计算理论笔记（五）Grammar Function

Grammar(unrestricted grammar)

• Context-free grammar: $A\to w$
• Grammar: $uAv\to w$

Definition

• A grammar is a 4-tuple $G=(V,\Sigma ,S,R)$
  • $V$ is an alphabet
  • $\Sigma \subseteq V$ is the set of terminals
    • $V-\Sigma$ is the set of non-terminals
  • $S\in V-\Sigma$ is the start symbol
  • $R$, the set of rules, is a finite subset of $(V^{\ast }(V-\Sigma )V^{\ast })\times V^{\ast }$
• $(u,v)\in R,u\to v$
• derives in one step, $u\underset{G}{\Rightarrow }v$
• derives ,$u{\underset{G}{\Rightarrow }}^{\ast }v$
• $G$ generates $w\in {\Sigma }^{\ast }$ if $S\underset{G}{\Rightarrow }w$
• L(G)={w∈Σ∗L(G)=\{w\in\Sigma^*L(G)={w∈Σ∗: $S\underset{G}{\Rightarrow }{w}^{\ast }\}$, $G$ generates $L(G)$

Example

• The following grammar generates {anbncn:n≥1}\{a^nb^nc^n:n\ge1\}{anbncn:n≥1}
  • $S\to ABCS$
  • $BA\to AB$, $CA\to AC$, $CB\to BC$
  • $S\to T_c$, $C{T}_c\to T_{c}c$
  • $B{T}_c\to B{T}_b$, $B{T}_b\to T_{b}b$
  • $A{T}_b\to A{T}_a$, $A{T}_a\to T_{a}a$
  • $T_a\to e$

Theorem

• A language is generated by a grammar iff it is semidecided by a TM
• $G\to M$
  • $M=$ on input $w$
    1. layer $0$ is {S}\{S\}{S}
    2. for $i=1,2,3,\dots$
    3. construct layer $i$ from layer $i-1$
    4. if $w$ in layer $i$, halts
• $M\to G$
  • WLOG, assume that
    • $K\cap \Sigma =\varnothing$
    • H={h}H=\{h\}H={h}, and if $M$ halts, it is always in configuration $(h,⊳\underline{\bigsqcup })$
  • $G=(V,{\Sigma }^{\prime },S,R)$
    • V=K∪Σ∪{S,⊲}V=K\cup\Sigma\cup\{S,\lhd\}V=K∪Σ∪{S,⊲}
    • Σ′=Σ−{⊳,⊔}\Sigma'=\Sigma-\{\rhd,\sqcup\}Σ′=Σ−{⊳,⊔}
    • $R=$
      1. $S\to ⊳\bigsqcup h⊲$
      2. $⊳\bigsqcup S\to e$, $⊲\to e$
      3. writing: if $M$ has $\delta (q,a)=(p,b)$, $G$ has $bp\to aq$
      4. moving right: if $M$ has $\delta (q,a)=(p,\to )$, $G$ has $abp\to aqb$
      5. moving left: if $M$ has $\delta (q,a)=(p,\leftarrow )$, $G$ has $pa\to aq$

Undecidable problems about grammar

• G1 Problem
  • AG={A_G=\{AG​={“$G$”“$w$”: $G$ is a grammar that generates $w\}$
    • $f($“$M$”“$w$”$)=$“$G$”“$w$”
    • $H⪯A_G$

Rice’s Theorem

• Suppose $\mathcal{L}$ is a proper, non-empty subset of all R.E. languages, the following is undecidable: given a grammar $G$, is $L(G)\in \mathcal{L}$?

Theorem

• Language: NDCFG={ND_\mathrm{CFG}=\{NDCFG​={"$G_1$"“$G_2$”: $G_1$ and $G_2$ are two CFGs with $L(G_1)\cap L(G_2)\ne \varnothing \}$
• Reduction by computation history

Numerical function

• We say a TM $M$ computes a numerical function $f:{\mathbb{N}}^k\to \mathbb{N}$ if for any $n_1,n_2,\dots ,n_k\in \mathbb{N}$, $M(\mathrm{b}\mathrm{i}\mathrm{n}(n_1),\mathrm{b}\mathrm{i}\mathrm{n}(n_2),\dots ,\mathrm{b}\mathrm{i}\mathrm{n}(n_k))=\mathrm{b}\mathrm{i}\mathrm{n}(f(n_1,n_2,\dots ,n_k))$

