Weighted Regular Tree Grammars with Storage

TL;DR

This paper introduces weighted regular tree grammars with storage, combining concepts from regular tree grammars, storage, and weighted automata, and explores their properties, logical characterizations, and simplifications.

Contribution

It presents a novel framework for weighted regular tree grammars with storage, including their characterization, support generation, and logical description.

Findings

Support of weighted tree languages can be generated by unweighted grammars.

Weighted tree languages are characterized by a composition of basic concepts.

Results on elimination of chain rules and finite storage types.

Abstract

We introduce weighted regular tree grammars with storage as combination of (a) regular tree grammars with storage and (b) weighted tree automata over multioperator monoids. Each weighted regular tree grammar with storage generates a weighted tree language, which is a mapping from the set of trees to the multioperator monoid. We prove that, for multioperator monoids canonically associated to particular strong bi-monoids, the support of the generated weighted tree languages can be generated by (unweighted) regular tree grammars with storage. We characterize the class of all generated weighted tree languages by the composition of three basic concepts. Moreover, we prove results on the elimination of chain rules and of finite storage types, and we characterize weighted regular tree grammars with storage by a new weighted MSO-logic.

Equations

$$A\in N$$

$$\mathrm{wt}(d)={\mathrm{wt}}^{\prime }(d)$$

$$[[\mathcal{G}]](\xi )$$

$$=\sum _{A\in Z}\sum _{d^{\prime }\in D_{{\mathcal{G}}^{\prime }}(A,\xi ,c_0)}{\mathrm{wt}}^{\prime }(d^{\prime })$$

$$\displaystyle=\sum_{A\in Z}\sum_{d^{\prime}\in D_{\mathcal{G}^{\prime}}(A,\xi,c_{0})}\mathrm{wt}^{\prime}(Z_{0}(\mathrm{true}_{C})\to A(\mathrm{id}_{C}))\Bigl{(}\mathrm{wt}^{\prime}(d^{\prime})\Bigr{)}$$

$$=\sum _{d^{\prime \prime }\in D_{{\mathcal{G}}^{\prime }}(Z_0,\xi ,c_0)}{\mathrm{wt}}^{\prime }(d^{\prime \prime })=[[{\mathcal{G}}^{\prime }]](\xi ).\text{qed}$$

$$L=\{\sigma ({\delta }^n(\alpha ),{\delta }^n(\alpha ))\mid n\ge 0\}$$

$$\tau =(A(p)\to \sigma (A_1(f_1),A_2(f_2)))$$

Full text

• •

For each $A\in Z$, the rule $r=(Z_{0}(\mathrm{true}_{C})\to A(\mathrm{id}_{C}))$ is in $R^{\prime}$ with $\mathrm{wt}^{\prime}(r)=\mathrm{id}_{K}$.

• •

Each rule $r\in R$ is in $R^{\prime}$ with $\mathrm{wt}^{\prime}(r)=\mathrm{wt}(r)$.

It is clear that,

[TABLE]

We can prove that $[\![\mathcal{G}]\!]=[\![\mathcal{G}^{\prime}]\!]$ as follows. Let $\xi\in T_{\Sigma}$. Then

[TABLE]

(a) For every $(S,\Sigma,M_{K})$-rtg $\mathcal{G}$, there is an $(S,\Sigma)$-rtg $\mathcal{G}^{\prime}$ such that $L(\mathcal{G}^{\prime})=\mathrm{supp}([\![\mathcal{G}]\!])$. (b) Moreover, if $(K,{\cdot},1)$ has a decidable ZGP, then $\mathcal{G}^{\prime}$ can be constructed effectively. 2. 2.

Assume that $\lvert\Sigma^{(1)}\rvert\geq 2$. If there is an effective construction of a $\Sigma$-rtg which generates $\operatorname{supp}([\![\mathcal{G}]\!])$ from any given $(\Sigma,M_{K})$-rtg $\mathcal{G}$, then $(K,{\cdot},1)$ has a decidable ZGP.

• •

Drop the sentence “By Lemma 3.2 we may assume that $Z\in N$.”.

• •

Replace “$Z^{\prime}=\{(Z,\bar{z})\mid\bar{z}\in T^{n},[\![\bar{z}]\!]_{\bar{a}}\neq 0\}$” by

“$Z^{\prime}=\{(A,\bar{z})\mid A\in Z,\bar{z}\in T^{n},[\![\bar{z}]\!]_{\bar{a}}\neq~{}0\}$”.

• •

Replace the last paragraph of the proof of $\mathrm{supp}([\![\mathcal{G}]\!])\subseteq L(\mathcal{G^{\prime}})$ by

“Now let $\xi\in\mathrm{supp}([\![\mathcal{G}]\!])$. Then there is some $A\in Z$ and a derivation tree $d\in D_{\mathcal{G}}(A,\xi,c_{0})$ such that $\mathrm{wt}(d)\neq 0$. By Statement (*) there are $\bar{y}\in\mathbb{N}^{n}$ and $d^{\prime}\in D_{\mathcal{G}^{\prime}}((A,\lfloor\bar{y}\rfloor_{\mathrm{dg}(\bar{a})}),\xi,c_{0})$ such that $[\![\bar{y}]\!]_{\bar{a}}=\mathrm{wt}(d)$. Since $[\![\bar{y}]\!]_{\bar{a}}\neq 0$, also $[\![\lfloor\bar{y}\rfloor_{\mathrm{dg}(\bar{a})}]\!]_{\bar{a}}\neq 0$ (by Lemma 4.2). Thus, $(A,\lfloor\bar{y}\rfloor_{\mathrm{dg}(\bar{a})})\in Z^{\prime}$ and $\xi$ is in $L(\mathcal{G}^{\prime})$.”.

