"""solve 包测试。""" from __future__ import annotations import numpy as np import pytest _NOT_IMPLEMENTED = "未实现：见 docs/03-roadmap.md L0/L1" def _normal_batch(rng, size, **kwargs): """`monte_carlo` 的 sampler 契约：收下实现传来的 `size`，一次抽够。""" return rng.normal(0, 1, size) @pytest.mark.skip(reason=_NOT_IMPLEMENTED) def test_monte_carlo_integral_converges_to_analytic_value(): """∫₀¹ x² dx = 1/3。n=2e5 时误差应 < 0.01。""" from mcm_toolkit.solve import monte_carlo_integral r = monte_carlo_integral(lambda X: X[:, 0] ** 2, bounds=[(0, 1)], n=200_000, random_state=0) assert abs(r.data - 1 / 3) < 0.01 def test_monte_carlo_probability_of_standard_normal(): """P(Z > 0) = 0.5，n=5000 时误差应 < 0.05。""" from mcm_toolkit.solve import monte_carlo_probability r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=5000, random_state=0) assert r.success assert abs(r.metrics["probability"] - 0.5) < 0.05 def test_monte_carlo_is_reproducible(): from mcm_toolkit.solve import monte_carlo_probability a = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=1000, random_state=9) b = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=1000, random_state=9) assert a.metrics["probability"] == b.metrics["probability"] def test_monte_carlo_returns_n_results_and_documented_metrics(): from mcm_toolkit.solve import monte_carlo r = monte_carlo(_normal_batch, lambda x: x**2, n=2000, random_state=0) assert r.success assert r.data.shape == (2000,) assert set(r.metrics) == {"mean", "std", "median", "p05", "p95", "ci_low", "ci_high"} def test_monte_carlo_calls_simulator_only_once(): """`simulator` 只被调用一次 —— 这是"必须向量化"的可执行版本。""" from mcm_toolkit.solve import monte_carlo counter = {"calls": 0} def simulator(params): counter["calls"] += 1 return params r = monte_carlo(_normal_batch, simulator, n=1000, random_state=0) assert r.success assert counter["calls"] == 1 def test_monte_carlo_data_is_reproducible_with_same_seed(): from mcm_toolkit.solve import monte_carlo a = monte_carlo(_normal_batch, lambda x: x, n=2000, random_state=11) b = monte_carlo(_normal_batch, lambda x: x, n=2000, random_state=11) assert np.array_equal(a.data, b.data) def test_monte_carlo_follows_set_seed_when_random_state_omitted(): from mcm_toolkit.core import set_seed from mcm_toolkit.solve import monte_carlo set_seed(5) a = monte_carlo(_normal_batch, lambda x: x, n=500) set_seed(5) b = monte_carlo(_normal_batch, lambda x: x, n=500) assert np.array_equal(a.data, b.data) def test_monte_carlo_estimates_known_expectation(): """E[X²] = 1（X ~ N(0, 1)）；n = 2e5 时误差应 < 0.02。""" from mcm_toolkit.solve import monte_carlo r = monte_carlo(_normal_batch, lambda x: x**2, n=200_000, random_state=0) assert abs(r.metrics["mean"] - 1.0) < 0.02 def test_monte_carlo_docstring_example_runs(): """docstring 的 Examples 必须真的能跑通（此前那一版是坏的）。""" from mcm_toolkit.solve import monte_carlo draw = lambda rng, size, **k: rng.normal(0, 1, size) # noqa: E731 r = monte_carlo(draw, lambda x: x**2, n=10000, random_state=0) assert r.success assert abs(r.metrics["mean"] - 1.0) < 0.1 def test_monte_carlo_accepts_numpy_integer_n(): """§4.7：`np.int64` 不是 `int` 的子类，不能漏。""" from mcm_toolkit.solve import monte_carlo r = monte_carlo(_normal_batch, lambda x: x, n=np.int64(100), random_state=0) assert r.success assert r.data.shape == (100,) def test_monte_carlo_counts_boolean_outcomes(): """布尔结果按 0/1 统计 —— 事件概率是 MC 最常见的用法。""" from mcm_toolkit.solve import monte_carlo r = monte_carlo( lambda rng, size, **k: rng.normal(0, 1, size) > 0, lambda x: x, n=5000, random_state=0, ) assert