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[Solved]: Undecidability of REGULAR_TM (Detail within Proof)

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Problem Detail: 

I'm reading through Sipser's Intro to the Theory of Computation for a class, and I'm having trouble understanding one of the examples in the book.

The example shows how $REGULAR_{TM}$, defined as the problem of determining if a Turing machine recognizes a regular language, is undecidable. The way they do it is by showing a reduction from $A_{TM}$, which is the acceptance problem for Turing Machines (the acceptance problem for some computational model is the task of determining whether or not it accepts a given input string).

From what I understand, all that needs to be done is show that if $REGULAR_{TM}$ was decidable by some TM $R$, then another TM could use $R$ to decide $A_{TM}$, which would be a contradiction, since $A_{TM}$ is undecidable.

They construct a TM $S$ that can be used to decide $A_{TM}$. Where I'm lost is through their use of an intermediate TM, $M_2$. Here is the full description:

  • $S$ = "On input $\langle M, w\rangle$, where $M$ is a TM and $w$ is a string:
    1. Construct the following TM $M_2$.
      • $M_2$ = "On input $x$:
        1. If $x$ has the form $0^{n}1^{n}$, accept.
        2. If $x$ does not have this form, run $M$ on input $w$ and accept if $M$ accepts $w$."
    2. Run $R$ on input $\langle M_2 \rangle$.
    3. If $R$ accepts, accept; if $R$ rejects, reject."

My two main questions are:

  1. Why are we allowed to do step 2 in the construction of $M_2$? If we could just say "run $M$ on $w$ and accept if $M$ accepts", then wouldn't that be a way to show that $A_{TM}$ is decidable?
  2. What exactly is the role of step 1 in the construction of $M_2$? The book says that the purpose of this TM is not to be run, but just to feed its description into $R$. The TM recognizes $\{0^n1^n\mid n \ge 0\}$ if $M$ does not accept $w$, and $\Sigma^*$ if it does, but I don't see how it does that from the description. Also, I don't see how $A_{TM}$ can be decided by using this (I'm thinking it's that the TM $M_2$ only outputs a regular language if the TM $M$ accepts $w$, but I don't understand how it can do that).

Any help is greatly appreciated, thanks!

Asked By : Andrew DiNunzio

Answered By : Ariel

This proof assumes, for the purpose of contradiction, that there exists a Turing machine $R$ deciding $R_{TM}=\left\{\langle M\rangle | \hspace{1mm} L(M) \text{ is regular}\right\}$, and uses the existence of $R$ to construct a Turing machine deciding $A_{TM}=\left\{\langle M,w\rangle | \hspace{1mm} M \text{ accepts $w$}\right\}$.

$R$ halts on every input (since it decides $R_{TM}$), so there is no problem running it (it will finally halt and either accept or reject the input). This avoids the problem of running $M$, some arbitrary Turing machine, on some input $w$, since we do not know that $M$ will finally halt.

Now, Sipser constructs a new Turing machine, $M_2$, such that:

$L(M_2) \text{ is regular} \iff M \text{ accepts $w$}$

If the above holds, then you would have $\langle M_2\rangle\in L(R) \iff M \text{ accepts $w$}$, so you could find out if $M$ accepts $w$ simply by running $R$ (which always halts) on $M_2$.

To see why this property holds, note that $M_2$ always accepts strings of the form $0^n1^n$, and it accepts any string not of this form iff $M$ accepts $w$. This means that $L(M_2)=\left\{0^n1^n | n\ge 0\right\}$ if $M$ does not accept $w$, and $\Sigma^*$ (which is regular) otherwise.

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Question Source : http://cs.stackexchange.com/questions/57580

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