Decision trees for regular factorial languages

Let $\omega=\{0,1,2,\ldots\}$ be the set of nonnegative integers and $\Sigma$ be a finite alphabet with at least two letters. By $\Sigma^{\ast}$, we denote the set of all finite words over the alphabet $\Sigma$, including the empty word $\lambda$. A word $w\in\Sigma^{\ast}$ is called a factor of a word $u\in\Sigma^{\ast}$ if $u=v_{1}wv_{2}$ and $v_{1},v_{2}\in\Sigma^{\ast}$. A language $L\subseteq\Sigma^{\ast}$ is called factorial if it contains all factors of its words. A word $w\in\Sigma^{\ast}$ is called a minimal forbidden word for $L$ if $w\notin L$ and all proper factors of $w$ belong to $L$. We denote by $\mathit{MF}(L)$ the language of minimal forbidden words for $L$. It is known [1] that a factorial language $L$ is regular if and only if the language $\mathit{MF}(L)$ is regular. In particular, a factorial language $L$ with a finite set of minimal forbidden words $\mathit{MF}(L)$ is regular. In this paper, we study arbitrary nonempty regular factorial languages.

A source over the alphabet $\Sigma$ is a triple $I=(G,q_{0},Q)$, where $G$ is a finite directed graph, possibly with multiple edges and loops, in which each edge is labeled with a letter from $\Sigma$ and edges leaving each node are labeled with pairwise different letters, $q_{0}$ is a node of $G$ called initial, and $Q$ is a nonempty set of the graph $G$ nodes called terminal. Note that the source $I$ can be interpreted as a partial deterministic finite automaton.

A path of the source $I$ is an arbitrary sequence $\xi=v_{1},d_{1},\ldots,v_{m},d_{m},v_{m+1}$ of nodes and edges of $G$ such that the edge $d_{i}$ leaves the node $v_{i}$ and enters the node $v_{i+1}$ for $i=1,\ldots,m$. We now define a word $w(\xi)$ from $\Sigma^{\ast}$ in the following way: if $m=0$, then $w(\xi)=\lambda$. Let $m>0$ and let $\delta_{j}$ be the letter attached to the edge $d_{j}$, $j=1,\ldots,m$. Then $w(\xi)=\delta_{1}\cdots\delta_{m}$. We say that the path $\xi$ generates the word $w(\xi)$. Note that different paths which start in the same node generate different words.

We denote by $\Xi(I)$ the set of all paths of the source $I$ each of which starts in the node $q_{0}$ and finishes in a node from $Q$. Let

We say that the source $I$ generates the language $L_{I}$. It is well known that $L_{I}$ is a regular language.

The source $I$ is called everywhere defined over the alphabet $\Sigma$ if exactly $|\Sigma|$ edges leave each node of $G$. Note that these edges are labeled with pairwise different letters from $\Sigma$. The source $I$ is called reduced if, for each node of $G$, there exists a path from $\Xi(I)$, which contains this node. It is known [3] that, for each regular language over the alphabet $\Sigma$, there exists an everywhere defined over the alphabet $\Sigma$ source, which generates this language. Therefore, for each nonempty regular language, there exists a reduced source, which generates this language.

Let $L$ be a regular factorial language and $I=(G,q_{0},Q)$ be a reduced source that generates the language $L$. Since the language $L$ is factorial, we can assume additionally that each node of the graph $G$ is terminal – it will not change the language generated by $I$. The source $I$ will be called t-reduced if it is reduced and each node of the graph $G$ is terminal. Further we will assume that a considered regular factorial language $L$ is nonempty and it is given by a t-reduced source, which generates this language.

