view mercurial/ancestor.py @ 4370:4ddc6d374265

localrepository.status: only acquire wlock if actually needed. This speeds up the common case of not needing to update the dirstate, and avoids the need to reload and parse the dirstate "just in case".
author Bryan O'Sullivan <bos@serpentine.com>
date Tue, 24 Apr 2007 11:05:39 -0700
parents eb0b4a2d70a9
children
line wrap: on
line source

# ancestor.py - generic DAG ancestor algorithm for mercurial
#
# Copyright 2006 Matt Mackall <mpm@selenic.com>
#
# 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