State Complexity of Binary Coded Regular Languages

Abstract

For the given non-unary input alphabet $\varSigma$, a maximal prefix code h mapping strings over $\varSigma$ to binary strings, and an optimal deterministic finite automaton (DFA) $\mathcal {A}$ with n states recognizing a language $\mathcal {L}$ over $\varSigma$, we consider the problem of how many states we need for an automaton $\mathcal {A}'$ that decides membership in $h(\mathcal {L})$, the binary coded version of $\mathcal {L}$. Namely, $\mathcal {A}'$ accepts binary inputs belonging to $h(\mathcal {L})$ and rejects binary inputs belonging to $h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}})$, where $\mathcal {L}^{\scriptscriptstyle \mathrm {C}}$ is the complement of $\mathcal {L}$. The outcome on inputs that are not valid binary codes for any string in $\varSigma ^{*}$ can be arbitrary: $\mathcal {A}'$ may accept, reject, or halt in a “don’t care” state. We show that any optimal deterministic don’t care finite automaton (dcDFA) $\mathcal {A}'$ solving this promise problem uses at most $(\Vert {\varSigma }\Vert -1){\cdot }n$ states but at least n states. We also show that, for each non-unary input alphabet $\varSigma$, there exists a maximal binary prefix code h such that, for each $n\ge 2$ and for each N in range from n to $(\Vert {\varSigma }\Vert -1){\cdot }n$, there exists a language $\mathcal {L}$ over $\varSigma$ such that the optimal DFA recognizing $\mathcal {L}$ uses exactly n states and any optimal dcDFA for solving the above promise problem uses exactly N states. Thus, we have the complete state hierarchy for deciding membership in the binary coded version of $\mathcal {L}$, with no magic numbers in between the lower and upper bounds.

Supported by the Slovak grant contract VEGA 1/0177/21.

1 Introduction

One of the earliest results in automata theory is the subset construction [16]: every n-state nondeterministic finite automaton (NFA) can be replaced by an equivalent deterministic finite automaton (DFA) using at most $2^n$ states. This raised later the question of whether it is possible, for a given number n, to find some $N\in \{n,\ldots ,2^n\}$ such that there is no optimal DFA with exactly N states, equivalent to some optimal NFA with exactly n states [10]; such numbers were named “magic”. The problem was solved in [11], showing that there are no magic numbers for ternary languages, contrary to the unary languages [5]. Since then, the magic numbers were studied for language operations, e.g., in [9], it was shown that, for the intersection of two languages, given by two DFAs with n and m states, we have no magic numbers in $\{1,\ldots ,n{\cdot }m\}$. Such state hierarchies were studied for other operations as well [4, 7, 9, 11].

From a different starting point, we are going to land in yet another complete state hierarchy with no magic numbers. Our initial motivation was the fact that most present-day computers store data in a binary coded form. This raises the following natural question: given a standard DFA $\mathcal {A}$ with n states for a regular language $\mathcal {L}$ over an input alphabet $\varSigma$, how many states we need to recognize $h(\mathcal {L})$, the binary coded version of $\mathcal {L}$? Clearly, the answer depends also on $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$, the binary code in use. In most cases, we can work with the assumption that the code is a homomorphism, such that $h(\alpha _1)= h(\alpha _2)$ implies $\alpha _1= \alpha _2$, so that each encoded string can be unambiguously decoded back. Since it is well known that regular languages are closed under any homomorphism (not necessarily a code—see e.g. [8, Sect. 4.2.3]), the situation seems clear at first glance:¹ construct an optimal DFA for $h(\mathcal {L})$.

However, if the automaton for $h(\mathcal {L})$ receives only inputs that are valid binary images of strings in $\varSigma ^{*}$, the outcome on inputs that are not valid images can be quite arbitrary, which allows us to save some states. This brings us to a modified problem: given a code $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ and a standard DFA $\mathcal {A}$ with n states for $\mathcal {L}\subseteq \varSigma ^{*}$, how many states we need for an automaton $\mathcal {A}'$ that accepts each $\beta \in h(\mathcal {L})$ and rejects each $\beta \in h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}})$? Here $\mathcal {L}^{\scriptscriptstyle \mathrm {C}}$ denotes the complement of $\mathcal {L}$.

This approach is not completely new: in general, we are given a pair of disjoint languages $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$ over the same alphabet $\varSigma$, called a promise problem, and we decide whether $w\in \mathcal {L}^{{\scriptscriptstyle \oplus }}$ or $w\in \mathcal {L}^{{\scriptscriptstyle \ominus }}$ by the use of a don’t care deterministic finite automaton (dcDFA) which, besides accepting and rejecting states, may also use neutral or “don’t care” states, otherwise it behaves like a standard DFA (see e.g. [6, 13]). In our settings, $\mathcal {L}^{{\scriptscriptstyle \oplus }}= h(\mathcal {L})$ and $\mathcal {L}^{{\scriptscriptstyle \ominus }}= h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}})$, where $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ is a code. We shall concentrate on the most common binary codes used in practice, that allow decoding by one-way deterministic finite-state transducers in real time and minimize $\sum _{a\in \varSigma }|{h(a)}|$, the sum of lengths of codewords. Such codes are called maximal prefix codes in literature [1, 3]. (See Definition 1.)

This paper shows that, for each maximal prefix code $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ and each optimal DFA $\mathcal {A}$ with n states recognizing some $\mathcal {L}$ over the alphabet $\varSigma$, the binary promise problem $\langle {}h(\mathcal {L}),h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ can be solved by a dcDFA $\mathcal {A}'$ using at most $(\Vert {\varSigma }\Vert -1){\cdot }n$ states, but at least n states.² We also show that, for each non-unary input alphabet $\varSigma$, there exists a maximal binary prefix code h such that, for each $n\ge 2$ and each $N\in \{n,\ldots ,(\Vert {\varSigma }\Vert -1){\cdot }n\}$, there exists a language $\mathcal {L}\subseteq \varSigma ^{*}$ such that the optimal DFA recognizing $\mathcal {L}$ uses exactly n states and any optimal dcDFA for solving $\langle {}h(\mathcal {L}), h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ uses exactly N states.

2 Preliminaries

Here we shall fix some basic definitions, notation, and preliminary properties. For more details, we refer the reader to [6, 8], or any other standard textbooks.

Definition 1

A homomorphism between strings over two alphabets is a mapping $h\,{:}\,\varSigma _1^{*}{\rightarrow }\varSigma _2^{*}$ preserving concatenation, i.e., $h(\alpha _1{\cdot }\alpha _2)= h(\alpha _1){\cdot }h(\alpha _2)$, for each $\alpha _1,\alpha _2\in \varSigma _1^{*}$. The image of a language $L\subseteq \varSigma _1^{*}$ is $h(L)= \{h(\alpha )\,{:}\,\alpha \in L\}\subseteq \varSigma _2^{*}$.

If $h(\alpha _1)= h(\alpha _2)$ implies that $\alpha _1= \alpha _2$, then h is called a code.

h is a prefix code, if no string in $h(\varSigma _1)= \{h(a)\,{:}\,a\in \varSigma _1\}$ is a proper prefix of another one. The code h is maximal, if there is no other code $h'\,{:}\,{\varSigma '_1}^{*}\,{\rightarrow }\,\varSigma _2^{*}$ (for some $\varSigma '_1\supseteq \varSigma _1$) such that $h'(\varSigma '_1)$ is a proper superset of $h(\varSigma _1)$.

Each homomorphism h is completely determined by the strings in $h(\varSigma _1)$, since $h(a_1{\cdots }a_{\ell })= h(a_1){\cdots }h(a_{\ell })$, for each $a_1,\ldots ,a_{\ell }\in \varSigma _1$. In addition, if h is a code, each $\beta \in h(\varSigma _1^{*})$ has a unique factorization into $\beta =\beta _1{\cdots }\beta _{\ell }$, where $\ell \ge 0$ and $\beta _1,\ldots ,\beta _{\ell }\in h(\varSigma _1)$. For examples of codes, see Fig. 1.

Fig. 1.