Let $L$ be a regular factorial language over the alphabet $\Sigma$. For any natural $n$, denote $L(n)=L\cap\Sigma^{n}$, where $\Sigma^{n}$ is the set of words over the alphabet $\Sigma$, which length is equal to $n$. We consider two problems related to the set $L(n)$. The problem of recognition: for a given word from $L(n)$, we should recognize it using attributes (queries) $l_{1}^{n},\ldots,l_{n}^{n}$, where $l_{i}^{n}$, $i\in\{1,\ldots,n\}$, is a function from $\Sigma^{n}$ to $\Sigma$ such that $l_{i}^{n}(a_{1}\cdots a_{n})=a_{i}$ for any word $a_{1}\cdots a_{n}\in\Sigma^{n}$. The problem of membership: for a given word from $\Sigma^{n}$, we should recognize if this word belongs to the set $L(n)$ using the same attributes. To solve these problems, we use decision trees over $L(n)$.

A decision tree over $L(n)$ is a marked finite directed tree with root, which has the following properties:

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  The root and the edges leaving the root are not labeled.

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  Each node, which is not the root nor terminal node, is labeled with an attribute from the set $\{l_{1}^{n},\ldots,l_{n}^{n}\}$.

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  Each edge leaving a node, which is not a root, is labeled with a letter from the alphabet $\Sigma$.

A decision tree over $L(n)$ is called deterministic if it satisfies the following conditions:

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  Exactly one edge leaves the root.

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  For any node, which is not the root nor terminal node, the edges leaving this node are labeled with pairwise different letters.

Let $\Gamma$ be a decision tree over $L(n)$. A complete path in $\Gamma$ is any sequence $\xi=v_{0},e_{0},\ldots,v_{m},e_{m},v_{m+1}$ of nodes and edges of $\Gamma$ such that $v_{0}$ is the root, $v_{m+1}$ is a terminal node, and $v_{i}$ is the initial and $v_{i+1}$ is the terminal node of the edge $e_{i}$ for $i=0,\ldots,m$. We define a subset $\Sigma(n,\xi)$ of the set $\Sigma^{n}$ in the following way: if $m=0$, then $\Sigma(n,\xi)=\Sigma^{n}$. Let $m>0$, the attribute $l_{i_{j}}^{n}$ be attached to the node $v_{j}$, and $b_{j}$ be the letter attached to the edge $e_{j}$, $j=1,\ldots,m$. Then

Let $L(n)\neq\emptyset$. We say that a decision tree $\Gamma$ over $L(n)$ solves the problem of recognition for $L(n)$ nondeterministically if $\Gamma$ satisfies the following conditions:

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  Each terminal node of $\Gamma$ is labeled with a word from $L(n)$.

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  For any word $w\in L(n)$, there exists a complete path $\xi$ in the tree $\Gamma$ such that $w\in\Sigma(n,\xi)$.

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  For any word $w\in L(n)$ and for any complete path $\xi$ in the tree $\Gamma$ such that $w\in\Sigma(n,\xi)$, the terminal node of the path $\xi$ is labeled with the word $w$.

We say that a decision tree $\Gamma$ over $L(n)$ solves the problem of recognition for $L(n)$ deterministically if $\Gamma$ is a deterministic decision tree, which solves the problem of recognition for $L(n)$ nondeterministically.

We say that a decision tree $\Gamma$ over $L(n)$ solves the problem of membership for $L(n)$ nondeterministically if $\Gamma$ satisfies the following conditions:

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  Each terminal node of $\Gamma$ is labeled with a number from the set $\{0,1\}$.

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  For any word $w\in\Sigma^{n}$, there exists a complete path $\xi$ in the tree $\Gamma$ such that $w\in\Sigma(n,\xi)$.

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  For any word $w\in\Sigma^{n}$ and for any complete path $\xi$ in the tree $\Gamma$ such that $w\in\Sigma(n,\xi)$, the terminal node of the path $\xi$ is labeled with the number $1$ if $w\in L(n)$ and with the number $0$, otherwise.

We say that a decision tree $\Gamma$ over $L(n)$ solves the problem of membership for $L(n)$ deterministically if $\Gamma$ is a deterministic decision tree which solves the problem of membership for $L(n)$ nondeterministically.

