decidable but not recognizable The unary separability problem is decidable for linear sets, as shown in [1], but we will not need this fact in the sequel. Therefore, (by Theorem 4. Languages decided by a TM are called decidable. Run M on w. Using the characterizations of Corollaries 4. 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. is not recognizable. Let A and B be Turing-recognizable languages such that A S B = . 1. Turing Decidable. , if there exists a Turing machine which will enumerate all valid strings of the language. Decidable and Semi-decidable Languages (Score: _____ out of 20 points) Let Sigma be an alphabet. • Theorem 2: If L is Turing-decidable then L is Turing-recognizable. L is Turing recognizable. 1. Turing-decidable language but not context-free language. A decider that recognizes language L is said to decide language L •How about languages that are not recognizable? The language . Theorem 4. That is, for every input string x2 on show, for each condition, that leaving it out makes P decidable for some P. There are some sets which are NOT-RECOGNIZABLE ( even members cannot be identified) . Here we show that there exists a language that is not even recognizable. 6) Which of the following statement is true about the mapping of integers on the set of TMs described in this week's Proof: 𝐴TM is T-recognizable but also undecidable decidable. So, it is natural to expect that A TM is NOT decidable. A Turing Machine is a decider if it halts on The complement of HALT is not CE. Class 3: Decidable but not Context-free. Therefore ATM is decidable. Consider M decidable. This is recognizable but not decidable (the enumerator for Hcan be used to enumerate L, and a decider for Lcould be used to decide H). is a Turing machine that accepts at least 42 different strings . That is, all words in the language are accepted by the TM. g (b) L 2 = f(hM 1i;hM 2i) : TM M 1 accepts input hM 2ior TM M 2 accepts input hM 1i(or both). 18 % \item [3. 2 Show that the following language is not Turing-recognizable: L C = fhM;kijM is a TM which accepts some string of length k but M loops on some (other) string of which is not even Turing-recognizable. (9 points) In order to arrive at a contradiction, suppose P is decidable, i. Unfortunately, there is still one thing that we must do before we can demonstrate a Turing-unrecognizable language. The TM Mon input w: 1. On any given input w, it might instead loop. (d)There is a recognizable but not decidable language. TM is Turing-recognizable (HW 8, problem 4) and A TM is not Turing-recognizable (Corollary 4. (Sipser 1. Argue why they are not recog-nizable To get a correct argument, let H be a language that is recognizable but not decidable. Q3 : Show that if A is Turing-recognizable and A m A, then A is decidable. The reasons for this are historical, and we will not So every decidable language is recognizable, but the converse is not true. We will show that A TM, the complement of A TM, is not Turing-recognizable. We will use RE to name this set. Turing-recognizable (and not Turing-recognizable) co-Turing-recognizable (and not co-Turing-recognizable) decidable (and undecidable) enumerator the Church-Turing thesis computable function mapping reduction / mapping reduciblility the halting problem Make sure you understand how your knowledge in this course ts together. By deﬁnition, there exists a Turing Machine M such that M accepts every word w ∈ L, but it might either reject or loop on any word w ∈ L. Proof: we know that HALT is CE but not decidable if complement of HALT wereCE, then HALT is CE and co-CE hence decidable. If xis an integer, then let [x] 2 denote its binary representation. (ii) (5 points) L = f0p: p is a primeg. 5. Let A be a Turing-recognizable language consisting of descriptions of Turing machines, fhM1i;hM2i;:::g, where every Mi is a decider. Since X does not exist, but Y could be used to make X, then Y must not exist. 1This is an interesting property in the following situation: imagine two undecidable, disjoint languages A and B. Concluded that the complement of A TM is not recognizable. 22 Language A Y -* is Turing-decidable Ù that is not Turing-recognizable. A language is Turing-recognizable (or recursively enumerable) if it is recognized by a TM. PROOF We know that ATM is Turing-recognizable. Pf: By contradiction. 1. Thus, if a string belongs to a a language that is Turing recognizable, there exists some Turing Machine which will accept that string. 22) A is decidable since A is both Turing-recognizable and co-Turing-recognizable. Recognizable and co-recognizable • Same method for arbitrary L: – L is recognizable but not decidable => L is not recognizable – Class of recognizable languages is not closed under complement • Same construction for arbitrary L: – L is recognizable AND co-recognizable => L is decidable 13 Turing-recognizable languages. If This is not a context free language, but surely this is recognizable by a Turing Machine that checks if the length N of the input is not divisible by any number between 2 and N-1 . If language recognized by TM, M, is {w/ Mlw)-accept} = {w/ M accepts a} This week till give a chance to test you uploading-pmba@blyviagradesape7Ch4iAcceptTmi. Problem 5. In fact, Theorem 1 better explains the situation. L ∈ R iff L is decidable Computable is a synonym of decidable. For each i from 0 to L, Create a finite automaton Ai that recognize strings that represent i-tuples of numbers that make phi(i) true. –L s iTuring-decidable if there is some TM that decides L. b/ L1 is Turing decidable but not recognizable. acs-07: Decidability Informatik Theorie II (A) WS2009/10 22 Some languages are not Turing- recognizable (T 4. Give examples of three different languages that are recognizable but not decidable. 4 it is sufficient to prove that the FPP for CU NA and CU N, and the condition (2) are decidable. Hence, by Theorem 4. L is Turing recognizable but not decidable. set A such that $\emptyset <_T A sets such as E are called DECIDABLE ( both members and non-members are identified) sets such as G are RECOGNIZABLE ( members can be identified , but non-members cannot be) Note that if a set and its complement are both RECOGNIZABLE , then the set is DECIDABLE. By Theorem 4. If M ever enters its accept state, accept; if M ever enters its reject state, reject. M may reject some strings not in L but it is also possible that 3. 2. Context-free language but not regular language. Exercise 2 (compulsory) 1. A language Lis called semi-decidable (also called recursively enumerable, r. IIt is possible for a TM to never reach a halting con guration. Note that Sipser, in his book, uses “recognizable” instead of “semi-decidable”. 10. Since A = A, A is Turing-recognizable and co-Turing-recognizable. (e. Note that step 2 is guaranteed to terminate, because Mdecides (not just recognizes) L. PSPACE=NPSPACE. does not belong to A TM (namely, ( DH, DH)) but H runs forever on the input ( DH, DH) So H cannot decide A TM Given any program that recognizes the Acceptance Problem, we can efficiently construct an input where the program hangs! Theorem: ATM is recognizable but NOT decidable Corollary: ¬¬¬ATM is not recognizable! A language is decidable if some TM decides it (chapter 3). 1. set is Turing-below the halting problem). , X = a TM that can decide ATM ) 2. 15) (T4. decidable is whether there are languages that are not Turing-recognizable. 