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3h
comment What is the right notion of security to use for file encryption?
Yes. ${}{}{}\;$
3h
revised What is the right notion of security to use for file encryption?
provided a more complete reference
4h
answered What is the right notion of security to use for file encryption?
12h
comment Fast PKI for embedded device
I saw [the restriction (involving $\overline{\pi}$) on the moduli] and the heuristic primality test (rather than Bernstein's zero-error test or Miller-Rabin). $\:$ I haven't yet looked further into the paper you linked to. $\;\;\;\;$
13h
comment Fast PKI for embedded device
The efficiency of these can be greatly improved if the signer and verifier(s) stay in sync (i.e., verify messages in the same order as they were signed). $\;$
13h
comment Fast PKI for embedded device
@MaartenBodewes : $\:$ Sections 4 and 7 of the paper linked to at the top of my answer each describe a reason why the scheme linked to in this answer is fast. $\;\;\;\;$
14h
comment Fast PKI for embedded device
Note that at least some of that paper is trading off correctness and security for things other than verification speed. $\:$ In the other direction, depending on how long those devices might remain in use, it may be worth using a 1280-bit modulus instead of a 1536-bit modulus. $\:$ (With that sort of idea in mind, it might even be worth having a dedicated public key for signing each device's files, to reduce the effect that factoring one of their moduli would have.) $\;\;\;\;$
15h
comment Fast PKI for embedded device
The files being "specific for this device" does not rule out interaction. $\:$ (However, if the pendrive is its only means of communication then interaction would require moving the pendrive from the device to a standard computer and back to the device.) $\;\;\;\;$
16h
answered Fast PKI for embedded device
1d
comment Asymmetric cipher with multiple public keys
(I didn't notice that you had also answered there, and) I don't know whether a ping would work for a user who had edited the question but not done anything else on a page. $\;$
1d
comment Operation modes of block ciphers how are used?
Do you know what $\oplus$ means? $\;$
1d
comment Non-interactive zero-knowledge proof for discrete logarithm without random oracle
www0.cs.ucl.ac.uk/staff/J.Groth/NIZKJournal.pdf $\;$
1d
comment Does the encryption algorithm DES perform random permutations? If so how is information not lost?
seperately $\mapsto$ separately $\:$ ? $\;\;\;\;$
1d
revised Why is public key cryptography not widely used in governments?
inserted question mark
2d
comment Would this be a plausible authenticated encryption scheme using nested encryption?
@Anon2000 : $\;\;\;$ Your minimal variant is close to CBC mode (you switched the order of E and xor), and subjects the block cipher to key-dependent data. $\:$ I'll be going to bed soon, so I probably won't get anything on it tonight. $\;\;\;\;\;\;\;\;$
2d
comment Would this be a plausible authenticated encryption scheme using nested encryption?
@Anon2000 : $\;\;\;$ Actually, my original proof happened to apply to your second variant but not this page's original proposal. $\:$ I fixed it so it would apply to both of those. $\;\;\;\;\;\;\;\;$
2d
revised Would this be a plausible authenticated encryption scheme using nested encryption?
fixed proof
2d
comment Would this be a plausible authenticated encryption scheme using nested encryption?
@Anon2000 : $\:$ That's what I'm trying to figure out right now. $\;\;\;\;$
2d
comment Would this be a plausible authenticated encryption scheme using nested encryption?
The flaw I was utilizing is that the random number generator is exposed to related seeds. $\:$ The only thing I can think of which would fix that is using a PRF instead. $\;\;\;\;$
2d
comment Is HMAC-MD5 still secure for commitment or other common uses?
If $\: 1 < L \:$ and there are at most $2^{\hspace{.02 in}L}$ permitted keys (in particular, if the key [must have length less than L] or [must have length exactly L]), then the probability of a random setup string being non-binding is less then $1\hspace{-0.02 in}\big/\hspace{-0.05 in}\left(\hspace{-0.02 in}2^{\hspace{.02 in}385-(2\cdot L)}\hspace{-0.05 in}\right)$. $\;\;\;$ That guarantee becomes meaningless slightly above $\: L = 192 \;$. $\;\;\;\;\;\;\;\;$