feat: disable two-word adjective+noun search by default

The new Wiktionary dictionary already includes 31k multi-word nouns,
making the combinatorial adjective+noun phase redundant. Add two_word
parameter (default False) to solve_word_dictionary to opt in when needed.

Co-Authored-By: Claude Opus 4.6 <[email protected]>

tests/test_word_solve.py

@@ -94,7 +94,7 @@ class TestSolveWordDictionaryTwoWord:
         target = font.getlength("large house")
         results = list(solve_word_dictionary(
             font, target, tolerance=1.0,
-            casing="lowercase", known_start="large",
+            casing="lowercase", known_start="large", two_word=True,
         ))
         texts = [r.text for r in results]
         assert "large house" in texts
@@ -104,7 +104,7 @@ class TestSolveWordDictionaryTwoWord:
         target = font.getlength("Large House")
         results = list(solve_word_dictionary(
             font, target, tolerance=1.0,
-            casing="capitalized", known_start="Large",
+            casing="capitalized", known_start="Large", two_word=True,
         ))
         texts = [r.text for r in results]
         assert "Large House" in texts
@@ -115,6 +115,7 @@ class TestSolveWordDictionaryTwoWord:
         results = list(solve_word_dictionary(
             font, target, tolerance=1.0,
             casing="lowercase", ensure_plural=True, known_start="large",
+            two_word=True,
         ))
         texts = [r.text for r in results]
         assert "large houses" in texts
@@ -124,7 +125,7 @@ class TestSolveWordDictionaryTwoWord:
         target = font.getlength("large house")
         results = list(solve_word_dictionary(
             font, target, tolerance=1.0,
-            casing="lowercase", known_start="la",
+            casing="lowercase", known_start="la", two_word=True,
         ))
         texts = [r.text for r in results]
         assert "large house" in texts
@@ -135,18 +136,30 @@ class TestSolveWordDictionaryTwoWord:
         target = font.getlength("large house")
         results = list(solve_word_dictionary(
             font, target, tolerance=1.0,
-            casing="lowercase", known_end="se",
+            casing="lowercase", known_end="se", two_word=True,
         ))
         texts = [r.text for r in results]
         assert "large house" in texts
         assert all(t.lower().endswith("se") for t in texts)
 
+    def test_two_word_disabled_by_default(self):
+        """Without two_word=True, only dictionary nouns are searched."""
+        font = _get_test_font()
+        target = font.getlength("large house")
+        results = list(solve_word_dictionary(
+            font, target, tolerance=1.0,
+            casing="lowercase", known_start="large",
+        ))
+        texts = [r.text for r in results]
+        # "large house" is not a dictionary noun, so it should NOT appear
+        assert "large house" not in texts
+
     def test_no_duplicates_across_phases(self):
         font = _get_test_font()
         target = font.getlength("dog")
         results = list(solve_word_dictionary(
             font, target, tolerance=1.0,
-            casing="lowercase",
+            casing="lowercase", two_word=True,
         ))
         texts = [r.text for r in results]
         assert len(texts) == len(set(texts))

unredact/pipeline/dictionary.py

@@ -276,11 +276,13 @@ def solve_word_dictionary(
     known_end: str = "",
     ensure_plural: bool = False,
     vocab_size: int = 0,
+    two_word: bool = False,
 ):
-    """Search English nouns (single-word) and adj+noun phrases (two-word).
+    """Search English nouns and optionally adj+noun phrases.
 
-    Yields results as found for SSE streaming. Phase 1 searches single nouns,
-    phase 2 searches two-word combinations using binary search.
+    Yields results as found for SSE streaming. Phase 1 searches nouns
+    (including multi-word nouns already in the dictionary). Phase 2
+    (disabled by default) searches adjective+noun combinations.
 
     Word lists are frequency-sorted (most common first). vocab_size > 0 slices
     to the N most common words; 0 means no limit.
@@ -327,6 +329,9 @@ def solve_word_dictionary(
         if error <= tolerance:
             yield SolveResult(text=display, width=float(width), error=float(error))
 
+    if not two_word:
+        return
+
     # Phase 2: two-word search (word1 + " " + noun)
     import bisect