计算理论笔记（四）Undecidability Enumerator

Halting Problem

• H={H=\{H={“$M$”“$w$”: $M$ is a TM that halts on $w\}$
• $H$ is R.E.
• If $H$ is recursive, then all R.E. languages are recursive
• $H$ is not recursive

Lemma: $H$ is recursively enumerable

• Universal TM: $U=$ on input “$M$”“$w$”
  1. run $M$ on “$w$”
  2. if $M$ halts on “$w$”, halts on “$M$”“$w$”
• $U$ halts on “$M$”“$w$” $⟺$ $M$ halts on $w⟺$ “$M$”“$w$”$\in H$
• $U$ semidecides $H$

Lemma: If $H$ is recursive, so are all R.E. languages

• Also: $H$ is complete for the class of R.E. languages
• Let $A$ be a R.E. language and $M$ be a TM that semidecides $A$
• Is $w\in A$? $⟺$ Does $M$ halt on $w$? $⟺$ “$M$”“$w$”$\in H$
• If $H$ is recursive, there exists a TM $M_H$ that decides $H$
• $M_{\mathrm{A}}=$ on input $w$
  1. run $M_{\mathrm{H}}$ on “$M$”“$w$”
  2. if $M_{\mathrm{H}}$ accepts “$M$”“$w$”, accept $w$
  3. else reject

Theorem: $H$ is not recursive

• H={H=\{H={“$M$”“$w$”: $M$ is a TM that halts on $w\}$
• H′={H'=\{H′={“$M$”: $M$ is a TM that halts on “$M$”$\}$
• Hd={H_d=\{Hd​={“$M$”: $M$ is a TM that does not halt on “$M$”$\}$
• Claim: If $H$ is recursive, then $H_d$ is also recursive
  • If $H$ is recursive, there exists a TM $M_H$ that decides $H$
  • $M_{\mathrm{d}}=$ on input “$M$”
    1. run $M_{\mathrm{H}}$ on “$M$”““$M$””
    2. if $M_{\mathrm{H}}$ accepts “$M$”““$M$””, reject “$M$”
    3. else accept
• Claim: $H_d$ is not R.E.
  • Suppose $H_d$ is R.E., there exists a TM $D$ semidecides $H_d$
  • $D$ on input “$M$”
    • halts if “$M$”$\in H_d$, therefore $M$ does not halt on “$M$”
    • does not halt if “$M$”$\notin H_d$, therefore $M$ halts on “$M$”
  • Barber Dilemma: $D$ on input “$D$”
    • halts if $D$ does not halt on “$D$”
    • does not halt if $D$ halts on “$D$”

Lemma: $L$ is recursive iff $L$ and $\overline{L}$ is both R.E.

• sufficiency: recursive under complement
  • $L$ and $\overline{L}$ are both recursive, therefore also R.E.
• necessity: $M_L$ and $M_{\overline{L}}$ semidecides $L$ and $\overline{L}$, respectively
  • $M^{\ast }=$ on input $w$
    1. run $M_{\mathrm{L}}$ and $M_{\overline{\mathrm{L}}}$ on $w$ in parallel
    2. if $M_{\mathrm{L}}$ halts, accept $w$
    3. else if $M_{\overline{\mathrm{L}}}$ halts, reject $w$

Theorem: The class of R.E. languages is not closed under complement

• $H$ and $\overline{H}$ are both R.E., then $H$ is recursive

Definition

• Let $A$ and $B$ be two languages over ${\Sigma }_A$ and ${\Sigma }_B$, respectively
• $A$ reduction from $A$ to $B$ is a computable function $f:{\Sigma }_A^{\ast }\to {\Sigma }_B^{\ast }$

Theorem

• Suppose that there is a reduction $f$ from $A$ to $B$, if $B$ is recursive, so is $A$
• $M_{\mathrm{A}}=$ on input $w$
  1. compute $f(w)$
  2. run $M_{\mathrm{B}}$ on $f(w)$
  3. if $M_{\mathrm{B}}$ accepts $f(w)$, accept $w$
  4. else reject
• $A⪯B$

