Recognizable but undecidable language. Decidability:Decidable Languages Chuck Cusack Based on M.

Recognizable but undecidable language The trick is to form two "opposite" languages that are based on unions, such that Then any undecidable language L0(and we know that undecidable languages exist — e. I have seen the two being used interchangeably, however I believe it is possible for a language to be recognizable but not decidable, so was hoping for C2. Viewed 4k times 1 $\begingroup$ If language one is undecidable then language three would have to be infinite, right? And what about the lengths of the strings in language three? Reduce undecidable language to decidable language? Ask Question Asked 6 years, 9 months ago. languages and co-r. An enumerator can be constructed as Is the Halting problem in the class of undecidable problems, or it is just in the set of unrecognizable problems? I understand that if it is undecidable, then it is also unrecognizable. 16 Robb T. In either case is there a better method for example is there a general way for proving that a language is not recognizable like for undecidability we try to reduce an undecidable problem to the current problem ? computability; turing-machines; semi-decidability; Share. 15 and 3. The document discusses the limits of computation and reducibility. Also suppose that you have an operation · which takes two of these objects and returns another object (the number two is just an example). 2. The machine can reject a word from or run forever in a loop. Forming a language that is recognizable but not co-recognizable. Any problem that involves Language of TM’s and is NOT trivial is going to be undecidable. 11. An exercise that was in a past session is the following: Prove that there exists an undecidable subset of $\{1\}^*$ This exercise looks very strange to me, because I think that all subsets are . So the simple answer to your question is that it's false, and an example, as in the comments, is $\Sigma^*$, which is decidable, but contains an undecidable language (indeed, every undecidable language). To consider uncountable languages we have to look at infinite strings in place of finite strings. It introduces different classes of languages based on whether they are decidable, recognizable, co-decidable, etc. Why is "decidable" included in "Turing-recognizable"? Hot Network Questions A language is recognizable if and only if we can build a Turing machine that accepts every string in the language, and does not accept any string not in the language. and provides examples like EMPTY, ACCEPT, and ACCEPT_ε. However, a recognizable language may or may not be decidable. Yuval Filmus commented that a pair of "random" languages should probably work. (a) Closed under union. Is the union of undecidable languages not Turing-recognizable? 2. For . To be undecidable, there must be individual strings is T-recognizable but also undecidable . Theorem : A language is decidable i it is Turing-recognizable and co-Turing-recognizable. After reading about it in the textbook and in the web, i was wondering about the "turing recognizable" concept. (Hampden-Sydney College)The Acceptance Problem - Undecidable Languages Fri, Nov 7, 2014 23 / 25. [ Choose ]reduce it to an undecidable language. For any All we know is that if a language is countable than it must be recognizable. 5 We will prove that two particular problems Let be a Turing recognizable language L Let be the Turing Machine that accepts M L L We will prove that is also decidable:L we will build a decider forL 10 NO L YES M s accept L Generally speaking, the existence of such languages follows from the implication you state -- the complement of enumerable yet undecidable languages can not be enumerable -- by the existence of precisely enumerable yet undecidable languages. We begin by provid-ing a diagram summarizing the languages we’ve seen in lecture, and whether they are decidable, recognizable, or Undecidable language-– A decision problem P is said to be undecidable if the language L of all yes instances to P is not decidable or a language is undecidable if it is not If no such Turing machine exists, L L is undecidable. In the Theory of Computation, problems can be classified into decidable and undecidable categories based on whether they can be solved using an algorithm. We can classify families of languages based on how complicated they are, and there are many results in this area, though they tend to be somewhat cumbersome The problem is known to be undecidable (but semidecidable). If I have two languages that aren't Turing-recognizable, is the union between them always not T-recognizable? Why? As an example, take a language where it and its complement are undecidable. In other words, if A TM was decidable, then every Turing-recognizable language would also be decidable. There's nothing special about the string abb. Prove that the following language is decidable / Whether a language is decidable or a language is decided by a TM is an entirely different although closely related concept. This can be considered impolite to answerers. Because L is undecidable lang we know one of the following: there is a word w1 from L that there is no TM that stop for this word As a result all languages whether decidable or undecidable are countable. The problem is that I cannot create a machine that accepts an undecidable language to put inside M' for this reduction. See the Encyclopedia of Mathematics for more on recognizable and undecidable languages (specifically Next I did some demonstrations to show how T-Recognizable languages are closed for Union, Intersection, Concatenation and Kleene Star. Mapping reducibility of Turing Machine to two state Turing Machine. We could clearly construct a decider for Bby running M A TM on hM B;wi. (c) Closed under complement. This would make me think that decidable languages include Turing-recognizable languages, and not viceversa. 