GATE Overflow - Recent questions tagged rice-theorem
https://gateoverflow.in/tag/rice-theorem
Powered by Question2AnswerMichael Sipser Edition 3 Exercise 5 Question 30 (Page No. 241)
https://gateoverflow.in/324077/michael-sipser-edition-3-exercise-5-question-30-page-no-241
<p>Use Rice’s theorem, to prove the undecidability of each of the following languages.</p>
<ol start="1" style="list-style-type:lower-alpha">
<li>$INFINITE_{TM} = \{\langle M \rangle \mid \text{M is a TM and L(M) is an infinite language}\}$.</li>
<li>$\{\langle M \rangle \mid \text{M is a TM and }\:1011 \in L(M)\}$.</li>
<li>$ ALL_{TM} = \{\langle M \rangle \mid \text{ M is a TM and}\: L(M) = Σ^{\ast} \}$.</li>
</ol>Theory of Computationhttps://gateoverflow.in/324077/michael-sipser-edition-3-exercise-5-question-30-page-no-241Sun, 20 Oct 2019 12:11:47 +0000Michael Sipser Edition 3 Exercise 5 Question 29 (Page No. 241)
https://gateoverflow.in/324076/michael-sipser-edition-3-exercise-5-question-29-page-no-241
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 machine descriptions where $P$ fulfills two conditions. First, $P$ is nontrivial—it contains some, but not all, $TM$ descriptions. Second, $P$ is a property of the $TM’s$ language—whenever $L(M_{1}) = L(M_{2})$, we have $\langle M_{1}\rangle \in P$ iff $\langle M_{2}\rangle \in P$ . Here, $M_{1}$ and $M_{2}$ are any $TMs$. Prove that $P$ is an undecidable language.<br />
<br />
Show that both conditions are necessary for proving that $P$ is undecidable.Theory of Computationhttps://gateoverflow.in/324076/michael-sipser-edition-3-exercise-5-question-29-page-no-241Sun, 20 Oct 2019 12:05:26 +0000Michael Sipser Edition 3 Exercise 5 Question 28 (Page No. 241)
https://gateoverflow.in/324075/michael-sipser-edition-3-exercise-5-question-28-page-no-241
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 machine descriptions where $P$ fulfills two conditions. First, $P$ is nontrivial—it contains some, but not all, $TM$ descriptions. Second, $P$ is a property of the $TM’s$ language—whenever $L(M_{1}) = L(M_{2})$, we have $\langle M_{1}\rangle \in P$ iff $\langle M_{2}\rangle \in P$ . Here, $M_{1}$ and $M_{2}$ are any $TMs$. Prove that $P$ is an undecidable language.Theory of Computationhttps://gateoverflow.in/324075/michael-sipser-edition-3-exercise-5-question-28-page-no-241Sun, 20 Oct 2019 12:03:03 +0000Ullman (TOC) Edition 3 Exercise 9.3 Question 4 (Page No. 400)
https://gateoverflow.in/317248/ullman-toc-edition-3-exercise-9-3-question-4-page-no-400
<p>We know by Rice's theorem that none of the following problems are decidable. However are they recursively enumerable,or non-RE?</p>
<ol start="1" style="list-style-type:lower-alpha">
<li>Does $L(M)$ contain at least two strings?</li>
<li>Is $L(M)$ infinite?</li>
<li>Is $L(M)$ a context-free language?</li>
<li>Is $L(M) = (L(M))^{R}$?</li>
</ol>Theory of Computationhttps://gateoverflow.in/317248/ullman-toc-edition-3-exercise-9-3-question-4-page-no-400Sun, 21 Jul 2019 15:57:07 +0000SELF DOUBT_ RICE THEOREM
https://gateoverflow.in/286327/self-doubt_-rice-theorem
L = {<M1, M2>M1 and M2 are two TMs, and ε ∈ L(M1) \ L(M2) }.<br />
<br />
is it RECURSIVE OR RECURSIVE ENUMERABLE OR NOT EVEN RECURSIVE ENUM.Theory of Computationhttps://gateoverflow.in/286327/self-doubt_-rice-theoremSat, 29 Dec 2018 17:25:16 +0000Rice theorem
https://gateoverflow.in/279786/rice-theorem
<p>1.{<M>| M is a TM accepts any string starting with 1}</p>
<p>2.{<M>| M is TM accept exactly 20 strings}</p>
<p> </p>
<p>Please guide </p>
<p>I don’t know how to apply rice theorem.</p>
<p>for 1. Is Tyes = { string starting with 1} Tno = { all strings – strings starting with 1}</p>
<ol start="2" style="list-style-type:decimal" type="1">
<li>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 .</li>
</ol>Theory of Computationhttps://gateoverflow.in/279786/rice-theoremMon, 17 Dec 2018 11:45:11 +0000Halting problem of TM which recognize recursive languages is undecidable?
