Day 5 of this year’s Advent of Code involves scanning “the chemical composition of [Santa’s new suit] material. We “discover that it is formed by extremely long polymers (one of which is available as your puzzle input).” We have to take these “polymers” (strings of upper and lowercase characters) and work through their reaction.

I’ll just paste from the challenge for us:

For example:

• In aA, a and A react, leaving nothing behind.
• In abBA, bB destroys itself, leaving aA. As above, this then destroys itself, leaving nothing.
• In abAB, no two adjacent units are of the same type, and so nothing happens.
• In aabAAB, even though aa and AA are of the same type, their polarities match, and so nothing happens.

Now, consider a larger example, dabAcCaCBAcCcaDA:

• dabAcCaCBAcCcaDA The first ‘cC’ is removed.
• dabAaCBAcCcaDA This creates ‘Aa’, which is removed.
• dabCBAcCcaDA Either ‘cC’ or ‘Cc’ are removed (the result is the same).
• dabCBAcaDA No further actions can be taken.

After all possible reactions, the resulting polymer contains 10 units. How many units remain after fully reacting the polymer you scanned?

In solving it, I wrote a react function that takes one of these polymer strings as a Vector of characters (Vec<char> is the Rust function signature). The first version of this react function contained two nested loops: the inner loop iterated through the polymer characters one character at a time, looking for characters that “cancel” or react. To test if any two given characters react, I wrote a separate function called do_these_two_chars_cancel.

If we indeed find a canceling pair, we remove them from the vector and break out of the inner for loop to an outer loop, where we begin all over, starting at the first two characters.

You can find my current solution here, but I wanted to walk us through some of the refactoring and optimizations two of the key functions went through.

Let’s start by looking at my original versions of these two functions. This code both compiles successfully and solves the puzzle, but running cargo build --bin day05 --release & time cargo run --bin day05 --release shows that solving part 2 of the challenge (which basically involves running through 26 long polymers) takes my Oryx Pro a whopping 2 minutes and 20 seconds.

fn react(mut p_vec: Vec<char>) -> Vec<char> {
let mut p_vec_len = p_vec.len();
loop {
let mut previous_c: char = p_vec[0];
let mut indexes_to_remove: Vec<usize> = vec![];
for c in 1..p_vec_len {
if do_these_two_chars_cancel(p_vec[c], previous_c) {
// "found a pair: {} and {}", previous_c, p_vec[c]
p_vec.remove(c);
p_vec.remove(c - 1);
break;
} else {
previous_c = p_vec[c];
}
}
if p_vec.len() == p_vec_len {
break;
} else {
p_vec_len = p_vec.len();
}
}
// let s: String = p_vec.into_iter().collect();
// println!("polymer is now {:?}", s);// }
p_vec
}

fn do_these_two_chars_cancel(a: char, b: char) -> bool {
if a.is_uppercase() && b.is_lowercase() && a.to_lowercase().to_string() == b.to_string() {
true
} else if a.is_lowercase() && b.is_uppercase() && a.to_uppercase().to_string() == b.to_string()
{
true
} else {
false
}
}

Can you spot ways to make this code more efficient? Turns out there are quite a few. The first one I implemented was actually submitted as a pull request by a Fediverse friend named Daniel.

## Don’t break so often

He changed the inner loop from a for loop to a while loop. The key change here is that the inner loop does NOT break when it finds a canceling pair – rather, it simply removes them, keeps its iterator index at its same value, so that on the next loop it’ll look at the very next pair.

fn react(mut p_vec: Vec<char>) -> Vec<char> {
let mut p_vec_len = p_vec.len();
loop {
let mut previous_c: char = p_vec[0];
let mut indexes_to_remove: Vec<usize> = vec![];
let mut index = 1;
while index < p_vec_len {
previous_c = p_vec[index - 1];
if do_these_two_chars_cancel(p_vec[index], previous_c) {
p_vec.remove(index);
p_vec.remove(index - 1);
index -= 1;
p_vec_len -= 2;
}
index += 1;
}
break;
}
}
p_vec
}

This led to a huge boost in efficiency… when doing part 2 of the challenge (which effectively requires you react/solve 26 long polymers), Daniel says his run time went from ~30 minutes to ~7 seconds! I didn’t do a speed test myself, but I saw a tremendous improvement.