1 1. An acceptor for is the following nondeterministic TM ; encode. A is decidable A is also Turing-recognizable Decidable and Undecidable Languages 37/38-6 RE is NOT Closed Under Complement accept reject input string w Suppose M accepts L and M accepts L . Since we know that is not decidable by Theorem 4. TM Variants 2/24/2021 CS332 ‐Theory of Computation 14. Theorem A TM is undecidable. It remains to show the forwards implication. Theorem: A language is decidable if and only if it is both Turing If B is decidable then A is decidable too. e. - the halting problem HALTTM is not decidable, but it is Turing-recognizable - undecidable languages: ATM, ETM, EQCFG, EQTM, etc. Decidable languages are often called also recursive languages. The reducing (D) decidable, (R) recognizable but not decidable, and indicate which undecidable examples follow from Rice’s Theorem. Run on for steps. Complement of T-recognizable =co-T-recognizable. Which of the following languages will not be accepted by this DFA? start q0 q 1q2 0 1 1 0 0 A {00}•{1}⇤ B {00}•{1⇤00}⇤ C {0}•{1}⇤ •{0}•{0}⇤ •{1} D {100}•{100}⇤ E {0}•{10 state (not equal to q) is reachable from q, or if there is a loop from q to itself. However, our proof was not constructive, so we have yet to see a concrete example of an unrecognizable language. is recognizable, however the complement of A TM is not recognizable. If and Turing recognizable then is decidable Let and be TMs for and Define : On input AAA AA A MM A A M w Decidable and recognizable languages are closed under but ∩ ∪ ° * not complement. Let and be two recognizable but not decidable languages. We will now show that if a language and its complement are Turing-recognizable, then the language is decidable. Note that, L might be recognized by other TM M’ that does not always halt. L =ATM ' A coTuringrecognizable LTuringrecognizableA Turingrecognizable TM TM ⇒ − − − ⇔ − _ _ '_ To prove a language recognizable, find a Turing Machine that accepts all strings of the language and does not accept any strings out of the language. Then T is A language is Turing-decidable (or decidable) if some Turing Machine decides it. We could clearly construct a decider for Bby running M A TM on hM B;wi. Given an input hM,wi, we want to decide if Mhalts on wor not. Regular ⊆ context free ⊆ decidable ⊆ Turing-recognizable ⊆ All languages Let's identify an example language for each set that is not in the next smaller set Summary: Decidable Languages and Their Machines (c) decidable but not in NP (and prove it) (d) Turing recognizable but not decidable (e) not Turing recognizable (f) undecidable and unary (unary means L ⊆ {1} *) Problem 14. g. e. Turing recognizable languages are closed under union and intersection. and the rst problem is therefore not decidable. It remains to show the forwards implication. UC Davis 20,780 not in the language A language is called Turing-recognizable or recursively enumerable, (or r. not in the language A language is called Turing-recognizable or recursively enumerable, (or r. • Not all languages are Turing-recognizable – There are some languages cannot be recognized by a TM. e. For example, if a language is both recognizable and decidable, but not context-free, circle decidable. Context-free language but not regular language. 3 (Chubb, Miller, and Solomon [CMS]). 2. The key observation is that no matter So far, all of the languages presented above are recognizable, even though some of them are not decidable. Any language recognized by a total deterministic TA is by de nition decidable. Score: 0 Accepted Answers: L is not Tunng recognizable. 2, Theorem 5. Decidable • A language L is Turing recognizable if some Turing machine recognizes it. Ask the oracle for A TM if M accepts w. (b)Prove or disprove: The language fhMi: M is a PDA that is duplicatedgis Turing recognizable. We say a language is co-Turing-recognizable if it is the decidable languages is decidable. Fact 1: A TM is Turing-recognizable (why?). On any given input w, it might instead loop. Conclude L1 is decidable in any general programming language. Introduction to language ATM, the halting problem; Universal Turing machines show that ATM is recognizable. For string w, if w ∈L,then M halts in ﬁnal state. Or A language is recursive if there is a membership algorithm for it. ¥ F or an y A and A . (halts always!). Conclude Lis not decidable in any programming language. Examples of decidable languages that are not context-sensitive are more difficult to describe. On words not belonging Recognizable vs. Run M 1 and M 2 on input win parallel. interventionism in the developing world Political economy of human rights Propaganda role of corporate media Noam Chomsky A TM A TM 0 n1n0 0 Semi- Decidable Problems – Semi-Decidable problems are those for which a Turing machine halts on the input accepted by it but it can either halt or loop forever on the input which is rejected by the Turing Machine. Examples – We will now consider few important Decidable problems: This known non-Turing-recognizable language can be any language for which non-Turing- recognizability has been proved in the textbook, in lectures, in class handouts, or in homework problems (but you should cite the appropriate reference). ", also known as Turing-recognizable (but not necessarily decidable)) sets that are undecidable and strictly Turing-below the Halting problem (note that any c. 1 Solution Let MA,MB, and MC be TMs that recognize languages A,B, and C respectively. a/ L1 is Turing recognizable but not decidable. Turing recognizable languages are closed under union and intersection. Let S be the set of inputs for which M loops forever. is a Turing Machine that has at least 42 states . UNSOLVED! Close. No, the answer is incorrect. 2. -ATM is Turing-recognizable, but it is not co-Turing-recognizable -ETM is not Turing-recognizable, but it is co-Turing-recognizable -EQCFG is not Turing-recognizable, but it is co-Turing-recognizable -Turing recognizable languages are not closed under complement. 𝑻𝑸𝑩𝑭 𝑬𝑸𝑹𝑬𝑿↑ 𝟑𝑺𝑨𝑻. Proved that HALT TM ("the halting problem") is not decidable (via a reduction from A TM). Justify your answer using methods from class. Statement 3 is true as Turing decidable languages (REC languages) are closed under intersection and complementation. (b) Show that Bis not recognizable. Undecidability (i)A proof technique by diagonalization (ii)Via reductions (iii)Rice’s theorem 3. • Turing-decidable languages: – For a , exists a M such that M decidesL – “Decide” means halting: either accept or reject • Turing-decidable Turing-recognizable – Halting Problem is Turing-recognizable, but not decidable. Turing Recognizable and Decidable Languages De nition A language L is Turing{recognizable (recursively enumerable) if there is a Turing machine M such that L(M) = L. To prove the language not decidable, use a reduction from a recognizable problem that is not decidable. regular context-free decidable recognizable not recognizable (e) {hMi : M is a Turing machine and there is some input string that causes M to run forever A language is Turing recognizable if it is recognized by some Turing Machine. There are some sets which are NOT-RECOGNIZABLE ( even members cannot be identified) . or semi-decidable) if some TM recognizes it A language is called decidable or recursive if some TM decides it [(ab)∗] is level-regular but not recognizable. Unknown D. c/ L1 is Turing recognizable and decidable. 