# Tag Info

1

Since Alice encrypts the message $m$ she knows the plaintext. Now Bob computes F with the public key of Alice.Alice knows the secret key of the underlying homomorphic scheme and decrypts $C'$ and obtains the underlying values. This is how homomorphic schemes operate

5

This concept is called targeted malleability: http://crypto.stanford.edu/~dabo/pubs/abstracts/reshom.html The abstract and introduction of that paper give a good overview of the ideas. In brief, the goal is to ensure that a homomorphic evaluator can only produce a ciphertext by evaluating a function from an "approved" class of functions. It is trivial to ...

3

it seems that you are looking for this paper of mine http://libeccio.dia.unisa.it/Papers/CHES/ Here we introduce the concept of a Controlled Homomorphic Encryption Scheme (CHES). In a CHES, for every circuit $C$, it is possible to homomorphically create a ciphertext of $C(m)$ from a ciphertext of $m$ (so far it is like a regular FHE) but only if a special ...

1

(even number mod even number) is always even number and (odd number mod even number) is always odd number. If $m=0$, the result of modular $x_0$ will be always even number. If $m=1$, the result will be odd number. So $x_0$ should be a odd number.

2

Let's look at the ciphertext: $c \leftarrow m + 2r + 2 \sum\limits_{i \in S} x_i \bmod x_0 = m + 2r + 2\sum\limits_{i \in S} (pq_i + r_i) + k(pq_0 + r_0)$ for some $k$. When we do modular reduction by $p$, what remains is $m + 2r + 2\sum\limits_{i \in S} r_i + k r_0$. This (without $m$) is called the noise - as long as it doesn't wrap around $\bmod ... 1 Just expand it out. You get$4\tau \cdot 2^\rho + 3 \cdot 2^\rho$. Then,$\tau \cdot 4 \cdot 2^\rho = \tau \cdot 2^2 \cdot 2^\rho = \tau \cdot 2^{\rho + 2}$and$3\cdot 2^\rho < 4 \cdot 2^\rho = 2^{\rho+2}$. Put it back together,$\tau \cdot 2^{\rho + 2} + 2^{\rho + 2} < \tau (2^{\rho + 2}+2^{\rho + 2}) = \tau 2^{\rho+3}\$

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