Gobbel2000

joined 1 year ago
[โ€“] Gobbel2000@programming.dev 2 points 7 months ago (5 children)

Rust

First part is solved by making a regex of the available towels, like ^(r|wr|bg|bwu|rb|gb|br)*$ for the example. If a design matches it, then it can be made. This didn't work for the second part, which is done using recursion and memoization instead. Again, it was quite surprising to see such a high solution number. 32 bits were not enough (thanks, debug mode overflow detection).

Solution

use regex::Regex;
use rustc_hash::FxHashMap;

fn parse(input: &str) -> (Vec<&str>, Vec<&str>) {
    let (towels, designs) = input.split_once("\n\n").unwrap();
    (towels.split(", ").collect(), designs.lines().collect())
}

fn part1(input: String) {
    let (towels, designs) = parse(&input);
    let pat = format!("^({})*$", towels.join("|"));
    let re = Regex::new(&pat).unwrap();
    let count = designs.iter().filter(|d| re.is_match(d)).count();
    println!("{count}");
}

fn n_arrangements<'a>(
    design: &'a str,
    towels: &[&str],
    cache: &mut FxHashMap<&'a str, u64>,
) -> u64 {
    if design.is_empty() {
        return 1;
    }
    if let Some(n) = cache.get(design) {
        return *n;
    }
    let n = towels
        .iter()
        .filter(|t| design.starts_with(*t))
        .map(|t| n_arrangements(&design[t.len()..], towels, cache))
        .sum();
    cache.insert(design, n);
    n
}

fn part2(input: String) {
    let (towels, designs) = parse(&input);
    let sum: u64 = designs
        .iter()
        .map(|d| n_arrangements(d, &towels, &mut FxHashMap::default()))
        .sum();
    println!("{sum}");
}

util::aoc_main!();

Also on github

[โ€“] Gobbel2000@programming.dev 2 points 7 months ago

Rust

Naive approach running BFS after every dropped byte after 1024. Still runs in 50ms. This could be much optimized by using binary search to find the first blocked round and using A* instead of BFS, but I didn't feel like doing more today.

Solution

use std::collections::VecDeque;

use euclid::{default::*, vec2};

fn parse(input: &str) -> Vec<Point2D<i32>> {
    input
        .lines()
        .map(|l| {
            let (x, y) = l.split_once(',').unwrap();
            Point2D::new(x.parse().unwrap(), y.parse().unwrap())
        })
        .collect()
}

const BOUNDS: Rect<i32> = Rect::new(Point2D::new(0, 0), Size2D::new(71, 71));
const START: Point2D<i32> = Point2D::new(0, 0);
const TARGET: Point2D<i32> = Point2D::new(70, 70);
const N_BYTES: usize = 1024;
const DIRS: [Vector2D<i32>; 4] = [vec2(1, 0), vec2(0, 1), vec2(-1, 0), vec2(0, -1)];

fn adj(
    field: &[[bool; BOUNDS.size.width as usize]],
    v: Point2D<i32>,
) -> impl Iterator<Item = Point2D<i32>> + use<'_> {
    DIRS.iter()
        .map(move |&d| v + d)
        .filter(|&next| BOUNDS.contains(next) && !field[next.y as usize][next.x as usize])
}

fn find_path(field: &[[bool; BOUNDS.size.width as usize]]) -> Option<u32> {
    let mut seen = [[false; BOUNDS.size.width as usize]; BOUNDS.size.height as usize];
    let mut q = VecDeque::from([(START, 0)]);
    seen[START.y as usize][START.x as usize] = true;
    while let Some((v, dist)) = q.pop_front() {
        for w in adj(field, v) {
            if w == TARGET {
                return Some(dist + 1);
            }
            if !seen[w.y as usize][w.x as usize] {
                seen[w.y as usize][w.x as usize] = true;
                q.push_back((w, dist + 1));
            }
        }
    }
    None
}

fn part1(input: String) {
    let bytes = parse(&input);
    let mut field = [[false; BOUNDS.size.width as usize]; BOUNDS.size.height as usize];
    for b in &bytes[..N_BYTES] {
        field[b.y as usize][b.x as usize] = true;
    }
    println!("{}", find_path(&field).unwrap());
}

fn part2(input: String) {
    let bytes = parse(&input);
    let mut field = [[false; BOUNDS.size.width as usize]; BOUNDS.size.height as usize];
    for (i, b) in bytes.iter().enumerate() {
        field[b.y as usize][b.x as usize] = true;
        // We already know from part 1 that below N_BYTES there is a path
        if i > N_BYTES && find_path(&field).is_none() {
            println!("{},{}", b.x, b.y);
            break;
        }
    }
}

util::aoc_main!();

