From mboxrd@z Thu Jan 1 00:00:00 1970 From: erik quanstrom Date: Sun, 16 May 2010 21:17:48 -0400 To: 9fans@9fans.net Message-ID: <425b6f8428135ee0c1c1358eb08750a9@kw.quanstro.net> In-Reply-To: <932700d9bb1c81831186e7aeb809fd9f@swcp.com> References: <6aaf2d79af665bf1905db13e44e194e5@quanstro.net> <3c68655ad1dadf393d44b4a945abbd7a@swcp.com> <26f3b3b7fc6f7e8e8d90094305925bdd@kw.quanstro.net> <932700d9bb1c81831186e7aeb809fd9f@swcp.com> MIME-Version: 1.0 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit Subject: Re: [9fans] nupas update Topicbox-Message-UUID: 25fdc416-ead6-11e9-9d60-3106f5b1d025 > > there is no 64 bit kernel. > > Will there ever be? Or is that even an appropriate question? i think it's a good question but lacking time travel or a working 64-bit kernel, this question is unknowable. :-) > > 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. - erik