Let $\Gamma$ be a decision tree over $L(n)$. We denote by $h(\Gamma)$ the maximum number of nodes in a complete path in $\Gamma$ that are not the root nor terminal node. The value $h(\Gamma)$ is called the depth of the decision tree $\Gamma$.

We denote by $h_{L}^{ra}(n)$ ($h_{L}^{rd}(n)$) the minimum depth of a decision tree over $L(n)$, which solves the problem of recognition for $L(n)$ nondeterministically (deterministically). If $L(n)=\emptyset$, then $h_{L}^{ra}(n)=h_{L}^{rd}(n)=0$.

We denote by $h_{L}^{ma}(n)$ ($h_{L}^{md}(n)$) the minimum depth of a decision tree over $L(n)$, which solves the problem of membership for $L(n)$ nondeterministically (deterministically). If $L(n)=\emptyset$, then $h_{L}^{ma}(n)=h_{L}^{md}(n)=0$.

Let $I=(G,q_{0},Q)$ be a t-reduced source over the alphabet $\Sigma$. A path of the source $I$ is called a cycle of the source $I$ if there is at least one edge in this path, and the first node of this path is equal to the last node of this path. A cycle of the source $I$ is called elementary if nodes of this cycle, with the exception of the last node, are pairwise different.

The source $I$ is called simple if every two different elementary cycles of the source $I$ do not have common nodes. Let $I$ be a simple source and $\xi$ be a path of the source $I$. The number of different elementary cycles of the source $I$, which have common nodes with $\xi$, is denoted by $cl(\xi)$ and is called the cyclic length of the path $\xi$. The value

is called the cyclic length of the source $I$.

Let $I$ be a simple source, $C$ be an elementary cycle of the source $I$, and $v$ be a node of the cycle $C$. Beginning with the node $v$, the cycle $C$ generates an infinite periodic word over the alphabet $\Sigma$. This word will be denoted by $W(I,C,v)$. We denote by $r(I,C,v)$ the minimum period of the word $W(I,C,v)$. The source $I$ is called dependent if there exist two different elementary cycles $C_{1}$ and $C_{2}$ of the source $I$, nodes $v_{1}$ and $v_{2}$ of the cycles $C_{1}$ and $C_{2}$, respectively, and a path $\pi$ of the source $I$ from $v_{1}$ to $v_{2}$, which satisfy the following conditions: $W(I,C_{1},v_{1})=W(I,C_{2},v_{2})$ and the length of the path $\pi$ is a number divisible by $r(I,C_{1},v_{1})$. If the source $I$ is not dependent, then it is called independent. Next theorem follows immediately from Theorem 2.1 [5].

Let $L$ be a nonempty regular factorial language over the alphabet $\Sigma$. For any natural $n$, we define a parameter $T_{L}(n)$ of the language $L$. If $L(n)=\emptyset$, then $T_{L}(n)=0$. Let $L(n)\neq\emptyset$, $w=a_{1}\cdots a_{n}\in L(n)$, and $J\subseteq\{1,\ldots,n\}$. Denote $L(w,J)=\{b_{1}\cdots b_{n}\in L(n):b_{j}=a_{j},j\in J\}$ (if $J=\emptyset$, then $L(w,J)=L(n)$) and $M_{L}(n,w)=\min\{|J|:J\subseteq\{1,\ldots,n\},|L(w,J)|=1\}$. Then

Note that, for any word $w\in L(n)$, $M_{L}(n,w)$ is the minimum number of letters of the word $w$, which allow us to distinguish it from all other words belonging to $L(n)$. One can show that $h_{L}^{ra}(n)=T_{L}(n)$.

Note that in general case (when we consider not only factorial languages) the classification of reduced sources depending on the minimum depth of decision trees solving the problem of recognition nondeterministically is more complicated [4]. In particular, there exists a dependent simple reduced source $I_{0}$ (see Fig. 1) with the initial node labeled with the symbol $+$ and the unique terminal node labeled with the symbol $*$ that generates the regular language $L_{0}=\{0^{i}10^{j}:i,j\in\omega\}$ over the alphabet $\{0,1\}$, which is not factorial and for which $H_{L_{0}}^{ra}(n)=O(1)$.