Examples of homomorphisms establishing some binary prefix codes. Each homomorphism is displayed as a tree in which each leaf represents some $a_i$, a letter of the original input alphabet; the edges are labeled so that the path from the root to $a_i$ gives the corresponding string $h(a_i)$. Internal nodes of the tree are related to prefixes of strings in $\{h(a)\,{:}\,a\in \varSigma \}$. The code h (left), defined by $h(a_0)=1$,  $h(a_1)=01$,  $h(a_2)=001$,  $h(a_3)=0001$,  $h(a_4)=00001$, and $h(a_5)=00000$, is maximal, while the code $\tilde{h}$ (right), with $\tilde{h}(a_0)=1$,  $\tilde{h}(a_1)=00011$, and $\tilde{h}(a_2)=00010$, is not—it can be extended, e.g., by defining $\tilde{h}(a_3)=01$.

Prefix codes allow easy decoding by a one-way deterministic finite-state machine such that, for the given $\beta \in h(\varSigma _1^{*})$, it computes the factorization of $\beta$ into $h(a_1){\cdots }h(a_{\ell })$ and prints $a_1{\cdots }a_{\ell }$ on the output in real time [1, Prop. 5.1.6].

Maximal codes minimize $\sum _{a\in \varSigma }|{h(a)}|$ and do not have “gaps” in images: each $\beta \in \varSigma _2^{*}$ can be extended to an image of some $\alpha \in \varSigma _1^{*}$, that is, to $\beta {\cdot }\beta '= h(\alpha )$, for some $\beta '\in \varSigma _2^{*}$ and some $\alpha \in \varSigma _1^{*}$, not excluding $\beta '= \varepsilon$.

Since we shall deal with binary codes only, we are going to simplify notation and write $\varSigma$ instead of $\varSigma _1$ and fix $\varSigma _2= \{0,1\}$.

Definition 2

A don’t care deterministic finite automaton (dcDFA) is a 6-tuple $\mathcal {A}=\langle {}Q,\varSigma , q_{\scriptscriptstyle \mathrm {I}}, f, F^{{\scriptscriptstyle \oplus }},F^{{\scriptscriptstyle \ominus }}{}\rangle$, in which Q is a finite set of states; $\varSigma$ is a finite input alphabet; $q_{\scriptscriptstyle \mathrm {I}}\in Q$ is the initial state; $f:Q\times \varSigma \rightarrow Q$ is a transition function; $F^{{\scriptscriptstyle \oplus }}\subseteq Q$ is the set of accepting states; and $F^{{\scriptscriptstyle \ominus }}\subseteq Q$ the set of rejecting states, $F^{{\scriptscriptstyle \oplus }}\cap F^{{\scriptscriptstyle \ominus }}=$ Ø. The remaining states are called neutral or “don’t care” states.

A (standard) deterministic finite automaton (DFA) is a 5-tuple $\mathcal {A}=\langle {}Q,\varSigma , q_{\scriptscriptstyle \mathrm {I}}, f,F{}\rangle$, with $F\subseteq Q$ denoting the set of accepting states and $Q{\setminus }F$ the set of rejecting states; the remaining components have the same meaning as above.

The transition function f can be extended to $f^{*}\,{:}\,Q\times \varSigma ^{*}\rightarrow Q$ in a natural way, taking by definition $f^{*}(q,\varepsilon )=q$ and $f^{*}(q,\alpha a)=f(f^{*}(q,\alpha ),a)$, for each $q\in Q$, $\alpha \in \varSigma ^{*}$, and $a\in \varSigma$. To simplify notation, $f^{*}(q,\alpha )= q'$ shall sometimes be displayed in a more compact form $q\mathop {\longrightarrow }\limits ^{\alpha }q'$.

A promise problem (see e.g. [6]) is a pair of disjoint languages $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$ over the same alphabet $\varSigma$. The promise problem is solved by a dcDFA $\mathcal {A}$, if $\mathcal {A}$ accepts each $w\in \mathcal {L}^{{\scriptscriptstyle \oplus }}$ (that is, $f^{*}(q_{\scriptscriptstyle \mathrm {I}},w)\in F^{{\scriptscriptstyle \oplus }}$) and rejects each $w\in \mathcal {L}^{{\scriptscriptstyle \ominus }}$ ($f^{*}(q_{\scriptscriptstyle \mathrm {I}},w)\in F^{{\scriptscriptstyle \ominus }}$). We do not have to worry about the outcome on inputs belonging neither to $\mathcal {L}^{{\scriptscriptstyle \oplus }}$ nor to $\mathcal {L}^{{\scriptscriptstyle \ominus }}$: on such inputs, $\mathcal {A}$ may accept, reject, or halt in a neutral state.

$\mathcal {A}$ is optimal for $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$, if it solves $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$ and there is no dcDFA $\mathcal {A}'$ that solves $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$ with fewer states than does $\mathcal {A}$.

If $\mathcal {L}^{{\scriptscriptstyle \oplus }}\cup \mathcal {L}^{{\scriptscriptstyle \ominus }}= \varSigma ^{*}$, then $\mathcal {A}$ has no neutral reachable states and can be viewed as a standard DFA; we have the standard language recognition and say that $\mathcal {L}^{{\scriptscriptstyle \oplus }}$ is recognized by $\mathcal {A}$. The language $\mathcal {L}^{{\scriptscriptstyle \oplus }}$ is then usually denoted by $\mathcal {L}(\mathcal {A})$ and its complement $\mathcal {L}^{{\scriptscriptstyle \ominus }}$ by $\mathcal {L}(\mathcal {A})^{\scriptscriptstyle \mathrm {C}}$. In this case, the concept of optimal dcDFA coincide with the concept of minimal DFA for $\mathcal {L}^{{\scriptscriptstyle \oplus }}$.

Note that if a promise problem $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$ can be solved by a dcDFA $\mathcal {A}$ with n states, it can also be solved by a standard dcDFA $\mathcal {A}'$ with n states, none of which is neutral: any neutral state could be set as accepting or rejecting, without affecting $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$. This leaves us some degree of freedom, leading to different machines. Namely, if $\mathcal {A}$ uses k neutral reachable states, we obtain $2^k$ different automata solving the same promise problem, all of them of size at most n. These automata do not agree in acceptance/rejection on inputs not belonging to $\mathcal {L}^{{\scriptscriptstyle \oplus }}\cup \mathcal {L}^{{\scriptscriptstyle \ominus }}$. Thus, the given dcDFA can also be viewed as a more concise template representing these $2^k$ DFAs.

This is related to the following separation problem: given DFAs $\mathcal {A}^{{\scriptscriptstyle \oplus }}$ and $\mathcal {A}^{{\scriptscriptstyle \ominus }}$ for two disjoint languages $\mathcal {L}^{{\scriptscriptstyle \oplus }}$ and $\mathcal {L}^{{\scriptscriptstyle \ominus }}$, find a DFA $\mathcal {A}'$ with minimal number of states, such that $\mathcal {L}^{{\scriptscriptstyle \oplus }}\subseteq \mathcal {L}(\mathcal {A}')$ and $\mathcal {L}^{{\scriptscriptstyle \ominus }}\subseteq \mathcal {L}(\mathcal {A}')^{\scriptscriptstyle \mathrm {C}}$. In general, this problem is NP-complete [13, Thm. 9]. This was shown by a simple application of NP-completeness for a slightly different computational model (in which some transitions f(q, a) may be undefined), presented in [14, 15].

