LCA Tree Merging

There are 2 ways that you get LCA merge resolution in bzr. First, if you use bzr merge --lca, the content of files will be resolved using a Least Common Ancestors algorithm. That is described in <lca-merge.html> not here.

This document describes how we handle merging tree-shape when there is not a single unique ancestor (criss-cross merge). With a single LCA, we use simple 3-way-merge logic.

When there are multiple possible LCAs, we use a different algorithm for handling tree-shape merging. Described here.

As a simple example, here is a revision graph which we will refer to often:

.    BASE
.  /      \
. LCA1   LCA2
. |   \ /   |
. |    X    |
. |   / \   |
. THIS  OTHER

In this graph, THIS and OTHER both have LCA1 and LCA2 in their ancestry but neither is an ancestor of the other, so we have 2 least common ancestors. The unique common ancestor is BASE. (It should be noted that in this text we will talk directly about LCA1 and LCA2, but the algorithms are designed to cope with more than 2 LCAs.)

Scalars

Definition

I’m defining scalar values as ones that cannot be ‘merged’ on their own. For example, the name of a file is “scalar”. If one person changes “foo.txt” to “foo.c” and someone else changes “foo.txt” to “bar.txt” we don’t merge the changes to be “bar.c”, we simply conflict and expect the user to sort it out.

We use a slightly different algorithm for scalars.

Resolution Algorithm

(This can be seen as breezy.merge.Merge3Merger._lca_multi_way`

  1. If THIS and OTHER have the same value, use it. There is no need to inspect any other values in this case. Either nothing was changed (all interesting nodes would have the same value), or we have “accidental convergence” (both sides made the same change.).

  2. Find the values from LCA1 and LCA2 which are not the same as BASE. The idea here is to provide a rudimentary “heads” comparison. Often, the whole tree graph will have a criss-cross, but the per-file (per-scalar) graph would be linear, and the value in one LCA strictly dominates the other. It is possible to construct a scenario where one side dominates the other, but the dominated value is not BASE, but a second intermediate value. Most scalars are rarely changed, so this is unlikely to be an issue. The trade-off is having to generate and inspect the per-scalar graph.

    If there are no LCA values that are different from BASE, we use a simple 3-way merge with BASE as the base value.

  3. Find the unique set of LCA values that do not include the BASE value. If there is only one unique LCA value, we again use three-way merge logic using that unique value as the base.

  4. At this point, we have determined that we have at least 2 unique values in our LCAs which means that THIS and OTHER would both have to resolve the conflict. If they resolved it in the same way, we would have caught that in step 1. So they either both picked a different LCA value, or one (or both) chose a new value to use.

    If OTHER and THIS both picked a different LCA value, we conflict.

    If OTHER and THIS both have values that are not LCA values, we also conflict. (Same as 3-way, both sides modified a value in different ways.)

  5. (optional) The only tricky part is this: if OTHER has a LCA value, but THIS does not, then we go with THIS, and conversely if THIS has an LCA value, but OTHER does not, then we go with OTHER. The idea is that THIS and OTHER may have resolved things in the same way, and then later changed the value to something newer. (They could have also resolved it differently, and then one side updated again.)

InventoryEntry.revision

The last-modified revision for an entry gets treated differently. This is because how it is generated allows us to infer more information. Specifically, any time there is a change to an entry (rename, or content change) the last modified revision is updated. Further, if we are merging, and both sides updated the entry, then we update the last-modified revision at the merge point.

For a picture example:

.   A
.  / \
. B   C
.  \ /
.   D

For a single entry, the last modified revision in D is:

  1. A if neither B or C modified it
  2. B if B modified and C did not
  3. C if C modified and B did not
  4. D if B and C modified it

This means that if the last modified revision is the same, there have been no changes in the intermediate time. If OTHER also has the same last modified revision as any LCA, then we know that all other LCAs’ last-modified revisions are in the ancestry of that value. (Otherwise, when OTHER would need to create a new last modified revision as part of the merge.)