For a regular factorial language $L$ over the alphabet $\Sigma$, we denote by $L^{C}$ its complementary language $\Sigma^{\ast}\setminus L$. The notation $|L|=\infty$ means that $L$ is an infinite language, and the notation $|L|<\infty$ means that $L$ is a finite language.

In this section, we assume that each regular factorial language over the alphabet $\Sigma$ is given by a t-reduced source $I$, which generates the considered language denoted by $L_{I}$. To study all possible types of joint behavior of functions $H_{L_{I}}^{rd}$, $H_{L_{I}}^{ra}$, $H_{L_{I}}^{md}$, and $H_{L_{I}}^{ma}$, we consider five classes of regular factorial languages $\mathcal{F}_{1},\ldots,\mathcal{F}_{5}$ described in the columns 2–4 of Table 1. In particular, $\mathcal{F}_{1}$ consists of all regular factorial languages $L_{I}$ for which the source $I$ is an independent simple source and $cl(I)=0$. It is easy to show that the complexity classes $\mathcal{F}_{1},\ldots,\mathcal{F}_{5}$ are pairwise disjoint, and each regular factorial language $L_{I}$ belongs to one of these classes. The behavior of functions $H_{L_{I}}^{rd}$, $H_{L_{I}}^{ra}$, $H_{L_{I}}^{md}$, and $H_{L_{I}}^{ma}$ for languages from these classes is described in the last four columns of Table 1. For each class, the results considered in Table 1 for the functions $H_{L_{I}}^{rd}$ and $H_{L_{I}}^{ra}$ follow directly from Theorems 1 and 2.

We now consider the behavior of the functions $H_{L_{I}}^{md}$ and $H_{L_{I}}^{ma}$ for each of the classes $\mathcal{F}_{1},\ldots,\mathcal{F}_{5}$. Let $I=(G,q_{0},Q)$ be a t-reduced source over the alphabet $\Sigma$, which generates a regular factorial language.

Let $L_{I}\in\mathcal{F}_{1}$. Since $cl(I)=0$, $G$ is a directed acyclic graph, and the language $L_{I}$ is finite. Using Theorem 3 we obtain $H_{L_{I}}^{md}(n)=O(1)$ and $H_{L_{I}}^{ma}(n)=O(1)$.

Let $L_{I}\in\mathcal{F}_{2}$. Since $cl(I)=1$, $G$ is a graph containing a cycle, and the language $L_{I}$ is infinite. By Lemma 4.2 [5], $|L_{I}(n)|=O(1)$. Therefore $L_{I}^{C}\neq\emptyset$. Using Theorem 3 we obtain $H_{L_{I}}^{md}(n)=\Theta(n)$ and $H_{L_{I}}^{ma}(n)=\Theta(n)$.

Let $L_{I}\in\mathcal{F}_{3}$. Since $cl(I)\geq 2$, $G$ is a graph containing a cycle, and the language $L_{I}$ is infinite. By Lemma 4.2 [5], $|L_{I}(n)|=O(n^{cl(I)})$. Therefore $L_{I}^{C}\neq\emptyset$. Using Theorem 3 we obtain $H_{L_{I}}^{md}(n)=\Theta(n)$ and $H_{L_{I}}^{ma}(n)=\Theta(n)$.

Let $L_{I}\in\mathcal{F}_{4}$. Since $I$ is not an independent simple source, $G$ is a graph containing a cycle, and the language $L_{I}$ is infinite. We know that $L_{I}^{C}\neq\emptyset$. Using Theorem 3 we obtain $H_{L_{I}}^{md}(n)=\Theta(n)$ and $H_{L_{I}}^{ma}(n)=\Theta(n)$.

