š Christmas Coding with Elixir, Day 1
Advent Of Code 2018
Advent of Code is upon us again! Iām going to attempt to solve each puzzle with Elixir this year and highlight any interesting things Iāve learned.
The Puzzle: Part 1
After feeling like youāve been falling for a few minutes, you look at the deviceās tiny screen. āError: Device must be calibrated before first use. Frequency drift detected. Cannot maintain destination lock.ā Below the message, the device shows a sequence of changes in frequency (your puzzle input). A value like +6 means the current frequency increases by 6; a value like -3 means the current frequency decreases by 3.
For example, if the device displays frequency changes of +1, -2, +3, +1, then starting from a frequency of zero, the following changes would occur:
Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.
In this example, the resulting frequency is 3.Here are other example situations:
+1, +1, +1 results in 3
+1, +1, -2 results in 0
-1, -2, -3 results in -6Starting with a frequency of zero, what is the resulting frequency after all of the changes in frequency have been applied?
Solving It
So part 1 is pretty simple. We need to parse a list of positive and negative integers and then sum them together.
Hereās our input:
And hereās some code to get our answer. Letās break it down line by line.
I like to be able to stream my puzzle inputs into my code, so we start with IO.read/2
ā this reads in from stdin so we can execute the code with
cat input.txt | ./advent1.1.exs
We pipe the string data into String.split/2
, which gives us a list of strings in the form [ā+16ā, ā-15ā, ā-2ā, ...]
.
Next, we pipe this list into Enum.map/2
and pass it the String.to_integer/1
function, which will return a list of parsed integers in the form [16, -15, -2, ...]
.
Now all that remains is to call Enum.sum/1
to sum all those integers together, and a call to IO.inspect/1
to dump out the result.
$ cat input.txt | ./advent1.1.exs
439
The Puzzle: Part 2
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, using the same list of changes above, the device would loop as follows:
Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.
(At this point, the device continues from the start of the list.)
Current frequency 3, change of +1; resulting frequency 4.
Current frequency 4, change of -2; resulting frequency 2, which has already been seen.
In this example, the first frequency reached twice is 2. 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.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.What is the first frequency your device reaches twice?
Solving It
Ok! So now it gets a little trickier. In Ruby, Iād use Enumerable#cycle
to give an infinite stream of frequencies and then use return
to break out early when I found one thatād been seen before.
But Elixir is functional, and early returns are very much an imperative approach. So how to solve it?
Well, after a lot of digging around, I found Enum.reduce_while/3
.
We start the solution in the same way as part 1, only this time we use Stream.cycle/1
to create an infinite stream of cycled frequencies. We pipe this stream into Enum.reduce_while/3
, which takes an enumerable, an accumulator, and a reducer function.
The reducer function either returns a tuple with the first element as :cont
and the second element as the updated accumulator, or it returns :halt
as the first element and the end result as the second element.
Using this, we can maintain a set of frequencies seen before and test for membership ā thatās pretty handy!
$ cat input.txt | ./advent1.2.exs
124645
Summary
So that was fun, and a little trickier than I was expecting for day 1! Iāll be filing away Enum.reduce_while/3
for future reference š
Iāll be adding my Ruby and Elixir solutions to Github if youād like to read further: https://github.com/seanhandley/adventofcode2018
Onwards to Day Two!