3. Vote. " The "lemma" used in the proof is the (apparently self-evident!) statement that " the complement of a decidable language is also decidable. Q2 : Show that no computable function reduces A TM to E TM. Option D says P1≤P3. regular context-free decidable recognizable not recognizable (d) {hMi : M is a Turing machine that accepts ε}. = { , | is a CFG that generates input string } 4. The (c) decidable but not in NP (and prove it) (d) Turing recognizable but not decidable (e) not Turing recognizable (f) undecidable and unary (unary means L ⊆ {1} *) Problem 14. (The reader should also note that in some of the literature, the term “recursive” is used instead of “decidable”, and the term “recursively enumerable” is used instead of “semi-decidable”. is a Turing Machine that runs for at least steps when started with a blank input tape, where -- between Recognizable and decidable If L is decidable, then L is recognized by a TM M that halts on all inputs. Run M on w. However, strings that do not belong to the language may run forever. Decidability Theorem If Lis Turing-recognizable and Lis Turing-recognizable, then Lis Turing-decidable. Since 0 1 is decidable, A is decid-able by Theorem 5. 4. H AL TT M is not T uring-decidable. In other words, no infinite looping is allowed. To decide a language, the Turing Machine must halt on all inputs. recognized by a deterministic queue automaton i the language is Turing-recognizable. Recall that a language Lis Turing recognizable if there is a Turing machine that accepts exactly the words in L, but can either reject or loop inde nitely on an input that’s not in L. (c) If two languages L1 and L2 over the same alphabet are decidable, then L14L2 is decidable. That is, a decider T is guaranteed to either accept, or reject, and never fall into an infinite loop. If any such paths or loops are found, reject. A TM is however Turing-recognizable. But it is not given in question about decidability of P3. 3. All computations of a decider TM must halt. If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. True B. Let FIN= f<M>: L(M) is nite g We will show this is not recognizable by reducing it to HP. 18]Show that a language is decidable iff some enumerator enumerates the language in the standard string order. of a language that is Turing recognizable A language is decidable if and only if it is both Turing-recogni co-Turi zable a ng recogni nd co-Turing recog za niz ble Theorem 12 able 1. As adjectives the difference between computable and decidable is that computable is capable of being computed while decidable is capable of being decided. Undecidable problems tend to be pretty hard, so you don’t usually think of an example right off the top of your head. If A is decidable, then A and A are both recognizable. A CFG Decidable Problems Concerning CFLs A CFG ={< G,w >:G is a CFG that generates string w}. W e construct TM S that decides A T M as follo ws: On input < M ,w > where M is a TM and w is a string, S Þrst run TM R on < M ,w > ,if R rejects, rejects. So option (D) is not correct. Two situations Deciders always say yes or no But some languages have (only) a TM that says yes, sooner or later, and may not say no all the time. 2 and 4. 16]Show that the collection of Turing-recognizable languages is closed under the operation of \begin {enumerate} \item [a. For full credit, your answer should include a brief argument supporting your answer (but a detailed proof is not needed). Def: Semi‐decidable language 48 A semi‐decidable (recursively enumerable) language is a language for which there exists a Turing Machine so that • if , then halts and accepts • if , then either does not halt, or does not accept Examples of semi‐decidable, but not decidable problems: • Halting problem: Does the program terminate? (b) There are in nitely many Turing-recognizable languages. Turing decidable languages are closed under intersection and complementation. theory. If w ∈/L,then M halts in non-ﬁnal state. Show that there exists a language over a unary alphabet that is Turing-recognizable but not Turing-decidable. Classify the following problems as in P, in NP, NP-complete, decidable, undecidable, Turing recognizable, not Turing recognizable. B It is not decidable but is recognizable C It is context-free D It is not context-free but is decidable E It can be recognized with a PDA 10. Proof that ATM is not decidable by a self-reference argument. e. We study the following decision problem: is the language recognized by a quantum finite automaton empty or non-empty? We prove that this problem is decidable or undecidable depending on whether \item [3. A Turing Machine is a decider if it halts on QED Corollary: L'' is not recognizable, not decidable, ~L'' is not decidable Theorem: ~A_TM =m ~L'' Proof (hint): f()= where M' on input x 1) Becomes M on input w (this may never return, but that's OK) 2) if M accepts then The goal today is to give a language that is not decidable, and it has to do with universal Turing machines: A TM = fhM;wijM is a TM and M accepts wg: Recall from the de nition of universal Turing machines that A TM is Turing-recognizable. The algorithm which, given inputs P and x, runs P(x) until it halts and then accepts, recognizes the halting problem. e. Hint: There is at least one language that is neither recognizable nor co-recognizable. 3. • Supposedly, somebody in the 60’s at MIT wrote a very complicated thesis about some class of languages showing all its great properties. L1 ={p|pis a polynomial with integer coeﬃcients∧phas one variable∧ phas an integer root} L1 is Turing-recognizable and decidable. In other words, if A TM was decidable, then every Turing-recognizable language would also be decidable. 4. 4. is recognizable. Now suppose A and A are Turing-recognizable by M 1 and M 2 respectively. Thus, if a string belongs to a a language that is Turing recognizable, there exists some Turing Machine which will accept that string. Show that L is decidable. Moreover, it will follow from our stronger decidability result about the We assume that P1 is non-RE but P2 is RE. L is not Turing recognizable. EXPTIME. If any of these derivations gen erate ; if not, "w n 0, w, accept reject. Thm The EQUIVALENCE problem is not recognizable ; Pf We show that is not recognizable by showing that (which means that ). Thoughtful comments (below) by Travis Service and Alex ten Brink suggest (in effect) that in Q1 the phrase "undecidable" is not synonymous with "not verifiably decidable" and that the answers to Q2–5 may depend upon this distinction. e. Then #is not Turing-recognizable. Then L is decidable (by M’ below)! M M’ accept r ejct M accept reject So if L is semi-decidable+ (L in RE - Dec), then L can’t be in RE! Important detail: M and M must be run in parallel, not Lis Turing-recognizable, but not decidable. Key idea: need to run both machines without committing to one. According to theorem 1 discussed, if P3 is decidable then P1 is also decidable. Decidable and Recognizable Languages IButnot all languages are decidable! We will show: I A tm = fhM;wijM is a TM and M accepts wgis undecidable IHowever A tm isTuring-recognizable! Proposition There are languages which are recognizable, but not decidable Turing- recognizable Decidable Context-free Regular 0n 0n1n 0n1n2n What We Know We already know of languages in several of the classes in our diagram We have shown that ATM is Turing-recognizable but not decidable. Assume both Aand Aare Turing recognizable by M 1 and M 2 respectively. Let L be some language that is not-decidable but recognizable (yes, such languages are there and we will study them soon) and let M be some DTM that accepts L. Here is our central deﬁnition: Deﬁnition 3. 