meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 6 months |
| seen | 18 mins ago | |
| stats | profile views | 9 |
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Dec 9 |
accepted | Why is OCB-AES mode not becoming a standard for authenticated encryption? |
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Dec 9 |
asked | Why is OCB-AES mode not becoming a standard for authenticated encryption? |
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Dec 9 |
revised |
Public key cryptography - public key encrypts and cannot decrypt? deleted 12 characters in body |
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Dec 7 |
answered | Public key cryptography - public key encrypts and cannot decrypt? |
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Dec 5 |
comment |
How to prove membership of a list without disclosing the list members? Yes you are correct. Another possibility would be to use modified bloom filters to support the security you want. This comes with a cost of false positive replies by the bloom filter for membership queries. fkerschbaum.org/dbsec11.pdf |
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Dec 3 |
answered | How to prove membership of a list without disclosing the list members? |
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Nov 27 |
answered | Is the AES encryption scheme CPA secure? |
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Nov 26 |
awarded | Yearling |
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Nov 26 |
comment |
Looking for examples for “proof by reduction” I may run the danger to be a bit out of topic but this is an interesting approach on rigorous proofs by reductions usually employed by cryptographers. It criticizes the too much effort on proofs by reductions which sometimes doesn't take into account all the attacker's window and consequently "tends to kick dust to eyes" |
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Nov 23 |
comment |
Mapping between subgroups and the integers Can you also explain why since $q$ is a big prime, it is odd, therefore $p = 3 \mod 4$ and that implies that $x$ is a quadratic residue? |
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Nov 19 |
accepted | Does a break in a collision resistance property of a hash function by definition implies an attack at the first pre-image attack? |
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Nov 19 |
comment |
Does a break in a collision resistance property of a hash function by definition implies an attack at the first pre-image attack? @CodesInChaos break collision resistance means to find two messages m1,m2 which are not equal s.t: h(m1)=h(m2). Break first preimage resistance means break the one-wayness, recover m1 from h(m1) |
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Nov 19 |
asked | Does a break in a collision resistance property of a hash function by definition implies an attack at the first pre-image attack? |
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Nov 19 |
comment |
Using CBC with a fixed IV and a random first plaintext block That's why it is a comment and not an answer. Since you are using a source of randomness i can't see why not use it to produce the initial IV. In terms of efficiency this will increase linear the size of the ciphertext compared with the standard way |
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Nov 19 |
comment |
Using CBC with a fixed IV and a random first plaintext block Why to do that?Is like CTR mode with 1 counting step, but in CTR the IV+the counter are encrypted with AES and this is XORed with the plaintext. If the IV is not random it must be unique, that's why we use a counter |
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Nov 19 |
comment |
Is it safer to encrypt twice with RSA? @D.W. Is there a quantitative assessment in how large this partial plaintext recovering should be in order to be conjectured as CPA? |
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Nov 18 |
answered | Relative merits of AES ECB and CBC modes for securing data at rest |
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Nov 17 |
accepted | Why does OAEP have 2 rounds with 2 random oracles? |
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Nov 16 |
comment |
Is it safer to encrypt twice with RSA? Does the fact that a tiny amount of information is discovered for the plaintext from the ciphertext violates the IND-CPA security? I am wondering as IND-CPA refers to fully recover the plaintext |
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Nov 15 |
asked | Why does OAEP have 2 rounds with 2 random oracles? |
