Re: Why triple encryption instead of split+encrypt?

[John Douceur <johndo@microsoft.com>]

-----BEGIN PGP SIGNED MESSAGE-----

>From:  <solman@MIT.EDU>
>Date: Tuesday, July 19, 1994 8:37AM

>> So now the meet-in-the-middle attack regains its earlier applicability:
>> A known-plaintext attack would encrypt P with the splitter, decrypt
>> C0 with DES, and attempt to meet in the middle to discover key K0;
>> similarly, decrypting with C1 to get K1.

>I don't believe this is true. You have C0 and C1, but you can not figure
>out P0 and P1 without the hash of the concatenation of both keys. Without
>this you can not do a meet in in the middle attack, right?

Wrong.  (sorry to sound so authoritative; just wanted to make my position
clear.)  If you knew how to perform the split, there would be no need for
a meet-in-the-middle attack; you could just attack each of the DES
encryptions of the split data separately.

Recall that a meet-in-the-middle attack is a method for cryptanalyzing a
message that has been doubly encrypted, as the following:

	I = E0_K0(P)
	C = E1_K1(I)

By this nomenclature, I mean to imply that not only the keys but also the
algorithms may be different between the first and second encryptions.
Meet-in-the-middle works by encrypting from P towards I, decrypting from
C towards I, and attempting to meet in the middle.  For algorithms with
large keyspaces, this attack requires so much memory for storing
intertext as to be almost absurd in today's world, but it is a valuable
theoretical technique for demonstrating that double encryption provides
little more computational security than single encryption.

I am claiming that your technique:

	P0, P1, P2, ... Pn = S_KS(P)

	C0 = E_K0(P0)
	C1 = E_K1(P1)
	C2 = E_K2(P2)
	.   .   .
	Cn = E_Kn(Pn)

Can be decomposed into parallel double encryptions, and is therefore just
as vulnerable to a meet-in-the-middle attack as double DES (or more so,
if your splitting algorithm is less secure than DES).  NB:  When I use
the term "double encryption" here, I am not referring to your use of DES
multiple times after the split; I am referring to the splitting itself as
the first encryption, and the DES as the second encryption.  Let us
define the function Sx_KS(P) as the portion of the splitting algorithm
which produces Px:

	P0 = S0_KS(P)
	P1 = S1_KS(P)
	.   .   .

We now have a parallel set of double encryptions as follows:

	P0 = S0_KS(P)
	C0 = E_K0(P0)

	P1 = S1_KS(P)
	C1 = E_K1(P1)

	.   .   .

Each of these double encryptions is vulnerable to a known-plaintext
meet-in-the-middle attack from P to Cx.

>Don't concatenate the negation of the two key hash to the hash. The
>point of that step was to split the cipher into two equal sized parts,
>but there is no reason to require that. In fact the possibility of
>different sized parts would add to the confussion. (The probability
>of an extreme imbalance in the size of the ciphers is extremelly
>small.)

>I think that multiplexing based on the hash of the concatenated keys
>is as secure as the one way hash function is, no?

In my above argument, I assumed a splitting key which is completely
independent of the DES keys.  This will be more secure than a splitting
key which is *any* function of the DES keys, since it increases the size
of the keyspace.

>> the security of this scheme is significantly less
>> than that of triple DES.

>Well I don't believe that this is the case,

Perhaps you do now?

>but there is one way to find out
>:). I believe that for messages longer than a couple of K, my algorithm
>provides substantially more security than its DES analog and is quicker.
>I'll write up a version of this that splits into 4 parts and post it here
>some time over the next week. I think that splitting into four parts should
>be about as quick as double DES while providing substantially more security
>than triple DES (which I will time it against).

If you still maintain this position, then either you have not understood my
argument above, or I seriously misunderstand your algorithm.  If you have
not yet been convinced that you have not eliminated the meet-in-the-middle
attack as triple encryption does, then I welcome your algorithm in code, so
that I may see if I am missing something fundamental in your approach.
However, I strongly suggest that you review meet-in-the-middle attacks as
described by Merkle and Hellman and judge for yourself their applicability
to and effectiveness against your algorithm.

>The question of the security of the split is difficult to resolve so I would
>like some help with it. Is multiplexing based on the hash of the concat of
>the keys as secure as the hash?

The security of the generation of the splitting key from the DES keys is
almost irrelevant.  You can guarantee that the splitting key is completely
uninferable from the DES keys by making them independent, yet the
split+encrypt algorithm is still as weak as (or weaker than) double DES.

JD

-----BEGIN PGP SIGNATURE-----
Version: 2.6

iQCVAgUBLiwC4EGHwsdH+oN9AQFfIQP+MoNBMzrrZiTJYdF2eIuwLiprxTLeqBpR
pxNfOrQ190Ugw+BGcjgbb7r1HZkpPtvNaXEtS/n0jBDasMalnwnPbNDM1rpl0ZkY
qWsGcLXhb5MQr/sCN9E5Bud8QCRD1eF+OL3jLUxIq3fKVuECA1zk+4osE2bTw2Fv
shX6vT8xZjg=
=COAe
-----END PGP SIGNATURE-----