Is offline secure two party computation possible?

Question

Is there a solution for the following problem?

• We have a function f(x,y) where the inputs are secret an cannot be known by anyone besides the party that submitted them.
• The output of f is public and can be known by anyone.
• An adversary has access to any machine in the middle (not the devices of the two submitting parties).
• The computation has has to happen on a third party device without interacting with any of the owners of x and y more than once to get an input eg. Enc(x).

To me this sounds pretty much impossible, but since I cannot come up with a proof and the first time I heard of MPC all of it sounded impossible, let's give it a try.

Answer 1

I'm not sure I understand your example exactly but I think you are talking about homomorphic encrpytion:

a form of encryption that permits users to perform computations on its encrypted data without first decrypting it.

It's a pretty nascent field and very inefficient compared to standard calcuations.

Answer 2

One way hashes make it possible for f() = x == y, x contains y etc (although now the owner of x knows y and vice-versa, does this count?)

Technically functions which discard information like abs(x) + abs(y) can also work, as you can send x or -x and get the same result, the attacker would have have to make 4 guesses. But I'm not sure how secure this would be considered?

A generic solution for any f is impossible as shown by f() = x

Answer 3

It is possible for some f, impossible for other f.

An example f for which it is possible would be bitwise xor:

Let f(x,y) := xor(x,y)
And enc(x) := xor(x,s) with s being a secret.
dec(x) would be the same as enc(x).

f(enc(x), enc(y)) = xor(xor(x,s), xor(y,s)) = xor(xor(xor(x,y),s),s) = xor(x,y)

An example for which it would be possible:
Let f(x,y) := (x,y)
This is trivial, because you cannot know (x,y) without knowing x and y