Also on github

[โ€“] Gobbel2000@programming.dev 2 points 7 months ago (1 children)

Rust

First part was straightforward (the divisions are actually just right shifts), second part not so much. I made some assumptions about the input program, namely that in the end register 8 is divided by 8, then an output is made, then everything starts from the beginning again (if a isn't 0). I found that the output always depends on at most 10 bits of a, so I ran through all 10-bit numbers and grouped them by the first generated output. At that point it's just a matter of chaining these 10-bit numbers from the correct groups so that they overlap on 7 bits. The other 3 bits are consumed each round.

Solution

use rustc_hash::FxHashMap;

fn parse(input: &str) -> Option<Program> {
    let mut lines = input.lines();
    let a = lines.next()?.split_once(": ")?.1.parse().ok()?;
    let b = lines.next()?.split_once(": ")?.1.parse().ok()?;
    let c = lines.next()?.split_once(": ")?.1.parse().ok()?;
    lines.next()?;
    let program = lines
        .next()?
        .split_once(": ")?
        .1
        .split(',')
        .map(|s| s.parse())
        .collect::<Result<Vec<u8>, _>>()
        .ok()?;
    Some(Program {
        a,
        b,
        c,
        out: vec![],
        program,
        ip: 0,
    })
}

#[derive(Debug, Clone, Default)]
struct Program {
    a: u64,
    b: u64,
    c: u64,
    out: Vec<u8>,
    program: Vec<u8>,
    ip: usize,
}

impl Program {
    fn run(&mut self) {
        while self.step() {}
    }

    // Returns true if a step was taken, false if it halted
    fn step(&mut self) -> bool {
        let Some(&[opcode, operand]) = &self.program.get(self.ip..self.ip + 2) else {
            return false;
        };
        self.ip += 2;
        match opcode {
            0 => self.adv(self.combo(operand)),
            1 => self.bxl(operand),
            2 => self.bst(self.combo(operand)),
            3 => self.jnz(operand),
            4 => self.bxc(),
            5 => self.out(self.combo(operand)),
            6 => self.bdv(self.combo(operand)),
            7 => self.cdv(self.combo(operand)),
            _ => panic!(),
        }
        true
    }

    fn combo(&self, operand: u8) -> u64 {
        match operand {
            0..=3 => operand as u64,
            4 => self.a,
            5 => self.b,
            6 => self.c,
            _ => unreachable!(),
        }
    }

    fn adv(&mut self, x: u64) {
        self.a >>= x
    }

    fn bxl(&mut self, x: u8) {
        self.b ^= x as u64
    }

    fn bst(&mut self, x: u64) {
        self.b = x % 8
    }

    fn jnz(&mut self, x: u8) {
        if self.a != 0 {
            self.ip = x as usize
        }
    }

    fn bxc(&mut self) {
        self.b ^= self.c
    }

    fn out(&mut self, x: u64) {
        self.out.push((x % 8) as u8)
    }

    fn bdv(&mut self, x: u64) {
        self.b = self.a >> x
    }

    fn cdv(&mut self, x: u64) {
        self.c = self.a >> x
    }
}

fn part1(input: String) {
    let mut program = parse(&input).unwrap();
    program.run();
    if let Some(e) = program.out.first() {
        print!("{e}")
    }
    for e in program.out.iter().skip(1) {
        print!(",{e}")
    }
    println!()
}

