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For all $\ell\in\mathbb N$, the cardinality of the set $\{0,1\}^\ell$ is $2^\ell$. This can be obtained inductively: We have $\{0,1\}^0=\{()\}$ (the empty tuple) which has cardinality $1=2^0$. If $\lvert\{0,1\}^\ell\rvert=2^\ell$, the elements of $\{0,1\}^{\ell+1}$ can be partitioned into two classes: The sequences that start with $0$ and those that start ...


...wouldn't key still get repeated every few hours or so - i.e. you come to the end of the PRG(K)... This is where you are mistaken. Modern cryptographic PRGs simply do no repeat within any conceivable time frame. That is, starting from a seed, a well-constructed PRG (and this is true even when they are not so well constructed, like RC4) will simply ...


An encryption algorithm does not need a keyspace. By definition, however, it has one. It sound to me like your confusion is mainly terminological. In cryptography, the "keyspace" of an encryption system is defined simply as the set of all possible (distinct) keys that the algorithm can accept. For example, let's say that we're back in the days of the ...


A keyspace is the set of all possible keys that an algorithm can use. So the keyspace for AES-256 for example is $2^{256}$ possible keys. The key chosen by the user exists in this keyspace.


You do supply your own key. This key has a certain format; the keyspace is just the set of valid keys.


You're right in that there's little chance you can break the logarithm in a well-chosen 512 bit group (using a home computer, in reasonable time — as pointed out by SEJPM, it is possible investing some time and a good amount of money). However, in your case, the parameters are bad: The order of $(\mathbb Z/p\mathbb Z)^\ast$, that is $p-1$, is a smooth ...


I know that all the subkeys $k_i$ are derived from the main key $K$, but how? However the cipher designer feel like. The Feistel design gives guidance as to how the block is processed (and in a way to make inverting the cipher easy), however it gives no guidance as to actually generate the subkeys. The designers can do anything they like, and still ...

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