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Jul
5
comment How to choose McEliece's parameters?
The paper Daniel J. Bernstein, Tung Chou, and Peter Schwabe: McBits: fast constant-time code-based cryptography lists a bunch of parameterizations as well. Since the paper is from 2013, the security estimates should take the attacks you mentioned into account.
Jul
5
comment space complexity of quantum collision search
Related paper Daniel J. Bernstein. "Cost analysis of hash collisions: Will quantum computers make SHARCS obsolete?" Workshop Record of SHARCS'09: Special-purpose Hardware for Attacking Cryptographic Systems.
Jul
5
comment Increasing the diffusion of the AES-CBC encryption algorithm in pycrypto for python
SIV mode might be interesting for you.
Jul
5
comment What informal indicators exist for estimating the computational infeasibility of cryptographic problems?
Interesting essay: Boaz Barak - Structure vs. Combinatorics in Computational Complexity
Jul
5
comment What informal indicators exist for estimating the computational infeasibility of cryptographic problems?
Related question How can we reason about the cryptographic capabilities of code-breaking agencies like the NSA or GCHQ?
Jul
5
comment What are the advantages of using OFB (Output Feedback Mode)?
Could you add a reference for CTR having better security bounds?
Jul
4
comment What are the advantages of using OFB (Output Feedback Mode)?
CBC and CFB decryption are parallelizable, so the speed advantage of CTR should only apply to encryption, not decryption.
Jul
4
comment Logjam-style attack on Factoring?
There is an technique called Batch NFS but I don't know about the details.
Jul
4
comment Parallel authentication of encrypted data. What AE type to choose?
If you have large files, why don't you split the file into chunks and encrypt these independently? That has the added advantage that you don't need to decrypt terabytes of data, only to throw them away at the end if the MAC doesn't validate.
Jul
3
comment Does XTS provide random write access?
@elisae__ The attacker learns which blocks (at the same position) changed between versions. If blocks consist a single bit (xor based streamciphers, including CTR mode), they learn which bits are identical, which is often enough to recover the message. If you use AES-XTS they learn which 16 byte blocks are identical. If you use some kind of wide block cipher construction to get 4096 byte blocks, they only learn if such a large block changed. But since you need to rewrite a whole block at once, larger blocks limit random access operations.
Jul
3
comment Can passwords be stored securely so that a similarity comparison can be made?
But it doesn't prevent somebody from alternating between two dissimilar passwords.
Jul
3
comment Can passwords be stored securely so that a similarity comparison can be made?
Of course this only works for the immediately prior password and not for comparing with older passwords.
Jul
3
comment Does XTS provide random write access?
How are you protecting integrity? A merkle tree of hashes/MACs?
Jul
2
comment How often does RSA-OAEP have a leading zero?
@TBridges42 Depends on the most significant byte of the public key. The probability should be approximately p(MSB(ciphertext)=0) = 1/MSB(public key).
Jul
1
comment How often does RSA-OAEP have a leading zero?
If this is about the most significant byte of the ciphertext being zero, it's possible. Padding affects the plaintext.
Jul
1
comment Is this client-side password hash scheme secure?
For code spanning several lines, I recommend indenting it with 4 spaces, instead of using a <code> html tag.
Jul
1
comment Building a combined encryption scheme from two encryption schemes that's secure if at least on of them is secure
@Gordon While you can't delete this question, you can contact stackexchange asking to disassociate it from your account.
Jul
1
comment How much computing resource is required to brute-force RSA?
Related question: How many RSA keys before a collision?
Jun
30
comment Double the key in block cipher - which approach is better?
@nopenope If you look at the decryption operation $D^\prime_2$ instead of the decryption operation, you get the same formula as for for $E^\prime_1$, except that $c$ and $m$ are swapped. Since an attacker knows both $m$ and $c$, they don't care if they're attacking $E^\prime_1$ or $D^\prime_2$.
Jun
30
comment Ideas for non duplicate cryptographically secure numbers
If 64 bit values are fine, you could encrypt a counteer using a 64 bit blockcipher, like blowfish or 3DES.