Let $L_{I}\in\mathcal{F}_{5}$. Then $L_{I}^{C}=\emptyset$. Using Theorem 3 we obtain $H_{L_{I}}^{md}(n)=O(1)$ and $H_{L_{I}}^{ma}(n)=O(1)$.

We now show that the classes $\mathcal{F}_{1},\ldots,\mathcal{F}_{5}$ are nonempty. For simplicity, we assume that $\Sigma=E$, where $E=\{0,1\}$. It is easy to generalize the considered examples to the case of an arbitrary finite alphabet $\Sigma$ with at least two letters. In the examples of sources, the initial node is labeled with the symbol $+$, and all nodes are terminal.

Denote by $I_{1}$ the source over the alphabet $E$ depicted in Fig. 2. One can show that $I_{1}$ is an independent simple t-reduced source and $cl(I_{1})=0$. This source generates the language $L_{I_{1}}=\{\lambda,0\}$, which is factorial. Therefore $L_{I_{1}}\in\mathcal{F}_{1}$.

Denote by $I_{2}$ the source over the alphabet $E$ depicted in Fig. 3. One can show that $I_{2}$ is an independent simple t-reduced source and $cl(I_{2})=1$. This source generates the language $L_{I_{2}}=\{0^{i}:i\in\omega\}$, which is factorial. Therefore $L_{I_{2}}\in$ $\mathcal{F}_{2}$.

Denote by $I_{3}$ the source over the alphabet $E$ depicted in Fig. 4. One can show that $I_{3}$ is an independent simple t-reduced source and $cl(I_{1})=2$. This source generates the language $L_{I_{3}}=\{0^{i}1^{j}:i,j\in\omega\}$, which is factorial. Therefore $L_{I_{3}}\in$ $\mathcal{F}_{3}$.

Denote by $I_{4}$ the source over the alphabet $E$ depicted in Fig. 5. One can show that $I_{4}$ is a dependent simple t-reduced source generating the language $L_{I_{4}}=\{0^{i}1^{j}0^{k}:i,k\in\omega,j\in\{0,1\}\}$, which is factorial. It is clear that $L_{I_{4}}^{C}\neq\emptyset$. Therefore $L_{I_{4}}\in$ $\mathcal{F}_{4}$.

Denote by $I_{5}$ the source over the alphabet $E$ depicted in Fig. 6. One can show that $I_{5}$ is a t-reduced source that is not simple. This source generates the language $L_{I_{5}}=E^{\ast}$, which is factorial. It is clear that $L_{I_{5}}^{C}=\emptyset$. Therefore $L_{I_{5}}\in$ $\mathcal{F}_{5}$.

A regular factorial language $L$ can have different t-reduced sources, which generate it. However, for each of such sources $I$, the language $L_{I}=L$ will belong to the same complexity class. Let us assume the contrary: there exist a regular factorial language $L$ and two t-reduced sources $I_{1}$ and $I_{2}$, which generate it and for which languages $L_{I_{1}}$ and $L_{I_{2}}$ belong to different complexity classes. Then, for some pair $bc\in\{rd,ra,md,ma\}$, the functions $H_{L_{I_{1}}}^{bc}$ and $H_{L_{I_{2}}}^{bc}$ have different behavior, but this is impossible since $H_{L_{I_{1}}}^{bc}(n)=H_{L_{I_{2}}}^{bc}(n)$ for any natural $n$.

Let $E=\{0,1\}$, $\alpha\in E^{\ast}$, and $\alpha\neq\lambda$. We denote by $L(\alpha)$ the language over the alphabet $E$, which consists of all words from $E^{\ast}$ that does not contain $\alpha$ as a factor. This is a regular factorial language with $\mathit{MF}(L(\alpha))=\{\alpha\}$. The following theorem indicates for each nonempty word $\alpha\in E^{\ast}$ the complexity class $\mathcal{F}_{i}$ to which the language $L(\alpha)$ belongs.