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Lets say we are given BFV encryption of x, let this encryption is represented as E(x). In FHE, the client can decrypt and get the value of x but what if we dont want the client to learn x. This paper suggests that:

  • the servers samples random r and perform E(x)+r and send that to client
  • The client decrypts and get x+r
  • The paper suggests that this is like a secret sharing of x. More concretely Server share is r and client share is x+r mod p. (p is plaintext mod)

My question is which ring r is sampled from ? is it from R_p or R_q

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The vector $\mathbf{r}$ serves as a one-time pad. So, it is sufficient to use a uniform $\mathbf{r}$ sampled from a vector space where the plaintext $\mathbf{x}$ is in. What really matters is how to add up the HE ciphertext $[\mathbf{x}]$ and the plaintext $\mathbf{r}$ to get a new HE ciphertext $[\mathbf{x} + \mathbf{r}]$. This operation is easy for any linear HE.

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  • $\begingroup$ I understand that this vector is going to server as OTP. I am worried about the following issue, if encryption is a, as+e+m \mod q then after addition it will become a, as+e+m+r mod q the client who holds secret key could remove as so she gets m+r+e mod q I am not sure how this is OTP in mod q ? or am I missing something (I know m+r mod p is secure as its uniform ) $\endgroup$
    – LWE-13
    Jul 20, 2022 at 16:04
  • $\begingroup$ I think that $\vec{m} + \vec{r} + \vec{e} \in R_q$ is only an intermediate result in the decryption of HE ciphertext $[\vec{m} + \vec{r}]$. For non-approximate HE schemes, the error should be then removed using the plaintext modulus. It may be helpful to view $(\vec{a}, \vec{a} \vec{s} + (\vec{m} + \vec{r}) + \vec{e})$ as the HE ciphertext for the plaintext $\vec{m} + \vec{r} \in R_p$ (which is the information can be learned by the client at most). Now, the OTP $\vec{r}$ works. $\endgroup$
    – X. G.
    Jul 21, 2022 at 0:54
  • $\begingroup$ oh I see yea that makes it indeed easier $\endgroup$
    – LWE-13
    Jul 21, 2022 at 1:25

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