MESSAGE
DATE | 2015-02-27 |
FROM | prmarino1@gmail.com
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SUBJECT | Re: [NYLXS - HANGOUT] Read the FUCKING NOTES
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From owner-hangout-outgoing-at-mrbrklyn.com Fri Feb 27 19:20:47 2015 Return-Path: X-Original-To: archive-at-mrbrklyn.com Delivered-To: archive-at-mrbrklyn.com Received: by mrbrklyn.com (Postfix) id 07E5F1612FA; Fri, 27 Feb 2015 19:20:47 -0500 (EST) Delivered-To: hangout-outgoing-at-mrbrklyn.com Received: by mrbrklyn.com (Postfix, from userid 28) id EA0B41612FD; Fri, 27 Feb 2015 19:20:46 -0500 (EST) Delivered-To: hangout-at-nylxs.com Received: from mail-qa0-f43.google.com (mail-qa0-f43.google.com [209.85.216.43]) by mrbrklyn.com (Postfix) with ESMTP id 0BCD41612FA; Fri, 27 Feb 2015 19:20:44 -0500 (EST) Received: by mail-qa0-f43.google.com with SMTP id bm13so15363107qab.2; Fri, 27 Feb 2015 16:20:44 -0800 (PST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=content-type:mime-version:content-transfer-encoding:message-id:date :subject:from:in-reply-to:references:to; bh=WsE0GdXyVn0pObdIc7UHHFp9cF3kUyFaGCty9VQX2Kk=; b=ZYrWmGs2L+mJ4K4P5WF4KF57XIkCoDMilveaM29ms0DHIu38cei7khzG6Cu20MUfn4 DLpb+4EES63xEg4kVry9dWjb6oZKi8gbOye8pyu7zjLdsz2LCTSK10WT2RTMv3nf+RCe RBn0PY16oR3ENhUeD54YKgCIam7wVQrxKD2Zw3ioT4CQg0quNR88e+SUr6OXZFCZF1vk HgLaHVseeF2nJ4yUPl8O1r/wxo81TFFjmB35LF7hRTJzMMsRNNQYJfR5QvzCRj9q83Hj Y7xt9XAxhSZZ643ZkY8WFjcIF1KlasxMih++GreySTQbt7bOceIF2LaaRJ6di3y9XOXU Fkeg== X-Received: by 10.140.101.22 with SMTP id t22mr33082419qge.9.1425082844090; Fri, 27 Feb 2015 16:20:44 -0800 (PST) Received: from [127.0.0.1] ([172.56.34.187]) by mx.google.com with ESMTPSA id 88sm153556qkz.1.2015.02.27.16.20.42 (version=TLSv1 cipher=ECDHE-RSA-RC4-SHA bits=128/128); Fri, 27 Feb 2015 16:20:43 -0800 (PST) Content-Type: text/plain; charset="utf-8" MIME-Version: 1.0 Content-Transfer-Encoding: quoted-printable X-Mailer: BlackBerry Email (10.2.1.3442) Message-ID: <20150228002042.5894291.22419.3653-at-gmail.com> Date: Fri, 27 Feb 2015 19:20:42 -0500 Subject: Re: [NYLXS - HANGOUT] Read the FUCKING NOTES From: prmarino1-at-gmail.com In-Reply-To: <20150228000724.5894291.2924.3651-at-gmail.com> References: <54F0E220.5010209-at-panix.com> <20150228000724.5894291.2924.3651-at-gmail.com> To: hangout-at-nylxs.com, learn-at-nylxs.com, hangout Sender: owner-hangout-at-mrbrklyn.com Precedence: bulk Reply-To: hangout-at-nylxs.com [NYLXS: HANGOUT] X-BeenThere: hangout-at-nylxs.com X-Mailing-list: hangout-at-nylxs.com Precedence: list List-Id: NYLXS General Discussion Forum List-Unsubscribe: List-Archive: List-Post: List-Help: List-Subscribe:
Correction it sounds like a might be a parity calculation=E2=80=8E. Again o= ver thinking it only makes sense if you are dealing with 4 or more disks. M= athematically it still sort of works if you can utilize decimals as disk co= unts but=C2=A0
Sent from my BlackBerry 10 smartphone. =C2=A0 Original Message =C2=A0 From: prmarino1-at-gmail.com Sent: Friday, February 27, 2015 19:07 To: hangout-at-nylxs.com; learn-at-nylxs.com; hangout Subject: Re: [NYLXS - HANGOUT] Read the FUCKING NOTES
Umm correct me if I'm wrong =E2=80=8Ebut 2 to the power of one is 2 so to s= implify the first equation 2-1=3D1. So n=3D1 is a really bad example.
Beyond that =E2=80=8Eoff the top of my head it looks like you may be dealin= g with RAID stripe, and stride equations. In that case I think you are over= thinking the problem.
Sent from my BlackBerry 10 smartphone. =C2=A0 Original Message =C2=A0 From: Ruben Safir Sent: Friday, February 27, 2015 16:31 To: learn-at-nylxs.com; hangout Reply To: hangout-at-nylxs.com Subject: [NYLXS - HANGOUT] Read the FUCKING NOTES
The minimum number of moves required to solve the Towers of Hanoi puzzle with n disks is ***2^n =E2=88=92 1***.
Proof is by mathematical induction: Basis: n =3D 1. Number of moves required is 2^1 =E2=88=92 1 =3D 1. Inductive step: Assume result true for n disks. Suppose now n + 1 disks are given. In order to move the largest disk from its initial location at the bottom of pile 1 to the final location at the bottom of pile 3, we clearly need to first have moved all other disks to pile 2, which by inductive hypothesis requires at least 2^n =E2=88=92 1 moves. Then one move is required to move the largest = disk from pile 1 to pile 3. Finally at least 2n =E2=88=92 1 moves are required to mov= e the remaining disks from pile 2 to pile 3. So the total number of moves required is at least (2^n =E2=88=92 1) + 1 + (2^n =E2=88=92 1) =3D ***2^(n+1) =E2=88=92 1***.
What don't i understand hear
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