Both-halves phrases may still tie into a third family

four-inch and door hinge join the orange/storage/porridge clan even
though four/door and inch/hinge each rhyme on their own — a phrase
whose target group differs from both halves' groups carries new
information. The full orange verse is now a test.

Co-Authored-By: Claude Opus 4.8 (1M context) <noreply@anthropic.com>

app.py

@@ -556,30 +556,37 @@ def analyze(draft: Draft):
             for key in keys:
                 by_multi[(t["sid"], key)].append(t)
 
-    # a phrase only competes when no word inside it already rhymes
+    # which group holds each already-rhyming word, per line
     grouped_spans = defaultdict(list)
-    for t in tokens:
-        if id(t) in grouped:
-            grouped_spans[t["line"]].append((t["start"], t["end"]))
+    for gi, g in enumerate(raw_groups):
+        for t in g["toks"]:
+            if " " not in t["word"]:
+                grouped_spans[t["line"]].append((t["start"], t["end"], gi))
     for p in phrases:
         if p["weak"]:
             continue
-        # a phrase yields only when BOTH its halves already rhyme (pure
-        # redundancy); if one half is new, the phrase carries information
-        # (beast mode / sleep though: though already rhymes, beast doesn't)
-        spans = grouped_spans[p["line"]]
-        a_taken = any(s <= p["start"] < e for s, e in spans)
-        b_taken = any(s < p["end"] <= e for s, e in spans)
-        if a_taken and b_taken:
-            continue
         vs = p["vowels"]
         if len(vs) >= 3 or (len(vs) == 2 and vs[1] not in REDUCED):
             key = _multi_key(vs)
         else:
             # V+schwa phrases must bring consonant support
             key = _m2_key(p["rime"].split())
-        if key:
-            attach_or_collect(p, key, by_multi, group_by_multi)
+        if not key:
+            continue
+        spans = grouped_spans[p["line"]]
+        a_gi = next((gi for s, e, gi in spans if s <= p["start"] < e), None)
+        b_gi = next((gi for s, e, gi in spans if s < p["end"] <= e), None)
+        if a_gi is not None and b_gi is not None:
+            # both halves already rhyme — the phrase only matters if it
+            # ties into a THIRD family (four-inch joining the orange/
+            # storage clan while four sits with door and inch with hinge)
+            gi = group_by_multi.get((p["sid"], key))
+            if gi is not None and gi != a_gi and gi != b_gi:
+                raw_groups[gi]["toks"].append(p)
+                p["slant"] = True
+                grouped.add(id(p))
+            continue
+        attach_or_collect(p, key, by_multi, group_by_multi)
 
     # biggest buckets claim first (a token may sit in several via its
     # anchors); distinctness by anchor word, so the phrase "fire burns"

tests/test_rhymes.py

@@ -356,3 +356,13 @@ def test_phrase_joins_slant_group_via_coda_consensus():
             "It's by the door hinge\n"
             "eating pourage")
     group_with(text, "orange", "door hinge", "pourage")
+
+
+def test_the_full_orange_verse():
+    # the Eminem demonstration: every line ties back to orange
+    text = ("I put my orange\nfour-inch\ndoor hinge\nin storage,\n"
+            "and ate porridge with George")
+    group_with(text, "orange", "storage", "porridge",
+               "four-inch", "door hinge")
+    group_with(text, "four", "door", "george")
+    group_with(text, "inch", "hinge")