Of course, this change would likely increase the overall program’s efficiency no matter what programming language I used. But the next two are (relatively) particular to Rust.

## Don’t make a String when sticking with &strs will do

Not wanting to be outdone, I took a look at optimizing the helper function do_these_two_chars_cancel further. The original version of the function has two calls to .to_string(), which I remembered are particularly costly, as Strings in Rust live on the heap. If I could use slices of strings (&str), I figured it’d lead to an increase in efficiency and thus a speed improvement.

When I was first trying to solve the challenge, I had written the seemingly logical if a.to_lowercase == b, however this results in an error:

error[E0369]: binary operation `==` cannot be applied to type `std::char::ToLowercase`
= note: an implementation of `std::cmp::PartialEq` might be missing for `std::char::ToLowercase`

I dropped the .to_string() calls in there out of necessity: if a.is_uppercase() && b.is_lowercase() && a.to_lowercase().to_string() == b.to_string()

But with time to go back, I found a similar method that did work on chars: to_ascii_lowercase(). I also took the time to clean up the conditional logic and got:

fn do_these_two_chars_cancel(a: char, b: char) -> bool {
a.to_ascii_lowercase() == b.to_ascii_lowercase() && a.is_uppercase() == b.is_lowercase()
}

As I wrote in a comment, while two <char>.to_lowercase() can’t be compared for equality, two <char>.to_ascii_lowercase()s can be. Informally, I found that this version of the function was about 9x faster than my original.

In order to be a little less verbose, I ended up using eq_ignore_ascii_case(), so it became just:

fn do_these_two_chars_cancel(a: char, b: char) -> bool {
a.eq_ignore_ascii_case(&b) && a.is_uppercase() == b.is_lowercase()
}

## One drain is faster than two removes

Then, Daniel submitted another pull request, this time with an even simpler change: substituting the two remove() calls with a splice(). (The reason I couldn’t do it in one remove() call is that you can only remove one leemnt at a time with remove.

I ended up using drain(), at the suggestion of another helper, as it fits a bit better semantically.

fn react(mut p_vec: Vec<char>) -> Vec<char> {
let mut p_vec_len = p_vec.len();
loop {
let mut previous_c: char = p_vec[0];
let mut indexes_to_remove: Vec<usize> = vec![];
let mut index = 1;
while index < p_vec_len {
previous_c = p_vec[index - 1];
if do_these_two_chars_cancel(p_vec[index], previous_c) {
p_vec.drain((index-1)..=index);
index -= 1;
p_vec_len -= 2;
}
index += 1;
}
break;
}
}
p_vec
}

## Removing the outer loop

Ever since I saw Daniel’s first pull request showing that we didn’t need to break out of the inner loop, I had a sneaking suspicion that we could do away with the outer loop all together. The trick would be managing the index such that no canceling pairs were missed, paying particular attention to those iterations where a pair was removed.

Below is how it worked out. I needed the clunky index = if index > 1 { index - 1 } else { index }; line to avoid a negative index on the very first pair. Otherwise it’s not terrible.

So here is, finally, the current version of those two functions:

fn react(mut p_vec: Vec<char>) -> Vec<char> {
let mut p_vec_len = p_vec.len();
let mut previous_c: char;
let mut index = 1;
while index < p_vec_len {
previous_c = p_vec[index - 1];
if do_these_two_chars_cancel(p_vec[index], previous_c) {
// Found a pair that react. Let's remove them!
// Use drain rather than remove. Drain is also a little semantically preferable
// to `splice`
p_vec.drain((index - 1)..=index);
p_vec_len -= 2;
// and, if we can, shift c back one
index = if index > 1 { index - 1 } else { index };
} else {
// these two weren't a pair. Move on to the next pair
// by shifting the iterator forward one character
index += 1;
}
}
p_vec
}

fn do_these_two_chars_cancel(a: char, b: char) -> bool {
a.eq_ignore_ascii_case(&b) && a.is_uppercase() == b.is_lowercase()
}

## How much faster?

With this version of the two functions, solving part 2 (again running cargo build --bin day-05 --release && time cargo run --bin day05 --release), on my Oryx Pro takes only about 1.1 seconds, about 127x faster than the original version.