Prove unrecognizable language Lecture 17: Proving Undecidability 14 Reference from here If a Language is Non-Recognizable then what about its complement? There exist complementary languages of unrecognizable languages that are recognizable, and there exist complementary languages of unrecognizable languages that are undecidable. 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 One of my final exam questions asked me to give a Turing-unrecognizable language over an unary alphabet, which I wasn't able to complete, but I ended using the Busy Beaver function (BB(k)) which is known to be something intractable or uncomputable so that I constructed the language as a set of strings 1^BB(k) for natural number k=1,2, So we can show that L is not decidable, because a TM U that receive L as a input need to test all elements over {1}* and then decide to accept in case of M rejected all of them, so it will never halt and it means that L is not decidable, implies that the empty Language is Let L1 and L2 be two languages such that there exist no string w that belongs to both L1 and L2. E is a language, to accept language E we construct a Turing Machine. Harry Porter; www. <M> denotes an encoding of a Turing machine M. Here, M is the As for what some properties of these languages would be, it becomes harder and harder to say as we move up the complexity hierarchy. Koether (Hampden-Sydney College)The Acceptance Problem - Undecidable Languages Fri, Nov 7, 2014 15 / 25 I convinced myself that this problem is decidable, but I am having trouble proving so. And we'll introduce the method for doing that called the diagonalization method. EM will be provided as input the encoding of another Turing machines, If that inputted machine M accepts an empty language then it will be a member of language E, else it will be not a member of language. So, if I were to prove either one of them as unrecognizable, then I should be good. Recall, the di erence is that a language is recognizable if a turing machine accepts every string in the language, but has $\begingroup$ @YuvalFilmus so how would you recommend to show the language is unrecognizable? im taking about the language not the turing machine, sorry for the confusion $\endgroup$ – marcelomo Commented Apr 21, 2021 at 12:58. Suppose that w accepts itself. 5. Let's consider the following two languages: L1 = {0n1n | n ≥ 0} (the language of all strings of the form ETM = {<T> | T is a TM, and L(T) = 0} 3. OK, so that is the plan-- proving A,TM is undecidable. I am not quite sure what you mean by algebraic. If w i 2=L, then the ith bit of s is 0. $\endgroup$ – You want to prove that the language of deciders—the language of Turing machines whose language is decidable—is not recognizable. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. So L consists of all Turing machines that satisfy the property described above. The good news is that we need not do this for every undecidable or unrecognizable language; we have a strategy that we can use to place languages in these categories. w7l?!;YIw} is undecidable (halt means reaches either qaccept or 9reject) ③ If L is undecidable and recognizable show that Lamp = { * LL is unrecognizable How to show a language is not computably enumerable? Related. Robb T. Here is one common example of an unrecognizable language: The language $\overline{A}_{TM}$ of all pairs (M,w) where M is a Turing machine that runs forever on input w. Check-in 8. I see three languages in this problem: The original recognized language; The set of strings in which the original recognizer halts in the reject state; The strings on which the recognizer loops; $\begingroup$ Indeed, the main idea here is the fact that the recognizable languages are closed under union, since their recognizers can be run in parallel. Just take the machine that always halts. ~A TM is the canonical example of a Turing-unrecognizable language. There are undecidable (and unrecognizable) languages over {0,1} {Turing Machines} {0,1}* {Sets of strings of 0s and 1s} {Languages over {0,1}} Set L Set of all subsets of L: 2L {Recognizable languages over {0,1}} There are (many) unrecognizable languages In the early 1900’s, logicians were trying to define consistent foundations for mathematics. This means there does not exist a Turing Machine which will accept the set of all machine-string pairs such that M does NOT halt when run This means that if we take any two unrecognizable languages L1 and L2, their union L1 ∪ L2 should also be unrecognizable. In Theorem 3. Given this fact, assume that L is also semi-decidable. The claim you want to prove or disprove is: The union of any undecidable language and any finite language is undecidable. 