State Complexity of Prefix Distance of Subregular Languages

Abstract

The neighbourhood of a regular language of constant radius with respect to the prefix distance is always regular. We give upper bounds and matching lower bounds for the size of the minimal deterministic finite automaton (DFA) needed for the radius k prefix distance neighbourhood of an n state DFA that recognizes, respectively, a finite, a prefix-closed and a prefix-free language. For prefix-closed languages the lower bound automata are defined over a binary alphabet. For finite and prefix-free regular languages the lower bound constructions use an alphabet that depends on the size of the DFA and it is shown that the size of the alphabet is optimal.

1 Introduction

The neighbourhood of radius r of a language L consists of all strings that are within distance at most r from some string of L. A distance measure d is said to be regularity preserving if the neighbourhood of any regular language with respect to d is regular. Calude et al. [2] have shown that additive distances are regularity preserving. Additivity requires, roughly speaking, that the distance is compatible with concatenation of words in a certain sense and best known examples of additive distances include the Levenshtein distance and the Hamming distance [2, 5].

The prefix distance of two words u and v is the sum of the lengths of the suffixes of u and v that begin after the longest common prefix of u and v. The suffix distance and the factor distance are defined analogously in terms of the longest common suffix (respectively, factor) of two words. It is known that the prefix, suffix and factor distance preserve regularity [4].

By the state complexity of a regularity preserving distance we mean the worst-case size of the minimal deterministic finite automaton (DFA) needed to recognize radius r neighbourhood of an n state DFA language (as a function of n and r). Tight bounds for the state complexity of prefix distance were recently obtained by the authors [14].

Worst-case state complexity bounds for general regular languages typically cannot be matched by finite languages, as first observed by Câmpeanu et al. [3], and the same holds for other proper sub-families of the regular languages. Relations between different sub-regular language families have been investigated recently by Holzer and Truthe [11]. Bordihn et al. [1] have studied the state complexity of determinization of automata for the different sub-regular language families and further recent work on the state complexity of sub-regular language families has been done by Holzer et al. [8, 10].

Here we study the state complexity of prefix distance for finite languages. Additionally, we concentrate on the classes of prefix-closed and prefix-free regular languages because their corresponding restricting properties can be viewed to be related to the definition of the prefix distance measure. We give tight state complexity bounds for the prefix distance of finite, prefix-closed and prefix-free regular languages. In the case of finite languages and prefix-free languages the lower bound construction uses an alphabet that depends linearly on the size of the DFA. We establish that the general upper bound cannot be matched by languages defined over an alphabet of smaller size.

2 Preliminaries

We briefly recall some definitions and notation used in the paper. For all unexplained notions on finite automata and regular languages the reader may consult the textbook by Shallit [15] or the survey by Yu [16]. A survey of distances is given by Deza and Deza [5]. Recent surveys on descriptional complexity of regular languages include [6, 9, 13].

In the following $\varSigma$ is always a finite alphabet, the set of strings over $\varSigma$ is $\varSigma ^*$ and $\varepsilon$ is the empty string. The reversal of a string $x \in \varSigma ^*$ is $x^R$. The set of nonnegative integers is $\mathbb {N}_0$. The cardinality of a finite set S is denoted |S| and the powerset of S is $2^S$. A string $w \in \varSigma ^*$ is a substring or factor of x if there exist strings $u,v \in \varSigma ^*$ such that $x = uwv$. If $u = \varepsilon$, then w is a prefix of x. If $v = \varepsilon$, then w is a suffix of x.

A nondeterministic finite automaton (NFA) is a 5-tuple $A = (Q,\varSigma ,\delta ,Q_0,F)$ where Q is a finite set of states, $\varSigma$ is an alphabet, $\delta$ is a multi-valued transition function $\delta :Q \times \varSigma \rightarrow 2^Q$, $Q_0 \subseteq Q$ is a set of initial states, and $F \subseteq Q$ is a set of final states. We extend the transition function $\delta$ to a function $Q \times \varSigma ^* \rightarrow 2^Q$ in the usual way. A string $w \in \varSigma ^*$ is accepted by A if, for some $q_0 \in Q_0$, $\delta (q_0,w) \cap F \ne \emptyset$ and the language recognized by A consists of all strings accepted by A. An $\varepsilon$-NFA is an extension of an NFA where transitions can be labeled by the empty string $\varepsilon$ [15, 16], i.e., $\delta$ is a function $Q \times (\varSigma \cup \{ \varepsilon \}) \rightarrow 2^Q$. It is known that every $\varepsilon$-NFA A has an equivalent NFA without $\varepsilon$-transitions and with the same number of states as A. An NFA $A = (Q, \varSigma , \delta , Q_0, F)$ is a deterministic finite automaton (DFA) if $|Q_0| = 1$ and, for all $q \in Q$ and $a \in \varSigma$, $\delta (q, a)$ either consists of one state or is undefined. Two states p and q of a DFA A are equivalent if $\delta (p,w) \in F$ if and only if $\delta (q,w) \in F$ for every string $w \in \varSigma ^*$. A DFA A is minimal if each state $q \in Q$ is reachable from the initial state, a final state is reachable from each state q, and no two states are equivalent.

