Mercurial > hg
view mercurial/ancestor.py @ 15880:02b135558756
mq: make qsave implementation more explicit
It wasn't obvious from the code how qsave mocked around with .hg/patches and
.hg/patches.? and what was going on.
This makes it more explicit so it will survive future refactorings.
author | Mads Kiilerich <mads@kiilerich.com> |
---|---|
date | Wed, 11 Jan 2012 02:29:55 +0100 |
parents | 1ffeeb91c55d |
children | 0b03454abae7 |
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 version 2 or any later version. import heapq def ancestor(a, b, pfunc): """ Returns the common ancestor of a and b that is furthest from a root (as measured by longest path) or None if no ancestor is found. If there are multiple common ancestors at the same distance, the first one found is returned. pfunc must return a list of parent vertices for a given vertex """ if a == b: return a a, b = sorted([a, b]) # find depth from root of all ancestors # depth is stored as a negative for heapq 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: # -(maximum distance of parents + 1) 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 True: 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