// Some assumptions on the input:
// * There is exactly one jump instruction at the end of the program, jumping to 0
// * Right before that, an output is generated
// * Right before that, register a is shifted right by 3: adv(3)
//
// Each output depends on at most 10 bits of a (it is used with a shift of at most 7).
// Therefore we look at all 10-bit a's and group them by the first number that is output.
// Then we just need to combine these generators into a chain that fits together.
fn number_generators(mut program: Program) -> [Vec<u16>; 8] {
    let mut out = [const { vec![] }; 8];
    for a in 1..(1 << 10) {
        program.a = a as u64;
        program.out.clear();
        program.ip = 0;
        program.run();
        let &output = program.out.first().unwrap();
        out[output as usize].push(a);
    }
    out
}

fn part2(input: String) {
    let mut program = parse(&input).unwrap();
    let generators = number_generators(program.clone());

    let output = program.program.clone();
    // a_candidates maps from 7-bit required prefixes to the lower bits of a that
    // generate the required numbers so far.
    let mut a_candidates: FxHashMap<u8, u64> = generators[output[0] as usize]
        .iter()
        .rev() // Collects the values for each prefix
        .map(|&a| ((a >> 3) as u8, a as u64 % 8))
        .collect();
    let len = output.len();
    for (i, x) in output.iter().enumerate().skip(1) {
        let mut next_candidates = FxHashMap::default();
        for (prefix, val) in generators[*x as usize]
            .iter()
            .filter(|&a| {
                // Take only short candidates in the end to ensure that not too many numbers are generated
                let max_bits = (len - i) * 3;
                (*a as u64) < (1u64 << max_bits)
            })
            // Only use generators that match any required prefix
            .filter(|&a| a_candidates.contains_key(&((a % (1 << 7)) as u8)))
            .map(|&a| {
                let prefix = (a >> 3) as u8;
                let val = a as u64 % 8;
                let prev = a_candidates[&((a % (1 << 7)) as u8)];
                (prefix, (val << (i * 3)) | prev)
            })
        {
            // Only insert first (smallest) encountered value
            next_candidates.entry(prefix).or_insert(val);
        }
        a_candidates = next_candidates;
    }
    println!("{}", a_candidates[&0]);

    // Verify result
    program.a = a_candidates[&0];
    program.run();
    assert_eq!(program.out, program.program);
}

util::aoc_main!();

Also on github

[โ€“] Gobbel2000@programming.dev 3 points 7 months ago (1 children)

Dijkstra's algorithm can fairly simply be modified to work for part 2. In the relaxation step you just need to also handle the case that the distances of two joining paths are equal.

[โ€“] Gobbel2000@programming.dev 2 points 7 months ago

Rust

Dijkstra's algorithm. While the actual shortest path was not needed in part 1, only the distance, in part 2 the path is saved in the parent hashmap, and crucially, if we encounter two paths with the same distance, both parent nodes are saved. This ensures we end up with all shortest paths in the end.

Solution

use std::cmp::{Ordering, Reverse};

use euclid::{default::*, vec2};
use priority_queue::PriorityQueue;
use rustc_hash::{FxHashMap, FxHashSet};

const DIRS: [Vector2D<i32>; 4] = [vec2(1, 0), vec2(0, 1), vec2(-1, 0), vec2(0, -1)];

type Node = (Point2D<i32>, u8);

fn parse(input: &str) -> (Vec<Vec<bool>>, Point2D<i32>, Point2D<i32>) {
    let mut start = None;
    let mut end = None;
    let mut field = Vec::new();
    for (y, l) in input.lines().enumerate() {
        let mut row = Vec::new();
        for (x, b) in l.bytes().enumerate() {
            if b == b'S' {
                start = Some(Point2D::new(x, y).to_i32());
            } else if b == b'E' {
                end = Some(Point2D::new(x, y).to_i32());
            }
            row.push(b == b'#');
        }
        field.push(row);
    }
    (field, start.unwrap(), end.unwrap())
}

fn adj(field: &[Vec<bool>], (v, dir): Node) -> Vec<(Node, u32)> {
    let mut adj = Vec::with_capacity(3);
    let next = v + DIRS[dir as usize];
    if !field[next.y as usize][next.x as usize] {
        adj.push(((next, dir), 1));
    }
    adj.push(((v, (dir + 1) % 4), 1000));
    adj.push(((v, (dir + 3) % 4), 1000));
    adj
}