2011 Theorem 4. We have already proven that there are languages that are not Turing-recognizable. Turing Machines that halt in all inputs are called deciders. 22 in Sipser. Bottom line: For every \strictly semi-decidable language", its complement cannot be semi-decidable. If some language C decidable separates them, then C can help give | decidable! | hints for strings of A and B, A language is Turing-decidable (simplified to just say decidable) if a TM exists that decides it. If a language Lis of Type NTR, give a proof that it is not Turing-recognizable. Note that “Turing-Decidable” is a stronger condition than “Turing-Recognizable”, because, if a language is Turing-Decidable then its corresponding Turing Machine never runs forever. Proof: we know that HALT is CE but not decidable if complement of HALT wereCE, then HALT is CE and co-CE hence decidable. The Halting Problem is Recognizable but Not Decidable - Duration: 1:19:07. Since ATM is undecidable, it must be the case that our assumption that T is decidable is false, so T is undecidable. Given a Turing machine M, is L(M) recursive (Turing-decidable)? You can assume without proof that the halting problem for Turing machines is undecidable. A decider that recognizes language L is said to decide language L Language is Turing decidable, or just decidable, if some Turing machine Statement 2 is false as Turing recognizable languages (RE languages) are not closed under complementation. 3. ‣ 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 an input not in L. w} = {〈 M, w 〉: M. krchowdhary TOC 2/11 provided. C. Answer: We will show that A TM reduces to A" TM. M),-. But first let’s look at a definition and a theorem. Decidabilit results for P A are mainly theoretical, and they help delineate some imp ortan t fron tiers in the eld of v eri cation is not decidable and is recognizable is not decidable and is not recognizable. (b)Any subset of a recognizable set is recognizable. WaitforEtoprintastrings. 11, this contradiction establishes the result. So the thesis was bogus. not sync hronize. \textbf {Solution:} \alreadyanswered \end {enumerate} % 3. A. Construct the new NTM N: On input w: 1. 20 Non-Recognizable via Reductions. 1 Decidable Languages Let A,B,C be recognizable languages over an alphabet Σ, such that A ∪ B ∪ C = Σ∗ and A ∩ B = ∅, A ∩ C = ∅, and B ∩ C = ∅. This is a contradiction with Theorem F and the counter-assumption does not hold. 7 If A is Turing-recognizable and A ≤m Ā, then (by Theorem 5. e. Assume Y exists. 11a (not in the book as theorem, but stated on page 174), ATM is Turing A TM = { (M, w) | M is a TM that accepts string w } A ConcreteUndecidable Problem: The Acceptance Problem for TMs Theorem [Turing ‘30s]: A TM is recognizable but NOT decidable 1 Decidable Languages Let A,B,C be recognizable languages over an alphabet Σ, such that A ∪ B ∪ C = Σ∗ and A ∩ B = ∅, A ∩ C = ∅, and B ∩ C = ∅. Theorem: A language is decidable if and only if it is both Turing-recognizable and co-Turing-recognizable. Let T be a c. 3. IThese are the three options for any input: accept, reject, loop. Show that A is not only recognizable, but also decidable. Turing Decidable ; Turing Recognizable (We haven't yet proved the relation between CF and TD langs) What are some examples in each class (and not in smaller class)? First three are easy: Regular, CF, decidable ; Are there languages that are recognizable but NOT decidable? Are the languages that are NOT recognizable? 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. Which of the following languages will not be accepted by this DFA? start q0 q 1q2 0 1 1 0 0 A {00}•{1}⇤ B {00}•{1⇤00}⇤ C {0}•{1}⇤ •{0}•{0}⇤ •{1} D {100}•{100}⇤ E {0}•{10 Recognizable = Decidable. Let L be a recursive language and M the Turing Machine that accepts (i. Classify each of the following languages and give a brief (1-2 lines) reason. First construct A(L) directly, iterate i from L to 0 The construct A(i-1) using Ai Check A0 accepts empty string if it does, phi is true and algorithm accepts Statement 3 is true as Turing decidable languages (REC languages) are closed under intersection and complementation. If 𝑇𝑀 also were Turing-recognizable, ATM would be decidable. (20 points) Show that the set of decidable languages is closed under intersection. regular. The TM M on input w: 1. ” CFG is co-Turing-recognizable. NFA are decidable, A TM is not. E) and now also pis Turing-recognizable. 12 Assume it is decidable to determine if a single-tape Turing machine ever writes a regular con text-free decidable recognizable not recognizable (f ) {h M i : M is a T uring machine and for all n ≥ 0, M accepts some string of length n } regular con text-free decidable recognizable not recognizable Turing-Recognizable and Decidable Languages Recall: Recursively Enumerable = Recognizable I Enumerate all strings in language I Compare with given input; if equal, accept Recall: Decidable ( Recognizable Both: willacceptall strings in language Decide: willrejectall strings not in language Recognize: may loop forever on strings not in language answer. • Non-recognizable Turing-recognizable languages and A ( B, then machine(A) is not decidably separable from machine(B). • Recall that the complement of a language L is the language L consisting of all strings that are not in the language L. " It is just stated as fact. Recursion Theorem A TM can obtain and execute its own description. Argue why the languages are recognizable but not decidable. Turing recognizable languages are closed under union and complementation. Such problems are termed as Turing Recognisable problems. 11 tells us that ATM is not decidable, so 𝑇𝑀 must not be Turing-recognizable. 2. So the halting problem is recognizable but not decidable. Every decidable language is Turing-Acceptable. Class 2: Context-free but not Regular. Show that A is not only recognizable, but also decidable. Fact 2: A TM is undecidable. [20 + 10 = 30 points] Solution: (a) We show that if Bis decidable, then we can construct a routine for deciding HALT TM which will be a contradiction. Theorem 12 A language is decidable if and only if it is Turing-recognizable and co-Turing-recognizable. Let L be some language that is not-decidable but recognizable (yes, such languages are there and we will study them soon) and let M be some DTM that accepts L. T decides a language L if T recognizes L, and halts in all inputs. A language is Turing-recognizable if there is a TM that accepts the language and may or may not halt on some of the strings NOT in the language. • Obviously. This table (below Introduction to language ATM, the halting problem; Universal Turing machines show that ATM is recognizable. rejects or loops on input . If the reachability problem for H is decidable in all presentations of H then we say that the problem is intrinsically decidable. Now construct an algorithm to reduce P1 to P2, but by this algorithm, P2 will be recognized. Theorem Theory of Languages and Automata Prof. If M 1 accepts, then accept; and if M 2 B. Assume both A and A