import { inputDay, munge } from "./lib/utilities.js"

const parser = inp =>
  inp
    .replaceAll(/[+=]/g, " ")
    .split("\n\n")
    .map(munge)
    .map(t => [t[0][3], t[0][5], t[1][3], t[1][5], t[2][2], t[2][4]])
const test = parser(`Button A: X+94, Y+34
Button B: X+22, Y+67
Prize: X=8400, Y=5400

Button A: X+26, Y+66
Button B: X+67, Y+21
Prize: X=12748, Y=12176

Button A: X+17, Y+86
Button B: X+84, Y+37
Prize: X=7870, Y=6450

Button A: X+69, Y+23
Button B: X+27, Y+71
Prize: X=18641, Y=10279
`)
display(test)

const input = await inputDay(13, { parser })

part 1

Our goal is to minimize such that

That's 2 unknowns () in 2 equations, so our problem should be solvable! Let's start by isolating in the first equation

Now we can substitute that into the second equation and solve for . My algebra is incredibly rusty, but let's give it a college try

Let's see if it works!

The only change I made from directly implementing the equation above is to check for non-integral results, and return [0, 0] in that case; that makes it easier for us to sum up the results later on.

function intersect(x1, y1, x2, y2, x, y) {
  if ((x1 * y - x * y1) % (x1 * y2 - x2 * y1) !== 0) return [0, 0]
  const b = (x1 * y - x * y1) / (x1 * y2 - x2 * y1)

  if ((x - b * x2) % x1 !== 0) return [0, 0]
  const a = (x - b * x2) / x1

  return [a, b]
}
display(intersect(94, 34, 22, 67, 8400, 5400))

Woo, it seems like it. Now we can sum up the intersections in the way specified by the problem, and we're done

function total(testcases) {
  return testcases
    .map(t => intersect.apply(null, t))
    .reduce((acc, [a, b]) => acc + (3 * a + b), 0)
}
display(total(test))
display(total(input))

part 2

Let's find out if our javascript integers overflow:

display(
  total(
    input.map(t => [
      t[0],
      t[1],
      t[2],
      t[3],
      10000000000000 + t[4],
      10000000000000 + t[5],
    ]),
  ),
)

Since my submitted answer was correct, it seems like they don't, so we don't need to rewrite our functions in terms of BigInts.

Day 13

prelude

part 1

part 2