The next theorem will play the same role for don’t care automata as the fooling set technique [2] for standard automata:

Theorem 1

Let $\mathcal {L}^{{\scriptscriptstyle \oplus }}$, $\mathcal {L}^{{\scriptscriptstyle \ominus }}$ be two disjoint languages over the same alphabet $\varSigma$. Suppose there exist m-tuple $X= \langle {}x_{e}{}\rangle _{e=1}^{m}$ and ${m\atopwithdelims ()2}$-tuple $Y= \langle {}y_{e,\tilde{e}}{}\rangle _{e,\tilde{e}=1,\,e<\tilde{e}}^{m}$ consisting of strings in $\varSigma ^{*}$ such that, for each $e,\tilde{e}\in \{1,\ldots ,m\}$, $e<\tilde{e}$,

(I):

both $x_{e}{\cdot }y_{e,\tilde{e}}$ and $x_{\tilde{e}}{\cdot }y_{e,\tilde{e}}$ are in $\mathcal {L}^{{\scriptscriptstyle \oplus }}\cup \mathcal {L}^{{\scriptscriptstyle \ominus }}$,

(II):

$x_{e}{\cdot }y_{e,\tilde{e}}\in \mathcal {L}^{{\scriptscriptstyle \oplus }}$ if and only if $x_{\tilde{e}}{\cdot }y_{e,\tilde{e}}\in \mathcal {L}^{{\scriptscriptstyle \ominus }}$.

Then any dcDFA solving the promise problem $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$ uses at least m states.

Proof

Let $\mathcal {A}=\langle {}Q,\varSigma , q_{\scriptscriptstyle \mathrm {I}}, f, F^{{\scriptscriptstyle \oplus }},F^{{\scriptscriptstyle \ominus }}{}\rangle$, satisfying $F^{{\scriptscriptstyle \oplus }}\cap F^{{\scriptscriptstyle \ominus }}=$ Ø, be an arbitrary dcDFA for solving $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$. Suppose that $\mathcal {A}$ uses less than m states.

Now, for the given m-tuple $X= \langle {}x_{1},x_{2},\ldots ,x_{m}{}\rangle$, let $q_{1},q_{2},\ldots ,q_{m}$ denote the respective states reached by $\mathcal {A}$ on these inputs, that is, $q_{\scriptscriptstyle \mathrm {I}}\mathop {\longrightarrow }\limits ^{x_e}q_{e}$, for each $e\in \{1,\ldots ,m\}$. But then, using a pigeonhole argument, some state must be repeated, i.e., we have $q_{e}=q_{\tilde{e}}$, for some $1\le e<\tilde{e}\le m$. This implies that the computations on inputs $x_{e}{\cdot }y_{e,\tilde{e}}$ and $x_{\tilde{e}}{\cdot }y_{e,\tilde{e}}$ must end in the same state, denoted here by r. That is, we have the following computations: $q_{\scriptscriptstyle \mathrm {I}}\mathop {\longrightarrow }\limits ^{x_e}q_{e}\mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{y_{e,\tilde{e}}}r$ and $q_{\scriptscriptstyle \mathrm {I}}\mathop {\longrightarrow }\limits ^{x_{\tilde{e}}}q_{\tilde{e}}=q_{e}\mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{y_{e,\tilde{e}}}r$. There are now two possibilities:

First, let $x_{e}{\cdot }y_{e,\tilde{e}}\in \mathcal {L}^{{\scriptscriptstyle \oplus }}$. Then, using (II), we also have that $x_{\tilde{e}}{\cdot }y_{e,\tilde{e}}\in \mathcal {L}^{{\scriptscriptstyle \ominus }}$. This implies that the computation on $x_{e}{\cdot }y_{e,\tilde{e}}$ ends in $r\in F^{{\scriptscriptstyle \oplus }}$ and, at the same time, the computation on $x_{\tilde{e}}{\cdot }y_{e,\tilde{e}}$ in $r\in F^{{\scriptscriptstyle \ominus }}$. But this is a contradiction: $F^{{\scriptscriptstyle \oplus }}\cap F^{{\scriptscriptstyle \ominus }}=$ Ø.

Second, let $x_{e}{\cdot }y_{e,\tilde{e}}\notin \mathcal {L}^{{\scriptscriptstyle \oplus }}$. Then, using (II), we have $x_{\tilde{e}}{\cdot }y_{e,\tilde{e}}\notin \mathcal {L}^{{\scriptscriptstyle \ominus }}$. Now, by (I), we see that $x_{e}{\cdot }y_{e,\tilde{e}}\in \mathcal {L}^{{\scriptscriptstyle \ominus }}$ and $x_{\tilde{e}}{\cdot }y_{e,\tilde{e}}\in \mathcal {L}^{{\scriptscriptstyle \oplus }}$. This leads to the same kind of contradiction as above, swapping the roles of $x_{e}{\cdot }y_{e,\tilde{e}}$ and $x_{\tilde{e}}{\cdot }y_{e,\tilde{e}}$.    $\square$

Note that, apart from providing the lower bound on the number of states, the statement of the above theorem does not deal with states in any dcDFA solving $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$ but, rather, with strings in $\varSigma ^{*}$. However, if there does exist a dcDFA $\mathcal {A}$ with m states for $\langle {}\mathcal {L}^{{\scriptscriptstyle \oplus }},\mathcal {L}^{{\scriptscriptstyle \ominus }}{}\rangle$, that is, if the lower bound provided by Theorem 1 matches the upper bound, then one can establish a one-to-one correspondence between states in $\mathcal {A}$ and strings in $X= \langle {}x_{1},x_{2},\ldots ,x_{m}{}\rangle$, with $Y= \langle {}y_{e,\tilde{e}}{}\rangle _{e,\tilde{e}=1,\,e<\tilde{e}}^{m}$ giving the pairwise distinguishability of states in $\mathcal {A}$. The standard fooling set technique (for DFAs) uses $y_{e,{\tilde{e}_1}}= y_{e,{\tilde{e}_2}}= \ldots ,$ with the condition (I) satisfied automatically.

3 Upper and Lower Bounds

We are now going to show that $(\Vert {\varSigma }\Vert -1){\cdot }n$ states are sufficient but n states necessary for a dcDFA that decides whether the given binary input is in $h(\mathcal {L})$ or in $h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}})$, that is, for a dcDFA solving $\langle {}h(\mathcal {L}),h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$, the binary promise-problem version of $\mathcal {L}$. Here $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ is a maximal prefix code and $\mathcal {A}$ is an optimal DFA with n states, recognizing a language $\mathcal {L}$ over a non-unary alphabet $\varSigma$.

For the given code h, let us begin by fixing some additional notation for the images of letters and proper prefixes:

$$\begin{aligned} \begin{array}{rcll} H &{}=&{} \{h(a)\,{:}\,a\in \varSigma \}, \\ P &{}=&{} \{\pi \,{:}\,\pi {\cdot }\beta \in H,~\text {for some}~\beta \in \{0,1\}^{+}\}. \end{array} \end{aligned}$$

(1)

Recall that h is a maximal prefix code. Thus, P includes the empty string $\varepsilon$, but no string from H. Next, if $\pi \in P$, then $\pi {\cdot }0,\pi {\cdot }1\in P\cup H$ (see also Fig. 1). The next two theorems provide the upper and lower bounds.

Theorem 2

Let $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ be a maximal binary prefix code and let $\mathcal {L}$ be a language over the non-unary alphabet $\varSigma$. Then, if $\mathcal {L}$ can be recognized by a DFA $\mathcal {A}=\langle {}Q,\varSigma , q_{\scriptscriptstyle \mathrm {I}}, f,F{}\rangle$ with n states, the binary promise problem $\langle {}h(\mathcal {L}),h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ can be solved by a dcDFA $\mathcal {A}'$ with at most $n'\le (\Vert {\varSigma }\Vert -1){\cdot }n$ states.

Proof

(a sketch). The idea of the construction is to remember $q\in Q$, the current state of $\mathcal {A}$ at the moment when $\mathcal {A}$ has read $a_1{\cdots }a_{\ell }\in \varSigma ^{*}$, and $\pi \in P$, the prefix of a code for the next input symbol $a_{\ell +1}$, not completed yet. This leads to $Q'= Q{\times }P$, with $q'_{\scriptscriptstyle \mathrm {I}}= \langle {}q_{\scriptscriptstyle \mathrm {I}},\varepsilon {}\rangle$, $F^{{\scriptscriptstyle \oplus }}= F{\times }\{\varepsilon \}$, and $F^{{\scriptscriptstyle \ominus }}= (Q{\setminus }F){\times }\{\varepsilon \}$. Transitions in $\mathcal {A}'$ are defined as follows, for each $q\in Q$, $\pi \in P$, and $b\in \{0,1\}$:

(I):

$f'(\langle {}q,\pi {}\rangle ,b)= \langle {}q,\pi b{}\rangle$, provided that $\pi {\cdot }b\in P$,

(II):

$f'(\langle {}q,\pi {}\rangle ,b)= \langle {}f(q,a),\varepsilon {}\rangle$, provided that, for some $a\in \varSigma$,  $\pi {\cdot }b= h(a)\in H$.

