Re: Why triple encryption instead of split+encrypt?

[solman@MIT.EDU]

> 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).

We thus far agree. Vulnerability is dependent on splitting it into
parallel problems.

> 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.

AH! I hadn't been looking at it that way. I wish I had thought of it like
that. You are then quite correct that meet-in-the-middle attacks can be
done, but the key to the first encryption (the hashing multiplex) is 112
bits (for the split into two parts version) which would require 2^112
stored messages, substantially more than could possibly be stored by
anybody ever (well, I guess ever is a bad word to use in this context). 

> 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.

When I am multiplexing based just on the hash of the keys and not
hash followed by negated hash, the cryptanalyst does not know how
to derive Ci (i=1...n) from C. This is even more true if I interleave
the cipher texts instead of sending them one after the other (which makes
more sense if I am doing them in parallel anyway). Of course this only
increases security by a few powers of two (about n-2 where the length of
the hash is 2^n and we constrain the keys slightly to avoid lopsided
splits) if the opponent has the memory available to do a meet in the middle 
attack for n=2. For n=4 this increased security becomes substantial
however. (Combinations of numbers that add up to the size of the hash as
constrained by the binomial distribution and splits that the program
determines to be acceptable.) It is still far less security than is
provided by the rest of the algorithm, however. So I suppose I should
consider this to negligible (even if it is around 2^10) and concede the
point.

> >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.

Certainly, but I figure that if using the hash of the keys stands up, then
the stronger totally seperate version certainly will.

> >> 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?

Your point is unquestionably valid, but I still believe that the security
of the scheme, even when just splitting into two parts and using the hash
of the keys to multiplex the split, is much worse (by more than a couple
of factors of two) than DES.

I suppose I have merely created a new hash based symetric cipher. I will
have to look up the other hash based symetric ciphers and see how they
compare.
 
> >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.

I don't think that meet in the middle attacks are relevant because nobody
has 2^112 memory. Its just alot. Schneier claims that at 128 bits there
probably isn't enough matter in the universe to meet an algorithm using
IDEA in the middle. I would say that 112 bits is nearly as solid a line of
defense.