view mercurial/ancestor.py @ 7104:9514cbb6e4f6

bdiff: normalize the diff (issue1295)

When the common part of a diff can be moved forward, move it forward.
Otherwise we don't get deterministic results (it would depends on the way we
split for the recursion).
That way we get identical hunks when doing the same change, it helps to solve
issue1295 (inconsistent diffs on different side during a merge).

# ancestor.py - generic DAG ancestor algorithm for mercurial
#
# Copyright 2006 Matt Mackall <[e-mail]>
#
# This software may be used and distributed according to the terms
# of the GNU General Public License, incorporated herein by reference.

import heapq

def ancestor(a, b, pfunc):
    """
    return the least common ancestor of nodes a and b or None if there
    is no such ancestor.

    pfunc must return a list of parent vertices
    """

    if a == b:
        return a

    # find depth from root of all ancestors
    visit = [a, b]
    depth = {}
    while visit:
        vertex = visit[-1]
        pl = pfunc(vertex)
        if not pl:
            depth[vertex] = 0
            visit.pop()
        else:
            for p in pl:
                if p == a or p == b: # did we find a or b as a parent?
                    return p # we're done
                if p not in depth:
                    visit.append(p)
            if visit[-1] == vertex:
                depth[vertex] = min([depth[p] for p in pl]) - 1
                visit.pop()

    # traverse ancestors in order of decreasing distance from root
    def ancestors(vertex):
        h = [(depth[vertex], vertex)]
        seen = {}
        while h:
            d, n = heapq.heappop(h)
            if n not in seen:
                seen[n] = 1
                yield (d, n)
                for p in pfunc(n):
                    heapq.heappush(h, (depth[p], p))

    def generations(vertex):
        sg, s = None, {}
        for g, v in ancestors(vertex):
            if g != sg:
                if sg:
                    yield sg, s
                sg, s = g, {v:1}
            else:
                s[v] = 1
        yield sg, s

    x = generations(a)
    y = generations(b)
    gx = x.next()
    gy = y.next()

    # increment each ancestor list until it is closer to root than
    # the other, or they match
    try:
        while 1:
            if gx[0] == gy[0]:
                for v in gx[1]:
                    if v in gy[1]:
                        return v
                gy = y.next()
                gx = x.next()
            elif gx[0] > gy[0]:
                gy = y.next()
            else:
                gx = x.next()
    except StopIteration:
        return None