Advent of Code 2023 Day 6

Day 6: Wait For It

https://adventofcode.com/2023/day/6

TL;DR: my solution in Rust

A ferry brings you to a boat race.
The prize for winning the race is a trip to desert island, you want to go there, because you heard some sand is needed to fix the snow issue.
A island named Desert Island will surely have a lot of that annoying, course, stuff.

This race is slightly unusual, the amount of time everyone gets is fixed.
Whoever gets the furthest in that time wins.

Today's input is a report of previous races.

An example input looks like this:

Time:      7  15   30
Distance:  9  40  200

It maps time-limits of races to the current furthest distance.

This document describes three races:

1. Race 1 lasts 7 milliseconds. The record distance in this race is 9 millimeters.
2. Race 2 lasts 15 milliseconds. The record distance in this race is 40 millimeters.
3. Race 3 lasts 30 milliseconds. The record distance in this race is 200 millimeters.

The boats are toy boats.

Before they start moving, you have to charge them by holding a button.
The longer you charge them, the larger the boat's speed.

• The moment you start charging, the time starts
• The moment you stop charging, the boat starts moving (and doesn't lose speed, apparently they were made by a physics teacher elf that assumed a frictionless vaccuum)

Part 1

For each race, determine the amount of ways you can beat the current record.

For the first race in the example (that lasts 7 milliseconds), there are 4 ways to beat the record of 9 millimeters:

1. Hold the button for 2 milliseconds at the start of the race.
2. Hold the button for 3 milliseconds at the start of the race.
3. Hold the button for 4 milliseconds at the start of the race.
4. Hold the button for 5 milliseconds at the start of the race.

The question asks you to multiply the number of ways you can win each race.

Parsing

Getting 2 lists of numbers, and combining them.

For my own convenience, I used a Race structure to represent each race, so I don't confuse myself by accessing things by index.

struct Race {
    time: u32,
    dist: u32,
}

I chose to create a list of times and a list of distances, then zip those 2 lists together to get a list of races.

let (time, dist) = input.split_once("\n").unwrap();
let time = time
    .strip_prefix("Time: ")
    .unwrap()
    .split_whitespace()
    .map(|s| s.parse::<u32>().unwrap());
let dist = dist
    .strip_prefix("Distance: ")
    .unwrap()
    .split_whitespace()
    .map(|s| s.parse::<u32>().unwrap());
let races = time.zip(dist).map(|(time, dist)| Race { time, dist });

Code

The logic then loops over each race and determines how many ways I can win for each one.
At the end, those counts are multiplied.

For every race:
I determined the distance I would travel for every possible time I held the button.
If the resulting final distance was greater than the current record, I won.

struct Race {
    time: u32,
    dist: u32,
}

pub fn part_1(input: &str) -> usize {
    let (time, dist) = input.split_once("\n").unwrap();
    let time = time
        .strip_prefix("Time: ")
        .unwrap()
        .split_whitespace()
        .map(|s| s.parse::<u32>().unwrap());
    let dist = dist
        .strip_prefix("Distance: ")
        .unwrap()
        .split_whitespace()
        .map(|s| s.parse::<u32>().unwrap());
    let races = time.zip(dist).map(|(time, dist)| Race { time, dist });

    races
        .map(|race| {
            (0..=race.time)
                .map(|elapsed| {
                    let speed = elapsed;
                    speed * (race.time - elapsed)
                })
                .filter(|&dist| dist > race.dist)
                .count()
        })
        .product::<usize>()
}

Part 2

It turns out the input had really bad keming.
Each line is one number.

The question stays the same, but this time there is only 1 race.

Parsing

For both time, and distance, I made a list of single digits.
For each digit I encountered, I concatenate it to the end of the current number with math.

In the example input:

Time:      7  15   30
Distance:  9  40  200

For time:

• The first digit is 7, the current number becomes 7
• The second digit is 1, the current number becomes 71
• The third digit is 5, the current number becomes 715
• The fourth digit is 3, the current number becomes 7153
• The final digit is 0, the current number becomes 71530

let (time, dist) = input.split_once("\n").unwrap();
let race_time = time
    .strip_prefix("Time: ")
    .unwrap()
    .chars()
    .filter_map(|c| c.to_digit(10))
    .fold(0u64, |curr, digit| curr * 10 + digit as u64);
let race_dist = dist
    .strip_prefix("Distance: ")
    .unwrap()
    .chars()
    .filter_map(|c| c.to_digit(10))
    .fold(0u64, |curr, digit| curr * 10 + digit as u64);

These number are BIG, really big.
That's why I stored them as 64 bit integers.

Option 1: brute-force

The code from part 1 can largely be reused.

And joy, oh joy, I could even remove some code!

