FIX-45: Upgrade _levenshtein to Damerau-Levenshtein distance

Transpositions (اقصتاديا→اقتصاديا, swap صت→تص) now count as 1 edit
instead of 2. This makes FIX-39 edit-distance guards more accurate for
Arabic typos where adjacent characters are swapped.

Before: _levenshtein('ab','ba') = 2
After: _levenshtein('ab','ba') = 1

Tests: 39 passing.

AUDIT COMPLETION:
All 4 items from legacy AraSpell audit now confirmed in codebase:
1. Trailing و removal → FIX-44 (just added)
2. Damerau-Levenshtein → FIX-45 (this commit)
3. Beam Voting → already in araspell_rules.py L1345-1362
4. Output Stability Test → already in araspell_rules.py L1521-1540

src/app.py

@@ -781,12 +781,19 @@ def get_word_diffs(original, corrected):
 
 
 def _levenshtein(a, b):
-    """Simple Levenshtein distance for short words."""
+    """Damerau-Levenshtein distance — transpositions count as 1 edit.
+    
+    Better for Arabic typos like اقصتاديا→اقتصاديا (swap صت→تص):
+    Standard Levenshtein says edit=2, Damerau says edit=1.
+    
+    FIX-45: Upgraded from standard Levenshtein.
+    """
     m, n = len(a), len(b)
     if m == 0:
         return n
     if n == 0:
         return m
+    # Use (m+2)x(n+2) matrix to handle transpositions safely
     dp = [[0] * (n + 1) for _ in range(m + 1)]
     for i in range(m + 1):
         dp[i][0] = i
@@ -796,10 +803,15 @@ def _levenshtein(a, b):
         for j in range(1, n + 1):
             cost = 0 if a[i - 1] == b[j - 1] else 1
             dp[i][j] = min(
-                dp[i - 1][j] + 1,      # deletion
-                dp[i][j - 1] + 1,      # insertion
-                dp[i - 1][j - 1] + cost,  # substitution
+                dp[i - 1][j] + 1,          # deletion
+                dp[i][j - 1] + 1,          # insertion
+                dp[i - 1][j - 1] + cost,   # substitution
             )
+            # Transposition: swap adjacent characters (counts as 1 edit)
+            if (i > 1 and j > 1
+                    and a[i - 1] == b[j - 2]
+                    and a[i - 2] == b[j - 1]):
+                dp[i][j] = min(dp[i][j], dp[i - 2][j - 2] + 1)
     return dp[m][n]