Edit-Distance Between Visibly Pushdown Languages

Abstract

We study the edit-distance between two visibly pushdown languages. It is well-known that the edit-distance between two context-free languages is undecidable. The class of visibly pushdown languages is a robust subclass of context-free languages since it is closed under intersection and complementation whereas context-free languages are not. We show that the edit-distance problem is decidable for visibly pushdown languages and present an algorithm for computing the edit-distance based on the construction of an alignment PDA. Moreover, we show that the edit-distance can be computed in polynomial time if we assume that the edit-distance is bounded by a fixed integer k.

Appendix

Context-free grammar (CFG). A context-free grammar (CFG) G is a four-tuple $G = (V, \varSigma , R, S)$, where V is a set of variables, $\varSigma$ is a set of terminals, $R \subseteq V \times (V\cup \varSigma )^*$ is a finite set of productions and $S\in V$ is the start variable. Let $\alpha A \beta$ be a word over $V \cup \varSigma$, where $A \in V$ and $A \rightarrow \gamma \in R$. Then, we say that A can be rewritten as $\gamma$ and the corresponding derivation step is denoted $\alpha A \beta \Rightarrow \alpha \gamma \beta$. A production $A \rightarrow t \in R$ is a terminating production if $t \in \varSigma ^*$. The reflexive, transitive closure of $\Rightarrow$ is denoted by and the context-free language generated by G is . We say that a variable $A \in V$ is nullable if .

Proposition 3 . Let $L \subseteq \varSigma ^*$ and $L' \subseteq \varSigma ^*$ be the languages over $\varSigma$. Then,

$$d(L,L') \le \max \{ \mathrm{lsw}(L), \mathrm{lsw}(L')\}$$

holds.

Proof

Assume that $\mathrm{lsw}(L) = m$ and $\mathrm{lsw}(L') = n$ where $n \le m$. It is easy to see that the edit-distance between two shortest words can be at most m since we can substitute all characters of the shortest word of length n with any subsequence of the longer word and insert the remaining characters.    $\square$

Proposition 5  (Han et al. [5]). Given a PDA $P = (Q, \varSigma , \varGamma , \delta , s, Z_0, F_P)$, we can obtain a shortest word in L(P) whose length is bounded by $2^{m^2 l + 1}$ in $O(n \cdot 2^{m^2 l})$ worst-case time and compute its length in $O(m^4 nl)$ worst-case time, where $m = |Q|$, $n = |\delta |$ and $l = |\varGamma |$.

Proof

Recall that we can convert a PDA into a CFG by the triple construction [6]. Let us denote the CFG obtained from P by $G_P$. Then, $G_P$ has $|Q|^2\cdot |\varGamma | +1$ variables and $|Q|^2 \cdot |\delta |$ productions. Moreover, each production of $G_P$ is of the form $A \rightarrow \sigma BC, A \rightarrow \sigma B, A \rightarrow \sigma$ or $A \rightarrow \lambda$, where $\sigma \in \varSigma$ and $A,B,C \in V$. Since we want to compute the shortest word from $G_P$, we can remove the occurrences of all nullable variables from $G_P$. Then, we pick a variable A that generates the shortest word $t \in \varSigma ^*$ among all variables and replace its occurrence in $G_P$ with t. We can compute the shortest word of L(P) by iteratively removing occurrences of such variables. We describe the algorithm in Algorithm 1.

Since a production of $G_P$ has at most one terminal followed by two variables, the length of the word to be substituted is at most $2^m -1$ when we replace mth variable. Since we replace at most $|Q|^2 \cdot |\varGamma |$ variables to have the shortest word, the length of the shortest word in L(P) can be at most $2^{|Q|^2 \cdot |\varGamma |+1}$. Since there are at most 2|R| occurrences of variables in R and |V| variables, we replace $\frac{2|R|}{|V|}$ occurrences of a given variable on average. Therefore, the worst-case time complexity for finding a shortest word is $O(n \cdot 2^{m^2 l})$. We also note that we can compute only the length of the shortest word in $O(m^4 nl)$ worst-case time by encoding a shortest word to be substituted with a binary number.    $\square$

Theorem 6 Given two VPAs $A_i = (\varSigma , \varGamma _i, Q_i, s_i, F_i, \delta _{i,c}, \delta _{i,r}, \delta _{i,l})$ for $i = 1,2,$ we can compute the edit-distance between $L(A_1)$ and $L(A_2)$ in $O( (m_1 m_2)^5 \cdot n_1n_2 \cdot (l_1l_2)^{10k})$ worst-case time, where $m_i = |Q_i|, n_i = |\delta _{i,c}| + |\delta _{i,r}| + |\delta _{i,l}|, l_i = |\varGamma _i|$ for $i =1,2$ and $k = \max \{ \mathrm{lsw}(L(A_1)), \mathrm{lsw}(L(A_2))\}.$

Proof

In the proof of Lemma 4, we have shown that we can construct an alignment PDA $\mathcal{A}(A_1,A_2) = (Q_E ,\varOmega , \varGamma _E, s_E, F_E, \delta _E)$ that accepts all possible alignments between two VPAs $A_1$ and $A_2$ of length up to k. From Proposition 5, we can compute the edit-distance in $O(m^4 n l)$ time, where $m = |Q_E|$, $n = |\delta _E|$ and $l = |\varGamma _E|$. Recall that

$$m = m_1m_2 \cdot \Biggl (\sum _{i=0}^{2k} l_1^{i}\Biggr ) \cdot \Biggl (\sum _{i=0}^{2k} l_2^{i}\Biggr ), \;\;n = m_1m_2 \cdot n_1n_2 \cdot \Biggl (\sum _{i=0}^{2k} l_1^{i}\Biggr ) \cdot \Biggl (\sum _{i=0}^{2k} l_2^{i}\Biggr ),$$

and $l = l_1l_2$. Note that

$$ \Biggl (\sum _{i=0}^{2k} l_1^{i}\Biggr ) \in O(l_1^{2k}) \text { and }\Biggl (\sum _{i=0}^{2k} l_2^{i}\Biggr ) \in O(l_2^{2k}) $$

if $l_1, l_2 > 0$.

Therefore, the time complexity of computing the edit-distance between two VPAs $A_1$ and $A_2$ is

$$ (m_1 m_2)^5 \cdot n_1 n_2 \cdot \Biggl (\sum _{i=0}^{2k} l_1^{i}\Biggr )^5 \cdot \Biggl (\sum _{i=0}^{2k} l_2^{i}\Biggr )^5 \cdot l_1l_2 \in O( (m_1 m_2)^5 \cdot n_1 n_2 \cdot (l_1l_2)^{10k}), $$

where k is the maximum of the length of the two shortest words from $L(A_1)$ and $L(A_2)$.    $\square$

Corollary 8. Given two VCAs $A_1,A_2$ and a positive integer $k \in \mathbb {N}$ in unary such that $d(L(A_1),L(A_2)) \le k$, we can compute the edit-distance between $L(A_1)$ and $L(A_2)$ in polynomial time.

Proof

If $l_1 = l_2 = 1$,

$$ \Biggl (\sum _{i=0}^{2k} l_1^{i}\Biggr ) \in O(k) \text { and }\Biggl (\sum _{i=0}^{2k} l_2^{i}\Biggr ) \in O(k). $$

If we replace $l_1^{2k}$ and $l_2^{2k}$ by k from the time complexity, we obtain the time complexity $O( (m_1 m_2)^5 \cdot n_1 n_2 \cdot k^{10})$ which is polynomial in the size of input.