1. 22) 9 . TM was decidable, but that some other undecidable language Bwas Turing-recognizable. No infinite loops. (9 points) (a) (5 points) Let L1 = {hM,wi| Turing machine M Indeed - that's why we use the relatively complicated notion of reductions between languages, rather than showing containment. Understanding Turing Machines: Recognizable and Decidable langauges. To make a language r. Proof idea: the elements of any infinite set could be put into lexicographic order, in which case there is a bijection with the language of all strings over the alphabet; because there are such recognizable but undecidable languages over the set of all strings, so too must there be recognizable but undecidable languages over this set of strings. Koether Homework Review Closure Properties of Decidable Languages Intersection Union Closure Properties of Recognizable Languages Intersection Union Assignment Closure Properties of Decidable and Recognizable Languages Lecture 28 Problems 3. Notice that by definition, LD(T M) ⊊ LR(T M). I also can't think of how it could be co-recognizable either, since I can't think of a way that you could decide that a given input string is not a substring of any word A language is recognizable if there’s a Turing machine that accepts all the words . Rice guarantees this language is undecidable. L(M) –“language recognized by M” is set of strings M accepts Language is Turing recognizable if some Turing machine recognizes it •Also called “recursively enumerable” Machine that halts on all inputs is a decider. It Is the union of undecidable languages not Turing-recognizable? 1. What is difference between "recognizable" and "decidable" in context of Turing machines? A language is Recognizable iff there is a Turing Machine which will halt and accept In this tutorial, we’ll study recognizable, co-recognizable, and decidable languages. Discover learning materials by subject, university or textbook. There are two main techniques for doing so: the first is a technique called diagonalization, and the second is called reductions . On input . 7. For any w ∉ (ℒ M), M does not Being co-recognizable means the language ${w\in \Sigma^*, w\not\in L}$ (or, in english, the set of all the words that are not in $L$, i. If the complement of a recognizable language is also recognizable, the language is, in fact, decidable. (d) Closed under concatenation. Viewed 8k times 0 $\begingroup$ The question ans its answer is given in the following picture: Is the union of undecidable languages not Turing-recognizable? 0. Corollary The complement H¯ of the halting problem H isnot Turing-recognizable. 5. (e) Closed under star. L is said to beTuring-recognizable(or simply recognizable) if there Do two CFGs generate the same language ? undecidable Acceptance Problem for Turing Machines A TM = fhM;w i j M is a TM that accepts string w g A TM is Turing-recognizable : simulate M on complement of a Turing-recognizable language. 4 C2. Classify a language as Turing-recognizable or co-Turing-recognizable. $\endgroup$ I can't seem to understand whether this would be unrecognizable or simply Turing undecidable since for inputs 0000 and 00 I think it would be in the same state but for inputs 0000w and 00w where w = 111, the first one should be accepted while the second one should not. A decider that recognizes language L is said to decide language L Language is Turing decidable, or just decidable, if TM is undecidable How shall we prove this? Let me count the ways 1. Ask Question Asked 6 years, 10 months ago. Proof semidecidable languages. . ) I know that a language is Turing-decidable if there is an algorithm to decide membership. Think how we can use this fact to construct an undecidable language. I'm having trouble coming up with a language with these properties. Partially decidable or Semi-Decidable Language-– A decision problem P is said to be semi-decidable (i. Your language L is indeed undecidable. There are languages which are recognizable, but not decidable Recognizing A tm $\begingroup$ ALLCFG is not recognizable - by reduction from $\overline{ATM}$ - but its complement is recognizable, using the algorithm deciding whether a word belongs to the language of a CFG. Only infinite languages can be undecidable. undecidable)? This thread is archived New comments cannot be posted and votes cannot be cast comments A few things, It's hard to find what your proof attempt is trying to do. If you don't care about the computably enumerable part, then I would say that your question is simply a duplicate of one of the similar questions. Follow edited Oct 3, 2017 at 15:40. Undecidable language problems. Skip to main content there are undecidable and unrecognizable languages. Decidable Languages: Decidable and Recognizable Languages Language Decidability Examples StudySmarterOriginal! Find study content Learning Materials. All of these statements are well-known and proofs can Is there an undecidable language A that is mapping reducible to its complement? If it is possible, since A is an undecidable language, so A's complement must also be an undecidable language. I. Now suppose that every language is recognizable. We say that C is closed under · when: given two objects x and y in C, applying the operator to them gives an object which is Every infinite Turing-recognizable language has an infinite decidable subset. 