#!/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