   $\square$

The above construction can be updated so that it works for prefix codes that are not maximal. Then $\pi \in P$ does not imply that $\pi {\cdot }b$ is in $P\cup H$. In such cases, we can define $f'(\langle {}q,\pi {}\rangle ,b)= q'_{\scriptscriptstyle \mathrm {E}}$, where $q'_{\scriptscriptstyle \mathrm {E}}$ is an additional trap state, in which we scan the rest of the input—the input can no longer be extended to a string in $h(\varSigma _1^{*})= h(H^{*})$. However, for such codes, $\Vert {P}\Vert$ is not bounded by $\Vert {\varSigma }\Vert -1$.

Theorem 3

Let $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ be a binary prefix code (not necessarily maximal) and let $\mathcal {L}$ be a language over the alphabet $\varSigma$ (not necessarily non-unary). Then, if the binary promise problem $\langle {}h(\mathcal {L}),h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ can be solved by a dcDFA $\mathcal {A}'= \langle {}Q',\{0,1\}, q'_{\scriptscriptstyle \mathrm {I}}, f', F^{{\scriptscriptstyle \oplus }},F^{{\scriptscriptstyle \ominus }}{}\rangle$ with $n'$ states, the language $\mathcal {L}$ can be recognized by a standard DFA $\mathcal {A}= \langle {}Q,\varSigma , q_{\scriptscriptstyle \mathrm {I}}, f,F{}\rangle$ with at most $n\le n'$ states.

Proof

If, for some $q,q'\in Q'$ and $a\in \varSigma$, the original $\mathcal {A}'$ has a path beginning in q, ending in $q'$, and reading from the input the string $h(a)\in H$ (introduced by (1)), we shall add the transition $q\mathop {\longrightarrow }\limits ^{a}q'$ to $\mathcal {A}$. Recall that $\mathcal {A}'$ is deterministic and $h(a)= b_1{\cdots }b_m$ is unique for each $a\in \varSigma$. But then $q'= {f'}^{*}(q,h(a))$ is also unique for each $q\in Q'$ and each $a\in \varSigma$, and hence $\mathcal {A}$ will be deterministic.

In addition, we can restrict the set of finite control states in $\mathcal {A}$ to states that can be reached from $q'_{\scriptscriptstyle \mathrm {I}}$ by reading some $\beta \in h(\varSigma ^{*})= H^{*}$, that is, to

$$\begin{aligned} \begin{array}{rcll} R= & {} \{{f'}^{*}(q'_{\scriptscriptstyle \mathrm {I}},\beta )\,{:}\,\beta \in H^{*}\}. \end{array} \end{aligned}$$

(2)

Thus, $Q= R$,  $q_{\scriptscriptstyle \mathrm {I}}= q'_{\scriptscriptstyle \mathrm {I}}$, and $F= R\cap F^{{\scriptscriptstyle \oplus }}$. Clearly, $R\subseteq F^{{\scriptscriptstyle \oplus }}\cup F^{{\scriptscriptstyle \ominus }}$, since no $\mathcal {A}'$ solving $\langle {}h(\mathcal {L}),h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ halts in a neutral state on any $h(\alpha )\in h(\varSigma ^{*})$. Finally, let

• $f(q,a)= {f'}^{*}(q,h(a))$, for each $q\in R$ and each $a\in \varSigma$.

It is easy to see, for each $q,q'\in R$ and each $a\in \varSigma$, that $\mathcal {A}$ has a transition from q to $q'$ reading the symbol $a\in \varSigma$ if and only if $\mathcal {A}'$ has a path connecting the same states and reading the string $h(a)\in H$. By a straightforward induction on the length of $\alpha \in \varSigma ^{*}$, we have $q\mathop {\longrightarrow }\limits ^{\alpha }q'$ in $\mathcal {A}$ if and only if $q\mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{h(\alpha )}q'$ in $\mathcal {A}'$.

Thus, if $\alpha \in \mathcal {L}$, then $h(\alpha )\in h(\mathcal {L})$ must be accepted by dcDFA $\mathcal {A}'$, which gives $q'_{\scriptscriptstyle \mathrm {I}}\mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{h(\alpha )}q'$ for some $q'\in F^{{\scriptscriptstyle \oplus }}$. Moreover, since $h(\alpha )\in H^{*}$,  $q'$ must also belong to R. But then, for $\mathcal {A}$, we have $q_{\scriptscriptstyle \mathrm {I}}= q'_{\scriptscriptstyle \mathrm {I}}\mathop {\longrightarrow }\limits ^{\alpha }q'$ with $q'\in R\cap F^{{\scriptscriptstyle \oplus }}= F$, and hence $\alpha$ is accepted by $\mathcal {A}$. Similarly, if $\alpha \notin \mathcal {L}$, then $\mathcal {A}'$ has a path $q'_{\scriptscriptstyle \mathrm {I}}\mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{h(\alpha )}q'$ ending in $q'\in R\cap F^{{\scriptscriptstyle \ominus }}$. For $\mathcal {A}$, this gives $q_{\scriptscriptstyle \mathrm {I}}= q'_{\scriptscriptstyle \mathrm {I}}\mathop {\longrightarrow }\limits ^{\alpha }q'$ ending in $q'\in R{\setminus }F^{{\scriptscriptstyle \oplus }}= R{\setminus }(R\cap F^{{\scriptscriptstyle \oplus }})= Q{\setminus }F$, and hence $\alpha$ is rejected by $\mathcal {A}$.

Summing up, $\mathcal {A}$ is a standard DFA recognizing $\mathcal {L}$, with $n\le n'$ states.    $\square$

By combining Theorems 2 and 3, we get:

Theorem 4

Let $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ be a maximal binary prefix code and let $\mathcal {L}$ be a language over the non-unary alphabet $\varSigma$. Then, if the optimal DFA $\mathcal {A}$ recognizing $\mathcal {L}$ uses n states, any optimal dcDFA $\mathcal {A}'$ solving the binary promise problem $\langle {}h(\mathcal {L}),h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ uses at least n states and at most $(\Vert {\varSigma }\Vert -1){\cdot }n$ states.

Proof

The upper bound is a direct consequence of Theorem 2: this theorem does not necessarily produce a dcDFA that is optimal, however, it does guarantee $(\Vert {\varSigma }\Vert -1){\cdot }n$ states for $\mathcal {A}'$. The lower bound follows from Theorem 3: suppose that $\langle {}h(\mathcal {L}),h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ can be solved by $\mathcal {A}'$ with $n'<n$ states. But then, by this theorem, we could obtain a standard DFA recognizing $\mathcal {L}$ with fewer states than n, which is a contradiction, since $\mathcal {A}$ is optimal.    $\square$

4 The Hierarchy

We are now ready to establish the complete state hierarchy and provide a witness language for each N between n and $(\Vert {\varSigma }\Vert -1){\cdot }n$. First, for the given

$$\begin{array}{rcll}\varSigma= & {} \{a_0,a_1,\ldots ,a_d\},&\text {where}~d= \Vert {\varSigma }\Vert -1\ge 2, \end{array}$$

define a maximal binary prefix code $h_{{\scriptscriptstyle \varSigma }}\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ as follows:

$$\begin{aligned} \begin{array}{rcll} h_{{\scriptscriptstyle \varSigma }}(a_j) &{}=&{} 0^{j}1, &{}\text {for}~j\in \{0,\ldots ,d-1\}, \\ h_{{\scriptscriptstyle \varSigma }}(a_d) &{}=&{} 0^{d}. \end{array} \end{aligned}$$

(3)

Second, for the given $\varSigma$ and any given

$$\begin{aligned} \begin{array}{rcll}&n\ge 2,\ \ g\in \{0,\ldots ,n\},\ \ k\in \{0,\ldots ,d-2\} , \end{array} \end{aligned}$$