2 or 2. Though it might be worth reading about The Arithmetical Hierarchy. A language is said to be Decidable if there is a Machine that will accept strings in the language and reject strings not in the language. A decidable problem is one for which a solution can be found in a finite amount of time, meaning there exists Show that Turing recognizable languages are closed under intersection. This is much harder to prove, however; it was the first example of a priority argument. Any regular language is recognizable and, since regular languages are closed under complement, its complement is also recognizable. g. (AFAIK, having an infinite alphabet is not very interesting and doesn't correspond to a realistic model of computation by itself. On input x: 2 Decidability:Decidable Languages Chuck Cusack Based on M. Check whether 2 languages are Turing 2 Recall: Recognizable vs. The answer: Is the union of undecidable languages not Turing-recognizable? 1. As sepp2k points out, a* is a regular language, hence decidable. Given 3 disjoint Turing-recognizable languages prove that one of them is decidable. Turing Machine That Accepts Machines With Undecidable Languages. the property of being a decidable language). Then the function f : M!Ldefined by f : M 7!L(M) is onto. W. For completeness, we have the following diagram that Here we show that undecidable languages are NOT closed under concatenation. Is this language recognizable (Turing machines) 0. ‣Some strings not in L may cause the TM to loop ‣Turing recognizable = recursively enumerable (RE) • A language L is Turing decidable if some Turing machine decides it ‣To decide is to return a definitive answer; the TM must halt on all inputs Corollary (the acceptance problem): The language AP of all strings (v,w) where v is an algorithm accepting the word w is recognizable but undecidable. The raw information is readily available -- to the point that we have to assume the asker either has the information at hand or the question is homework -- but the asker seems to have some problems with basic definitions. 1 The fast way Skip ahead to section 2. Otherwise, the diagonalization technique described in answers to similar questions would have provided simple counterexamples. $\begingroup$ I am asking for better intuition on the steps needed to reduce a language to another to prove it is not recognizable. Since L' is in NP by definition there exist a TM ( NTM but since they are equivalent in power I write TM ) M' such that decides L'. A reduces to B means can use TM f\൯r B as subroutine in TM for A What about the undecidable languages? I think they are close under complementation, but not under concatenation, union and intersection. Created Date: 11/15/2006 6:20:00 PM The empty language is recognizable and its complement (the language containing all the words) is also recognizable. A Turing machine is under no obligation to read its entire input. Then by the theorem that \a language is decidable i it is both recognizable and corecognizable", it must be that E TM is not decidable. turing-machines; decision-problem; Share. The halting problem H is Turing-recognizable but not decidable. its complementary) is recognizable. However, that doesn't tell you that every regular language is a subset of every decidable language: it tells you that every regular language is a decidable language. Ask Question Asked 3 years, 2 months ago. } Some languages are Turing-recognizable, but not decidable. However, I am not sure where to start with this proof. $\endgroup$ – D. It is Turing recognizable if in the latter case the C program simply never halts. ) Why is this language Turing recognizable and not not-Turing recognizable. Citation from Wikipedia: . The same is not true of recognizable languages. Let me quote the definition in the book introduction to the theory of computation by Michael Sipser. If a language is not even partially decidable , then there exists no Turing machine for that language. Improve this answer A language which is Turing Recognizable if there is a Machine that will halt and accept only the strings in that language and not in that language, then that TM either rejects, or does not halt at all. But I don't know how to prove it My prove "Lets say L is undecidable lang. Now I'm trying to answer a question to show why the classe of T-Recognizable languages are not closed for the operation of Complementation, but I cannot understand it. Yes, in fact one can show that there are uncountably many languages (say by representing a language as a sequence of the form $\{0, 1\}^\mathbb N$ which can be interpreted as the binary form of some real number in $[0, 1]$) while there are only countably many TMs (as any TM can be represented by a finite string), so almost all languages are undecidable, i. 