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H��H�Y4����L��A\�H��H���sequence expectedOOOreplacedeleteinsertsnnnnequal|iOOnnnnOiPatienceSequenceMatcher_c_patiencediff_cunique_lcs_crecurse_matches_cPatienceSequenceMatcherget_matching_blocksget_opcodesget_grouped_opcodesC implementation of PatienceSequenceMatcher;����d���t��TT������@���xD��������X����h$����hT��4�DzRx�$p��PFJw�?;*3$"D���`\p��vB�F�E �B(�A0�A8�I`(
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���o ]6FVfv��������&6FVf`C implementation of PatienceSequenceMatcherIsolate change clusters by eliminating ranges with no changes.

Return a list of groups with upto n lines of context.
Each group is in the same format as returned by get_opcodes().

>>> from pprint import pprint
>>> a = map(str, range(1,40))
>>> b = a[:]
>>> b[8:8] = ['i']     # Make an insertion
>>> b[20] += 'x'       # Make a replacement
>>> b[23:28] = []      # Make a deletion
>>> b[30] += 'y'       # Make another replacement
>>> pprint(PatienceSequenceMatcher(None,a,b).get_grouped_opcodes())
[[('equal', 5, 8, 5, 8), ('insert', 8, 8, 8, 9), ('equal', 8, 11, 9, 12)],
 [('equal', 16, 19, 17, 20),
  ('replace', 19, 20, 20, 21),
  ('equal', 20, 22, 21, 23),
  ('delete', 22, 27, 23, 23),
  ('equal', 27, 30, 23, 26)],
 [('equal', 31, 34, 27, 30),
  ('replace', 34, 35, 30, 31),
  ('equal', 35, 38, 31, 34)]]
Return list of 5-tuples describing how to turn a into b.

Each tuple is of the form (tag, i1, i2, j1, j2).  The first tuple
has i1 == j1 == 0, and remaining tuples have i1 == the i2 from the
tuple preceding it, and likewise for j1 == the previous j2.

The tags are strings, with these meanings:

'replace':  a[i1:i2] should be replaced by b[j1:j2]
'delete':   a[i1:i2] should be deleted.
               Note that j1==j2 in this case.
'insert':   b[j1:j2] should be inserted at a[i1:i1].
               Note that i1==i2 in this case.
'equal':    a[i1:i2] == b[j1:j2]

>>> a = "qabxcd"
>>> b = "abycdf"
>>> s = PatienceSequenceMatcher(None, a, b)
>>> for tag, i1, i2, j1, j2 in s.get_opcodes():
...    print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" %
...           (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2]))
 delete a[0:1] (q) b[0:0] ()
  equal a[1:3] (ab) b[0:2] (ab)
replace a[3:4] (x) b[2:3] (y)
  equal a[4:6] (cd) b[3:5] (cd)
 insert a[6:6] () b[5:6] (f)
Return list of triples describing matching subsequences.

Each triple is of the form (i, j, n), and means that
a[i:i+n] == b[j:j+n].  The triples are monotonically increasing in
i and in j.

The last triple is a dummy, (len(a), len(b), 0), and is the only
triple with n==0.

>>> s = PatienceSequenceMatcher(None, "abxcd", "abcd")
>>> s.get_matching_blocks()
[(0, 0, 2), (3, 2, 2), (5, 4, 0)]
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