are Turing recognizable by M 1 and M 2 respectively. To show this involves a discussion of the halting problem for Turing machines, and an appeal to the notion of a "universal" Turing machine. not in the language A language is called Turing-recognizable or recursively enumerable, (or r. Class 4: Recognizable but not Decidable. Which of the following statements are false? is a recognizable language is a recognizable language is a decidable 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. “Decidable” Language is Turing recognizable if some Turing machine recognizes it •Also called “recursively enumerable” Machine that halts on all inputs is a decider. ]union. • Later it was shown this class of languages was empty. Check-in 8. We also prove that, in general, its FO[Reach] theory is not decidable. d/ L1 is neither Turing recognizable nor decidable. Let’s look at a di erent one: De nition 4. M should stop and accept if either one of M. yes. Suppose for contradiction that A" TM is decidable, and let R be a TM that 1. Prove that some decidable language D is not decided by any decider Mi whose description appears in A. Although it might take a staggeringly long time, M will eventually accept or reject w. 1 Solution Let MA,MB, and MC be TMs that recognize languages A,B, and C respectively. ) 2. We know X does not exist. Consider the language A" TM = fhMijM is a TM that accepts "g: Show that A" TM is undecidable. The set R is the set of all decidable languages. Basically, we want to run M. L1 ={p|pis a polynomial with integer coeﬃcients∧phas one variable∧ phas an integer root} L1 is Turing-recognizable and decidable. Any language recognized by a total deterministic TA is by de nition decidable. If L is recognizable, then there might be such TM M that recognizes L but run forever, rather than rejecting, some inputs not in L. Regular language. is a Turing Machine that has at least 42 states . (5 points) Prove that there exists a subset of {1}∗ which is not Turing-recognizable. Please give complete explanation as well. Show how to use Y to make X. If M 1 accepts, then accept; and if M recognizable. Every decidable language is Turing-recognizable, but not vice-versa - stay tuned! If A is decidable, then both A and A are Turing-recognizable: Any decidable language is Turing-recognizable, and the complement of a decidable language also is decidable. We Proved that A TM is not decidable (via diagonalization). UTMs define what a computer is in the way that TM This makes sense even if B is not decidable! We say A is recognizable inB if there is an oracle TM M with oracle B that recognizes A We say A is decidable inB if there is an oracle TM M with oracle B that decides A HALT TM is decidable in ATM On input (M,w), decide if M halts on w as follows: 1. Turing-recognizable language but not Turing-decidable language. Suppose is Turing-recognizable. Proof: ( <== ) If both L and complement of L are Turing recognizable, we let Machine M1 be the Turing recognizer for L and M2 be the Counter-assumption: nis Turing-recognizable • Then by Theorem B, n ýERR = pis Turing-recognizable. Note that it is ﬁrst required to remove all the states (actually, just accepting states) not reachable from q 0 as these states cannot lead to any string being in the language. ¥ If A and A are both T uring-recognizable, then A is T uring-decidable. There are many languages in CFG are not decidable! ) string describing a pair of (1) a (D) decidable, (R) recognizable but not decidable, (C) co-recognizable but not decidable, or (N) neither recognizable nor co-recognizable, and indicate which undecidable examples follow from Rice’s Theorem. For the graphs in Examples 1 and 2, reachability is intrinsically decidable. So this is also an example of a problem that is Turing-recognizable, but not Turing-decidable. Decidable • A language L is Turing recognizable if some Turing machine recognizes it. , nis not Turing-recognizable. If A m B and A is not Turing-recognizable then B is not Turing-recognizable. decidable Turing recognizable • D is a DFA that accepts w • N is an NFA that accepts w • D is a DFA that accepts a non-empty language • A, B are DFAs and L(A) = L(B) • C is a CFL that accepts w • C is a CFL that accepts a non-empty language • But, A, B are CFLs and L(A) = L(B) not decidable • When the model of comp increases in Decidable ⇒ Recognizable (converse not true). g. Rao, CSE 322 4 The Chomsky Hierarchy – Then & Now… CFLs Decidable T-recognizable Not T-recognizable Then (1950s) Now U. Let L = fhAi: A is a DFA and L(A) = g. e. languages Proposition (Decidable languages and total DTAs) A language is decidable if and only if its recognized by a total DTA. Contradiction. For any language, if it is decidable, then it is also recognizable. e. (a) L 1 = f(hMi;w) : TM M accepts all strings shorter than w. Applications: showing (un)decidability of other problems (i)A string matching problem: Post’s Correspondance Problem I'm referring to the proof of theorem 4. Show that there of a language that is Turing recognizable A language is decidable if and only if it is both Turing-recogni co-Turi zable a ng recogni nd co-Turing recog za niz ble Theorem 12 able 1. Question 2 : Let L be a language and L’ be its complement. 23: (the complement of ) is not Turing-recognizable. Fix any decidable language L and assume that L is recognized by Specifically, Emil Post posed the question: are there computably enumerable (aka "c. Conclude Lis not decidable in any programming language. Turing-recognizable language. Note that, L might be recognized by other TM M’ that does not always halt. Dec = Recursive (Turing-Decidable) Languages CFL = Context-Free Languages anbn wwR anbncn ww semi-decidable+ decidable Machine = all languages described by a non-looping TM. CFL. Answers: Let M 1 be a TM that accepts A TM (see Sipser’s book), which is a Turing-recognizable (recursively enumerable) language. 5. (i) (5 points) f[2p] 2: p is a prime numberg. \\ sets such as E are called DECIDABLE ( both members and non-members are identified) sets such as G are RECOGNIZABLE ( members can be identified , but non-members cannot be) Note that if a set and its complement are both RECOGNIZABLE , then the set is DECIDABLE. Class 3: Decidable but not context-free. Proof. That is, show that if L1 and L2 are decidable languages, then L1 intersection L2 is a decidable language. Which of the following statements are false? is a recognizable language is a recognizable language is a decidable Lis Turing-recognizable, but not decidable. • is Turing ‐recognizable • is not decidable (1949‐70) 2/24/2021 CS332 ‐Theory of Computation 13. Statement 4 is true as Turing recognizable languages (RE languages) are closed under union and intersection. Theorem: A language is decidable if and only if it is both Turing regular context-free decidable recognizable not recognizable (c) {0n: n ∈ IN}. (Solved) : Problem Concerns Languages Decidable Recognizable Fivetm M Tm L M 5 Decidable Hint Basmat Q30165678 . A language Lis called decidable (or recursive) if 9a Turing machine Msuch that L(M) = L, and additionally, Mhalts on all inputs x2 . Let S be the set of inputs for which M loops forever. 1 and M 2 does. 3. According to Thm 4. Bottom line: For every \strictly semi-decidable language", its complement cannot be semi-decidable. 