6k 1 1 gold badge 39 39 silver badges 67 But for strings not in the language (the first given machine cannot generate all the strings the second one can), our machine may halt and reject, or may never halt. An unrecognizable language is a language that is not Turing-recognizable. a language is decidable if and only if it is Is a language Learning which is a CFL, regular: This is an Undecidable Problem as we can not find from the production rules of the CFL whether it is regular or not. 9. Shaull Shaull. Consider the following machine M0: 1. Modified 6 years, 9 months ago. The reason for this is that if a language is decidable, then its complement must be decidable as well. A Turing Machine recognizes the language, but it will loop infinitely on some inputs } Some languages are not Turing-recognizable This makes the language recognizable but the machine may go on forever if executed. Then their union is decided by a TM that always says yes. , have a semi-algorithm) if the Why aren't recursively enumerable languages undecidable. We have that A is Turing-recognizable, and thus we are not sure we have a machine that decides it (and thus halts). Later, I have written a proof to show that Turing Recognizable languages are closed under union. Our first known undecidable language is HALT = { M,w |M halts on w} 1 Some Closure Recall the definition of decidable and recognizable languages. 22: Hierarchy of undecidable languages. e (recursive enumerable), in your terms co-recognizable we need a aditional property: Termination properties of your TM For the recursive languages, we need this also, because REC is a subclass of RE. 3. The most well-known undecidable problem is the Halting Problem, which asks whether a given program will stop running (halt) or continue running forever for a particular input. This known undecidable language can be any language for which undecidability has been that non-Turing-recognizable language to the given language). Follow answered Dec 1, 2016 at 17:09. Therefore, if you know decidable languages must be recognizable, its contrapositive readily follows: languages that are not recognizable can not be decidable. Why is "decidable" included in "Turing-recognizable"? 0. The Halting Problem H is Undecidable A Language that is not Turing-recognizable We have the following results: A language L is decidable iff bothL and L¯ are Turing-recognizable. This will prove that that Etm is NP - hard. Prove the intersection of two Turing-decidable languages is Turing-decidable. It is actually known that recursive languages are the intersection of r. Simple concrete example of a language that L(N) => I could think of a turing machine that takes as an input 10 strings of the language and accept them, if the turing machine accepts them then it's turing recognizable but not turing decidable (given Rice's Theorem) But I can't figure it out a proof for L(M) even though I know it's undecidable also fro Rice's Theorem One of the most applicable basic laws of basic logic is the law of contraposition: $(P \rightarrow Q) \leftrightarrow (\neg Q \rightarrow \neg P)$. A Turing machine M is said todecidea language L if L = L (M ) and M halts on every input. Koether Homework Review Universal Turing Machines The Acceptance Problem for Turing Machines Turing-Unrecognizable Languages Assignment Turing-Unrecognizable Languages Proof of the theorem. Some infinite languages are regular. } Some languages are Turing-decidable A Turing Machine will halt on all inputs (either accepting or rejecting). You are given a TM M and an input x and an oracle Q for the problem you describe. But the other is not, namely: What I don't understand is why the author assumed that M will halt. [ Choose ]show that its complement is decidable A TM halting on infinitely many cases does not imply that the recognized language is decidable. RE accept pipe For every language L in Dec, there is a deciding machine M that for an input string w is guaranteed to deliver a ball to either the accept pipe or reject pipe. $ would also be, and conversely. ) : A decidable ) A To show that a language is NOT recognizable, one could Group of answer choicesshow that its complement is recognizable but not decidable. Construct an algorithm recognizing AP as follows: when presented with a string (v,w), check if v is an algorithm for recognizing languages or not. Nothing is implied about the complement of . There are no regular languages which are undecidable and therefore there are no finite languages which are undecidable. Can a promblem be undecidable but yet be fully Turing recognizable? For example - if language is Turing recognizable OR it's complement is Turing recognizable - so it still undecidable? And another question - does Atm is np or np hard? Does a list of all the configuration that accepts the word w can be verifier? Thanks! $\begingroup$ Please don't delete your question after receiving an answer. Prove whether this language is decidable and recognizable. All semi-decidable+ languages are undecidable, but we’ll see there are undecidable languages that aren’t semi-decidable+! Decidable and Undecidable Languages 32-3 Dec vs. [ Choose ]reduce it to an unrecognizable language. We can prove that, say, ${\overline{A_{TM}}}$ is not turing-recognizable, because that would make ${{A_{TM}}}$ Indeed, the language of all Turing-machine