Files
kicad-zone-resistance/tests/test_quadtree.py
T
janik 14cdf4c7ee Add balanced quadtree grid engine (adaptive cells, phase 1)
fill_resistance/quadtree.py decomposes a layer's fine copper mask into
2:1-balanced power-of-two leaves: boundary and keep-fine cells stay at
the fine size, interiors coarsen with their Chebyshev distance to the
nearest feature (guard factor, default 4), and an explicit enforcement
pass splits any leaf more than twice an edge-adjacent neighbor. Face
conductances use the series-half-cell rule, which reduces to the
production harmonic mean for equal sizes and EXACTLY to sigma in the
uniform limit - verified edge-for-edge against solver.build_edges and
to rel 1e-12 in R against run_solve, so the exact-value test suite
stays authoritative for this engine.

Measured (feature-dense 120x120 plate, h=50um, production AMG solver):
uniform 5.58M unknowns ~35s; adaptive guard=4 823k / ~8s at -1.1%;
guard=8 1.74M / ~12s at -0.47%. tools/adaptive_proto.py now benchmarks
the engine itself.

Not yet wired into the pipeline: phase 2 ports electrodes, barrels,
1D chains, buildup and field output onto leaves.

Co-Authored-By: Claude Fable 5 <noreply@anthropic.com>
2026-07-15 17:05:31 +07:00

