On Fri, 22 Feb 2013 14:47:14 -0300 Sergio Lerner <[email protected]> wrote:
> BUFFER1[0]=IVK1 > BUFFER1[i] =Hash(BUFFER1[i-1]) > > BUFFER2[0]=IVK2 > BUFFER2[i] =Hash(BUFFER2[i-1]) > > Encryption: C = AES(EK,BUFFER1[i] XOR BUFFER2[i]) XOR P And also the security of the hash concatenation in this case provides no greater security/entropy then the highest hash...in particular the xor in this case reduces effective randomness of the hash, by create a seperate function f(i) = hash(yi) xor hash(zi), where y and z are dependant values...so why two hash buffers? The value of xor of two non-random data values, further decreases the entropy? _______________________________________________ OTR-dev mailing list [email protected] http://lists.cypherpunks.ca/mailman/listinfo/otr-dev