encodings which accept the empty language is: undecidable, since there is no TM which answers yes for strings in the language and no for strings not in the language; The language L that consists of all Turing Machine descriptions M, for which the language accepted by M is finite. Therefore one way of showing that a language is decidable is by describing a Turing machine that accepts it. 1. Share. Modified 6 years, 10 months ago. A language L L is recognizable (or, recursively enumerable, abbreviated r. $\begingroup$ @templatetypedef, echadromani: it's not about the level, it's about the SE-badness of the question. Modified 3 years, 2 months ago. The first direction is OK. $\endgroup$ – Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site L is a recognizable undecidable language ,M is a Turing machine that recognizes L, does M reject or infinitely loop for s belonging to L-complement? Ask Question Asked 1 year, 9 months ago Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site If the complementary language of an recognizable language is a non-recognizable language, is the recognizable language a non-decidable language? 3 Can I reduce a non semi decidable and undecidable language to a semi decidable and undecidable langauge? many-one reduction What is the easiest and the most straightforward way to find whether a given language is decidable? For example, how do we know if the following languages are decidable or not? if you could decide your problem X then you would be able to decide some other problem Y which is already known to be undecidable. Turing recognizable languages are closed under union and complementation. Note all languages that sit outside of the region of "TM decidable" is "TM undecidable". By definition, a language is decidable if there exists a Turing machine that accepts it, that is, halts on all inputs, and answers "Yes" on words in the language, "No" on words not in the language. 3 . e. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site The question of whether all languages are Turing recognizable is a fundamental one in the field of computational complexity theory and the theory of computation. It covers three key points: 1. Simulate M1 on w. Sipser, “Introduction to the Theory of Computation,” Second Edition, Thomson/Course Technology, 2006, Chapter 4. Suppose is recognizable and recognizes . Koether Created Date: Recognizable Languages Robb T. Then there would exist a machine M B that recognized (but did not decide) B. To prove that a given language is undecidable: Construct a (mapping) reduction from another language already We say a L language is recognizable (L ∈ LR(T M)1) if there exists a Turing machine M such that. languages, i. Ask Question Asked 2 years, 3 months ago. 3 H is Undecidable C2. can the closure of an undecidable language be decidable? Also, can we generalize such undecidable classes? formal-languages; undecidability; closure-properties; Share. 4 Reprise: Type-0 Languages C2. David Richerby Why isn't the class of Turing-Recognizable languages closed under Complement? 10. For recognizable language, we can specify Closure for Recognizable Languages Turing-Recognizable languages are closed under ∪, °, *, and ∩ (but not complement! We will see this in the final lecture) Example: Closure under ∩ Let M1 be a TM for L1 and M2 a TM for L2 (both may loop) A TM M for L1 ∩L2: On input w: 1. Run on for steps. I know you're stuck, but you should at least have a strategy of what you want to do. ) if there exists a Turing machine M How to Prove Undecidability or Non-Turing-Recognizability in This Course. Note that every Turing machine (like every integer) has a finitely-long description, so the language is a set (an infinite set, in this case) of finite strings. By definition there is a Turing machine M such that L(M) = L. undecidable languages •We first introduce the diagonalization method, which is a powerful tool to show a language is undecidable •Afterwards, we give examples of undecidable languages that are –Turing recognizable but not decidable –Non-Turing recognizable Objectives Is the union of undecidable languages not Turing-recognizable? Hot Network Questions Meaning and source of names of maidservants Understanding the benefit of non principal repayment loan Do we know when a cohomology theory "comes from a site"? In particular, is compactly supported cohomology a sheaf cohomology on a site? I have written a proof to show that a Turing Decidable languages are closed under union (amongst other things). A recursive language is a formal language for which there exists a Turing machine that, when presented with any finite input string, halts and accepts if the string is in the language, and Now suppose that every language is recognizable. There are two equivalent major definitions for the concept of a recursive (also decidable) language: 2. ♦ I'm not sure I understand the question properly but every finite language is regular. { langle M, w rangle mid M text{ is a Turing machine and } M(w) text{ halts} } ). We showed that HALT 6∈ LD(T M) but it turns The language of a Turing machine M, denoted ℒ(M), is the set of all strings that M accepts: ℒ(M) = { w ∈ Σ* | M accepts w} For any w ∈ (ℒ M), M accepts w. Showing that Turing-recognizable languages are closed under union. Viewed 538 times Turing Machine That Accepts Machines With Undecidable