Theorem: The following problems are undecidable

• A={A=\{A={“$M$”: $M$ is a TM that halts on $e\}$
  • Given a TM $M$ and a string $w$
  • $M_w=$ on input $w$
    1. run $M$ on $u$
    2. if $M$ halts on $u$, halts
  • $M_w$ halts on $e$ (“$M_w$”$\in A$) $⟺$ $M_w$ halts on every input $⟺M_w$ halts on $w$ (“$M$”“$w$”$\in H$)
  • Notice: “$M_w$” is used as input of $A$
  • $H⪯A$
• B={B=\{B={“$M$”: $M$ is a TM that halts on some strings$\}$
  • $f($“$M$”“$w$”$)=$“$M_w$”
  • $H⪯B$
• C={C=\{C={“$M$”: $M$ is a TM that halts on every string$\}$
  • $f($“$M$”“$w$”$)=$“$M_w$”
  • $H⪯C$
• D={D=\{D={"$M_1$"“$M_2$”: $M_1$ and $M_2$ are two TMs that halts on the same string$\}$
  • Let $M^{\ast }$ be a TM with $S\in H⟺M^{\ast }$ halts on every input
  • $f($“$M$”$)=$“$M$”“$M^{\ast }$”
  • $C⪯D$
• E1={E_1=\{E1​={“$M$”: $M$ is a TM that $L(M)$ is regular$\}$
• E2={E_2=\{E2​={“$M$”: $M$ is a TM that $L(M)$ is context-free$\}$
• E3={E_3=\{E3​={“$M$”: $M$ is a TM that $L(M)$ is recursive$\}$
  • Given a TM $M$ and a string $w$
  • $f($“$M$”“$w$”$)=$“$M_w$”
  • $M_w=$ on input $v$
    1. run $M$ on $w$
    2. run $U$ on “$M$”“$v$”
  • if “$M$”“$w$”$\notin H$, does not halt in step 1, $L(M_w)=\varnothing$
  • if “$M$”“$w$”$\in H$, $L(M_w)=L(U)=H$
  • $L(M_w)$ is regular/context-free/recursive iff “$M$”“$w$”$\notin H$
  • $H⪯\overline{E_1},\overline{E_2},\overline{E_3}$
• There is a certain fixed TM $M$, for which the following problem is undecidable:
  • Given a string $w$, does $M$ halt on $w$?
  • Language: $L(M)$
  • Run $U$ on “$M$”“$w$”

Rice’s Theorem

• Suppose $\mathcal{L}$ is a proper, non-empty subset of all R.E. languages, the following is undecidable: given a TM $M$, is $L(M)\in \mathcal{L}$?
• S={S=\{S={“$M$”: $M$ is a TM that $L(M)\in \mathcal{L}\}$
• Proof:
  • Assume $\varnothing \notin \mathcal{L}$, otherwise let $\mathcal{L}$ be its complement
  • Let a language $A\in \mathcal{L}$ and a TM $M_{\mathrm{A}}$ that semidecides $A$
  • $f($“$M$”“$w$”$)=$ on input $u$
    1. run $M$ on $w$
    2. run $M_{\mathrm{A}}$ on $u$
  • if “$M$”“$w$”$\notin H$, $L(f($“$M$”“$w$”$))=\varnothing$
  • if “$M$”“$w$”$\in H$, $L(f($“$M$”“$w$”$))=L(M_A)=A\in \mathcal{L}$
  • $H⪯S$
• Given a TM $M$, does $M$ halt on $e$?
  • L={L\mathcal{L}=\{LL={L: $L$ is R.E. and $e\in L\}$
• Given a TM $M$, does $M$ halt on some input?
  • L={L\mathcal{L}=\{LL={L: $L$ is R.E. and $L\ne \varnothing \}$
• Given a TM $M$, does $M$ halt on every input?
  • L={Σ∗}\mathcal{L}=\{\Sigma^*\}L={Σ∗}
• Given a TM $M$, is $L(M)$ regular/context-free/recursive?
  • L={L\mathcal{L}=\{LL={L: $L$ is regular/context-free/recursive$\}$

Theorem: A language $L$ is R.E. iff it is Turing enumerable

• Assume $L$ is infinite
• sufficiency:
  • Let $M$ be a TM that enumerates $L$
  • Construct a TM $M^{\prime }$ to semidecide $L$
  • $M^{\prime }=$ on input $w$
    1. run $M$ to enumerate $L$
    2. every time $M$ outputs a string, compare it with $w$, and halt if they match
• necessity:
  • Let $M$ be a TM that semidecides $L$
  • Construct a TM $M^{\prime }$ to enumerate $L$
  • Name all strings over $\Sigma$ as $s_1,s_2,\dots$ in lexicographical order
  • $M^{\prime }=$ on input $w$
    1. for $i=1,2,\dots$
    • run $M$ on $s_1,s_2,\dots ,s_i$ at most $i$ steps
    2. if $M$ hatls on $s_j$ in $i$ steps, output $s_j$

Definition

• Let $M$ be a TM enumerates $L$, we say $M$ lexicographically enumerate $L$ if whenever $(q,⊳\underline{\bigsqcup }w){⊢}_M^{+}(q,⊳\underline{\bigsqcup }{w}^{\prime })$
  • $w^{\prime }$ is after $w$ in lexico order

Theorem: A language is recursive iff it is lexico Turing enumerable

• sufficiency:
  • Let $M$ be a TM that decides $L$
  • Construct a TM $M^{\prime }$ to lexico enumerate $L$
  • $M^{\prime }=$ on input $w$
    1. generate all strings voer $\Sigma$ in lexico order
    2. run $M$ on each string $w$ to see if $w\in L$
    3. if $M$ accepts $w$, accept
    4. else reject
• necessity:
  • Let $M$ be a TM that lexico enumerates $L$
  • Construct a TM $M^{\prime }$ to decide $L$
  • $M^{\prime }=$ on input $w$
    1. run $M$ to enumerate $L$ until $M$ outputs a string is after $w$ in lexico order
    2. if $w$ is among strings that output, accept $w$
    3. else reject