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3

That is the general idea of it yes. Some modes of operation (eg CTR) work in such a way that only known values are ever encrypted, forming a stream of pseudo-random data that is then combined with the plaintext by a keyless reversible operation (often xor) to form the ciphertext. Other modes (eg CBC) directly encrypt secret (ie plaintext) values, meaning ...


2

I'm not sure if your understanding is correct. However, to explain it a bit to be sure you’re on the right track… Some mode of operation only use an encryption function because it is used to generate something to XOR with the plaintext. There is no point decrypt the generated bytes. To decrypt the ciphertext, you just need the same stream of bytes.


4

Just enumerate all powers of your primitive element. There are only 15 of them. That will give you a lookup table that allows you to translate from field element to its expression as a power of the primitive element.


3

Remark: in One Time Pad the pad is used once, thus this is not OTP, since here $k$ is reused. Hint for part 1: Write the relations between $k$, the message blocks $m_i$, the ciphertext blocks $\small C_i$ with the convention $\small\text{IV}=\small C_0$. Then, find equations that allow computing the desired $m_3⊕m_4$ from known quantities. Hint for part 2, ...


3

In fact the equation is not used directly. If you work in the field of integers modulo $p$, both the $x$ and $y$ coordinates are integers in the $0$ to $p-1$ range, so there are $p^2$ possible points. The equation tells you which points are part of the curve (the $(x,y)$ such that $y^2 = x^3 + ax + b$) and which points are not. In that sense, the equation ...


2

Generally, no. There is no calculus in the design and implementation of the RSA. There are two things that could add some analysis in the whole scheme of things: Maybe involving the Prime Number Theorem, which explains why we won't run out of prime numbers to choose from. Some proofs of the theorem use complex analysis (including the simplest known proof, ...


4

This is purely a counting problem. We want the number $N(n,m)$ of possible permutations of $n$ things, with the constraint that only $m$ among these things can map to themselves ($0\le m\le n$). The values in the question are $n=11$, $m=5$. It holds that $N(n,n)=n!$ (that's the number of permutations of $n$ elements). It holds that ...


3

Yes, we need symmetric cryptosystems, for many reasons; to give three of these: We need a hash function to make most asymmetric cryptosystems secure (e.g. we simply do not have a secure signature system based on RSA without a hash), and current hash functions are (or are built from) symmetric cryptosystems. All asymmetric encryption cryptosystems are bound ...


1

Scroll to the end for tl;dr. Regarding your contrived example: Alice doesn't have Bob's keypair, but sends a message in such a way that only Bob can read it, eg. puts it in a dead-drop. So she takes out her pen and writes Hey Bob, could you sign and send me $X$ along with your public key? Here's mine: $P_A$ Signed, Alice Bob has no way of knowing if ...



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