发表机构
School of Mathematics, Fudan University（复旦大学数学学院）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文证明q元至多三种码字长度的免固定码的3/4猜想，通过矩阵方法和插值定理给出确定性构造，将二元结果推广到任意有限字母表。
AI 中文摘要
我们证明了对于任意整数q≥2，具有至多三种不同码字长度的q元免固定码的3/4猜想。每个具有Kraft和至多3/4的给定长度分布都可以通过确定性构造实现。我们引入了一种基于禁止前缀扩展与后缀残差之间重叠的精确恒等式的矩阵方法。自然数值顺序固定最短层，剩余的选择问题通过行和列计数来表达。一个固定基数的插值定理在嵌套端点之间提供每个中间基数的可行集合，前提是它们的差异满足单侧唯一性以及无环性或整数松弛。逆序选择直接处理均匀情况；在其余情况下，层完成和对齐组为插值提供端点。这些方法共同将二元三长度结果推广到任意有限字母表，并给出了构造该码的确定性程序。
英文摘要
We prove the \(3/4\) conjecture for \(q\)-ary fix-free codes with at most three distinct codeword lengths, for every integer \(q\geq2\). Every prescribed length distribution with Kraft sum at most \(3/4\) is realized by a deterministic construction. We introduce a matrix approach based on an exact identity for the overlap between forbidden prefix extensions and suffix residuals. Natural numerical order fixes the shortest layer, and the remaining selection problem is expressed through row and column counts. A fixed-cardinality interpolation theorem supplies feasible sets of every intermediate cardinality between nested endpoints, provided their differences satisfy one-sided uniqueness and either acyclicity or integral slack. Reverse-order selection handles the uniform cases directly; in the remaining cases, layer completion and aligned groups provide endpoints for interpolation. Together, these methods extend the binary three-length result to arbitrary finite alphabets and give a deterministic procedure for constructing the code.
Comments33 pages, 4 figures, 1 table