142 lines
5.6 KiB
Python

"""Quadtree grid engine tests (phase 1): exact uniform limit against the
production graph and solver, partition/alignment/balance invariants, and
adaptive-vs-fine R agreement."""
import numpy as np
import pytest
from scipy import ndimage
from fill_resistance import quadtree, raster, solver
from tests.util import NM, make_problem, strip_problem
def _plate_with_holes(n=5, size_mm=40.0):
holes = []
pitch = size_mm / n
for i in range(n):
for j in range(n):
x, y = pitch * (i + 0.4), pitch * (j + 0.4)
holes.append([(x, y), (x + 1, y), (x + 1, y + 1), (x, y + 1)])
outline = [(0, 0), (size_mm, 0), (size_mm, size_mm), (0, size_mm)]
return make_problem([(outline, holes)],
rect1_mm=(0, 15, 2, 25),
rect2_mm=(size_mm - 2, 15, size_mm, 25))
def _solve_on_leaves(problem, stack, e1, e2, grid):
"""Equipotential mini-solve on the leaf graph, using the production
assembly and linear solver."""
ia, ib, g = quadtree.leaf_edges(grid, problem.sigma_s(0))
state = np.ones(grid.n, dtype=np.uint8)
for e, code in ((e1[0], 2), (e2[0], 3)):
ids = grid.id_grid[e]
state[ids[ids >= 0]] = code
edges = solver.Edges(a=ia, b=ib, w=g,
via_index=np.full(len(ia), -1, dtype=np.int32))
A, rhs, _ = solver._assemble(state, edges, None)
x, _ = solver.solve_system(A, rhs)
V = np.zeros(grid.n)
V[state == 2] = 1.0
V[state == 1] = x
Ie = g * (V[ia] - V[ib])
sa, sb = state[ia], state[ib]
I1 = float(Ie[sa == 2].sum() - Ie[sb == 2].sum())
I2 = float(Ie[sb == 3].sum() - Ie[sa == 3].sum())
return 1.0 / (0.5 * (I1 + I2))
def test_uniform_limit_graph_identical():
"""max_block=1: one leaf per cell, and the edge list matches the
production in-plane graph exactly (same pairs, conductance sigma)."""
p = _plate_with_holes()
stack = raster.rasterize_stack(p, 0.5 * NM)
grid = quadtree.build_leaves(stack.masks[0], max_block=1)
assert int(stack.masks.sum()) == grid.n
assert (grid.size == 1).all()
ny, nx = stack.shape2d
flat_of_leaf = grid.y0.astype(np.int64) * nx + grid.x0
ia, ib, g = quadtree.leaf_edges(grid, p.sigma_s(0))
ours = np.sort(np.stack([
np.minimum(flat_of_leaf[ia], flat_of_leaf[ib]),
np.maximum(flat_of_leaf[ia], flat_of_leaf[ib])], axis=1), axis=0)
edges = solver.build_edges(stack, p, [p.sigma_s(0)])
ref = np.sort(np.stack([np.minimum(edges.a, edges.b),
np.maximum(edges.a, edges.b)], axis=1), axis=0)
assert ours.shape == ref.shape
assert np.array_equal(np.sort(ours.view("i8,i8"), order=["f0", "f1"],
axis=0),
np.sort(ref.view("i8,i8"), order=["f0", "f1"],
axis=0))
assert np.allclose(g, p.sigma_s(0), rtol=0, atol=0)
def test_uniform_limit_R_matches_production():
p = strip_problem(length=50, width=10, e_len=5)
stack = raster.rasterize_stack(p, 0.25 * NM)
e1, e2 = raster.electrode_masks(stack, p)
ref = solver.run_solve(p, stack, e1, e2, 1.0,
contact_model="equipotential")
stack2 = raster.rasterize_stack(p, 0.25 * NM)
e1b, e2b = raster.electrode_masks(stack2, p)
grid = quadtree.build_leaves(stack2.masks[0], max_block=1)
R = _solve_on_leaves(p, stack2, e1b, e2b, grid)
assert R == pytest.approx(ref.R_ohm, rel=1e-12)
def test_partition_alignment_and_balance():
p = _plate_with_holes()
stack = raster.rasterize_stack(p, 0.1 * NM)
mask = stack.masks[0]
grid = quadtree.build_leaves(mask, max_block=32)
# exact partition of the copper
assert int((grid.size.astype(np.int64) ** 2).sum()) == int(mask.sum())
assert (grid.id_grid >= 0).sum() == int(mask.sum())
assert not (grid.id_grid[~mask] >= 0).any()
counts = np.bincount(grid.id_grid[grid.id_grid >= 0], minlength=grid.n)
assert np.array_equal(counts, grid.size.astype(np.int64) ** 2)
# power-of-two sizes, aligned to their own size
assert np.array_equal(grid.size & (grid.size - 1),
np.zeros_like(grid.size))
assert (grid.y0 % grid.size == 0).all()
assert (grid.x0 % grid.size == 0).all()
# 2:1 balance and real coarsening (max size is geometry-limited by
# the guard distance, not by max_block, on this feature-dense plate)
assert quadtree.balanced(grid)
assert grid.n < 0.5 * int(mask.sum())
assert grid.size.max() >= 4
def test_boundary_and_keep_fine_stay_fine():
p = _plate_with_holes()
stack = raster.rasterize_stack(p, 0.1 * NM)
mask = stack.masks[0]
keep = np.zeros_like(mask)
keep[50:60, 50:60] = True
grid = quadtree.build_leaves(mask, keep_fine=keep, max_block=32)
boundary = mask & ndimage.binary_dilation(~mask)
assert (grid.size[grid.id_grid[boundary]] == 1).all()
assert (grid.size[grid.id_grid[keep & mask]] == 1).all()
def test_adaptive_R_close_to_fine():
"""Adaptive leaves reproduce the fine-uniform R within 1% on the
holey plate (features everywhere - the adversarial case)."""
p = _plate_with_holes()
stack = raster.rasterize_stack(p, 0.1 * NM)
e1, e2 = raster.electrode_masks(stack, p)
ref = solver.run_solve(p, stack, e1, e2, 1.0,
contact_model="equipotential")
stack2 = raster.rasterize_stack(p, 0.1 * NM)
e1b, e2b = raster.electrode_masks(stack2, p)
grid = quadtree.build_leaves(stack2.masks[0],
keep_fine=(e1b[0] | e2b[0]))
R = _solve_on_leaves(p, stack2, e1b, e2b, grid)
assert grid.n < 0.4 * ref.solve_info.n_unknowns
assert R == pytest.approx(ref.R_ohm, rel=0.01)