If we consider a set of numbers say a set $s=\{a,b,c,d\}$ , where $a,b,c,d>1$ and the numbers $a, b, c, d$ do not share any relation between them , i.e. for any two numbers, $n_1,n_2\in s$ the following relations are guaranteed to hold: $n_1+n_2\neq 0$ and $n_1*n_2\neq 1$.
Is it possible to break the fully homomorphic encryption scheme on these set of numbers?
Note
This is a followup to this question which does not restrict the set $s$ is at all.
This is an interesting question. If the input set does not have any identity element (for both mul and add operations) then one cannot construct a field and subsequently there cannot be homomorphic operations. More clearly, if the input plaintext elements do not form a field (and their respective ciphertext elements do not form a field either) then there cannot be homomorphisms of operations. This is by definition of Field.
So if the input set contains identity elements (and other properties needed to become a field) then while the given set members may not have any obvious relation. There would be some constant $c$ for which $enc(n_1) + c * enc(n_2) = enc(0) $ subsequently finding which attacks mentioned in previous question apply !
Also an attacker could try to find random complex relations among them. For example is $ enc(a) + enc(a) = enc(b) $ or is $ enc(a) *enc( b) = enc(c) $ etc. revealing inferences.
If your fully homomorphic scheme is asymmetric and deterministic, then, it is already broken.
One could just encrypt the value one, $c_1 = E(pk, 1)$, then, adds $c_1$ to itself until
$c_1 + c_1 + ... + c_1 == E(pk, a)$
or
$c_1 + c_1 + ... + c_1 == E(pk, b)$
etc...
