guzhujushi
cd6f613568
两条口径由用户第二轮改定，Codex 顺手补工程侧。 - 用户：恢复 start 的入口类型校验；n_steps/n_paths 注解放宽为 int | np.integer - Codex：start 报错文案"必须是整数" -> "必须是数值"（这条校验本来就收 float，原文案自相矛盾）， 类型名统一 type(x).__name__；n_paths 默认值补空格；docstring 的 start 说明与 Raises 同步 - tests：start="a" 恢复断言 TypeError；补回 np.int64 的 n_steps/n_paths；solve 段 42 条 - 验证：209 收集 / 178 passed / 31 skipped / 0 xfail，ruff 0，pyright 0 - docs：STATUS 决策 55（方阵歧义两条路）+ 验证计数；CHANGELOG
444 lines
16 KiB
Python
444 lines
16 KiB
Python
"""solve 包测试。"""
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from __future__ import annotations
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import numpy as np
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import pytest
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_NOT_IMPLEMENTED = "未实现：见 docs/03-roadmap.md L0/L1"
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def _normal_batch(rng, size, **kwargs):
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"""`monte_carlo` 的 sampler 契约：收下实现传来的 `size`，一次抽够。"""
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return rng.normal(0, 1, size)
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def test_monte_carlo_integral_converges_to_analytic_value():
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"""∫₀¹ x² dx = 1/3。n=2e5 时误差应 < 0.01。"""
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from mcm_toolkit.solve import monte_carlo_integral
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r = monte_carlo_integral(lambda X: X[:, 0] ** 2, bounds=[(0, 1)], n=200_000, random_state=0)
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assert r.success
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assert abs(r.metrics["mean"] - 1 / 3) < 0.01 # 体积必须是正的
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assert len(r.data) == 200_000 # data 是 n 个样本各自的贡献
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def test_monte_carlo_integral_odd_dimension_sign():
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"""∫∫∫ x dV over [0,1]³ = 1/2 —— 奇数维最容易暴露"体积符号写反"。"""
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from mcm_toolkit.solve import monte_carlo_integral
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r = monte_carlo_integral(lambda X: X[:, 0], bounds=[(0, 1)] * 3, n=200_000, random_state=0)
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assert abs(r.metrics["mean"] - 0.5) < 0.01
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def test_monte_carlo_integral_signed_bounds():
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"""∫₀² x dx = 2 —— 边界带宽度时，体积算错和符号写反都会现形。"""
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from mcm_toolkit.solve import monte_carlo_integral
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r = monte_carlo_integral(lambda X: X[:, 0], bounds=[(0, 2)], n=200_000, random_state=0)
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assert abs(r.metrics["mean"] - 2.0) < 0.02
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def test_monte_carlo_integral_2d_and_params():
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from mcm_toolkit.solve import monte_carlo_integral
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bounds = [(0, 1), (0, 1)]
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r = monte_carlo_integral(lambda X: X[:, 0] * X[:, 1], bounds=bounds, n=200_000, random_state=0)
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assert abs(r.metrics["mean"] - 0.25) < 0.01
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assert r.name == "monte_carlo_integral"
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assert r.params["dimension"] == 2
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assert r.params["bounds"] == bounds
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def test_monte_carlo_integral_rejects_bad_bounds():
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"""四类非法 bounds 都要在入口抛 ValueError：区间反了 / 漏括号 / 空 / 参差不齐。"""
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from mcm_toolkit.solve import monte_carlo_integral
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for bad in ([(1, 0)], [0, 1], [], [(0, 1), (0, 2, 3)]):
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with pytest.raises(ValueError):
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monte_carlo_integral(lambda X: X[:, 0], bounds=bad, n=100, random_state=0)
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def test_monte_carlo_integral_failure_message_stays_failure():
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"""失败时 message 不能被包装层改写成"成功"（易错点 22）。"""
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from mcm_toolkit.solve import monte_carlo_integral
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r = monte_carlo_integral(lambda X: 1.0, bounds=[(0, 1)], n=100, random_state=0)
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assert r.success is False
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assert "成功" not in r.message
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assert "失败" in r.message
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assert r.params["dimension"] == 1 # 失败也要留下追溯信息
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def test_monte_carlo_integral_accepts_numpy_integer_n():
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"""§4.7：`np.int64` 不是 `int` 的子类，注解与运行期都要收（易错点 1）。"""
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from mcm_toolkit.solve import monte_carlo_integral
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r = monte_carlo_integral(lambda X: X[:, 0], bounds=[(0, 1)], n=np.int64(1000), random_state=0)
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assert r.success
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assert len(r.data) == 1000
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def test_monte_carlo_probability_of_standard_normal():