Note that our definition of a DFA allows some transitions to be undefined, that is, by a DFA we mean an incomplete DFA. It is well known that, for a regular language L, the sizes of the minimal incomplete and complete DFAs differ by at most one. The constructions used in this paper are more convenient to formulate using incomplete DFAs but our results would not change in any significant way if we were to require that all DFAs are complete. The (incomplete deterministic) state complexity of a regular language L, ${{\mathrm{sc}}}(L)$, is the size of the minimal DFA recognizing L.

We define ${{\mathrm{pref}}}(L)$ to be the language of all prefixes of words belonging to L,

$$\begin{aligned} {{\mathrm{pref}}}(L) = \{ u \in \varSigma ^* \mid (\exists v \in \varSigma ^*) \; uv \in L \}. \end{aligned}$$

A language L is prefix-closed if $L = {{\mathrm{pref}}}(L)$. A language L is prefix-free if no word $u \in L$ is a proper prefix of any other word in L. A DFA A is non-exiting if a final state of A has no outgoing transitions. The minimal DFAs recognizing a prefix-free language have always the following property.

Lemma 1

([7]). If A is minimal and L(A) is prefix-free, then A is non-exiting.

To conclude this section, we recall definitions of the distance measures used in the following. Generally, a function $d : \varSigma ^* \times \varSigma ^* \rightarrow [0,\infty )$ is a distance if it satisfies for all $x,y,z \in \varSigma ^*$, the conditions $d(x,y) = 0$ if and only if $x = y$, $d(x,y) = d(y,x)$, and $d(x,z) \le d(x,y) + d(y,z)$. The neighbourhood of a language L of radius k with respect to a distance d is the set

$$\begin{aligned} E(L,d,k) = \{w \in \varSigma ^* \mid (\exists x \in L) \; d(w,x) \le k \}. \end{aligned}$$

Let $x, y \in \varSigma ^*$. The prefix distance of x and y counts the number of symbols which do not belong to the longest common prefix of x and y [4]. Formally, it is defined by

$$\begin{aligned} d_p(x,y) = |x| + |y| - 2 \cdot \max _{z \in \varSigma ^*} \{ |z| \mid x,y \in z \varSigma ^* \}. \end{aligned}$$

The state complexity of prefix distance was established in [14].

Theorem 1

([14]). For $n > k \ge 0$, if ${{\mathrm{sc}}}(L) = n$ then

$$\begin{aligned} {{\mathrm{sc}}}(E(L,d_p,k)) \le n \cdot (k+1) - \frac{k(k+1)}{2} \end{aligned}$$

and this bound can be reached in the worst case.

To conclude this section we recall from [14] the construction of a DFA that recognizes the prefix-distance neighbourhood of a regular language.

Let $A = (Q,\varSigma ,\delta ,q_0,F)$ be a DFA and $\varphi _A: Q \rightarrow \mathbb N_0$ be a function defined by

$$\begin{aligned} \varphi _A(q) = \min _{w \in \varSigma ^*} \{ |w| \mid \delta (q,w) \in F \} \end{aligned}$$

The function $\varphi _A(q)$ gives the length of the shortest path from a state q to the closest reachable final state. Note that if $q \in F$, then $\varphi _A(q) = 0$.