fn shortest_path_length(field: &[Vec<bool>], start: Node, end: Point2D<i32>) -> u32 {
    let mut dist: FxHashMap<Node, u32> = FxHashMap::default();
    dist.insert(start, 0);
    let mut pq: PriorityQueue<Node, Reverse<u32>> = PriorityQueue::new();
    pq.push(start, Reverse(0));
    while let Some((v, _)) = pq.pop() {
        for (w, weight) in adj(field, v) {
            let dist_w = dist.get(&w).copied().unwrap_or(u32::MAX);
            let new_dist = dist[&v] + weight;
            if dist_w > new_dist {
                dist.insert(w, new_dist);
                pq.push_increase(w, Reverse(new_dist));
            }
        }
    }
    // Shortest distance to end, regardless of final direction
    (0..4).map(|dir| dist[&(end, dir)]).min().unwrap()
}

fn part1(input: String) {
    let (field, start, end) = parse(&input);
    let distance = shortest_path_length(&field, (start, 0), end);
    println!("{distance}");
}

fn shortest_path_tiles(field: &[Vec<bool>], start: Node, end: Point2D<i32>) -> u32 {
    let mut parents: FxHashMap<Node, Vec<Node>> = FxHashMap::default();
    let mut dist: FxHashMap<Node, u32> = FxHashMap::default();
    dist.insert(start, 0);
    let mut pq: PriorityQueue<Node, Reverse<u32>> = PriorityQueue::new();
    pq.push(start, Reverse(0));
    while let Some((v, _)) = pq.pop() {
        for (w, weight) in adj(field, v) {
            let dist_w = dist.get(&w).copied().unwrap_or(u32::MAX);
            let new_dist = dist[&v] + weight;
            match dist_w.cmp(&new_dist) {
                Ordering::Greater => {
                    parents.insert(w, vec![v]);
                    dist.insert(w, new_dist);
                    pq.push_increase(w, Reverse(new_dist));
                }
                // Remember both parents if distance is equal
                Ordering::Equal => parents.get_mut(&w).unwrap().push(v),
                Ordering::Less => {}
            }
        }
    }
    let mut path_tiles: FxHashSet<Point2D<i32>> = FxHashSet::default();
    path_tiles.insert(end);

    // Shortest distance to end, regardless of final direction
    let shortest_dist = (0..4).map(|dir| dist[&(end, dir)]).min().unwrap();
    for dir in 0..4 {
        if dist[&(end, dir)] == shortest_dist {
            collect_tiles(&parents, &mut path_tiles, (end, dir));
        }
    }
    path_tiles.len() as u32
}

fn collect_tiles(
    parents: &FxHashMap<Node, Vec<Node>>,
    tiles: &mut FxHashSet<Point2D<i32>>,
    cur: Node,
) {
    if let Some(pars) = parents.get(&cur) {
        for p in pars {
            tiles.insert(p.0);
            collect_tiles(parents, tiles, *p);
        }
    }
}

fn part2(input: String) {
    let (field, start, end) = parse(&input);
    let tiles = shortest_path_tiles(&field, (start, 0), end);
    println!("{tiles}");
}

util::aoc_main!();

Also on github

[โ€“] Gobbel2000@programming.dev 2 points 7 months ago

Rust

Part 2 was a bit tricky. Moving into a box horizontally works mostly the same as for part 1, for the vertical case I used two recursive functions. The first recurses from the left and right side for each box just to find out if the entire tree can be moved. The second function actually does the moving in a similar recursive structure, but now with the knowledge that all subtrees can actually be moved.

Lots of moving parts, but at least it could very nicely be debugged by printing out the map from the two minimal examples after each round.

Solution

use euclid::{default::*, vec2};