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 Proof that there are languages which are not even recognizable. (c) Closed under complement. reading: automata constructions, 1. Prove that the language H of all halting algorithms is There are four major theorems (and their uses) that we will study during this course, providing complete proofs: the pumping Lemma for regular languages, used to show that there are languages that are not regular; the existence of a Universal Turing Machine; undecidability of the Halting problem; and Cook's theorem that NP-complete problems exist. $\endgroup$ – R. By First, what is a "simple" example of a language which is not Turing recognizable: If H H is the halting problem, then I claim Hc H c (that is, the complement of H H) is not Turing-Unrecognizable Languages Proof of the theorem. Citation from Wikipedia: . A language is decidable if there is a TM for which this is the case. Rao, CSE 322 7 Review of Chapters 0-1 See Midterm Review Slides ¼Emphasis on: Sets, strings, and languages Operations on strings/languages (concat, *, union, etc) Lexicographic ordering of strings DFAs and NFAs: definitions and how they work Regular languages and properties Regular expressions and GNFAs (see lecture slides) Pumping lemma for regular language recognizable? By a similar counting argument, we know that there are uncountably many languages and only countably many recognizable languages, so most languages are unrecognizable. If a language is unrecognizable, then it must be undecidable. Is there a notable unrecognizable language, in the same sense that HALT is a notable undecidable language? Lets prove two theorems about closure to show the Turing-Unrecognizable Languages Lemma The set Bof all inﬁnite binary strings is uncountable. Math Mode. Show that this language is not recognizable. Introduced The Reducibility Method to prove undecidability and T-unrecognizability. Unfortunately your proof is not correct. Now we demonstrate a language which is not even Turing-recognizable. . If more than n of the running instances of M ever halt-accept, then halt-accept. episolon, 0, 00, etc). 1. Encoding input for TMs Sipser p. And except that I can not even come up with a language so it and its complement are unrecognizable. Note that if a language of any number of 1s is unrecognizable, then it is I need to prove that every L ∈ A contains an unrecognizable subset. Here I would like to address the following question. • We have seen a language that is undecidable. 5. Visit Stack Exchange 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. is T-recognizable but also undecidable . SOL- Let HP={ A,s :HP={ A,s : A is a TM that halts on s }} and HP¯={ A,s :HP¯={ A,s : A is a TM that does not halt on s }} Suppose RR is a recognizer for LL. Prove that this language is undecidable. If M1 accepts, then ACC w and halt; if M2 accepts, REJ w and halt. "Use diagonalization in a proof of undecidability. A better way to phrase my question is: Is there a language of any number of 1s over $\Sigma=\{0,1\}$ that you can make unrecognizable?. Or: Construct a I have a problem giving "intuitive" explanation to turing-unrecognizable languages. Suppose we create a Turing EM for the language E. The bad news is that this process is A Turing-unrecognizable language. 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 One way to prove a language is decidable is to produce a decider. Rice’s Theorem. I know that hint (and the whole hierarchy of languages). 1or when the context is clear just L(TM) 16: Undecidability by Reduction-1 undecidable, since there is no TM which answers yes for strings in the language and no for strings not in the language; co-recursively-enumerable in that there is a TM which answers no for strings not in the language (using dovetailing, try all strings on all TMs and you will eventually answer no for any string not in the language) the first). We say a language A is Turing-reducible to a language B, written A ≤ T B, if given an oracle that decides B, there exists a decider for A. ) Notes: 1. Koether Homework Review Closure Properties of Decidable Languages Intersection Union Closure Properties of Recognizable Languages Intersection Union Assignment Homework Review Exercise 3. If M does not accept w, then U may halt in its reject state or it may Some common patterns are those addressed by the Rice's theorem (proving a set undecidable) and Rice-Shapiro theorem (proving a set unrecognizable). First, you should show that if there exists an enumerator E which enumerates all strings in the language L, we can construct a recognizer for this language L. For (a) I know that EQ_tm and comp(EQ_tm) are unrecognizable, but I don't know how that can be proven. The Halting Problem - Undecidable Languages Robb T. pdx/~harry TM) and one unrecognizable language (A TM). Is the difference between an unrecognizable language and a finite language decidable? recognizable? Hot Network Questions Short story(?) Boys and Girls live apart, and meet for procreation on a space station. is T-unrecognizable . Though it might be worth reading about The Arithmetical Hierarchy. Recognizable languages. The machine can reject a word from or run forever in a loop. Nothing is implied about the complement of . Theorem 2. For L to be RE, we would need to be able to tell that TM always accepts w^R when it accepts w. For example, if L ∈ A, it is decidable but not regular. Then, by the definition of L, w is not in L. Show $\overline{L}$ is in $RE$ by Simulate M1 and M2 on w one step at a time, alternating between them. Some unrecognizable languages \overline{A_{TM}} = \{\langle M, w \rangle \; | \; M \text{ is a Turing machine and } M \text{ does not accept } w \}. If a Language is Non-Recognizable then what about its complement? 