We construct a DFA $A' = (Q',\varSigma ,\delta ', q_0', F')$ for the neighbourhood $E(L(A),d_p,k)$, $k \in \mathbb {N}$, as follows. We define the state set

$$\begin{aligned} Q' = ((Q-F) \times \{1, \dots , k+1\}) \cup F \cup \{p_1,\dots ,p_k\}. \end{aligned}$$

(1)

The initial state $q_0'$ is defined by

$$\begin{aligned} q_0' = {\left\{ \begin{array}{ll} q_0, &{} \text { if }q_0 \in F; \\ (q_0,\varphi _A(q_0)) &{} \text { if }q_0 \not \in F \text { and } \varphi _A(q_0) \le k; \\ (q_0,k+1) &{} \text { if } q_0 \not \in F \text { and } \varphi _A(q_0) > k. \end{array}\right. } \end{aligned}$$

The set of final states is given by

$$\begin{aligned} F' = ((Q - F) \times \{1,\dots , k\}) \cup F \cup \{p_1,\dots ,p_k\}. \end{aligned}$$

Let $q_{i,a} = \delta (i,a)$ for $i \in Q$ and $a \in \varSigma$, if $\delta (i,a)$ is defined. Then for all $a \in \varSigma$, the transition function $\delta '$ is defined for states $i \in F$ by

$$\begin{aligned} \delta '(i,a) = {\left\{ \begin{array}{ll} (q_{i,a},1), &{} \text { if } q_{i,a} \in Q - F; \\ q_{i,a}, &{} \text { if } q_{i,a} \in F; \\ p_1, &{} \text { if }\delta (i,a) \,\text {is undefined.} \end{array}\right. } \end{aligned}$$

For states $(i,j) \in Q - F \times \{1, \dots , k+1 \}$, $\delta '$ is defined

$$\begin{aligned} \delta '((i,j),a) = {\left\{ \begin{array}{ll} q_{i,a}, &{} \text { if }q_{i,a} \in F; \\ (q_{i,a},\min \{j+1,\varphi _A(q_{i,a})\}), &{} \text { if } \varphi _A(q_{i,a}) \text { or } j+1 \le k; \\ (q_{i,a},k+1), &{} \text { if } \varphi _A(q_{i,a}) \text { and } j+1 > k; \\ p_{j+1}, &{} \text { if } \delta (i,a) \,\text {is undefined.} \end{array}\right. } \end{aligned}$$

Finally, we define $\delta '$ for states $p_\ell$ for $\ell = 1,\dots ,k-1$ by $\delta '(p_\ell ,a) = p_{\ell +1}$.

The following Proposition 1 follows from the proof of Proposition 2 of [14]. Note that Proposition 2 of [14] establishes a stronger claim and the statement of the below proposition includes only the parts that we need in the later sections.

Proposition 1

([14]). (a) The DFA $A'$ recognizes the neighbourhood $E(L(A), d_p, k)$.

(b) The elements of the set $S_{ur} = \{ (q,j) \mid q \in Q-F, 1 \le j \le k+1, j > \varphi _A(q) \}$ are all unreachable as states of the DFA $A'$.

3 Neighbourhoods of Finite Languages

We first consider the state complexity of neighbourhoods of finite languages with respect to the prefix distance.

Proposition 2

Let L be a finite language recognized by a minimal DFA $A = (Q, \varSigma , \delta , q_0, F)$ with n states. Then

$$\begin{aligned} {{\mathrm{sc}}}(E(L,d_p,k)) \le (n-2)\cdot (k+1) - k^2 + 2. \end{aligned}$$

Proof

We know that the neighbourhood of L of radius k with respect to the prefix distance is recognized by a DFA $A' = (Q', \varSigma , \delta ', q_0'. F')$ obtained from A as in Proposition 1 where, furthermore, all elements of the set $S_{ur}$ are unreachable. We show that there are more unreachable states in the case of finite languages.

Since A is acyclic, the number and length of words that reach each state $q \in Q$ is bounded. For $q \in Q$, let $w_q$ denote the longest word that reaches q from the initial state $q_0$ without passing through a final state. Then for all states q with $|w_q| \le k$, the states $(q,j) \in Q'$ with $j > |w_q|$ are unreachable as states of $A'$ (where the set of states of $A'$ is as in (1). That is, all states in the set

$$\begin{aligned} R_{ur} = \{ (q,j) \mid q \in Q-F, 1 \le j \le k+1, j > |w_q| \} \end{aligned}$$

are unreachable in $A'$. By Proposition 1 (b) all elements of the set $S_{ur} = \{ (q,j) \mid q \in Q-F, 1 \le j \le k+1, j > \varphi _A(q) \}$ are also unreachable in $A'$. We note that increasing the number of final states of A by one decreases the cardinality of $Q'$ by k and decreases the cardinality of $S_{ur}$ and $R_{ur}$ by at most k. However, we observe that A must have at least two final states to reach the bound. The last state of A, with no outgoing transitions, must be a final state since, otherwise, there are useless states. But this cannot be the only final state, since otherwise, for every state $q \in Q$ with $\varphi _A(q) > k$, only $(q,k+1)$ is reachable. Thus, the initial state $q_0$ must also be a final state.