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"""P(Z > 0) = 0.5，n=5000 时误差应 < 0.05。"""
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from mcm_toolkit.solve import monte_carlo_probability
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r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=5000, random_state=0)
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assert r.success
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assert abs(r.metrics["probability"] - 0.5) < 0.05
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def test_monte_carlo_is_reproducible():
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from mcm_toolkit.solve import monte_carlo_probability
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a = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=1000, random_state=9)
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b = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=1000, random_state=9)
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assert a.metrics["probability"] == b.metrics["probability"]
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def test_monte_carlo_returns_n_results_and_documented_metrics():
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from mcm_toolkit.solve import monte_carlo
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r = monte_carlo(_normal_batch, lambda x: x**2, n=2000, random_state=0)
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assert r.success
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assert r.data.shape == (2000,)
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assert set(r.metrics) == {"mean", "std", "median", "p05", "p95", "ci_low", "ci_high"}
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def test_monte_carlo_calls_simulator_only_once():
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"""`simulator` 只被调用一次 —— 这是"必须向量化"的可执行版本。"""
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from mcm_toolkit.solve import monte_carlo
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counter = {"calls": 0}
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def simulator(params):
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counter["calls"] += 1
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return params
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r = monte_carlo(_normal_batch, simulator, n=1000, random_state=0)
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assert r.success
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assert counter["calls"] == 1
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def test_monte_carlo_data_is_reproducible_with_same_seed():
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from mcm_toolkit.solve import monte_carlo
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a = monte_carlo(_normal_batch, lambda x: x, n=2000, random_state=11)
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b = monte_carlo(_normal_batch, lambda x: x, n=2000, random_state=11)
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assert np.array_equal(a.data, b.data)
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def test_monte_carlo_follows_set_seed_when_random_state_omitted():
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from mcm_toolkit.core import set_seed
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from mcm_toolkit.solve import monte_carlo
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set_seed(5)
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a = monte_carlo(_normal_batch, lambda x: x, n=500)
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set_seed(5)
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b = monte_carlo(_normal_batch, lambda x: x, n=500)
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assert np.array_equal(a.data, b.data)
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def test_monte_carlo_estimates_known_expectation():
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"""E[X²] = 1（X ~ N(0, 1)）；n = 2e5 时误差应 < 0.02。"""
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from mcm_toolkit.solve import monte_carlo
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r = monte_carlo(_normal_batch, lambda x: x**2, n=200_000, random_state=0)
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assert abs(r.metrics["mean"] - 1.0) < 0.02
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def test_monte_carlo_docstring_example_runs():
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"""docstring 的 Examples 必须真的能跑通（此前那一版是坏的）。"""
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from mcm_toolkit.solve import monte_carlo
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draw = lambda rng, size, **k: rng.normal(0, 1, size) # noqa: E731
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r = monte_carlo(draw, lambda x: x**2, n=10000, random_state=0)
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assert r.success
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assert abs(r.metrics["mean"] - 1.0) < 0.1
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def test_monte_carlo_accepts_numpy_integer_n():
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"""§4.7：`np.int64` 不是 `int` 的子类，不能漏。"""
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from mcm_toolkit.solve import monte_carlo
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r = monte_carlo(_normal_batch, lambda x: x, n=np.int64(100), random_state=0)
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assert r.success
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assert r.data.shape == (100,)
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def test_monte_carlo_counts_boolean_outcomes():
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"""布尔结果按 0/1 统计 —— 事件概率是 MC 最常见的用法。"""
