One of the conditions of FHE is hop correctness. Why the stages in hops are orthogonal to each other.
During circuit evaluation some inputs are ignored, maybe to reduce the noise growth rate. If yes how we can ensure about the weak inputs.
We have C1 that is encryption of m and we have to compute f(C1) in real word but due to recryption we are computing f(C3). As we know C3 = C1+C2 (where C2 is also known as refreshed ciphertext encryption of secret key) . So it is necessary that f(C1) and f(C3) will produce the same result or if not then the difference between both the results is negligible ?
Could you explain the context of your questions? Are you referring to a specific formal description? What do you mean by a hop here?
Regarding question 3, there is a difference between exact, e.g., BGV and BFV, and approximation FHE, e.g., CKKS. In the exact case, f(C1) and f(C3) will produce the same result as long as the decryption failure probability is negligible. In the approximate FHE case, the result will be different by design (with some approximation error).
Hop refers as: An i-hop homomorphic encryption where Evaluate can be called on it’s on output up to i times (while still being able to decrypt the result).
And number of hops supported by a scheme is somewhat orthogonal to its homomorphic capacity.
So, why they are orthogonal.
I see. So you are asking about the definition in i-Hop Homomorphic Encryption and Rerandomizable Yao Circuits The high-level idea there is to achieve circuit privacy at each hop so that if the result is decrypted at the next hop, no information about the functions used at previous hops is learned. The number of hops is orthogonal because bootstrapping can be invoked at the end of computation at each hop. This will reset the homomorphic capacity.