1. But if you mean by just using the definition in terms of recognizing morphisms in such a manner that the proof would easily generalize to other monoids, then this would not be possible. This recognizer works with an input w ( a string ) and runs the E inside. e. A mathematical proof would be of great help since im unable to think of any way to prove this. Oftentimes, we can use a decider that exists for a somewhat different language as a component of our decider. $\overline{One}$ I'll define as the set that doesn't contain 1 in it (i. To prove that a given language is non-Turing-recognizable: Either do both of these: Prove that its complement is Turing-recognizable. Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. Except one explanation from math. Introduction to reductions. • Recall that the complement of a language L is the language L consisting of all strings that are not in the language L. • We say that a language is co-Turing-recognizable if it is the The idea is, that we want to show that ATM is undecidable, i. To do a proof by contradiction, you need to take the opposite: The union of some undecidable language and some finite language is decidable. stackexchang. The proof for both of these was a bit involved, and required us to generate a para-dox. D. So I have to show a translation of any given PCP-problem into some form that makes it appearent that it actually is or can be viewed as a problem of 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. Since we know ATM is undecidable, we can show a new language B is undecidable if a machine that can decide B could be used to build a machine that can decide ATM. To prove this, we need to use a diagonalization argument, which is based on proof by contradiction. For any 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. (This is the complement of the halting A few answers has addressed the confusion about the length of a word being infinite. Is this language decidable, recognizable, or unrecognizable? 1. ! TM is T-unrecognizable. 1. (a) Closed under union. Koether Turing-Unrecognizable Languages Proof of the theorem. 6, page 160. There are two equivalent major definitions for the concept of a recursive (also decidable) language: 2. 3. 12 Stack Exchange Network. Proof: ! TM. (Reduce from ~A TM. More than one of the above. 1 Unrecognizable languages The contradiction we just showed demonstrates that the language L halt is not decidable. When proving closure of the class of decidable languages under a given operation the obvious choice is an assumed decider for a given decidable language. 3 Lecture 17: Proving Undecidability 13 Acceptance Language A TM= { < M, w> | M is a TM description and M accepts input w} We proved ATM is undecidable last class. - I can take both recognizers and run them in parellel, simulating a step on each Build a universal Turing machine U and use it to simulate M on the input w. Prove the intersection of two Turing-decidable languages is Turing-decidable. Show that there exists a language that is Turing-recognizable but not decidable. 4. We can prove that, say, ${\overline{A_{TM}}}$ is not turing-recognizable, because that would make In case your language is in $Co-RE$ you can do one of the following: Show $\overline{L}$ is in $RE$ by describing a TM that accepts $\overline{L}$. (d) Closed under concatenation. 2. If we can find an unrecognizable languageA that is mapping-reducible to L (that is, A ≤ m L), then L is unrecognizable. A language is RE if a TM can halt-accept on all strings in the language. uvixyiz Let $L$, $L_1$, $L_2$ $\subseteq \Sigma^*$ such that $L = L_1 \cup L_2$, and $L_2$ is decidable. The bad news is that this process is Next I did some demonstrations to show how T-Recognizable languages are closed for Union, Intersection, Concatenation and Kleene Star. Then the function f : M!Ldeﬁned by f : M 7!L(M) We can easily prove this language is semi-decidable. This is a contradiction. Instead, we have to prove We will show $\mathcal{T}$ is countable by showing it is encodable. 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. Prove that the language L = { <T> : T is a Turing machine that runs in polynomial time } is not Turing-recognizeable Hot Network Questions How to reconstruct a lost VeraCrypt keyfile? 1. Clearly, this is a one-to-one correspondence between Show language and its complement are both recognizable: If is T-recog and ̅ is T-recog then is T-decidable. 