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from mcm_toolkit.solve import monte_carlo
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r = monte_carlo(
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lambda rng, size, **k: rng.normal(0, 1, size) > 0,
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lambda x: x,
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n=5000,
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random_state=0,
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)
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assert r.success
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assert r.data.dtype == float
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assert abs(r.metrics["mean"] - 0.5) < 0.05
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def test_monte_carlo_rejects_nan_results():
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"""结果里出现 NaN 就判失败，不静默给出一列 nan 指标。"""
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from mcm_toolkit.solve import monte_carlo
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r = monte_carlo(
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lambda rng, size, **k: np.r_[np.zeros(size - 1), np.nan],
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lambda x: x,
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n=10,
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random_state=0,
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)
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assert not r.success
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assert "NaN" in r.message
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def test_monte_carlo_rejects_bad_arguments():
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from mcm_toolkit.solve import monte_carlo
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with pytest.raises(TypeError):
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monte_carlo(123, lambda x: x) # type: ignore[arg-type] # 就是要传错
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with pytest.raises(TypeError):
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monte_carlo(_normal_batch, 123) # type: ignore[arg-type] # 就是要传错
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with pytest.raises(TypeError):
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monte_carlo(_normal_batch, lambda x: x, n=True)
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with pytest.raises(TypeError):
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monte_carlo(_normal_batch, lambda x: x, n=10.0) # type: ignore[arg-type] # 就是要传错
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with pytest.raises(ValueError):
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monte_carlo(_normal_batch, lambda x: x, n=0)
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def test_monte_carlo_probability_returns_documented_metrics():
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from mcm_toolkit.solve import monte_carlo_probability
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r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=20000, random_state=0)
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assert r.success
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assert set(r.metrics) == {"probability", "ci_low", "ci_high", "std_error"}
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assert r.data.shape == (20000,)
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assert 0.0 <= r.metrics["ci_low"] <= r.metrics["probability"]
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assert r.metrics["probability"] <= r.metrics["ci_high"] <= 1.0
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def test_monte_carlo_probability_std_error_matches_binomial_formula():
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"""`std_error` 就是比例的标准误 √(p(1-p)/n)。"""
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from mcm_toolkit.solve import monte_carlo_probability
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n = 20000
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r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=n, random_state=0)
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p = r.metrics["probability"]
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assert np.isclose(r.metrics["std_error"], np.sqrt(p * (1 - p) / n))
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def test_monte_carlo_probability_ci_narrows_with_n():
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"""区间宽度随 n 增大而收窄（O(1/√n)）。"""
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from mcm_toolkit.solve import monte_carlo_probability
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def width(n):
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r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=n, random_state=0)
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return r.metrics["ci_high"] - r.metrics["ci_low"]
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assert width(100_000) < width(1_000)
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def test_monte_carlo_probability_higher_confidence_is_wider():
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from mcm_toolkit.solve import monte_carlo_probability
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def width(confidence):
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r = monte_carlo_probability(
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lambda z: z > 0, _normal_batch, n=5000, random_state=0, confidence=confidence
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)
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return r.metrics["ci_high"] - r.metrics["ci_low"]
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assert width(0.99) > width(0.90)
