pack-bitmap-write: build fewer intermediate bitmaps

The bitmap_writer_build() method calls bitmap_builder_init() to
construct a list of commits reachable from the selected commits along
with a "reverse graph". This reverse graph has edges pointing from a
commit to other commits that can reach that commit. After computing a
reachability bitmap for a commit, the values in that bitmap are then
copied to the reachability bitmaps across the edges in the reverse
graph.

We can now relax the role of the reverse graph to greatly reduce the
number of intermediate reachability bitmaps we compute during this
reverse walk. The end result is that we walk objects the same number of
times as before when constructing the reachability bitmaps, but we also
spend much less time copying bits between bitmaps and have much lower
memory pressure in the process.

The core idea is to select a set of "important" commits based on
interactions among the sets of commits reachable from each selected commit.

The first technical concept is to create a new 'commit_mask' member in the
bb_commit struct. Note that the selected commits are provided in an
ordered array. The first thing to do is to mark the ith bit in the
commit_mask for the ith selected commit. As we walk the commit-graph, we
copy the bits in a commit's commit_mask to its parents. At the end of
the walk, the ith bit in the commit_mask for a commit C stores a boolean
representing "The ith selected commit can reach C."

As we walk, we will discover non-selected commits that are important. We
will get into this later, but those important commits must also receive
bit positions, growing the width of the bitmasks as we walk. At the true
end of the walk, the ith bit means "the ith _important_ commit can reach
C."

MAXIMAL COMMITS
---------------

We use a new 'maximal' bit in the bb_commit struct to represent whether
a commit is important or not. The term "maximal" comes from the
partially-ordered set of commits in the commit-graph where C >= P if P
is a parent of C, and then extending the relationship transitively.
Instead of taking the maximal commits across the entire commit-graph, we
instead focus on selecting each commit that is maximal among commits
with the same bits on in their commit_mask. This definition is
important, so let's consider an example.

Suppose we have three selected commits A, B, and C. These are assigned
bitmasks 100, 010, and 001 to start. Each of these can be marked as
maximal immediately because they each will be the uniquely maximal
commit that contains their own bit. Keep in mind that that these commits
may have different bitmasks after the walk; for example, if B can reach
C but A cannot, then the final bitmask for C is 011. Even in these
cases, C would still be a maximal commit among all commits with the
third bit on in their masks.

Now define sets X, Y, and Z to be the sets of commits reachable from A,
B, and C, respectively. The intersections of these sets correspond to
different bitmasks:

 * 100: X - (Y union Z)
 * 010: Y - (X union Z)
 * 001: Z - (X union Y)
 * 110: (X intersect Y) - Z
 * 101: (X intersect Z) - Y
 * 011: (Y intersect Z) - X
 * 111: X intersect Y intersect Z

This can be visualized with the following Hasse diagram:

	100    010    001
         | \  /   \  / |
         |  \/     \/  |
         |  /\     /\  |
         | /  \   /  \ |
        110    101    011
          \___  |  ___/
              \ | /
               111

Some of these bitmasks may not be represented, depending on the topology
of the commit-graph. In fact, we are counting on it, since the number of
possible bitmasks is exponential in the number of selected commits, but
is also limited by the total number of commits. In practice, very few
bitmasks are possible because most commits converge on a common "trunk"
in the commit history.

With this three-bit example, we wish to find commits that are maximal
for each bitmask. How can we identify this as we are walking?

As we walk, we visit a commit C. Since we are walking the commits in
topo-order, we know that C is visited after all of its children are
visited. Thus, when we get C from the revision walk we inspect the
'maximal' property of its bb_data and use that to determine if C is truly
important. Its commit_mask is also nearly final. If C is not one of the
originally-selected commits, then assign a bit position to C (by
incrementing num_maximal) and set that bit on in commit_mask. See
"MULTIPLE MAXIMAL COMMITS" below for more detail on this.

Now that the commit C is known to be maximal or not, consider each
parent P of C. Compute two new values:

 * c_not_p : true if and only if the commit_mask for C contains a bit
             that is not contained in the commit_mask for P.

 * p_not_c : true if and only if the commit_mask for P contains a bit
             that is not contained in the commit_mask for P.

If c_not_p is false, then P already has all of the bits that C would
provide to its commit_mask. In this case, move on to other parents as C
has nothing to contribute to P's state that was not already provided by
other children of P.

We continue with the case that c_not_p is true. This means there are
bits in C's commit_mask to copy to P's commit_mask, so use bitmap_or()
to add those bits.

If p_not_c is also true, then set the maximal bit for P to one. This means
that if no other commit has P as a parent, then P is definitely maximal.
This is because no child had the same bitmask. It is important to think
about the maximal bit for P at this point as a temporary state: "P is
maximal based on current information."

In contrast, if p_not_c is false, then set the maximal bit for P to
zero. Further, clear all reverse_edges for P since any edges that were
previously assigned to P are no longer important. P will gain all
reverse edges based on C.

The final thing we need to do is to update the reverse edges for P.
These reverse edges respresent "which closest maximal commits
contributed bits to my commit_mask?" Since C contributed bits to P's
commit_mask in this case, C must add to the reverse edges of P.

If C is maximal, then C is a 'closest' maximal commit that contributed
bits to P. Add C to P's reverse_edges list.

Otherwise, C has a list of maximal commits that contributed bits to its
bitmask (and this list is exactly one element). Add all of these items
to P's reverse_edges list. Be careful to ignore duplicates here.

