Decidable but not recognizable Engineering; Computer Science; Computer Science questions and answers; To show that a language is NOT recognizable, one could show that its complement is recognizable but not decidable. –If w ∉ L, M enters qReject. 4) It might be that A and its complement are both not recognizable. 4) Intersection. Bottom line: For every \strictly semi-decidable language", its complement cannot be semi-decidable. Are there countably many symbols (in the context of computation theory)? 0. But one idea I had was this: 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 Main Ideas in Section 4. 3) A might not be recognizable but its complement has to be recognizable. Halting (on empty input tape) for an infinite subset of all Turing machines. Improve this question. , A TM. On the other side, if S has no bound, it contains elements larger than any number. Stack Exchange Network. Note: it’s not enough to just design a TM that loops on some input. $\begingroup$ I am asking for better intuition on the steps needed to reduce a language to another to prove it is not recognizable. An undecidable language maybe a partially decidable language or something else but not decidable. reduce it to an undecidable language. ; T executes one step of 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 clarification. If M A accepts x, T accepts x. It is actually known that recursive languages are the intersection of r. These are also called theTuring-decidableor decidable languages. So, there are languages that are recognizable, but not decidable. ’ Solution (a) Let M be the Turing machine that loops inde nitely on all inputs, which has language L = ?. 103k 6 6 gold badges 73 73 silver badges 121 121 bronze badges To show that a language is NOT recognizable, one could. $4. Languages recognized by a TM are called recognizable. The set of all languages that are recognizable and co-recognizable are the decidable languages, so if the halting problem were co-recognizable then it would be decidable. (c) False, by Corollary 4. If L is undecidable, and show its complement is recognizable, then L not recognizable. So I'm stuck. 2(Recognizable Language). WewanttobuildaTMM thatdecidesA REX. Thus, Lc is Turing-recognizable. Which of the following language is/are Turing decidable? But the issue here is L(G) is not closed under complement. (Th 4. The 3-TM T (a Turing machine with 3 tapes) that decides A can be described as follows: T: input x. As we’ve seen, this may just be a poor approach, and a better (TM decider) approach may exist. But clearly, anything is decidable, is recognizable. 12/2/2016. 18) ; We prove that A Alin Tomescu, 6. We define the language A TM and prove: ; A TM is recognizable but NOT decidable; A TM ' is NOT recognizable; The proof that A TM is recognizable uses machine U, the Universal Turing Machine ; Simulates the action of any Turing Machine on any input ; We prove there are languages that are not Turing-recognizable (Corollary 4. Turing decidable means it halts in an accepting state if the input word is in the language, and halts in a rejecting state if the word is not in the language, Turing recognizable means it halts in an accepting state if the word is in the language, and in a rejecting state or fails to halt if the word First, what is a "simple" example of a language which is not Turing recognizable: If $\mathcal{H}$ is the halting problem, then I claim $\mathcal{H} Understanding Turing Machines: Recognizable and Decidable langauges. Let M L be a TM accepting L, and let M L be a TM accepting How can you prove a language is decidable ? Lecture 17: Proving Undecidability 3 What Decidable Means A language L is decidable if there exists a TM M such that for all strings w: –If w ∈ L,M enters qAccept. 3) Kleen closure. For any problem P, let Q be the problem of determining whether P is decidable or not. My thinking is, if M halts, then you can test the output and accept or 1 Semi-decidable vs. recognizable but not decidable. (2) If L and Lc are Turing-recognizable, L is decidable. T decides a language L if T recognizes L, and halts in all inputs. 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). We will show some specific language 0. 279k 27 27 gold badges 316 316 silver badges 512 512 bronze badges $\endgroup$ 2 $\begingroup$ But I don't understand how are you defining Y $\endgroup$ Turing-recognizable languages. Contradiction. Classification Table: Now we will classify most commonly asked problems as Decidable, Semi-decidable and Undecidable. 840 Theory of Computation (Fall 2013), taught by Prof. We will use this is recognizable but not decidable. (2 points) Must L2−L1 be Turing-recognizable? Prove your answer. Definition 1. bigger languages are not necessarily ‘harder. "-" denotes set subtraction. Conclusion. Well, we already showed above that ATM is not decidable, but we showed that it IS recognizable. ) I know that a language is Turing-decidable if there is an algorithm to decide membership. 6. Can decide this by the following TM: M = “On input hN,Ri, where N is an NFA and Ris a regular expression: 1. We saw in 1 that A is co-Turing-recognizable and in 2 that A is Turing-recognizable. Yuval Filmus Yuval Filmus. Given the string not in the co-recognizable language, Turing machine can eventually confirm that the string indeed does not belong to the language. We construct a Turing TM is not decidable. (Hint: first find a computable “one-to-one correspondence” from {1}* to {0,1}, sending each unary string in to some binary string sn, and A language is decidable iff it is Turing-recognizable and co-Turing-recognizable. A decidable language and a Turing recognizable but not decidable language are two distinct concepts in the field of computational complexity theory, specifically in relation to Turing machines. 