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29

I am struggling to understand what is meant by "standard cryptographic assumption". ‘Standard assumption’ broadly means an assumption that has withstood the scrutiny of many smart cryptanalysts for a long time. Examples: We think that, for uniform random 1024-bit primes $p$ and $q$, solving $y = x^3 \bmod pq$ for uniform random $x$ is hard given $pq$ and $...


7

There is no formal definition of standard assumption, but we usually say that an assumption is standard if it has already been used in several cryptographic schemes and if it is well-accepted in the crypto community. It usually also implies that several researchers tried to solve the problem and were not able to find efficient ways of doing so, therefore, ...


1

It is exactly $2^{64}$. You need to look for all possible keys for a successful brute-force. One cannot guarantee that the key will be in the half, $2^{63}$, that you are going to search. In, CS there is an adversary argument, that the adversary can always force you to the worst case. Tell him how you the search, he will produce a case that you are not going ...


1

Firstly consider the simple case where we know that there is only a simple x-or relation between two key bits. $k_i \oplus k_j = 1$ and $i \neq j$ and call it $R1$. If we consider the x-or table we will see that only half of the rows for $k_i$ and $k_j$ will satisfies this equation. \begin{array}{cc|c} k_i & k_j & k_i\oplus k_j \\ \hline 0 & ...


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