Prefix Sorting DFAs: a Recursive Algorithm

In the setting of the previous works on the topic [1, 8, 21], the problem that we want to solve is defined as follows. Consider a deterministic finite automaton (DFA) such that (i) all edges entering the same state have the same label, (ii) each state is reachable from the initial state, (iii) each state is co-reachable, that is, it is either final or it allows reaching a final state, (iv) the initial state has no incoming edges. Then, determine the min/max partition of the set of states (see Section 2.2 for the formal definition of min/max partition, and see Figure 1 for an example).

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  Assumptions (ii) and (iii) are standard assumptions in automata theory, because all states that do not satisfy these assumptions can be removed without changing the accepted language.

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  In this setting, all the non-initial states have an incoming edge, but the initial state has no incoming edges. This implies for some state $u$ it may hold $I_{u}=\emptyset$ (remember that $I_{u}$ is the set of all infinite strings that can be read starting from $u$ and following edges in a backward fashion, see the introduction), so Kim et al. [21] need to perform a tedious case analysis which also takes finite strings into account in order to define the min/max-partition (in particular, the minimum and maximum strings reaching the initial state are both equal to the empty string). However, we can easily avoid this complication by means of the same trick used in [5]; we can add a self-loop to the initial state, and the label of the self-loop is a special character $\#$ smaller than any character in the alphabet. Intuitively, $\#$ plays the same role as the termination character in the Burrows-Wheeler transform of a string, and since $\#$ is the smallest character, adding this self-loop does not affect the min/max-partition (see [21] for details).

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  Notice that the initial state and the set of all final states play no role in the definition of the min/max partition; this explains why, more generally, it will be expedient to consider deterministic graphs rather than DFAs (otherwise we would need to artificially add an initial state and add a set of final states when we recursively build a graph in our algorithm). Equivalently, one may assume to work with semiautomata in which the transition function is not necessarily total. This justifies the assumptions that we will make in Section 2.2.

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  Some recent papers [6, 7] have shown that assumptions (i) and (iv) can be removed. The partial order $\preceq_{\mathcal{A}}$ is defined analogously, and all the algorithms for building $\preceq_{\mathcal{A}}$ that we have mentioned still work. Indeed, if a state $u$ is reached by edges with the distinct labels, we need to only consider all edges with the smallest label when computing $\min_{u}$ and all edges with the largest label when computing $\max_{u}$; moreover, we once again assume that the initial state has a self loop labeled $\#$. The only difference is that assumption (i) implies that $m\leq n^{2}$ ($n$ being the number of states, $m$ being the number of edges) because each state can have at most $n$ incoming edges, but this is no longer true if we remove assumption (i). As a consequence, the running time of our algorithm is no longer $O(n^{2})$ but $O(m+n^{2})$ (and the running time of the $O(n^{2}\log n)$ algorithm in [21] becomes $O(m+n^{2}\log n)$) because we still need to process all edges in the DFA.

To sum up, all the algorithms for computing $\preceq_{\mathcal{A}}$ work on arbitrary DFAs.

Let $\Sigma$ be a finite alphabet. We consider finite, edge-labeled graphs $G=(V,E)$, where $V$ is the set of all nodes and $E\subseteq V\times V\times\Sigma$ is the set of all edges. Up to taking a subset of $\Sigma$, we assume that all $c\in\Sigma$ label some edge in the graph. We assume that all nodes have at least one incoming edge, and all edges entering the same node $u$ have the same label $\lambda(u)$ (input consistency). This implies that an edge $(u,v,a)\in E$ can be simply denoted as $(u,v)$, because it must be $a=\lambda(v)$. In particular, it must be $|E|\leq|V|^{2}$ (and so an $O(|E|)$ algorithm is also a $O(|V|^{2})$ algorithm). If we do not know the $\lambda(u)$’s, we can easily compute them by scanning all edges. In addition, we always assume that $G$ is deterministic, that is, for every $u\in V$ and for every $a\in\Sigma$ there exists at most one $v\in V$ such that $(u,v)\in E$ and $\lambda(v)=a$.

