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  • Committer: John Arbash Meinel
  • Date: 2007-12-05 18:47:51 UTC
  • mfrom: (3079 +trunk)
  • mto: This revision was merged to the branch mainline in revision 3080.
  • Revision ID: john@arbash-meinel.com-20071205184751-58fi7xoj8ddox6ch
[merge] bzr.dev 3079

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bzrlib/graph.py
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            order if they need it.
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        """
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        candidate_heads = set(keys)
 
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        if revision.NULL_REVISION in candidate_heads:
 
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            # NULL_REVISION is only a head if it is the only entry
 
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            candidate_heads.remove(revision.NULL_REVISION)
 
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            if not candidate_heads:
 
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                return set([revision.NULL_REVISION])
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        if len(candidate_heads) < 2:
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            return candidate_heads
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        searchers = dict((c, self._make_breadth_first_searcher([c]))
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            common_walker.start_searching(new_common)
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        return candidate_heads
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    def find_unique_lca(self, left_revision, right_revision):
 
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    def find_unique_lca(self, left_revision, right_revision,
 
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                        count_steps=False):
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        """Find a unique LCA.
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        Find lowest common ancestors.  If there is no unique  common
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        Note that None is not an acceptable substitute for NULL_REVISION.
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        in the input for this method.
 
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        :param count_steps: If True, the return value will be a tuple of
 
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            (unique_lca, steps) where steps is the number of times that
 
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            find_lca was run.  If False, only unique_lca is returned.
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        """
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        revisions = [left_revision, right_revision]
 
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        steps = 0
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        while True:
 
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            steps += 1
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            lca = self.find_lca(*revisions)
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            if len(lca) == 1:
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                return lca.pop()
 
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                result = lca.pop()
 
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                if count_steps:
 
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                    return result, steps
 
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                else:
 
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                    return result
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            if len(lca) == 0:
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                raise errors.NoCommonAncestor(left_revision, right_revision)
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            revisions = lca
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        smallest number of parent looksup to determine the ancestral
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        relationship between N revisions.
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        """
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        if revision.is_null(candidate_ancestor):
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            return True
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        if revision.is_null(candidate_descendant):
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            # if candidate_descendant is NULL_REVISION, then only
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            # candidate_ancestor == NULL_REVISION is an ancestor, but we've
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            # already handled that case.
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            return False
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        return set([candidate_descendant]) == self.heads(
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            [candidate_ancestor, candidate_descendant])
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