Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S1422984AbWJUBdp (ORCPT ); Fri, 20 Oct 2006 21:33:45 -0400 Received: (majordomo@vger.kernel.org) by vger.kernel.org id S1751104AbWJUBdp (ORCPT ); Fri, 20 Oct 2006 21:33:45 -0400 Received: from omx1-ext.sgi.com ([192.48.179.11]:65479 "EHLO omx1.americas.sgi.com") by vger.kernel.org with ESMTP id S1750934AbWJUBdo (ORCPT ); Fri, 20 Oct 2006 21:33:44 -0400 Date: Fri, 20 Oct 2006 18:33:24 -0700 From: Paul Jackson To: "Siddha, Suresh B" Cc: nickpiggin@yahoo.com.au, akpm@osdl.org, mbligh@google.com, menage@google.com, Simon.Derr@bull.net, linux-kernel@vger.kernel.org, dino@in.ibm.com, rohitseth@google.com, holt@sgi.com, dipankar@in.ibm.com, suresh.b.siddha@intel.com, clameter@sgi.com Subject: Re: [RFC] cpuset: add interface to isolated cpus Message-Id: <20061020183324.b29caa37.pj@sgi.com> In-Reply-To: <20061020135944.B8481@unix-os.sc.intel.com> References: <20061019092607.17547.68979.sendpatchset@sam.engr.sgi.com> <453750AA.1050803@yahoo.com.au> <20061019105515.080675fb.pj@sgi.com> <4537BEDA.8030005@yahoo.com.au> <20061019115652.562054ca.pj@sgi.com> <4537CC1E.60204@yahoo.com.au> <20061019203744.09b8c800.pj@sgi.com> <453882AC.3070500@yahoo.com.au> <20061020130141.b5e986dd.pj@sgi.com> <20061020135944.B8481@unix-os.sc.intel.com> Organization: SGI X-Mailer: Sylpheed version 2.2.4 (GTK+ 2.8.3; i686-pc-linux-gnu) Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: 7bit Sender: linux-kernel-owner@vger.kernel.org X-Mailing-List: linux-kernel@vger.kernel.org Content-Length: 1016 Lines: 25 Suresh wrote: > Its just a corner case issue that Nick didn't consider while doing a quick > patch. Nick meant to partition the sched domain at the top > exclusive cpuset and he probably missed the case where root cpuset is marked > as exclusive. This makes no sense. If P is a partition of S, that means that P is a set of subsets of S such that the intersection of any two members of P is empty, and the union of the members of P equals S. If P is a partition of S, then adding S itself to P as another member makes P no longer a partion, for then every element of S is in two elements of P, not one. -- I won't rest till it's the best ... Programmer, Linux Scalability Paul Jackson 1.925.600.0401 - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/