AI 中文总结
本文证明无限制静态输出反馈镇定问题在单输入多输出情形下是NP难的，通过从Betweenness问题多项式时间归约构造整数矩阵，并利用Routh行列式与频率分离技术实现，同时得出仿射参数化有理多项式族的Hurwitz成员判定也是NP难的。
AI 中文摘要
我们证明无限制的静态输出反馈镇定是NP难的，即使在单输入多输出情形下也是如此。更精确地，我们给出了从NP完全的Betweenness问题到该问题的多项式时间多一归约。给定一个包含$n$个元素和$m$个Betweenness约束的实例，该归约构造整数矩阵$A\in\mathbb Z^{D\times D}$，$B\in\mathbb Z^{D\times1}$和$C\in\mathbb Z^{n\times D}$，使得该实例可满足当且仅当存在一个无限制的实行向量$K\in\mathbb{R}^{1\times n}$使得$A+BKC$是Hurwitz的。状态维数为$D=6(m+n)$，且每个矩阵条目在输入规模下具有对数比特长度。该归约使用一个通用的三次源多项式，其Routh行列式强制实施编码Betweenness的符号条件，而固定的锚定值提供了一种排除任意大幅值的虚假镇定增益的机制。然后，频率分离将同时的源条件压缩为一个公共的仿射多项式，该多项式直接实现为$A+BKC$的特征多项式。作为副产品，判定一个仿射参数化的有理多项式族在有理盒上是否包含一个Hurwitz成员是NP难的。
英文摘要
We prove that unrestricted static output-feedback stabilization is NP-hard, already in the single-input multiple-output case. More precisely, we give a polynomial-time many-one reduction from the NP-complete Betweenness problem. Given an instance with $n$ elements and $m$ betweenness constraints, the reduction constructs integer matrices $A\in\mathbb Z^{D\times D}$, $B\in\mathbb Z^{D\times1}$, and $C\in\mathbb Z^{n\times D}$ such that the instance is satisfiable if and only if there exists an unrestricted real row vector $K\in\mathbb{R}^{1\times n}$ for which $A+BKC$ is Hurwitz. The state dimension is $D=6(m+n)$, and every matrix entry has logarithmic bit length in the input size. The reduction uses one universal cubic source polynomial whose Routh determinant enforces the sign condition encoding betweenness, while fixed anchor values provide a mechanism for excluding spurious stabilizing gains of arbitrarily large magnitude. Frequency separation then compresses the simultaneous source conditions into one common affine polynomial, which is realized directly as the characteristic polynomial of $A+BKC$. As a by-product, deciding whether an affinely parameterized rational polynomial family contains a Hurwitz member on a rational box is NP-hard.
CommentsSubmitted