// Common type for both parts. In part 1 all boxes are BoxL.
#[derive(Clone, Copy)]
enum Spot {
    Empty,
    BoxL,
    BoxR,
    Wall,
}

impl From<u8> for Spot {
    fn from(value: u8) -> Self {
        match value {
            b'.' | b'@' => Spot::Empty,
            b'O' => Spot::BoxL,
            b'#' => Spot::Wall,
            other => panic!("Invalid spot: {other}"),
        }
    }
}

fn parse(input: &str) -> (Vec<Vec<Spot>>, Point2D<i32>, Vec<Vector2D<i32>>) {
    let (field_s, moves_s) = input.split_once("\n\n").unwrap();
    let mut field = Vec::new();
    let mut robot = None;
    for (y, l) in field_s.lines().enumerate() {
        let mut row = Vec::new();
        for (x, b) in l.bytes().enumerate() {
            row.push(Spot::from(b));
            if b == b'@' {
                robot = Some(Point2D::new(x, y).to_i32())
            }
        }
        field.push(row);
    }

    let moves = moves_s
        .bytes()
        .filter(|b| *b != b'\n')
        .map(|b| match b {
            b'^' => vec2(0, -1),
            b'>' => vec2(1, 0),
            b'v' => vec2(0, 1),
            b'<' => vec2(-1, 0),
            other => panic!("Invalid move: {other}"),
        })
        .collect();
    (field, robot.unwrap(), moves)
}

fn gps(field: &[Vec<Spot>]) -> u32 {
    let mut sum = 0;
    for (y, row) in field.iter().enumerate() {
        for (x, s) in row.iter().enumerate() {
            if let Spot::BoxL = s {
                sum += x + 100 * y;
            }
        }
    }
    sum as u32
}

fn part1(input: String) {
    let (mut field, mut robot, moves) = parse(&input);
    for m in moves {
        let next = robot + m;
        match field[next.y as usize][next.x as usize] {
            Spot::Empty => robot = next, // Move into space
            Spot::BoxL => {
                let mut search = next + m;
                let can_move = loop {
                    match field[search.y as usize][search.x as usize] {
                        Spot::BoxL => {}
                        Spot::Wall => break false,
                        Spot::Empty => break true,
                        Spot::BoxR => unreachable!(),
                    }
                    search += m;
                };
                if can_move {
                    robot = next;
                    field[next.y as usize][next.x as usize] = Spot::Empty;
                    field[search.y as usize][search.x as usize] = Spot::BoxL;
                }
            }
            Spot::Wall => {} // Cannot move
            Spot::BoxR => unreachable!(),
        }
    }
    println!("{}", gps(&field));
}

// Transform part 1 field to wider part 2 field
fn widen(field: &[Vec<Spot>]) -> Vec<Vec<Spot>> {
    field
        .iter()
        .map(|row| {
            row.iter()
                .flat_map(|s| match s {
                    Spot::Empty => [Spot::Empty; 2],
                    Spot::Wall => [Spot::Wall; 2],
                    Spot::BoxL => [Spot::BoxL, Spot::BoxR],
                    Spot::BoxR => unreachable!(),
                })
                .collect()
        })
        .collect()
}

// Recursively find out whether or not the robot can move in direction `dir` from `start`.
fn can_move_rec(field: &[Vec<Spot>], start: Point2D<i32>, dir: Vector2D<i32>) -> bool {
    let next = start + dir;
    match field[next.y as usize][next.x as usize] {
        Spot::Empty => true,
        Spot::BoxL => can_move_rec(field, next, dir) && can_move_rec(field, next + vec2(1, 0), dir),
        Spot::BoxR => can_move_rec(field, next - vec2(1, 0), dir) && can_move_rec(field, next, dir),
        Spot::Wall => false,
    }
}

// Recursively execute a move for vertical directions into boxes.
fn do_move(field: &mut [Vec<Spot>], start: Point2D<i32>, dir: Vector2D<i32>) {
    let next = start + dir;
    match field[next.y as usize][next.x as usize] {
        Spot::Empty | Spot::Wall => {}
        Spot::BoxL => {
            do_move(field, next, dir);
            do_move(field, next + vec2(1, 0), dir);
            let move_to = next + dir;
            field[next.y as usize][next.x as usize] = Spot::Empty;
            field[next.y as usize][next.x as usize + 1] = Spot::Empty;
            field[move_to.y as usize][move_to.x as usize] = Spot::BoxL;
            field[move_to.y as usize][move_to.x as usize + 1] = Spot::BoxR;
        }
        Spot::BoxR => {
            do_move(field, next - vec2(1, 0), dir);
            do_move(field, next, dir);
            let move_to = next + dir;
            field[next.y as usize][next.x as usize - 1] = Spot::Empty;
            field[next.y as usize][next.x as usize] = Spot::Empty;
            field[move_to.y as usize][move_to.x as usize - 1] = Spot::BoxL;
            field[move_to.y as usize][move_to.x as usize] = Spot::BoxR;
        }
    }
}

