/brz/remove-bazaar

#!/usr/bin/env python

# Copyright (C) 2005 Bram Cohen, Copyright (C) 2005, 2006 Canonical Ltd

#

# This program is free software; you can redistribute it and/or modify

# it under the terms of the GNU General Public License as published by

# the Free Software Foundation; either version 2 of the License, or

# (at your option) any later version.

#

# This program is distributed in the hope that it will be useful,

# but WITHOUT ANY WARRANTY; without even the implied warranty of

# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the

# GNU General Public License for more details.

#

# You should have received a copy of the GNU General Public License

# along with this program; if not, write to the Free Software

# Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA

from bisect import bisect

import difflib

from bzrlib.trace import mutter

__all__ = ['PatienceSequenceMatcher', 'unified_diff', 'unified_diff_files']

def unique_lcs_py(a, b):

    """Find the longest common subset for unique lines.

    :param a: An indexable object (such as string or list of strings)

    :param b: Another indexable object (such as string or list of strings)

    :return: A list of tuples, one for each line which is matched.

            [(line_in_a, line_in_b), ...]

    This only matches lines which are unique on both sides.

    This helps prevent common lines from over influencing match

    results.

    The longest common subset uses the Patience Sorting algorithm:

    http://en.wikipedia.org/wiki/Patience_sorting

    """

    # set index[line in a] = position of line in a unless

    # a is a duplicate, in which case it's set to None

    index = {}

    for i in xrange(len(a)):

        line = a[i]

        if line in index:

            index[line] = None

        else:

            index[line]= i

    # make btoa[i] = position of line i in a, unless

    # that line doesn't occur exactly once in both,

    # in which case it's set to None

    btoa = [None] * len(b)

    index2 = {}

    for pos, line in enumerate(b):

        next = index.get(line)

        if next is not None:

            if line in index2:

                # unset the previous mapping, which we now know to

                # be invalid because the line isn't unique

                btoa[index2[line]] = None

                del index[line]

            else:

                index2[line] = pos

                btoa[pos] = next

    # this is the Patience sorting algorithm

    # see http://en.wikipedia.org/wiki/Patience_sorting

    backpointers = [None] * len(b)

    stacks = []

    lasts = []

    k = 0

    for bpos, apos in enumerate(btoa):

        if apos is None:

            continue

        # as an optimization, check if the next line comes at the end,

        # because it usually does

        if stacks and stacks[-1] < apos:

            k = len(stacks)

        # as an optimization, check if the next line comes right after

        # the previous line, because usually it does

        elif stacks and stacks[k] < apos and (k == len(stacks) - 1 or

                                              stacks[k+1] > apos):

            k += 1

        else:

            k = bisect(stacks, apos)

        if k > 0:

            backpointers[bpos] = lasts[k-1]

        if k < len(stacks):

            stacks[k] = apos

            lasts[k] = bpos

        else:

            stacks.append(apos)

            lasts.append(bpos)

    if len(lasts) == 0:

        return []

    result = []

    k = lasts[-1]

    while k is not None:

        result.append((btoa[k], k))

        k = backpointers[k]

    result.reverse()

    return result

def recurse_matches_py(a, b, alo, blo, ahi, bhi, answer, maxrecursion):

    """Find all of the matching text in the lines of a and b.

    :param a: A sequence

    :param b: Another sequence

    :param alo: The start location of a to check, typically 0

    :param ahi: The start location of b to check, typically 0

    :param ahi: The maximum length of a to check, typically len(a)

    :param bhi: The maximum length of b to check, typically len(b)

    :param answer: The return array. Will be filled with tuples

                   indicating [(line_in_a, line_in_b)]

    :param maxrecursion: The maximum depth to recurse.

                         Must be a positive integer.

