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RSA is almost always used in hybrid mode, where AES (or another symmetric cipher) is used to encrypt the data itself, and RSA is then used to encrypt the random data key. That way RSA has only a static overhead: the modulus size (which is also the key size) in bytes. So for RSA-1024 that would mean an overhead of 128 bytes + whatever overhead is required for ...


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In theory, you appear to be right. A combination of HE and FE for that particular use case, where FE computes the non-linear operations, might be useful. However, you need an appropriate FE scheme that can operate correctly on HE-encrypted data, or a protocol to switch between HE and FE-compatible scheme. Nowadays such scheme doesn't exist - and currently ...


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The main difference between FE and HE is that: In HE, for the decryption you need to interact with the owner of the secret-key, but in FE everybody who has access to the ciphertext and functional key can decrypt the massage. While HE can efficiently support different kinds of computation over encrypted data, FE supports limited class of functionality (in ...


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What you're describing is called a known-plaintext attack, and any secure encryption algorithm should prevent this. If we take an encryption function with no IV $C = ENC(M, K)$ we should get a system where knowing both $M$ and $C$ reveals nothing about $K$. We can expand this principal to $C = ENC(M \oplus IV, K)$ and simplify it back to the example without ...


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