fn part2(input: String) {
    let (field1, robot1, moves) = parse(&input);
    let mut field = widen(&field1);
    let mut robot = Point2D::new(robot1.x * 2, robot1.y);
    for m in moves {
        let next = robot + m;
        match field[next.y as usize][next.x as usize] {
            Spot::Empty => robot = next, // Move into space
            Spot::BoxL | Spot::BoxR if m.y == 0 => {
                let mut search = next + m;
                let can_move = loop {
                    match field[search.y as usize][search.x as usize] {
                        Spot::BoxL | Spot::BoxR => {}
                        Spot::Wall => break false,
                        Spot::Empty => break true,
                    }
                    search += m;
                };
                if can_move {
                    robot = next;
                    // Shift boxes by array remove/insert
                    field[next.y as usize].remove(search.x as usize);
                    field[next.y as usize].insert(next.x as usize, Spot::Empty);
                }
            }
            Spot::BoxL | Spot::BoxR => {
                if can_move_rec(&field, robot, m) {
                    do_move(&mut field, robot, m);
                    robot = next;
                }
            }
            Spot::Wall => {} // Cannot move
        }
    }
    println!("{}", gps(&field));
}

util::aoc_main!();

Also on github

[โ€“] Gobbel2000@programming.dev 2 points 7 months ago (1 children)

Very cool, taking a statistical approach to discern random noise from picture.

[โ€“] Gobbel2000@programming.dev 2 points 7 months ago

Rust

Part 2 was very surprising in that it had a very vague requirement: "Find christmas tree!". But my idea of finding the first round where no robots overlap turned out to just work when printing the map, so that was nice. I'm glad I did not instead start looking for symmetric patterns, because the christmas tree map is not symmetric at all.

Solution

use euclid::default::*;
use regex::Regex;

fn parse(input: &str) -> Vec<(Point2D<i32>, Vector2D<i32>)> {
    let re = Regex::new(r"p=(\d+),(\d+) v=(-?\d+),(-?\d+)").unwrap();
    re.captures_iter(input)
        .map(|cap| {
            let (_, [p0, p1, v0, v1]) = cap.extract();
            (
                Point2D::new(p0.parse().unwrap(), p1.parse().unwrap()),
                Vector2D::new(v0.parse().unwrap(), v1.parse().unwrap()),
            )
        })
        .collect()
}

const ROOM: Size2D<i32> = Size2D::new(101, 103);
const TIME: i32 = 100;

fn part1(input: String) {
    let robots = parse(&input);
    let new_pos: Vec<Point2D<i32>> = robots.iter()
        .map(|&(p, v)| (p + v * TIME).rem_euclid(&ROOM))
        .collect();

    assert_eq!(ROOM.width % 2, 1);
    assert_eq!(ROOM.height % 2, 1);
    let mid_x = ROOM.width / 2;
    let mid_y = ROOM.height / 2;
    
    let mut q = [0u32; 4];
    for p in new_pos {
        use std::cmp::Ordering::*;
        match (p.x.cmp(&mid_x), p.y.cmp(&mid_y)) {
            (Less, Less) => q[0] += 1,
            (Greater, Less) => q[1] += 1,
            (Less, Greater) => q[2] += 1,
            (Greater, Greater) => q[3] += 1,
            _ => {}
        };
    }
    let prod = q[0] * q[1] * q[2] * q[3];
    println!("{prod}");
}

fn print_map(map: &[Vec<bool>]) {
    for row in map {
        for p in row {
            if *p { print!("#") } else { print!(".") }
        }
        println!();
    }
    println!();
}


fn part2(input: String) {
    let mut robots = parse(&input);
    let mut map = vec![vec![false; ROOM.width as usize]; ROOM.height as usize];
    for i in 1.. {
        let mut overlap = false;
        for (p, v) in &mut robots {
            *p = (*p + *v).rem_euclid(&ROOM);
            if map[p.y as usize][p.x as usize] {
                // Found two robots on the same spot,
                // which is used as a heuristic for detecting the easter egg.
                overlap = true;
            } else {
                map[p.y as usize][p.x as usize] = true;
            }
        }
        if !overlap {
            print_map(&map);
            println!("Round: {i}");
            break;
        }
        for row in &mut map {
            row.fill(false);
        }
    }
}

util::aoc_main!();

