--- a/mercurial/revset.py Tue Jun 14 11:05:36 2016 +0100
+++ b/mercurial/revset.py Mon Jun 13 18:20:00 2016 +0100
@@ -1887,6 +1887,204 @@
raise error.ParseError(_("unknown sort key %r") % fk)
return baseset([c.rev() for c in ctxs])
+def groupbranchiter(revs, parentsfunc, firstbranch=()):
+ """Yield revisions from heads to roots one (topo) branch at a time.
+
+ This function aims to be used by a graph generator that wishes to minimize
+ the number of parallel branches and their interleaving.
+
+ Example iteration order (numbers show the "true" order in a changelog):
+
+ o 4
+ |
+ o 1
+ |
+ | o 3
+ | |
+ | o 2
+ |/
+ o 0
+
+ Note that the ancestors of merges are understood by the current
+ algorithm to be on the same branch. This means no reordering will
+ occur behind a merge.
+ """
+
+ ### Quick summary of the algorithm
+ #
+ # This function is based around a "retention" principle. We keep revisions
+ # in memory until we are ready to emit a whole branch that immediately
+ # "merges" into an existing one. This reduces the number of parallel
+ # branches with interleaved revisions.
+ #
+ # During iteration revs are split into two groups:
+ # A) revision already emitted
+ # B) revision in "retention". They are stored as different subgroups.
+ #
+ # for each REV, we do the following logic:
+ #
+ # 1) if REV is a parent of (A), we will emit it. If there is a
+ # retention group ((B) above) that is blocked on REV being
+ # available, we emit all the revisions out of that retention
+ # group first.
+ #
+ # 2) else, we'll search for a subgroup in (B) awaiting for REV to be
+ # available, if such subgroup exist, we add REV to it and the subgroup is
+ # now awaiting for REV.parents() to be available.
+ #
+ # 3) finally if no such group existed in (B), we create a new subgroup.
+ #
+ #
+ # To bootstrap the algorithm, we emit the tipmost revision (which
+ # puts it in group (A) from above).
+
+ revs.sort(reverse=True)
+
+ # Set of parents of revision that have been emitted. They can be considered
+ # unblocked as the graph generator is already aware of them so there is no
+ # need to delay the revisions that reference them.
+ #
+ # If someone wants to prioritize a branch over the others, pre-filling this
+ # set will force all other branches to wait until this branch is ready to be
+ # emitted.
+ unblocked = set(firstbranch)
+
+ # list of groups waiting to be displayed, each group is defined by:
+ #
+ # (revs: lists of revs waiting to be displayed,
+ # blocked: set of that cannot be displayed before those in 'revs')
+ #
+ # The second value ('blocked') correspond to parents of any revision in the
+ # group ('revs') that is not itself contained in the group. The main idea
+ # of this algorithm is to delay as much as possible the emission of any
+ # revision. This means waiting for the moment we are about to display
+ # these parents to display the revs in a group.
+ #
+ # This first implementation is smart until it encounters a merge: it will
+ # emit revs as soon as any parent is about to be emitted and can grow an
+ # arbitrary number of revs in 'blocked'. In practice this mean we properly
+ # retains new branches but gives up on any special ordering for ancestors
+ # of merges. The implementation can be improved to handle this better.
+ #
+ # The first subgroup is special. It corresponds to all the revision that
+ # were already emitted. The 'revs' lists is expected to be empty and the
+ # 'blocked' set contains the parents revisions of already emitted revision.
+ #
+ # You could pre-seed the <parents> set of groups[0] to a specific
+ # changesets to select what the first emitted branch should be.
+ groups = [([], unblocked)]
+ pendingheap = []
+ pendingset = set()
+
+ heapq.heapify(pendingheap)
+ heappop = heapq.heappop
+ heappush = heapq.heappush
+ for currentrev in revs:
+ # Heap works with smallest element, we want highest so we invert
+ if currentrev not in pendingset:
+ heappush(pendingheap, -currentrev)
+ pendingset.add(currentrev)
+ # iterates on pending rev until after the current rev have been
+ # processed.
+ rev = None
+ while rev != currentrev:
+ rev = -heappop(pendingheap)
+ pendingset.remove(rev)
+
+ # Seek for a subgroup blocked, waiting for the current revision.
+ matching = [i for i, g in enumerate(groups) if rev in g[1]]
+
+ if matching:
+ # The main idea is to gather together all sets that are blocked
+ # on the same revision.
+ #
+ # Groups are merged when a common blocking ancestor is
+ # observed. For example, given two groups:
+ #
+ # revs [5, 4] waiting for 1
+ # revs [3, 2] waiting for 1
+ #
+ # These two groups will be merged when we process
+ # 1. In theory, we could have merged the groups when
+ # we added 2 to the group it is now in (we could have
+ # noticed the groups were both blocked on 1 then), but
+ # the way it works now makes the algorithm simpler.
+ #
+ # We also always keep the oldest subgroup first. We can
+ # probably improve the behavior by having the longest set
+ # first. That way, graph algorithms could minimise the length
+ # of parallel lines their drawing. This is currently not done.
+ targetidx = matching.pop(0)
+ trevs, tparents = groups[targetidx]
+ for i in matching:
+ gr = groups[i]
+ trevs.extend(gr[0])
+ tparents |= gr[1]
+ # delete all merged subgroups (except the one we kept)
+ # (starting from the last subgroup for performance and
+ # sanity reasons)
+ for i in reversed(matching):
+ del groups[i]
+ else:
+ # This is a new head. We create a new subgroup for it.
+ targetidx = len(groups)
+ groups.append(([], set([rev])))
+
+ gr = groups[targetidx]
+
+ # We now add the current nodes to this subgroups. This is done
+ # after the subgroup merging because all elements from a subgroup
+ # that relied on this rev must precede it.
+ #
+ # we also update the <parents> set to include the parents of the
+ # new nodes.
+ if rev == currentrev: # only display stuff in rev
+ gr[0].append(rev)
+ gr[1].remove(rev)
+ parents = [p for p in parentsfunc(rev) if p > node.nullrev]
+ gr[1].update(parents)
+ for p in parents:
+ if p not in pendingset:
+ pendingset.add(p)
+ heappush(pendingheap, -p)
+
+ # Look for a subgroup to display
+ #
+ # When unblocked is empty (if clause), we were not waiting for any
+ # revisions during the first iteration (if no priority was given) or
+ # if we emitted a whole disconnected set of the graph (reached a
+ # root). In that case we arbitrarily take the oldest known
+ # subgroup. The heuristic could probably be better.
+ #
+ # Otherwise (elif clause) if the subgroup is blocked on
+ # a revision we just emitted, we can safely emit it as
+ # well.
+ if not unblocked:
+ if len(groups) > 1: # display other subset
+ targetidx = 1
+ gr = groups[1]
+ elif not gr[1] & unblocked:
+ gr = None
+
+ if gr is not None:
+ # update the set of awaited revisions with the one from the
+ # subgroup
+ unblocked |= gr[1]
+ # output all revisions in the subgroup
+ for r in gr[0]:
+ yield r
+ # delete the subgroup that you just output
+ # unless it is groups[0] in which case you just empty it.
+ if targetidx:
+ del groups[targetidx]
+ else:
+ gr[0][:] = []
+ # Check if we have some subgroup waiting for revisions we are not going to
+ # iterate over
+ for g in groups:
+ for r in g[0]:
+ yield r
+
@predicate('subrepo([pattern])')
def subrepo(repo, subset, x):
"""Changesets that add, modify or remove the given subrepo. If no subrepo