Not really. At minimum it's 2^128 bytes. But it's probably closer to 2^160.
On Thu, Sep 4, 2014 at 12:37 AM, Givon Zirkind <[email protected]> wrote: > forgive me for jumping into the middle of the conversation. but, if > memory serves, such a scheme is limited to 2^128 bytes. then, the counter > repeats. which causes redundancy, weakness in the hash. > > > On 9/3/2014 7:18 PM, Ryan Carboni wrote: > > isn't the simplest solution would be to concatenate or XOR a counter? > > Thus H[0] = Hash(input) > H[N] = Hash(H[N-1]+CTR) > > considering that hashes from MD4 to SHA-2 all have block sizes of 512 > bits, much larger than their outputs, one could simply concatenate a > 128-bit counter. > > > > > _______________________________________________ > cryptography mailing > [email protected]http://lists.randombit.net/mailman/listinfo/cryptography > > >
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