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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...


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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 ...



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