_frequency, it will delegate the finding duplicate job to accumulate_and_find/3 and check if the return result is a found duplicate or not.
    • If there is no duplicate found, it will call itself recursively, with the same frequencies, to start the next iteration.
    • The initial frequency will be replaced with the last frequency from the previous iteration.
    • The appeared frequencies from the previous iteration will also be pass forward to the next iteration.
  4. In accumulate_and_find/3, we use Map.reduce_while/3 to map through each elements in the frequencies and keep track of the previous frequency and appeared frequencies.
    • reduce_while/3 is used so that we can control when to halt the looping.
    • It accepts an accumulator, which in this case is our { initial, past_frequencies }, which indicates the state we want to keep track of while mapping through the elements
    • It also accepts a function, which in this case, will be used to calculate the new frequency, by new_frequency = x + prev_frequency and check if it is a duplicate using is_duplicate?/2.
  5. In is_duplicate?/2, it basically check if the element already existed by Enum.member/2?.
    • If it exists, it return :halt and the duplicate to the reduce_while code block. Thus, ending the loop.
    • Else, it return :cont with the current frequency and the new appeared frequencies to the reduce_while/3 code block.

Performance:

$ time elixir part-2.ex

real	0m17.047s
user	0m16.978s
sys	0m0.184s

At first, I thought I am having infinity loop and was kind of disappointed as it was my forth time fixing the solution. Then, I jrealized, it just take a long time to compute the answer.

After a few hours, I started to think about how to improve the performance. Then, I realized why it took so long.

Improved Solution

It is beacuse of the use of List. Enum.member?/2 will iterate through every single element until it find the member. So, when there is a new element added to appeared_frequency, it has to loop through additional element in the Enum.member?/2.

Looping through the frequencies itself is already an O(n) operation. With, Enum.member?/2, that’s another O(n) operation.Thus, resulting in a O(n^2) solution.

defmodule Finder do

  def is_duplicate?(frequency, appeared_frequencies) do
    # Use Map.has_key?/2 to check if the frequency already existed
    if Map.has_key?(appeared_frequencies, frequency) do
      {:halt, frequency}
    else
      # Use Map.put/3 to add the new frequency in to the map.
      appeared_frequencies = Map.put(appeared_frequencies, frequency, 0)
      {:cont, {frequency, appeared_frequencies}}
    end
  end

  def accumulate_and_find(frequencies, initial, past_frequencies) do
    frequencies
    |> Enum.reduce_while({initial, past_frequencies},
      fn (x, {prev_frequency, appeared_frequencies})  ->
        new_frequency = x + prev_frequency
        is_duplicate?(new_frequency, appeared_frequencies)
      end)
  end

  def find_first_duplicate_frequency(frequencies, initial, past_frequencies) do
    case accumulate_and_find(frequencies, initial, past_frequencies) do
      {result, appeared_frequencies} ->
        find_first_duplicate_frequency(frequencies, result, appeared_frequencies)
      result -> result
    end
  end
end


case File.read("input.txt") do
  {:ok, content} ->
    frequencies = content
                  |> String.split()
                  |> Enum.map(&String.to_integer/1)

    # Subsitute [0] to %{0 => 0} to represent the appeared frequencies.
    Finder.find_first_duplicate_frequency(frequencies, 0, %{0 => 0})
    |> IO.inspect
  {:error, _} -> IO.puts "Error opening files"
end

The second solution is just to switch the data structure we used to store the appeared_frequencies. Instead of List, we use a Map. Map has a constant look up time O(1) to access the element. Hence, it improve the performance of the solution.

By using the right data structure, with a change of 3 lines of code, we successfully improve the execution speed of our program by 34x faster.

Performance:

$ time elixir part-2.ex

real	0m0.517s
user	0m0.481s
sys	0m0.188s

Day 1 Part 2, Done.


Footnote

  1. At first, I didn’t notice this and submitted the wrong answer.
  2. Then I submitted the wrong answer again, as I didn’t use the last frequency from the previous iteration while starting the next iteration…
  3. I replace the data structure to keep track of the appeared frequencies (which required lookup) from List to Map.