https://gateoverflow.in/276798/halting-problem-recognize-recursive-languages-undecidable
Halting problem of Turing machines which recognize recursive languages is undecidable. (True / False)Theory of Computationhttps://gateoverflow.in/276798/halting-problem-recognize-recursive-languages-undecidableMon, 10 Dec 2018 14:46:59 +0000TOC-Undecidability
https://gateoverflow.in/269304/toc-undecidability
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=11452310310240732429"></p>
<p>Here is my analysis.</p>
<p>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.</p>
<p>So, P1 is Decidable or REC.</p>
<p>P2: Here I can have two TM for this say $T_{yes}=\{\epsilon,a\}$ and $T_{no}=\{aaa\}$. This is Undecidable, but since we cannot have $T_{yes}$ such that it should be a proper subset of $T_{no}$, this is Recursively Enumerable.So, this RE but not REC.</p>
<p>P3:I can fix the moves of TM to 99, and only $| \sum|^{99}$ inputs are possible. I can run TM on such inputs in an interleaved way and hence I can decide P3.</p>
<p>Hence, P3 is decidable.->REC.</p>
<p>So, I think here answer must be 1.</p>
<p>Please let me know what's right.</p>Theory of Computationhttps://gateoverflow.in/269304/toc-undecidabilityFri, 23 Nov 2018 12:32:18 +0000SELF DOUBT _ RICE THEOREM
https://gateoverflow.in/264943/self-doubt-_-rice-theorem
<blockquote>
<p>L1 = { <M> | M is a TM and | L (M) <=1 }</p>
</blockquote>
<p>L2= { <M> | M is a TM and | L (M) >=1 }</p>
<p><span class="marker">NOW QUESTION IS WHICH ARE RECURSIVE ENUMERABLE AND WHICH ARE NOT ????</span></p>
<p>I JUST READ BASICS OF rice theorem DONT PRACTICE MUCH QUESTIONS ON THIS</p>
<p>I SOLVED ABOVE QUESTION BY THIS... IS IT CORRECT ?????if any mistake plzz point out</p>
<blockquote>
<p> </p>
<p>statement 1 :
<br>
<br>
means that for all turing machines T (m).. each turing ,machine has one language which have atmost 1 string ......
<br>
<br>
now by use rice theorem .......
<br>
<br>
2 properties of theorem .......
<br>
<br>
and by using 2nd property monotonic one ------->
<br>
<br>
any non monotonic property of recursively enumerable languages is not even semi decidable..........
<br>
<br>
what is non monotonic property ??? ------> means if set having property is a proper subset of any set not having the property......
<br>
<br>
and we also want REL_ yes to be a proper subset of REL_no.....
<br>
<br>
statement 1
<br>
<br>
REL_ yes means any language of one string i.e. suppose l = { a}.
<br>
<br>
no take REL-no means which dont have one string so suppose L = {a,b,aa }
<br>
<br>
see here REL_ yes is a proper subset of rel _No
<br>
<br>
so its not rel.....
<br>
<br>
for 2nd statement ......
<br>
<br>
<br>
<br>
again i can use monotonic property ...... REL_no as string length >= 0...... and REL_yes as string length >= 1 .
