Mercurial > hg
view mercurial/ancestor.py @ 17777:af7c6bc48d8d
run-tests: alternative way of handling \r on Windows
After f71d60da58fb all \r was stripped from output on Windows, and the places
where a \r explicitly was expected it was accepted that it was missing. Ugly
hack.
Instead we now accept that an extra \r might appear at the end of lines on
Windows. That is more to the point and less ugly.
author | Mads Kiilerich <mads@kiilerich.com> |
---|---|
date | Mon, 15 Oct 2012 02:33:12 +0200 |
parents | 1ffeeb91c55d |
children | 0b03454abae7 |
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# 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