37) Consider the language Fixing the construction for Union. The set A TM is mapping reduced to J by the function g(x) = 0x. Class 4: Recognizable but not Decidable. A TM Definition:A language is called semi-decidable(or recognizable) if there exists an algorithm that accepts a given string if and only if the string belongs to that language. 25 1 recognizable, But then, according to Theorem 23, the acceptance problem would be decidable. 2. Proof : ⇐ Assume A ≤m ATM. Proof. Non-deterministically divide winto two “Turing recognizable” vs. Simulate M1 on w. Option (B) says P3≤P2’. Find out whether the following problem is decidable is not Turing-recognizable. Paul Goldberg Intro to Foundations of CS; slides 3, 2017-18 24/42 Turing Recognizable and Decidable Languages De nition A language L is Turing{recognizable (recursively enumerable) if there is a Turing machine M such that L(M) = L. Theor em : A and A : ¥ If A is T uring-decidable, so is A . Show that if S is finite, then there exists a deterministic TM N that uses S to decide L. , do not just cite a theorem that establishes this without a proof. Now, let’s assume there is a TM Fthat U (Turing-recognizable languages or recursively enumerable languages, Type-0 grammars), described by unrestricted grammars, recognized by Turing machines (but not guaranteed to halt thus not necessarily decidable). If the complement of a recognizable language is also recognizable, the language is, in fact, decidable. L15: Proof by diagonalization that ATM (Halting problem) is not decidable UC Davis Academics. (0. Let L be a recognizable language of A * X B*. Chubb, Miller, and Solomon proved: Theorem 1. If and Turing recognizable then is decidable Let and be TMs for and Define : On input AAA AA A MM A A M w L is decidable. Also, prove that Lis not decidable. Decidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. What algorithm could we write that would recognize this language? Diagonalization. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). 2. If s = w, accept 3. Accep ted An sw ers: is decidable and is recognizable Let and be two decidable languages. This problem is not decidable, but it is Turing recognizable. D. From the deﬁnition of B that recognized (but did not decide) B. (a) L 1 = f(hMi;w) : TM M accepts all strings shorter than w. In particular, recognizable languages would coincide with decidable languages (which is verycoincide with decidable languages (which is very unrealistic!). Call a language L ⊆ Σ* decidable if there is a TM M such that for all w ∈ Σ*, M accepts w iff w ∈ L, and which is not even Turing-recognizable. A. Proof for L being decidable. From what we’ve learned, which closure properties can we prove for the class of T-recognizable languages? Choose all that apply. g (b) L 2 = f(hM 1i;hM 2i) : TM M 1 accepts input hM 2ior TM M 2 accepts input hM 1i(or both). Here is a decider for . Problem 5. Let L 1, L 2 be two recognizable languages and M 1, M 2 be two TMs that recognize L 1, L 2 respectively. HamPath If Ais decidable, then both Aand Aare Turing-recognizable: Any decidable language is Turing-recognizable, and the complement of a decidable language also is decidable. If is decidable then and Turing recognizable Trivial 2. , the Equivalence problem for NFA’s. Theorem 4. I have chosen the mapping reduction approach to get the result directly. 03/03/2010 20 Decidable and r. • We say that a language is co-Turing-recognizable if it is the complement of a Turing-recognizable language. Otherwise, it is undecidable. Given M and w, we design a TM M 2 that behaves like M 3. Turing-recognizability If A m B and B is Turing-recognizable, then A is Turing-recognizable. P A giv es rise to in nite-state systems and is quite expressiv e. 3. We have already seen that the universal Turing machine is a recognizer for \(\atm\). (iii) f[2p In mathematics, logic and computer science, a formal language is called recursively enumerable (also recognizable, partially decidable, semidecidable, Turing-acceptable or Turing-recognizable) if it is a recursively enumerable subset in the set of all possible words over the alphabet of the language, i. If any of these derivations generate w, accept; if not, reject. 22, it follows that is decidable. esystem (. 2. If you can figure out a systematic way (an algorithm) to answer the question correctly, then the problem is called decidable. e. Turing Machines Recognizable/Decidable Proofs. E. We will use Decto name this set. Department of Software Systems 157 OHJ-2306 Introduction to Theoretical Computer Science, Fall 2011 13. Q4 : Show that the Post Correspondence Problem is decidable over the unary alphabet = f1g. Suppose a language L is enumerated in lexicographic order by an enumer-ator E. Assume TM R decides H AL TT M. 15) Some languages are not Turing recognizable There is a countable number of Turing Machines (Each Turing Machine can be encoded in a string; the set of all strings over a finite alphab Decidable and r. A CFG String description of Therefore, is decidable But there is a Turing-Acceptable language which is undecidable Contradiction!!!! Since is chosen arbitrarily, every Turing-Acceptable language is decidable END OF PROOF We have shown: Decidable Undecidable We can actually show: Decidable Turing-Acceptable Turing machine that accepts : 1. L is Turing recognizable. A language that is recognizable but not decidable is the Halting problem. Turing-decidable language but not context-free language. X (e. However, strings that do not belong to the language may run forever. Reading: 4. Recognizable vs. decidable. , the Equivalence problem for NFA’s. (ii) f[2n] 2: n‚0g. UTMs define what a computer is in the way that TMs define what algorithms are. If L is nite, then of course it’s decidable, so we suppose that L is in nite. • The classes of Turing-recognizable and Turing-decidable languages are different. Languages recognized by a TM are called recognizable. False C. Give brief explanations. \(\mathcal{P}(\mathcal{P}(\Sigma^*))\), may have unrestricted grammars or no grammar at all, not necessarily Turing-recognizable. 22: Prove that A is Turing-recognizable iff A ≤m ATM. In each part below, if you need to prove that the given language Lis decidable, undecidable, or not Turing-recognizable, you must give an explicit proof of this; i. Turing Decidable. How do we show that a language \(L\) is recognizable? Similar to how we show that a language is decidable, we need only write a recognizer for \(L\). we ha ve one of the follo wing possibilities: (1) Both are T uring-decidable; (2) Neither is (a) Show that Bis not decidable. D. Fix any decidable language L and assume that L is recognized by A language is co-Turing-recognizable if it is the complement of a Turing-recognizable language The complement of a language is the language consisting of all strings that are not in the language. , Y = a TM that can decide B) Y 3. A decision problem P is decidable if the language L of all yes instances to P is decidable. Conclude L1 is decidable in any general programming language. In case the string does not belong to the language, the algorithm either rejects No. ibb. To prove that a given language is Turing-recognizable: If A m B and B is decidable, then A is decidable. Using B we can construct a Turning machine A that accepts the language {(M,w)| M is the description of a Turing machine that accepts the string w}. Which one of the following is NOT a viable possibility? Enumerator Homework/classwork Provide formal definition of enumerator Decidable vs Turing Recognizable A language is decidable if there is a TM that accepts the language and always halts. A TM M which decides L works as follows: M="On input w 1. If M1 halts and accepts w, go to step 2. Y 4. P. 1 and M 2 in parallel, but our computational model does not explicitly support parallelism, so simulate it by alternating steps of the two on two tapes: (iii)Decidable and Turing recognizable languages (iv)Church-Turing Hypothesis 2. 