After inspecting all parents P for a commit C, we can clear the
commit_mask for C. This reduces the memory load to be limited to the
"width" of the commit graph.

Consider our ABC/XYZ example from earlier and let's inspect the state of
the commits for an interesting bitmask, say 011. Suppose that D is the
only maximal commit with this bitmask (in the first three bits). All
other commits with bitmask 011 have D as the only entry in their
reverse_edges list. D's reverse_edges list contains B and C.

COMPUTING REACHABILITY BITMAPS
------------------------------

Now that we have our definition, let's zoom out and consider what
happens with our new reverse graph when computing reachability bitmaps.
We walk the reverse graph in reverse-topo-order, so we visit commits
with largest commit_masks first. After we compute the reachability
bitmap for a commit C, we push the bits in that bitmap to each commit D
in the reverse edge list for C. Then, when we finally visit D we already
have the bits for everything reachable from maximal commits that D can
reach and we only need to walk the objects in the set-difference.

In our ABC/XYZ example, when we finally walk for the commit A we only
need to walk commits with bitmask equal to A's bitmask. If that bitmask
is 100, then we are only walking commits in X - (Y union Z) because the
bitmap already contains the bits for objects reachable from (X intersect
Y) union (X intersect Z) (i.e. the bits from the reachability bitmaps
for the maximal commits with bitmasks 110 and 101).

The behavior is intended to walk each commit (and the trees that commit
introduces) at most once while allocating and copying fewer reachability
bitmaps. There is one caveat: what happens when there are multiple
maximal commits with the same bitmask, with respect to the initial set
of selected commits?

MULTIPLE MAXIMAL COMMITS
------------------------

Earlier, we mentioned that when we discover a new maximal commit, we
assign a new bit position to that commit and set that bit position to
one for that commit. This is absolutely important for interesting
commit-graphs such as git/git and torvalds/linux. The reason is due to
the existence of "butterflies" in the commit-graph partial order.

Here is an example of four commits forming a butterfly:

   I    J
   |\  /|
   | \/ |
   | /\ |
   |/  \|
   M    N
    \  /
     |/
     Q

Here, I and J both have parents M and N. In general, these do not need
to be exact parent relationships, but reachability relationships. The
most important part is that M and N cannot reach each other, so they are
independent in the partial order. If I had commit_mask 10 and J had
commit_mask 01, then M and N would both be assigned commit_mask 11 and
be maximal commits with the bitmask 11. Then, what happens when M and N
can both reach a commit Q? If Q is also assigned the bitmask 11, then it
is not maximal but is reachable from both M and N.

While this is not necessarily a deal-breaker for our abstract definition
of finding maximal commits according to a given bitmask, we have a few
issues that can come up in our larger picture of constructing
reachability bitmaps.

In particular, if we do not also consider Q to be a "maximal" commit,
then we will walk commits reachable from Q twice: once when computing
the reachability bitmap for M and another time when computing the
reachability bitmap for N. This becomes much worse if the topology
continues this pattern with multiple butterflies.

The solution has already been mentioned: each of M and N are assigned
their own bits to the bitmask and hence they become uniquely maximal for
their bitmasks. Finally, Q also becomes maximal and thus we do not need
to walk its commits multiple times. The final bitmasks for these commits
are as follows:

  I:10       J:01
   |\        /|
   | \ _____/ |
   | /\____   |
   |/      \  |
   M:111    N:1101
        \  /
       Q:1111

Further, Q's reverse edge list is { M, N }, while M and N both have
reverse edge list { I, J }.

PERFORMANCE MEASUREMENTS
------------------------

Now that we've spent a LOT of time on the theory of this algorithm,
let's show that this is actually worth all that effort.

To test the performance, use GIT_TRACE2_PERF=1 when running
'git repack -abd' in a repository with no existing reachability bitmaps.
This avoids any issues with keeping existing bitmaps to skew the
numbers.

Inspect the "building_bitmaps_total" region in the trace2 output to
focus on the portion of work that is affected by this change. Here are
the performance comparisons for a few repositories. The timings are for
the following versions of Git: "multi" is the timing from before any
reverse graph is constructed, where we might perform multiple
traversals. "reverse" is for the previous change where the reverse graph
has every reachable commit.  Finally "maximal" is the version introduced
here where the reverse graph only contains the maximal commits.

      Repository: git/git
           multi: 2.628 sec
         reverse: 2.344 sec
         maximal: 2.047 sec

      Repository: torvalds/linux
           multi: 64.7 sec
         reverse: 205.3 sec
         maximal: 44.7 sec

So in all cases we've not only recovered any time lost to switching to
the reverse-edge algorithm, but we come out ahead of "multi" in all
cases. Likewise, peak heap has gone back to something reasonable:

      Repository: torvalds/linux
           multi: 2.087 GB
         reverse: 3.141 GB
         maximal: 2.288 GB

While I do not have access to full fork networks on GitHub, Peff has run
this algorithm on the chromium/chromium fork network and reported a
change from 3 hours to ~233 seconds. That network is particularly
beneficial for this approach because it has a long, linear history along
with many tags. The "multi" approach was obviously quadratic and the new
approach is linear.

Helped-by: Jeff King <peff@peff.net>
Signed-off-by: Derrick Stolee <dstolee@microsoft.com>
Helped-by: Johannes Schindelin <Johannes.Schindelin@gmx.de>
Signed-off-by: Taylor Blau <me@ttaylorr.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>

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