Tag Info

New answers tagged

0

Not always, it depends on the particular encryption scheme. Strictly speaking, the proofs only say that breaking indistinguishability is equivalent to breaking the hardness assumption they are based on. There are some cryptosystems, like Rabin's, where the security of the key is equivalent to the security of the ciphertexts, i.e. factoring <=> key ...


3

If an attacker can choose the points $P_i$, than this system is not semantically secure. For example, they may choose $P_2=2P_1$, and the corresponding encryption $Q_2$ would be equal to $2Q_1$. If the points are chosen at random, this system is semantically secure if decisional Diffie-Hellman assumption holds for the curve. This assumption is presumed to ...



Top 50 recent answers are included