Graph-walking automata (GWA) traverse graphs by moving between the nodes following the edges, using a finite-state control to decide where to go next. It is known that every GWA can be transformed to a GWA that halts on every input, to a GWA returning to the initial node in order to accept, and to a reversible GWA. This paper establishes lower bounds on the state blow-up of these transformations, as well as closely matching upper bounds. It is shown that making an $n$-state GWA traversing $k$-ary graphs halt on every input requires at most $2nk+1$ states and at least $2(n-1)(k-3)$ states in the worst case; making a GWA return to the initial node before acceptance takes at most $2nk+n$ and at least $2(n-1)(k-3)$ states in the worst case; Automata satisfying both properties at once have at most $4nk+1$ and at least $4(n-1)(k-3)$ states in the worst case. Reversible automata have at most $4nk+1$ and at least $4(n-1)(k-3)-1$ states in the worst case.

中文翻译：

---

停止，返回和可逆图行走自动机的状态复杂性

图形行走自动机（GWA）通过使用有限状态控制来决定下一步去向，在沿着边缘的节点之间移动来遍历图形。众所周知，每个GWA都可以转换为在每个输入上停止的GWA，还可以转换为返回到初始节点以便接受的GWA，还可以转换为可逆GWA。本文确定了这些变换的状态爆炸的下限，以及紧密匹配的上限。结果表明，在每个输入上使$ n $状态的GWA遍历$ k $ -ary图时，最多需要$ 2nk + 1 $个状态和至少$ 2（n-1）（k-3）$个状态。最坏的情况下; 在最坏的情况下，在接受之前，GWA返回初始节点最多需要$ 2nk + n $，并且至少需要$ 2（n-1）（k-3）$个状态；同时满足这两个属性的自动机在最坏的情况下最多具有$ 4nk + 1 $和至少$ 4（n-1）（k-3）$个状态。在最坏的情况下，可逆自动机的状态最多为$ 4nk + 1 $，状态至少为$ 4（n-1）（k-3）-1 $。