# Tag Info

## New answers tagged known-plaintext-attack

1

Let $k=\left(\begin{array}{cc}k_0 & k_1\\k_2 & k_3\end{array}\right)$ be the key. And I'll assume the transformation $a=0$, $b=1$, and so on. So you know $k\left(\begin{array}{c}0\\1\end{array}\right)=\left(\begin{array}{c}c_0\\c_1\end{array}\right)$ where you know $c_0$ and $c_1$. Thus $c_0=0k_0+1k_1$ and $c_1=0k_2+1k_3$. So that is two equations, ...

2

I'm assuming Bob uses the standard format, OpenPGP aka RFC4880, to store his private key in a private keyfile. If the adversary Eve somehow obtains Bob's private keyfile, and the passphrase on that keyfile is so short and weak that the adversary breaks it with a dictionary attack or brute force, then -- Things are very bad for Bob. Bob's best choice of ...

4

No, this does not weaken ed25519 in any way. Known plaintext will not have any effect on a signature algorithm, if it did it would make that algorithm completely useless.

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