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In Fully Homomorphic Encryption scheme represented here Fully Homomorphic Encryption over the Integers

In the Evaluate process (see section “3.1 The Construction” of the paper): $$Evaluate(pk, C, c1, \dots , ct):$$

Now…

  1. If I want to do some operations on an encrypted data, should I convert that operation to a binary circuit, then do the operation, and re-encrypt the final result?

  2. If I have this simple binary circuit below, how could I evaluate the process?

    Simple binary circuit to compute

  3. What is the data that will be passed to the Input A and input B?

  4. And how can I determine the number of addition and multiplication operations that will be used?

I did not understand how to do it from the Paper.

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  • $\begingroup$ Hi, this link is dead $\endgroup$ Commented Dec 13, 2021 at 20:11

1 Answer 1

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It's actually straight-forward; we'll assume that all the inputs are either encrypted versions of 0, or encrypted version of 1; then:

  • We can replace an AND gate with just an FHE multiplication of the two inputs:

    $$AND(x,y) = x*y$$

    Where $*$ is our Homeomorpic multiplcation operation. This obviously evaluates to an encrypted 1 if both of the inputs are encryped 1's; and an encrypted 0 if either of the inputs are encryped 0's

  • To replace an exclusive or gate, we encrypt the constants 1 and -1, and then compute:

    $$XOR(x,y) = x*(1 + (-1 * y)) + y*(1 + (-1 * x))$$

    Where + is our homeomorphic addition operation, and 1 and -1 stand for our encrypted constants.

  • To replace an or gate, we take our encrypted 1 and -1 constants, and compute:

    $$OR(x,y) = 1 + -1 * (-1 + x) * (-1 + y)$$

  • To replace a NOT gate (that doesn't appear in your circuit, but does come up in others), we compute:

    $$NOT(x) = 1 + -1 * x$$

It is easy to see that, in all cases, if the inputs are restricted to encrypted 0 and encrypted 1, the result is either an encrypted 0 or an encrypted 1 (and which will be the logical result of the operation). Obviously, things can be simplified somewhat if our FHE operation includes a subtraction operation.

As for the number of operations used, we just translate each gate and count them up. On the other hand, this construction is really a proof that just addition and multiplication suffice to be complete operations (that is, given just those two operations, we can compute anything); when you look at operation count, it turns out to be quite inefficient.

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  • $\begingroup$ The fact that addition and multiplication can't turn 0s into a 1 shows that they are not complete operations. (However, addition and multiplication and $-1$ suffice to be complete operations.) $\endgroup$
    – user991
    Commented Sep 23, 2014 at 17:57
  • $\begingroup$ @RickyDemer: with FHE, it is assumed that we can take any value, and generate an encrypted form of it with the public key. I assumed that as part of my demonstration; if those aren't available, the problem could include a fixed encryption of -1. $\endgroup$
    – poncho
    Commented Sep 23, 2014 at 18:08
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    $\begingroup$ $XOR(x,y)$ and $OR(x,y)$ can be simplified to $\:x+y+((-2)*x*y)$ $\hspace{2.15 in}$ and $\:x+y+((-1)*x*y)\:$ respectively. $\;\;\;$ $\endgroup$
    – user991
    Commented Sep 23, 2014 at 18:28
  • $\begingroup$ @RickyDemer: cool! $\endgroup$
    – poncho
    Commented Sep 23, 2014 at 18:32

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