this post was submitted on 13 Dec 2024
18 points (90.9% liked)

Advent Of Code

920 readers
60 users here now

An unofficial home for the advent of code community on programming.dev!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

Solution Threads

M T W T F S S
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 18 20 21 22
23 24 25

Rules/Guidelines

Relevant Communities

Relevant Links

Credits

Icon base by Lorc under CC BY 3.0 with modifications to add a gradient

console.log('Hello World')

founded 1 year ago
MODERATORS
 

Day 13: Claw Contraption

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

you are viewing a single comment's thread
view the rest of the comments
[โ€“] [email protected] 2 points 4 days ago

Rust

This problem is basically a linear system, which can be solved by inverting the 2x2 matrix of button distances. I put some more detail in the comments.

Solution

use std::sync::LazyLock;

use regex::Regex;

#[derive(Debug)]
struct Machine {
    a: (i64, i64),
    b: (i64, i64),
    prize: (i64, i64),
}

impl Machine {
    fn tokens_100(&self) -> i64 {
        for b in 0..=100 {
            for a in 0..=100 {
                let pos = (self.a.0 * a + self.b.0 * b, self.a.1 * a + self.b.1 * b);
                if pos == self.prize {
                    return b + 3 * a;
                }
            }
        }
        0
    }

    fn tokens_inv(&self) -> i64 {
        // If [ab] is the matrix containing our two button vectors: [ a.0 b.0 ]
        //                                                          [ a.1 b.1 ]
        // then prize = [ab] * x, where x holds the number of required button presses
        // for a and b, (na, nb).
        // By inverting [ab] we get
        //
        // x = [ab]โปยน * prize
        let det = (self.a.0 * self.b.1) - (self.a.1 * self.b.0);
        if det == 0 {
            panic!("Irregular matrix");
        }
        let det = det as f64;
        // The matrix [ a b ] is the inverse of [ a.0 b.0 ] .
        //            [ c d ]                   [ a.1 b.1 ]
        let a = self.b.1 as f64 / det;
        let b = -self.b.0 as f64 / det;
        let c = -self.a.1 as f64 / det;
        let d = self.a.0 as f64 / det;
        // Multiply [ab] * prize to get the result
        let na = self.prize.0 as f64 * a + self.prize.1 as f64 * b;
        let nb = self.prize.0 as f64 * c + self.prize.1 as f64 * d;

        // Only integer solutions are valid, verify rounded results:
        let ina = na.round() as i64;
        let inb = nb.round() as i64;
        let pos = (
            self.a.0 * ina + self.b.0 * inb,
            self.a.1 * ina + self.b.1 * inb,
        );
        if pos == self.prize {
            inb + 3 * ina
        } else {
            0
        }
    }

    fn translate(&self, tr: i64) -> Self {
        let prize = (self.prize.0 + tr, self.prize.1 + tr);
        Machine { prize, ..*self }
    }
}

impl From<&str> for Machine {
    fn from(s: &str) -> Self {
        static RE: LazyLock<(Regex, Regex)> = LazyLock::new(|| {
            (
                Regex::new(r"Button [AB]: X\+(\d+), Y\+(\d+)").unwrap(),
                Regex::new(r"Prize: X=(\d+), Y=(\d+)").unwrap(),
            )
        });
        let (re_btn, re_prize) = &*RE;
        let mut caps = re_btn.captures_iter(s);
        let (_, [a0, a1]) = caps.next().unwrap().extract();
        let a = (a0.parse().unwrap(), a1.parse().unwrap());
        let (_, [b0, b1]) = caps.next().unwrap().extract();
        let b = (b0.parse().unwrap(), b1.parse().unwrap());
        let (_, [p0, p1]) = re_prize.captures(s).unwrap().extract();
        let prize = (p0.parse().unwrap(), p1.parse().unwrap());
        Machine { a, b, prize }
    }
}

fn parse(input: String) -> Vec<Machine> {
    input.split("\n\n").map(Into::into).collect()
}

fn part1(input: String) {
    let machines = parse(input);
    let sum = machines.iter().map(|m| m.tokens_100()).sum::<i64>();
    println!("{sum}");
}

const TRANSLATION: i64 = 10000000000000;

fn part2(input: String) {
    let machines = parse(input);
    let sum = machines
        .iter()
        .map(|m| m.translate(TRANSLATION).tokens_inv())
        .sum::<i64>();
    println!("{sum}");
}

util::aoc_main!();

Also on github