From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: from localhost (localhost [127.0.0.1]) by mail.saout.de (Postfix) with ESMTP id C983E91A3 for ; Tue, 4 Aug 2009 15:55:28 +0200 (CEST) Received: from mail.saout.de ([127.0.0.1]) by localhost (mail.saout.de [127.0.0.1]) (amavisd-new, port 10024) with ESMTP id CIeKlAv5WUQZ for ; Tue, 4 Aug 2009 15:55:23 +0200 (CEST) Received: from qw-out-2122.google.com (qw-out-2122.google.com [74.125.92.26]) by mail.saout.de (Postfix) with ESMTP for ; Tue, 4 Aug 2009 15:55:23 +0200 (CEST) Received: by qw-out-2122.google.com with SMTP id 9so2224416qwb.55 for ; Tue, 04 Aug 2009 06:55:22 -0700 (PDT) Date: Tue, 4 Aug 2009 16:55:09 +0300 From: Moji Message-ID: <20090804165509.6fa226ec@gmail.com> In-Reply-To: References: <20090803125342.CF87216440B5@mail.absint.com> <20090804004626.4a811f96@gmail.com> Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: quoted-printable Subject: Re: [dm-crypt] 1,5 TB partition: use cbc-essiv or xts-plain? List-Id: List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , To: dm-crypt@saout.de On Tue, 04 Aug 2009 15:27:20 +0200 Henrik Theiling wrote: > >From the wording of the Wikipedia article, however, it is not =20 > completely clear to me how serious the watermarking attack on CBC is. > The IV function is known, so can two blocks be easily constructed in > such a way that their cbc-essiv:sha256 encryption (with whatever main > algorithm) is identical? You'd need to know the sector for that plus > break SHA256, because ESSIV uses the hash of the encryption key plus > the sector number to generate the IV, right? If I understood that > correctly, then I can safely get back to relaxing, enjoying the summer > and drinking beer instead of thinking about this any longer. =46rom Clemens Fruhwirth: "ESSIV E(Sector|Salt) IV, short ESSIV, derives the IV from key material via encryp= tion of the sector number with a hashed version of the key material, the sa= lt. ESSIV does not specify a particular hash algorithm, but the digest size= of the hash must be an accepted key size for the block cipher in use. As t= he IV depends on a none public piece of information, the key, the sequence = of IV is not known, and the attacks based on this can't be launched." This covers watermarks, I hope this provides for drinking much beer. -MJ