Also on github

[โ€“] Gobbel2000@programming.dev 2 points 7 months 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

[โ€“] Gobbel2000@programming.dev 2 points 7 months ago

Rust

Areas are found by flooding, in the meantime whenever the adjacent plot would be outside the region (or out of bounds) the edge (inside plot, outside plot) is saved in a perimeter list. Part 1 takes just the size of that list, in part 2 we remove fence parts and all entries directly next to it on both sides.

Solution

use std::collections::{HashSet, VecDeque};

use euclid::{default::*, point2, vec2};

type Fences = HashSet<(Point2D<i32>, Point2D<i32>)>;
const DIRS: [Vector2D<i32>; 4] = [vec2(0, -1), vec2(1, 0), vec2(0, 1), vec2(-1, 0)];

fn parse(input: &str) -> Vec<&[u8]> {
    input.lines().map(|l| l.as_bytes()).collect()
}

fn price(field: &[&[u8]], start: (usize, usize), visited: &mut [Vec<bool>]) -> (u32, Fences) {
    let crop = field[start.1][start.0];
    let width = field[0].len();
    let height = field.len();
    let mut area_visited = vec![vec![false; width]; height];
    let mut area = 0;
    let mut fences: Fences = HashSet::new();

    area_visited[start.1][start.0] = true;
    visited[start.1][start.0] = true;
    let start = point2(start.0 as i32, start.1 as i32);
    let bounds = Rect::new(Point2D::origin(), Size2D::new(width, height).to_i32());
    let mut frontier = VecDeque::from([start]);
    while let Some(p) = frontier.pop_front() {
        area += 1;
        for dir in DIRS {
            let next = p + dir;
            if bounds.contains(next) {
                let next_u = next.to_usize();
                if area_visited[next_u.y][next_u.x] {
                    continue;
                }
                if field[next_u.y][next_u.x] == crop {
                    visited[next_u.y][next_u.x] = true;
                    area_visited[next_u.y][next_u.x] = true;
                    frontier.push_back(next);
                    continue;
                }
            }
            fences.insert((p, next));
        }
    }
    (area, fences)
}

fn part1(input: String) {
    let field = parse(&input);
    let width = field[0].len();
    let height = field.len();
    let mut visited = vec![vec![false; width]; height];
    let mut total_price = 0;
    for y in 0..height {
        for x in 0..width {
            if !visited[y][x] {
                let (area, fences) = price(&field, (x, y), &mut visited);
                total_price += area * fences.len() as u32;
            }
        }
    }
    println!("{total_price}");
}

fn count_perimeter(mut fences: Fences) -> u32 {
    let list: Vec<_> = fences.iter().copied().collect();
    let mut perimeter = 0;
    for (v, w) in list {
        if fences.contains(&(v, w)) {
            perimeter += 1;
            let dir = w - v;
            let orth = dir.yx();
            let mut next = v + orth;
            while fences.remove(&(next, next + dir)) {
                next += orth;
            }
            let mut next = v - orth;
            while fences.remove(&(next, next + dir)) {
                next -= orth;
            }
        }
    }
    perimeter
}

fn part2(input: String) {
    let field = parse(&input);
    let width = field[0].len();
    let height = field.len();
    let mut visited = vec![vec![false; width]; height];
    let mut total_price = 0;
    for y in 0..height {
        for x in 0..width {
            if !visited[y][x] {
                let (area, fences) = price(&field, (x, y), &mut visited);
                total_price += area * count_perimeter(fences);
            }
        }
    }
    println!("{total_price}");
}

util::aoc_main!();

Also on github

[โ€“] Gobbel2000@programming.dev 3 points 8 months ago

Rust

Part 2 is solved with recursion and a cache, which is indexed by stone numbers and remaining rounds and maps to the previously calculated expansion size. In my case, the cache only grew to 139320 entries, which is quite reasonable given the size of the result.

