B.4 Key as initialization vector

The des-cbc-crc algorithm (see Cryptographic Overview) uses the DES key as the initialization vector. This is problematic in general (see below⁵), but may be mitigated in Kerberos by the CRC checksum that is also included.

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
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References: <4h39li$33o@gaia.ns.utk.edu>
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In article <4h39li$33o@gaia.ns.utk.edu>,
Nair Venugopal <venu@mars.utcc.utk.edu> 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.