Tag Info

Hot answers tagged

2

If you were using $e=3$, then there is a well known attack by Bleichenbacher that enables the trivial generation of a signature that passes verification. This attack was never published, but is described here. Note that this attack appeared in a real vulnerability in Kindle (and some versions of Android). In any case, the attack does not work for $e=65536$. ...


1

One property that this unpadded system is that it is homomorphic; if $A^d = X$ and $B^d = Y$, then we know that $(AB)^d = XY$, and it doesn't matter if we don't know what $d$ is. More generally, if we have a collection of $H_1, H_2, H_3, ... H_n$, and a collection of signatures $S_1, S_2, S_3, ..., S_n$, then for any set of integers $e_1, e_2, e_3, ..., ...


1

To obtain the tag, OCB xors the plaintext blocks and encrypts them. Thus a sufficient condition for a forgery is finding another plaintext with the same xor as an existing plaintext. Consider a known plaintext attack where the attacker obtained (plaintext, ciphertext) pairs for two messages encrypted using the same key and nonce. The attacker picks between ...



Only top voted, non community-wiki answers of a minimum length are eligible