3 . Otherwise, build a NDTM which recognizes it. But by the definition of w recognizing L, w must be in L. Show transcribed image text Stack Exchange Network. Part of the theorem: "A language is Turing-recognizable if and only if some Lecture 23: Union of recognizable languages is recognizable. From what we’ve learned, which closure properties can we prove for the class of T-recognizable languages? Choose all that apply. To prove that this assumption is false, we need to find two unrecognizable languages whose union is recognizable. Recognizable; Non-Recognizable; May be Recognizable, May be Non-recognizable. I. 2 EQTM is Turing-unrecognizable Prove that the following language is Turing-unrecognizable by showing that ATM ≤m EQTM EQTM = {(T1, T₂) | T₁ and T₂ are TMs and L(T₁)=L(T₂)} PROOF: For B is Turing recognizable since we can interleave executions of M on all possible input tapes. , no Turing machine can semi-decide L′). A language is coRE if a TM can halt-reject on all strings not in the language. Proof that the union of recognizable languages is recognizable. Definition: A language is a set of strings (i. For example, let . 2 Decidability and Undecidability Recall that: A Turing Machine Mdecides a language Lif it halts on every input and Lis the set of words for which Mhas output 1. I will also show the complement of A,TM is Turing-unrecognizable. ) I know that a language is Turing-decidable if there is an algorithm to decide membership. For L to be coRE, we would need to be able to tell that TM accepts some w but not the corresponding I claim that L is an unrecognizable language. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company • The set of all languages over a finite alphabet,P(Σ∗) • The set of all undecidable languages • The set of all unrecognizable languages 2 Turing Reductions and Undecidability Definition 2. Any suggestions ? I need to prove that the language L(EVEN) = { M : |L(M)| is even } is undecidable. I am not sure how to formally construct such a subset or prove this rigorously. ms/Lm. I am struggling on How to Prove that, if L1 and L2 are both co-Turing-recognizable, there exists a Is this language decidable, recognizable, or unrecognizable? 2. The second statement gives us a strategy to prove thatL is unrecognizable. To prove that a given language is non-Turing-recognizable: Either do both of these: • Prove that its complement is Turing-recognizable. So given a language A that we already know to be unrecognizable, we could prove the un- A,TM is not decidable, which we promised we will prove today. So we did a quick example of that last time. I suspect that I can construct a subset L′⊆L L′ is unrecognizable (i. For every i 1, If w i 2L, then the ith bit of s is 1. I said L is a decidable language because I can just run M on a function D(M) that returns false if there exists a loop somewhere between start and accept state of M, and returns true otherwise. How to prove that the provided. 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 language L that consists of all Turing Machine descriptions M, for which the language accepted by M is finite. In other words, the language L(EVEN) is the set of all Turing Machines which accept some language of even cardinality. Defined mapping reducibility as a type of reducibility. We're going to go over that Proving additional languages are not decidable, by using reductions. Visit Stack Exchange Thm: There are unrecognizable languages Assuming the Church-Turing Thesis, this means there are problems that NO computing device will ever solve! The proof will be very NON-CONSTRUCTIVE: We will prove there is no onto function from the set of all Turing Machines to the set of all languages over {0,1}. 159 C. In fact, \most" languages are both undecidable and unrecognizable. In order to show that every decidable language is recognizable, you take the Turing machine deciding the language and modify it to recognize the language (exercise). cs. If M accepts w, then U will halt in its accept state. ≤ You have to prove both parts since it includes "if and only if" phrase. Show that there exists a language that is not Turing-recognizable. ~ATM is the canonical example of a Turing-unrecognizable language. "Theory of Computation"; Portland State University: Prof. Each Turing-recognizable language is associated with a TM, so there can be no more Turing-recognizable languages than TMs. Could someone please explain this? Question: [3 marks] Prove that the set of all unrecognizable languages is not closed under union. Now suppose that every language is recognizable. Creating a turing machine that simply goes right and accepts if we reach end without hitting a one, rejects if we do hit a one seems to work for this case. L1 and L2 decidable ==> INTERLACE(L1, L2) decidable. !$ TM and !