<br>
<br>
REL_YES is a proper subset of REL _NO . so we can say that it is also non re</p>
</blockquote>Theory of Computationhttps://gateoverflow.in/264943/self-doubt-_-rice-theoremWed, 14 Nov 2018 07:20:26 +0000Undecidability
https://gateoverflow.in/259120/undecidability
L1:{<M> | there exist a Turing machine M' such that <M>$\neq$<M'> and L(M) = L(M')}<br />
<br />
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 or string may get accepted by some Turing machine and not by some other Turing machines and hence it becomes undecidable. Hence it becomes undecidable when a given Turing machine it will accept a given string or not as it becomes non-trivial property.<br />
<br />
For every non deterministic TM M1 there exists an equivalent deterministic TM M2 recognizing the same language, in this case we will have 2 different machines and both will accept same language, is this description holds true?? 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.Theory of Computationhttps://gateoverflow.in/259120/undecidabilityTue, 30 Oct 2018 05:22:43 +0000undecidability
https://gateoverflow.in/245306/undecidability
<blockquote>
<p><strong>Writes Non Blank: </strong>Given a turing machine T, does it ever writes a non-blank symbol on its tape, when started with a blank tape.</p>
</blockquote>
<p>how the above problem is solvable?</p>
<p>somewhere i got this explanation:</p>
<blockquote>
<p>Let the machine only writes blank symbol. Then its number of configurations in the com computation on w is q × 2, where q is the number of states of M; the factor 2 is for the choices re. the direction of heads movement; there is no factor for the written symbol because that is always blank. So the problem is decidable, decided by the following machine: input <M,w>, run M on w for q × 2 steps; if it M ever writes a non blank symbol, stop with yes answer; if M never writes a non blank symbol, stop with no answer.</p>
</blockquote>
<p>why is it that q*2 is taken as an upper bound for writing a non blank symbol? there can be some machine which can write non blank on more than q*2 moves.</p>
<p> </p>
<p>in fact we can apply rice's theorem here.. as this is a non-trivial property of turing machine and every non trivial property of turing machine is undecidable, so this is also undecidable.</p>Theory of Computationhttps://gateoverflow.in/245306/undecidabilityFri, 21 Sep 2018 07:04:37 +0000TOC- Undecidability
https://gateoverflow.in/241325/toc-undecidability
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=7657023284748237473"></p>Theory of Computationhttps://gateoverflow.in/241325/toc-undecidabilitySat, 08 Sep 2018 19:50:26 +0000Rice's Theorem
https://gateoverflow.in/198800/rices-theorem
What is monotonic and non-monotonic property. Please explain the second postulate of Rice's Theorem.Theory of Computationhttps://gateoverflow.in/198800/rices-theoremTue, 23 Jan 2018 07:29:35 +0000Undecidability Confusion
https://gateoverflow.in/194029/undecidability-confusion
<p>I was Studying About Undecidability on GateCSE. I am facing a doubt that :</p>
<ol>
<li><strong>L = {<M> | M accepts "1"}</strong>
<br>
L is set of String & each String is an Encoding of TM & TM accepts 1</li>
<li><strong>L = {<M> | L(M) = {1}}</strong>
<br>
Given a Input Program we have to see it Accepts 1 & nothing Else.
<br>
<br>
<em><strong>What Does this Input Program is ?
<br>
What does Language here specifically means (Does it mean the language accepted by that particular selected encoding ?)
<br>
What really is difference between both of them ?