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def test_monte_carlo_probability_ci_matches_statsmodels_wilson():
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"""区间必须对得上独立实现（statsmodels 的 wilson）—— `^` 写成异或那次就靠它拦住。"""
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from statsmodels.stats.proportion import proportion_confint
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from mcm_toolkit.solve import monte_carlo_probability
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n = 5000
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r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=n, random_state=0)
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count = int(r.data.sum())
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lo, hi = proportion_confint(count, n, alpha=0.05, method="wilson")
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assert np.isclose(r.metrics["ci_low"], lo)
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assert np.isclose(r.metrics["ci_high"], hi)
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def test_monte_carlo_probability_rejects_bad_confidence():
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from mcm_toolkit.solve import monte_carlo_probability
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for bad in (0.0, 1.0, -0.5, 1.5):
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with pytest.raises(ValueError):
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monte_carlo_probability(lambda z: z > 0, _normal_batch, n=100, confidence=bad)
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def test_random_walk_shape_and_start():
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"""`data` 形状 (n_paths, n_steps + 1)，第 0 列是起点。"""
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from mcm_toolkit.solve import random_walk
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r = random_walk(n_steps=100, n_paths=5, start=2.5, random_state=0)
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assert r.success
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assert r.data.shape == (5, 101)
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assert np.all(r.data[:, 0] == 2.5)
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def test_random_walk_default_steps_are_plus_minus_one():
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"""默认是对称随机游走：每一步都恰好是 ±1。"""
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from mcm_toolkit.solve import random_walk
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r = random_walk(n_steps=200, n_paths=3, random_state=0)
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assert set(np.unique(np.diff(r.data, axis=1))) == {-1.0, 1.0}
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def test_random_walk_is_reproducible():
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from mcm_toolkit.solve import random_walk
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a = random_walk(n_steps=50, n_paths=4, random_state=7)
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b = random_walk(n_steps=50, n_paths=4, random_state=7)
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assert np.array_equal(a.data, b.data)
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def test_random_walk_accepts_custom_step_sampler():
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"""换成高斯增量就是带漂移的高斯游走 —— 建模差异全在 step_sampler 里。"""
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from mcm_toolkit.solve import random_walk
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r = random_walk(
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n_steps=100,
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n_paths=3,
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step_sampler=lambda rng, size: rng.normal(0.5, 1.0, size),
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random_state=0,
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)
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assert r.success
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assert r.data.shape == (3, 101)
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assert abs(r.data[:, -1].mean() - 50.0) < 20.0 # 100 步 × 漂移 0.5
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def test_random_walk_rejects_bad_arguments():
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"""§4.7 的分工：类型不对抛 TypeError，取值不对（<= 0）抛 ValueError。"""
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from mcm_toolkit.solve import random_walk
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with pytest.raises(TypeError):
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random_walk(n_steps=2.5) # type: ignore[arg-type] # 就是要传错
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with pytest.raises(TypeError):
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random_walk(n_steps=True)
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with pytest.raises(TypeError):
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random_walk(n_steps=10, n_paths=1.5) # type: ignore[arg-type] # 就是要传错
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with pytest.raises(ValueError):
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random_walk(n_steps=0)
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with pytest.raises(ValueError):
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random_walk(n_steps=10, n_paths=0)
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with pytest.raises(TypeError):
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random_walk(n_steps=10, start="a") # type: ignore[arg-type] # 就是要传错
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def test_random_walk_failure_returns_result_not_exception():
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"""失败也要返回带 params 的 Result，不能把异常抛给调用方 —— 包装层引用未赋值变量那次就靠它拦住。"""
