We investigate the complexity of the containment problem “Does \(L(\mathcal {A})\subseteq L({\mathscr{B}})\) hold?” for register automata and timed automata, where \({\mathscr{B}}\) is assumed to be unambiguous and \(\mathcal {A}\) is arbitrary. We prove that the problem is decidable in the case of register automata over \((\mathbb N,=)\), in the case of register automata over \((\mathbb Q,<)\) when \({\mathscr{B}}\) has a single register, and in the case of timed automata when \({\mathscr{B}}\) has a single clock. We give a 2-EXPSPACE algorithm in the first case, whose complexity is a single exponential in the case that \({\mathscr{B}}\) has a bounded number of registers. In the other cases, we give an EXPSPACE algorithm.

我们调查了“ （\（L（\ mathcal {A}）\ subseteq L（{\ mathscr {B}}）\）是否成立？”这个容纳问题的复杂性。对于寄存器自动机和定时自动机，其中\（{\ mathscr {B}} \）假定是明确的，而\（\ mathcal {A} \）是任意的。我们证明，这个问题是在寄存器自动机上的情况下，可判定\（（\ mathbb N，=）\） ，在寄存器自动机上的情况下\（（\ mathbb Q，<）\）时\（{\ mathscr {B}} \）有一个寄存器，如果是定时自动机，则\（{\ mathscr {B}} \）有一个时钟。在第一种情况下，我们给出了2-EXPSPACE算法，在\（{\ mathscr {B}} \）的情况下，其复杂度为单指数有一定数量的寄存器。在其他情况下，我们给出一个EXPSPACE算法。