As in [14], we note that the cardinality of $S_{ur}$ is minimized when exactly one non-final state has a shortest path of length i that reaches $q_f$. From the above it then follows that reaching the upper bound requires exactly two final states, one of which must be the initial state and the other which must have no outgoing transitions. Since A is acyclic, the initial state cannot have any incoming transitions, so the states in $S_{ur}$ consist of those that can reach the non-initial final state, giving $\frac{k(k+1)}{2}$ unreachable states. Similarly, the cardinality of $R_{ur}$ is minimized when exactly one non-final state has a longest word of length i which reaches it from $q_0$, giving $\frac{k(k+1)}{2}$ unreachable states.

Thus, the number of states of the minimal DFA for $E(L,d_p,k)$ is upper bounded by

$$\begin{aligned} (n-2)(k+1) + 2 + k - 2 \cdot \frac{k(k+1)}{2} = (n-2)(k+1) - k^2 + 2. \end{aligned}$$

   $\square$

Next we give a lower bound construction that matches the upper bound of Proposition 2.

Lemma 2

There exists a finite language recognized by a DFA with n states such that $E(L(A),d_p,k)$ requires at least $(n-2)(k+1) - k^2 + 2$ states.

Proof

Let $A_n = (Q_n,\varSigma _n,\delta _n,q_0,F_n)$ where $Q_n = \{0,\dots ,n-1\}$, $\varSigma _n = \{a_1,\dots ,a_{n-3}\}$, $q_0 = 0$, $F_n = \{0,n-1\}$, and the transition function is defined by

• $\delta _n(0,a_i) = i$ for $1 < j \le n-3$,

• $\delta _n(i,a_{i+1}) = i+1$ for $0 \le i < n-3$,

• $\delta _n(i,a_1) = i+1$ for $i = n-3,n-2$.

The DFA $A_n$ is depicted in Fig. 1.

Fig. 1.

The DFA $A_n$.

Let $A_n' = (Q_n',\varSigma _n,\delta _n',q_0',F_n')$ be the DFA constructed from $A_n$ as in Proposition 1. First, we show that $(n-2)(k+1) - k^2 + 2$ states are reachable. States of the form $p_i$ with $1 \le i \le k$ are reachable from states $0 \le i \le k$ on symbols $a_j$ with $j \ne i+1$. For states of the form $(i,j) \in (Q_n - F_n) \times \{1,\dots ,k+1\}$, with $\varphi _{A_n}(i) > k$ and $j \le i$, each (i, j) is reachable on the word $a_{i-j} a_{i-j+1} \cdots a_i$. However, states (i, j) with $j > \varphi _{A_n}(i)$ are unreachable by definition of $A_n'$ and states (i, j) with $i < j \le k$ are unreachable. Thus the number of unreachable states in $(Q_n - F_n) \times \{1,\dots ,k+1\}$ is

$$\begin{aligned}&\sum _{i=n-k}^{n-1} |\{i\} \times \{\varphi _{A_n}(i)+1,\dots ,k+1\}| + \sum _{i=1}^k |\{i+1,\dots ,k+1\}| \\ =&\; 2 \cdot \sum _{i=1}^k |\{i=1,\dots ,k+1\}| = 2 \cdot \sum _{i=1}^k i = 2 \cdot \frac{k(k+1)}{2}. \end{aligned}$$

Thus the number of reachable states is

$$\begin{aligned} (n-2)(k+1) - 2 + k - 2 \cdot \frac{k(k+1)}{2} = (n-2)(k+1) - k^2 + 2. \end{aligned}$$

Now, we show that all reachable states are pairwise inequivalent.

• For states of the form $p_i$ and $p_j$, $i < j$, the word $a_1^{k-i}$ takes the machine from state $p_i$ to $p_k$ and is accepted. However, from state $p_j$, the word $a_1^{k-i}$ reaches state $p_k$ on the prefix $a_1^{k-j}$ with no further transitions to read $a_1^{j-i}$ and thus, the word is not accepted.