<br>
Can you definition of 2 in words like " L is set of String ............"</strong></em>
<br>
</li>
</ol>Theory of Computationhttps://gateoverflow.in/194029/undecidability-confusionSat, 13 Jan 2018 03:18:36 +0000undecidability
https://gateoverflow.in/191944/undecidability
Define languages L0 and L1 as follows :<br />
<br />
L0={⟨M,w,0⟩∣M halts on w}<br />
<br />
L1={⟨M,w,1⟩∣M does not halt on w}<br />
<br />
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.<br />
<br />
Let L= L0 ∪ L1. Which of the following is true?<br />
<br />
L is not even recursively enumerable as we cannot even design an acceptor for L as even when L0 is RE L1 is not RE<br />
<br />
but can anyone explain me what about L COMPLEMENT what is the language ??Theory of Computationhttps://gateoverflow.in/191944/undecidabilityTue, 09 Jan 2018 06:30:01 +0000Doubt in Rice's Theorem
https://gateoverflow.in/181185/doubt-in-rices-theorem
<p>I have a doubt while understanding step 2 in proof of Rice's Theorem-</p>
<p>According to my understanding,proof of Rice's theorem as follows ( Please suggest If something is wrong in my understanding)</p>
<table border="1" cellpadding="1" style="width:500px; border-spacing: 1px;">
<tbody>
<tr>
<td>
<p><strong>P is a property of languages of TM which is non-trivial and semantic. We have to prove Lp={<M>| L(M) is element of P} is un-decidable</strong></p>
<p>Proof:</p>
<p>Show that $A_{TM} \leq_{m} L_{P}$</p>
<p><strong>Step 1 :</strong>There are Turing machines M1 and M2 L(M1) is element of P and L(M2) is not element of P(P is non trivial)</p>
<p><strong>Step 2: </strong>Construct a TM M' : M' takes input as <M,w> M' simulates TM M on string w if M accepts w then return <M2> else return <M1></p>
<p><strong>Step 3: M<sub>p </sub></strong>(A decider which decides that given<M> has property P or not) takes input as <M1> or <M2> if Mp accepts then accept else reject</p>
</td>
</tr>
</tbody>
</table>
<p>Step 2 says , we can construct a Turing Machine M' which simulate M on w and If M accepts w or reject w take the decision and feed input to the hypothetical decider. But how does M' decides if w is accepted by M or not since M can be looping forever and may not halt at all(Same problem as A<sub>TM</sub>). <strong>Can M' take decision in finite time.</strong></p>
<p>Please give me some insights to I can understand this point.</p>Theory of Computationhttps://gateoverflow.in/181185/doubt-in-rices-theoremThu, 14 Dec 2017 21:33:18 +0000Rice theorem G is a CFG. Is L(G)= Σ ∗ L(G)=Σ∗ .What is it - Decidable / Undeciable .How?
https://gateoverflow.in/175823/rice-theorem-is-cfg-is-%CF%83%E2%88%97-what-is-it-decidable-undeciable-how
Theory of Computationhttps://gateoverflow.in/175823/rice-theorem-is-cfg-is-%CF%83%E2%88%97-what-is-it-decidable-undeciable-howFri, 01 Dec 2017 02:50:34 +0000Rice theorem problem
https://gateoverflow.in/175771/rice-theorem-problem
Problem : It is undecidable whether an arbitrary Turing Machines halt within 10 steps?<br />
<br />
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.<br />
<br />
@arjun sir ,@bikram sir or @othersTheory of Computationhttps://gateoverflow.in/175771/rice-theorem-problemFri, 01 Dec 2017 01:10:01 +0000Can any one proof decidability/undecidability of following on Rice Theorem ?
https://gateoverflow.in/175756/proof-decidability-undecidability-following-rice-theorem
<p>Rice theorem :</p>
<p>1. Any non-trivial <strong>property</strong> of the LANGUAGE recognizable by a Turing machine <strong>is</strong> undecidable.</p>
<p>2. Any non-monotonic <strong>property</strong> of the LANGUAGE recognizable by a Turing machine <strong>is</strong> unrecognizable</p>
<p>While solving please describe non-trivial/trivial and non-monotonic /monotonic property of particular statement .</p>
<p>Which of the following problems are undecidable?</p>
<ol style="list-style-type:upper-alpha">
<li>Membership problem in context-free languages.</li>
<li>Whether a given context-free language is regular.</li>
<li>Whether a finite state automation halts on all inputs.</li>
<li>Membership problem for type 00 languages.</li>
</ol>Theory of Computationhttps://gateoverflow.in/175756/proof-decidability-undecidability-following-rice-theoremFri, 01 Dec 2017 00:44:45 +0000Rice's Theorem
https://gateoverflow.in/163559/rices-theorem
<p>Example# 3 from Part-1 Rice's Theorem from <a href="https://gatecse.in/rices-theorem/" rel="nofollow" target="_blank">https://gatecse.in/rices-theorem/</a> states as follows
<br>
</p>
<p><em>(3) L(M) is recognized by a TM having even number of states</em></p>
<p><em>Sol: This is a trivial property. This set equals the set of recursively enumerable languages.</em></p>
<p><em>According to the previous examples, if i can find a Turing Machine which says Yes and a Turing Machine which says No, so i'll choose T<sub>NO </sub>as <strong>PHI </strong>(Empty String) and T<sub>YES</sub> as a TM which accepts even number of states, then is it true that my L(M) is undecidable as the property is non-trivial, is my Reasoning correct ?</em></p>
<p>It seems like i can find T<sub>YES</sub> and T<sub>NO</sub> for every property then?
