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1h
reviewed Leave Open “AND” concatenation of two secure PRGs
1d
reviewed Leave Open Bouncy Castle and Salsa 20
1d
reviewed Leave Open Simplicity and precision in cryptography papers
1d
reviewed Close How to multiply two field elements generated by bilinear paring map?
1d
reviewed No Action Needed ElGamal scheme attack when one message is known and ciphertexts are intercepted
1d
reviewed No Action Needed Source for PKCS#11 Header Files
1d
reviewed No Action Needed Double the key in block cipher - which approach is better?
1d
reviewed No Action Needed Extracting a WinRAR password
1d
comment Ideas for non duplicate cryptographically secure numbers
How large do the numbers need to be (in terms of bits)? They need to be at least 36 bits (so that there are at least 50 billion distinct possibilities)
1d
comment “AND” concatenation of two secure PRGs
It looks like you answered your own question...
1d
comment What happens if no final subtraction is done in Montgomery multiplication?
Stupid question: why are you using Montgomery Multiplication for $2^{255}-19$ in the first place? The reason we do Montgomery Multiplication is to make the modulo operation easier; however with $N=2^{255}-19$, it's already awfully easy, as $a \cdot 2^{255} + b \equiv a \cdot 19 + b \pmod{2^{255}-19}$
1d
comment “AND” concatenation of two secure PRGs
Hint: what's the definition of Advantage?
1d
revised Adding tweak to a block cipher
added 84 characters in body
1d
answered Adding tweak to a block cipher
Jun
28
comment Security of MSS
@dylan7: yes, Grover's algorithm is the best known against today's hash function (say, SHA-256). What it does is, given a function with $m$ possible inputs, find the input where the function returns "yes" in $O(\sqrt{m})$ time (and when you use it to find a hash preimage, you make the function one that computes the hash, and then compare the result to the desired value). As for PRNGs, there is no known attack against most PRNGs better than what's implied by Grover's algorithm.
Jun
28
answered Security of MSS
Jun
26
comment Simple protocol for 1-out-of-2 oblivious transfer
Important note on the RSA protocol; Alice needs to generate a fresh RSA key for every transfer; if not, that is, if Alice uses the same RSA key for two exchanges, then Bob can cheat and for his second transfer, learn the different between a secret from the first transfer (that he didn't pick) and a secret from the second. That is, from the first exchange with secrets $(m_0, m_1)$, he learns $m_0$, from the second exchange with secrets $(m'_0, m'_1)$, he can learn $m'_1 - m_1$
Jun
26
reviewed Approve cryptdb tag wiki excerpt
Jun
26
reviewed Approve cryptdb tag wiki
Jun
26
reviewed No Action Needed How can a lattice attack be applied to ECDSA signatures?