Re: An hash-Encryption algorithm

Kristian Gjøsteen <kristiag+news@xxxxxxxxxxxx> wrote in
news:uo7qa3-j1m1.ln1@xxxxxxxxxxxxxxxxxxxxx:

> Neo <neoscandal@xxxxxxxxx> wrote:
>>What if from n=2 onwards K(n) = MD5( K(n-1) + C(n-1) + user entered
>>key). theres no way the attacker can obtain the key.. even given both
>>P1 and C1.
>
> So this would be K(n) = MD5(K(n-1), C(n-1), K(0)). At first sight,
> this looks ok. I think MD5(C(n-1),K(0)) (CFB mode) and
> MD5(K(n-1),K(0)) (OFB mode) would work just as well.
>
>>Or have i missed something again.
>
> You are getting somewhere. Now you are just missing a somewhat
> subtle point: Encrypt two messages P1(1,...,n) and P2(1,..m) with
> the same key K(0). Find the expression for the first ciphertext
> blocks C1(1) and C2(1) and compute their exclusive ors. What do you
> find?
>

Hmm.. according to my design that would imply ( consider pa,ca,pb,cb)...

ca xor cb = (pa xor (ka(0) + salt1)) xor (pb xor (kb(0) + salt2))

|_______________________| |______________________|
ciphertext1 block 1 ciphertext2 block 1

I dont see how this may be simplified...

Note that salts are generated at the time of encryption .. and
thoeretically no two salts would match. Thus no two encryptions would
use similar keys during the first round.


> The solution is to use a nonce or initialization vector (iv).
>
> You would probably enjoy reading the chapter on block cipher modes of
> operation in any cryptography book and translating CTR, OFB and CFB
> into MD5-like constructions.
>

Will do.

thank you
.

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