• For states of the form (i, j) and $p_\ell$ with $\ell < k$, we consider the word $z = w_i a_2^k$ with

  $$\begin{aligned} w_i = a_{n-i+1} a_{n-i+2} \cdots a_{n-3} a_1 a_1. \end{aligned}$$

  The prefix $w_i$ takes the machine from state (i, j) to state $n-1$ and on the rest of the word $a_2^k$, the machine moves from $n-1$ to $p_k$ and is accepted. However, from state $p_\ell$, the computation on z reaches $p_k$ before all of z is read, since $|z| = n - i + k > k - \ell$ and it is rejected.

• For states of the form (i, j) and $(i',j')$ with $i < i'$ the states can be distinguished by $z = w_i a_2^k$ as above. For $i = i'$ and $j < j'$, let $z = a_i a_1^{k-j}$. From (i, j), the machine reads $a_i$ and is taken to $p_j$, while from $(i,j')$, the machine is taken to $p_{j'}$. From above, $p_j$ and $p_{j'}$ are distinguishable by $a_1^{k-j}$.

Thus, we have shown that there are $(n-2)(k+1) - k^2 + 2$ reachable states and that all reachable states are pairwise inequivalent.

   $\square$

Proposition 2 and Lemma 2 now yield a tight state complexity bound for the prefix distance neighbourhoods of regular languages.

Theorem 2

Let L be a finite language. For $n > 2k \ge 0$, if ${{\mathrm{sc}}}(L) = n$, then

$$\begin{aligned} {{\mathrm{sc}}}(E(L,d_p,k)) \le (n-2)\cdot (k+1) - k^2 + 2, \end{aligned}$$

and this bound can be reached in the worst case.

The lower bound construction of Lemma 2 uses, for a DFA with n states, an alphabet of cardinality $n-3$. To conclude this section we show that the construction is optimal in the sense that the upper bound of Theorem 2 cannot be reached with an alphabet of cardinality less than $n-3$.

Proposition 3

Let A be a DFA recognizing a finite language with n states. If the state complexity of $E(L(A),d_p,k)$ equals $(n-2)(k+1) - k^2 + 2$, then the alphabet of A needs at least $n-3$ letters.

Proof

Let $A = (Q,\varSigma ,\delta ,q_0,F)$ with $|Q| = n$. Let $A' = (Q',\varSigma ,\delta ', q_0' F')$ be the DFA recognizing $E(L(A),d_p,k)$ constructed in Proposition 1. Recall from the proof of Proposition 2 that in order for $A'$ to have the maximal number of states $(n-2)(k+1) - k^2 + 2$, a necessary condition is that $F = \{q_0,q_f\}$ and that there can be only one state $q_1$ with $\varphi _A(q_1) = 1$.

Now for all $q \in Q - \{q_0,q_f,q_1\}$, $\varphi _A(q) \ge 2$. By definition of the transition function $\delta '$, if $\varphi _A(q) \ge 2$, the state (q, 1) can only be reached by a direct transition from a final state. Since $q_f$ does not have any outgoing transitions, $q_0$ must have $n-3$ outgoing transitions—one for each state q.

Furthermore, since A contains a final state $q_f$ with no outgoing transitions, no additional symbols are required to reach $p_1$, as it can be reached from $q_f$ via a direct transition on any symbol.

Since A is a DFA and $q_0$ has at least $n-3$ outgoing transitions, the cardinality of the alphabet must be at least $n-3$.    $\square$

4 Neighbourhoods of Prefix-Closed and Prefix-Free Languages

Next, we consider the state complexity of neighbourhoods of prefix-closed and prefix-free regular languages with respect to the prefix distance.

Theorem 3

Let L be a prefix-closed regular language recognized by an n-state DFA A. Then there is a DFA $A'$ that recognizes the neighbourhood $E(L,d_p,k)$ with at most $n+k$ states and this bound is reachable.

Proof

Since L is prefix-closed, every state of A must be an accepting state [12]. If A has n states, this means that the DFA $A'$ constructed in Proposition 1 for the radius k neighbourhood has $n+k$ states.

We now define a prefix-closed regular language $L_n$ such that a DFA recognizing $E(L_n,d_p,k)$ requires at least $n+k$ states. Let $L_n = \{a^i \mid 0 \le i \le n \}$. Then we define $A_n = (Q_n,\{a,b\},\delta _n,q_0,F_n)$ where $Q_n = F_n = \{0,\dots ,n-1\}$, $q_0 = 0$, and the transition function $\delta _n$ is defined by $\delta _n(i,a) = i+1$ for $0 \le i \le n-1$.