Basic functions

• for any $k\ge 0$, the $k$-ary zero function is $\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}(n_1,\dots ,n_k)=0$
• for any $k,j\ge 0$, the $j$-th $k$-ary identity function is ${\mathrm{i}\mathrm{d}}_{k,j}(n_1,\dots ,n_k)=n_j$
• for any $n\in \mathbb{N}$, the successor function $\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{c}(n)=n+1$

Operations

• Composition: given $g:{\mathbb{N}}^k\to \mathbb{N}$ and $h_1,h_2,\dots ,h_k:{\mathbb{N}}^t\to \mathbb{N}$, the composition of $g$ with $h_1,h_2,\dots ,h_k$ is the $t$-ary function $f(n_1,n_2,\dots ,n_t)=g(h_1(n_1,n_2,\dots ,n_t),h_2(n_1,n_2,\dots ,n_t),\dots ,h_k(n_1,n_2,\dots ,n_t))$
• Recursive definition: given $g:{\mathbb{N}}^k\to \mathbb{N}$ and $h:{\mathbb{N}}^{k+2}\to \mathbb{N}$, the function defined recursively by $g$ and $h$ is the $(k+1)$-ary function $f(n_1,n_2,\dots ,n_k,0)=g(n_1,n_2,\dots ,n_k),f(n_1,n_2,\dots ,n_k,m+1)=h(n_1,n_2,\dots ,n_k,m,f(n_1,n_2,\dots ,n_k,m))$
• Given $f:{\mathbb{N}}^k\to \mathbb{N}$, there exists $g:{\mathbb{N}}^{t+k}\to \mathbb{N}$ such that $g(m_1,\dots ,m_t,n_1,\dots ,n_k)=f(n_1,\dots ,n_k)$

Example: $\mathrm{p}\mathrm{l}\mathrm{u}\mathrm{s}(m,n)=m+n$

• $\mathrm{p}\mathrm{l}\mathrm{u}\mathrm{s}(m,0)={\mathrm{i}\mathrm{d}}_{2,1}(m,0)=m$
• $\mathrm{p}\mathrm{l}\mathrm{u}\mathrm{s}(m,n+1)=\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{c}({\mathrm{i}\mathrm{d}}_{3,3}(m,n,\mathrm{p}\mathrm{l}\mathrm{u}\mathrm{s}(m,n)))$

Definition

• The primitive recursive functions are all the basic functions and all the functions obtained from the basic function by composition and recursive definition
• Collary
  • Compositions of primitive functions are primitive recursive
  • Functions defined recursively by primitive recursive functions are primitive recursive

Example

• $\mathrm{p}\mathrm{l}\mathrm{u}{\mathrm{s}}_2(n)=n+2$
• $\mathrm{p}\mathrm{l}\mathrm{u}\mathrm{s}(m,n)=m+n$
• $\mathrm{m}\mathrm{u}\mathrm{l}(m,n)=m\cdot n$
• $\exp (m,n)=m^n$
• Constant function: $f(n_1,\dots ,n_k)=c$
• Sign function: $\mathrm{s}\mathrm{g}\mathrm{n}(0)=0,\mathrm{s}\mathrm{g}\mathrm{n}(n+1)=1$
• Predessor function: $\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}(0)=0,\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}(n)=n-1$
  • $\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}(0)=0$
  • $\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}(n+1)=\mathrm{s}\mathrm{u}\mathrm{c}\mathrm{c}({\mathrm{i}\mathrm{d}}_{2,1}(n,\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}(n)))$
• Non-negative substraction: m∼n=max⁡{m−n,0}m\sim n=\max\{m-n,0\}m∼n=max{m−n,0}

Definition

• A primitive recursive function is a primitive recursive predicate if it only takes values of $0$ and $1$

Example

• Is-zero function: $\mathrm{i}\mathrm{s}\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}(0)=1,\mathrm{i}\mathrm{s}\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}(n+1)=0$
• Is-positive function: $\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{o}\mathrm{s}(n)=\mathrm{s}\mathrm{g}\mathrm{n}(n)$
• Great-or-equal function: $\mathrm{g}\mathrm{e}\mathrm{q}(m,n)=\mathrm{i}\mathrm{s}\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}(n\sim m)$