Let $\Sigma^{*}$ be the set of all finite strings on $\Sigma$, and let $\Sigma^{\omega}$ be the set of all (countably) right-infinite strings on $\Sigma$. If $\alpha\in\Sigma^{*}\cup\Sigma^{\omega}$ and $i\geq 1$, we denote by $\alpha[i]\in\Sigma$ the $i^{\text{th}}$ character of $\alpha$ (that is, $\alpha=\alpha[1]\alpha[2]\alpha[3]\dots$). If $1\leq i\leq j$, we define $\alpha[i,j]=\alpha[i]\alpha[i+1]\dots\alpha[j-1]\alpha[j]$, and if $j<i$, then $\alpha[i,j]$ is the empty string $\epsilon$. If $\alpha\in\Sigma^{*}$, then $|\alpha|$ is length of $\alpha$; for every $0\leq j\leq|\alpha|$ the string $\alpha[1,j]$ is a prefix of $\alpha$, and if $0\leq j<|\alpha|$ it is a strict prefix of $\alpha$; analogously, one defines suffixes and strict suffixes of $\alpha$. An occurrence of $\alpha\in\Sigma^{*}$ starting at $u\in V$ and ending at $u^{\prime}\in V$ is a sequence of nodes $u_{1},u_{2},\dots,u_{|\alpha|+1}$ of $V$ such that (i) $u_{1}=u$, (ii) $u_{|\alpha|+1}=u^{\prime}$, (iii) $(u_{i+1},u_{i})\in E$ for every $1\leq i\leq|\alpha|$ and (iv) $\lambda(u_{i})=\alpha[i]$ for every $1\leq i\leq|\alpha|$. An occurrence of $\alpha\in\Sigma^{\omega}$ starting at $u\in V$ is a sequence of nodes $(u_{i})_{i\geq 1}$ of $V$ such that (i) $u_{1}=u$, (ii) $(u_{i+1},u_{i})\in E$ for every $i\geq 1$ and (iii) $\lambda(u_{i})=\alpha[i]$ for every $i\geq 1$. Intuitively, a string $\alpha\in\Sigma^{*}\cup\Sigma^{\omega}$ has an occurrence starting at $u\in V$ if we can read $\alpha$ on the graph starting from $u$ and following edges in a backward fashion.

In the paper, occurrences of strings in $\Sigma^{\omega}$ will play a key role, while occurrences of strings in $\Sigma^{*}$ will be used as a technical tool. For every $u\in V$, we denote by $I_{u}$ the set of all strings in $\Sigma^{\omega}$ admitting an occurrence starting at $u$. Since every node has at least one incoming edge, then $I_{u}\not=\emptyset$.

A total order $\leq$ on a set $V$ if a reflexive, antisymmetric and transitive relation on $V$. If $u,v\in V$, we write $u<v$ if $u\leq v$ and $u\not=v$.

Let $\preceq$ be a fixed total order on $\Sigma$. We extend $\preceq$ to $\Sigma^{*}\cup\Sigma^{\omega}$ lexicographically. It is easy to show that in every $I_{u}$ there is a lexicographically smallest string $\min_{u}$ and a lexicographically largest string $\max_{u}$ (for example, it follows from [21, Observation 8]).

We will often use the following immediate observation. Let $u\in V$, and let $(u_{i})_{i\geq 1}$ be an occurrence of $\min_{u}$. Fix $i\geq 1$. Then, $(u_{j})_{j\geq i}$ is an occurrence of $\min_{u_{i}}$, and $\min_{u}=\min_{u}[1,i-1]\min_{u_{i}}$.

