Advent of Code 2020

I was not planning on doing anything specific this December, but a lot of folks around me (on Twitter, at work) have chosen this Advent of Code to pick up Rust, and I’ve got big FOMO energy, so, let’s see where this goes.

I’ll be doing all of these on Linux, so there may be some command-line tools involved, but don’t worry about them - the code itself should run on all platforms no problem.

Day 2, Day 2! Woo!

The Advent of Code 2020, Day 2 problem talks about passwords. Sounds familiar.

Basically, our input looks like this:

1-3 a: abcde 1-3 b: cdefg 2-9 c: ccccccccc

Each line contains a “password policy” and a “password”. For the first line, the policy is that the password must contain between 1 and 3 (inclusive) times the letter “a”.

Hello all, and welcome back to Advent of Code 2020, featuring Cool Bear.

Hey y’all!

Let’s get right to it.

The problem statement for Day 3 is as follows: we’re given a map, that looks like this:

..##....... #...#...#.. .#....#..#. ..#.#...#.# .#...##..#. ..#.##..... .#.#.#....# .#........# #.##...#... #...##....# .#..#...#.#

And we imagine that it repeats infinitely to the right, like so:

It’s time for Day 4 of the Advent of Code 2020!

Now, I’ve already had a look at the problem statement, at least for part 1, and I’m not particularly excited.

But it will allow me to underline some of the points I’ve recently been *trying to make about types and correctness.

The problem is to parse passports, with fields like these:

Time for another day of Advent of Code 2020.

For Day 5, we’re going to have to do…

So there’s an airline that uses binary space partitioning when referring to seats - there’s 128 rows and 8 columns. The first 7 characters are either F (Front, for the lower half) and B (back, for the upper half), and the last 3 are L (Left, for the lower half) or R (Right, for the upper half).

The end of Advent of Code 2020 is fast approaching, and we’re nowhere near done. Time to do Day 6!

The problem statement here is a little contrived, as uh, as the days that came before it, but that won’t stop us.

Basically, the input looks like this:

abc a b c ab ac a a a a b

Each line represents one person, and “groups of persons” are separated by blank lines.

Another day, another Advent of Code 2020 problem.

That one seems fun! For some nerdy values of fun.

Our input is a set of rules:

light red bags contain 1 bright white bag, 2 muted yellow bags. dark orange bags contain 3 bright white bags, 4 muted yellow bags. bright white bags contain 1 shiny gold bag. muted yellow bags contain 2 shiny gold bags, 9 faded blue bags. shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags. dark olive bags contain 3 faded blue bags, 4 dotted black bags. vibrant plum bags contain 5 faded blue bags, 6 dotted black bags. faded blue bags contain no other bags. dotted black bags contain no other bags.

Time for another Advent of Code 2020 problem!

That one sounds like it’s going to be fun. Our input is pretty much assembly, like this:

nop +0 acc +1 jmp +4 acc +3 jmp -3 acc -99 acc +1 jmp -4 acc +6

So, the first thing we’re going to do is write down some types.

There’s more than one way to approach this problem, but let’s go with this:

#[derive(Debug, Clone, Copy)] enum InstructionKind { Nop, Acc, Jmp, } #[derive(Debug, Clone, Copy)] struct Instruction { kind: InstructionKind, operand: isize, } type Program = Vec<Instruction>;

Day 9’s problem statement is convoluted - the “ah maybe that’s why I don’t usually do Advent of Code” kind of convoluted, but let’s give it a go anyway.

So, we have a series of numbers, like so:

35 20 15 25 47 40 62 55 65 95 102 117 150 182 127 219 299 277 309 576

And uh the first N numbers are a “preamble” and every number that comes after that must be the sum of any two of the numbers that come before it.

Day, 10! Day, 10!

Okay, Day 10.

Again, the problem statement is very confusing - but what it all boils down to is this. We have a list of numbers:

16 10 15 5 1 11 7 19 6 12 4

To which we need to add 0 and whatever the maximum was, plus three:

16 10 15 5 1 11 7 19 6 12 4 0 22

From there on, if we take them in order, we’ll have gaps of 1 and gaps of 3:

Another day, another problem.

This time the problem looks suspiciously like Conway’s Game of Life, or, I guess, any old Cellular automaton.

We have a map like so:

L.LL.LL.LL LLLLLLL.LL L.L.L..L.. LLLL.LL.LL L.LL.LL.LL L.LLLLL.LL ..L.L..... LLLLLLLLLL L.LLLLLL.L L.LLLLL.LL

And for each iteration:

• L symbols turn into # if there’s no # in any of the 8 adjacent cells

Time for the Day 12 problem!

In this problem, we have a ship. And we have navigation instructions:

• Action N means to move north by the given value.
• Action S means to move south by the given value.
• Action E means to move east by the given value.
• Action W means to move west by the given value.
• Action L means to turn left the given number of degrees.
• Action R means to turn right the given number of degrees.

In the Day 13 problem, we’re trying to take the bus.

Our input looks like this:

939 7,13,x,x,59,x,31,19

The first line indicates the earliest minute we can leave from the bus terminal, and the second line indicates the “identifier” of the buses that are active.

Each bus departs every “bus ID” minutes - bus 7 leaves at minute 0, minute 7, minute 14, minute 21, etc. The question is: which bus can we take first (apparently they either all go to the same destination, or we don’t really care where we’re going), and how long do we have to wait for it?

It’s time for the Day 14 problem!

After the hassle that was Day 13, I hope this time we’ll have a relatively chill time. And, at least for Part 1, that is true.

Our input looks something like this:

mask = XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X mem[8] = 11 mem[7] = 101 mem[8] = 0

mem is our memory. Our addresses are 36-bit wide, but as you’ll see, that doesn’t matter much.