Refactor 2023 for new system.

src/holt59/aoc/2023/day24.py

@@ -1,63 +1,68 @@
-import sys
+from typing import Any, Iterator
 
 import numpy as np
 from sympy import solve, symbols
 
-lines = sys.stdin.read().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()
+from ..base import BaseSolver
 
 
-answer_1 = count
-print(f"answer 1 is {answer_1}")
+class Solver(BaseSolver):
+    def solve(self, input: str) -> Iterator[Any]:
+        lines = input.splitlines()
 
-# 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
-#
+        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]
+        )
 
-# 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
+        # part 1
+        low, high = (
+            [7, 27] if len(positions) <= 10 else [200000000000000, 400000000000000]
+        )
 
-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)
+        count = 0
+        for i1, (p1, v1) in enumerate(zip(positions, velocities)):
+            p, r = p1[:2], v1[:2]
 
-r = solve(equations, [x, y, z, vx, vy, vz] + list(ts), dict=True)[0]
+            q, s = positions[i1 + 1 :, :2], velocities[i1 + 1 :, :2]
 
-answer_2 = r[x] + r[y] + r[z]
-print(f"answer 2 is {answer_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]