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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 ...


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Looks like Marvin32 is a part of a patent :-) And Microsoft really do believe it is resistant to Hash Flood attack. http://www.google.com/patents/US20130262421


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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 ...


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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?


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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.


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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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