fix: require both horizontal AND vertical overlap for word dedup

Previous version only checked X overlap, causing false positives for
short words like "=" and "I" that appear at similar X positions in
different rows. Now requires >=50% overlap in both dimensions.

Co-Authored-By: Claude Opus 4.6 <noreply@anthropic.com>

klausur-service/backend/ocr_pipeline_api.py

@@ -3359,11 +3359,14 @@ def _merge_paddle_tesseract(paddle_words: list, tess_words: list) -> list:
     return merged_all
 
 
-def _deduplicate_words(words: list, overlap_ratio: float = 0.4) -> list:
+def _deduplicate_words(words: list) -> list:
     """Remove duplicate words with same text at overlapping positions.
 
     PaddleOCR can return overlapping phrases (e.g. "von jm." and "jm. =")
     that produce duplicate words after splitting.  This pass removes them.
+
+    A word is a duplicate only when BOTH horizontal AND vertical overlap
+    exceed 50% — same text on the same visual line at the same position.
     """
     if not words:
         return words
@@ -3374,19 +3377,25 @@ def _deduplicate_words(words: list, overlap_ratio: float = 0.4) -> list:
         if not wt:
             continue
         is_dup = False
+        w_right = w["left"] + w.get("width", 0)
+        w_bottom = w["top"] + w.get("height", 0)
         for existing in result:
             et = existing.get("text", "").lower().strip()
             if wt != et:
                 continue
-            # Check horizontal overlap
-            ol = max(w["left"], existing["left"])
-            or_ = min(
-                w["left"] + w.get("width", 0),
-                existing["left"] + existing.get("width", 0),
-            )
-            ow = max(0, or_ - ol)
+            # Horizontal overlap
+            ox_l = max(w["left"], existing["left"])
+            ox_r = min(w_right, existing["left"] + existing.get("width", 0))
+            ox = max(0, ox_r - ox_l)
             min_w = min(w.get("width", 1), existing.get("width", 1))
-            if min_w > 0 and ow / min_w >= overlap_ratio:
+            if min_w <= 0 or ox / min_w < 0.5:
+                continue
+            # Vertical overlap — must also be on the same line
+            oy_t = max(w["top"], existing["top"])
+            oy_b = min(w_bottom, existing["top"] + existing.get("height", 0))
+            oy = max(0, oy_b - oy_t)
+            min_h = min(w.get("height", 1), existing.get("height", 1))
+            if min_h > 0 and oy / min_h >= 0.5:
                 is_dup = True
                 break
         if not is_dup: