Non turing recognisable language

Non turing recognisable language. Turing recognizable languages are closed under union and intersection. If M accepts, then remove that string from B and put it in a list, R. Recursive language (REC) – A language ‘L’ is said to be recursive if there 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. intersection a. P is non-trivial: There exist TMs M1 and M2 such that hM1i ∈ P but hM2i 6∈P . –A deciderthat recognizesa language decides it. We construct a TM M0that recognizes the union L 1 [L 2: M0 = \On input string w: 1. A non-mechanical way to get an infinite decidable subset of a Turing-recognizable language? 2 Non Turing-recognizable language. union. 16 Show that the collection of Turing-recognizable languages is closed under the op eration of d. P is a property of (only) a TMs language: For any TMs M1 and M2 such that L(M1) = L(M2), hM1i ∈ P if and only if hM2i ∈ P . •Halting is not required for ?∉6, just non-acceptance •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. Consequently, A' is decidable. 2. De nition 2. A Turing machine could loop on a specific instance. Explanation : Statement 1:For every non-deterministic Turing machine, there exists an equivalent deterministic Turing machine. a non-empty language). If anyone has any ideas on how to approach both of these problems, I'd be very thankful. We construct a TM TM M′ M ′ that recognize the union of L1 L 1 and L2 L 2: On input w w: This is not a context free language, but surely this is decidable by a Turing Machine that checks if the length N of the input is not divisible by any number between 2 and N-1 . Rice Theorem - Rice theorem states that any non-trivial semantic property of a language which is recognized by a Turing machine is undecidable. 1. One can construct a Turing Machine T that simulates D. 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. Nov 12, 2019 · What is an example of a Turing-recognizable infinite word, which is not Turing-decidable? 1 Why is this language Turing recognizable and not not-Turing recognizable Nov 3, 2017 · This question is taken from an exam of a Computer Theory Course. Run M 1 and M 2 alternately on w, one step at a time. It is co-recognizable if there is a Turing machine that enumerates all words not in the language. For every non-deterministic Turing machine, there exists an equivalent deterministic Turing machine. (a)A language is regular i it is recognised by a nite automaton (FA). Feb 19, 2018 · $\begingroup$ @SumeetSingh Under your proposed definition, every language is recognizable. A regular expression that generates words of 0’s and 1’s that begin and end with the same symbol. Jan 26, 2021 · Prove the languages |L<M>| = 2 and |L<M>| $\not=$ 2 to be non-Turing recognizable or non-recursively enumerable 1 Turing recognizable language between languages that aren't recognizable? A Turing Machine decides a language if it rejects every string it doesn’t accept – i. There exist non-Turing-recognisable languages. This Statement is true. – Peter Leupold. To prove Turing Key point. However, I also know that Turing-recognizable languages are not closed under complement, and A¯ ∪B¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯ = A ∩ B = C A ¯ ∪ B ¯ ¯ = A ∩ B = C, which seems to suggest that Turing-recognizable languages are not closed under intersection. After a lot of searching I came to the conclusion that yes, there are languages that are not even Turing Recognisable, but I can't get good examples which are simple to understand. I want to prove that their symmetric difference is Turing Recognizable. If a match exists, it will be found. Just one example for each . 0. 5. You need to prove that using those algorithms you can obtain a new one that decides membership of strings in the intersection. Solution 1. Nov 25, 2015 · There's a famous theorem that every infinite Turing-recognizable language has an infinite decidable subset. * Otherwise, leave it in B. It's a essentially a Turing-complete language, with one (major) caveat: every loop must contain a bound on the number of iterations. Perhaps the language of strings x. $\endgroup$ – Recognizers and Deciders. e. Programming languages can be used to create programs that control the behavior of a machine and/or to express algorithms precisely. Just take the machine that always halts. In this article, we have explored the idea of Non Turing Complete Programming Languages and listed all examples of Non Turing Complete Programming Languages along with advantages of such languages. Proof: ⇒. Turing decidable languages are closed under intersection and complementation. 28) is given at the back of Chapter 5 of Sipser’s book. The turing machine accepts all the language even though they are recursively enumerable. Here is my thinking, Since B is Turing Recognizable => There is some TM which accepts all the words of language B => There is a TM which accepts (all the words of language A + some other words) => There is a TM which accepts all the words of language A => A is Turing Recognizable. We will use Dec to name this set Jan 25, 2023 · Truly Turing-complete languages - ones that require the full computing power of Turing machines - are pretty rare. A language is Turing-decidable is there is a Turing machine that always stops and answers YES or NO correctly. – D. How to show that this language is Turing recognizable. $\endgroup$ – Gyro Gearloose. For example, if L1 is regular and L2 is recognisable, then is L1 intersection L2 decidable? My answer is that L1 is also recognisable and therefore L1 intersection L2 is recognisable. 1 Show that single-tape TMs that cannot write on the portion of the tape containing the input string recognizes only regular languages. The proof of Rice’s Theorem (Prob. To prove decidability or recognisability, it’s often easiest to provide a Turing machine with the desired attributes. Describe how a NON-Deterministic Turing Machine with two tapes recognize the language generated from the grammar: $ S \\rightarrow Apr 14, 2015 · I'm studying Turing Machines and I've already showed how Turing-Decidable is closed for the operations of Union, Intersection, Concatenation, Complement and Kleene Star. These are also called theTuring-decidable or decidable languages. I'm very new to this topic and didn't find any similar problems online, so I'm looking for any kind of tips to solve these problems :) A function that maps an instance ATM = {<M,w>| M accepts input w} to a N such that. Because of the Church-Turing thesis, any programming language is as powerful as a Turing machine. $\endgroup$ – Apr 3, 2013 · Every regular language is Turing-decidable and therefore Turing acceptable / recognisable (but note that Turing acceptable does not imply Turing decidable). Mar 4, 2017 · Turing machines. Turing recognizable languages are closed under union. Turing recognizable languages are known as RE languages whose complement may not 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; co-recursively-enumerable in that there is a TM which answers no for strings not in the language (using dovetailing, try all strings on Feb 26, 2018 · Show that Turing recognizable languages are closed under intersection. $\endgroup$ – Jul 26, 2020 · 1. RE R E is countable, by the same logic, since there is a TM semi-deciding each RE R E language. LaTeX is Turing-complete, despite being a text-processing language. CS. , all programs that can be written in some given programming language that is general enough to be equivalent to a Turing machine. The following TM U recognizes A. After a lot of searching ME was to the close that absolutely, thither become languages that are doesn even Turing Recognisable, although I can't get good examples which are simple to understand. Infinite loops are not allowed. The standard proof of this result works by constructing an enumerator for the Turing-recognizable language, then including the first enumerated string in the decidable language, then the first string that comes after it lexicographically Mar 2, 2024 · 1. y. 4. The set of non- RE R E languages, RE¯ ¯¯¯¯¯¯¯ R E ¯, is not countable, then by A programming language is an artificial language designed to communicate instructions to a machine, particularly a computer. This means that our Turing Machine is Recognizable, but it is not decidable. However, infinite languages (Turing recognizable or not) always have subsets which are not Turing recognizable, simply because they have uncountably many subsets, but there are only countably many Turing recognizable languages. @SamB. Turing decidable languages are closed under intersection. 4 Non-Recognizable Languages Theorem 16. See this thread for lots of examples. Mar 25, 2016 · No, finite languages don't have subsets which aren't Turing recognizable. If Mar 6, 2015 · The question: Show that the collection of Turing-recognizable languages is closed under the operation of union. The answer: For any two Turing-Recognizable languages L1 L 1 and L2 L 2, let M1 M 1 and M2 M 2 be the TM TM s that recognize them. If you can solve a problem with a non-deterministic Turing Machine, then you can definitely find a deterministic Turing Machine to solve that problem. All languages are subset of $\Sigma^{*}$ and hence this set contains all languages including all recursively enumerable languages. T's states will be similar to D's. Jun 18, 2022 · The answer to "Can we sometimes recognise the intersection between a not-recognisable language and a recognisable one?" is "yes". Note that it need not be possible to make a machine that halts and rejects all strings not in the language - the machine may loop. TM is Turing-recognizable. However, this comment thread probably isn't the right place for that discussion. It is the set of all strings that are members of both languages. See Answer. Recognizers and Deciders. Continuing the previous example, the language composed of all {strings giving Combining, there must be some language which is non-Turing recognizable, as each TM can recognize only one language. Jul 16, 2019 · The interpretation of the notation, and the meanings of such terms as "the language decided by" a given Turing machine should be covered by your course materials in some detail. There exist undecidable languages. Thus, L L is a Turing Recognizable Language (since the TM M M recognizes it). @Hellnar: No, for example a language that contains all words consisting of n-times 0 followed by n-times 1 (e. Nov 11, 2021 · 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; 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 there exists a Turing Machine such that when encountering a string in that language, the machine terminates and accepts that string then we can say that type of language is a Turing Dec 21, 2014 · 1. Now, consider a Turing Machine M M and a language L L (over input alphabet Σ Σ) that is recognized by M M. A basic connection between Turing-recognizable and Turing-decidable languages: Theorem 4: L is Turing decidable if and only if L and Lcare both Turing-recognizable. , it never loops The recursive languages = the set of all languages that are decided by some Turing Machine = all languages described by a non-looping TM. They've been discussed in the past, and you can find discussions using search -- or you could post a new question and start a new discussion. Question: 3. Turing recognizable languages are closed under union and complementation. –A language is I'm studying Turing Machines and I've already showed how Turing-Decidable is closed for the operations of Union, Intersection, Concatenation, Complement and Kleene Star. There exist non-regular languages. Therefore, A is decidable. Aug 3, 2023 · Partially decidable or Semi-Decidable Language -– A decision problem P is said to be semi-decidable (i. Jun 14, 2017 · Those languages for which there is a Turing machine that will always halt and accept in a finite amount of time for any string in the language are called Turing recognizable languages. For example, let the recognisable one be the empty language. To prove non-Turing-recognizability: If A m Band Ais non-Turing-recognizable, then B is non-Turing-recognizable. So in my courses, it was perfectly acceptable to provide an algorithm in a programming language or pseudo code. Also, I wanna know that what are the properties of a language which is non-Turing Recognisable? Jun 26, 2012 · 0. Now, for C - TM for A should accept which would as its acceptor and TM for B should reject as it shouldnt belong to it. For any string , a recognizer of should switch between its states based on the counts of , , and in . The TM also accepts the computable functions, such as addition, multiplication, subtraction, division, power function, and many more. A decider of a language is a machine that decides that language. Mar 22, 2024 · Which of the following statements is/are FALSE? 1. You need to represent the number n in some way, but that's not possible using finite number of states. A Decider also halts if the string is not in the language. Prove that a language L is decidable if and only if there is some enumerator that enumerates the language in lexicographic order. 01, 0011, 000111, ) is not recognizable by finite state machines. We can clearly recognize these with a Turing Machine, since we can just make a machine that always says "YES" for the first case, or make a machine that always says "NO" for the second. Answer: For any two Turing-recognizable languages L 1 and L 2, let M 1 and M 2, respectively, be TMs that recognize them. Jun 15, 2015 · A language L J, K of arrows from J to K is recognizable in C if it is the language of some (J, K)-automaton. This is a standard fact that probably doesn't need to be proven here. Any language "generated" by a Turing machine that always halts the way you mentioned (i. A closure proof provides an answer to the question, "If I have a class of languages, and do [blah] to a language in it, is the new language still in the class?” So, to prove that the set of Turing recognisable languages is closed under Kleene star and concatenation, it is sufficient to show that there exits a Non-deterministic Turning Machine that can recognise the language. 6. A decider that recognizes language L is said to decide language L Language is Turing decidable, or just decidable, if some Turing machine decides it 2 Example non-halting machine Nov 8, 2021 · $\begingroup$ More generally: whenever faced with a question of the form "is there any language such that" or "are all languages such that", always try the empty language and the universal language (of all strings) first. Nothing is implied about the complement of . The difference between data and computation is tricky. Dec 5, 2015 · First you need to define what the intersection is. Then P is undecidable. Recall two de nitions from last class: De nition 1. , have a semi-algorithm) if the language L of all yes instances to P is RE. See the Encyclopedia of Mathematics for more on recognizable and undecidable languages (specifically For every non-deterministic Turing machine, there exists an equivalent deterministic Turing machine. Every language has a one-to-one correspondence to the natural numbers, simply because the number of possible words is countable. As a result, the Halting Problem can be solved for BlooP programs. In the following section we will study the languages of semi-automata and show under which conditions they are recognizable. If something is computable then there exists a DTM(Deterministic Turing Machine) for it. So both the properties you have identified are non Apr 10, 2017 · Therefore, A is Turing-recognizable. b. A language is recognizable if there is a Turing machine that enumerates all words in the language. To prove NP-completeness: If A P B and Ais NP-complete (and B 2NP), then B is NP-complete. Statement 2 : Turing recognizable languages are closed under union and complementation. An enumerator can be constructed as TM. If Nov 22, 2019 · $\begingroup$ The question asks for Two non Turing decidable language A and B such that A cannot be Turing reduces to B, B cannot be Turing reduced to A. That is, R =R. This basically comes from the definition where recursive languages reside in both RE and coRE. Both types of machine halt in the Accept state on strings that are in the language. Simulate M1 on w. $\endgroup$ BlooP (short for B ounded loop) is an interesting non-Turing-complete language. Requiring a TM to halt on all inputs restricts the kinds of languages that it can recognize. Second part is, if L is Turing recognizable then there must be a Turing Machine M that recognizes L. Consider (a^n)(b^n)(c^n); a simple Turing machine for this language can run back and forth over the tape, removing one of each symbol in a pass, until all symbols are removed or it runs out of one kind of symbol before another. Language is Turing recognizable if some Turing machine recognizes it •Also called “recursively enumerable” Machine that halts on all inputs is a decider. Then L is Turing-recognizable, by Theorem 2. 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). This set is not recognizable in $\mathbb N^2$. Oct 28, 2020 · Intersection of Turing-recognizable language and regular language. Many properties of regular languages can be shown for recognizable languages in a straightforward way; see [4 (b)Show that the class of Turing-recognizable languages is closed under union. Jun 16, 2021 · A Language is called Turing Recognizable if some Turing Machine recognizes it. A language is Turing-recognizable if there exists a Turing machine which halts in an accepting state i its input is in the language. Mar 18, 2023 · A language is recognizable if there’s a Turing machine that accepts all the words . = “on input <M,w>, where M is a TM and w is a string: Simulate M on input w. Note that we sort of waved our hands there when we "split up" stuff not in L1 or L2 into two infinite languages, one co-Turing-recognizable and the other co-Turing Jun 25, 2012 · Mar 24, 2010 at 16:21. The machine can reject a word from or run forever in a loop. Since A TM is undecidable and it can be reduced to Lp, Lp is also undecidable. Also, Lcis Turing-decidable, by Theorem 3. (hence recognizers are more powerful than deciders). Is this a correct solution? Deciable Languages are subsets of Recognizable Languages. Consider the set of strings that are not in L L (we call it L¯¯¯¯ L ¯ ). Turing machines can be encoded as strings, and other Turing machines can read those strings to peform \simulations". Prove that NE is Turing-recognizable (show an algorithm to recognize it, and argue the correctness of your algorithm). and complementation. Plenty of non-context-free languages are recursive. The question is asking for a false statement. z where x is a number, y is a Turing-machine and z is an initial tape configuration, and y halts on z in fewer than x! steps - perhaps that qualifies (though it would need to be shown!) A common Some Languages are Not Turing-recognizable Proof: – The set ∑* is countable: there are only a finite number of strings of each length, we may form a list of ∑* by writing down all strings of length 0, length 1, length 2, etc. TM . 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. The problem is to determine, given a program and an input to the program, whether Jun 9, 2017 · I am trying to understand closure properties between different language classes. They help establish the hierarchy of computational complexity classes, demonstrate undecidability and intractability results, and provide insights into the limits of computation. , I encourage you to take discussion of those issues to Meta. Nov 12, 2019 · Add a comment. Is your question maybe: Is every enumerable language that can be enumerated in non-decreasing order decidable? Then the answer is yes. R R (the family of recursive langauges) is countable, because there the set of Turing Machines is countable, and there is at least one TM for each R R language. It is clear that the language is recognizable since one could systematically enumerate sequences of dominos, checking each one to to see if it is a match. Suppose that L is Turing-decidable. Thus L is both recognizable and co-recognizable. A recognizer of a language is a machine that recognizes that language. The way to recognise the language S S is to use the Turing machine which does the following: Machine A: Turing machine to Nov 12, 2017 · A property is non-trivial unless it contains no languages, or contains all Turing-recognizable languages. May 1, 2022 · There is another similar problem, but instead of the language being recognizable, it's decidable. May 31, 2018 · The reason is that recognizable languages in general are not closed under Kleene star. First we observe that A. Aug 2, 2023 · Non-Turing recognizable languages hold great significance in computational complexity theory, particularly in the field of cybersecurity. Also, I wanna k Aug 6, 2015 · If the input string is equal to one of these generated strings then ACCEPT. A non-turing complete DSL could easily meet all of these requirements. Recursive means repeating the same set of rules for any number of times and enumerable means a list of elements. A property, P, is the language of all Mar 11, 2018 · That question asks two questions, one (in the title) is "is the class of Turing-recognizable languages closed under homomorphism", and the other is "is my proof correct". If by Turing-acceptable you mean that there exists a Turing machine M such that M always halts and accepts on words in the language, then yes, all recursive languages are Turing-acceptable, or Turing-recognisable. This is also why language non-emptiness is recognizable - you can perform a BFS on the infinite configuration graph. Recursive languages are also recursively enumerable. Then M recognizes L, and the Turing machine which halts with the opposite output of M recognizes L¯. 