i want to tell you, Holmes: Dr. Watson’s voice was enthusiastic, That your recent activities in…

i want to tell you, Holmes: Dr. Watson’s voice was enthusiastic, That your recent activities in network security have increased my interest in cryptography. And just yesterday I found a way to make one-time pad encryption practical’ ‘Oh, really?” Holmes’ face lost its sleepy look. ‘Yes, Holmes. The idea is quite simple. For a given one-way function F, I generate a long pseudorandom sequence of elements by applying F to some standard sequence of arguments. The cryptanalyst is assumed to know F and the general nature of the sequence, which may be as simple as S, S + 1, S + 2, …, but not secret S. And due to the one-way nature of F, no one is able to extract S given F(S + i) for some i, thus even if he somehow obtains a certain segment of the sequence, he will not be able to determine the rest’
? am afraid, Watson, that your proposal isn’t without flaws and at least it needs some additional conditions to be satisfied by F. Let’s consider, for instance, the RSA encryption function, that is F(M) = MK mod N, K is secret. This function is believed to be one-way, but I wouldn’t recommend its use, for example, on the sequence M = 2, 3, 4, 5, 6, … ” ‘But why, Holmes?” Dr. Watson apparently didn’t understand. °Why do you think that the resulting sequence 2K mod N, 3K mod N, 4K mod N, … is not appropriate for one-time pad encryption if K is kept secret?” ‘Because it is—at least partially—predictable, dear Watson, even if K is kept secret. You have said that the cryptanalyst is assumed to know F and the general nature of the sequence. Now let’s assume that he will obtain somehow a short segment of the output sequence. In crypto circles, this assumption is generally considered to be a viable one. And for this output sequence, knowledge of just the first two elements will allow him to predict quite a lot of the next elements of the sequence, even if not all of them, thus this sequence can’t be considered to be cryptographically strong. And with the knowledge of a longer segment he could predict even more of the next elements of the sequence. Look, knowing the general nature of the sequence and its first two elements 2K mod N and 3K mod N, you can easily compute its following elements.”
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