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Here is the document: More Undecidable Problems

For a given property $P$ of languages, define $L_P$ as the set of all Turing machines (resp. their encodings) that accept languages with $P$, that is

$\qquad \displaystyle L_{p} = \{ ⟨M⟩ \mid \mathcal{L}(M) \text{ has property } P \}$

If $P$ is a trivial property, that is if $P$ holds for all or no language, $L_P$ is decidable, too (as $L_P=\emptyset$ or $L_P = \{\langle M\rangle \mid M \text{ Turing machine }\}$. If $P$ is not trivial, $L_p$ is undecidable (by Rice's theorem), which means the strings in this language (or with such property) cannot be determined if it can be halt.

We determine a language $M$ has property $P$ by reduce $M$ to another $M'$, and check that if $M'$ accepts the reduced string $x$ from the initial string $w$, we can conclude that $L(M)$ and $L(M')$ have property $P$.

However, as the title suggests, we only reduce a single string $w$ to $x$ and if $x$ is accepted, but we conclude the whole $L(M')$ has property $P$. Thus, $M'$ is obviously part of $L_p$. What if some random strings in $L(M')$ do not have property $P$?

What if M accepts some string, but the reduction of those strings are not accepted by M'?

A reduction from language $L$ to language $L’$ is an algorithm (TM that always halts) that takes a string $w$ and converts it to a string $x$, with the property that: $x$ is in $L’$ if and only if $w$ is in $L$.

Does this imply the reduced language $L'$ will contain every property $P$ from $L$? Since we can conclude that if $L'$ is decidable, then $L$ is decidable as well and vice verse. Can we conclude the same thing to property $P$?

Raphael
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Amumu
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  • the question is not clear. What is $L_p$? is it the union of $L(M_0)$, $L(M_1)$, etc? or is it the encodings of those machines? – Ran G. Jun 10 '12 at 20:22
  • @RanG. Typo error. Fixed. – Amumu Jun 11 '12 at 02:19
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    Hmmm... so it is a union? why not writing it as $L(M_0) \cup L(M_1) \cdots$? I am still confused. There is a mix between languages (sets of words) and properties (sets of languages). Check http://cs.stackexchange.com/q/2293/157 for a similar definition and try to make it clearer. – Ran G. Jun 11 '12 at 02:55
  • I put it as a set, since according to the definition, Lp is a set of all RE languages from all TMs.

    Regardless, I answered it myself. Pleas have a look if you're interest.

     – Amumu Jun 11 '12 at 03:32
  • usually $L_p$ is a language of encodings of TMs. The string $\langle M \rangle$ describes the TM $M$. Then, $L_p$ is a languague - a set that contains strings $L_p={ \langle M_0 \rangle, \langle M_1 \rangle, \ldots }$. – Ran G. Jun 11 '12 at 04:15
  • Your definitions of $L_p$ seem to contradict each other. I have problems parsing the question as well. – Raphael Jun 11 '12 at 10:44
  • @Raphael I wrote it according to the provided document. You can look it up in slide 4 (slide number is at the corner). Language L with property P contains sets of strings accepted by Turing Machines, and a TM M is in L with property P if its language has property P as well. That's what I understand. – Amumu Jun 11 '12 at 12:47
  • @Raphael But you can look at the slide 14 for the picture, and the slides 13-17 for the design of M'. Or the examples from this page: http://www.cs.uiuc.edu/class/fa07/cs273/Handouts/reductions/more-examples.html . All of the examples only use a single string w to determine a language is decidable or not. I also can't make sense of why we have to reduce from a non-existence undecidable language to prove our language isnot decidable.

    What I wanted to do was just trying to make sense of it, by reading other resources and post what I learned in my own answer.

     – Amumu Jun 11 '12 at 12:51
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    @Amumu: The (informal) definitions on slides 3 and 4 are not even close to what you write. I'll edit that. Now a misunderstanding has appeared: "$L_p$ is undecidable, which means the strings in this language (or with such property) cannot be determined if it can be halt." -- that does not make sense. In general, properties have nothing to do with halting. Please look up what "decidability of a language" means. – Raphael Jun 11 '12 at 13:11
  • @Raphael As I understand, a decidable problem is the problem which has an algorithm (Turing Mach) to answer it.An algorithm is a TM that halts on all inputs, accepted or not. You can look here for the definition of decidable problem (slide 13) :http://spark-public.s3.amazonaws.com/automata/slides/18_tm3.pdf – Amumu Jun 11 '12 at 13:34
  • @Raphael Rereading what you edited, I can see that if it exists a Turing Machine that can accept all the strings of L (L is language with property P), then the L is decidable. Otherwise, it's not. Initially, I thought each string encodes a TM, and if there exists a TM not in L, then L is not decidable. My line of thinking (each string is a TM) derives from Wikipedia definition: http://en.wikipedia.org/wiki/Rice's_theorem. (Btw, if we really want to construct language with Turing Machines like I described, we can just do just, can't we?) – Amumu Jun 11 '12 at 13:59
  • @Raphael But then again, if you look at my own answer, at the picture of Rice's Theorem, you can see what I meant. Each word in L with property P encodes a TM. It's the lecture note from UIUC. So what definition is correct? I'm so confused. – Amumu Jun 11 '12 at 14:35

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I think I got it. Here is Rice's Theorem (Rice Theorem and Turing machine b ehavior prop erties):

Rice's Theorem

I looked it up here: Decidability for the definition of a problem in the context of Turing Machine (slides 9 and 11):

  • Formally, a problem is a language.
  • Each string encodes some instance.
  • The string is in the language if and only if the answer to this instance of the problem is “yes.”

A problem is decidable if there is an algorithm to answer it.

  • Recall: An “algorithm,” formally, is a TM that halts on all inputs, accepted or not.
  • Put another way, “decidable problem” = “recursive language.”

Otherwise, the problem is undecidable

That answers why I only need one string w and reduced x from w to prove a language L with property P is undecidable.

According to Rice's Theorem, if language L with property P has Turing Machines that does not accept language of L, L is undecidable. Thus:

$$L_{p} = \{ L(M_{0}), L(M_{1}), L(M_{2}),..., L(M_{n})\}$$ $$L'_{NotInp} = \{ L(M_{0}), L(M_{1}), L(M_{2}),..., L(M_{n}) \}$$

$$M_{i}\mspace{8mu} \text{is}\mspace{8mu} \text{an}\mspace{8mu} i\text{th}\mspace{8mu} \text{Turing} \mspace{8mu} \text{machine},\mspace{8mu} i ≤ n. $$

If the language L' (which contains TMs not in P) is not an empty set, the above L sub P is undecidable. Recall that language with property P is decidable if and only if it is trivial, which means either it contains every RE languages or not at all. Since if a language L with property P is decidable, every Turing Machines accept every string in L, so we don't need to test the acceptance of input to a given TM.

Also, according to the definition of a problem in the context of Turing Machine, if we have the answer (an algorithm) to every instance of the language L sub P (obviously, since L sub P contains every TM), L sub P is decidable. Otherwise, it's not.

Let's be more concrete with an example from here: More Reduction Examples

Language L3

Aside from the proof given on the page, let's look at it more intuitively. We may have Turing Machine M which accepts the three strings. Other than that, the input w can be anything which is not in M, but belong to other M'. Input w can be "Not UIUC", "Not Iowa", or "Not Michigan" or any other strings which do not belong to M. Since we do not have the answer to every instance of the problem with M (because if we do, M will accept every string since every string has property P. In this case, it's not). The language is not trivial, thus it is undecidable.

This is my reasoning. What do you think?

Amumu
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