(4)

define a DFA $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}= \langle {}Q,\varSigma , q_{\scriptscriptstyle \mathrm {I}}, f,F{}\rangle$ with the state set $Q= \{0,\ldots ,n-1\}$, $q_{\scriptscriptstyle \mathrm {I}}=0$,  $F=\{0\}$, and the following transitions:

$$\begin{aligned} \begin{array}{rcll} f(i,a) &{}=&{} (i+1){\,\bmod \,}n, &{}\quad \text {if}~i<g~\text {and}~a\in \varSigma , \\ f(g,a) &{}=&{} (g+1){\,\bmod \,}n, &{}\quad \text {if}~g\le n-2~\text {and}~a\in \{a_0,\ldots ,a_k\} \cup \{a_d\}, \\ &{}=&{} (g+2){\,\bmod \,}n, &{}\quad \text {if}~g\le n-2~\text {and}~a\in \{a_{k+1},\ldots ,a_{d-1}\}, \\ f(i,a) &{}=&{} (i+1){\,\bmod \,}n, &{}\quad \text {if}~i>g~\text {and}~a\in \varSigma {\setminus }\{a_d\}, \\ &{}=&{} i, &{}\quad \text {if}~i>g~\text {and}~a=a_d. \end{array} \end{aligned}$$

(5)

There are two special cases. The first one is $g= n-1$, with no states $i\in Q$ satisfying $i>g$. Moreover, this case differs in one value, namely, in $f(g,a_d)$:

$$\begin{aligned} \begin{array}{rcll} f(g,a) &{}=&{} (g+1){\,\bmod \,}n =0, &{}\text {if}~g= n-1~\text {and}~a\in \{a_0,\ldots ,a_k\}{,} \\ &{}=&{} (g+2){\,\bmod \,}n =1, &{}\text {if}~g= n-1~\text {and}~a\in \{a_{k+1},\ldots ,a_d\}. \end{array} \end{aligned}$$

(6)

The second special case is $g= n$, with no states $i\in Q$ satisfying $i>g$ or $i=g$. Thus, the condition $i<g$ is satisfied automatically for each $i\in Q$, which reduces (5)–(6) above to $f(i,a)= (i+1){\,\bmod \,}n$ for each $i\in Q$ and each $a\in \varSigma$.

Examples of $h_{{\scriptscriptstyle \varSigma }}$, $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$, and subsequent $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ are displayed in Fig. 1 (left), Fig. 2 (bottom), and Fig. 2 (top), respectively, for $d=5$,  $n=7$,  $g=3$, and $k=2$. The special case of $g=n-1$ is illustrated by Fig. 3.

Fig. 2.

Examples of a DFA $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ (bottom) and the corresponding dcDFA $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ (top), if $\varSigma = \{a_0,\ldots ,a_5\}$, $d=5$, $n=7$, $g=3$, and $k=2$. Accepting states are tagged by “$+$”, rejecting states by “−”, and neutral don’t care states by no sign. To simplify notation for $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$, the ordered pairs “$\langle {}i,j{}\rangle$” are displayed here in the form “ij”.

Finally, for the given $\varSigma ,n,g,k$ satisfying (4), consider a dcDFA $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}= \langle {}Q',\{0,1\}, q'_{\scriptscriptstyle \mathrm {I}}, f', F^{{\scriptscriptstyle \oplus }},F^{{\scriptscriptstyle \ominus }}{}\rangle$, not constructed for $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ by the use of Theorem 2, but defined as follows. First, let $Q'= Q_0 \cup \ldots \cup Q_{n-1}$, where

$$\begin{aligned} \begin{array}{rcll} Q_i &{}=&{} \{ \langle {}i,0{}\rangle ,\ldots ,\langle {}i,d-1{}\rangle \}, &{}\text {for}~i\in \{0,\ldots ,g-1\}, \\ Q_g &{}=&{} \{ \langle {}g,0{}\rangle ,\ldots ,\langle {}g,k{}\rangle \},\\ Q_i &{}=&{} \{ \langle {}i,0{}\rangle \}, &{}\text {for}~i\in \{g+1,\ldots ,n-1\}. \end{array} \end{aligned}$$

(7)

There are no sections $Q_i$ with $i>g$, if $g= n-1$, and no section $Q_g$, if $g= n$, in accordance with the two special cases for $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$.

Now, let $q'_{\scriptscriptstyle \mathrm {I}}=\langle {}0,0{}\rangle$, $F^{{\scriptscriptstyle \oplus }}= \{\langle {}0,0{}\rangle \}$, and $F^{{\scriptscriptstyle \ominus }}= \{\langle {}1,0{}\rangle ,\langle {}2,0{}\rangle ,\ldots ,\langle {}n-1,0{}\rangle \}$. Transitions are defined as follows:

$$\begin{aligned} \begin{array}{rclll} f'(\langle {}i,j{}\rangle ,1) &{}=&{} \langle {}(i+1){\,\bmod \,}n,0{}\rangle , &{}&{}\text {for each}~\langle {}i,j{}\rangle \in Q', \\ f'(\langle {}i,j{}\rangle ,0) &{}=&{} \langle {}i,j+1{}\rangle , &{}&{}\text {if}~i<g~\text {and}~j<d-1, \\ &{}=&{} \langle {}(i+1){\,\bmod \,}n,0{}\rangle , &{}&{}\text {if}~i<g~\text {and}~j=d-1, \\ f'(\langle {}g,j{}\rangle ,0) &{}=&{} \langle {}g,j+1{}\rangle , &{}&{}\text {if}~j<k, \\ &{}=&{} \langle {}(g+1){\,\bmod \,}n,0{}\rangle &{}=\langle {}g+1,0{}\rangle , &{}\text {if}~g\le n-2~\text {and}~j=k, \\ &{}=&{} \langle {}(g+1){\,\bmod \,}n,j+1{}\rangle &{}=\langle {}0,k+1{}\rangle , &{}\text {if}~g= n-1~\text {and}~j=k, \\ f'(\langle {}i,0{}\rangle ,0) &{}=&{} \langle {}i,0{}\rangle , &{}&{}\text {if}~i>g. \end{array} \end{aligned}$$

(8)

Note that also here the case of $g=n-1$ is different. (See also Fig. 3).

Fig. 3.

Examples of $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ and $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ for the special case of $g= n-1$, namely, for $\varSigma = \{a_0,\ldots ,a_5\}$, $d=5$, $n=7$,  $g= n-1= 6$, and $k=2$ (graph rotated, so that the state $g=n-1$ is not displayed at the right end).

Lemma 1

Let $h_{{\scriptscriptstyle \varSigma }}$, $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$, and $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ be the binary code, DFA, and dcDFA defined above. Then $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ solves the binary promise problem $\langle {}h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})), h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})^{\scriptscriptstyle \mathrm {C}}){}\rangle$.

Proof

First, it is not too hard to see that if $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ has a transition $i\mathop {\longrightarrow }\limits ^{a}i'$, then $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ has the corresponding computation path $\langle {}i,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{h_{{\scriptscriptstyle \varSigma }}(a)}\langle {}i',0{}\rangle$. This can be shown by consulting (3) and by comparing all transitions presented by (5), (6), and (8), for each $i\in Q$ and each $a\in \varSigma$:

• $f(i,a)= (i+1){\,\bmod \,}n$, if $i<g$ and $a\in \varSigma$ (which covers the case of $g=n$):

  • $\langle {}i,0{}\rangle \mathop {\longrightarrow }\limits ^{0^j} \langle {}i,j{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}(i+1){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^j1$, $j\in \{0,\ldots ,d-1\}$,

  • $\langle {}i,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{d-1}} \langle {}i,d-1{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}(i+1){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^d$.

• $f(g,a)= (g+1){\,\bmod \,}n$, if $g\le n-2$ and $a\in \{a_0,\ldots ,a_k\}\cup \{a_d\}$:

  • $\langle {}g,0{}\rangle \mathop {\longrightarrow }\limits ^{0^j} \langle {}g,j{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}(g+1){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^j1$, $j\in \{0,\ldots ,k\}$,

  • $\langle {}g,0{}\rangle \mathop {\longrightarrow }\limits ^{0^k} \langle {}g,k{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}g+1,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{d-k-1}} \langle {}g+1,0{}\rangle = \langle {}(g+1){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^d$.

