19,161 reputation
12676
bio website
location Paris, France
age
visits member for 2 years, 11 months
seen 1 hour ago

I'm an engineer with experience in applied cryptography, in particular in Smart Card systems.


Jul
9
comment Is calculating a hash code for a large file in parallel less secure than doing it sequentially?
As rightly pointed by CodesInChaos, Merkle-Damgård hashes are not second-preimage-resistant to $2^n$ effort (see John Kelsey and Bruce Schneier's Second Preimages on $n$-bit Hash Functions for Much Less than $2^n$ Work), so it would seem that the proposed construction does NOT weaken SHA-256 after all.
Jul
9
revised Is calculating a hash code for a large file in parallel less secure than doing it sequentially?
Generalize to any number of separately hashed blocks; use TeX
Jul
9
comment Client authentication on limited hardware
RSA is very fast for the purpose of authenticating something (rather than: authenticate w.r.t. something). For 2048-bit public modulus, public exponent $e=3$, and a CPU with 32x32 bit multiplication, in the order of 17000 multiply-and-accumulate are enough, with textbook algorithms and straightforward loops. This can be about halved with Rabin, while still having standard-conformance, e.g to ISO/IEC 9796-2, and then there's DjB's work.
Jul
9
revised Is calculating a hash code for a large file in parallel less secure than doing it sequentially?
Fix typos
Jul
9
comment RSA-based authentication and key-agreement protocol
Ah, right, I did not read up to the KGC-free certificate-based variant (page 24), sorry about that; I do see it now, thanks for your patience! $\;$ Still, the public-key certificates need more parameters than in RSA, and it seems non-trivial to implement the protocol on top of RSA primitives: I think we need modular squaring and/or inverse, and either is a nightmare to implement in the subset of Java available in a Java Card Classic Smart Card.
Jul
8
revised How is this affine function a pair wise independent permutation?
Polish
Jul
8
revised How is this affine function a pair wise independent permutation?
Add missing step
Jul
8
revised How is this affine function a pair wise independent permutation?
Polish and simplify
Jul
8
revised How is this affine function a pair wise independent permutation?
Shorten
Jul
8
revised How is this affine function a pair wise independent permutation?
I'm more confortable with a finite field after all.
Jul
8
revised How is this affine function a pair wise independent permutation?
Polish
Jul
8
answered How is this affine function a pair wise independent permutation?
Jul
8
revised How is this affine function a pair wise independent permutation?
Added definition requested
Jul
8
revised How is this affine function a pair wise independent permutation?
Exact quote (based on the preprint)
Jul
8
comment How is this affine function a pair wise independent permutation?
Are you reading the actual paper, or what appears to be the Nov 1997 preprint, which does contain (nearly) the sentence now in the second paragraph of the question?
Jul
8
comment An unpredictable PRG is secure (Theorem Yao'82)
I think a proof sketch using contraposition might go: Assume $G$ is not a secure PRG. There is thus an algorithm that breaks $G$. Define $G_i$ which substitutes true random to bits at position $j\ge i$. The same algorithm breaks $G_n$ (since that's $G$) but not $G_0$ (since that's random). So there must be a $i$ such that it breaks $G_{i+1}$ but not $G_i$. Now you can build a predictor for position $i$.
Jul
7
answered Adi Shamir's secret database of all primes
Jul
7
comment Generate Finite Field power of g
Explaining the $(0010)$ notation could come earlier in the response. Kudos for the rest.
Jul
7
comment Adi Shamir's secret database of all primes
That database is to cryptography venues what the Dahu is to French summer camps. Also see the answers to this. $\;$ The three other 'future work' items in the presentation are in the same vein.
Jul
4
comment Getting 88bytes cipher output from 48bytes input in AES
The paper considers that a block cipher's output is wider than its input, likely because the authors confused with some mode of operation of a block cipher complete with IV and padding; that's both for DES and AES.