Preliminaries. Let $n,m,i,j,p,c \in \mathbb{N}$ with $n,m,i,j,p,c \geq 1$. Let our alphabet be $\{0,1\}$, with non-empty languages $ L_i \subseteq \Sigma^n$ and $ L'_i \subseteq \Sigma^m$. The other preliminaries are the same as a previous question:

We follow the standard definition for deterministic finite-state automata except that we allow the state-transition function $\delta$ to be a partial function. In other words, an FSM has a finite number of states with transitions between them. We define the depth of a state $s$ as the length of the shortest path from the start state (at depth zero) to $s$.

A state $q$ is considered accessible if there is a path from the start state to $q$. A state $q$ is called co-accessible if there is a path from $q$ to a final state. Finally, an automaton is called trim if all its states are both accessible and co-accessible. This is defined here.

Question: Consider the minimal trim deterministic finite-state automaton $A$ for the language

$$\cup_{i=1}^{p} L_i \circ L'_i$$

We observe that this language is finite. Can we conclude that the number of states in $A$ at level $n$ is $p$? We are given three additional constraints:

  1. All of the $L'_i$ are mutually disjoint, so for $i \neq j$, we have $L'_i \cap L'_j = \emptyset$.
  2. Same for the $L_i$: for $i \neq j$, $L_i \cap L_j = \emptyset$.
  3. The upper limit $p$ is polynomially bounded by $n$: $p = O(n^c)$ for some fixed $c$.

Argument: By Myhill-Nerode, the only distinguishing extensions come from the $L'_i$, and there are only $p$ mutually disjoint sets of those, so there are only $p$ states at level $n$ of the automaton. All strings not in any of the $L_i$ will land in the sink state, which is trimmed out of the minimal trim automaton.


1 Answer 1


Yes, the number of states in $A$ at level $n$ is $p$, even without the 3rd constraint. The following is a proof.

Assume $A$ is constructed as in this answer.

As observed in question, it is straightforward to verify that each $L_i$ is an equivalence class of the Nerode congruence for the given language $L=\cup_{i=1}^{p} L_i \circ L'_i$. Recall that for each equivalence class $e$, the label of any path from the initial state to state $q_e$ is in $e$. Since $A$ is a trim automaton for $L$, all paths from the initial state of length $n$ end at $q_{L_i}$ for some $i$.

It is enough to prove that $q_{L_i}$ is of depth $n$ for all $i$.

Fix an arbitrary $i$.

Let $f\in L_i$ and $g\in L_i'$. There is a path from the initial state to $q_{L_i}$ with label $f$ as well as a path from $q_{L_i}$ to an accept state with label $g$. The former path implies the depth of $q_{L_i}$ is at most $n$.

Towards a contradiction, suppose the depth of $q_{L_i}$ is $<n$, i.e. there is a path of length $<n$ from the initial state to $q_{L_i}$. Let $h$ be the label of this path.

Both $fg$ and $hg$ are in $L$ since each of them is the label of a path from the initial state to an accept state. The length of $fg$ is different from that of $hg$. However, all strings in $L$ are of the same length. This contradiction completes our proof.

  • $\begingroup$ Thank you so much! $\endgroup$
    – ShyPerson
    May 11, 2023 at 3:02
  • 1
    $\begingroup$ You are welcome! $\endgroup$
    – John L.
    May 11, 2023 at 3:06

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