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0

Yes you can. This is because hash can be made using block ciphers and we know that block ciphers can be made from stream ciphers. Thus, by the transitive property we can observe that hash can be made from stream ciphers. I hope this helps!


3

I would think these numbers would have been put on the google search engine, and yield (probably) many hits. This assumption is wrong. Certificate serial numbers are not indexed by common search engines, nor are they typically posted to any HTML site. Frankly, I'm not sure why you would assume they'd be indexed. The Wordpress certificate is used for ...


5

You can generate a random string $s_1$ as long as the plaintext. Then XOR this value with the plaintext generating $s_2$. Now encrypt both parts using $\mathrm{Enc}_1$ and $\mathrm{Enc}_2$. You need to decrypt both to XOR the two parts together again. This is similar to secret sharing where you need two parts of a key to decrypt. If $\mathrm{Gen}_1$ and ...


3

For any $n \in \mathbf{N}$, let $X_n$ be a random variable which always equals $n$, and $Y_n$ be a random variable which equals $n$ or $n+1$ each with probability $1/2$. Then the probability ensembles $X = \{X_n\}_{n\in \mathbf{N}}$ and $Y = \{Y_n\}_{n\in \mathbf{N}}$ are not computationally indisinguishable. A possible distinguisher is an algorithm $D$ ...


0

Yes, a function $f$ is said to be negligible if for every polynomial function $p(n)$ there exits some constant $N$ such that $f(n) < \frac{1}{p(n)}$ for all $n > N$. If $\frac{1}{n!} < \frac{1}{p(n)}$ then $n! > p(n)$, for all polynomials $p(n)$ and suitable $N$ such that $n>N$. Thus, you'd only have to prove the second segment of the ...


2

Hint: you can notice that $n! > 2^n$ (except for very small $n$).


1

$A$ has outsourced it's private data to the server, right? In that case $A$'s private data has already been leaked to the adversary when the server was corrupted. So there is no added leakage in $B$ also learning this data. In fact since only one of $B$ and the server can be corrupted (assuming static corruptions) we can conclude that $B$ is honest, and ...


2

Assuming that the probability distributions of $\pi_{k_1}$ and $\pi_{k_2}$ are both uniform (that is, each permutation can take on any particular setting with probability $1/n!$), then no, adversary does not have enough information to learn anything about the original positions. This remains true even if we assume the adversary can perform unbounded ...



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