$ TM are T-unrecognizable. Because L and its complement are both semi-decidable, we have an effective procedure for deciding each: run the TMs for L and its complement alternately, and one will eventually list the input. a subset of many languages and only countably many recognizable languages, so most languages are unrecognizable. The Chomsky Hierarchy – Then & Now 1. Counting proof that there are unrecognizable languages. e cant comment. Prove that its complement is undecidable. This means there does not exist a Turing Machine which will accept the So I have to show that if I could decide L then I could decide PCP. • The set of all languages over a finite alphabet,P(Σ∗) • The set of all undecidable languages • The set of all unrecognizable languages 2 Turing Reductions and Undecidability Definition 2. First, based on Remark (Equivalence of Turing machines), we will normalize TMs and assume the set of states $Q$ is always \(\{0,1,2,\ldots, k-1\} \cup If my L, and its complement, aren't recongnizable, I need to show it through a reduction with another known unrecognizable language (which is, in my case, Atm and it complement, halting problem, EQtm and its complement). Recall, the di erence is that a language is recognizable if a turing machine accepts every string in the language, but has the first). A ≤ m B ⇐⇒A ≤ m B. We know that A cannot be Turing-recognizable because, if it were, the language B' = {<N> | N is a TM and L(N) contains no more than n strings } would be Turing-recognizable (we could interleave 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 TM) and one unrecognizable language (A TM). So given a language A that we already know to be unrecognizable, we could prove the un- 1. 2 Decidable Languages Decidable Problems Concerning Regular Languages Decidable Problems Concerning Context-free Languages 3 Undecidability The Diagonalization Method An Undecidable Language A Turing-unrecognizable Language Sipser's Theory of Computation, Third edition, chapter three asks me to prove this. (But the proof will work for any finite Σ) Languages Robb T. that there is no TM H, that when given any another TM M and any other input w to this TM M as input can decide if the TM M will accept input w. And we will do so. To prove that a given language is Turing-recognizable: unrecognizable languages. (b) Closed under intersection. Note: it’s not enough to just design a TM that loops on some input. ÷. ! TM is undecidable. Does anyone know how to do this reduction, or if it is possible to prove that the language of TMs that accept every input string is unrecognizable using a reduction from a different unrecognizable language? Answer to how to prove there is an unrecognizable language A. In other words: if L1, L2 are unrecognizable, then L1 UL2 is not necessarily unrecognizable. It is Turing recognizable if in the latter case the C program simply never halts. 21 we showed that a language is Turing-recognizable iff some enumerator enumerates it. Even though A,TM itself is recognizable, the complement is not. A language is recognizable if there’s a Turing machine that accepts all the words . Is there a notable unrecognizable language, in the same sense that HALT is a notable undecidable language? Lets prove two theorems about closure to show the answer is yes. Proof. "Use counting arguments to prove the existence of unrecognizable (undecidable) languages. that given ( M, w) can " simulate " M's computation on input w ② Given that Atm is undecidable show that HALT. These are particularly Given the following language: Lf = { p(m) | The language of M is finite } Is Lf recognizable? if not, prove using reduction. I'm kinda sure that Lf can not be recognized but I ain't sure how to prove it. $\endgroup$ – Jens Bossaert Commented May 27, 2015 at 15:45 Breakout Room Problems ①Give a more detailed description of a universal TM iie. However, I am not sure where to start with this proof. We can actually prove that there exist languages that aren’t even recognizable. (Given algorithms to decide each language, describe an algorithm to determine if a string belongs to the intersection. Definition of a recognizable language. $\begingroup$ @SumeetSingh Under your proposed definition, every language is recognizable. Therefore one way of showing that a language is decidable is So using reducibilities to prove problems are undecidable, or unrecognizable, and the basic way that works is we're going to leverage another some problem we already know is undecidable say, or unrecognizable to prove other problems are unrecognizable. A rough outline of my answer would be : Assume we have a TM T that has only one halting state, and if it wants to go through all the states it needs to pass through this halt state and we somehow need to show how they cycle through all the states as such. Prove that if $L$ is not recognizable then $L_1$ is not recognizable. For assume that there is an algorithm, represented as a word w, that recognizes L. As we’ve seen, this may just be a poor approach, and a better (TM decider) approach may exist. Is the complement of a Non-Recognizable language. qgkei ckvu pcztfdby avdq wqmv jtngriv vvgfxc ewld tfxd pnygi bvsk ruj jvxsl wfu jlsl

Prove unrecognizable language. (b) Closed under intersection.