Lecture 9¶

Example

Show that $R_{TM} = \{\langle M \rangle : M \text{ is a TM with } L(M) \text{ being regular}\}$ is not recursive.

Proof

Suppose $M_1$ decides on $\overline{R_{TM}}$, then we construct $M_2$ runs on $H$.

$M_2$ = on input $\langle M, w \rangle$,

construct a TM $M'$ = on input $x$
1. run $M$ on $w$.
2. run $U$ on $x$. ($U$ is the universal TM)
\[ L(M') = \left\{ \begin{aligned} & \phi, && \text{if $M$ does not halt on $w$}, \\ & L(U) = H, && \text{if $M$ halts on $w$}. \end{aligned} \right. \]

NOTE that $\phi$ is regular, but $L(U) = H$ is not regular (because it's even not recursice).
run $M_1$ on $\langle M' \rangle$.
return the result of $M_1$.

Since $M_2$ can not decide $H$, thus $M_1$ can not decide $\overline{R_{TM}}$. Therefore $\overline{R_{TM}}$ is not recursive and so is $R_{TM}$.

Example

Show that $CF_{TM} = \{\langle M \rangle : M \text{ is a TM with } L(M) \text{ being context-free}\}$ is not recursive.

Proof

The same reduction pattern of $R_{TM}$. Since $\phi$ is context-free and $H$ not.

Example

Show that $REC_{TM} = \{\langle M \rangle : M \text{ is a TM with } L(M) \text{ being recursive}\}$ is not recursive.

From the examples above, we can conclude some pattern of not recursive langauges.

Rice's Theorem¶

Theorem

Suppose $\mathcal{L}(P)$ is a non-empty proper subset of all recursively enumarable languages, then the following language is NOT recursive.

\[ R(P) = \{\langle M \rangle : M \text{ is a TM with } L(M) \in \mathcal{L}(P)\}. \]

Proof

Case 1. $\phi \notin \mathcal{L}(P)$

There exists $A \in \mathcal{L}(P)$, so there exists a TM $M_A$ that semidecides $A$. Suppose $M_R$ decides $R(P)$, then we construct $M_H$ on $H$.

$M_H$ = on input $\langle M, w \rangle$,

construct a TM $M^*$ = on input $x$
1. run $M$ on $w$
2. run $M_A$ on $x$
\[ L(M^*) = \left\{ \begin{aligned} & \phi, && \text{if $M$ does not halt on $w$}, \\ & L(M_A) = A, && \text{if $M$ halts on $w$}. \end{aligned} \right. \]
run $M_R$ on $\langle M^* \rangle$.
return the result of $M_R$.

Case 2. $\phi \in \mathcal{L}(P)$

Then $\phi \notin \overline{\mathcal{L}(P)}$, we can reduce $H$ to $\overline{R(P)}$ like in case 1. And thus $R(P)$ is not recursive.

Summary¶

To prove a language $A$ recursive, we can prove

By definition: construct a Turing machine.
By reduction: $A \le \text{known recursive language}$.

To prove a language $A$ not recursive, we can prove

By reduction: $\text{known recursive language} \le A$.
- The mostly used language is $H$.

To prove a language $A$ recursively enumerable, we can prove

By definition: construct a Turing machine.
By reduction: $A \le \text{known recursively enumerable language}$.

To prove a language $A$ not recursively enumerable, we can prove

By reduction: $\text{known non-recursively enumerable language} \le A$.
By the following theorem.

Theorem

If $A$ and $\overline{A}$ are recursively enumerable, then $A$ is recursive.

Proof

Since $A$ and $\overline{A}$ are recursively enumerable, then there exists $M_1$ and $M_2$ that semidecides $A$ and $\overline{A}$ respectively.

Then we construct a $M_3$ that decides $A$.

$M_3$ = on input $x$

run $M_1$ and $M_2$ on $x$ in parallel.
if $M_1$ halts on $x$,
accept $x$.
if $M_2$ halts on $x$,
reject $x$.

Example

Since $H$ is recursively enumerable and not recursive, then $\overline{H}$ is not recursively enumerable.

Example

Show that $A = \{\langle M \rangle : M \text{ is a TM that halts on some input}\}$ is recursively enumerable.

Proof

$M_A$ = on input $\langle M \rangle$

for $i$ = $1, 2, 3, \dots$
for $s$ = $s_1, s_2, \dots, s_i$ ($s_i \in L$)
run $M$ on $s$ for $i$ steps,
if $M$ halts on $s$ within $i$ steps,
halt.

Closure Property¶

Theorem

If $A$ and $B$ are recursive langauges, then so as

$A \cup B$.
$A \cap B$.
$\overline{A}$.
$A \circ B$.
$A^*$.

If $A$ and $B$ are recursively enumerable langauges, then so as

$A \cup B$.
$A \cap B$.
$A \circ B$.
$A^*$.

NOTE not so as $\overline{A}$, a counter example is $H$.

Enumerator¶

There are some extra properties of Turing machine.

Definition

A Turing machine $M$ enumerates a language $L$ if for some state $q$,

\[ L = \{w | (s, \triangleright \underline{\sqcup}) \vdash_M^* (q, \triangleright \underline{\sqcup}w) \}. \]

And $L$ is Turing enumerable.

$M$ has no input.

Theorem

A language $L$ is Turing enumerable iff it's recursively enumerable.

Proof

If $L$ is finite, it's trivial. Assume $L$ is infinite,

$\Rightarrow$. There exists a TM $M$ that enumerates $L$, we construct $M'$.

$M'$ = on input $x$

run $M$ to enumerate $L$.
every time $M$ outputs a string $w$,
if $w$ equals to $x$,
halt.

Then $M'$ semidecides $L$ and $L$ is recursively enumerable.

$\Leftarrow$. There exists a TM $M$ that semidecides $L$, we construct $M'$.

$M'$ =

for $i$ = $1, 2, 3, \dots$
for $s$ = $s_1, s_2, \dots, s_i$ ($s_i \in L$)
run $M$ on $s$ for $i$ steps.
if $M$ halts on $s$ within $i$ steps.
output $s$.

Here the "some state $q$" of $M'$ is the halting state.

Definition

Let $M$ be a TM that enumerates $L$, then $M$ is lexicographically enumerates $L$ if whenever

\[ (q, \triangleright \underline{\sqcup} w) \vdash_M^+ (q, \triangleright \underline{\sqcup} w') \]

we have $w'$ after $w$ in lexicographical order.

Theorem

$L$ is lexicographically enumerable iff it's recursive.

Proof

$\Leftarrow$. There exists $M$ decides $L$, we construct $M'$.

$M'$ on input $x$

enumerate all strings in lexicographical order.
for each string $s$,
run $M$ on $s$,
if $M$ accepts $s$,
output $s$.

$\Rightarrow$. There exists $M$ lexicographically enumerate $L$, we construct $M'$.

$M'$ = on input $x$

run $M$ to enumerate $L$.
every time $M$ outputs a string $w$,
if $w$ equals to $x$,
accept $x$.
reject $x$.

最后更新: 2024.01.29 15:36:20 CST
创建日期: 2023.12.04 16:11:01 CST