Minor cleaning 2023.

src/holt59/aoc/2023/day24.py

@@ -1,11 +1,8 @@
 import sys
-from collections import defaultdict
-from dataclasses import dataclass
+
 import numpy as np
-
 from sympy import solve, symbols
 
-
 lines = sys.stdin.read().splitlines()
 
 positions = [
@@ -19,7 +16,7 @@ velocities = [
 low, high = [7, 27] if len(positions) <= 10 else [200000000000000, 400000000000000]
 count = 0
 for i, (p1, v1) in enumerate(zip(positions, velocities)):
-    for p2, v2 in zip(positions[i + 1:], velocities[i + 1:]):
+    for p2, v2 in zip(positions[i + 1 :], velocities[i + 1 :]):
         p, r = p1[:2], v1[:2]
         q, s = p2[:2], v2[:2]
 
@@ -31,7 +28,7 @@ for i, (p1, v1) in enumerate(zip(positions, velocities)):
             continue
         else:
             assert rs != 0
-            t = np.cross((q - p), s) / rs 
+            t = np.cross((q - p), s) / rs
             u = np.cross((q - p), r) / rs
 
             if t >= 0 and u >= 0:
@@ -50,14 +47,16 @@ print(f"answer 1 is {answer_1}")
 #   p3 + t3 * v3 == p0 + t3 * v0
 #      ...
 #   pn + tn * vn == p0 + tn * v0
-# 
+#
 
-# we can solve with only 3 lines since each lines contains 3 
+# 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)))
+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 i, ti in zip(range(n), ts):
     for p, d, pi, di in zip((x, y, z), (vx, vy, vz), positions[i], velocities[i]):