Let $V^{\prime}\subseteq V$. Let $\mathcal{A}$ be the unique partition of $V^{\prime}$ and let $\leq$ be the unique total order on $\mathcal{A}$ such that, for every $I,J\in\mathcal{A}$ and for every $u\in I$ and $v\in J$, (i) if $I=J$, then $\min_{u}=\min_{v}$ and (ii) if $I<J$, then $\min_{u}\prec\min_{v}$. Then, we say that $(\mathcal{A},\leq)$, or more simply $\mathcal{A}$, is the min-partition of $V^{\prime}$. The max-partition of $V^{\prime}$ is defined analogously. Now, consider the set $V^{\prime}\times\{\min,\max\}$, and define $\rho((u,\min))=\min_{u}$ and $\rho((u,\max))=\max_{u}$ for every $u\in V^{\prime}$. Let $\mathcal{B}$ be the unique partition of $V^{\prime}\times\{\min,\max\}$ and let $\leq$ be the unique total order on $\mathcal{B}$ such that, for every $I,J\in\mathcal{B}$ and for every $x\in I$ and $y\in J$, (i) if $I=J$, then $\rho(x)=\rho(y)$ and (ii) if $I<J$, then $\rho(x)\prec\rho(y)$. Then, we say that $(\mathcal{B},\leq)$, or more simply $\mathcal{B}$, is the min/max-partition of $V^{\prime}$.

The main result of this paper will be proving that the min/max partition of $V$ can be determined in $O(|V|^{2})$ time.

Let $G=(V,E)$ be a graph. We will first show how to build the min-partition of $V$ in $O(n^{2})$ time, where $n=|V|$ (Section 4); then, we will show how the algorithm can be adapted so that it builds the min/max-partition in $O(n^{2})$ time (Section 5).

In order to build a min-partition of $V$, we will first classify all minima into three categories (Section 3), so that we can split $V$ into three pairwise-disjoint sets $V_{1},V_{2},V_{3}$. Then, we will show that in $O(n^{2})$ time:

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  we can compute $V_{1},V_{2},V_{3}$ (Section 4.1);

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  we can define a graph $\bar{G}=(\bar{V},\bar{E})$ having $|V_{3}|$ nodes (Section 4.2);

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  assuming that we have already determined the min-partition of $\bar{V}$, we can determine the min-partition of $V$ (Section 4.3).

Analogously, in $O(n^{2})$ time we can reduce the problem of determining the min-partition of $V$ to the problem of determining the min-partition of the set of all nodes of a graph having $|V_{1}|$ (not $|V_{3}|$) nodes (Section 4.4). As a consequence, since $\min\{|V_{1}|,|V_{3}|\}\leq|V|/2=n/2$, we obtain a recursive algorithm whose running time is given by the recurrence:

and we conclude that the running time of our algorithm is $O(n^{2})$.

Let us first show how to compute all $u\in V$ such that $\tau(u)=1$.

Corollary 4.4 yields an algorithm to decide whether $u\in V$ is such that $\tau(u)=1$.

Now, let us show how to determine all $u\in V$ such that $\tau(u)=2$. We can assume that we have already determined all $u\in V$ such that $\tau(u)=1$.

From Corollary 4.5 and Corollary 4.7 we immediately obtain the following result.

Let us sketch the general idea to build a smaller graph for the recursive step. We consider each $u\in V$ such that $\tau(u)=3$, and we follow edges in a backward fashion, aiming to determine a prefix of $\min_{u}$. As a consequence, we discard edges through which no occurrence of $\min_{u}$ can go, and by Corollary 3.3 we can restrict our attention to the nodes $v$ such that $\tau(v)$ is minimal. We proceed like this until we encounter nodes $v^{\prime}$ such that $\tau(v^{\prime})=3$.

Let us formalize our intuition. We will first present some properties that the occurrences of a string $\min_{u}$ must satisfy.

The previous results allow us to give the following definition.