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 thing I understand is that the complement of every Turing recognizable(but not decidable) language is non Turing recognizable. (%): from above, decidable 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. Halts for all rejects, but might accept/loop otherwise? 2. Use Diagonalizaton or 2. 0. L is decidable (the Turing machine that rejects all inputs is a decider for L), but M is not a Decidable Turing-recognizable • We proved Regular ⊆ Context-free since we can convert a FA into a CFG • We just proved that every Context-free language is decidable • From the definitions in Chapter 3 it is clear that every Decidable language is trivially Turing-recognizable. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). We will use Decto name this set. Because there are more languages than TMs. Find an infinite recognizable not decidable subset of an infinite decidable Decidable Languages 4 •decidable problems concerning regular languages •computational problems concerning finite automata •represent computational problems through languages •already have a framework and terminology •e. Show transcribed image text. Total views 100+ University of California, San Diego. So, BUT since, B is an Turing recognizable(not decider) only (assuming from your query), it means it might never halt for the input which doesnt belongs to it. It is important to note that these regular expressions 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 Prove the intersection of two Turing-decidable languages is Turing-decidable. We can combine machines for language one and the complement of language 2 in a single no deterministic Turing machines that is a decider. If is not decidable, then or ̅ is not Turing-recognizable. 1(Decidable Language). Are there problems that cannot be solved by any algorithm? Consider the language: ATM = {<M,w> | M is a TM and M accepts w} NOTE: <A,B, > is just a string encoding the objects A, B, In particular, <M,w> is a string listing all the components of TM Turing Recognizable: Let's check it: If L is CSL : It is turing recognizable as well as turing decidable If L is RE : It is turing recognizable but may not be turing decidable. Intersection of 2 recursively enumerable languages is recursively recognizable language is decidable. Therefore, if we recognize that w is not in L2 and decide it's not in R2, we have recognized that w is in A. If L 3 was not decidable, no possible Turing machine could decide L 3. reduce an undecidable language to it. Therefore, A is decidable. Gabriele R oger (University of Basel) Theory of Computer Science April 25/May 2, 2022 20 / 27. answered Oct 19 $\begingroup$ I know that intersection of two decidable is decidable and intersection of two recognizable is recognizable. Tautologically, either P is decidable, or it isn't. So we dont know what kind of language this is. 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. If A m B and A is undecidable then B is undecidable. Partially decidable or Semi-Decidable Language-– A decision problem P is said to be semi-decidable (i. t or f. In conclusion, decidable and undecidable problems highlight the boundaries of what computers can and cannot solve. Let’s look at a di erent one: De nition 4. Languages decided by a TM are called decidable. "Theory of Computation"; Portland State University: Prof. Otherwise, the class of decidable languages would equivalent to the class of recognizable languages, which we’ve just shown to be false. Let's call M A, M B and M C the Turing machines that recognize A, B and C, respectively. Proof. Share Cite Turing-recognizable language Answer: A language A that is recognized by a Turing machine; i. Rice’s Theorem (the version I use): ﬁLet C be a proper, nonempty subset of RE. Although it might take a staggeringly long time, M will eventually accept or reject w. Statement 4 is true as Turing recognizable languages (RE languages) are closed under union and intersection. Understanding the difference between these two types of languages is important in the realm of cybersecurity, as it has implications for the solvability and computability of problems. If 𝐽 is undecidable and 𝐽≤𝑚𝐽, then both ̅ 𝐽 and 𝐽 ̅are not Turing-recognizable. We would like to show you a description here but the site won’t allow us. computability; undecidability; closure-properties; sets; $\begingroup$ I think the problem is recognizable but not decidable. A language •Turing-recognizable languagesare sometimes referred as “recursively enumerable languages” •A TM which recognizes a language 6is called a recognizerfor 6 –A Turing machine that halts on allinputs is a decider. to L, alternatively via reduction of other undecidable language to L) Not Recognizable: 1. (d) False. ) 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. Are there languages that are NOT Turing-recognizable? Lemma If language L is Turing-recognizable and its complement L is also Turing-recognizable, then L is decidable. Follow asked Oct 27, 2020 at 7:52. Decidable languages. Turing-recognizability If A m B and B is Turing-recognizable, then A is Turing-recognizable. rvsrx lzecax vbh izneon eulel bkboa jaqg lfawsr bxezh ygzte hixos atqis cruqa qpfn ehvsv