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16h
comment To prove $r_2$ is a uniformly at random value in $Z_n$, where $r_2=r_1 . m$
By computation while they are at the server, you mean the RSA public or private operations? If so, well, what you're doing is known as RSA Blinding; it's a well known technique to allow someone to do the RSA operations without leaking what they're operating on. It's known to be safe.
16h
comment To prove $r_2$ is a uniformly at random value in $Z_n$, where $r_2=r_1 . m$
The answer to that would depend on what you mean by 'security'. You may find it helpful to write down exactly what you mean by that; what security properties are you expecting?
20h
comment To prove $r_2$ is a uniformly at random value in $Z_n$, where $r_2=r_1 . m$
@user153465: the answer to question 1 is "no"; if $m$ is an arbitrary value, consider the case that $m=0$; obviously in that case, $r_2$ is not a uniform random value. For the answer to question 1 to be "yes", we need to put restrictions on the distribution $m$ is chosen from (specifically, that probabilities that $m$ is a multiple of $p$, $q$ and $pq$ must be $1/p$, $1/q$ and $1/pq$)
2d
comment Does the transposition cipher have a network application?
@jake.toString: I suspect it'd be unlikely that anyone would write "we never use transposition ciphers in network applications" unless someone specifically asked.
2d
comment problem with “one time pad”
possible duplicate of How does one attack a two-time pad (i.e. one time pad with key reuse)?
2d
comment Does the transposition cipher have a network application?
I'm not sure how to answer this other than saying that I've never seen a transposition cipher within a network security protocol, and I have a good deal of experience in that area. It's not clear to me why anyone would choose to use a transposition cipher as opposed to one we are confident is semantically secure.
Oct
10
comment Is it possible to demonstrate that md5(x) != x for any x?
One minor nit: the computation about a fixed point assumes a random function. MD5 isn't a random function, and might not behave as one (we don't know any specific way that it doesn't, at least, any way that appears relevant to the question), however that may be due more to our ignorance rather than the internals of MD5. However, it is an appropriate plausibility argument to say "there might be one, and there might not"
Oct
9
comment Do Export Restrictions Still Apply To The Key Length of RC4?
@CodesInChaos: unless you're exporting to North Korea, Syria, or an handful of other countries. If you want to export to those countries, it's a lot more than "fill out some paperwork"
Oct
9
comment How do I generate a number for a lottery and later proves its existence
@KentMuntheCaspersen: I missed where you said that
Oct
8
comment How costly is to find millions of large prime numbers for RSA?
@fgrieu: you're right; I was worried that if the step you take between sieved elements was too small, the discovered primes would be too close. However, if you're looking for primes for this purpose, you'd select a step of circa $2^{900}$ (say) for 1024 bit primes; the resulting primes would have plenty of spread. (To user153465): Doing this would mean that if someone found two primes, factoring the rest would be a lot easier; that's why fgrieu asked if that would be a problem
Oct
8
comment How costly is to find millions of large prime numbers for RSA?
@fgrieu: one requirement is that the product of two of those primes be hard to factor. The problem with those known algorithms is that they generate primes which are numerically close to each other, which means that their product may be easy to factor (say, via Fermat factorization)
Oct
8
comment How costly is to find millions of large prime numbers for RSA?
Mind if I ask about the problem you're solving? If you're trying to generate a large number of distinct RSA keys, well, there're actually faster approaches...
Oct
7
comment Can adding nonces make challenge response authentications weaker?
@PlasmaHH: someone who tweaks a crypto design without knowing what they are doing is like someone who tweaks the shape of an airplane wing without knowing what they are doing -- what they do might be safe, but I wouldn't want to trust it...
Oct
5
comment Why don't we use bcrypt and scrypt together?
@ddddavidee: no, it wouldn't, as bcrypt (the outer transform) cannot be inverted.
Oct
2
comment RSA given q, p and e?
I'm saying that what you have labeled as the 'key' doesn't make sense as an RSA modulus.
Oct
2
comment RSA given q, p and e?
Also, it's obvious that, in the above example, $p \times q \neq key$
Oct
2
comment Generating child keys for a hill climb algorithm
That makes sense -- however, it strikes me that the combination of that scoring function and a single letter swap is likely to be prone to local maximum. Consider the case where it guesses a common quadrigram is "tion" (which gets a great score), but the actual decryption is "ould"(which is also quite common); it's hard to see how the system would move from "tion" to "ould" without having a bad intermediate score.
Oct
2
comment Generating child keys for a hill climb algorithm
What is the scoring function you are using? One problem you might be running into is that you fall into a local maximum, where any single step you can make the score wrose, and so your mechanism doesn't want to do anything. You need to consider your scoring mechanism when selecting a mutation method (and, in particular, so that your mutation method as a decent probability of finding it's way out of a local maximum)
Oct
1
comment A variant of Shamir secret sharing
@user153465: if my edits to the question were inaccurate, please go ahead and reedit the question.
Oct
1
comment A variant of Shamir secret sharing
I tried to make your question more understandable; if I garbled it, or misinterpreted it, feel free to roll back my changes.