From: EBo <ebo@sandien.com>
To: Fans of the OS Plan 9 from Bell Labs <9fans@9fans.net>
Subject: Re: [9fans] nupas update
Date: Sun, 16 May 2010 19:52:09 -0600 [thread overview]
Message-ID: <0666415404bd7d49e9a122f364b96fbd@swcp.com> (raw)
In-Reply-To: <425b6f8428135ee0c1c1358eb08750a9@kw.quanstro.net>
> i think it's a good question but lacking time travel or a working
> 64-bit kernel, this question is unknowable. :-)
;-) After thinking about it I think amd might have been a better example
>> > please, no use flags. we can't test what we've got. use
>> > flags make the problem go factorial. (gentoo for example
>> > doesn't work if you set the profile use flag.)
>>
>> Now we are getting to the heart of a very important matter. I agree
that
>> use flags causes the dependency graph to go factorial -- but pruned to
>> the
>> number of use flags implemented in each ebuild (so it is not factorial
to
>> the number of accepted use flags).
>
> if each package has only n use flags, then you still have
> 2^n^m options, where m is the number of packages.
>
> proof: each use flag may be on or off. if we order the use
> flags for a package arbitrarly, we can think of them as binary
> digits and there would be 2^n possible values. since there
> are m packages, we can consider this an m digit number with
> each digit taking the value 0 ... 2^n-1.
>
> if they don't all have the same use flags, we can redo this.
> if for package k, there are k_n use flags we would have
> 2^{k_0}2^{k_1} ... =
> 2^(sum k_i}
>
> which i'll grant isn't factorial. but it's still 2^{sum of use flags
> per package}. :-)
>
> i'll give you that this isn't factorial. :-) but on the other hand,
> if a package might not be installed at all, that's like another use
> flag.
and without use flags I end up having k*m packages instead of m. So the
question still comes to do I write it to allow 2^n^m possible combinations
and document the two most common scenarios, or write 2*m package variants
and leave it to the interested to populate any of the remaining 2^{k-2}
permutations. I'm still undecided, but I know then kinds of bugs that
creep up when splitting trees like that. (I guess like splitting hairs ;-)
EBo --
next prev parent reply other threads:[~2010-05-17 1:52 UTC|newest]
Thread overview: 49+ messages / expand[flat|nested] mbox.gz Atom feed top
2009-09-03 2:16 erik quanstrom
2009-09-03 2:29 ` David Leimbach
2009-09-03 2:37 ` erik quanstrom
2010-05-15 23:17 ` Akshat Kumar
2010-05-15 23:45 ` erik quanstrom
2010-05-16 2:36 ` ron minnich
2010-05-16 4:15 ` erik quanstrom
2010-05-16 4:28 ` Akshat Kumar
2010-05-16 4:35 ` ron minnich
2010-05-16 4:39 ` erik quanstrom
2010-05-16 4:51 ` ron minnich
2010-05-16 4:57 ` Akshat Kumar
2010-05-16 5:44 ` ron minnich
2010-05-16 13:42 ` EBo
2010-05-16 14:03 ` erik quanstrom
2010-05-16 14:51 ` Ethan Grammatikidis
2010-05-16 15:37 ` EBo
2010-05-16 15:53 ` Ethan Grammatikidis
2010-05-16 16:02 ` ron minnich
2010-05-16 17:10 ` Ethan Grammatikidis
2010-05-16 17:19 ` EBo
2010-05-16 15:21 ` EBo
2010-05-16 15:42 ` Ethan Grammatikidis
2010-05-16 17:34 ` EBo
2010-05-16 17:47 ` hiro
2010-05-16 18:58 ` Corey
2010-05-16 19:29 ` EBo
2010-05-18 21:35 ` Georg Lehner
2010-05-18 22:07 ` ron minnich
2010-05-16 15:46 ` Jorden M
2010-05-16 15:59 ` Ethan Grammatikidis
2010-05-16 15:42 ` Jorden M
2010-05-16 18:24 ` erik quanstrom
2010-05-16 18:49 ` EBo
2010-05-16 20:44 ` erik quanstrom
2010-05-16 21:44 ` EBo
2010-05-17 1:17 ` erik quanstrom
2010-05-17 1:52 ` EBo [this message]
2010-05-17 1:58 ` erik quanstrom
2010-05-17 1:35 ` Akshat Kumar
2010-05-17 4:09 ` ron minnich
2010-05-17 4:19 ` erik quanstrom
2010-05-17 4:56 ` ron minnich
2010-05-18 23:50 ` Federico G. Benavento
2010-05-18 23:59 ` ron minnich
2010-05-19 0:40 ` Jorden M
2010-05-19 1:40 ` Robert Ransom
-- strict thread matches above, loose matches on Subject: below --
2009-08-06 15:23 erik quanstrom
2008-10-23 0:08 erik quanstrom
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