Then we define the DFA recognizing $E(L_n,d_p,k)$ by $A' = (Q_n',\{a,b\},\delta _n',q_0,F_n')$ where $Q_n' = F_n' = Q_n \cup \{p_1,\dots ,p_k\}$ and the transition function defined by

• $\delta _n'(i,a) = i+1$ for $0 \le i < n-1$,

• $\delta _n'(n-1,a) = p_1$,

• $\delta _n'(i,b) = p_1$ for $0 \le i < n-1$,

• $\delta _n'(p_i,a) = \delta _n'(p_i,b) = p_{i+1}$ for $1 \le i < k$.

Every state i, $0 \le i \le n-1$, is reachable on the word $a^i$ and every state $p_i$, $1 \le i \le k$ is reachable on the word $b^i$. The states $0 \le i,i' \le n-1$ are distinguished by the word $b^{k-i}$ and the states $p_i, p_i'$, $1 \le i,i' \le k$ are also distinguished by the word $b^{k-i}$. The states i, $0 \le i \le n-1$ and $p_j$, $1 \le j \le k$ are distinguished by the word $a^{n-j} b^k$. Thus, there are $n+k$ reachable states and they are all pairwise distinguishable.    $\square$

Proposition 4

Let L be a prefix-free regular language recognized by a minimal n-state DFA $A = (Q, \varSigma , \delta , q_0, F)$. Then there is a DFA B with at most $(n-1) k +2 - \frac{k(k-1)}{2}$ states that recognizes the neighbourhood $E(L,d_p,k)$.

Proof

Let $A' = (Q', \varSigma , \delta , q_0', F')$ be the DFA constructed for the neighbourhood $E(L,d_p,k)$ as in Proposition 1. Since L is prefix-free, A must be non-exiting. That is, A has a single final state with no outgoing transitions. This property creates additional unreachable states in the DFA $A'$ for $E(L,d_p,k)$.

For all non-final states $q \in Q - F$, the state (q, 1) is reachable only if either $\varphi _A(q) = 1$ or there is a transition from a final state to q. However, since A is non-exiting, no final states may have any outgoing transitions, so the only states q where (q, 1) is reachable are those with $\varphi _A(q) = 1$. However, for all such states q, the states (q, i) with $2 \le i \le k+1$ are unreachable. Thus, to reach the upper bound on the number of states, the number of states q with $\varphi _A(q) = 1$ must be minimized if $k \ge 2$. If $k = 1$, then for each state $q \in Q - F$, either (q, 1) is reachable or $(q,k+1)$ is reachable, so the number of states with $\varphi _A(q) = 1$ need not be minimized.

By Proposition 1 (b) elements of the set $S_{ur} = \{(q,j) \mid q \in Q-F, 2 \le j \le k+1, j > \varphi _A(q) \}$ are unreachable as states of $A'$ (even without assuming that L(A) is prefix-free. Let $q_f$ be the sole final state of A. The set $S_{ur}$ is minimized when exactly one non-final state $q_i$ in the DFA A for each $1 \le i \le k$ has a shortest path of length i that reaches $q_f$. In this case, we have $|S_{ur}| = \frac{k(k-1)}{2}$.

Thus, in order to maximize the number of reachable states of $A'$, the DFA A has a single final state and a single state $q_1$ with $\varphi _A(q_1) = 1$ if $k \ge 2$, giving us at most $(n-2) k + k + 2 - \frac{k(k-1)}{2} = (n-1) k + 2 - \frac{k(k-1)}{2}$ states of $A'$ which are reachable.    $\square$

Next we present a lower bound construction that matches the bound of Proposition 4.

Lemma 3

There exists a DFA A with n states recognizing a prefix-free regular language such that a DFA recognizing the neighbourhood $E(L(A),d_p,k)$ requires at least $(n-1) k + 2 - \frac{k(k-1)}{2}$ states.

Proof

We define a DFA $A_n = (Q_n,\varSigma _n,\delta _n,q_0,F)$, shown in Fig. 2, by choosing

$$\begin{aligned} Q_n = \{0,\dots ,n-1\}, \varSigma _n = \{a_1,\dots ,a_{n-3},b\}, \end{aligned}$$

$q_0 = 0$, $F = \{n-1\}$, and the transition function $\delta _n$ is given by

• $\delta _n(0,a_i) = i$ for $i = 1,\dots ,n-3$,

• $\delta _n(i,a_i) = i$ for $i = 1,\dots ,n-3$,

• $\delta _n(i,a_{i+1}) = i+1$ for $i = 1,\dots ,n-4$,

• $\delta _n(n-3,b) = n-2$, $\delta _n(n-2,b) = 0$, $\delta _n(0,b) = n-1$.

