Advent of Code 2023 Day 6
Nicky Meuleman
Posted on December 6, 2023
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:
- Race 1 lasts 7 milliseconds. The record distance in this race is 9 millimeters.
- Race 2 lasts 15 milliseconds. The record distance in this race is 40 millimeters.
- 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:
- Hold the button for 2 milliseconds at the start of the race.
- Hold the button for 3 milliseconds at the start of the race.
- Hold the button for 4 milliseconds at the start of the race.
- 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
}
Posted on December 6, 2023
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