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I would like to know if this process is feasible under homomorphic encryption, ideally under paillier or any other additive scheme

  1. Apply a mask X to obfuscate a message A ie. Am = A (op) X where (op) can be +, x, ...
  2. Homomorphically encrypt Am => E(Am)
  3. Remove X from E(Am) in order to obtain E(A) without any decryption (without knowing the private key)

A can be a matrix, vector, ...etc.

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It sounds like you're looking for a way to do blinded encryption (that is, encryption in such a way that the encryptor does not know what he is encrypting).

It is certainly possible with Paillier. Here is one way:

  • The (op) will be multiplication modulo $n$ the Paillier modulus (which is in the public key).

  • To mask, we pick a random $r$ relatively prime to $n$, and compute $Am = A \cdot r \bmod n$. We can see that, if we selected $r$ uniformly, then $Am$ is distributed independently of $A$ (except for values not relatively prime to $n$)

  • We ask the encryptor to take $Am$ to compute $E(Am)$

  • To unmask, we compute $rinv = r^{-1} \bmod n$, and then homomorphically compute $E(rinv \cdot Am)$; this Paillier, this is done by computing $E(Am)^{rinv} \bmod n = E(A)$.

One drawback to this is that it doesn't actually mask the value $A=0$; is that a concern?

If it is, you can still do it by asking the encryptor to perform two masked encrypted operations.

First, you generate two random values $r, s$, making sure that neither is $0$

Then, you have the encryptor encrypt the values $A-s$ and $r\cdot s$, producing $E(A-s)$ and $E(r \cdot s)$. Then, you use the above procedure to recover $E(s)$ from $E(r \cdot s)$. Then, you homomorphically add $E(s)$ to $E(A-s)$, producing $E(A)$

You could just generate $E(s)$ yourself with $s$ and the pubic key. However, if you could perform encryption yourself, you could just generate $E(A)$ from $A$, and it would appear you wanted to avoid that...

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  • $\begingroup$ Thanks, and r can be of any dimension ? $\endgroup$ – witdev Apr 8 at 10:51
  • $\begingroup$ @witdev: For Paillier, it would be a value between 1 and $n-1$, If the value $A$ is a vector, then you would have a vector of $r$'s, and blind encrypt each element of the vector separately. $\endgroup$ – poncho Apr 8 at 12:48
  • $\begingroup$ Thanks again, 1. Could it be a vector with a different $r$ in each of its elements or you mean a vector with the same $r$ repeated ? 2. I found in shorturl.at/jZ026 that the equivalence of $E(a [mult] b)$ is $E(a)^{E(b)}$, while you used $E(Am)^{rinv} mod n$ .. what is the difference ? $\endgroup$ – witdev Apr 8 at 16:34
  • $\begingroup$ @witdev: no, using the same $r$ would leak information; use a different random $r$ each time. $\endgroup$ – poncho Apr 8 at 18:33
  • $\begingroup$ I have no idea what that url is - it doesn''t resolve for me $\endgroup$ – poncho Apr 9 at 3:06

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