feat(places): registry of towns and authorities

Two namespaces because 67 town names collide with an authority name and
neither set contains the other — postal towns cross authority boundaries, so
Bedford the town holds 104 schools against the authority's 86.

Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_015mWQnpye9F299NVRCCSRvj

docs/superpowers/plans/2026-08-21-w2-location-layer.md

@@ -132,9 +132,12 @@ def _df(rows: list[dict]) -> pd.DataFrame:
     return pd.DataFrame([{**base, **r} for r in rows])
 
 
-def _town(n: int, town: str, la: str, **kw) -> list[dict]:
+def _town(n: int, town: str, la: str, start: int = 100000, **kw) -> list[dict]:
+    """`start` offsets the URNs so two calls can describe different schools —
+    the Bedford case needs two authorities' worth of distinct URNs in one
+    town."""
     return [
-        {"urn": 100000 + i, "school_name": f"{town} School {i}",
+        {"urn": start + i, "school_name": f"{town} School {i}",
          "town": town, "local_authority": la, **kw}
         for i in range(n)
     ]
@@ -164,7 +167,7 @@ def test_town_and_authority_of_the_same_name_are_separate_places():
     # 104 schools, Bedford the authority 86, because postal towns cross
     # authority boundaries.
     rows = (_town(MIN_SCHOOLS, "Bedford", "Bedford")
-            + _town(MIN_SCHOOLS, "Bedford", "Central Bedfordshire"))
+            + _town(MIN_SCHOOLS, "Bedford", "Central Bedfordshire", start=200000))
     reg = build_place_registry(_df(rows))
     town, authority = reg["town:bedford"], reg["authority:bedford"]
     assert set(town.urns) != set(authority.urns)