• •

Replace the third paragraph of the proof of $L(\mathcal{G^{\prime}})\subseteq\mathrm{supp}([\![\mathcal{G}]\!])$ by

“Now let $\xi\in L(\mathcal{G}^{\prime})$. Then there is a derivation tree $d^{\prime}\in D_{\mathcal{G}^{\prime}}((A,\bar{z}),\xi,c_{0})$ for some $A\in Z$ and $\bar{z}\in T^{n}$ with $[\![\bar{z}]\!]_{\bar{a}}\neq 0$. By Statement (**) there are $\bar{y}\in\mathbb{N}^{n}$ and $d\in D_{\mathcal{G}}(A,\xi,c_{0})$ such that $[\![\bar{y}]\!]_{\bar{a}}=\mathrm{wt}(d)$ and $\bar{z}=\lfloor\bar{y}\rfloor_{\mathrm{dg}(\bar{a})}$. Since $[\![\bar{z}]\!]_{\bar{a}}\neq 0$ also $[\![\bar{y}]\!]_{\bar{a}}\neq 0$ (by Lemma 4.2). Thus, since $K$ is zero-sum free, $\xi$ is in $\mathrm{supp}([\![\mathcal{G}]\!])$.”.

∎

• (i)

$s$ is $(S,\Sigma,K)$-regular.

• (ii)

$s=\mathcal{B}_{\Delta};h(\cal L(\mathcal{H}))$ for some

• –

finite sets $P^{\prime}\subseteq P$, $F^{\prime}\subseteq F$, ranked alphabet $\Delta$ corresponding to $\Sigma$, $P^{\prime}$, and $F^{\prime}$,

• –

ranked alphabet $\Theta$, unambiguous and chain-free $\Theta$-rtg $\mathcal{H}$, and

• –

alphabetic mapping $h\colon T_{\Theta}\to K[T_{\langle\Delta,\Sigma\rangle}]$ such that $h_{1}(\Theta^{(1)})\subseteq\Omega^{(1)}\times\langle\Delta,\Sigma\rangle^{(1)}$.

[TABLE]

is in $\mathrm{Reg}(\operatorname{\mathrm{P}}-op^{1},\Sigma)$, because it can be generated by the $(\operatorname{\mathrm{P}}-op^{1},\Sigma)$-grammar which we obtain from $\mathcal{G}$ of Example 3.1 by dropping its weight structure and weight function (cf. [Gue83]).

On the other hand, we can show by contradiction that $L\not\in\mathrm{Reg}_{\text{nc}}(S,\Sigma)$ for any storage type $S$. For this, we assume that there is a storage type $S$ and a chain-free $(S,\Sigma)$-rtg $\mathcal{G}=(N,Z,R)$ such that $\mathcal{L}(\mathcal{G})=L$. Since $\mathcal{G}$ is chain-free, the root of each derivation tree of $\mathcal{G}$ has to be a $\sigma$-rule, i.e., a rule of the form

[TABLE]

from $R$ for some $A\in Z$, $A_{1},A_{2}\in N$, $p\in P$, and $f_{1},f_{2}\in F$. As $L$ is infinite and, in contrast, $R$ is finite, there have to be two different integers $m,m^{\prime}\in\mathbb{N}$ and a $\sigma$-rule $\tau\in R$ such that $\tau$ is the root of some $d\in D_{\mathcal{G}}(\sigma(\delta^{m}(\alpha),\delta^{m}(\alpha)))$ and also of some $d^{\prime}\in D_{\mathcal{G}}(\sigma(\delta^{m^{\prime}}(\alpha),\delta^{m^{\prime}}(\alpha)))$. Assume that $\tau$ is of the above form. Then $p(c_{0})=1$, $f_{1}(c_{0})$ and $f_{2}(c_{0})$ are defined (where $c_{0}$ is the initial storage configuration of $S$), and there are derivation trees

• •

$d_{1}\in D_{\mathcal{G}}(A_{1},\delta^{m}(\alpha),f_{1}(c_{0}))$ and $d_{1}^{\prime}\in D_{\mathcal{G}}(A_{1},\delta^{m^{\prime}},f_{1}(c_{0}))$, and

• •

$d_{2}\in D_{\mathcal{G}}(A_{2},\delta^{m}(\alpha),f_{2}(c_{0}))$ and $d_{2}^{\prime}\in D_{\mathcal{G}}(A_{2},\delta^{m^{\prime}},f_{2}(c_{0}))$

such that $d=\tau(d_{1},d_{2})$ and $d^{\prime}=\tau(d_{1}^{\prime},d_{2}^{\prime})$. But then $\tau(d_{1},d_{2}^{\prime})\in D_{\mathcal{G}}(\sigma(\delta^{m}(\alpha),\delta^{m^{\prime}}(\alpha)))$ and, hence, also $\sigma(\delta^{m}(\alpha),\delta^{m^{\prime}}(\alpha))\in\mathcal{L}(\mathcal{G})$. This is a contradiction to the assumption $\mathcal{L}(\mathcal{G})=L$. ∎