Solution

use std::collections::HashMap;

fn parse(input: String) -> Vec<u64> {
    input
        .split_whitespace()
        .map(|w| w.parse().unwrap())
        .collect()
}

fn part1(input: String) {
    let mut stones = parse(input);
    for _ in 0..25 {
        let mut new_stones = Vec::with_capacity(stones.len());
        for s in &stones {
            match s {
                0 => new_stones.push(1),
                n => {
                    let digits = s.ilog10() + 1;
                    if digits % 2 == 0 {
                        let cutoff = 10u64.pow(digits / 2);
                        new_stones.push(n / cutoff);
                        new_stones.push(n % cutoff);
                    } else {
                        new_stones.push(n * 2024)
                    }
                }
            }
        }
        stones = new_stones;
    }
    println!("{}", stones.len());
}

fn expansion(s: u64, rounds: u32, cache: &mut HashMap<(u64, u32), u64>) -> u64 {
    // Recursion anchor
    if rounds == 0 {
        return 1;
    }
    // Calculation is already cached
    if let Some(res) = cache.get(&(s, rounds)) {
        return *res;
    }

    // Recurse
    let res = match s {
        0 => expansion(1, rounds - 1, cache),
        n => {
            let digits = s.ilog10() + 1;
            if digits % 2 == 0 {
                let cutoff = 10u64.pow(digits / 2);
                expansion(n / cutoff, rounds - 1, cache) +
                expansion(n % cutoff, rounds - 1, cache)
            } else {
                expansion(n * 2024, rounds - 1, cache)
            }
        }
    };
    // Save in cache
    cache.insert((s, rounds), res);
    res
}

fn part2(input: String) {
    let stones = parse(input);
    let mut cache = HashMap::new();
    let sum: u64 = stones.iter().map(|s| expansion(*s, 75, &mut cache)).sum();
    println!("{sum}");
}

util::aoc_main!();

Also on github

[โ€“] Gobbel2000@programming.dev 2 points 8 months ago

Rust

This was a nice one. Basically 9 rounds of Breadth-First-Search, which could be neatly expressed using fold. The only difference between part 1 and part 2 turned out to be the datastructure for the search frontier: The HashSet in part 1 unifies paths as they join back to the same node, the Vec in part 2 keeps all paths separate.

Solution

use std::collections::HashSet;

fn parse(input: &str) -> Vec<&[u8]> {
    input.lines().map(|l| l.as_bytes()).collect()
}

fn adj(grid: &[&[u8]], (x, y): (usize, usize)) -> Vec<(usize, usize)> {
    let n = grid[y][x];
    let mut adj = Vec::with_capacity(4);
    if x > 0 && grid[y][x - 1] == n + 1 {
        adj.push((x - 1, y))
    }
    if y > 0 && grid[y - 1][x] == n + 1 {
        adj.push((x, y - 1))
    }
    if x + 1 < grid[0].len() && grid[y][x + 1] == n + 1 {
        adj.push((x + 1, y))
    }
    if y + 1 < grid.len() && grid[y + 1][x] == n + 1 {
        adj.push((x, y + 1))
    }
    adj
}

fn solve(input: String, trailhead: fn(&[&[u8]], (usize, usize)) -> u32) -> u32 {
    let grid = parse(&input);
    let mut sum = 0;
    for (y, row) in grid.iter().enumerate() {
        for (x, p) in row.iter().enumerate() {
            if *p == b'0' {
                sum += trailhead(&grid, (x, y));
            }
        }
    }
    sum
}

fn part1(input: String) {
    fn score(grid: &[&[u8]], start: (usize, usize)) -> u32 {
        (1..=9)
            .fold(HashSet::from([start]), |frontier, _| {
                frontier.iter().flat_map(|p| adj(grid, *p)).collect()
            })
            .len() as u32
    }
    println!("{}", solve(input, score))
}

fn part2(input: String) {
    fn rating(grid: &[&[u8]], start: (usize, usize)) -> u32 {
        (1..=9)
            .fold(vec![start], |frontier, _| {
                frontier.iter().flat_map(|p| adj(grid, *p)).collect()
            })
            .len() as u32
    }
    println!("{}", solve(input, rating))
}

util::aoc_main!();

Also on github

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