Consider the problem of distinguishing between polynomially many samples of either \begin{equation} (x, b, As + e) ~~\text{or}~~\left(x, b, ~Ax + b\cdot(As + e) + e'\right). \end{equation}

Here, $A$ is a public matrix and $s$ is a secret vector chosen uniformly at random. $e$ and $e'$ are Gaussian errors. $x$ and $b$ are sampled uniformly at random.

The dimensions of different objects are:

\begin{align} b &\in \{0, 1\}, \\ x &\in \mathbb{Z}_{q}^{n}, \\ s &\in \mathbb{Z}_{q}^{n}, \\ A &\in \mathbb{Z}_{q}^{m \times n}, \\ e, e' &\in \mathbb{Z}_{q}^{m}, \\ \end{align}

$q \geq 2$ is a prime integer.

Are these two cases (computationally) indistinguishable, when we are given polynomially many samples? I think they are, but I could not tie them to a conjecture.

Note that by LWE,

\begin{equation} (x, b, As + e) ~~\text{and}~~\left(x, b, u\right), \end{equation} are computationally indistinguishable and so are \begin{equation} (x, b, ~Ax + b\cdot(As + e) + e') ~~\text{and}~~\left(x, b, ~Ax + b\cdot u + e'\right). \end{equation}

$u$ is a uniformly random sample. However, I could not reduce my case to LWE.


1 Answer 1


One can trivially distinguish $(x,0,u)$ from $(x,0,Ax+e’)$ by subtracting $Ax$ from the third entry and seeing if the entries look uniform or Gaussian.

Distinguishing $(x,1,u)$ from $(x,1,A(x+s)+(e+e’))$ is a standard LWE problem (noting that the variance of $e+e’$ is the sum of the variances of $e$ and $e’$.

Thus samples with $b=0$ are trivial and samples with $b=1$ Are presumably not. Taking polynomially many samples is virtually certain to give at least one with $b=0$ and so allow us to distinguish trivially.

  • $\begingroup$ Just to sanity check, if there were polynomially many samples from either \begin{equation} (x, b_1, b_2, \ldots, b_k, ~Ax + b_1\cdot(As_1 + e_1) + b_2\cdot(As_2 + e_2) + \cdots + b_k\cdot(As_k + e_k) + e') ~~\text{or}~~\left(x, b_1, b_2, \ldots, b_k, u \right), \end{equation} for $b_i \in \{0, 1\}$, for a polynomially large $k$ and for secret vectors $s_1, \ldots, s_k$, then these will be indistinguishable, is that right? $\endgroup$
    – Morbius
    Commented Apr 15, 2022 at 0:39
  • $\begingroup$ The argument is just that when every $b_i$ is $0$, they are easily distinguishable, but such a case is exponentially unlikely. For every other case, for any choice of $k$, we can reduce it to LWE. $\endgroup$
    – Morbius
    Commented Apr 15, 2022 at 0:42
  • $\begingroup$ No, the argument is that if at least $b_i$ is 0, the set is easily distinguishable $\endgroup$
    – Daniel S
    Commented Apr 15, 2022 at 5:52
  • $\begingroup$ How can that be true? Let's say only $b_1 = 0$. Then, we are essentially distinguishing between $A(x + s_2 + s_3 + \cdots + s_k) + (e_1 + e_2 + \cdots + e_k) + e'$ and $u$. Isn't that just a variant of LWE? So, do we not need every $b_i$ to be $0$ for the samples to be distinguishable and not just at least one $b_i$ to be $0$? $\endgroup$
    – Morbius
    Commented Apr 16, 2022 at 5:36
  • $\begingroup$ *We are distinguishing between $A(x + s_2 + \cdots + s_k) + (e_2 + \cdots + e_k + e')$ and $u$.... $\endgroup$
    – Morbius
    Commented Apr 16, 2022 at 5:46

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