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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.
Oct
1
answered A variant of Shamir secret sharing
Oct
1
comment A variant of Shamir secret sharing
Why do you want to use this? Is there any specific reason? In other words, what is your application scenario.
Oct
1
answered Hardness proof?
Oct
1
revised what are DP and DQ in Encryption by RSA in c#
edited body
Oct
1
answered what are DP and DQ in Encryption by RSA in c#
Oct
1
comment Why are we not using multiple ciphers per message?
@Einar, not sure. Don't have the time to read and digest all that at the moment. Could be a useful question to ask on this site (or possibly another SE site).
Oct
1
comment Is Attribute based encryption sub-class of Identity based encryption
To add to what @DrLecter says, given an ABE scheme, you can construct an IBE scheme. You can't go the other way. The way to construct IBE from ABE is to give every participant in the system a single attribute. The attribute value is their identity.
Sep
30
comment Why are we not using multiple ciphers per message?
You can't compute in parallel things that depend on the output of each other.
Sep
30
revised Homomorphic encryption based on XOR
added 118 characters in body
Sep
30
comment Homomorphic encryption based on XOR
@PaŭloEbermann re second comment, for $+$, I was talking about the typical way we do addition in finite fields of characteristic 2.
Sep
30
comment Homomorphic encryption based on XOR
@PaŭloEbermann re first comment, you mean $=(a\oplus b)\neq(a\oplus b)\oplus d$ right? I was more going along the question of the OP "can we derive b+a from any combination of c1 and c2". In this case, we can derive $a\oplus b$ from some combination of c1 and c2.