Consider the following modification to the Short Integer Solution (SIS) problem:

Let $n$ be an integer and $\alpha=\alpha(n),\beta=\beta(n),m=m(n)>\Omega(n\log \alpha)$ be functions of $n$. Sample a uniform $A\gets[-\alpha,\alpha]^{n\times m}$. The task is to compute "short" vector $e\in\mathbb{Z}^m$ in the kernel of $A$. That is:

  1. $|e| < \beta$.
  2. $A.e=0^n$. Here, equality holds over the integers

The usual version of SIS is the same as above, except where $A.e=0^n$ holds mod $q$, and $q=2\alpha+1$ (so that $A$ is uniform in $\mathbb{Z}_q^{n\times m}$). This variant removes the modulus.

Question: Are there any non-trivial hardness/easiness results for this version of SIS? What settings of parameters are easy, and which (if any) can be proved hard based on worse-case lattice problems, as in the usual version of SIS?

Trivial attack: There is an trivial algorithm in the case where $\beta$ is huge. You can compute a kernel vector over the integers by taking minors of the matrix $A$. These minors, and hence the kernel vector, can be easily upper bounded by $(\alpha n)^{O(n)}$. So in the regime $\beta= (\alpha n)^{O(n)}$, there is a trivial attack.

What I'm most curious about is the case where $\alpha,\beta$ are polynomial in $n$. Are there any attacks here, or can any hardness be shown?

I picked the distribution for $A$ above to give a concrete problem. But I'd also be interested in other distributions on $A$. For example, what if the entries of $A$ are discrete Gaussians, etc?

One can also consider an inhomogeneous version of this SIS variant, where $A.e=u$, for some vector $u$ (again, without a modulus). We have to be careful, though as for large $u$ there won't be a solution. Maybe we restrict to random $u\in\{0,1\}$, or in $[-\gamma,\gamma]^n$. I'd also be interested if anything can be said about this problem as well, besides the straightforward adaptation of the trivial attack from above.

  • $\begingroup$ I would be very careful with an assumption like this, namely because LWE without the modulus is easy. $\endgroup$
    – Mark Schultz-Wu
    Jul 9, 2021 at 6:22
  • $\begingroup$ I certainly agree that it would be dangerous to assume hardness without any formal justification. At the same time, I'm not aware of any actual attacks, besides the trivial attack mentioned. $\endgroup$
    – AAA
    Jul 9, 2021 at 16:12
  • $\begingroup$ I have linked to an attack on a closely related problem in the same setting. It would not be surprising to me if one could potentially extend the attack to SIS, which is why I linked you the paper. $\endgroup$
    – Mark Schultz-Wu
    Jul 9, 2021 at 21:34
  • $\begingroup$ The LWE hardness reduction constrains the modulus as $q \le 2^{O(n)}$, whereas the SIS reduction constrains it as $q \ge \beta \cdot O(n)$. Sufficiently large $q$ will be equivalent to the problem over the integers, I imagine. $\endgroup$ Jul 11, 2021 at 16:53
  • $\begingroup$ @Samuel Neves: the problem is that SIS is usually defined where the matrix is random over $\mathbb{Z}_q$. So as $q$ scales up, so do the entries of $A$. This means $A.e$ will almost certainly have wrap-around mod $q$. So I don't immediately see how to use this for my problem $\endgroup$
    – AAA
    Jul 12, 2021 at 5:44

1 Answer 1


It turns out some version of the problem is actually as hard as SIS. Concretely, I claim that the version where $A$ is a random binary matrix and $\beta$ is polynomial will be hard, assuming SIS is hard for an appropriate choice of parameters.

Let $q=2^\ell$ be a power of 2 that is sufficiently larger than $\beta$. Let $n'=n/\ell$ (we assume $n$ divisible by $\ell$ for simplicity). Then consider a SIS instance with parameters $n',m,q,\beta$: given a random matrix $A\in\mathbb{Z}_q^{n'\times m}$, the goal is to find a non-zero vector $e\in\mathbb{Z}^m$ such that $A\cdot e\equiv 0\pmod q$ and $|e|<\beta$.

We reduce to the stated modulus-free problem as follows. Let $A_i\in\{0,1\}^{n'\times m}$ be the matrix where we replace each entry in $A$ by the $i$th bit of that entry. Then let $A'\in\{0,1\}^{n\times m}$ be the matrix obtained by stacking all of the $A_i$ on top of each other.

If we could solve modulus-free SIS for $A'$, this would give us a vector $e\neq 0$ such that $A'\cdot e=0$ (over the integers) and $|e|<\beta$. But then I claim that $A\cdot e = 0$. Indeed, each entry of $A$ is just a linear combination of entries in the corresponding column of $A'$. Therefore, each entry of $A\cdot e$ is just a linear combination of the entries in $A'\cdot e$, and is therefore 0.


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