A recent eprint paper claims to bound $\lambda_1(\Lambda^\perp(\mathbf{A}))$ for $\mathbf{A}\in\mathbb{Z}^{n\times m}$, a uniformly random matrix, by $O(1)$, specifically by $4$. This has applications to solving $\mathsf{SIS}_{n,m,q,4}$ in $\mathsf{P}$.

I'm no expert in this area, but it seems to me this contradicts the common thought that $\lambda_1(\Lambda^\perp(\mathbf{A})) = \Omega(\sqrt{n\log q})$ (see, for example, section 2.4.2 of this paper).

As getting into the details of the recent paper is likely off topic for this forum, I'm interested instead in another question --- what concrete/experimental evidence has been collected for the asymptotics of $\lambda_1(\Lambda^\perp(\mathbf{A}))$? Looking into data such as this would be an easy way to gain intuition for whether we're in an $O(1)$ regime or $\Omega(\sqrt{n\log q})$ one, and I had assumed that someone had papers along these lines, but don't actually recall seeing any of it myself.

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    $\begingroup$ I contacted the authors, one of whom (Jia Huiwen in particular) replied: (after translation) Currently there isn't a specific algorithm, it's only deduction on theoretical level, also there's a problem with the proof in the article which we're fixing. $\endgroup$
    – DannyNiu
    Commented Jan 17, 2019 at 5:51
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    $\begingroup$ @DannyNiu Their claim that $\lambda_1(\Lambda^\perp(A)) = O(1)$ shouldn't be dependent on any particular algorithm though. For a particular distribution over $A$ (with fixed $m,n,q$), we have that $\lambda_1(\Lambda^\perp(A))$ is simply a statistic, and should be able to be estimated computationally. Then, by repeating this for various $m,n,q$, you should be able to gain some intuition for the asymptotic properties of $\lambda_1(\Lambda^\perp(A))$. I'm curious if anyone has done this computational exercise, and made some paper out of it (for example), as this seems plausible. $\endgroup$
    – Mark Schultz-Wu
    Commented Jan 17, 2019 at 6:18
  • $\begingroup$ @DannyNiu That being said, I appreciate your update with regards to that paper! $\endgroup$
    – Mark Schultz-Wu
    Commented Jan 17, 2019 at 6:19
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    $\begingroup$ It looks like the bug is in the proof of the (false) inequality labeled "on the other hand" on page 7. E.g., a random lattice has $bl \approx \sqrt{n}$ but $\rho_1(L^*) \approx 2$. So, they're inequality says $2 \gtrsim \sqrt{n}$. Their proof of this inequality seems to boil down to an argument that the dual lattice $L^*$ always has a vector of length $1/bl$, which is false in general. It has a vector of length $1/\|\tilde{b}_n\|$, but typically not a vector of length, e.g., $1/\|\tilde{b}_1\|$. $\endgroup$ Commented Jan 17, 2019 at 16:32

2 Answers 2


The first inequality at the bottom of page 3 of the paper is false. For example, Conway and Thompson proved the existence of "self-dual" $n$-dimensional lattices $L$ (i.e., $L^* = L$) where $\lambda_1(L) = \lambda_1(L^*) = \Omega(\sqrt{n})$, hence $\eta_{2^{-n}}(L) = O(1)$ but $\tilde{bl}(L) \geq \lambda_1(L) = \Omega(\sqrt{n})$.

The statement and proof of Conway and Thompson's result can be found as Theorem 9.5 in https://www.springer.com/us/book/9783642883323 .


Independently of the algorithmic claim, I indeed have serious doubts about Theorem 2. Here is a counterargument (using standard techniques) cooked up with Yang Yu and Wessel van Woerden:

Suppose that $q$ is prime, and suppose that the minimum distance in the 2-norm is indeed bounded by $b=O(1)$. There are at most $Binom[m,b] \cdot (2b+1)^b = poly(m)$ elements x in $\mathbb Z^m \setminus \{0\}$ with $||x||^2_2 \leq b$. But every such vector has only probability $q^{-n}$ to be part of the lattice over the randomness of $A$. By a union bound this means that the probability that any such vector is part of the lattice over the randomness of $A$ is at most $poly(m) \cdot q^{-n}$ which is negligible unless m itself is exponential in $n$.

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    $\begingroup$ You are right and this can be turned into a simple proof (see Micciancio's lecture notes, Exercise 4). So without looking at the paper in detail, it seems there must be something wrong with the proof. $\endgroup$
    – blowfish
    Commented Jan 17, 2019 at 9:27

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