Let's assume I have datapiece A which, after being put through a model or neural network, has a known output X in the unencrypted space. When I move datapiece A into an encrypted space, and put it through an encrypted model, it becomes datapiece e-A. Now, it has output e-X. In this way, I am using a known (unencrypted) data to map X into the encrypted space (e-X).
The model in the unencrypted space has only two possible outputs: X and Y. When I encrypt this model, I'd like there to still be only two possible outputs: e-X and e-Y.
When someone else sends me an encrypted input data, datapiece e-Unknown, I plug it into the function, and I also get output e-X. Since datapiece e-unknown produces e-X in the encrypted space, can it be assumed that datapiece-Unknown also produces output X in the unencrypted space? In other words, does the map from the first paragraph enable deciphering outputs for the anonymized input datapiece-Unknown? Assume datapiece unknown and datapiece A are different (i.e. not exactly the same).
Eval(Enc(f),Enc(dA))=Eval(Enc(f),Enc(dU))⇒f(dA)=f(dU)
where Eval(f,c) stands to "apply f homomorphically to the ciphertext c" .... "f" is the model, "dA" is datapiece-A, "dU" is datapiece-unknown.
If this can be done without encrypting the model, that would also work.
Eval(f,Enc(dA))=Eval(f,Enc(dU))⇒f(dA)=f(dU)
For an unusual use case, I want this to be true. How can I make it work?
Does this hold true for homomorphic encryption? Since h.e. is probabilistic, are there any tricks to make it hold true for homomorphic encryption (e.g. re-seeding with each new input)? If not, why? I am asking specifically about H.E. because a number of libraries already exist that make it easy to use.
Does this hold true for functional encryption? I have heard it does, but there are not many libraries available for functional encryption.