advent-of-code/src/holt59/aoc/2023/day24.py
2024-12-26 15:26:36 +01:00

69 lines
2.1 KiB
Python

from typing import Any, Iterator
import numpy as np
from sympy import solve, symbols
from ..base import BaseSolver
class Solver(BaseSolver):
def solve(self, input: str) -> Iterator[Any]:
lines = input.splitlines()
positions = np.array(
[[int(c) for c in line.split("@")[0].strip().split(", ")] for line in lines]
)
velocities = np.array(
[[int(c) for c in line.split("@")[1].strip().split(", ")] for line in lines]
)
# part 1
low, high = (
[7, 27] if len(positions) <= 10 else [200000000000000, 400000000000000]
)
count = 0
for i1, (p1, v1) in enumerate(zip(positions, velocities)):
p, r = p1[:2], v1[:2]
q, s = positions[i1 + 1 :, :2], velocities[i1 + 1 :, :2]
rs = np.cross(r, s)
q, s, rs = q[m := (rs != 0)], s[m], rs[m]
t = np.cross((q - p), s) / rs
u = np.cross((q - p), r) / rs
t, u = t[m := ((t >= 0) & (u >= 0))], u[m]
c = p + np.expand_dims(t, 1) * r
count += np.all((low <= c) & (c <= high), axis=1).sum()
yield count
# part 2
# equation
# p1 + t1 * v1 == p0 + t1 * v0
# p2 + t2 * v2 == p0 + t2 * v0
# p3 + t3 * v3 == p0 + t3 * v0
# ...
# pn + tn * vn == p0 + tn * v0
#
# we can solve with only 3 lines since each lines contains 3
# equations (x / y / z), so 3 lines give 9 equations and 9
# variables: position (3), velocities (3) and times (3).
n = 3
x, y, z, vx, vy, vz, *ts = symbols(
"x y z vx vy vz " + " ".join(f"t{i}" for i in range(n + 1))
)
equations = []
for i1, ti in zip(range(n), ts):
for p, d, pi, di in zip(
(x, y, z), (vx, vy, vz), positions[i1], velocities[i1]
):
equations.append(p + ti * d - pi - ti * di)
r = solve(equations, [x, y, z, vx, vy, vz] + list(ts), dict=True)[0]
yield r[x] + r[y] + r[z]