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seen Dec 16 at 17:33

Oct
1
comment finding collision for truncated SHA1 hash output
@CodesInChaos Thanks, fixed. I had to hand-repair several lines when I pasted into this post but I missed those. Not sure why crypto.SE is behaving this way suddenly (normally it preserves indentation for me).
Sep
22
comment Best way to reduce chance of hash collisions: Multiple hashes, or larger hash?
Thomas: you switched from SHA-256 in your answer to MD5 at the end. As you know, these are two different animals (neither animal, being a gorilla) since MD5 outputs only 128 bits and it has known collisions. Finding a collision by brute force (a suddenly very apt term in the present context) for a 128 bit hash is going to take "only" $2^{64}$ expected trials, making it roughly on par with gorilla attacks.
Sep
10
comment What NIST protocol was allegedly backdoored by NSA in 2006?
It wouldn't be the first time the media got something wrong; even the NYT makes technical mistakes occasionally. More likely, there is a wrong perception among researchers about the history of SP 800-90.
Aug
31
comment Recent attacks on RSA
Seems that Schneier (and other cryptographers) agree with me: schneier.com/blog/archives/2013/08/the_cryptopocal.html
Aug
30
comment Do ciphertexts leak information about their algorithmic creators?
So if I say "I have 1024 bits of ciphertext" you can definitively ascertain whether that is an output from Rijndael-1024, or it's three blocks of AES-CBC with a random IV? I can't.
Aug
25
comment Do ciphertexts leak information about their algorithmic creators?
@Ninveh You're confusing block cipher encipherment with full-blown encryption again. The concept of an "IV" does not apply to AES, Twofish, Serpent, etc. That's an idea used with modes in order to (typically) implement randomization. So you can't say "IV structure in some ciphers". In any case, the normal approach to security in crypto is to NOT try to hide which algorithms have been used at all (unless you are the NSA).
Aug
3
comment Recent attacks on RSA
Elliptic curve cryptanalysis has had 1/100th the attention of RSA and yet these guys want to run to it? I'm guessing if we start to convert to ECC, the recommended key lengths will quickly increase in the next few years.
Aug
3
comment Recent attacks on RSA
Thanks for posting this. I am not an algebraist, but breakthroughs in discrete log for small-characteristic finite fields doesn't seem particularly related to factoring in zero-characteristic rings like $\mathbb{Z}$. Of course it may be relevant to RSA (which is on the ring $\mathbb{Z}_N$) and despite many wrong proclamations in the cited articles, breaking RSA does not require factoring. In any case, in my opinion the presenters have overstepped in declaring a state of emergency with respect to RSA. Remember 10 years ago all the factorization papers (Bernstein and Shamir mostly)?
Aug
2
comment Recent attacks on RSA
That post refers to the Blackhat speakers as "experts in cryptography." I have been a professional cryptographer for 20 years... I have never heard of these guys...
Jan
24
comment Why is MixColumns omitted from the last round of AES?
@PaĆ­loEbermann Yes, but you need a separate set of precomputed table just for the final round. This is onerous in environments where memory is tight.
Jul
29
comment Can I jettison MAC if I already have SHA1(M)?
@D.W. I'm not sure why I would care about perturbing the ciphertext if the underlying message is immutable; padding attacks rely on being able to change the message (in order to decrypt it based on padding-validity rules). Anyway, thanks for your input.
Jul
29
comment Can I jettison MAC if I already have SHA1(M)?
@D.W. My assertion that adversarial messages will (almost certainly) be rejected is based on the assumption that any perturbation to a string whose SHA1-digest is fixed-and-known is effectively immutable. Gave you give an example where this is false (choose any padding scheme you like).
Jul
24
comment Can I jettison MAC if I already have SHA1(M)?
It's standard in CCA security to give the adversary a decryption oracle (that's what every definition of CCA security does, that I've seen). You of course don't give the adversary credit for decrypting a message he's encrypted with the corresponding encryption oracle. (en.wikipedia.org/wiki/…)
Jul
22
comment Can I jettison MAC if I already have SHA1(M)?
Yeah, I know the padding attacks well.
Jul
21
comment Can I jettison MAC if I already have SHA1(M)?
Thanks for the reply. Your "secure" above works even without encryption(!). My intent was to preserve privacy even in some "reasonable" attack model, meaning the adversary cannot decrypt C=E_K(M) even with access to a decryption oracle and subject to the usual complexity-theoretic limits.
Jul
21
comment Can I jettison MAC if I already have SHA1(M)?
I thought about padding oracles, by the way, but any adversarial message is overwhelmingly likely to just be rejected (the same effect a MAC would cause). Another concern is extension attacks, but I think I've ruled those out as well.
Jul
21
comment Can I jettison MAC if I already have SHA1(M)?
Thanks. Actually in my application there is a digest for every 4MiB, so the resend-problem isn't a problem. I wrote my question in a simplified form to focus on the essential issue.
Jun
7
comment Is there a simple hash function that one can compute without a computer?
The 6k$\pm$1 rule doesn't help much: every integer is between -2 and 3 mod 6. Half of these are even and therefore obviously composite; one is an odd divisible by 3, which is quickly found out. The last 2 satisfy the 6k$\pm$1 rule, so it tells you nothing further.
May
15
comment Exposing RSA private-key data… bad?
The $d$ is the RSA private exponent. Have a look at any exposition on RSA encryption for more details.
Feb
22
comment Exposing RSA private-key data… bad?
Thanks. I had never seen this attack on $n$ when we know $e$ and $d$.