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I'm trying to understand reduction, this is from my textbook and is not a homework problem or even any exercise, just trying to understand an example they present.

This is the reduction they give:

PROOF We let R be a TM that decides REGULARTM and construct TM S to decide ATM . Then S works in the following manner.

S = “On input $M$, $w$ , 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 $M_{2}$ .

  3. If R accepts, accept ; if R rejects, reject .”

So, we start with a machine that decides whether a language of a TM is regular. And we want to use that to decide if a TM halts on a given input.

My question: What if $w$ does have the form $0^{n} 1^{n}$? Well, $M_{2}$ accepts that string just cause of the form. But we never actually run $M$ on $w$. So how can we say that it will accept or reject it? We have no idea what it does because we never ran it on $w$.

Raphael
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Zach
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  • Is that the whole proof? – Karolis Juodelė Dec 11 '13 at 20:37
  • Yes, that is all of the formal proof. They also state: "The idea is for S to take its input M, w and modify M so that the result- ing TM recognizes a regular language if and only if M accepts w. We call the modified machine M2 . We design M2 to recognize the nonregular language {0n 1n | n ≥ 0} if M does not accept w, and to recognize the regular language Σ∗ if M accepts w. We must specify how S can construct such an M2 from M and w. Here, M2 works by automatically accepting all strings in {0n 1n | n ≥ 0}. In addition, if M accepts w, M2 accepts all other strings." – Zach Dec 11 '13 at 20:42
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    So, as it's explained, M2 accepts a regular language only if M accepts w. Note that x and w are unrelated. M2 never runs on w. – Karolis Juodelė Dec 11 '13 at 20:51
  • I understand the logic between having M2 accept a regular language iff M accepts w. That makes a lot of sense and I can see why the reduction goes that way. What I don't see is how the formal description of M2 they give satisfies that. I don't see how M2 accepts a regular language iff M accepts w. Shouldn't it have something like : "If M accepts w, accept all strings", "Else, accept if of the form $0^{n} 1^{n}$? – Zach Dec 11 '13 at 20:54
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    M2 has exactly that, but written from the other end. It will accept 01 always, and accept anything else, if M accepts w. – Karolis Juodelė Dec 11 '13 at 21:02
  • But we could be accepting non-regular strings like $0^{n} 1^{n}$ even if M doesn't accept w? – Zach Dec 11 '13 at 21:06
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    Strings are not regular, languages are. $0^n1^n$ is in the language ${0^n1^n}$ and it is also in the language $\Sigma^*$. Why should it not be accepted? – Karolis Juodelė Dec 11 '13 at 21:52
  • Closely related question. – Raphael Apr 07 '16 at 20:22

1 Answers1

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Here's the key: work out what $L(M_2)$ is.

Notice that $M_2$ takes a string $x$ as input, so we are trying to identify which strings $x$ will cause $M_2$ to accept.

There are two cases:

  • Case 1: If $M$ accepts on input $w$. Then $L(M_2)=\Sigma^*$.

    In other words, in this case, it doesn't matter what $x$ is. $M_2$ will always accept on input $x$. If $x$ has the form $0^n 1^n$, then $M_2$ accepts (in step 1). If $x$ does not have the form $0^n 1^n$, then $M_2$ also accepts (in step 2). $M_2$ accepts either way; it accepts on every possible input.

  • Case 2: If $M$ does not accept on input $w$. Then $L(M_2) = \{0^n 1^n : n \in \mathbb{N}\}$.

    Let's look at why this is. If $x$ has the form $0^n 1^n$, then $M_2$ accepts in step 1. If $x$ does not have the form $0^n 1^n$, then $M_2$ runs step 2 and doesn't accept (since $M$ doesn't accept on input $w$).

So, $L(M_2)$ is one of two possibilities: it is either $\Sigma^*$ (everything) or $\{0^n 1^n : n \in \mathbb{N}\}$. Which of those two possibilities it is will depend upon $M$ and on $w$ (and in particular on whether $M$ accepts on $w$). Remember that implicitly $M_2$ is dependent upon $M$ and $w$: it's got them hard-coded into its code.

Moreover, $\Sigma^*$ is regular, but $\{0^n 1^n : n \in \mathbb{N}\}$ is not regular. So if we have a way to tell whether the language of a TM is regular or not, we have a way to tell whether $L(M_2)$ is $\Sigma^*$ or $\{0^n 1^n : n \in \mathbb{N}\}$ -- which in turn means we have a way to distinguish Case 1 from Case 2, or in other words, we have a way to tell whether $M$ accepts on input $w$ or not.

That means, if we have a way to tell whether the language of a TM is regular or not, we have a way to solve the halting problem. (It's not the same TM; it's a related TM. The way we tell whether $M$ accepts on $w$ is not by analyzing whether the language of $M$ is regular; we construct some new TM that is related to $M$ and $w$ in a crazy way, and then test whether the language of that new TM is regular or not.)

It's important to remember that $w$ is not an input to $M_2$. Instead, $w$ is just a constant that is hard-coded into the source code of $M_2$. It would be like hard-coding the value of pi as 3.1415927 in your code, except here the hardcoded constant is going to be whatever $w$ is. Each different $w$ corresponds to a different $M_2$.

P.S. I'll also repeat a crucial point from Karolis Juodelė: "Strings are not regular, languages are."

D.W.
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  • I suppose my confusion is because strings of the form $0^{n} 1^{n}$ are in the language even if $M$ does not accept $w$. – Zach Dec 11 '13 at 21:47
  • @Zach, one thing that might help you clear up your confusion is to be very explicit about which language you are talking about. The language $L(M_2)$ is completely different from the language $L(M)$. See my edited, extended answer. – D.W. Dec 11 '13 at 21:49