• $f(g,a)= (g+2){\,\bmod \,}n$, if $g\le n-2$ and $a\in \{a_{k+1},\ldots ,a_{d-1}\}$:

  • $\langle {}g,0{}\rangle \mathop {\longrightarrow }\limits ^{0^k} \langle {}g,k{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}g+1,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{j-k-1}} \langle {}g+1,0{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}(g+2){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^j1$, $j\in \{k+1,\ldots ,d-1\}$.

• $f(i,a)= (i+1){\,\bmod \,}n$, if $i>g$ and $a\in \varSigma {\setminus }\{a_d\}$:

  • $\langle {}i,0{}\rangle \mathop {\longrightarrow }\limits ^{0^j} \langle {}i,0{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}(i+1){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^j1$, $j\in \{0,\ldots ,d-1\}$.

• $f(i,a)= i$, if $i>g$ and $a=a_d$:

  • $\langle {}i,0{}\rangle \mathop {\longrightarrow }\limits ^{0^d} \langle {}i,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^d$.

The same can be seen for different transitions in the case of $g=n-1$:

• $f(g,a)= (g+1){\,\bmod \,}n = 0$, if $g= n-1$ and $a\in \{a_0,\ldots ,a_k\}$:

  • $\langle {}g,0{}\rangle \mathop {\longrightarrow }\limits ^{0^j} \langle {}g,j{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}(g+1){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^j1$, $j\in \{0,\ldots ,k\}$.

• $f(g,a)= (g+2){\,\bmod \,}n = 1$, if $g= n-1$ and $a\in \{a_{k+1},\ldots ,a_d\}$:

  • $\langle {}g,0{}\rangle \mathop {\longrightarrow }\limits ^{0^k} \langle {}g,k{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}0,k+1{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{j-k-1}} \langle {}0,j{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}1,0{}\rangle = \langle {}(g+2){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^j1$, $j\in \{k+1,\ldots ,d-1\}$,

  • $\langle {}g,0{}\rangle \mathop {\longrightarrow }\limits ^{0^k} \langle {}g,k{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}0,k+1{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{d-k-2}} \langle {}0,d-1{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}1,0{}\rangle =$ $\langle {}(g+2){\,\bmod \,}n,0{}\rangle$, for $h_{{\scriptscriptstyle \varSigma }}(a)=0^d$.

Now, by induction on the length of $\alpha = a_{i_1}\!{\cdots }a_{i_{\ell }}\in \varSigma ^{*}$, we easily obtain that the computation path $i\mathop {\longrightarrow }\limits ^{\alpha }i'$ in $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ implies the existence of the corresponding path $\langle {}i,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{h_{{\scriptscriptstyle \varSigma }}(\alpha )} \langle {}i',0{}\rangle$ for $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$, for each $i,i'\in Q$ and each $\alpha \in \varSigma$.

Thus, if $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ has a path $q_{\scriptscriptstyle \mathrm {I}}= 0\mathop {\longrightarrow }\limits ^{\alpha }0\in F$, then $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ has the corresponding path $q'_{\scriptscriptstyle \mathrm {I}}= \langle {}0,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{h_{{\scriptscriptstyle \varSigma }}(\alpha )} \langle {}0,0{}\rangle \in F^{{\scriptscriptstyle \oplus }}$, and hence $h_{{\scriptscriptstyle \varSigma }}(\alpha )$ is accepted by $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$, if $\alpha \in \mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})$. On the other hand, if this path in $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ halts in some $i'\ne 0$, that is, in some $i'\in Q{\setminus }F$, the corresponding path in $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ will halt in $\langle {}i',0{}\rangle \in F^{{\scriptscriptstyle \ominus }}$, and hence $h_{{\scriptscriptstyle \varSigma }}(\alpha )$ is rejected by $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$, if $\alpha \in \mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})^{\scriptscriptstyle \mathrm {C}}$.

Therefore, $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ is a valid dcDFA for solving the binary promise problem $\langle {}h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})), \,h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})^{\scriptscriptstyle \mathrm {C}}){}\rangle$.    $\square$

$\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ uses fewer states than dcDFA obtained by the use of Theorem 2, but it may accept/reject some binary inputs that are not images of any $\alpha \in \varSigma ^{*}$. That is, it does not necessarily halt in a neutral state on such inputs.

Lemma 2

Let $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ be the DFA defined above. Then $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ is optimal and uses exactly n states.

Proof

Using (5) and (6), we see that $f(i,a_0)= (i+1){\,\bmod \,}n$, for each $i\in Q= \{0,\ldots ,n-1\}$, not excluding the special cases of $g= n-1$ or $g= n$. Since $q_{\scriptscriptstyle \mathrm {I}}=0$ and $F=\{0\}$,  $a_0^e\in \mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})$ if and only if e is an integer multiple of n.

This implies that $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ cannot be replaced by an equivalent DFA using fewer states: such DFA would accept $a_0^n$ by a computation path $r_0\mathop {\longrightarrow }\limits ^{a_0}{} r_1\mathop {\longrightarrow }\limits ^{a_0}{} r_2\mathop {\longrightarrow }\limits ^{a_0}{} \cdots {} \mathop {\longrightarrow }\limits ^{a_0}{} r_n$ along which some state would be repeated, which gives a valid accepting computation path for some $a_0^e$ with $0<e<n$, a contradiction.    $\square$

Lemma 3

Let $h_{{\scriptscriptstyle \varSigma }}$, $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$, and $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ be the binary code, DFA, and dcDFA defined above. Then $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ is optimal for solving the binary promise problem $\langle {}h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})), h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})^{\scriptscriptstyle \mathrm {C}}){}\rangle$ and uses exactly $m= (\Vert {\varSigma }\Vert -1){\cdot }g+ (k+1)+ (n-g-1)$ states, if $g\le n-1$, but exactly $m= (\Vert {\varSigma }\Vert -1){\cdot }n$ states, if $g=n$.

Proof

By Lemma 1, $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ solves the given promise problem and, by (7), it uses $m= d{\cdot }g+ (k+1)+ (n-g-1)$ states, if $g\le n-1$. For $g=n$, all sections $Q_i$ are of equal size d in (7), which gives $m= d{\cdot }n$. Since $d= \Vert {\varSigma }\Vert -1$, the upper bound for $m= \Vert {Q'}\Vert$ follows.

We only have to show that this bound cannot be reduced. Let $\langle {}i,j{}\rangle ,\langle {}\tilde{\imath },\tilde{\jmath }{}\rangle$ represent two arbitrary—but different—states in $Q'$, that is, either $i<\tilde{\imath }$, or $i=\tilde{\imath }$ but $j<\tilde{\jmath }$. For $i=\tilde{\imath }$ we have two subcases, depending on whether $i<g$ or $i=g$. (There are no pairs of different states with $i=\tilde{\imath }>g$; this condition implies $j=\tilde{\jmath }=0$, by (7). We do not consider $i>\tilde{\imath }$, or $i=\tilde{\imath }$ with $j>\tilde{\jmath }$ either—the roles of $\langle {}i,j{}\rangle ,\langle {}\tilde{\imath },\tilde{\jmath }{}\rangle$ can be swapped). Let us now define the following binary strings:

$$\begin{aligned} \begin{array}{rcll} x_{\scriptscriptstyle \langle {}i,j{}\rangle } &{}=&{} 1^i{\cdot }0^j , &{}\text {for each}~\langle {}i,j{}\rangle \in Q', \\ y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle } &{}=&{} 1{\cdot }1^{n-i-1} , &{}\text {if}~0\le i<\tilde{\imath }\le n-1, \\ y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}i,\tilde{\jmath }{}\rangle } &{}=&{} 0^{d-\tilde{\jmath }}1{\cdot }1^{n-i-1} , &{}\text {if}~0\le i<g~\text {and}~0\le j<\tilde{\jmath }\le d-1, \\ y_{\scriptscriptstyle \langle {}g,j{}\rangle \langle {}g,\tilde{\jmath }{}\rangle } &{}=&{} 0^{k+1-\tilde{\jmath }}1{\cdot }1^{n-g-1} , &{}\text {if}~g\le n-1~\text {and}~0\le j<\tilde{\jmath }\le k. \end{array} \end{aligned}$$

(9)

It can be seen from (8) that each state $\langle {}i,j{}\rangle \in Q'$ is reached by reading $x_{\scriptscriptstyle \langle {}i,j{}\rangle }$ from the input: $q'_{\scriptscriptstyle \mathrm {I}}= \langle {}0,0{}\rangle \mathop {\longrightarrow }\limits ^{1^i} \langle {}i{\,\bmod \,}n,0{}\rangle = \langle {}i,0{}\rangle \mathop {\longrightarrow }\limits ^{0^j} \langle {}i,j{}\rangle$.