    :return: None, the return value is in the parameter answer, which

             should be a list

    """

    if maxrecursion < 0:

        mutter('max recursion depth reached')

        # this will never happen normally, this check is to prevent DOS attacks

        return

    oldlength = len(answer)

    if alo == ahi or blo == bhi:

        return

    last_a_pos = alo-1

    last_b_pos = blo-1

    for apos, bpos in unique_lcs_py(a[alo:ahi], b[blo:bhi]):

        # recurse between lines which are unique in each file and match

        apos += alo

        bpos += blo

        # Most of the time, you will have a sequence of similar entries

        if last_a_pos+1 != apos or last_b_pos+1 != bpos:

            recurse_matches_py(a, b, last_a_pos+1, last_b_pos+1,

                apos, bpos, answer, maxrecursion - 1)

        last_a_pos = apos

        last_b_pos = bpos

        answer.append((apos, bpos))

    if len(answer) > oldlength:

        # find matches between the last match and the end

        recurse_matches_py(a, b, last_a_pos+1, last_b_pos+1,

                           ahi, bhi, answer, maxrecursion - 1)

    elif a[alo] == b[blo]:

        # find matching lines at the very beginning

        while alo < ahi and blo < bhi and a[alo] == b[blo]:

            answer.append((alo, blo))

            alo += 1

            blo += 1

        recurse_matches_py(a, b, alo, blo,

                           ahi, bhi, answer, maxrecursion - 1)

    elif a[ahi - 1] == b[bhi - 1]:

        # find matching lines at the very end

        nahi = ahi - 1

        nbhi = bhi - 1

        while nahi > alo and nbhi > blo and a[nahi - 1] == b[nbhi - 1]:

            nahi -= 1

            nbhi -= 1

        recurse_matches_py(a, b, last_a_pos+1, last_b_pos+1,

                           nahi, nbhi, answer, maxrecursion - 1)

        for i in xrange(ahi - nahi):

            answer.append((nahi + i, nbhi + i))

def _collapse_sequences(matches):

    """Find sequences of lines.

    Given a sequence of [(line_in_a, line_in_b),]

    find regions where they both increment at the same time

    """

    answer = []

    start_a = start_b = None

    length = 0

    for i_a, i_b in matches:

        if (start_a is not None

            and (i_a == start_a + length)

            and (i_b == start_b + length)):

            length += 1

        else:

            if start_a is not None:

                answer.append((start_a, start_b, length))

            start_a = i_a

            start_b = i_b

            length = 1

    if length != 0:

        answer.append((start_a, start_b, length))

    return answer

def _check_consistency(answer):

    # For consistency sake, make sure all matches are only increasing

    next_a = -1

    next_b = -1

    for (a, b, match_len) in answer:

        if a < next_a:

            raise ValueError('Non increasing matches for a')

        if b < next_b:

            raise ValueError('Non increasing matches for b')

        next_a = a + match_len

        next_b = b + match_len

class PatienceSequenceMatcher_py(difflib.SequenceMatcher):

    """Compare a pair of sequences using longest common subset."""

    _do_check_consistency = True

    def __init__(self, isjunk=None, a='', b=''):

        if isjunk is not None:

            raise NotImplementedError('Currently we do not support'

                                      ' isjunk for sequence matching')

        difflib.SequenceMatcher.__init__(self, isjunk, a, b)

    def get_matching_blocks(self):

        """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)]

        """

        # jam 20060525 This is the python 2.4.1 difflib get_matching_blocks

        # implementation which uses __helper. 2.4.3 got rid of helper for

        # doing it inline with a queue.

        # We should consider doing the same for recurse_matches

        if self.matching_blocks is not None:

            return self.matching_blocks

        matches = []

        recurse_matches_py(self.a, self.b, 0, 0,

                           len(self.a), len(self.b), matches, 10)

        # Matches now has individual line pairs of

        # line A matches line B, at the given offsets

        self.matching_blocks = _collapse_sequences(matches)

        self.matching_blocks.append( (len(self.a), len(self.b), 0) )

        if PatienceSequenceMatcher_py._do_check_consistency:

            if __debug__:

                _check_consistency(self.matching_blocks)

        return self.matching_blocks