diff -r fe30f5434b51 -r 3616c0d7ab88 mercurial/hg.py
--- a/mercurial/hg.py	Sun Aug 14 12:23:36 2005 -0800
+++ b/mercurial/hg.py	Sun Aug 14 12:23:45 2005 -0800
@@ -1072,6 +1072,112 @@
     def heads(self):
         return self.changelog.heads()
 
+    # branchlookup returns a dict giving a list of branches for
+    # each head.  A branch is defined as the tag of a node or
+    # the branch of the node's parents.  If a node has multiple
+    # branch tags, tags are eliminated if they are visible from other
+    # branch tags.
+    #
+    # So, for this graph:  a->b->c->d->e
+    #                       \         /
+    #                        aa -----/
+    # a has tag 2.6.12                     
+    # d has tag 2.6.13
+    # e would have branch tags for 2.6.12 and 2.6.13.  Because the node
+    # for 2.6.12 can be reached from the node 2.6.13, that is eliminated
+    # from the list.
+    #
+    # It is possible that more than one head will have the same branch tag.
+    # callers need to check the result for multiple heads under the same
+    # branch tag if that is a problem for them (ie checkout of a specific
+    # branch).
+    #
+    # passing in a specific branch will limit the depth of the search
+    # through the parents.  It won't limit the branches returned in the
+    # result though.
+    def branchlookup(self, heads=None, branch=None):
+        if not heads:
+            heads = self.heads()
+        headt = [ h for h in heads ]
+        chlog = self.changelog
+        branches = {}
+        merges = []
+        seenmerge = {}
+
+        # traverse the tree once for each head, recording in the branches
+        # dict which tags are visible from this head.  The branches
+        # dict also records which tags are visible from each tag
+        # while we traverse.
+        while headt or merges:
+            if merges:
+                n, found = merges.pop()
+                visit = [n]
+            else:
+                h = headt.pop()
+                visit = [h]
+                found = [h]
+                seen = {}
+            while visit:
+                n = visit.pop()
+                if n in seen:
+                    continue
+                pp = chlog.parents(n)
+                tags = self.nodetags(n)
+                if tags:
+                    for x in tags:
+                        if x == 'tip':
+                            continue
+                        for f in found:
+                            branches.setdefault(f, {})[n] = 1
+                        branches.setdefault(n, {})[n] = 1
+                        break
+                    if n not in found:
+                        found.append(n)
+                    if branch in tags:
+                        continue
+                seen[n] = 1
+                if pp[1] != nullid and n not in seenmerge:
+                    merges.append((pp[1], [x for x in found]))
+                    seenmerge[n] = 1
+                if pp[0] != nullid:
+                    visit.append(pp[0])
+        # traverse the branches dict, eliminating branch tags from each
+        # head that are visible from another branch tag for that head.
+        out = {}
+        viscache = {}
+        for h in heads:
+            def visible(node):
+                if node in viscache:
+                    return viscache[node]
+                ret = {}
+                visit = [node]
+                while visit:
+                    x = visit.pop()
+                    if x in viscache:
+                        ret.update(viscache[x])
+                    elif x not in ret:
+                        ret[x] = 1
+                        if x in branches:
+                            visit[len(visit):] = branches[x].keys()
+                viscache[node] = ret
+                return ret
+            if h not in branches:
+                continue
+            # O(n^2), but somewhat limited.  This only searches the
+            # tags visible from a specific head, not all the tags in the
+            # whole repo.
+            for b in branches[h]:
+                vis = False
+                for bb in branches[h].keys():
+                    if b != bb:
+                        if b in visible(bb):
+                            vis = True
+                            break
+                if not vis:
+                    l = out.setdefault(h, [])
+                    l[len(l):] = self.nodetags(b)
+        return out
+
     def branches(self, nodes):
         if not nodes: nodes = [self.changelog.tip()]
         b = []