Languages. Why? Well this would let you solve the halting problem. If this language is infinite, then the recognizer may not halt, which is not a problem for a recognizer since it is not a decider. For example, determining if a string is a palindrome is a decision problem because there’s a binary output for every word (yes or no). That said, there are also non-enumerable languages whose complement isn't either. Review: Turing Here we show that the A_TM problem is undecidable and recognizable, which is asking if there is a decider for whether an arbitrary Turing Machine accepts an To prove that a given language is undecidable: Construct a (mapping) reduction from another language already known to be undecidable to the given language. Turing-recognizable languages are languages whice are accepted by a Turing machine; decidable languages are languages for which a Turing machines halts, i. If a language A is Turing-Recognizable and Undecidable, what can be said of the Turing-Machine that recognizes the complement of A? To my understanding this turing machines accept states, are all those that were of the rejection states in the first Turing Machine. If this problem is decidable, we'd expect the problem to be decidable for any arbitrary string. If accepts, accept. Such problems have a direct correspond language is undecidable or unrecognizable. If M1 halts and accepts w, go to step 2. On input x, M simulates (N, y) for length(x) steps. Example 3. Follow I need to prove that the language L(EVEN) = { M : |L(M)| is even } is undecidable. (Th 4. 1 Suppose that C is the class of all objects that have some property (e. Being decidable means you can build an If one language is polynomial time reducible to a language already known to have a polynomial time solution, we obtain a polynomial time solution to the original language, as in the following R. 3 and show that E TM is not recognizable. Check-in 8. I am supposed to identify why closing a Turing Recognizable language under some operation is trickier to prove than when dealing with Turing Decidable languages. However, the reduction given from D to L is to prove how ATM is not decidable. 2 By contradiction S’pose that E For any language L' that is in NP if we show that we can poly-time reduce to Etm. 7k 3 3 gold Given 3 disjoint Turing-recognizable languages prove that one of them is decidable. Is this a correct idea? A recognizable language has a computable verifier but the verifier doesn't necessarily run in polynomial time (and the time hierarchy theorem says that there I am trying to prove that language K is Turing-recognizable but undecidable using the diagonalization method. (5 points) Prove that there exists a subset of {1}∗ which is not Turing-recognizable. and develop the set of languages co-r. Modified 2 years, 3 months ago. Definition 1. Show that the collection of Turing-recognizable languages is closed under the operation of union. But is this language Turing-recognizable or co-Tur Does every Turing-recognizable undecidable language have a NP-complete subset? The question could be seen as a stronger version of the fact that every infinite Turing-recognizable language has an infinite decidable subset. Improve this question. 2 A classic example of an undecidable language is the set of (descriptions of) Turing machines which halt on every input. If 2. If L is undecidable, and show its complement is recognizable, then L not recognizable. 5 Summary Gabriele R oger (University of Basel) Theory of Computer Science April 25/May 2, 2022 2 / 27 De nition (Turing-recognizable Language) We call a languageTuring-recognizableif some deterministic Turing machine recognizes it. Follow answered Dec 21, 2014 at 13:48. Example 2. Decidable • A language L is Turing recognizable if some Turing machine recognizes it. 1 An Undecidable but Recognizable Language Decidable and Recognizable Languages But not all languages are decidable! In the next class we will see an example: { A tm = fhM;wijMis a TM and w2L(M)gis undecidable However A tm is Turing-recognizable! Proposition 2. Proving Ackermann's function is decidable through a Turing Machine. Since we know that Diag is not decidable In particular, why isn't the complement of decidable languages just the union of unrecognizable languages and recognizable languages that are undecidable (i. Usually this means reduction to the Halting Problem or Rice Can a Turing recognizable language be decidable if it is possible to enumerate its strings in non-decreasing length? I think it's not because you can go to infinity, and this will make it undecidable right? turing-machines; Share. 1(Decidable Language). Second part is, if L is Turing recognizable then there must be a Turing Machine M that recognizes L. If that were the case, we could define a reduction from the known undecidable language $$ HALT = \{(\langle M\rangle \mid M \text{ halts on input }w\} $$ to $\overline{L}$ by the My thinking for part A is currently,for a given Turing Machine the language will always halt because the input will either accept or reject the empty string by definition of DFA. I am thinking a decidable problem could be reduced to a recognizable, undecidable problem because a polynomial time verifier would work on a decidable problem. All finite languages are regular. [ Choose ]reduce an undecidable language to it. either accepts or rejects, but never loops. 