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from mcm_toolkit.solve import random_walk
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def transposed(rng, size):
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return rng.normal(0, 1, (size[1], size[0]))
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r = random_walk(n_steps=4, n_paths=6, step_sampler=transposed, random_state=0)
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assert r.success is False
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assert "形状" in r.message
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assert r.params["n_paths"] == 6 # 失败也要留下追溯信息
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assert r.data is None
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def test_random_walk_rejects_nan_steps():
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"""增量含 NaN 判失败 —— 与 monte_carlo 的 NaN 口径（决策 48）一致。"""
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from mcm_toolkit.solve import random_walk
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r = random_walk(
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n_steps=4,
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n_paths=2,
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step_sampler=lambda rng, size: np.full(size, np.nan),
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random_state=0,
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)
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assert r.success is False
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assert "NaN" in r.message
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def test_random_walk_accepts_numpy_integer():
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"""§4.7：`np.int64` 不是 `int` 的子类，注解与运行期都要收（易错点 1）。"""
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from mcm_toolkit.solve import random_walk
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r = random_walk(n_steps=np.int64(20), n_paths=np.int64(3), random_state=0)
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assert r.success
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assert r.data.shape == (3, 21)
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def test_random_walk_metrics_are_terminal_positions():
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"""终点位置的标准差按 √t 扩散（t = n_steps）—— 随机游走的核心性质。"""
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from mcm_toolkit.solve import random_walk
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n_steps = 100
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r = random_walk(n_steps=n_steps, n_paths=2000, start=5.0, random_state=0)
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assert abs(r.metrics["mean_terminal"] - 5.0) < 5.0 # 对称游走，期望还在起点附近
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assert abs(r.metrics["std_terminal"] - np.sqrt(n_steps)) < 5.0 # √t 扩散
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def test_random_walk_docstring_example_runs():
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from mcm_toolkit.solve import random_walk
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r = random_walk(n_steps=100, n_paths=5, random_state=0)
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assert r.data.shape == (5, 101)
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assert np.all(np.abs(np.diff(r.data, axis=1)) == 1)
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|
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|
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@pytest.mark.skip(reason=_NOT_IMPLEMENTED)
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def test_markov_steady_state_two_state_chain():
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"""P=[[0.9,0.1],[0.2,0.8]] 的平稳分布是 [2/3, 1/3]。"""
|
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from mcm_toolkit.solve import markov_steady_state
|
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|
||
r = markov_steady_state(np.array([[0.9, 0.1], [0.2, 0.8]]))
|
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assert np.allclose(r.data, [2 / 3, 1 / 3], atol=1e-6)
|
||
|
||
|
||
@pytest.mark.skip(reason=_NOT_IMPLEMENTED)
|
||
def test_knapsack_dp_known_optimum():
|
||
from mcm_toolkit.solve import knapsack_dp
|
||
|
||
r = knapsack_dp(weights=[1, 3, 4, 5], values=[1, 4, 5, 7], capacity=7)
|
||
assert np.isclose(r.metrics["total_value"], 9.0)
|
||
assert np.isclose(r.metrics["total_weight"], 7.0)
|
||
|
||
|
||
@pytest.mark.skip(reason=_NOT_IMPLEMENTED)
|
||
def test_linear_program_basic():
|
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from mcm_toolkit.solve import linear_program
|
||
|
||
r = linear_program(c=[1, 2], A_ub=[[1, 1]], b_ub=[4], bounds=[(0, None)] * 2)
|
||
assert r.success
|
||
assert np.isclose(r.metrics["objective"], 4.0)
|
||
|
||
|
||
@pytest.mark.skip(reason=_NOT_IMPLEMENTED)
|
||
def test_mm1_metrics_closed_form():
|
||
from mcm_toolkit.solve import mm1_metrics
|
||
|
||
r = mm1_metrics(arrival_rate=2, service_rate=3)
|
||
assert np.isclose(r.metrics["rho"], 2 / 3, rtol=1e-6)
|
||
assert np.isclose(r.metrics["L"], 2.0, rtol=1e-6)
|
||
|
||
|
||
@pytest.mark.skip(reason=_NOT_IMPLEMENTED)
|
||
def test_mm1_rejects_unstable_system():
|
||
"""λ >= μ 时系统不稳定，应 success=False 而不是返回 inf。"""
|
||
from mcm_toolkit.solve import mm1_metrics
|
||
|
||
assert mm1_metrics(arrival_rate=3, service_rate=3).success is False
|
||
|
||
|
||
@pytest.mark.skip(reason=_NOT_IMPLEMENTED)
|
||
def test_simulated_annealing_finds_quadratic_minimum():
|
||
from mcm_toolkit.solve import simulated_annealing
|
||
|
||
f = lambda v: (v[0] - 3.0) ** 2 + (v[1] + 1.0) ** 2 # noqa: E731
|
||
r = simulated_annealing(f, x0=[0.0, 0.0], n_iter=5000, random_state=0)
|
||
assert np.allclose(r.params["x"], [3.0, -1.0], atol=0.1)
|