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But the question is, can the Chinese remainder theorem in ECDSA be applied to the parameters in secp256k1? That precise attack doesn't work - we don't use the Chinese remainder theorem when computing with secp256k1 (as the group order is prime). On the other hand, there are certainly side channel attacks available against naïve implementations of ECDSA and ...


3

decryptions of Microsoft SEAL ciphertexts should be treated as private information only available to the secret key owner, as sharing decryptions of ciphertexts may in some cases lead to leaking the secret key. This was put in place as a response to the Li Micciancio attack on CKKS. The model Li Micciancio [LM] work in is traditional IND-CPA security ...


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The standard way is by using the LWE estimator (if it is simple or not it is debatable). It estimates the cost of known attacks against LWE instances given parameters $n$, $\alpha$, and $q$, where $\alpha$ represents the noise ratio and can be obtained from the parameter $\sigma$ from the discrete Gaussian using the formula $\alpha = \sqrt{2 \pi} \cdot \...


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The key step is in section 3 where the protocol is described. In step 2, Bob "scalarises" the $c_t$ values which converts them to a set of random $c_t'$ values where $\mathrm{Dec}(c_t)=\mathrm{Dec}(c'_t)$. Unfortunately, the authors have given the scalarised $c_t$ the same notation as the unscalarised, which is confusing. Alice can recreate the ...


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Regarding average, I was thinking about multiplying each entry by $2$ to make sure it is even. The number of all elements is known in my service so I can divide the encrypted numbers by non-encrypted values. Then the encrypted results can be sent to all companies and they can divide it by $2$ to get the final result, do you think it's fine? About the second ...


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My system knows only public key so it can calculate average salary for specific job title Actually, you can compute the sum; computing the average, that is, the value $\text{Encrypt}_k( \lfloor sum / n \rfloor )$ is rather trickier (and the floor operation is necessary if $sum$ is not necessarily a multiply of $n$ the number of values). This could be ...


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