<br>
Can someone give me an example for which T<sub>YES</sub> and T<sub>NO</sub> cannot be found and let me know if my approach is correct ?</p>Theory of Computationhttps://gateoverflow.in/163559/rices-theoremSat, 28 Oct 2017 08:25:16 +0000$\{\langle M \rangle \mid M$ is a TM and there exist an input whose length is less than 100, on which $M$ halts$\}$
https://gateoverflow.in/151057/langle-rangle-there-exist-input-whose-length-than-which-halts%24
<p><strong>My Question </strong></p>
<hr>
<p>$\{\langle M \rangle \mid M$ is a TM and there exist an input whose length is less than 100, on which $M$ halts$\}$</p>
<p>I have to check that it is Turing Recognizable or not (i.e R.E)</p>
<p><strong>My Approach/Doubt</strong></p>
<hr>
<p>Question taken from </p>
<p><a rel="nofollow" href="https://gateoverflow.in/7427/which-following-languages-recursively-enumerable-language">https://gateoverflow.in/7427/which-following-languages-recursively-enumerable-language</a></p>
<p>I tried solving it using Rice theorem -2,i.e showing <strong>non-monotonic property.</strong></p>
<p>My $T_{yes}=\left \{ \epsilon \right \}$//as length $=0$,it make sense</p>
<p>My $T_{no}=$Aceepting string of length at least $100$</p>
<p>$T_{yes}\subset T_{no}$.</p>
<p>Hence it is not RE.But in the solution it is R.E.</p>
Theory of Computationhttps://gateoverflow.in/151057/langle-rangle-there-exist-input-whose-length-than-which-halts%24Sat, 09 Sep 2017 12:26:33 +0000Turing Machine Rice's Theorem
https://gateoverflow.in/141892/turing-machine-rices-theorem
<p><span class="marker">L = {M|M is a TM that accepts all even numbers}</span></p>
<p>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. </p>
<p>But to show it is Not RE. What should be my Tno,so that it should be proper superset of Tyes?Can i take sigma* as my Tno?Although it will accepts the even numbers ,but still it has odd numbers also.Will that work as Tno machine?I am assuming here that the property of the language as "<span class="marker">Only all even numbers</span>",i guess the same has been given in the question.</p>
Theory of Computationhttps://gateoverflow.in/141892/turing-machine-rices-theoremSat, 05 Aug 2017 06:51:01 +0000Rice theorem Example from (https://www.cs.rice.edu/~nakhleh/COMP481/final_review_sp06_sol.pdf)
https://gateoverflow.in/138444/theorem-example-https-nakhleh-comp481-final_review_sp06_sol
<h1><a rel="nofollow" href="https://gateoverflow.in/105965/l-m-m-is-a-tm-that-accepts-all-even-numbers">L={<M>: M is a TM that accepts all even numbers }</a> Why it is not recursively enumerable language???</h1>Theory of Computationhttps://gateoverflow.in/138444/theorem-example-https-nakhleh-comp481-final_review_sp06_solTue, 18 Jul 2017 00:40:53 +0000http://www.cs.rice.edu/~nakhleh/COMP481/
https://gateoverflow.in/130674/http-www-cs-rice-edu-nakhleh-comp481
<p>Can anybody please explain this reduction and rice theorem.</p>
<p><a rel="nofollow" href="http://www.cs.rice.edu/~nakhleh/COMP481/">http://www.cs.rice.edu/~nakhleh/COMP481/</a></p>
<p>Thanks</p>
Theory of Computationhttps://gateoverflow.in/130674/http-www-cs-rice-edu-nakhleh-comp481Wed, 24 May 2017 06:28:04 +0000rice theorem problem
https://gateoverflow.in/110408/rice-theorem-problem
L(M)<br />
<br />
has at most 10 strings<br />
<br />
We can have Tyes<br />
<br />
for ϕ and Tno for Σ∗. Hence, L={M∣L(M) has at most 10 strings} is not Turing decidable (not recursive).<br />
