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There are 2 ways that you get LCA merge resolution in bzr. First, if you use
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``bzr merge --lca``, the *content* of files will be resolved using a Least Common
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Ancestors algorithm. That is described in <lca-merge.html> not here.
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This document describes how we handle merging tree-shape when there is not
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a single unique ancestor (criss-cross merge). With a single LCA, we use
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simple 3-way-merge logic.
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When there are multiple possible LCAs, we use a different algorithm for
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handling tree-shape merging. Described here.
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As a simple example, here is a revision graph which we will refer to often::
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In this graph, ``THIS`` and ``OTHER`` both have ``LCA1`` and ``LCA2`` in their
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ancestry but neither is an ancestor of the other, so we have 2 least common
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ancestors. The unique common ancestor is ``BASE``. (It should be noted that in
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this text we will talk directly about ``LCA1`` and ``LCA2``, but the algorithms
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are designed to cope with more than 2 LCAs.)
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I'm defining scalar values as ones that cannot be 'merged' on their own. For
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example, the name of a file is "scalar". If one person changes "foo.txt" to
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"foo.c" and someone else changes "foo.txt" to "bar.txt" we don't merge the
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changes to be "bar.c", we simply conflict and expect the user to sort it out.
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We use a slightly different algorithm for scalars.
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(This can be seen as ``bzrlib.merge.Merge3Merger._lca_multi_way```
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1. If ``THIS`` and ``OTHER`` have the same value, use it. There is no need to
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inspect any other values in this case. Either nothing was changed (all
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interesting nodes would have the same value), or we have "accidental
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convergence" (both sides made the same change.).
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2. Find the values from ``LCA1`` and ``LCA2`` which are not the same as
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``BASE``. The idea here is to provide a rudimentary "heads" comparison.
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Often, the whole tree graph will have a criss-cross, but the per-file
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(per-scalar) graph would be linear, and the value in one LCA strictly
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dominates the other. It is possible to construct a scenario where one side
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dominates the other, but the dominated value is not ``BASE``, but a second
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intermediate value. Most scalars are rarely changed, so this is unlikely to
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be an issue. The trade-off is having to generate and inspect the
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If there are no LCA values that are different from ``BASE``, we use a simple
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3-way merge with ``BASE`` as the base value.
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3. Find the unique set of LCA values that do not include the ``BASE`` value.
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If there is only one unique LCA value, we again use three-way merge logic
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using that unique value as the base.
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4. At this point, we have determined that we have at least 2 unique values in
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our LCAs which means that ``THIS`` and ``OTHER`` would both have to resolve
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the conflict. If they resolved it in the same way, we would have caught that
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in step 1. So they either both picked a different LCA value, or one (or
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both) chose a new value to use.
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If ``OTHER`` and ``THIS`` both picked a different LCA value, we conflict.
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If ``OTHER`` and ``THIS`` both have values that are not LCA values, we also
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conflict. (Same as 3-way, both sides modified a value in different ways.)
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5. (optional) The only tricky part is this: if ``OTHER`` has a LCA value, but
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``THIS`` does not, then we go with ``THIS``, and conversely if ``THIS`` has
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an LCA value, but ``OTHER`` does not, then we go with ``OTHER``. The idea is
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that ``THIS`` and ``OTHER`` may have resolved things in the same way, and
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then later changed the value to something newer. (They could have also
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resolved it differently, and then one side updated again.)
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``InventoryEntry.revision``
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---------------------------
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The last-modified revision for an entry gets treated differently. This is
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because how it is generated allows us to infer more information. Specifically,
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any time there is a change to an entry (rename, or content change) the last
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modified revision is updated. Further, if we are merging, and both sides
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updated the entry, then we update the last-modified revision at the merge
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For a picture example::
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For a single entry, the last modified revision in ``D`` is:
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1) ``A`` if neither ``B`` or ``C`` modified it
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2) ``B`` if ``B`` modified and ``C`` did not
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3) ``C`` if ``C`` modified and ``B`` did not
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4) ``D`` if ``B`` and ``C`` modified it
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This means that if the last modified revision is the same, there have been no
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changes in the intermediate time. If ``OTHER`` also has the same last modified
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revision as *any* LCA, then we know that all other LCAs' last-modified
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revisions are in the ancestry of that value. (Otherwise, when ``OTHER`` would
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need to create a new last modified revision as part of the merge.)
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doc/developers/lca_tree_merging.txt