Fig. 2.

The DFA $A_n$.

We transform $A_n$ into the DFA $A_n' = (Q_n',\varSigma _n,\delta _n',q_0',F')$ via the construction from Proposition 1. To determine the reachable states of $Q_n'$, we first note that the state (0, 1) is reachable as it is the initial state. Note that the initial state is (0, 1) since $\varphi _{A_n}(0) = 1$. The final state $n-1$ is reachable on the word b. Now consider states $p_1,\dots ,p_k$. The state $p_\ell$ is reachable on the word $b^{\ell +1}$ by first reading b to reach the final state and $b^\ell$ to reach the state $p_\ell$.

Now consider states of the form $(i,j) \in (Q_n - \{0,n-1\}) \times \{2,\dots ,k+1\}$. Recall that states (i, 1) are unreachable for any state $i \in Q_n$ with $\varphi _{A_n} > 1$. Then for states $i \in Q_n$ with $\varphi _{A_n} > k$ and each $2 \le j \le k+1$, we can reach state (i, j) from (0, 1) via the word $a_i^{j-1}$. For states $i \in Q_n$ with $\varphi _{A_n} \le k$, we can reach state (i, j) via the word $a_i^{j-1}$ for $j = 2,\dots ,\varphi _{A_n}(i)$ and states (i, j) with $j > \varphi _{A_n}(i)$ are unreachable by definition of $A_n'$.

Finally, we can reach state $(n-2,2)$ via the word $a_{n-3} b$ and states $(n-2,j)$ are unreachable for $j > 2$ since $\varphi _{A_n}(n-2) = 2$. Thus the number of unreachable states in $(Q_n - \{0,n-1\}) \times \{2,\dots ,k+1\}$ is

$$\begin{aligned} \sum _{i=n-k}^{n-2} |\{i\} \times \{\varphi _{A_n}(i) + 1, \dots , k+1\}| = \sum _{i=1}^k |\{i+1,\dots ,k+1\}| = \sum _{i=1}^k i = \frac{k(k-1)}{2}. \end{aligned}$$

Thus, the number of reachable states is

$$\begin{aligned} (n-2) \cdot k + 2 - \frac{k(k-1)}{2} + k = (n-1) \cdot k + 2 - \frac{k(k-1)}{2}. \end{aligned}$$

Now, we show that all reachable states are pairwise inequivalent. First, note that as a final state of A, $n-1$ is not equivalent to a state of the form (i, j) in $A'$. Next, we distinguish states of the form (i, j) from states of the form $p_\ell$. For each $1 \le i \le n-3$, reading the word $a_i^k$ from state (i, j) takes the machine to state $(i,\min \{\varphi _A(i),k+1\})$. Then subsequently reading $a_{i+1} a_{i+2} \cdots a_{n-3} bbb$ takes the machine to the final state $n-1$. However, for every state $p_\ell$, reading $a_i^k$ forces the machine beyond state $p_k$, after which there are no transitions defined. The state $(n-2,2)$ is distinguished from all $p_\ell$ by the word $b^{2+k}$, (0, 1) by $b^{1+k}$, and $n-1$ by $b^k$.

Next, without loss of generality, let $\ell < \ell '$ and consider states $p_\ell$ and $p_{\ell '}$. Choose $z = b^{k-\ell }$. The string z takes state $p_\ell$ to the state $p_k$, where it is accepted. However, the computation on string z from state $p_{\ell '}$ is undefined since ${\ell ' + k - \ell > k}$.

Finally, we consider states of the form (i, j). Let $i < i'$ and consider states (i, j) and $(i',j')$. Let $z = a_{i+1} a_{i+2} \cdots a_{n-3} bbb b^k$. From state (i, j), the word z goes to state $n-1$ on $a_{i+1} \cdots a_{n-3} bbb$. Then by reading $b^k$ from state $n-1$, we reach state $p_k$, an accepting state. However, when reading z from state $(i',j')$, we immediately reach state $p_{j'+1}$ on $a_{i+1}$, since the transition on $a_{i+1}$ is defined only for states (0, 1) and (i, j). Since the rest of the word z is of length greater than k, reading it takes us to state $p_k$ with no further defined transitions for the rest of the word.