3. e. . is recognizable but not decidable. Paul Goldberg Intro to Foundations of CS; slides 3, 2017-18 24/42 3. 2. This shows that A TM is mapping reducible to J and hence that J is not Turing-recognizable. (This is the negation of the halting problem. Movaghar 23 Decidable and Recognizable Languages Butnot all languages are decidable! In the next class we will see an example: A tm = fh M ;w ij M is a TM and M accepts w g is undecidable However A tm isTuring-recognizable! Proposition There are languages which are recognizable, but not decidable Agha-Viswanathan CS373 If a language is recognizable, but not decidable, then its complement cannot be recognizable. because by Theorem: if L is Turing Decidable if and only if both L and complement of L are Turing recognizer. 5 Recognizability vs. 4. True. Therefore, L is not Turing-recognizable. A. Decidable and Recognizable Languages Butnot all languages are decidable! In the next class we will see an example: A tm = fh M ;w ij M is a TM and M accepts w g is undecidable However A tm isTuring-recognizable! Proposition There are languages which are recognizable, but not decidable Agha-Viswanathan CS373 (D) decidable, (R) recognizable but not decidable, and indicate which undecidable examples follow from Rice’s Theorem. languages Proposition (Decidable languages and total DTAs) A language is decidable if and only if its recognized by a total DTA. • Recall that the complement of a language L is the language L consisting of all strings that are not in the language L. Classify each of the languages as described above. If is decidable then and Turing recognizable Trivial 2. Thus, J is not Turing-recognizable. Prove that the language it recognizes is equal to the given language and that the algorithm halts on all inputs. , the Equivalence problem for NFA’s. In the Silly Post Correspondence Problem, SPCP, in each pair the top string has . Posted by just now. recognizes) it. 4. -- between Recognizable and decidable If L is decidable, then L is recognized by a TM M that halts on all inputs. IIt is possible for a TM to never reach a halting con guration. Moreover since decidable languages are closed under complement, L is also Turing H AL TT M is T uring-recognizable since itcan be recognized by TM U . (Note: the symmetric di erence S4T of two sets S;T is de ned to be the set of elements belonging to S or T but not both. Closed under intersection. Run M 1 and M 2 on input w in parallel. S. 𝐴TM 𝐴TM Check-in 8. g (c) L Problem I. (Do not give redundant classifications. yes. B It is not decidable but is recognizable C It is context-free D It is not context-free but is decidable E It can be recognized with a PDA 10. Non-Turing Recognizable Language (example) Theorem: The complement of A TM is not Turing-recognizable. This means that our Turing Machine is Recognizable, but it is not decidable. 2. (b) Using Rice’s Theorem, prove that the following language is undecidable: ALLTM = {hMi|M is a TM and L(M) = Σ∗} 2. On input . Class 5: Not recognizable. As we know that one can convert an instance of w in P1 to an instance x in P2. 3. Decidable and Recognizable Languages IButnot all languages are decidable! We will show: I A tm = fhM;wijM is a TM and M accepts wgis undecidable IHowever A tm isTuring-recognizable! Proposition There are languages which are recognizable, but not decidable But it is not given in question about undecidability of P3. B. Atm is undecidable NOT ON EXAM. 1 1. “ • Here we describe algorithms to test whether a CFG generates a particular string and to test whether the language of a CFG is empty. The language A TM = (fhMi;w) : M accepts wg The question of whether A TM is decidable is precisely the question of whether one TM can predict the output of another TM! Question. The Turing Machine is allowed to loop forever instead of rejecting. 3. Although it might take a staggeringly long time, M will eventually accept or reject w. 22. False. w} The language . Problem 5. Turing-recognizable language but not Turing-decidable language. Show that the collection of recognizable languages is closed under the following operations 1. If a language is not recognizable, but its com-plement is, we call it co 5. 5 pts each) For each of the following languages, select one of A-E that best describes it. NP. Let’s prove this. I. (i) f[n] 2: n‚0g. is a Turing machine that accepts at least 42 different strings . • U is known to be Turing-recognizable (Th. • Let E DFA This is not a uniform procedure, because it depends on the choice of the element ain the model A. 3. T-recognizable. or semi-decidable) if some TM recognizes it A language is called decidable or recursive if some TM decides it •Corollary 4. So Th(!;S)is relatively decidable but not uniformly relatively decidable. 3. 2 Clearly mark if the language is decidable, recognizable, co-recognizable, or neither. or semi-decidable) if some TM recognizes it A language is called decidable or recursive if some TM decides it w L ? accept reject TM yes no w Σ* L is decidable (recursive) w L ? accept reject or no output TM yes no w Σ* L is semi-decidable L is Turing recognizable if there is a Turing machine M that recognizes L, that is, M should accept all strings in L and M should not accept any strings not in L. Turing Machines Recognizable/Decidable Proofs. CLIQUE. IThese are the three options for any input: accept, reject, loop. Let and be two recognizable but not decidable languages. The theorem: "A language is decidable iff it is Turning-recognizable and co-Turing-recognizable. These are also called theTuring-decidableor decidable languages. None of the above 3/17/2016 reachability is not absolute in terms of decidability. • Let Theorem 6: is a decidable language. Accep ted An sw ers: is decidable and is recognizable Let and be two decidable languages. e, or Turing-recognizable) if 9a Turing machine Msuch that L(M) = L. a. 3. A TM = { (M, w) | M is a TM that accepts string w } A ConcreteUndecidable Problem: The Acceptance Problem for TMs Theorem [Turing ‘30s]: A TM is recognizable but NOT decidable If a language is Turing-recognizable and its complement is Turing-recognizable then it is said to be co-Turing-recognizable. A language L is decidable if and only if both L and L are Turing recognizable. (a) f1p: p is a prime numberg regular context-free decidable recognizable not recognizable (b) f0n1n: n 0g regular context-free decidable recognizable not recognizable (c) fx 2f0;1g : x contains an even number of 0’sg regular context For each of these languages, say whether it is (i) decidable, (ii) recognizable but not decidable, or (iii) not recognizable. This is not the same as decidability because recognizability does not require that M actually reject strings not in L. But first let’s look at a definition and a theorem. y (Hint: You may ﬂnd it TM-recognizable but not decidable ; Pf It is easy to see that since ; So cannot be decidable. For each of these languages, say whether it is (i) decidable, (ii) recognizable but not decidable, or (iii) not recognizable. ‣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 “Turing recognizable” vs. UNSOLVED! https://i. when encountering a string not in that language, the machine terminates and rejects that string. Statement 4 is true as Turing recognizable languages (RE languages) are closed under union and intersection. Class 5: Not Recognizable. That is, we want a c. TM. 22. “Decidable” 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. C. Theorem 1. For . The FO theory of the suﬃx rewriting graph of a recognizable trace rewriting system with level-regular contexts (RTL graph) is thus decidable. 