If $g=n-1$, we get $y_{\scriptscriptstyle \langle {}g,j{}\rangle \langle {}g,\tilde{\jmath }{}\rangle }= 0^{k+1-\tilde{\jmath }}1$, if $g=n$, we do not define $y_{\scriptscriptstyle \langle {}g,j{}\rangle \langle {}g,\tilde{\jmath }{}\rangle }$.

We are now going to show that (i) both $x_{\scriptscriptstyle \langle {}i,j{}\rangle }{\cdot } y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }$ and $x_{\scriptscriptstyle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }{\cdot } y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }$ are valid binary images of some strings in $\varSigma ^{*}$, i.e., both of them are in $h_{{\scriptscriptstyle \varSigma }}(\varSigma ^{*})$, and that (ii) the computation of $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ on $x_{\scriptscriptstyle \langle {}i,j{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }$ starts in $q'_{\scriptscriptstyle \mathrm {I}}= \langle {}0,0{}\rangle$ and ends in $\langle {}0,0{}\rangle \in F^{{\scriptscriptstyle \oplus }}$, while the computation on $x_{\scriptscriptstyle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }$ starts in $q'_{\scriptscriptstyle \mathrm {I}}= \langle {}0,0{}\rangle$ and ends in some $\langle {}i',0{}\rangle \in F^{{\scriptscriptstyle \ominus }}$, with $i'\ne 0$. Taking into account (i), this gives that $x_{\scriptscriptstyle \langle {}i,j{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }\in h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}))$ and $x_{\scriptscriptstyle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }\in h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})^{\scriptscriptstyle \mathrm {C}})$. These statements can be shown by analyzing all cases, using (9), (3), and (8) (see also Figs. 2 and 3):

• If $0\le i<\tilde{\imath }\le n-1$, and hence $0<\tilde{\imath }-i\le n-1$, with $j,\tilde{\jmath }\in \{0,\ldots ,d-1\}$:

  • $x_{\scriptscriptstyle \langle {}i,j{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }= 1^i{\cdot }0^j{\cdot }1{\cdot }1^{n-i-1}= h_{{\scriptscriptstyle \varSigma }}(a_0^i{\cdot }a_j{\cdot }a_0^{n-i-1})$, $\langle {}0,0{}\rangle \mathop {\longrightarrow }\limits ^{1^i0^j} \langle {}i,j{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}(i+1){\,\bmod \,}n,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{1^{n-i-1}} \langle {}0,0{}\rangle$,

  • $x_{\scriptscriptstyle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }= 1^{\tilde{\imath }}{\cdot }0^{\tilde{\jmath }}{\cdot }1{\cdot }1^{n-i-1}= h_{{\scriptscriptstyle \varSigma }}(a_0^{\tilde{\imath }}{\cdot }a_{\tilde{\jmath }}{\cdot }a_0^{n-i-1})$, $\langle {}0,0{}\rangle \mathop {\longrightarrow }\limits ^{1^{\tilde{\imath }}0^{\tilde{\jmath }}} \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}(\tilde{\imath }+1){\,\bmod \,}n,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{1^{n-i-1}} \langle {}\tilde{\imath }-i,0{}\rangle \ne \langle {}0,0{}\rangle$.

• If $0\le i<g$ and $0\le j<\tilde{\jmath }\le d-1$, and hence $1\le j+d-\tilde{\jmath }\le d-1$:

  • $x_{\scriptscriptstyle \langle {}i,j{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}i,\tilde{\jmath }{}\rangle }= 1^i{\cdot }0^j{\cdot }0^{d-\tilde{\jmath }}1{\cdot }1^{n-i-1}= h_{{\scriptscriptstyle \varSigma }}(a_0^i{\cdot }a_{j+d-\tilde{\jmath }}{\cdot }a_0^{n-i-1})$, $\langle {}0,0{}\rangle \mathop {\longrightarrow }\limits ^{1^i0^j} \langle {}i,j{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{d-\tilde{\jmath }}} \langle {}i,j+d-\tilde{\jmath }{}\rangle \mathop {\longrightarrow }\limits ^{1} \langle {}(i+1){\,\bmod \,}n,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{1^{n-i-1}} \langle {}0,0{}\rangle$,

  • $x_{\scriptscriptstyle \langle {}i,\tilde{\jmath }{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}i,\tilde{\jmath }{}\rangle }= 1^i{\cdot }0^{\tilde{\jmath }}{\cdot }0^{d-\tilde{\jmath }}1{\cdot }1^{n-i-1}= h_{{\scriptscriptstyle \varSigma }}(a_0^i{\cdot }a_d{\cdot }a_0^{n-i})$, $\langle {}0,0{}\rangle \mathop {\longrightarrow }\limits ^{1^i0^{\tilde{\jmath }}} \langle {}i,\tilde{\jmath }{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{d-1-\tilde{\jmath }}} \langle {}i,d-1{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}(i+1){\,\bmod \,}n,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{1^{n-i}} \langle {}1,0{}\rangle$.

• If $g\le n-1$ and $0\le j<\tilde{\jmath }\le k$, and hence $1\le j+k+1-\tilde{\jmath }\le k$:

  • $x_{\scriptscriptstyle \langle {}g,j{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}g,j{}\rangle \langle {}g,\tilde{\jmath }{}\rangle }= 1^g{\cdot }0^j{\cdot }0^{k+1-\tilde{\jmath }}1{\cdot }1^{n-g-1}= h_{{\scriptscriptstyle \varSigma }}(a_0^g{\cdot }a_{j+k+1-\tilde{\jmath }}{\cdot }a_0^{n-g-1})$, $\langle {}0,0{}\rangle \mathop {\longrightarrow }\limits ^{1^g0^j} \langle {}g,j{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{k+1-\tilde{\jmath }}} \langle {}g,j+k+1-\tilde{\jmath }{}\rangle \!\mathop {\longrightarrow }\limits ^{1} \langle {}(g+1){\,\bmod \,}n,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{1^{n-g-1}} \langle {}0,0{}\rangle$,

  • $x_{\scriptscriptstyle \langle {}g,\tilde{\jmath }{}\rangle }{\cdot }y_{\scriptscriptstyle \langle {}g,j{}\rangle \langle {}g,\tilde{\jmath }{}\rangle }= 1^g{\cdot }0^{\tilde{\jmath }}{\cdot }0^{k+1-\tilde{\jmath }}1{\cdot }1^{n-g-1}= h_{{\scriptscriptstyle \varSigma }}(a_0^g{\cdot }a_{k+1}{\cdot }a_0^{n-g-1})$, $\langle {}0,0{}\rangle \mathop {\longrightarrow }\limits ^{1^g0^{\tilde{\jmath }}} \langle {}g,\tilde{\jmath }{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{0^{k-\tilde{\jmath }}} \langle {}g,k{}\rangle$, with two different ways to continue: if $g\le n-2$, then $\langle {}g,k{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}g+1,0{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{1^{n-g}} \langle {}(n+1){\,\bmod \,}n,0{}\rangle = \langle {}1,0{}\rangle$, if $g= n-1$, then $\langle {}g,k{}\rangle \mathop {\longrightarrow }\limits ^{0} \langle {}0,k+1{}\rangle \mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{1^{n-g}} \langle {}(n-g){\,\bmod \,}n,0{}\rangle = \langle {}1,0{}\rangle$.