r.success assert r.data.dtype == float assert abs(r.metrics["mean"] - 0.5) < 0.05 def test_monte_carlo_rejects_nan_results(): """结果里出现 NaN 就判失败，不静默给出一列 nan 指标。""" from mcm_toolkit.solve import monte_carlo r = monte_carlo( lambda rng, size, **k: np.r_[np.zeros(size - 1), np.nan], lambda x: x, n=10, random_state=0, ) assert not r.success assert "NaN" in r.message def test_monte_carlo_rejects_bad_arguments(): from mcm_toolkit.solve import monte_carlo with pytest.raises(TypeError): monte_carlo(123, lambda x: x) # type: ignore[arg-type] # 就是要传错 with pytest.raises(TypeError): monte_carlo(_normal_batch, 123) # type: ignore[arg-type] # 就是要传错 with pytest.raises(TypeError): monte_carlo(_normal_batch, lambda x: x, n=True) with pytest.raises(TypeError): monte_carlo(_normal_batch, lambda x: x, n=10.0) # type: ignore[arg-type] # 就是要传错 with pytest.raises(ValueError): monte_carlo(_normal_batch, lambda x: x, n=0) def test_monte_carlo_probability_returns_documented_metrics(): from mcm_toolkit.solve import monte_carlo_probability r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=20000, random_state=0) assert r.success assert set(r.metrics) == {"probability", "ci_low", "ci_high", "std_error"} assert r.data.shape == (20000,) assert 0.0 <= r.metrics["ci_low"] <= r.metrics["probability"] assert r.metrics["probability"] <= r.metrics["ci_high"] <= 1.0 def test_monte_carlo_probability_std_error_matches_binomial_formula(): """`std_error` 就是比例的标准误 √(p(1-p)/n)。""" from mcm_toolkit.solve import monte_carlo_probability n = 20000 r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=n, random_state=0) p = r.metrics["probability"] assert np.isclose(r.metrics["std_error"], np.sqrt(p * (1 - p) / n)) def test_monte_carlo_probability_ci_narrows_with_n(): """区间宽度随 n 增大而收窄（O(1/√n)）。""" from mcm_toolkit.solve import monte_carlo_probability def width(n): r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=n, random_state=0) return r.metrics["ci_high"] - r.metrics["ci_low"] assert width(100_000) < width(1_000) def test_monte_carlo_probability_higher_confidence_is_wider(): from mcm_toolkit.solve import monte_carlo_probability def width(confidence): r = monte_carlo_probability( lambda z: z > 0, _normal_batch, n=5000, random_state=0, confidence=confidence ) return r.metrics["ci_high"] - r.metrics["ci_low"] assert width(0.99) > width(0.90) def test_monte_carlo_probability_ci_matches_statsmodels_wilson(): """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。""" from statsmodels.stats.proportion import proportion_confint from mcm_toolkit.solve import monte_carlo_probability n = 5000 r = monte_carlo_probability(lambda z: z > 0, _normal_batch, n=n, random_state=0) count = int(r.data.sum()) lo, hi = proportion_confint(count, n, alpha=0.05, method="wilson") assert np.isclose(r.metrics["ci_low"], lo) assert np.isclose(r.metrics["ci_high"], hi) def test_monte_carlo_probability_rejects_bad_confidence(): from mcm_toolkit.solve import monte_carlo_probability for bad in (0.0, 1.0, -0.5, 1.5): with pytest.raises(ValueError): monte_carlo_probability(lambda z: z > 0, _normal_batch, n=100, confidence=bad) @pytest.mark.skip(reason=_NOT_IMPLEMENTED) def test_markov_steady_state_two_state_chain(): """P=[[0.9,0.1],[0.2,0.8]] 的平稳分布是 [2/3, 1/3]。""" from mcm_toolkit.solve import markov_steady_state r = markov_steady_state(np.array([[0.9, 0.1], [0.2, 0.8]])) 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(): 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)