Next up: the lobby of a resort on a tropical island. The Historians take a moment to admire the hexagonal floor tiles before spreading out.
Fortunately, it looks like the resort has a new arcade! Maybe you can win some prizes from the claw machines?
Visit the website for the full story and puzzle description.
We got a math problem today. The movements triggered by buttons A, B and the target position P form a linear equation that we can solve for the two unknowns: the number of button presses. Using math terms, if i and j marks the number of button presses, we are to solve A * i + B * j = P
. This is simple enough to do using Cramer's rule, especially with two dimensional vectors. In this case the determinats can be computed with a simple cross product. We should not forget about the special cases: A and B can be parallel (didn't occur in my input), and the solution needs to be non negative integer for i and j.
namespace AdventOfCode.Y2024.Day13;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
using Machine = (Vec2 a, Vec2 b, Vec2 p);
record struct Vec2(long x, long y);
[ProblemName("Claw Contraption")]
class Solution : Solver {
public object PartOne(string input) => Parse(input).Sum(GetPrize);
public object PartTwo(string input) => Parse(input, shift: 10000000000000).Sum(GetPrize);
long GetPrize(Machine m) {
var (a, b, p) = m;
// solve a * i + b * j = p for i and j using Cramer's rule
var i = Det(p, b) / Det(a, b);
var j = Det(a, p) / Det(a, b);
// return the prize when a non negative _integer_ solution is found
if (i >= 0 && j >= 0 && a.x * i + b.x * j == p.x && a.y * i + b.y * j == p.y) {
return 3 * i + j;
} else {
return 0;
}
}
long Det(Vec2 a, Vec2 b) => a.x * b.y - a.y * b.x;
IEnumerable<Machine> Parse(string input, long shift=0) {
var blocks = input.Split("\n\n");
foreach (var block in blocks) {
var nums =
Regex.Matches(block, @"\d+", RegexOptions.Multiline)
.Select(m => int.Parse(m.Value))
.Chunk(2).Select(p => new Vec2(p[0], p[1]))
.ToArray();
nums[2] = new Vec2(nums[2].x + shift, nums[2].y + shift);
yield return (nums[0], nums[1], nums[2]);
}
}
}
Please ☆ my repo if you like it!