Uiua

Pretty much just a transcription of my Dart solution.

Data  ← ⊜(⊜(⊜⋕⊸∈"0123456789")⊸≠@\n)⊸(¬⦷"\n\n")"Button A: X+94, Y+34\nButton B: X+22, Y+67\nPrize: X=8400, Y=5400\n\nButton A: X+26, Y+66\nButton B: X+67, Y+21\nPrize: X=12748, Y=12176\n\nButton A: X+17, Y+86\nButton B: X+84, Y+37\nPrize: X=7870, Y=6450\n\nButton A: X+69, Y+23\nButton B: X+27, Y+71\nPrize: X=18641, Y=10279"
IsInt ← <0.00001⌵-⁅.
AB    ← ÷°⊂≡(/-×⇌°⊟)⊏[0_1 2_1 0_2]
Cost  ← /+×IsInt.×3_1AB
&p /+≡Cost Data
&p /+≡(Cost⍜(⊡2|+1e13))Data

Nim

I’m embarrasingly bad with math. Couldn’t have solved this one without looking up the solution. =C

type Vec2 = tuple[x,y: int64]

const
  PriceA = 3
  PriceB = 1
  ErrorDelta = 10_000_000_000_000

proc isInteger(n: float): bool = n.round().almostEqual(n)
proc `+`(a: Vec2, b: int): Vec2 = (a.x + b, a.y + b)

proc solveEquation(a, b, prize: Vec2): int =
  let res_a = (prize.x*b.y - prize.y*b.x) / (a.x*b.y - a.y*b.x)
  let res_b = (a.x*prize.y - a.y*prize.x) / (a.x*b.y - a.y*b.x)
  if res_a.isInteger and res_b.isInteger:
    res_a.int * PriceA + res_b.int * PriceB
  else: 0

proc solve(input: string): AOCSolution[int, int] =
  let chunks = input.split("\n\n")
  for chunk in chunks:
    let lines = chunk.splitLines()
    let partsA = lines[0].split({' ', ',', '+'})
    let partsB = lines[1].split({' ', ',', '+'})
    let partsC = lines[2].split({' ', ',', '='})

    let a = (parseBiggestInt(partsA[3]), parseBiggestInt(partsA[6]))
    let b = (parseBiggestInt(partsB[3]), parseBiggestInt(partsB[6]))
    let c = (parseBiggestInt(partsC[2]), parseBiggestInt(partsC[5]))

    result.part1 += solveEquation(a,b,c)
    result.part2 += solveEquation(a,b,c+ErrorDelta)

Rust

Hardest part was parsing the input, i somehow forgot how regexes work and wasted hours.

Learning how to do matrix stuff in rust was a nice detour as well.

#[cfg(test)]
mod tests {
    use nalgebra::{Matrix2, Vector2};
    use regex::Regex;

    fn play_game(ax: i128, ay: i128, bx: i128, by: i128, gx: i128, gy: i128) -> i128 {
        for a_press in 0..100 {
            let rx = gx - ax * a_press;
            let ry = gy - ay * a_press;
            if rx % bx == 0 && ry % by == 0 && rx / bx == ry / by {
                return a_press * 3 + ry / by;
            }
        }
        0
    }

    fn play_game2(ax: i128, ay: i128, bx: i128, by: i128, gx: i128, gy: i128) -> i128 {
        // m * p = g
        // p = m' * g
        // |ax bx|.|a_press| = |gx|
        // |ay by| |b_press|   |gy|
        let m = Matrix2::new(ax as f64, bx as f64, ay as f64, by as f64);
        match m.try_inverse() {
            None => return 0,
            Some(m_inv) => {
                let g = Vector2::new(gx as f64, gy as f64);
                let p = m_inv * g;
                let pa = p[0].round() as i128;
                let pb = p[1].round() as i128;
                if pa * ax + pb * bx == gx && pa * ay + pb * by == gy {
                    return pa * 3 + pb;
                }
            }
        };
        0
    }

    #[test]
    fn day13_part1_test() {
        let input = std::fs::read_to_string("src/input/day_13.txt").unwrap();
        let re = Regex::new(r"[0-9]+").unwrap();

        let games = input
            .trim()
            .split("\n\n")
            .map(|line| {
                re.captures_iter(line)
                    .map(|x| {
                        let first = x.get(0).unwrap().as_str();
                        first.parse::<i128>().unwrap()
                    })
                    .collect::<Vec<i128>>()
            })
            .collect::<Vec<Vec<i128>>>();

        let mut total = 0;
        for game in games {
            let cost = play_game2(game[0], game[1], game[2], game[3], game[4], game[5]);
            total += cost;
        }
        // 36870
        println!("{}", total);
    }