Theorem

• The negation of a primitive recursive predicate is primitive recursive
  • $\neg p(m)=1\sim p(m)$
• The disjunction and conjunction of primitive recursive predicates are primitive recursive
  • $p(m)\vee q(m)=1\sim \mathrm{i}\mathrm{s}\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}(p(m)+q(m))$
  • $p(m)\wedge q(m)=1\sim \mathrm{i}\mathrm{s}\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}(p(m)\cdot q(m))$
• Let $g,h:{\mathbb{N}}^k\to \mathbb{N}$ be primitive recursive functions and $p:{\mathbb{N}}^k\to \mathbb{N}$ be a primitive recursive predicate, the following function is primitive recursive
  • $f(n_1,\dots ,n_k)=\{\begin{array}{l}g(n_1,\dots ,n_k),\mathrm{i}\mathrm{f}p(n_1,\dots ,n_k)\\ h(n_1,\dots ,n_k),\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$
  • $f=p\cdot g+(1\sim p)\cdot h$

Example

• Equal function: $\mathrm{e}\mathrm{q}(m,n)=\mathrm{g}\mathrm{e}\mathrm{q}(m,n)\wedge \mathrm{g}\mathrm{e}\mathrm{q}(n,m)$
• Remainder function: $\mathrm{r}\mathrm{e}\mathrm{m}(m,n)$ is the remainder of $m$ divided by $n$
  • $\mathrm{r}\mathrm{e}\mathrm{m}(0,n)=0$
  • $\mathrm{r}\mathrm{e}\mathrm{m}(m+1,n)=\{\begin{array}{l}0,\mathrm{i}\mathrm{f}\mathrm{e}\mathrm{q}(\mathrm{r}\mathrm{e}\mathrm{m}(m,n),\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}(n))\\ \mathrm{r}\mathrm{e}\mathrm{m}(m,n)+1,\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$
• Division function: $\mathrm{d}\mathrm{i}\mathrm{v}(m,n)$ is the quotient of $m$ divided by $n$
  • $\mathrm{d}\mathrm{i}\mathrm{v}(0,n)=0$
  • $\mathrm{d}\mathrm{i}\mathrm{v}(m+1,n)=\{\begin{array}{l}\mathrm{d}\mathrm{i}\mathrm{v}(m,n)+1,\mathrm{i}\mathrm{f}\mathrm{e}\mathrm{q}(\mathrm{r}\mathrm{e}\mathrm{m}(m,n),\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{d}(n))\\ \mathrm{d}\mathrm{i}\mathrm{v}(m,n),\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$
• Digit function: $\mathrm{d}\mathrm{i}\mathrm{g}\mathrm{i}\mathrm{t}(m,n,p)$ is the $m$-th digit of the base-$p$ representation of $n$
  • $\mathrm{d}\mathrm{i}\mathrm{g}\mathrm{i}\mathrm{t}(m,n,p)=\mathrm{d}\mathrm{i}\mathrm{v}(\mathrm{r}\mathrm{e}\mathrm{m}(n,p^m),p^{m-1})$
• If $f(m,n)$ is primitive recursive, so is ${\mathrm{s}\mathrm{u}\mathrm{m}}_f(m,n)={\sum }_{k=0}^{n}f(m,k)$
  • ${\mathrm{s}\mathrm{u}\mathrm{m}}_f(m,0)=f(m,0)$
  • ${\mathrm{s}\mathrm{u}\mathrm{m}}_f(m,n+1)={\mathrm{s}\mathrm{u}\mathrm{m}}_f(m,n)+f(m,n+1)$

Theorem

• If a function is primitive recursive, it is computable

Theorem

• There exists a function that is computable but not primitive recursive
• Claim: The set of all unary primitive recursive functions can be lexicographically enumerated by a TM
  • $M=$
    1. enumerate all strings in ${\Sigma }^{\ast }$ in lexico order
    2. every time a string is enumerated, output it if it is a valid encoding of a unary primitive recursive function
• Name all the unary primitive recursive functions as $f_0,f_1,f_2,\dots$
• Define $f^{\ast }(n)=f_n(n)+1$, for $\forall n\in \mathbb{N}$
  • $f^{\ast }\ne f_n$, therefore $f^{\ast }$ is not primitive recursive
• $f^{\ast }$ is computable
  • $M=$ on input $n$
    1. enumerat all the unary primitive recursive functions until it output $f_n(n)$
    2. output $f_n(n)+1$