##// END OF EJS Templates
revlog: faster hash computation when one of the parent node is null...
revlog: faster hash computation when one of the parent node is null Because we often compute sha1(nullid), it's interesting to copy a precomputed hash of nullid instead of computing everytime the same hash. Similarly, when one of the parents is null, we can avoid a < comparison (sort). Overall, this change adds a string equality comparison on each hash() call, but when p2 is null, we drop one string < comparison, and copy a hash instead of computing it. Since it is common to have revisions with only one parent, this change makes hash() 25% faster when cloning a big repository.

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ancestor.py
85 lines | 2.3 KiB | text/x-python | PythonLexer
# 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
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 = {}
while h:
d, n = heapq.heappop(h)
if n not in seen:
seen[n] = 1
yield (d, n)
for p in parentcache[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