    #[test]
    fn day12_part2_test() {
        let input = std::fs::read_to_string("src/input/day_13.txt").unwrap();
        let re = Regex::new(r"[0-9]+").unwrap();

        let games = input
            .trim()
            .split("\n\n")
            .map(|line| {
                re.captures_iter(line)
                    .map(|x| {
                        let first = x.get(0).unwrap().as_str();
                        first.parse::<i128>().unwrap()
                    })
                    .collect::<Vec<i128>>()
            })
            .collect::<Vec<Vec<i128>>>();

        let mut total = 0;
        for game in games {
            let cost = play_game2(
                game[0],
                game[1],
                game[2],
                game[3],
                game[4] + 10000000000000,
                game[5] + 10000000000000,
            );
            total += cost;
        }
        println!("{}", total);
    }
}

J

I think this puzzle is a bit of a missed opportunity. They could have provided inputs with no solution or with a line of solutions, so that the cost optimization becomes meaningful. As it is, you just have to carry out Cramer’s rule in extended precision rational arithmetic.

load 'regex'

data_file_name =: '13.data'
raw =: cutopen fread data_file_name
NB. a b sublist y gives elements [a..b) of y
sublist =: ({~(+i.)/)~"1 _
parse_button =: monad define
  match =. 'X\+([[:digit:]]+), Y\+([[:digit:]]+)' rxmatch y
  ". (}. match) sublist y
)
parse_prize =: monad define
  match =. 'X=([[:digit:]]+), Y=([[:digit:]]+)' rxmatch y
  ". (}. match) sublist y
)
parse_machine =: monad define
  3 2 $ (parse_button >0{y), (parse_button >1{y), (parse_prize >2{y)
)
NB. x: converts to extended precision, which gives us rational arithmetic
machines =: x: (parse_machine"1) _3 ]\ raw

NB. A machine is represented by an array 3 2 $ ax ay bx by tx ty, where button
NB. A moves the claw by ax ay, button B by bx by, and the target is at tx ty.
NB. We are looking for nonnegative integer solutions to ax*a + bx*b = tx,
NB. ay*a + by*b = ty; if there is more than one, we want the least by the cost
NB. function 3*a + b.

solution_rank =: monad define
  if. 0 ~: -/ . * }: y do. 0  NB. system is nonsingular
  elseif. */ (=/"1) 2 ]\ ({. % {:) |: y do. 1  NB. one equation is a multiple of the other
  else. _1 end.
)
NB. solve0 yields the cost of solving a machine of solution rank 0
solve0 =: monad define
  d =. -/ . * }: y
  a =. (-/ . * 2 1 { y) % d
  b =. (-/ . * 0 2 { y) % d
  if. (a >: 0) * (a = <. a) * (b >: 0) * (b = <. b) do. b + 3 * a else. 0 end.
)
NB. there are actually no machines of solution rank _1 or 1 in the test set
result1 =: +/ solve0"_1 machines

machines2 =: machines (+"2) 3 2 $ 0 0 0 0 10000000000000 10000000000000
NB. there are no machines of solution rank _1 or 1 in the modified set either
result2 =: +/ solve0"_1 machines2

C

“The cheapest way” “the fewest tokens”, that evil chap!

I’m on a weekend trip and thought to do the puzzle in the 3h train ride but I got silly stumped on 2D line intersection*, was too stubborn to look it up, and fell asleep 🤡

When I woke up, so did the little nugget of elementary algebra somewhere far in the back of my mind. Tonight I finally got to implementing, which was smooth sailing except for this lesson I learnt:

int64 friends don’t let int64 friends play with float32s.

*) on two parts:

1. how can you capture a two-dimensional problem in a linear equation (ans: use slopes), and
2. what unknown was I supposed to be finding? (ans: either x or y of intersection will do)

#include "common.h"

static int64_t
score(int ax, int ay, int bx, int by, int64_t px, int64_t py)
{
	int64_t a,b, x;
	double as,bs;

	as = (double)ay / ax;
	bs = (double)by / bx;

	/* intersection between a (from start) and b (from end) */
	x = (int64_t)round((px*bs - py) / (bs-as));

	a = x / ax;
	b = (px-x) / bx;

	return
	    a*ax + b*bx == px &&
	    a*ay + b*by == py ? a*3 + b : 0;
}

int
main(int argc, char **argv)
{
	int ax,ay, bx,by;
	int64_t p1=0,p2=0, px,py;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));
	
	while (scanf(
	    " Button A: X+%d, Y+%d"
	    " Button B: X+%d, Y+%d"
	    " Prize: X=%"SCNd64", Y=%"SCNd64,
	    &ax, &ay, &bx, &by, &px, &py) == 6) {
		p1 += score(ax,ay, bx,by, px,py);
		p2 += score(ax,ay, bx,by,
		    px + 10000000000000LL,
		    py + 10000000000000LL);
	}

	printf("13: %"PRId64" %"PRId64"\n", p1, p2);
	return 0;
}

https://github.com/sjmulder/aoc/blob/master/2024/c/day13.c