11 A language L is decidable iff L is recognizable and co-recognizable. , s ∈ S s ∈ S iff T(x) = s T ( x) = s for some input x x to the fixed Turing machine T T) will be Turing recognisable. The halting problem is a decision problem about properties of computer programs on a fixed Turing-complete model of computation, i. \co-R. Markup languages (used for describing data, not computation) like XML and JSON are not Turing complete. For example, let . Proof: Suppose L is decidable. Mar 27, 2020 · Nope! If you have a regular language, you can get a DFA for it, then convert that DFA into a Turing machine by slightly adjusting the transitions so that they mechanically move the tape head forward. My reasoning: B is finite, therefore the finite number of strings can be put through a DTM , M, such that L(M)= A. 3. A multitape Turing Machine has definitely more computational power than a single tape Turing Machine. E. May 6, 2020 · L=A^* is very decidable, and contains ervery non-decidable language. It is Turing-recognizable if there exists a Turing machine that either outputs YES or never terminates (there are other, equivalent definitions). . Jul 2, 2015 · Let A be a Turing Recognizable Language and B a finite Language. Other, less common, uses for mapping reductions: To prove decidability: If A m Band Bis decidable, then Ais decidable. Oct 4, 2022 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Jan 27, 2016 · Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Nov 27, 2012 · 1. W. Next I did some demonstrations to show how T-Recognizable languages are closed for Union, Intersection, Concatenation and Kleene Star. As a result, that language is also Turing-recognizable. This is my opinion. Suppose you are given a DFA D such that L = L(D). ♦. . We are given hint that we can choose one of the language to be the halting problem $\endgroup$ – 16. Apr 22, 2012 · Your discussion successfully confused me :( "Can A be not Turing-recognizable?" I feel A is always Turing-recognizable. We saw in 1 that A is co-Turing-recognizable and in 2 that A is Turing-recognizable. A Language is called Turing Decidable if some Turing Machine decides it. If M1 halts and accepts w, go to step 2. A language ‘L’ is partially decidable if ‘L’ is a RE but not REC language. A language is Turing recognizable iff some multitape Turing machine recognizes it Programming Turing machines is incredibly tedious and usually not enlightening: it's like programming in a particularly nasty and human-unfriendly programming languages. We use the fact that a language A is decidable if and only if A is Turing-recognizable and its Apr 14, 2015 · Regular languages - those that can be described as regular expressions - are not Turing complete. g. This can be done by Turing machine So, this statement is correct. A FA is a tuple hQ; ; ;q s;Q f iconsisting of a nite set of states Q, a nite set of input symbols , a transition function : Q !Q, a start state q Some languages not Turing-recognizable III One-to-one correspondence between B and L B is uncountable (like real numbers) Therefore, L is uncountable Each TM ⇒handles one language in L Set of TM is countable, but L is not Thus some languages cannot be handled by TM December 29, 20223/12 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. As both languages are turing decidable means that there exist such algorithm for each. Dec 1, 2015 · Case 3: Intersection of A and B is non-empty. However, this cannot guarantee that it is decidable. Or more specifically, in general we have rational sets that are not recognizable. Def : A Language is called Turing Recognizable if some Turing Machine recognizes it. Also, there is an additional step that is worth explaining: it's worth mentioning that any language that's recognizable by a NTM is also recognizable by a standard deterministic TM. If there is a path to an accepting state, you will find it eventually. – The set of all Turing Machines M is countable since each TM M has an encoding into a string <M> May 8, 2017 · A recognizable language CAN always be decidable. Part 2 (For some unrecognizable language) Any non-monotonic property of the LANGUAGE recognizable by a Turing machine (recursively enumerable language) is unrecognizable Nov 11, 2021 · The discussion includes a proof of undecidability by reduction from the language of encodings <M,w> where M is a Turing machine that accepts w. One particular simple example for Kleene star would be $(1,1)^* \subseteq \mathbb N^2$. if ua rm rj oo qv di mw mb bq