Let $C$ be an $[n, k]$ linear code chosen uniformly at random over a finite field $\mathbb{F}_q$ of size $q$. The following asymptotic probability of $C$ being maximum distance separable (MDS) as $q,n,k\to\infty$ is known: If $\frac{1}{q}\binom{n}{k} \to 0$, then $P(C\text{ is MDS}) \to 1$. We demonstrate that this growth rate is in fact a threshold by proving: If $\frac{1}{q}\binom{n}{k} \to \infty$, then $P(C\text{ is MDS}) \to 0$. A matrix is (\textit{contiguous}) \textit{super-regular} if all of its (contiguous) square submatrices are nonsingular. The above results imply that for any $k \times k$ matrix $A$ chosen uniformly at random over $\mathbb{F}_q$, the following hold: If $\frac{4^k/\sqrt{k}}{q} \to 0$, then $P(A \text{ is super-regular}) \to 1$. If $\frac{4^k/\sqrt{k}}{q} \to \infty$, then $P(A \text{ is super-regular}) \to 0$. We also obtain the following asymptotic probabilities for two variations of the above questions: If $\frac{1}{q}\binom{n}{k} \to λ\in (0,\infty)$ and $k/n \to 0$, then $P(C\text{ is MDS}) \to e^{-λ}$. If $\frac{k^3/3}{q} \to λ\in (0,\infty)$, then $P(A \text{ is contiguous super-regular}) \to e^{-λ}$. The number of contiguous super-regular $3 \times 3$ matrices is also a polynomial. Finally, for $4 \times 4$ matrices, we show that the number of super-regular matrices is not a polynomial, nor even a quasi-polynomial of period less than 7, whereas our experimental evidence suggests that the number of contiguous super-regular matrices is a polynomial.

翻译：设$C$为有限域$\mathbb{F}_q$上均匀随机选取的$[n, k]$线性码，其中$q$为域的大小。已知当$q,n,k\to\infty$时，$C$为极大距离可分（MDS）码的渐近概率如下：若$\frac{1}{q}\binom{n}{k} \to 0$，则$P(C\text{为MDS}) \to 1$。我们通过证明以下结论表明该增长率实际上是一个阈值：若$\frac{1}{q}\binom{n}{k} \to \infty$，则$P(C\text{为MDS}) \to 0$。若矩阵的所有（连续）子方阵均为非奇异矩阵，则称该矩阵为（连续）超正则矩阵。上述结果蕴含：对于$\mathbb{F}_q$上均匀随机选取的任意$k \times k$矩阵$A$，若$\frac{4^k/\sqrt{k}}{q} \to 0$，则$P(A \text{为超正则}) \to 1$；若$\frac{4^k/\sqrt{k}}{q} \to \infty$，则$P(A \text{为超正则}) \to 0$。我们还得到了上述问题两种变体的渐近概率：若$\frac{1}{q}\binom{n}{k} \to λ\in (0,\infty)$且$k/n \to 0$，则$P(C\text{为MDS}) \to e^{-λ}$；若$\frac{k^3/3}{q} \to λ\in (0,\infty)$，则$P(A \text{为连续超正则}) \to e^{-λ}$。此外，连续超正则$3 \times 3$矩阵的数量是一个多项式。最后，对于$4 \times 4$矩阵，我们证明超正则矩阵的数量既不是多项式，也不是周期小于7的拟多项式，而我们的实验证据表明连续超正则矩阵的数量是一个多项式。