Note that $\ell_{u}$ is well-defined, because (i) it cannot hold $\tau(u_{k})=1$ for every $k\geq 2$ by Lemma 4.10 (indeed, if $\tau(u_{2})=1$, then $(u_{i})_{i\geq 2}$ is an occurrence of $\min_{u_{2}}$ starting at $u_{2}$, and by Lemma 4.10 there exists $2\leq k\leq|V|+2$ such that $\tau(u_{k})\not=1$) and (ii) $\ell_{u}$ does not depend on the choice of $(u_{i})_{i\geq 1}$ by Lemma 4.9. In particular, it must be $\ell_{u}\leq|V|+1$ because $u_{1},u_{2},\dots,u_{\ell_{u}-1}$ are pairwise distinct ($u_{1}$ is distinct from $u_{2},\dots,u_{\ell_{u}-1}$ because $\tau(u_{1})=3$ and $\tau(u_{2})=\tau(u_{3})=\dots\tau(u_{\ell_{u}-1})=1$ by the minimality of $\ell_{u}$).

If $R\subseteq Q$ is a nonempty set of nodes such that for every $u,v\in R$ it holds $\lambda(u)=\lambda(v)$, we define $\lambda(R)=\lambda(u)=\lambda(v)$. If $R\subseteq Q$ is a nonempty set of nodes such that for every $u,v\in R$ it holds $\tau(u)=\tau(v)$, we define $\tau(R)=\tau(u)=\tau(v)$.

Let $R\subseteq Q$ be a nonempty set of states. Let $\mathtt{F}(R)=\operatorname*{arg\,min}_{u\in R^{\prime}}\tau(u)$, where $R^{\prime}=\operatorname*{arg\,min}_{v\in R}\lambda(v)$. Notice that $\mathtt{F}(R)$ is nonempty, and both $\lambda(\mathtt{F}(R))$ and $\tau(\mathtt{F}(R))$ are well-defined. In other words, $\mathtt{F}(R)$ is obtained by first considering the subset $R^{\prime}\subseteq\mathtt{F}(R)$ of all nodes $v$ such that $\lambda(v)$ is as small as possible, and then considering the subset of $R^{\prime}$ of all nodes $v$ such that $\tau(v)$ is as small as possible. This is consistent with our intuition on how we should be looking for a prefix of $\min_{u}$.

Define:

Notice that we also require that a node in $G_{i}(u)$ has not been encountered before. Intuitively, this does not affect our search for a prefix of $\min_{u}$ because, if we met the same node twice, then we would have a cycle where all edges are equally labeled (because by Lemma 4.12 labels can only decrease), and since $\tau(G_{i}(u))=1$ for every $2\leq i\leq\ell_{u}-1$, then no occurrence of the minimum can go through the cycle because if we remove the cycle from the occurrence we obtain a smaller string by Lemma 3.2.

The following technical lemma is crucial to prove that our intuition is correct.

Let $u\in V$ such that $\tau(u)=3$. We define:

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  $\gamma_{u}=\min_{u}[1,\ell_{u}]$;

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  $\mathtt{t}_{u}=\tau(G_{\ell_{u}}(u))\in\{2,3\}$

Now, in order to define the smaller graph for the recursive step, we also need a new alphabet $(\Sigma^{\prime},\preceq^{\prime})$, which must be defined consistently with the mutual ordering of the minima. The next lemma yields all the information that we need.

Now, let $\Sigma^{\prime}=\{(\gamma_{u},\mathtt{t}_{u})\;|\;u\in V,\tau(u)=3\}$, and let $\preceq^{\prime}$ be the total order on $\Sigma^{\prime}$ such that for every distinct $(\alpha,x),(\beta,y)\in\Sigma^{\prime}$, it holds $(\alpha,x)\prec^{\prime}(\beta,y)$ if and only if one of the following is true:

1. 1.

   $\alpha$ is not a prefix of $\beta$ and $\alpha\prec\beta$.

2. 2.

   $\alpha=\beta$, $x=2$ and $y=3$.

3. 3.

   $\beta$ is a strict prefix of $\alpha$.

It is immediate to verify that $\preceq^{\prime}$ is a total order: indeed, $\preceq^{\prime}$ is obtained (i) by first comparing the $\gamma_{u}$’s using the variant of the (total) lexicographic order on $\Sigma^{*}$ in which a string is smaller than every strict prefix of it and (ii) if the $\gamma_{u}$’s are equal by comparing the $\mathtt{t}_{u}$’s, which are elements in $\{2,3\}$.

