mercurial/ancestor.py

run-tests: sort options

# 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 version 2 or any later version.

import heapq

def ancestor(a, b, pfunc):
    """
    return a minimal-distance ancestor of nodes a and b, or None if there is no
    such ancestor. Note that there can be several ancestors with the same
    (minimal) distance, and the one returned is arbitrary.

    pfunc must return a list of parent vertices for a given vertex
    """

    if a == b:
        return a

    # find depth from root of all ancestors
    parentcache = {}
    visit = [a, b]
    depth = {}
    while visit:
        vertex = visit[-1]
        pl = pfunc(vertex)
        parentcache[vertex] = pl
        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 = set()
        while h:
            d, n = heapq.heappop(h)
            if n not in seen:
                seen.add(n)
                yield (d, n)
                for p in parentcache[n]:
                    heapq.heappush(h, (depth[p], p))

    def generations(vertex):
        sg, s = None, set()
        for g, v in ancestors(vertex):
            if g != sg:
                if sg:
                    yield sg, s
                sg, s = g, set((v,))
            else:
                s.add(v)
        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