<br />
<br />
<br />
problem :<br />
<br />
It should not b Tyes Σ∗ and Tno for ϕTheory of Computationhttps://gateoverflow.in/110408/rice-theorem-problemTue, 24 Jan 2017 10:39:29 +0000General Doubt In rice theorem
https://gateoverflow.in/108917/general-doubt-in-rice-theorem
<p>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.</p>
<p>Say,L={TM| L(TM) has atleast one string},</p>
<p>Now it is clearly Language property of RE and hence undecidable,But if we look on non-monotone property,then the Tno is phi as it is the only machine which will say NO as per me,and the langauge having atleast one string will never be proper subset of Tno,so non-monotone property doesn't hold,so is this sufficient to say that it is turing recogonizable?I mean satisfying Rice theorem (i) part and not (ii) will always mean theat RE but not REC.</p>
<p>P.S:- By (i) and (ii) ,i mean the definitions mentioned here.(<a rel="nofollow" href="http://gatecse.in/rices-theorem/">http://gatecse.in/rices-theorem/</a>)</p>Theory of Computationhttps://gateoverflow.in/108917/general-doubt-in-rice-theoremSun, 22 Jan 2017 03:38:33 +0000Rice theorem Clarification
https://gateoverflow.in/104918/rice-theorem-clarification
I need to understand when to apply RICE's theorem and when to not.<br />
<br />
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.<br />
<br />
But these are NON TRIVIAL properties?Some TM will make 5 moves and some will not,Some can halt on every input <50 and some can not?So if this is Non trivial property then why cant we apply RICE's theorem?<br />
<br />
I was reading Arjun's sir blog on GateCse,there will say that TM accepts atleast 10 strings is undecidable because some TM will say yes and some will say NO.Then why can't we use same concept on above metioned questions?<br />
<br />
Please helpTheory of Computationhttps://gateoverflow.in/104918/rice-theorem-clarificationFri, 13 Jan 2017 06:55:32 +0000Rice's Theorem
https://gateoverflow.in/100112/rices-theorem
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=14742691634759696563"></p>
<p>I am unable to understand when to apply Rice's theorem and when to not. How L2 is decidable.</p>Theory of Computationhttps://gateoverflow.in/100112/rices-theoremTue, 03 Jan 2017 15:29:46 +0000Rices theorem
https://gateoverflow.in/99369/rices-theorem
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=15475671492620916836"></p>
<p>I mean how it is a trivial property, we can write Tyes ( TM having even states) and Tno ( TM having odd states) for it.Can Turing machine can never have odd number of states? </p>Theory of Computationhttps://gateoverflow.in/99369/rices-theoremMon, 02 Jan 2017 06:03:55 +0000L1 = {<M> | M is a TM and L(M) ⊆ {00, 11}}
https://gateoverflow.in/97192/l1-m-%C2%A0-m-is-a-tm-and-l-m-%E2%8A%86-00-11
L1 = {<M> | M is a TM and L(M) ⊆ {00, 11}}<br />
<br />
R.E or not RE..??Theory of Computationhttps://gateoverflow.in/97192/l1-m-%C2%A0-m-is-a-tm-and-l-m-%E2%8A%86-00-11Tue, 27 Dec 2016 04:41:44 +0000#Selfdoubt Rice's Theorem
https://gateoverflow.in/97080/%23selfdoubt-rices-theorem
<p>What is the difference between <span class="marker">Non-trivial property</span> (in 1st theorem)and <span class="marker">Non-monotonic property</span> ( in 2nd theorem)?</p>
<p>I was going through Rice's theorem but unable to differentiate between these 2 properties.Please give the detail to differentiate both </p>
