ClaudeandClaude Opus 5.5
b31a6141b3
common_window requires every pair from each case's own K_case to K+2; a-priori domain criteria against the 60 D reference (separate from Table 1); bilinear sampling returns the original fluid corner weight sum; analysis time-averages bin samples, reports coverage/missing points and fails closed on an empty common profile set; cases run to the end of cycle 11. Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
65 lines
3.2 KiB
Python
65 lines
3.2 KiB
Python
"""CPU tests of the Dütsch protocol arithmetic (docs/validation/duetsch_re100_kc5_protocol.md §6a)."""
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import importlib.util
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import math
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from pathlib import Path
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import numpy as np
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import pytest
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_spec = importlib.util.spec_from_file_location(
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"duetsch_protocol", Path(__file__).resolve().parents[2] / "research/duetsch_re100_kc5/protocol.py")
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p = importlib.util.module_from_spec(_spec)
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_spec.loader.exec_module(p)
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T = 2520
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def test_phase_and_cycle_bookkeeping():
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assert p.phase(0, T) == 270.0 and p.cycle(0, T) == 0 # +A, at rest, phi = -90
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assert p.phase(T // 4, T) == 0.0 and p.cycle(T // 4, T) == 1 # mean position, u = -U_max
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assert p.position(0, T, 1.0) == 1.0
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assert p.cycle(T // 4 - 1, T) == 0 and p.cycle(T // 4 + T - 1, T) == 1
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assert p.cycle(T // 4 + T, T) == 2
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assert p.phase(T // 6, T) == pytest.approx(330.0) and p.cycle(T // 6, T) == 0 # startup quarter
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assert p.phase(7 * T // 6, T) == pytest.approx(330.0) and p.cycle(7 * T // 6, T) == 1
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for ph, frac in ((180.0, .75), (210.0, 5 / 6)):
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s = round(frac * T) + T
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assert p.phase(s, T) == pytest.approx(ph) and p.bin_phase(s, T) == ph
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assert p.bin_phase(round(.75 * T) + T - 1, T) is None
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assert sum(p.bin_phase(s, T) == 180.0 for s in range(T, 2 * T)) == T // 360
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def test_settling_and_common_window():
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passes = {k: True for k in range(3, 12)}
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assert p.settled_start(passes) == 7 # pairs 3->4, 4->5, 5->6
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late = dict(passes, **{5: False})
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assert p.settled_start(late) == 9 # pairs 6->7, 7->8, 8->9
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assert p.common_window([7, 9], [passes, passes]) == (9, 10, 11)
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assert p.common_window([7, 10], [passes, passes]) is None # K + 2 = 12 beyond the cap
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assert p.common_window([7, None], [passes, passes]) is None
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assert p.common_window([7, 7], [passes, dict(passes, **{8: False})]) is None # must stay settled
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# a case settled early must keep passing while waiting for the common K
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assert p.common_window([7, 9], [dict(passes, **{8: False}), passes]) is None
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def test_solid_aware_bilinear():
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field = np.arange(16.0).reshape(4, 4)
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fluid = np.ones((4, 4), bool)
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value, used, total = p.bilinear_fluid(field, fluid, 1.25, 1.5)
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assert total == pytest.approx(1.0) and value == pytest.approx(0.75 * 0.5 * (5 + 9) + 0.25 * 0.5 * (6 + 10))
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fluid[1, 1] = False # one solid corner
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value, used, total = p.bilinear_fluid(field, fluid, 1.25, 1.5)
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assert total == pytest.approx(1.0 - 0.75 * 0.5)
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assert all(not (i == 1 and j == 1) for i, j, _ in used)
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assert sum(w for *_, w in used) == pytest.approx(1.0)
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fluid[:] = False
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assert p.bilinear_fluid(field, fluid, 1.25, 1.5) == (None, (), 0.0)
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def test_interval_bases_and_morison_fit_recover_known_coefficients():
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b = np.array([p.interval_bases(n, T) for n in range(T // 4, T // 4 + 3 * T)])
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phi = 2 * np.pi * (np.arange(T // 4, T // 4 + 3 * T) + .5) / T - np.pi / 2
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assert np.allclose(b[:, 1], np.sin(phi), atol=1e-5) # midpoint ~ interval mean
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f = 2.09 * b[:, 0] - (math.pi ** 2 * 1.45 / p.KC) * b[:, 1]
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assert p.morison_fit(f, b[:, 0], b[:, 1]) == pytest.approx((2.09, 1.45), abs=1e-9)
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