arXiv 0812.10612008-12cs.FLcs.CC

多字母量子有限自动机等价性的判定

Decidability of the Equivalence of Multi-Letter Quantum Finite Automata

Daowen Qiu, Xiangfu Zou, Lvzhou Li, Paulo Mateus

备注：: 18 pages; this is a further revised version

摘要：

多字母{ \it量子有限自动机}（QFA）是经典{ \it单向多头有限自动机}的量子变体（J. Hromkovietum，Acta Informatica 19（1983）377-384），并且已经表明，这种新的单向QFA（多字母QFA）可以毫无错误地接受某些常规语言$(a+b)^{*}b$，这些常规语言是先前的单向QFA所不能接受的。本文研究了多字母QFA等价性的判定问题，主要技术贡献如下：（1）我们证明了在同一个输入字母表$Σ$上的任意两个自动机，一个$k_{1}$ -字母QFA ${\cal A}_1$和一个$k_{2}$ -字母QFA ${\cal A}_2$是等价的当且仅当它们是$(n^2m^{k-1}-m^{k-1}+k)$ -等价的，其中$m=|Σ|$是$Σ$的基数，$k=\max(k_{1},k_{2})$，和$n=n_{1}+n_{2}$，其中$n_{1}$和$n_{2}$分别是${\cal A}_{1}$和${\cal A}_{2}$的状态数。当$k=1$时，我们得到了文献中一次测量QFA等价性的可判定性。值得一提的是，我们的技术方法本质上不同于单输入字母情况下的可判定性（即，$m=1$）。(2)然而，如果我们通过检查长度不超过$ n^2m^{k-1}-m^{k-1}+k$的所有字符串来确定多字母QFA的等价性，那么最差的时间复杂度是指数的，即，$O(n^6m^{n^2m^{k-1}-m^{k-1}+2k-1})$。因此，我们设计了一个多项式时间的$O(m^{2k-1}n^{8}+km^kn^{6})$算法来确定任何两个多字母QFA的等价性。这里，时间复杂度与多字母QFA中的状态数有关，并且$k$被认为是常数。

英文摘要：

Multi-letter {\it quantum finite automata} (QFAs) were a quantum variant of classical {\it one-way multi-head finite automata} (J. Hromkovič, Acta Informatica 19 (1983) 377-384), and it has been shown that this new one-way QFAs (multi-letter QFAs) can accept with no error some regular languages $(a+b)^{*}b$ that are unacceptable by the previous one-way QFAs. In this paper, we study the decidability of the equivalence of multi-letter QFAs, and the main technical contributions are as follows: (1) We show that any two automata, a $k_{1}$-letter QFA ${\cal A}_1$ and a $k_{2}$-letter QFA ${\cal A}_2$, over the same input alphabet $Σ$ are equivalent if and only if they are $(n^2m^{k-1}-m^{k-1}+k)$-equivalent, where $m=|Σ|$ is the cardinality of $Σ$, $k=\max(k_{1},k_{2})$, and $n=n_{1}+n_{2}$, with $n_{1}$ and $n_{2}$ being the numbers of states of ${\cal A}_{1}$ and ${\cal A}_{2}$, respectively. When $k=1$, we obtain the decidability of equivalence of measure-once QFAs in the literature. It is worth mentioning that our technical method is essentially different from that for the decidability of the case of single input alphabet (i.e., $m=1$). (2) However, if we determine the equivalence of multi-letter QFAs by checking all strings of length not more than $ n^2m^{k-1}-m^{k-1}+k$, then the worst time complexity is exponential, i.e., $O(n^6m^{n^2m^{k-1}-m^{k-1}+2k-1})$. Therefore, we design a polynomial-time $O(m^{2k-1}n^{8}+km^kn^{6})$ algorithm for determining the equivalence of any two multi-letter QFAs. Here, the time complexity is concerning the number of states in the multi-letter QFAs, and $k$ is thought of as a constant.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

摘要：

英文摘要：

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