The most efficient way to test if repositories share the same objects

Hi, all:

What is the most efficient way to test if repoA and repoB share common commits? My goal is to automatically figure out if repoB can benefit from setting alternates to repoA and repacking. I currently do it by comparing the output of "show-ref --tags -s", but that does not work for repos without tags.

Best,

On Thu, Mar 22 2018, Konstantin Ryabitsev wrote:

If you're using show-ref already to get the tag tips, you can use for-each-ref to get all tips.

But of course that'll just give you the tips. You could then use `git cat-file --batch-check` on both ends to see what commits from the other they report knowing about, in case they have branches that are ahead/behind the other.

Ævar Arnfjörð Bjarmason <avarab@gmail.com> writes:

I am not sure how you are envisioning to use "cat-file --batch-check" here. Do you mean to take "rev-list --all" output from both and compare, or something?

I am not sure how Konstantin defines "the most efficient", but if it is "with the smallest number of bits exchanged between the repositories", then the answer would probably be to find the root commit(s) in each repository and if they share any common root(s). If there isn't then there is no hope to share objects between them, of course.

On Thu, Mar 22 2018, Junio C. Hamano wrote:

By doing something like this:

    (
        cd /tmp &&
        git clone http://github.com/gitster/git gitster-git;
        git clone http://github.com/avar/git avar-git;
        git -C gitster-git for-each-ref --format="%(object)" | git -C avar-git cat-file --batch-check|grep -E -o '(commit|missing)'|sort|uniq -c &&
        git -C avar-git for-each-ref --format="%(object)" | git -C gitster-git cat-file --batch-check|grep -E -o '(commit|missing)'|sort|uniq -c
    )

Which outputs:

    673 commit
    696 missing
    374 commit
    495 missing

Which of course, as noted, isn't going to be a good general solution in all cases, but it's blindingly fast, and since the original question is essentially that he's starting out with doing a rough equivalent of that, but for tags only, maybe it'll work for his use-case.

Yes, that would probably be much better:

    diff -ru <(git -C avar-git log --oneline --pretty=format:%H --max-parents=0) \
             <(git -C gitster-git log --oneline --pretty=format:%H --max-parents=0) &&
    do_stuff_because_they_have_stuff_in_common

On 03/22/18 15:35, Junio C Hamano wrote:

Hmm... so, this a cool idea that I'd like to use, but there are two annoying gotchas:

1. I cannot assume that refs/heads/master is meaningful -- my problem is
actually with something like
https://source.codeaurora.org/quic/la/kernel/msm-3.18 -- you will find
that master is actually unborn and there are 7700 other heads (don't get
me started on that unless you're buying me a lot of drinks).

2. Even if there is a HEAD I know I can use, it's pretty slow on large
repos (e.g. linux.git):

$ time git rev-list --max-parents=0 HEAD a101ad945113be3d7f283a181810d76897f0a0d6 cd26f1bd6bf3c73cc5afe848677b430ab342a909 be0e5c097fc206b863ce9fe6b3cfd6974b0110f4 1da177e4c3f41524e886b7f1b8a0c1fc7321cac2

real 0m6.311s user 0m6.153s sys 0m0.110s

If I try to do this for each of the 7700 heads, this will take roughly 12 hours.

My current strategy has been pretty much:

git -C msm-3.10.git show-ref --tags -s | sort -u > /tmp/refs1 git -C msm-3.18.git show-ref --tags -s | sort -u > /tmp/refs2

and then checking if an intersection of these matches at least half of refs in either repo:

---- #/usr/bin/env python import numpy

refs1 = numpy.array(open('/tmp/refs1').readlines()) refs2 = numpy.array(open('/tmp/refs2').readlines())

in_common = len(numpy.intersect1d(refs1, refs2))
if in_common > len(refs1)/2 or in_common > len(refs2)/2:
    print('Lots of shared refs')
else:
    print('None or too few shared refs')
----

This works well enough at least for those repos with lots of shared tags, but will miss potentially large repos where there's only heads that can be pointing at commits that aren't necessarily the same between two repos.

Thanks for your help!

Best,

Konstantin Ryabitsev <konstantin@linuxfoundation.org> writes:

Wouldn't it be more efficient to avoid doing so one-by-one? That is, wouldn't

	rev-list --max-parents=0 --all

be a bit faster than

	for-each-ref |
	while read object type refname
	do
		rev-list --max-parents=0 $refname
	done

I wonder?

On 03/22/18 17:44, Junio C Hamano wrote:

Yeah, you're right -- I forgot that we can pass --all. The check takes 30 seconds, which is a lot better than 12 hours. :) It's a bit heavy still, but msm kernel repos are one of the heaviest outliers, so let me try to run with this.

Thanks for the suggestion!

Best,

On Fri, Mar 23 2018, Konstantin Ryabitsev wrote:

Unless you have just one CPU core your script would probably benefit from being wrapped in GNU parallel. I.e.:

    parallel 'stuff-to-do {}' ::: $(list-of-repos)

Or something similar.