A simple, recently observed generalization of the classical Singleton bound to list-decoding asserts that rate $R$ codes are not list-decodable using list-size $L$ beyond an error fraction $\frac{L}{L+1} (1-R)$ (the Singleton bound being the case of $L=1$, i.e., unique decoding). We prove that in order to approach this bound for any fixed $L >1$, one needs exponential alphabets. Specifically, for every $L>1$ and $R\in(0,1)$, if a rate $R$ code can be list-of-$L$ decoded up to error fraction $\frac{L}{L+1} (1-R -\varepsilon)$, then its alphabet must have size at least $\exp(\Omega_{L,R}(1/\varepsilon))$. This is in sharp contrast to the situation for unique decoding where certain families of rate $R$ algebraic-geometry (AG) codes over an alphabet of size $O(1/\varepsilon^2)$ are unique-decodable up to error fraction $(1-R-\varepsilon)/2$. Our bounds hold even for subconstant $\varepsilon\ge 1/n$, implying that any code exactly achieving the $L$-th generalized Singleton bound requires alphabet size $2^{\Omega_{L,R}(n)}$. Previously this was only known only for $L=2$ under the additional assumptions that the code is both linear and MDS. Our lower bound is tight up to constant factors in the exponent -- with high probability random codes (or, as shown recently, even random linear codes) over $\exp(O_L(1/\varepsilon))$-sized alphabets, can be list-of-$L$ decoded up to error fraction $\frac{L}{L+1} (1-R -\varepsilon)$.

翻译：最近观察到的一个经典Singleton界到列表解码的简单推广表明：速率为$R$的码在列表大小$L$下，无法在超过错误率$\frac{L}{L+1} (1-R)$时进行列表解码（Singleton界对应$L=1$的特殊情况，即唯一解码）。我们证明，对于任意固定的$L>1$，要逼近该界需要指数级的字母表。具体而言，对于每个$L>1$和$R\in(0,1)$，如果一个速率为$R$的码能够以列表大小$L$在错误率$\frac{L}{L+1} (1-R -\varepsilon)$内被解码，则其字母表大小至少为$\exp(\Omega_{L,R}(1/\varepsilon))$。这与唯一解码的情况形成鲜明对比，在唯一解码中，某些速率为$R$的代数几何（AG）码族使用大小为$O(1/\varepsilon^2)$的字母表可以在错误率$(1-R-\varepsilon)/2$内实现唯一解码。我们的界甚至对亚常数$\varepsilon\ge 1/n$也成立，这意味着任何精确达到第$L$个广义Singleton界的码都需要字母表大小$2^{\Omega_{L,R}(n)}$。此前，这一结论仅在$L=2$且附加假设码是线性且MDS的情况下已知。我们的下界在指数中常数因子意义下是紧的——在大小为$\exp(O_L(1/\varepsilon))$的字母表上，高概率随机码（或如最近所示，甚至随机线性码）可以在错误率$\frac{L}{L+1} (1-R -\varepsilon)$内以列表大小$L$进行解码。