<p><strong>For a property to be non-trivial, there should exist at least two Turing machines, the property holding for the language of one (Tyes) and not holding for the language of other (Tno)</strong></p>
<p><strong>For a property to be non-monotonic, there should exist at least two Turing machines, the property holding for the language of one (Tyes) and not holding for the language of other (Tno) and </strong> <strong>L(Tyes)⊂L(Tno)</strong></p>
<p>Source: <a rel="nofollow" href="http://gatecse.in/rices-theorem/">http://gatecse.in/rices-theorem/</a></p>Theory of Computationhttps://gateoverflow.in/97080/%23selfdoubt-rices-theoremMon, 26 Dec 2016 18:11:43 +0000L={⟨M⟩|TM accepts exactly 154 strings}
https://gateoverflow.in/93397/l-%E2%9F%A8m%E2%9F%A9-tm-accepts-exactly-154-strings
<p><strong>L={⟨M⟩|TM accepts exactly 154 strings}</strong></p>
<p>--------------------------------------------------------------------------------------------------------</p>
<p>this language is not decidable but is this R.E??</p>
<p>by second rice theorem, T<sub>yes =</sub>{154 strings} and T<sub>no</sub> = more than 154 strings</p>
<p>hence, Tyes is subset of Tno.so,it is not even R.E.</p>
<p>am i right?</p>
<p>please correct me.</p>Theory of Computationhttps://gateoverflow.in/93397/l-%E2%9F%A8m%E2%9F%A9-tm-accepts-exactly-154-stringsFri, 16 Dec 2016 11:56:06 +0000L= {<TM> | TM halts on every input}
https://gateoverflow.in/93284/l-tm-tm-halts-on-every-input
$L=\left \{ <TM> | TM\ halts\ on\ every\ input\ \right \}$<br />
<br />
is above language Recursively enumerable or non recursively enumerable??Theory of Computationhttps://gateoverflow.in/93284/l-tm-tm-halts-on-every-inputFri, 16 Dec 2016 06:15:22 +0000Basic Doubt in Rice Theorem
https://gateoverflow.in/85964/basic-doubt-in-rice-theorem
<p>I read this blog <a rel="nofollow" href="http://gatecse.in/rices-theorem/">http://gatecse.in/rices-theorem/</a> and I have a doubt in the first property.</p>
<p>An example in this blog is L(M) = {0} and it's written that TM<sub>no </sub>=sigma* . My doubt is how can it be sigma* as sigma* can also contain {0} which should be accepted by TM.</p>
Theory of Computationhttps://gateoverflow.in/85964/basic-doubt-in-rice-theoremFri, 25 Nov 2016 11:49:39 +0000Rice theorem
https://gateoverflow.in/83068/rice-theorem
<p><img alt="" src="https://gateoverflow.in/?qa=blob&qa_blobid=5538089539405211230">s</p>
<p>Rice theorem says any non-trivial property of a language is Undecidable. Then how is above one REL.</p>
<p>Does it mean we cannot say anything for a trivial property?</p>Theory of Computationhttps://gateoverflow.in/83068/rice-theoremWed, 16 Nov 2016 21:43:39 +0000Please explain?
https://gateoverflow.in/30890/please-explain
Consider the following Languages:<br />
<br />
$L_{ne}=\{\langle M \rangle \mid L(M)\neq \phi \}$<br />
<br />
$L_{e}=\{\langle M \rangle \ \mid L(M)=\phi \}$<br />
<br />
where $\langle M \rangle$ denotes encoding of a Turing Machine $M$<br />
Then which of the following is true?<br />
<br />
(a) $L_{ne}$ is r.e. but not recursive and $L_{e}$ is not r.e.<br />
(b) Both are not r.e.<br />
(c) Both are recursive <br />
(d) $L_{e}$ is r.e. but not recursive and $L_{ne}$ is not r.e.Theory of Computationhttps://gateoverflow.in/30890/please-explainTue, 15 Dec 2015 02:10:46 +0000