# Tag Info

2

There is a way to generate forgeries for (EC)DSA when the hash function is not one-way: Let $n$ be the order of the group, $P$ a generator, and $Q = aP$ for some secret $a$; Pick arbitrary $\alpha$ and $\beta$ $\in \{0, \dotsc, n\}$; $r = x \bmod n$, where $(x, y) = \alpha P + \beta Q$; $s = r \beta^{-1} \bmod n$; $h = s \alpha \bmod n$; Invert $H(h)$ to ...

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 $e=65536$. ...

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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, ..., ...

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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 ...

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