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8

Yes. Such proofs are possible for El Gamal. It involves a zero knowledge proof of equality of a discrete log, together with the homomorphic property of El Gamal encryption. Recall that given $E(a)$ and $E(b)$, anyone can form $E(a/b)$ using the homomorphic property of El Gamal. Suppose $E(a/b)=(r,s)=(g^k,h^k a/b)$ (where $g$ is the generator and $h$ is ...


6

First, remark that the desired commutativity is incompatible with security under Chosen Plaintext Attack, which (under the name IND-CPA) is considered a requirement for modern encryption systems. Proof, expanded following tylo's comment, using the IND-CPA game as played for symmetric encryption (see the CPA indistinguishability experiment in section 3.5 of ...


4

In general, from a security analysis viewpoint, there isn't much difference between a peer-to-peer protocol and one using an untrusted server — and, insofar as there is any, the peer-to-peer case is strictly harder to secure. This is because, if we have a secure peer-to-peer protocol, we can always implement it over an untrusted server by simply ...


2

Unfortunately, if you want advice on a specific patent, I'm afraid you are most likely going to need to consult with a lawyer. The odds of getting reliable authoritative advice over the Internet are... not so great, in my experience. Generally speaking, this site focuses primarily on technical questions (which are answerable by many in the community here) ...


2

You are in a twist here: semantic security (equal to IND-CPA) can only be fulfilled by probabilistic encryption schemes. You need a deterministic encryption scheme for your drop-out tolerance. As it was pointed out previously, any homomorphic encryption allows you to proof in zero knowledge the equality of two ciphertexts: known: $c_0 = ...


1

With any convergent encryption algorithm E, it's easy for Alice to prove -- without revealing(*) a, b or the private key -- that a == b. In order for the data deduplication feature to work, convergent algorithms are specifically designed such that when Alice encrypts two messages a and b, such that x=E(a), y=E(b), then x == y whenever a == b. There's some ...



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