Summing up, we have constructed m-tuple $X= \langle {}x_{\scriptscriptstyle \langle {}i,j{}\rangle }{}\rangle _{\langle {}i,j{}\rangle \in Q'}$ and ${m\atopwithdelims ()2}$-tuple $Y= \langle {}y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }{}\rangle _ {\langle {}i,j{}\rangle {,}\langle {}\tilde{\imath },\tilde{\jmath }{}\rangle \in Q', \,\langle {}i,j{}\rangle \ne \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }$, where $m= \Vert {Q'}\Vert$, consisting of binary strings such that, for each pair $\langle {}i,j{}\rangle \ne \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle$,

• both $x_{\scriptscriptstyle \langle {}i,j{}\rangle }{\cdot } y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }$ and $x_{\scriptscriptstyle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }{\cdot } y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }$ are in $h_{{\scriptscriptstyle \varSigma }}(\varSigma ^{*})= h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}))\cup h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})^{\scriptscriptstyle \mathrm {C}})$,

• $x_{\scriptscriptstyle \langle {}i,j{}\rangle }{\cdot } y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }\in h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}))$ and $x_{\scriptscriptstyle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }{\cdot } y_{\scriptscriptstyle \langle {}i,j{}\rangle \langle {}\tilde{\imath },\tilde{\jmath }{}\rangle }\in h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})^{\scriptscriptstyle \mathrm {C}})$.

But then, by Theorem 1, any dcDFA solving the binary promise problem $\langle {}h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})), h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})^{\scriptscriptstyle \mathrm {C}}){}\rangle$ must use at least $m= \Vert {Q'}\Vert$ states, which gives that $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ is optimal.    $\square$

Theorem 5

For each non-unary input alphabet $\varSigma$, there exists a maximal binary prefix code $h\,{:}\,\varSigma ^{*}{\rightarrow }\{0,1\}^{*}$ such that, for each $n\ge 2$ and each value $N\in \{n,\ldots ,(\Vert {\varSigma }\Vert -1){\cdot }n\}$, there exists a language $\mathcal {L}\subseteq \varSigma ^{*}$ such that the optimal DFA recognizing $\mathcal {L}$ uses exactly n states and any optimal dcDFA for solving $\langle {}h(\mathcal {L}), h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$, the binary promise-problem version of $\mathcal {L}$, uses exactly N states.

Proof

Let us handle the pathological case of $\varSigma = \{a_0,a_1\}$ first. There are only two maximal prefix codes in this case, both of them are bijections from $\{a_0,a_1\}$ to $\{0,1\}$, and none of them can change the state complexity of any language. This corresponds to the fact that here $N\in \{n,\ldots ,(\Vert {\varSigma }\Vert -1){\cdot }n\}= \{n\}$.

Now, for the given $\varSigma$, with $\Vert {\varSigma }\Vert \ge 3$, let us fix $h=h_{{\scriptscriptstyle \varSigma }}$, as introduced by (3). Next, the witness language $\mathcal {L}$ depends on $\varSigma$ and on the given values n and N:

First, if $N\le (\Vert {\varSigma }\Vert -1){\cdot }n-1= d{\cdot }n -1$, we can take $\mathcal {L}= \mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})$, using the following parameters:

$$\begin{array}{rcll}g=\lfloor {(N-n)/(d-1)}\rfloor ,&\,&k=(N-n){\,\bmod \,}(d-1). \end{array}$$

Clearly, $g\le \lfloor {(d{\cdot }n-1-n)/(d-1)}\rfloor = \lfloor {n-\frac{1}{d-1}}\rfloor = n-1$ and $k\le d-2$. By Lemma 2, $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ is the optimal DFA for recognizing $\mathcal {L}$, using exactly n states. Similarly, by Lemma 3, $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ is optimal dcDFA for solving $\langle {}h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}), h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ and uses exactly $m= d{\cdot }g\,+\,(k\,+\,1)\,+\,(n\,-\,g\,-\,1)$ states. This gives $m= d{\cdot }g\,+\,(k\,+\,1)\,+\,(n\,-\,g\,-\,1) = (d\,-\,1){\cdot }g\,+\, k\,+\, n = (d-1){\cdot }\lfloor {(N\,-\,n)/(d\,-\,1)}\rfloor \,+\, (N\,-\,n){\,\bmod \,}(d\,-\,1)+ n = (N\,-\,n)+ n = N$ states, using the fact that $a{\cdot }\lfloor {b/a}\rfloor + b{\,\bmod \,}a= b$.

Second, if $N= (\Vert {\varSigma }\Vert -1){\cdot }n$, take $\mathcal {L}= \mathcal {L}(\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k})$ with $g=n$ and $k=0$. (Here $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ does not actually depend on k.) Again, by Lemma 2, $\mathcal {A}_{{\scriptscriptstyle \varSigma },n,g,k}$ is optimal and uses exactly n states and, by Lemma 3, $\mathcal {A}'_{{\scriptscriptstyle \varSigma },n,g,k}$ is optimal for solving $\langle {}h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}), h_{{\scriptscriptstyle \varSigma }}(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$, this time with exactly $m= (\Vert {\varSigma }\Vert -1){\cdot }n= N$ states.    $\square$

5 Concluding Remarks

By a more careful analysis of the construction in Theorem 3, we see that it does not increase the number of accepting or rejecting states. As a direct consequence, if the optimal DFA $\mathcal {A}$ recognizing $\mathcal {L}$ uses $n^{{\scriptscriptstyle \oplus }}$ accepting and $n^{{\scriptscriptstyle \ominus }}$ rejecting states (neither of these values can be reduced, since the optimal $\mathcal {A}$ is unique), then any optimal dcDFA $\mathcal {A}'$ solving the binary promise problem $\langle {}h(\mathcal {L}),h(\mathcal {L}^{\scriptscriptstyle \mathrm {C}}){}\rangle$ must use at least $n^{{\scriptscriptstyle \oplus }}$ accepting and at least $n^{{\scriptscriptstyle \ominus }}$ rejecting states. But all states in $\mathcal {A}'$ that cannot be reached from $q'_{\scriptscriptstyle \mathrm {I}}$ by reading some $\beta \in h(\varSigma ^{*})$ can be made neutral (see also (2) in the proof of Theorem 3). This will only change acceptance/rejection to “don’t care” answers on some inputs not belonging to $h(\varSigma ^{*})$.

This allows to establish some kind of pseudo-isomorphism between $\mathcal {A}$ and $\mathcal {A}'$. (Proof omitted here due to space constraints, to appear in a journal version.) Namely, there exists a bijective function $t\,{:}\,Q{\rightarrow }F^{{\scriptscriptstyle \oplus }}{\cup }F^{{\scriptscriptstyle \ominus }}$ that maps $q_{\scriptscriptstyle \mathrm {I}}$ to $q'_{\scriptscriptstyle \mathrm {I}}$, F to $F^{{\scriptscriptstyle \oplus }}$, and $Q{\setminus }F$ to $F^{{\scriptscriptstyle \ominus }}$, preserving the machine’s transitions, i.e., $t(f(q,a))= {f'}^{*}(t(q),h(a))$, for each $q\in Q$ and $a\in \varSigma$. However, such pseudo-isomorphism does not exclude, for some transition $q\mathop {\longrightarrow }\limits ^{a}q'$ in $\mathcal {A}$, passing through some state $t(q'')\in F^{{\scriptscriptstyle \oplus }}{\cup }F^{{\scriptscriptstyle \ominus }}$ along the corresponding path $t(q)\mathop {-\!\!\!-\!\!\!-\!\!\!\longrightarrow }\limits ^{h(a)}t(q')$ in $\mathcal {A}'$—even more than once in the meantime.³

There are more open questions in the related area than the known answers. As an example, we do not know the cost of binary coded intersection; the same holds for other basic operations with regular languages. It can be expected that answers may depend also on the code h in use, and we expect some anomalies for prefix codes that are not maximal.