70 lines
1.9 KiB
Python
70 lines
1.9 KiB
Python
import sys
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from collections import defaultdict
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from dataclasses import dataclass
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import numpy as np
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from sympy import solve, symbols
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lines = sys.stdin.read().splitlines()
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positions = [
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np.array([int(c) for c in line.split("@")[0].strip().split(", ")]) for line in lines
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]
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velocities = [
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np.array([int(c) for c in line.split("@")[1].strip().split(", ")]) for line in lines
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]
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# part 1
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low, high = [7, 27] if len(positions) <= 10 else [200000000000000, 400000000000000]
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count = 0
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for i, (p1, v1) in enumerate(zip(positions, velocities)):
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for p2, v2 in zip(positions[i + 1:], velocities[i + 1:]):
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p, r = p1[:2], v1[:2]
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q, s = p2[:2], v2[:2]
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rs = np.cross(r, s)
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qpr = np.cross((q - p), r)
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if rs == 0:
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# assume there are no colinear lines
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continue
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else:
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assert rs != 0
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t = np.cross((q - p), s) / rs
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u = np.cross((q - p), r) / rs
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if t >= 0 and u >= 0:
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c = p + t * r
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if low <= c[0] <= high and low <= c[1] <= high:
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count += 1
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answer_1 = count
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print(f"answer 1 is {answer_1}")
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# part 2
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# equation
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# p1 + t1 * v1 == p0 + t1 * v0
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# p2 + t2 * v2 == p0 + t2 * v0
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# p3 + t3 * v3 == p0 + t3 * v0
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# ...
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# pn + tn * vn == p0 + tn * v0
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#
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# we can solve with only 3 lines since each lines contains 3
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# equations (x / y / z), so 3 lines give 9 equations and 9
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# variables: position (3), velocities (3) and times (3).
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n = 3
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x, y, z, vx, vy, vz, *ts = symbols('x y z vx vy vz ' + ' '.join(f't{i}' for i in range(n + 1)))
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equations = []
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for i, ti in zip(range(n), ts):
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for p, d, pi, di in zip((x, y, z), (vx, vy, vz), positions[i], velocities[i]):
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equations.append(p + ti * d - pi - ti * di)
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r = solve(equations, [x, y, z, vx, vy, vz] + list(ts), dict=True)[0]
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answer_2 = r[x] + r[y] + r[z]
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print(f"answer 2 is {answer_2}")
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