From: Konstantin Ryabitsev <konstantin@linuxfoundation.org>
To: "Junio C Hamano" <gitster@pobox.com>,
"Ævar Arnfjörð Bjarmason" <avarab@gmail.com>
Cc: git@vger.kernel.org
Subject: Re: The most efficient way to test if repositories share the same objects
Date: Thu, 22 Mar 2018 17:32:40 -0400 [thread overview]
Message-ID: <906555df-e906-775a-0255-fbc71f7138f6@linuxfoundation.org> (raw)
In-Reply-To: <xmqqlgejlx8e.fsf@gitster-ct.c.googlers.com>
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On 03/22/18 15:35, Junio C Hamano wrote:
> 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.
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
Director, IT Infrastructure Security
The Linux Foundation
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next prev parent reply other threads:[~2018-03-22 21:32 UTC|newest]
Thread overview: 8+ messages / expand[flat|nested] mbox.gz Atom feed top
2018-03-22 16:48 The most efficient way to test if repositories share the same objects Konstantin Ryabitsev
2018-03-22 19:20 ` Ævar Arnfjörð Bjarmason
2018-03-22 19:35 ` Junio C Hamano
2018-03-22 19:53 ` Ævar Arnfjörð Bjarmason
2018-03-22 21:32 ` Konstantin Ryabitsev [this message]
2018-03-22 21:44 ` Junio C Hamano
2018-03-23 13:57 ` Konstantin Ryabitsev
2018-03-23 14:36 ` Ævar Arnfjörð Bjarmason
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