Next, consider the state (i, j) and $(i,j')$, where $j < j'$. First, consider the case when $\varphi _{A_n}(i) > k$. Then let $z = a_i^{k-j}$. Reading z from (i, j) takes us to state (i, k), which is a final state. However, from $(i,j')$, reading z brings us to state $(i,k+1)$ and so the computation is rejected.

Now, consider the case when $\varphi _{A_n}(i) \le k$. Let $z = b b^{k-j-1}$. From state (i, j), reading b takes the machine to state $p_{j+1}$ and reading $b^{k-j-1}$ puts the machine in the accepting state $p_k$. However, reading z from $(i,j')$ takes us to state $p_k$ with $b^{j'-j}$ still unread since $j'+k-j-1 > k$ and thus, with no further transitions available, the computation is rejected.

Thus, we have shown that there are $(n-1) \cdot k + 2 - \frac{k(k-1)}{2}$ reachable states and that all reachable states are pairwise inequivalent.    $\square$

Combining Proposition 4 and Lemma 3 we have:

Theorem 4

Let L be a prefix-free regular language. For $n > k \ge 0$, if ${{{\mathrm{sc}}}(L) = n}$, then

$$\begin{aligned} {{\mathrm{sc}}}(E(L,d_p,k)) \le (n-1)\cdot k + 2 - \frac{k(k-1)}{2}, \end{aligned}$$

and this bound can be reached in the worst case.

The construction of Lemma 3 that establishes the lower bound for Theorem 4 uses an alphabet of size $n-2$, where n is the number of states of the DFA. The below result establishes that the size of the alphabet cannot be reduced.

Proposition 5

Let A be a DFA recognizing a prefix-free regular language with n states. If the state complexity of $E(L(A),d_p,k)$ equals $(n-1) k + 2 - \frac{k(k-1)}{2}$, then the alphabet of A needs at least $n-2$ letters.

Proof

Let $A = (Q,\varSigma ,\delta ,q_0,F)$ with $|Q| = n$. Let $A' = (Q',\varSigma ,\delta ', q_0' F')$ be the DFA recognizing $E(L(A),d_p,k)$ constructed in Proposition 1. Recall that as an automaton recognizing a prefix-free regular language A must be non-exiting. That is, A has a single final state $q_f$ and it cannot have any outgoing transitions. Recall also from the proof of Proposition 4 that in order for $A'$ to have the maximal number of states $(n-1) k + 2 - \frac{k(k-1)}{2}$, a necessary condition is that there can be only one state $q_1$ with $\varphi _A(q_1) = 1$ and one state $q_2$ with $\varphi _A(q_2) = 2$.

Now for all $q \in Q - \{q_f,q_1,q_2\}$, $\varphi _A(q) \ge 3$. Recall that since the sole final state $q_f$ has no outgoing transitions, states (q, 1) are reachable only if $\varphi _A(q) = 1$. Then by definition of the transition function $\delta '$, if $\varphi _A(q) \ge 3$, the state (q, 2) can only be reached by a direct transition from a state q with $\varphi _A(q) = 1$. Thus, $q_1$ must have $n-2$ outgoing transitions—one for each state q with $\varphi _A(q) \ge 3$ and one additional transition to the final state $q_f$. Note that $q_2$ requires no direct transition from $q_1$ since $\varphi _A(q_2) = 2$ and thus $(q_2,2)$ is the only reachable state of the form $(q_2,j)$.

Furthermore, since A contains a final state $q_f$ with no outgoing transitions, no additional symbols are required to reach $p_1$, as it can be reached from $q_f$ via a direct transition on any symbol.

Since A is a DFA and $q_1$ has at least $n-2$ outgoing transitions, the cardinality of the alphabet must be at least $n-2$.    $\square$

5 Conclusion

We have given tight state complexity bounds for the prefix-distance neighbourhood of, respectively, finite, prefix-closed, and prefix-free languages. As can, perhaps, be expected the bound for prefix-closed languages is relatively easier to obtain and the matching lower bound construction uses a binary alphabet. The upper bound constructions for the finite and the prefix-free languages are more involved and the lower bound constructions use a variable size alphabet. Furthermore, we have shown that, in both cases, the alphabet size is optimal.

Since the reversal of a DFA is not, in general, deterministic, the state complexity bounds for suffix-distance (or factor-distance) neighbourhoods differ significantly from the corresponding bounds for prefix-distance neighbourhoods. Tight lower bounds are not known for suffix-distance neighbourhoods of general regular languages [14] or for various sub-regular language families. Such questions can be a topic for further research.