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Oct
6
comment Does AES CTR mode store header information in encrypted files?
Not sure. Haven't ever really looked.
Oct
6
answered Does AES CTR mode store header information in encrypted files?
Oct
6
answered How to guarantee every plaintext is in a rainbow table?
Oct
6
comment How to guarantee every plaintext is in a rainbow table?
Depends. What is the length and how big is your hard drive?
Oct
4
answered How can I do minus on plaintexts in the Paillier cryptosystem?
Oct
3
comment How can I do minus on plaintexts in the Paillier cryptosystem?
If I get a chance I'll write-up an answer here, but check out the wiki article from thep on negative numbers. Full disclosure: I wrote the article and thep.
Oct
3
comment Need to choose an ecnryption method
I'd implement RC4. It is not too hard to understand and yet not as easy to break as something like a classical cipher.
Oct
2
comment RSA given q, p and e?
I changed "key" so it would make more sense
Oct
2
revised RSA given q, p and e?
added 14 characters in body
Oct
2
comment Generating child keys for a hill climb algorithm
@Jamiec, you could do a simulated annealing type approach to child generation. At first the children you generate are more chaotic, more different from the parent. Over time, however, you "cool" it down, i.e., the children are more similar to the parent (only small changes). There are entire textbooks written on these sorts of things :)
Oct
2
comment Generating child keys for a hill climb algorithm
Are you only generating one child each iteration?
Oct
1
comment A variant of Shamir secret sharing
I just gave you explicit write access (should have happened automatically though). Try again. There should be an input dialog at the bottom of the screen.
Oct
1
comment A variant of Shamir secret sharing
Let us continue this discussion in chat.
Oct
1
comment A variant of Shamir secret sharing
When you multiply two polynomials, the coefficients of the resulting polynomial are no longer independent. Consider $(1+3x) \cdot (5+7x) = (1)(5) + (1)(7)(3)(5)x + (3)(7)x^2$. So the $5$ appears in the first 2 coefficients, the $7$ appears in the last two, etc. So the coefficients have interdependencies.
Oct
1
comment Hardness proof?
@TravisMayberry very good comment. You could even add it as an answer.
Oct
1
comment Triple DES in Firefox - in practice
Do you have any references we can look at. I have no idea how/where firefox uses 3DES.
Oct
1
comment A variant of Shamir secret sharing
What is the probability that all randomly generated coefficients are small enough to not need a modulo? What is the probability that a random share is a multiple of the secret?
Oct
1
comment A variant of Shamir secret sharing
The fact that this only works because we are not doing a modulo operation tells me that in practice it really isn't an attack. That is because the only time you won't do a modulo in practice is when all the coefficients are very small and the secret is very small (compared to the modulus).
Oct
1
comment A variant of Shamir secret sharing
@user153465 I think I see what you are saying. Since $2x+3$ becomes $18x+27$, all shares will be a multiple of $9$, so the adversary can figure that the secret is $9$.
Oct
1
comment A variant of Shamir secret sharing
@user153465 your example here assumes that the adversary knows that the constant term of the polynomial is $0$. Like I said in my answer, if the constant term of the polynomial is $0$ (or in this case, knows to be $0$ by the adversary) there is an issue. If the constant term is not necessarily $0$, the adversary does not know which secret the belongs to.