Abstract ancestor algorithm into generic function

Make depth calculation non-recursive
Add simple shortcut for linear ancestry

Convert context to use ancestor function
 make memoized parents function

Convert revlog to use ancestor function

mercurial/ancestor.py

@@ -0,0 +1,83 b''
+# 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, 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

mercurial/context.py

@@ -7,7 +7,7 b''
 
 from node import *
 from demandload import demandload
-demandload(globals(), "heapq")
+demandload(globals(), "ancestor")
 
 class changectx(object):
     """A changecontext object makes access to data related to a particular
@@ -161,103 +161,26 b' class filectx(object):'
         find the common ancestor file context, if any, of self, and fc2
         """
 
-        a, b = (self._path, self._filenode), (fc2._path, fc2._filenode)
-
-        if a == b:
-            return self
-
-        if a[0] == b[0]:
-            n = self._filelog.ancestor(a[1], b[1])
-            if n != nullid:
-                return filectx(self._repo, self._path,
-                               fileid=n, filelog=self._filelog)
-
-        # build a graph of all ancestors, crossing renames
-        ag = {}
-        fv = [a, b]
+        acache = {}
         flcache = {self._path:self._filelog, fc2._path:fc2._filelog}
-
-        while fv:
-            f,n = fv.pop()
-            try:
-                fl = flcache[f]
-            except KeyError:
+        def parents(vertex):
+            if vertex in acache:
+                return acache[vertex]
+            f, n = vertex
+            if f not in flcache:
                 flcache[f] = self._repo.file(f)
                 fl = flcache[f]
-            v = [n]
-            while v:
-                n = v.pop()
-                c = (f, n)
-                if c in ag:
-                    continue
-
-                pl = [ n for n in fl.parents(n) if n != nullid ]
-                v += pl
-                pl = [ (f, n) for n in pl ]
+            pl = [ (f,p) for p in fl.parents(n) if p != nullid ]
                 re = fl.renamed(n)
                 if re:
                     pl.append(re)
-                    if re not in ag:
-                        fv.append(re)
-                ag[c] = pl
-
-        dist = {}
-        def depth(node):
-            try:
-                return dist[node]
-            except KeyError:
-                pl = ag[node]
-                if not pl:
-                    dist[node] = 0
-                else:
-                    dist[node] = max([depth(p) for p in pl]) + 1
-                return dist[node]
-
-        # traverse ancestors in order of decreasing distance from root
-        def ancestors(vertex):
-            h = [(-depth(vertex), vertex)]
-            seen = {}
-            while h:
-                d, v = heapq.heappop(h)
-                if v not in seen:
-                    seen[v] = 1
-                    yield (-d, v)
-                    for p in ag[v]:
-                        heapq.heappush(h, (-depth(p), p))
+            acache[vertex]=pl
+            return pl
 
-        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()
+        a, b = (self._path, self._filenode), (fc2._path, fc2._filenode)
+        v = ancestor.ancestor(a, b, parents)
+        if v:
+            f,n = v
+            return filectx(self._repo, f, fileid=n, filelog=flcache[f])
 
-        # increment each ancestor list until it is closer to root than
-        # the other, or they match
-        try:
-            while 1:
-                if gx[0] == gy[0]:
-                    # find the intersection
-                    i = [ n for n in gx[1] if n in gy[1] ]
-                    if i:
-                        fp,fn = i[0]
-                        fl = flcache[fp]
-                        return filectx(self._repo, fp, fileid=fn, filelog=fl)
-                    else:
-                        gy = y.next()
-                        gx = x.next()
-                elif gx[0] < gy[0]:
-                    gy = y.next()
-                else:
-                    gx = x.next()
-        except StopIteration:
             return None

mercurial/revlog.py

@@ -13,7 +13,7 b' of the GNU General Public License, incor'
 from node import *
 from i18n import gettext as _
 from demandload import demandload
-demandload(globals(), "binascii changegroup errno heapq mdiff os")
+demandload(globals(), "binascii changegroup errno ancestor mdiff os")
 demandload(globals(), "sha struct util zlib")
 
 # revlog version strings
@@ -1016,78 +1016,10 b' class revlog(object):'
     def ancestor(self, a, b):
         """calculate the least common ancestor of nodes a and b"""
 
-        # start with some short cuts for the linear cases
-        if a == b:
-            return a
-        ra = self.rev(a)
-        rb = self.rev(b)
-        if ra < rb:
-            last = b
-            first = a
-        else:
-            last = a
-            first = b
-
-        # reachable won't include stop in the list, so we have to use a parent
-        reachable = self.reachable(last, stop=self.parents(first)[0])
-        if first in reachable:
-            return first
-
-        # calculate the distance of every node from root
-        dist = {nullid: 0}
-        for i in xrange(self.count()):
-            n = self.node(i)
-            p1, p2 = self.parents(n)
-            dist[n] = max(dist[p1], dist[p2]) + 1
+        def parents(node):
+            return [p for p in self.parents(node) if p != nullid]
 
-        # traverse ancestors in order of decreasing distance from root
-        def ancestors(node):
-            # we store negative distances because heap returns smallest member
-            h = [(-dist[node], node)]
-            seen = {}
-            while h:
-                d, n = heapq.heappop(h)
-                if n not in seen:
-                    seen[n] = 1
-                    yield (-d, n)
-                    for p in self.parents(n):
-                        heapq.heappush(h, (-dist[p], p))
-
-        def generations(node):
-            sg, s = None, {}
-            for g,n in ancestors(node):
-                if g != sg:
-                    if sg:
-                        yield sg, s
-                    sg, s = g, {n:1}
-                else:
-                    s[n] = 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
-        while 1:
-            #print "ancestor gen %s %s" % (gx[0], gy[0])
-            if gx[0] == gy[0]:
-                # find the intersection
-                i = [ n for n in gx[1] if n in gy[1] ]
-                if i:
-                    return i[0]
-                else:
-                    #print "next"
-                    gy = y.next()
-                    gx = x.next()
-            elif gx[0] < gy[0]:
-                #print "next y"
-                gy = y.next()
-            else:
-                #print "next x"
-                gx = x.next()
+        return ancestor.ancestor(a, b, parents) or nullid
 
     def group(self, nodelist, lookup, infocollect=None):
         """calculate a delta group