Request Comments: Transpose/XOR Hash

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Hi!

I'm working on a cipher that utilizes Cipher Block Chaining
and One-way hashing to create a sequence of pseudo-random
numbers to XOR against a plaintext.

I'd like to know if anyone has comments regarding the hash
method I came up with. It is a combination of Transposition
and XORing. Basically, it works like this:

Let's say K is the 8 character key that will be hashed. 
There are two binary accumulators M and L, which store the
Most and Least significant bits in each byte of K. After M and
L are both 8 bits long, they are XORed together and the 
resulting value replaces byte N in the Key. This is repeated
8 times, and each time, N is incremented by one.

	Any thoughts, comments?

		-olcay

- --
"For he who lives more lives than one, |) Olcay Cirit -- olcay@libtech.com
   more deaths than one must die"      (| http://www.libtech.com/olo2.html

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Olcay Cirit writes:
> I'd like to know if anyone has comments regarding the hash
> method I came up with. 

My primary suggestion is that you do some reading in a good crypto text like
Applied Cryptography (2nd ed.). The introduction to PRZ's manual for PGP is
also particularly germane, as is the sci.crypt FAQ.

> It is a combination of Transposition
> and XORing. Basically, it works like this:
> 
> Let's say K is the 8 character key that will be hashed. 
> There are two binary accumulators M and L, which store the
> Most and Least significant bits in each byte of K. After M and
> L are both 8 bits long, they are XORed together and the 
> resulting value replaces byte N in the Key. This is repeated
> 8 times, and each time, N is incremented by one.

OK, since I'm procrastinating doing some non-crypto work right now, I looked
at your algorithm for 10 minutes or so. Your description is rather vague, so
I'm not sure I understand exactly what you're proposing. My best guess is:

The hash has 8 rounds. The initial 64-bit digest value is H = K. For 
notational convenience, let H[i] denote the i-th byte of H, and H[i,j] denote 
the j-th bit of H[i]. Juxtaposition denotes concatenation. I'll assume bit 1 
is the MSB. You hash K by doing:

H = K
for k = 1 to 8 do
	M = H[1,1] H[2,1] ... H[8,1]
	L = H[1,8] H[2,8] ... H[8,8]
	H[k] = M xor L
od
return H

First of all, this isn't even a good checksum, since the output depends on 
only 16 of the 64 input bits (namely the MSBs and LSBs of each byte of K).

Many pairs of output bits are highly correlated (in fact, equal).
H[2,1] == H[3,1] == ... == H[8,1] because H[1] doesn't change after the first 
round. Similarly H[1,8] == H[2,8] == ... == H[8,8] because H[8] doesn't
change until the end of the algorithm. For each other choice of bit index j, 
there's a "before" value H[1,j] == ... == H[j,j] and an "after" value
H[j+1,j] == ... == H[8,j].

This tells us that there are at most 2^(1 + 7*2) == 2^15 possible hash output 
values. 

But it would be faster to take advantage of the observation that for each j,
H[1,j] == K[j,1] xor K[j,8]. We guess the 8 LSBs as g_1, g_2, ..., g_8, and
compute the corresponding 8 MSBs as m_j = g_j xor H[1,j], which gives us all
the information we need to compute a hash value.

So we can compute a preimage of an arbitrary hash value with at most 2^8 = 
256 guesses.

In any case, this is an extremely weak cryptographic hash.

Lewis			"...made my own pretty hate machine" (Tori Amos)
lmccarth@cs.umass.edu	http://www.cs.umass.edu/~lmccarth

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