From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S261346AbTDKSib (for ); Fri, 11 Apr 2003 14:38:31 -0400 Received: (majordomo@vger.kernel.org) by vger.kernel.org id S261380AbTDKSib (for ); Fri, 11 Apr 2003 14:38:31 -0400 Received: from [64.246.18.23] ([64.246.18.23]:10444 "EHLO ensim.2hosting.net") by vger.kernel.org with ESMTP id S261346AbTDKSi3 (for ); Fri, 11 Apr 2003 14:38:29 -0400 From: "Steve Lee" To: Cc: Subject: RE: [ANNOUNCE] udev 0.1 release Date: Fri, 11 Apr 2003 13:50:16 -0500 Message-ID: <001f01c3005b$362a1f60$0201a8c0@pluto> MIME-Version: 1.0 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit X-Priority: 3 (Normal) X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook, Build 10.0.4024 In-Reply-To: <200304112018.11931.freesoftwaredeveloper@web.de> X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 Importance: Normal Sender: linux-kernel-owner@vger.kernel.org X-Mailing-List: linux-kernel@vger.kernel.org Yes, 1998 3 into 1 splitters would be needed. The break down is as follows: Three of the original power connectors would need 400 splitters, while the remaining two would just need 399 splitters. This would result in 4001 power connectors, or 4001 disks. Because it's a Friday, here is a simple program that verified my findings. int main(int argc,char **argv) { int con_a,con_b,con_c,con_d,con_e; int pow_a,pow_b,pow_c,pow_d,pow_e; con_a=con_b=con_c=con_d=con_e=0; pow_a=pow_b=pow_c=pow_d=pow_e=1; while(1) { con_a++; pow_a=con_a*3-(con_a==1?0:con_a-1); if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000) break; con_b++; pow_b=con_b*3-(con_b==1?0:con_b-1); if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000) break; con_c++; pow_c=con_c*3-(con_c==1?0:con_c-1); if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000) break; con_d++; pow_d=con_d*3-(con_d==1?0:con_d-1); if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000) break; con_e++; pow_e=con_e*3-(con_e==1?0:con_e-1); if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000) break; } fprintf(stderr,"con_a = %d, pow_a = %d\n",con_a,pow_a); fprintf(stderr,"con_b = %d, pow_b = %d\n",con_b,pow_b); fprintf(stderr,"con_c = %d, pow_c = %d\n",con_c,pow_c); fprintf(stderr,"con_d = %d, pow_d = %d\n",con_d,pow_d); fprintf(stderr,"con_e = %d, pow_e = %d\n",con_e,pow_e); fprintf(stderr,"pow_a+pow_b+pow_c+pow_d+pow_e = %d\n",pow_a+pow_b+pow_c+pow_d+pow_e); return 0; } -----Original Message----- From: freesoftwaredeveloper@web.de [mailto:freesoftwaredeveloper@web.de] Sent: Friday, April 11, 2003 1:18 PM To: oliver@neukum.name Cc: Steve Lee; Desmet_Jochen@emc.com Subject: Re: [ANNOUNCE] udev 0.1 release On Friday 11 April 2003 20:15, Oliver Neukum wrote: > Am Freitag, 11. April 2003 20:03 schrieb Michael Buesch: > > On Friday 11 April 2003 19:46, John Bradford wrote: > > > [Puzzle] > > > Say the power supply had five 5.25" drive power connecters, how many 1 > > > into 3 power cable splitters would you need to connect all 4000 disks? > You are hereby hit. All drivers need to be connected to a splitter. > Thus any answer under 4000/3 must be wrong. Yea, but: "Say the power supply had five 5.25" drive power connecters" :) -- My homepage: http://www.8ung.at/tuxsoft fighting for peace is like fu**ing for virginity