4. (Given algorithms to decide each language, describe an algorithm to determine if a string belongs to the intersection. for $\sigma$ is $$\{1^{\sigma(n)}: n\in \Bbb N\}=\{\epsilon, 1, 1111, 111111, 1111111111111, \cdots\},$$ which is undecidable. language over {1} which is recognizable but not decidable? A Turing recognizable language is decidable or not? 1. The halting problem is known to be undecidable, meaning there is no Turing machine that can decide An Undecidable but Recognizable Language Complementation Decidable and Recognizable Languages Recall: De nition A Turing machine M is said torecognizea language L if L = L (M ). Can anyone give examples of languages A and B such that we can prove B is undecidable using A in a proof by contradiction but we A is not $\leq_m$ B. Most problems we deal with in computing are decision problems. the equality problem of CFG, or the acceptance problem of a TM) will satisfy the precondition of the claim Let Lbe a recognizable language. Why REC languages is undecidable under emptiness and finiteness? Hot Network Questions Transformation of skewed independent variables for GLMMs How could I have come up with a proof of this proposition about subgroups myself? Does the rolling resistance increase with decreased Then, we can take the complement of all problems in r. Decidable and Undecidable Languages: Decidable languages have a definitive decision process via Turing machines, whereas I think you'll find the answer is actually that it's undecidable. In other words, the language L(EVEN) is the set of all Turing Machines which accept some language of even cardinality. Follow You are probably thinking about undecidable languages which are computably enumerable. theory of computation: how can A be the language recognized by machine M1. (b) Using Rice’s Theorem, prove that the following language is undecidable: ALLTM = {hMi|M is a TM and L(M) = Σ∗} 2. Iff Undecidable TM accepts M', then M must accept x, else M' does not accept an undecidable language. nir shahar nir shahar. Can we solve the halting problem for M with input x using oracle Q? First, we connect a new TM N to the front of M. L1 and L2 decidable ==> INTERLACE(L1, L2) decidable. 0. Since the universe of strings over any finite alphabet is countable, every language can be mapped to a subset of the natural numbers. RE Turing Machine M for a language L in Dec accept pipe reject pipe input string w Turing Machine M for a language L in RE Undecidable Languages Robb T. Here, M is the encoding of some Turing Machine which would be passed in as input if there existed a decider for L(EVEN). Assume that the complement, is also recognizable and recognizes . Here is a decider for . Any language outside Dec is undecidable. I didn't understand his proof and think that a well-defined bijection between a recognizable but undecidable language and N will help. What I tried regarding proving K is undecidable using diagonalization: Suppose there exists a Turing machine H that decides K, denoted by H. ) Intuitively, a language is Turing-recognizable if there is some computer program that, given a string in the language, can confirm that the string is indeed within the language. Now, recall that if a language is both recognizable and co-recognizable, then it is decidable. Prove that A language is decidable iff it is Turing-recognizable and co-Turing-recognizable. But i don't know whether undecidable languages is closed under complement or not. An undecidable language is a language that is not Turing-decidable. So you just Today we expand on Turing’s work. You have to prove that the language is undecidable: that no Turing machine can always say "no" correctly, even if it can always say "yes" correctly. All semi-decidable+ languages are undecidable, but we’ll see there are undecidable languages that aren’t semi-decidable+! Decidable and Undecidable Languages 30-3 Dec vs. 9 As for what some properties of these languages would be, it becomes harder and harder to say as we move up the complexity hierarchy. If you want another kind of proof, consider the following. This can be shown by reducing the halting problem to L: For the halting problem instance (N, y), create a new machine M for the L problem. R. The complementary language of a recognizable undecidable language is not recognizable. From what we’ve learned, which closure properties can we prove for the class of T-recognizable languages? Choose all that apply. For example In this lecture, we will prove that certain languages are undecidable. (I'm unsure about why you think this would be the case, it would make the halting problem decidable, for instance. True or False: Any decidable language is also recognizable. Why is "decidable" included in "Turing-recognizable"? Hot Network Questions Sourdough starter- what is happening? Recap: Recognizable versus Decidable Languages A language L is called Turing-Recognizable if there exists a TM M such that L(M) = L ¼Note: M need not halt on all inputs but it should halt and accept all and only those strings that are in L; it can reject strings by Turing-recognizable languages are closed under intersection. Title: The Acceptance Problem - Undecidable Languages - Lecture 29 Section 4. 