des-cbc-crc algorithm (see Cryptographic Overview) uses
the DES key as the initialization vector. This is problematic in
general (see below1), but
may be mitigated in Kerberos by the CRC checksum that is also
From daw@espresso.CS.Berkeley.EDU Fri Mar 1 13:32:34 PST 1996 Article: 50440 of sci.crypt Path: agate!daw From: daw@espresso.CS.Berkeley.EDU (David A Wagner) Newsgroups: sci.crypt Subject: Re: DES-CBC and Initialization Vectors Date: 29 Feb 1996 21:48:16 GMT Organization: University of California, Berkeley Lines: 31 Message-ID: <firstname.lastname@example.org> References: <email@example.com> NNTP-Posting-Host: espresso.cs.berkeley.edu In article <firstname.lastname@example.org>, Nair Venugopal <email@example.com> wrote: > Is there anything wrong in using the key as the I.V. in DES-CBC mode? Yes, you're open to a chosen-ciphertext attack which recovers the key. Alice is sending stuff DES-CBC encrypted with key K to Bob. Mary is an active adversary in the middle. Suppose Alice encrypts some plaintext blocks P_1, P_2, P_3, ... in DES-CBC mode with K as the IV, and sends off the resulting ciphertext A->B: C_1, C_2, C_3, ... where each C_j is a 8-byte DES ciphertext block. Mary wants to discover the key K, but doesn't even know any of the P_j's. She replaces the above message by M->B: C_1, 0, C_1 where 0 is the 8-byte all-zeros block. Bob will decrypt under DES-CBC, recovering the blocks Q_1, Q_2, Q_3 where Q_1 = DES-decrypt(K, C_1) xor K = P_1 Q_2 = DES-decrypt(K, C_2) xor C_1 = (some unimportant junk) Q_3 = DES-decrypt(K, C_1) xor 0 = P_1 xor K Bob gets this garbage-looking message Q_1,Q_2,Q_3 which Mary recovers (under the chosen-ciphertext assumption: this is like a known-plaintext attack, which isn't too implausible). Notice that Mary can recover K by K = Q_1 xor Q_3; so after this one simple active attack, Mary gets the key back! So, if you must use a fixed IV, don't use the key-- use 0 or something like that. Even better, don't use a fixed IV-- use the DES encryption of a counter, or something like that.
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