From a10fc044f31bd1bc87ab2c0dab5f4d5b595d6a9b Mon Sep 17 00:00:00 2001 From: Jean-Philippe Bossuat Date: Thu, 26 Oct 2023 17:06:49 +0200 Subject: [PATCH] updated SECURITY.md --- SECURITY.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/SECURITY.md b/SECURITY.md index bd84b8be..208fd70a 100644 --- a/SECURITY.md +++ b/SECURITY.md @@ -21,7 +21,7 @@ If at any point of an application, decrypted values have to be shared with exter Estimating $PR[\epsilon < x] \leq 2^{-s}$, for $s$ a security parameter, of the circuit must be done carefully and we suggest the following process to do so: 1. Given a security parameter $\lambda$ and a circuit $C$ that takes as inputs length-_n_ vectors $\omega$ following a distribution $\chi$, select the appropriate parameters enabling the homomorphic evaluation of $C(\omega)$, denoted by $H(C(\omega))$, which includes the encoding, encryption, evaluation, decryption and decoding. 2. Sample input vectors $\omega$ from the distribution $\chi$ and record $\epsilon = C(\omega) - H(C(\omega))$ for each slots. The user should make sure that the underlying circuit computed by $H(C(\cdot))$ is identical to $C(\cdot)$; i.e., if the homomorphic implementation $H(C(\cdot))$ uses polynomial approximations, then $C(\cdot)$ should use them too, instead of using the original exact function. Repeat until until enough data points are collected to construct a CDF of $PR[\epsilon > x]$. - 3. Use the CDF to select the value $E[\epsilon]$ such that any given slot will fail with probability $2^{-2}$ to reach $\log_{2}(1/\epsilon)$ bits of precision. + 3. Use the CDF to select the value $E[\epsilon]$ such that any given slot will fail with probability $2^{-s}$ to reach $\log_{2}(1/\epsilon)$ bits of precision. 3. Use the encoder method $\textsf{DecodePublic}$ with the parameter $\log_{2}(1/\epsilon)$ to decode plaintexts that will be published. Note that, for composability with differential privacy, the variance of the error introduced by the rounding is $\text{Var}[x - \lfloor x \cdot \epsilon \rceil / \epsilon] = \tfrac{\epsilon}{12}$ and therefore $\text{Var}[x - \lfloor x/(\sigma\sqrt{12})\rceil\cdot(\sigma\sqrt{12})] = \sigma^2$.