17. (b) Closed under intersection. This theorem implies that a recognizable but undecidable language cannot have an increasing bijection with N. We can conclude that either the halting problem is decidable, or the halting problem for LBAs is undecidable or the thing that you're calling a mapping reduction isn't actually a mapping reduction. Thus, we can decide whether a machine M accepts input x. which is accompanied by Theorem 4. language over {1} which is recognizable but not decidable? Classify a language as Turing-recognizable or co-Turing-recognizable. If the complementary language of an recognizable language is a non-recognizable language, is the recognizable language a non-decidable language? I understand following about recognizable (aka recursively enumerable) and co-recognizable languages: Definition 1: Recognizable language is one which have one-to-one correspondence with the natural number with the additional property that we could specify an algorithm to enumerate the language elements. Proving that the set of deciders is not Turing-recognizable. 2 Author: Robb T. A language is co-Turing-recognizable if it is the complement of a Turing-recognizable language. Pass M' to Undecidable TM. ATM is very closely related to the language L I need to prove is not recognizable. Rao, CSE 322 5 One Last Concept: Reducibility How do we show a new problem B is undecidable? Idea: Show that ATM is reducible to the new problem B What does this mean and how do we show this? Show that if B was decidable, then you can use the decider for B as a subroutine to decide ATM Contradiction, therefore B must also be undecidable Since I don't think the "substring language" would help to decide such a language (and produce a contradiction), I'm not sure how to argue that such a language is undecidable. undecidable languages: Decidable Turing-AcceptableL L Lis Turing-Acceptable and undecidable. Languages (1) and (2) are, respectively, {0, 1}* and the empty language, both of which are decidable (so there are TMs that always halt that accept those languages). I've completed other problems Why are regular tree languages closed under intersection, but deterministic context free languages are not closed under intersection? 1 Give a class of languages which is closed under intersection and union, but not under complement Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site I have this language L = {&lt;M&gt; | M is a TM that accepts w whenever it accepts w^R} I was able to prove that this language is undecidable. Cite. Joey A language is Turing decidable if you can write a C program (replace C with your favorite programming language) that outputs YES if the input belongs to the language and outputs NO otherwise. If the simulation halted within that number of steps I have a problem giving "intuitive" explanation to turing-unrecognizable languages. We say a language L is decidable (L ∈L D(TM)) if there exists a Turing machine M such that • w ∈L ⇐⇒M accepts w TM was decidable, but that some other undecidable language Bwas Turing-recognizable. The bijection constructed from the proof of "every subset of natural numbers is countable" seems not If the input string is equal to one of these generated strings then ACCEPT. Rao, CSE 322 1 Are There Languages That Are Not Even Recognizable? Recall from last class: A TM = {<M,w> | M is a TM and M accepts w} A H = {<M,w> | M is a TM and M halts on w} A TM and A H are undecidable but Turing-recognizable ¼Are there languages that are not even Turing-recognizable? What happens if a language A and its complement A are both Turing Not all Recognizable languages are closed under complement. So DFA A will either accept the empty string or reject the empty string, but regardless the Turing Machine that represents the language will halt. so for instance, if i take a simple language like:"L = {< M > | M ACCEPTS < M >}", then it should be a turing recognizable language since there can be a turing machine that halts and accepts strings in it, and for strings not in that language it doesn't halt or just skip them. True or False: Any recognizable reduction of other undecidable language to L) Not Recognizable: 1. Follow answered Nov 18, 2022 at 18:28. Saying that the regular languages are a subset of the decidable languages (which is true) is just saying that every regular language is decidable: this is true. Prove that this language is undecidable. Koether Hampden-Sydney College Wed, Oct 28, 2009 An undecidable language maybe a partially decidable language or something else but not decidable. Improve this answer. Part of our mission is to build up an archive of high-quality questions and answers that will be useful not only to the original asker, but also to others in the future. Proof: That AP is recognizable is shown in the same way as for the language M above. A better result is that there are computably enumerable sets which are Turing incomparable (the Friedberg-Muchnik theorem). For example, let . Proving innumerable number of RE-hard languages. Note that we usually say "decidable", "recognizable", "co-recognizable" and "undecidable" simply without the prefix "TM". This means that our Turing Machine is Recognizable, but it is not decidable. ) This language contains all natural numbers, or; This language contains no natural numbers, or; This language contains all natural numbers greater than some natural number n. abjze vrtt evgwalu tthyb bysmj jexwdb grkqfb fksmswbt sjfl jvv