5. For one such example, some familiarity with mathematical logic is required: Presburger arithmetic is the first-order theory of the natural numbers with addition (but without multiplication). Namely, there is an algorithm that can solves it (for any input). (Do not give redundant classifications. Otherwise, accept. Proof. It is not at all clear (to me) which definitional choice would lead to the strongest theorems, and also, best is not decidable and is recognizable is not decidable and is not recognizable. Assume that the complement, is also recognizable and recognizes . False. L /∈ RE and L ∈ RE. TM. L ∈ R iff L is 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. If A m B and A is undecidable then B is undecidable. e. This claim is true. Thank you! So far, all of the languages presented above are recognizable, even though some of them are not decidable. 1. • Another interesting question about a regular language is whether or not it is empty. • But the other direction does not hold---there are languages that are Turing-recognizable but not Turing-decidable. Not all Recognizable languages are closed under complement. Share Cite TM is not decidable. To show that a language is decidable, we need to TM were decidable, the language of any TM would be decidable. Proof. Closed under union. Classify the following problems as in P, in NP, NP-complete, decidable, undecidable, Turing recognizable, not Turing recognizable. Consider the niteness problem. there is a halting Turning machine B that recognizes the descriptions of Turing machines that satisfy P. TM is recognizable, but NOT decidable! Given: code of a Turing machine M and an input w for that Turing machine, Decide: Does M accept w? A TM decidable ⇒There is an algorithm ALG which, given any code and input, ALG determines in finite time if the code will stop and accept the input The complement of HALT is not CE. 22, U is decidable. is a TM that does not accept . , A is co-Turing-recognizable. The following TM U recognizes A TM: U =“On input string hM;wi, where M is a TM and w is a string: 1. L1 ∪ L2 where L1 = {w | w has even number of 1's} and L2 = {w | w contains 001 as a substring} language L C E* is Turing-recognizable if there exists a TM M sot, La (M) = L, (also called recursive) Note: regular context-free decidable recognizable ote: a decidable L is also recognizable, ut not vice versa (also called recursively enumerable) Defn, A language L C E* is Turing-decidable if there exists a decider D sot, La (D) = L, Decidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. concatenation Solution: Proof. From the deﬁnition of not regular. Suppose is recognizable and recognizes . That means there will be a Turing machine that says 'yes' if the input is P2 but may or may not halt for the input which is not in P2. 3. e. A. A. Here we show that there exists a language that is not even recognizable. -Midterm Nov 3 Nov 5 Tu w Th spend 40 minutes 4-⑤ review answers to questions you may have ocfh midterm material recognized by M is regular. is not decidable in order to conclude that it is not recognizable. . EXPSPACE. We know that is Turing-recognizable. subset of decidable languages recognized by linear-bounded automata (LBA)) R. The Star Problem is decidable in Proof. Simulate M on input w 2. Show that if S is finite, then there exists a deterministic TM N that uses S to decide L. So option (C) is not correct. 2. Problem 8. 23), contradicting Theorem 5. 6 is a decidable language Relies on the following property : If is in Chomsky Normal Form, then any derivation of has length at mo CFG CFG A G w G w A G w = Theorem De nition 13. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the following diagram − Example 1. Recen tly, sev eral v eri cation problems ha e b een sho wn decidable for P A using a v ariet y of fairly in olv ed tec hniques 1. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). Since CA, C NA and N, are recognizable, we already know, from [13,24], that the FPP's arc decidable. Contradiction. Classify each of the following languages as: (R) Regular, (CF) Context-Free, but not regu-lar, (TD) Turing-Decidable, but not context-free, (TR) Turing-Recognizable, but not Turing-decidable, or (None) None of the above. If yes TM is admittedly a rather strange language, and it’s not obvious why we should really care that it’s not recognizable. Turing recognizable languages are closed under union and complementation. 4. . TM = {〈 M, w 〉: M. If P(x) halts, this algorithm will eventually accept, and if P(x) doesn’t halt, this algorithm runs forever. co that are not Turing-recognizable ( #is the complement of A): Theorem C Let A Y -* be a Turing-recognizable language that is not Turing-decidable. Fact 3: If complement of A TM is Turing-recognizable, then A TM is decidable. Give examples of three different languages that are not recognizable. g (c) L A language is Turing recognizable if it is recognized by some Turing Machine. Justify your answer using methods from class. this seems 3 Not all languages are recognizable Letus showanon-constructive proofthat Theor em : A T uring-decidable language is also T uring-recognizable, but not vice versa. Now form a language Lby replacing each string w2Hby w jw. 4. (c)There is a decidable but not recognizable language. language that is not Turing-recognizable. Proof in two directions: First, if A is decidable, show both A and its complement are Turing-recognizable. If L is decidable then it is Turing recognizable. 5. July 21, 2019 July 21, 2019 ASSIGNMENT Leave a comment This problem concerns what languages are decidable/recognizableor not. e. coNP. Otherwise, we could decide the halting problem for 2-counters Minsky machine. • We say that a language is co-Turing-recognizable if it is the complement of a Turing-recognizable language. If accepts, accept. e. 28) Ā is Turing-recognizable – i. Ch 3 includes (graph), LT. HP = f<M;x>: Mdoes not halt on xg) Obviously, HP is co-recognizable but not recognizable. • Not every Turing-recognizable recognizable PSPACE=NPSPACE EXPSPACE EXPTIME decidable P CFL is decidable, but not in polynomial time. We use the fact that A PDA is decidable. We’ll soon see examples of languages that are in REbut not in Dec. Turing decidable languages are closed under intersection and complementation. If x is an integer, let [x] 2 denote its binary representation. Proof: W e will reduce A T M to H AL TT M. 24 The set A TM is mapping reduced to J by the function f(y) = 1y. We have shown (sort of) that A TM is Turing-recognizable but not decidable. The set R is the set of all decidable languages. decidable but not recognizable