A T depends on every connection between its copper and its group

Anchoring a T only on the connection whose copper it touched missed the
connections that tie that copper to the T's own group. When one of those
was ripped, the T stayed routed but cut off: on charge-sense freeroute
counted 15 unrouted while KiCad found 27 opens. Routed connections of a net
form a forest, so the path back is unique; the T now depends on all of it.
KiCad's opens and freeroute's count agree again (18 against 17, the extra
one a GND pour island left to the fill).
This commit is contained in:
2026-10-05 12:58:58 -06:00
parent b91e44e360
commit 61f51129e1
2 changed files with 52 additions and 16 deletions
+29
View File
@@ -139,3 +139,32 @@ def test_occupancy_stays_honest(tree_nets):
for anchor, hanging in router._dependents.items():
for i in hanging:
assert conns[i].routed and conns[anchor].routed
# P1-P2 along the top, P3 far off to the lower right, P4 just under the
# middle of the P1-P2 leg.
CHAIN_TEE = TEE.replace(
""" (place P3 100000 -70000 front 0)
""",
""" (place P3 190000 -90000 front 0)
(place P4 100000 -40000 front 0)
""",
).replace("(pins P1-1 P2-1 P3-1)", "(pins P1-1 P2-1 P3-1 P4-1)")
def test_a_tee_depends_on_every_connection_back_to_its_group():
"""Joined as P1-P2, P2-P3, P3-P4, P4's connection tees onto P1-P2 copper,
which reaches P3's group only through P2-P3. Ripping P2-P3 must take the T
too, or it would be counted routed while cut off from P3."""
router = _router(text=CHAIN_TEE)
conns = router._conns = router._build_connections()
groups = {c.a.component_ref: c.a_group for c in conns}
groups.update({c.b.component_ref: c.b_group for c in conns})
for conn, (a, b) in zip(conns, (("P1", "P2"), ("P2", "P3"), ("P3", "P4")), strict=True):
conn.a_group, conn.b_group = groups[a], groups[b]
for conn in conns:
assert router._try_route(conn, conns)
first, second, third = conns
assert third.path[0] in first.copper # the T lands on the far leg
assert third.anchors == {first.index, second.index}
assert router._with_dependents([second], conns) == [second, third]