不可补全词与矩阵死亡性的二次界
Quadratic bounds for uncompletable words and matrix mortality
浏览论文内容
中文总结 AI 辅助
该研究为不可补全词和矩阵死亡性建立了二次长度上界，并给出多项式时间构造算法，在 Lean 中完成形式化证明。
中文摘要 AI 辅助
每个有限非空的不完整唯一可解码码，若其最大词长为 $k$，则存在长度至多 $4k^2-3k$ 的不可补全词。该界与码字数量及其总长度无关。删除一个完整的码字循环可得到一个有限的路径计数恒等式；Kraft 等式随后提供一个压缩质量不足的短词。循环平均与填充将其转化为不可补全词。条件期望使该构造为多项式时间，并同时判定完整性。首次返回词将该界推广至非负整数 $n\ imes n$ 矩阵的死亡族，其联合谱半径至多为 1，前提是每个强连通分量都有一个顶点与每个循环相交。这样的族存在长度至多 $4n^2-3n$ 的零乘积。具有 $2k-1$ 个状态的二元部分确定性族的最短零乘积长度为 $k^2+k-1$，确立了最优的二次阶。这些界以及显式码算法（包括其多项式工作量界）均在 Lean 中证明。
英文摘要
Every finite nonempty incomplete uniquely decipherable code with maximum word length $k$ has an uncompletable word of length at most $4k^2-3k$. The bound is independent of the number of codewords and their total length. Deleting a complete codeword cycle gives a finite path-counting identity; Kraft equality then supplies a short word of deficient compressed mass. Cyclic averaging and padding turn it into an uncompletable word. Conditional expectation makes the construction polynomial-time and also decides completeness. First-return words extend the bound to mortal families of nonnegative integer $n\times n$ matrices with joint spectral radius at most one, provided every strongly connected component has a vertex meeting every cycle. Such a family has a zero product of length at most $4n^2-3n$. A binary partial deterministic family with $2k-1$ states has shortest zero product of length $k^2+k-1$, establishing the optimal quadratic order. The bounds and the explicit-code algorithm, including its polynomial work bound, are proved in Lean.
发表机构
- Efficient Computation Inc.
- Rexion Intelligence
机构由 AI 辅助整理，请以论文原文为准。
