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EDIT: The answer below answers the RSA modulus case… There is some related discussion in the answers to the Crypto.SE question “Why is the following RSA PRNG cryptographically secure?”. Edit: As pointed out in comments, this answer is about the zRSA The consensus among the answers seems to be that the required security assumption is too strong (strong RSA ...


Looks like Marvin32 is a part of a patent :-) And Microsoft really do believe it is resistant to Hash Flood attack.


Randomness testing uses asymptotic properties. Thus, as the length of your input increases the effect you are concerned about will disappear. Specifically, there is no need to convert ternary $1$ or $2-$tuples to binary. If you let $n$ increase you cover the interval $\{0,1,\ldots,3^n-1\}$ and then you convert these integers to binary, paying at most a 1 ...


I won't give the answer to homework questions, but I will give a hint. Suppose you learn the tags for $m_1 || m_2$ and $m_3 || m_4$; what other messages could you deduce the tags for?


Yes. A secure PRF is a secure MAC. A secure MAC of a secure MAC is a secure MAC. Therefore, applying a PRF to a MAC still gives you a MAC. Depending on the length of the inner MAC and the PRF you may lose security bits, but if they are long enough it works.


The full $m-$sequence (periodically repeated to avoid modular addition in the subscript} with $$C_{xx}(\tau):=\sum_{k=0}^{2^n-2} (-1)^{x_k+x_{k+\tau}}$$ satisfies $C_{xx}(\tau)=-1+\delta(\tau)(2^n),$ where $\delta$ is the dirac delta function. Now, one might define an $m-$symbol partial correlation function, whose average is proportional to what you want, if ...

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