From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S261700AbTDKVRL (for ); Fri, 11 Apr 2003 17:17:11 -0400 Received: (majordomo@vger.kernel.org) by vger.kernel.org id S261719AbTDKVRL (for ); Fri, 11 Apr 2003 17:17:11 -0400 Received: from albireo.ucw.cz ([81.27.194.19]:55817 "EHLO albireo.ucw.cz") by vger.kernel.org with ESMTP id S261700AbTDKVRK (for ); Fri, 11 Apr 2003 17:17:10 -0400 Date: Fri, 11 Apr 2003 23:28:50 +0200 From: Martin Mares To: Mike Dresser Cc: John Bradford , linux-kernel@vger.kernel.org, linux-hotplug-devel@lists.sourceforge.net, message-bus-list@redhat.com Subject: Re: [ANNOUNCE] udev 0.1 release Message-ID: <20030411212850.GA13608@ucw.cz> References: <200304111746.h3BHk9hd001736@81-2-122-30.bradfords.org.uk> Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Disposition: inline In-Reply-To: User-Agent: Mutt/1.3.28i Sender: linux-kernel-owner@vger.kernel.org X-Mailing-List: linux-kernel@vger.kernel.org Hello, world!\n > Someone PLEASE tell me the simpler way to do this. If you have an arbitrary system of splitters and you hook another 3-way splitter to any of its outputs, you lose one output and gain 3 new outputs, so the total number of outputs increases by 2. Hence if you take K inputs and N 3-way splitters, the network has K+2N outputs, no matter how the splitters are connected (of course unless you create a cycle :-) ). So in our case, we are searching for the smallest possible N such that 5+2N >= 4000, which equals ceil((4000-5)/2) = 1998. Have a nice fortnight -- Martin `MJ' Mares http://atrey.karlin.mff.cuni.cz/~mj/ Faculty of Math and Physics, Charles University, Prague, Czech Rep., Earth God is real, unless declared integer.