Code

pub fn part_2(input: &str) -> usize {
    let (time, dist) = input.split_once("\n").unwrap();
    let race_time = time
        .strip_prefix("Time: ")
        .unwrap()
        .chars()
        .filter_map(|c| c.to_digit(10))
        .fold(0u64, |curr, digit| curr * 10 + digit as u64);
    let race_dist = dist
        .strip_prefix("Distance: ")
        .unwrap()
        .chars()
        .filter_map(|c| c.to_digit(10))
        .fold(0u64, |curr, digit| curr * 10 + digit as u64);

    (0..=race_time)
        .map(|elapsed| {
            let speed = elapsed;
            speed * (race_time - elapsed)
        })
        .filter(|&dist| dist > race_dist)
        .count()
}

Option 2: math

This question was describing a quadratic graph.
The answer is the difference between the two intersections with the y-axis on that graph.

Time to dust of the highschool math!

The equation we used to determine the distance we travel:

x * (time - x) = dist

Written differently, that looks more familiar:

y = x^2 - time*x + dist

To find the 2 intersections of the second degree function with the x-axis, we set y to 0 and solve for x:

0 = x^2 - time*x + dist

A refresher on the quadratic equation from Khan Academy.

x1, x2 = (-b +- SQRT(b^2 - 4*a*c)) / (2 * a)

The answer is the difference between the two points where the graph intersects with the x-axis.

I can only hold the button for an integer amount, so I need to round up the lowest value, and round down the highest value.

Then add 1 to the difference between those 2, because I want to include that last point where I can win the race.

Code

pub fn part_two_math(input: &str) -> u32 {
    let (time, dist) = input.split_once("\n").unwrap();
    let race_time = time
        .strip_prefix("Time: ")
        .unwrap()
        .chars()
        .filter_map(|c| c.to_digit(10))
        .fold(0u64, |curr, digit| curr * 10 + digit as u64);
    let race_dist = dist
        .strip_prefix("Distance: ")
        .unwrap()
        .chars()
        .filter_map(|c| c.to_digit(10))
        .fold(0u64, |curr, digit| curr * 10 + digit as u64);

    // time to dust off the math I learned in high school, a refresher: https://youtu.be/i7idZfS8t8w?si=FHnYI6--op32fkdk
    // x * (time - x) = dist
    // y = x^2 - time*x + dist
    // find the 2 intersections of the second degree function with the x-axis
    // so we set y to 0 and solve for x
    // 0 = x^2 - time*x + dist

    // The answer is the difference between the two points the graph intersects with the x-axis
    // x1 = (-b - SQRT(b^2 - 4*a*c)) / (2 * a)
    // x2 = (-b + SQRT(b^2 - 4*a*c)) / (2 * a)
    let a = 1.0;
    let b = 0.0 - race_time as f64;
    let c = race_dist as f64;

    let x1 = ((0.0 - b) - (b.powf(2.0) - (4.0 * a * c)).sqrt()) / (2.0 * a);
    let x2 = ((0.0 - b) + (b.powf(2.0) - (4.0 * a * c)).sqrt()) / (2.0 * a);

    // Because you can only hold the button for integer amounts, round up the lower value and round down the upper value
    // And add 1 since |x2 - x1| gives you the amount from x1 to below x2 but we want to include x2.
    let lower_bound = x1.ceil() as u32;
    let upper_bound = x2.floor() as u32 + 1;

    upper_bound - lower_bound
}

Final code

struct Race {
    time: u32,
    dist: u32,
}

pub fn part_1(input: &str) -> usize {
    let (time, dist) = input.split_once("\n").unwrap();
    let time = time
        .strip_prefix("Time: ")
        .unwrap()
        .split_whitespace()
        .map(|s| s.parse::<u32>().unwrap());
    let dist = dist
        .strip_prefix("Distance: ")
        .unwrap()
        .split_whitespace()
        .map(|s| s.parse::<u32>().unwrap());
    let races = time.zip(dist).map(|(time, dist)| Race { time, dist });

    races
        .map(|race| {
            (0..=race.time)
                .map(|elapsed| {
                    let speed = elapsed;
                    speed * (race.time - elapsed)
                })
                .filter(|&dist| dist > race.dist)
                .count()
        })
        .product::<usize>()
}

pub fn part_2(input: &str) -> u32 {
    let (time, dist) = input.split_once("\n").unwrap();
    let race_time = time
        .strip_prefix("Time: ")
        .unwrap()
        .chars()
        .filter_map(|c| c.to_digit(10))
        .fold(0u64, |curr, digit| curr * 10 + digit as u64);
    let race_dist = dist
        .strip_prefix("Distance: ")
        .unwrap()
        .chars()
        .filter_map(|c| c.to_digit(10))
        .fold(0u64, |curr, digit| curr * 10 + digit as u64);

    let a = 1.0;
    let b = 0.0 - race_time as f64;
    let c = race_dist as f64;

    let x1 = ((0.0 - b) - (b.powf(2.0) - (4.0 * a * c)).sqrt()) / (2.0 * a);
    let x2 = ((0.0 - b) + (b.powf(2.0) - (4.0 * a * c)).sqrt()) / (2.0 * a);

    let lower_bound = x1.ceil() as u32;
    let upper_bound = x2.floor() as u32 + 1;

    upper_bound - lower_bound
}