strip: make query to get new bookmark target cheaper The current query to get the new bookmark target for stripped revisions involves multiple walks up the DAG, and is really expensive, taking over 2.5 seconds on a repository with over 400,000 changesets even if just one changeset is being stripped. A slightly simplified version of the current query is max(heads(::<tostrip> - <tostrip>)) We make two observations here. 1. For any set s, max(heads(s)) == max(s). That is because revision numbers define a topological order, so that the element with the highest revision number in s will not have any children in s. 2. For any set s, max(::s - s) == max(parents(s) - s). In other words, the ancestor of s with the highest revision number not in s is a parent of one of the revs in s. Why? Because if it were an ancestor but not a parent of s, it would have a descendant that would be a parent of s. This descendant would have a higher revision number, leading to a contradiction. Combining these two observations, we rewrite the revset query as max(parents(<tostrip>) - <tostrip>) The time complexity is now linear in the number of changesets being stripped. For the above repository, the query now takes 0.1 seconds when one changeset is stripped. This speeds up operations that use repair.strip, like the rebase and strip commands.

--- a/mercurial/repair.py	Tue Dec 04 12:54:18 2012 -0800
+++ b/mercurial/repair.py	Wed Dec 05 14:33:15 2012 -0800
@@ -112,8 +112,10 @@
         saverevs.difference_update(descendants)
     savebases = [cl.node(r) for r in saverevs]
     stripbases = [cl.node(r) for r in tostrip]
-    newbmtarget = repo.revs('sort(heads((::%ld) - (%ld)), -rev)',
-                            tostrip, tostrip)
+
+    # For a set s, max(parents(s) - s) is the same as max(heads(::s - s)), but
+    # is much faster
+    newbmtarget = repo.revs('max(parents(%ld) - (%ld))', tostrip, tostrip)
     if newbmtarget:
         newbmtarget = repo[newbmtarget[0]].node()
     else: