# Recent questions tagged rice-theorem

1 vote
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Use Rice’s theorem, to prove the undecidability of each of the following languages. $INFINITE_{TM} = \{\langle M \rangle \mid \text{M is a TM and L(M) is an infinite language}\}$. $\{\langle M \rangle \mid \text{M is a TM and }\:1011 \in L(M)\}$. $ALL_{TM} = \{\langle M \rangle \mid \text{ M is a TM and}\: L(M) = Σ^{\ast} \}$.
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Rice's theorem. Let $P$ be any nontrivial property of the language of a Turing machine. Prove that the problem of determining whether a given Turing machine's language has property $P$ is undecidable. In more formal terms, let $P$ be a language consisting of Turing ... any $TMs$. Prove that $P$ is an undecidable language. Show that both conditions are necessary for proving that $P$ is undecidable.
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Rice's theorem. Let $P$ be any nontrivial property of the language of a Turing machine. Prove that the problem of determining whether a given Turing machine's language has property $P$ is undecidable. In more formal terms, let $P$ ... $M_{1}$ and $M_{2}$ are any $TMs$. Prove that $P$ is an undecidable language.
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We know by Rice's theorem that none of the following problems are decidable. However are they recursively enumerable,or non-RE? Does $L(M)$ contain at least two strings? Is $L(M)$ infinite? Is $L(M)$ a context-free language? Is $L(M) = (L(M))^{R}$?
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L = {<M1, M2>M1 and M2 are two TMs, and ε ∈ L(M1) \ L(M2) }. is it RECURSIVE OR RECURSIVE ENUMERABLE OR NOT EVEN RECURSIVE ENUM.
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1.{<M>| M is a TM accepts any string starting with 1} 2.{<M>| M is TM accept exactly 20 strings} Please guide I don’t know how to apply rice theorem. for 1. Is Tyes = { string starting with 1} Tno = { all strings – strings starting with 1} what is Tyes and Tno here? I only conclude by intution that when we provide strings as input some got into loop and some got accepts .
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Halting problem of Turing machines which recognize recursive languages is undecidable. (True / False)
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Here is my analysis. P1: When we bound the number of steps a turing machine can tape, the total number of input possible that can be taken by such turing machine becomes finite and by running TM in an interleaved mode I can decide whether TM M halts on x within k steps. So, ... hence I can decide P3. Hence, P3 is decidable.->REC. So, I think here answer must be 1. Please let me know what's right.
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L1 = { <M> | M is a TM and | L (M) <=1 } L2= { <M> | M is a TM and | L (M) >=1 } NOW QUESTION IS WHICH ARE RECURSIVE ENUMERABLE AND WHICH ARE NOT ???? I JUST READ BASICS OF rice theorem DONT PRACTICE MUCH QUESTIONS ON THIS I SOLVED ABOVE QUESTION BY THIS... ... string length >= 0...... and REL_yes as string length >= 1 . REL_YES is a proper subset of REL _NO . so we can say that it is also non re
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L1:{<M> | there exist a Turing machine M' such that <M>$\neq$<M'> and L(M) = L(M')} How this problem becomes trivial? and if it non-trivial then please explain why is that so. According to my understanding, non-trivial properties are the one where a language ... and if it is then is it okay to say M1=M2 because they are kind of same machine but other one is just with some non deterministic nature.
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Writes Non Blank: Given a turing machine T, does it ever writes a non-blank symbol on its tape, when started with a blank tape. how the above problem is solvable? somewhere i got this explanation: Let the machine only writes blank symbol. Then its ... this is a non-trivial property of turing machine and every non trivial property of turing machine is undecidable, so this is also undecidable.
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1 vote
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What is monotonic and non-monotonic property. Please explain the second postulate of Rice's Theorem.
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I was Studying About Undecidability on GateCSE. I am facing a doubt that : L = {<M> | M accepts "1"} L is set of String & each String is an Encoding of TM & TM accepts 1 L = {<M> | L(M) = {1}} Given a Input Program we have to see it Accepts 1 & nothing ... selected encoding ?) What really is difference between both of them ? Can you definition of 2 in words like " L is set of String ............"
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Define languages L0 and L1 as follows : L0={⟨M,w,0⟩∣M halts on w} L1={⟨M,w,1⟩∣M does not halt on w} Here ⟨M,w,i⟩is a triplet, whose first component M is an encoding of a Turing Machine, second component w is a string, and the third component i is a bit. Let ... even design an acceptor for L as even when L0 is RE L1 is not RE but can anyone explain me what about L COMPLEMENT what is the language ??
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I have a doubt while understanding step 2 in proof of Rice's Theorem- According to my understanding,proof of Rice's theorem as follows ( Please suggest If something is wrong in my understanding) P is a property of languages of TM which is non-trivial and semantic. We ... at all(Same problem as ATM). Can M' take decision in finite time. Please give me some insights to I can understand this point.
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Problem : It is undecidable whether an arbitrary Turing Machines halt within 10 steps? Let consider Two Turing machine in which first one it is halt in 10 steps while in other it is not , so as it is undecidable. @arjun sir ,@bikram sir or @others
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Rice theorem : 1. Any non-trivial property of the LANGUAGE recognizable by a Turing machine is undecidable. 2. Any non-monotonic property of the LANGUAGE recognizable by a Turing machine is unrecognizable While solving please describe non-trivial/trivial and non- ... context-free language is regular. Whether a finite state automation halts on all inputs. Membership problem for type 00 languages.
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Example# 3 from Part-1 Rice's Theorem from https://gatecse.in/rices-theorem/ states as follows (3) L(M) is recognized by a TM having even number of states Sol: This is a trivial property. This set equals the set of recursively enumerable languages. According to the ... property then? Can someone give me an example for which TYES and TNO cannot be found and let me know if my approach is correct ?
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My Question $\{\langle M \rangle \mid M$ is a TM and there exist an input whose length is less than 100, on which $M$ halts$\}$ I have to check that it is Turing Recognizable or not (i.e R.E) My Approach/Doubt Question taken from https://gateoverflow.in/7427/which-following ... $100$ $T_{yes}\subset T_{no}$. Hence it is not RE.But in the solution it is R.E.
1 vote
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L = {M|M is a TM that accepts all even numbers} For the above language i can have Tyes machine which has all even numbers.And Tno as machine whose language is empty.So i can say it is undecidable. But to show it is Not RE. What should be my Tno,so that ... Tno machine?I am assuming here that the property of the language as "Only all even numbers",i guess the same has been given in the question.
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L={<M>: M is a TM that accepts all even numbers } Why it is not recursively enumerable language???
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Can anybody please explain this reduction and rice theorem. http://www.cs.rice.edu/~nakhleh/COMP481/ Thanks
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L(M) has at most 10 strings We can have Tyes for ϕ and Tno for Σ∗. Hence, L={M∣L(M) has at most 10 strings} is not Turing decidable (not recursive). problem : It should not b Tyes Σ∗ and Tno for ϕ
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If we are not able to apply non-monote property ,then is it always true that it is RE but not REC,are there any scenarios where we can't apply non-monotone property but still language is NOT RE. Say,L={TM| L(TM) has atleast one string}, Now it is clearly Language property ... mean theat RE but not REC. P.S:- By (i) and (ii) ,i mean the definitions mentioned here.(http://gatecse.in/rices-theorem/)
1 vote
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I need to understand when to apply RICE's theorem and when to not. Questions like:- Turing machine makes at least five moves,It accepts a string input of length atleast five ,TM halts for every input on length <50 are all decidable. But these are NON TRIVIAL properties ... because some TM will say yes and some will say NO.Then why can't we use same concept on above metioned questions? Please help