I already asked a question about the feasibility of LWE when the matrix A is sparse or small here.

Let $q$ be a prime, let $\chi$ be a distribution of $\textit{small}$ elements over $\mathbb{Z}/q$, and let $n,m$ be two integers (dimensions of vectors that we work with). Sample $s \leftarrow \chi^n$, $e \leftarrow \chi^m$ and $A \leftarrow \mathbb{Z}/q^{m \times n}$, where elements of $A$ are sampled uniformly at random. Let $t=As + e$.

Given $A$ and $t$, find $s$.

This problem is believed to be hard under some reasonable assumptions on $q$, $\chi$, $n$, $m$.

My question is - what is $A$ is a (somewhat) sparse matrix, for example $n=m$, and $A$ has $\approx\sqrt{n}$ non-zero elements in each column, and every non-zero element is $\pm1$. Does the problem remain hard? Or does the problem become trivial? What if I increase $m$ from $m=n$ to something larger?

I was provided a good answer showing that when $A$ is small/sparse, the solution to the LWE problem is not unique, and a solution can be easily found.

My question is the following: can the exact solution of the above problem be found, if it is committed to via another regular LWE instance. In other words:

Let $s \leftarrow \chi^n$; $e,e' \leftarrow \chi^n$, let $A$ be a sparse matrix ($n=m$, $\sqrt{n}$ non-zero elements, each equal to $\pm 1$), and let $A'$ be a uniformly random matrix. Let $t=As+e$ and $t' = A's+e'$.

Given $t$ and $t'$ can we find $s$?

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    $\begingroup$ How about this paper eprint.iacr.org/2018/741? This paper says that if $A$ is binary, LWE with $A$ is broken. $\endgroup$ Commented Jul 26, 2022 at 9:12
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    $\begingroup$ This helps a lot. Thanks. It doesn't directly provide an answer to the problem, right? I'm considering an $n\times n$ matrix, for which I know there are many solutions in the sparse matrix case, and I want to find one specific solution committed to in a separate LWE instance. $\endgroup$
    – Kolja
    Commented Jul 29, 2022 at 3:19

1 Answer 1


This is not intended as an answer, but hopefully a helpful comment. (Apparently I don't have the reputation on this website to leave a comment -- apologies.)

Let's view this is an LWE with side information problem. That is, you're intending to attack $({\bf A}, {\bf b} = {\bf A}{\bf s} + {\bf e})$, and you're presented with auxiliary information $({\bf A'}, {\bf b}' = {\bf A}{\bf s} + {\bf e}')$.

Since solving for any one of the many solutions ${\bf s}'_1, ..., {\bf s}'_k$ is easy (since ${\bf A'}$ is not only sparse, but ternary), a naïve way to view the side information $({\bf A}', {\bf b}')$ is that you're presented with a auxiliary set $S = \{{\bf s}'_1, ..., {\bf s}'_k\}$ of possible values of ${\bf s}.$ Obviously, if $|S| = k$ is ${\sf poly}(n),$ there is a simple polynomial-time attack by enumerating the ${\bf s}_i'.$ Similarly -- in concrete analysis -- if enumerating $|S|$ ($2^{30}$? $2^{40}$? etc.) is within your computational bounds, the problem is immediately concretely easy. So, the more interesting situation is when $|S| = {\sf superpoly}(n).$ (As this is just meant to be a comment, I haven't bothered to check $|S|$ for this, or other possible, distribution(s) of the auxiliary $({\bf A}', {\bf b}')$ -- but for any problem of this form, that is the first step to perform, of course.)

However, more information is clearly present in the presented auxiliary value $({\bf A}', {\bf b}')$ than simply the set $S.$

For example, given ${\bf A}'$, for each candidate-solution ${\bf s}_i' \in S$, it is easy to derive the corresponding induced candidate-error term ${\bf e}_i'.$

Note that the ${\bf e}_i'$ will be in the support of the corresponding distribution $\chi^n$. However, the various ${\bf e}_i'$ will not occur with equal probability under i.i.d. sampling of $\chi^n$.

Intuitively, for each candidate ${\bf s}_i' \in S$, what I want to do is check whether ${\bf s}_i'$ is indeed the solution ${\bf s}$ of $({\bf A}, {\bf b})$. (You can do the obvious thing -- subtract ${\bf A}{\bf s}_i'$ off ${\bf b}$ and check if the result is short.)

However, if I could somehow cheaply order my enumeration through the ${\bf s}_i'$ according to the probability that the induced ${\bf e}_i'$ would be sampled from $\chi^n,$ I will clearly do much better in total attack cost in expectation than $\tilde O(|S|)$. (Such ordering is, of course, possible in principle -- finding a way to do it that gives a speed-up is the important question. Details left to the person who will actually answer this question. =))

Further, I wanted to point out that if there is a way to "shoe-horn" the auxiliary information provided by $({\bf A}', {\bf b}')$ into the format of, e.g., the Distorted Bounded Distance Decoding (DBDD) approach of Dachman-Soled et al. (cf. https://eprint.iacr.org/2020/292.pdf), then you should be able to very cleanly derive a good cost estimate of this problem (for many variant distributions of the sparse/short ${\bf A}'$) using that single DBDD framework.

Edit: An extra, parting thought-- An interesting approach might be to artificially add additional error (in some controlled way) to the $({\bf A}', {\bf b}')$ auxiliary information before attempting to enumerate a solution set $S$. Note that if I increase the error, it seems I might be able to 'naturally eliminate' potential ${\bf s}_i'$ candidates that induce error terms ${\bf e}_i'$ that are on the 'tails' of the $\chi^n$ distribution. An approach like this seems to naturally restrict the set $S$ toward solutions that are more likely on average to be the solution ${\bf s}$ to the primary instance $({\bf A}, {\bf b})$. Good luck!

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    $\begingroup$ I never considered the number of solutions ($|S|$) of $As+e$, is it indeed polynomial? If it is, and if all the solutions can be enumerated, the problem becomes trivial. $\endgroup$
    – Kolja
    Commented Aug 4, 2022 at 15:59
  • $\begingroup$ I can comment now - thanks! $|S|$ depends heavily on the distribution of $A, s$ and $e$. For natural choices, it is polynomial; for other natural choices, it is exponential. For some natural choices of dist that lead to polynomially-bounded $S$, finding one element of $S$ is hard; for other natural choices of dist that are polynomially-bounded, find every element of $S$ is easy. If we assume there is a good motivation for this question, then answering these questions is the content of a publishable paper (outside the scope of SE =)). I would first ask (yourself) - why do you care? =) $\endgroup$ Commented Aug 5, 2022 at 2:01
  • $\begingroup$ Example: In the typical LWE-crypto setting (let $A, s$ be uniform and $e \leftarrow \chi$ for "typical" $\chi,$) then $|S| = 1.$ (This is what leads to a correctly decryptable ciphertext.) -- Or, modify the above so that $A = I_n$ for the $n$-by-$n$ identity matrix $I_n.$ Now $|S| = |{\sf support}(\chi)| = B^n = \Omega(2^n)$ where $B$ is the number of valid coordinates per individual error term of $e$. $\endgroup$ Commented Aug 5, 2022 at 2:04
  • $\begingroup$ Maybe you misunderstood my comment, I was asking if $|S|$ was polynomial in size when $A$ is a sparse matrix with $\pm1$ as only non-zero values. The problem is related to a different more practical issue, but I preferred to explain it in simplest terms through a generic LWE instance. $\endgroup$
    – Kolja
    Commented Aug 5, 2022 at 16:05

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