Starting from $G=(V,E)$, we define a new graph $\bar{G}=(\bar{V},\bar{E})$ as follows:

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  $\bar{V}=\{\bar{u}\;|\;u\in V,\tau(u)=3\}$.

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  The new totally-ordered alphabet is $(\Sigma^{\prime},\preceq^{\prime})$.

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  For every $\bar{u}\in\bar{V}$, we define $\lambda(\bar{u})=(\gamma_{u},\mathtt{t}_{u})$.

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  $\bar{E}=\{(\bar{v},\bar{v})\;|\;\mathtt{t}_{v}=2\}\cup\{(\bar{u},\bar{v})\;|\;\mathtt{t}_{v}=3,u\in G_{\ell_{v}}(v)\}$.

Note that for every $\bar{v}\in\bar{V}$ such that $\mathtt{t}_{v}=3$ and for every $u\in G_{\ell_{v}}(v)$ it holds $\tau(u)=\tau(G_{\ell_{v}}(v))=\mathtt{t}_{v}=3$, so $\bar{u}\in\bar{V}$ and $(\bar{u},\bar{v})\in E$. Moreover, $\bar{G}=(\bar{V},\bar{E})$ satisfies all the assumptions about graphs that we use in this paper: (i) all edges entering the same node have the same label (by definition), (ii) every node has at least one incoming edge (because if $\bar{v}\in\bar{V}$, then $G_{\ell_{v}}(v)\not=\emptyset$ by Lemma 4.13) and (iii) $\bar{G}$ is deterministic (because if $(\bar{u},\bar{v}),(\bar{u},\bar{v}^{\prime})\in\bar{E}$ and $\lambda(\bar{v})=\lambda(\bar{v}^{\prime})$, then $\gamma_{v}=\gamma_{v^{\prime}}$ and $\mathtt{t}_{v}=\mathtt{t}_{v^{\prime}}$, so by the definition of $\bar{E}$ if $\mathtt{t}_{v}=\mathtt{t}_{v^{\prime}}=2$ we immediately obtain $\bar{v}=\bar{u}=\bar{v}^{\prime}$, and if $\mathtt{t}_{v}=\mathtt{t}_{v}^{\prime}=3$ we obtain $u\in G_{\ell_{v}}(v)\cap G_{\ell_{v^{\prime}}}(v^{\prime})$; since by Lemma 4.13 there exist two occurrences of $\min_{v}[1,\ell_{v}-1]=\gamma_{v}[1,\ell_{v}-1]=\gamma_{v^{\prime}}[1,\ell_{v^{\prime}}-1]=\min_{v^{\prime}}[1,\ell_{v^{\prime}}-1]$ starting at $v$ and $v^{\prime}$ and both ending at $u$, the determinism of $G$ implies $v=v^{\prime}$ and so $\bar{v}=\bar{v}^{\prime}$).

Notice that if $\bar{v}\in\bar{V}$ is such that $\mathtt{t}_{v}=2$, then $I_{\bar{v}}$ contains exactly one string, namely, $\lambda(\bar{v})^{\omega}$; in particular, $\min_{\bar{v}}=\max_{\bar{v}}=\lambda(\bar{v})^{\omega}$.

When we implement $G=(V,E)$ and $\bar{G}=(\bar{V},\bar{E})$, we use integer alphabets $\Sigma=\{0,1,\dots,|\Sigma|-1\}$ and $\Sigma^{\prime}=\{0,1,\dots,|\Sigma^{\prime}|-1\}$; in particular, we will not store $\Sigma^{\prime}$ by means of pairs $(\gamma_{u},\mathtt{t}_{u})$’s, but we will remap $\Sigma^{\prime}$ to an integer alphabet consistently with the total order $\preceq^{\prime}$ on $\Sigma^{\prime}$, so that the mutual order of the $\min_{\bar{u}}$’s is not affected.

Let us prove that we can use $\bar{G}=(\bar{V},\bar{E})$ for the recursive step. We will start with some preliminary results.