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$\textbf{Continuous LWE}$ : $(\overrightarrow{a}, b)\in \mathbb{Z}_q^n\times \mathbb{T}$, where $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, $b = \langle \overrightarrow{a},\overrightarrow{s}\rangle/q + e\mod 1$, where the error $e$ is sampled from $\Psi_\alpha(x) := \sum_{k=-\infty}^{\infty}\frac{1}{\alpha}\cdot exp(-\pi(\frac{x-k}{\alpha})^2), x\in [0,1)$ over the torus $\mathbb{T}$. The density function $\Psi_\alpha$ is just the Guassian function $\rho_\alpha(x) = \frac{1}{\alpha}exp(-\pi x^2/\alpha^2) \mod 1$.

$\textbf{The discretization}: $ transform the continuous sample $(\overrightarrow{a},b)$ to $(\overrightarrow{a}, \lfloor qb\rceil) \in \mathbb{Z}_q^{n+1} $, the $\lfloor qb\rceil = \langle \overrightarrow{a},\overrightarrow{s}\rangle + \lfloor qe \rceil \mod q$, therefore, the error in discretization is the distribution $q\cdot\Psi_\alpha$ over $\mathbb{Z}_q$.

$\textbf{The rounded Gaussian}:$ $\rho_\alpha(x) = \frac{1}{\alpha}exp(-\pi x^2/\alpha^2)$ which is the Gaussian distribution over $\mathbb{R}$, we transform it to $\mathbb{Z}_q$ by $\lfloor \rho_\alpha \rceil \mod q$, which means that we sample a real from $\rho_\alpha$, then round it to integer and modulo $q$, then $\lfloor \rho_\alpha \rceil \mod q$ is also a distribution over $\mathbb{Z}_q$..

$\textbf{My Question}:$

  1. Are the distribution in discretization $q\cdot \Psi_\alpha$ and the rounded Gaussian $\lfloor \rho_\alpha \rceil \mod q$ over $\mathbb{Z}_q$ the same?

  2. If we choose the $\lfloor \rho_\alpha \rceil \mod q$ or $\lfloor q\cdot \Psi_\alpha\rceil $ as the error distribution in discretization LWE, is it still hard?

I think the two distribution over $\mathbb{Z}_q$ are different. The $\lfloor q\cdot \Psi_\alpha\rceil $ is just the distribution in [Regev05] which has been proved hard. Then, how about the $\lfloor \rho_\alpha \rceil \mod q$ ?

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The distributions are the same. That is, rounding and modding (by any integer $q$) essentially commute: $\lfloor \rho_a \rceil \bmod q = \lfloor \rho_a \bmod q \rceil$, where on the right we are rounding $\mathbb{R}/q\mathbb{Z}$ to the closest element of $\mathbb{Z}/q\mathbb{Z}$ (so the result remains modulo $q$). This follows simply from the fact that $\lfloor x \rceil +kq =\lfloor x +kq \rceil$ for any integer $k$. So, for any $v \in \mathbb{Z}/q\mathbb{Z}$ its probability is the same under the two distributions.

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  • $\begingroup$ Thank you for your answer. I've tried to derive that: If the density function of $e$ is $\Psi_\alpha$, then density function of $qe$ is $\frac{1}{q}\Psi_\alpha(\frac{y}{q})=\sum_{k=-\infty}^{\infty}\frac{1}{\alpha q} exp(-\pi \frac{(y-kq)^2}{(\alpha q)^2})$, it should be a Gaussian distribution with parameter $\alpha q$, but for the $\rho_\alpha \mod q$, its parameter seems to be $\alpha$, not $\alpha q$. So I guess what you mean is the $qe$ is the same as $\rho_{\alpha q} \mod q$ ? $\endgroup$ – 2646jiaxing Jun 11 at 2:19
  • $\begingroup$ Of course, you need to scale both $\Psi_\alpha$ and $\rho_\alpha$ by the same $q$ factor, or they obviously won’t match. Then, the rounding commutes with the modding. $\endgroup$ – Chris Peikert Jun 11 at 2:22
  • $\begingroup$ Thank you very much for your reply！ $\endgroup$ – 2646jiaxing Jun 11 at 2:26

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