Language of emptiness of cfg is decidable If w ∉ L(B), then Decide whether two automata are equivalent, i. L2 = { a n b n c m d m | m >= 0 and n >= 0} is also context free. They are also known as Non-Recursively Enumerable Language. opt C:Emptiness problem is decidable with a finite state machine. I did get some info on list of Note, that if a grammar Ggenerates an empty language i the start ariablev S62V0, where V0is the set computed by the above algorithm. See this for details. Rao, CSE 322 8 The Chomsky Hierarchy – Then & Now Reg CFLs CSLs T-recognizable Not T-recognizable Then (1950s) Now U. 1998 ACM Subject Classification F. In other words, the question is whether there exists a string that is accepted by the DFA recognizing the language. 4 LetBbe the set of all infinite sequences over{0, 1}. A CFG . 1 is a decidable language ="On input , , where is a DFA and is a string: 1. that decides ! DFA. To determine the emptiness of a regular language, one needs to analyze the structure of the DFA and its states. CFG is decidable and that was used to prove that every contextprove that every context-free language is decidable (byfree language is decidable (by simulating a TM for A CFG). Why decidable problems would be semidecidable?Recursive languages encloses regular,CFL,CSL, and few other languages (some of which are recursively enumerable as well) but for all of this turing machine can output yes or no. • R does not have to simulate M on w. a nb c),a subset of decidable languages recognized by linear-bounded automata (LBA)) R. Follow edited Apr 20, 2021 at 1:28. To decide whether CFL is finite or not. ECFG = {hGi | G is a CFG and L(G) = ∅} is decidable. My question is that how can we say CFL regularity is undecidable Our start symbol S can become either a or b, so the language of the CFG is {a,b}. From the finiteness of the set of symbols it How could I prove that the following language is decidable? $\{\langle G\rangle \mid G\ \text{is a CFG over}\ \{0,1\}\ \text{and}\ 1^* \subseteq L(G)\}$ P. 3 of 18. Therefore, this is the regular language (ab + aabb)+. The answer depends on how the two languages are given to you; I will assume that the context-free language is given as a CFG, and that the regular language is given as a DFA/NFA. Mark any variable A where G has a rule A I am not sure how he reached the conclusion that it is equal to $\mathrm{ALL}_{\mathrm{CFG}}$. (2)Every TM is a recognizer for some language. We know that Emptiness problem is decidable: determine whether the start symbol of G is generating. Since converting a CFG into CNF is decidable,AǫCF Gis decidable. question asked if Can you guys list the problems that you are aware that are decidable for Context free Language and for Deterministic Context free languages. Prove that E CFG = { G : G is a CFG and L(G) = ∅} is a decidable language. One variable to two variables One variable to one terminal Epsilon only appears if part of the language derived with the CFG, and it should only appear directly from the Theorem 6 A CFG is a decidable language PROOF IDEA For CFG G and string w we from MATH 5510 at East Tennessee State University. Language A TM is defined as Emptiness of context free language: Given a context free language, Option 1 is whether a CFG is empty or not, this problem is decidable. Mark all terminal symbols in G. Non-emptiness of intersection for pair of grammars In this section, we investigate the computational complexity of the problem of deciding non-emptiness of the intersection of the languages generated by a pair of grammars, at least one of which is non-recursive (see Appendix A for the definition of non-recursive CFG). CFG Deciders. So we formally Then $\bar{C}$ is a CFG by closure under complement. Carl Mummert. Theorem 4. TM. We start by Decidable properties of regular languages: membership, emptiness, finiteness (and hence infiniteness), equality to , equality, inclusion (A B), Algorithms to demonstrate these properties are decidable. Encodings; examples of decidable languages Now that we have studied basic aspects of the Turing machine model, including DFA, a CFG, another DTM (maybe even a description of itself), or a list of objects of multiple types. Emptiness Testing for CFGs Theorem E CFG = If we try to decide EQ CFG using the same symmetric difference approach that Stack Exchange Network. , for proving A TM undecidable Reducing from A TM. Because there are more languages than TMs. S. Q: What Some language is not decidable. Study Resources. To prove that A CFG is decidable, we note the following two facts: Every CFG has an equivalent Chomsky Normal Form grammar language is HALT = { M,w |M halts on w} 1 Some Closure Recall the definition of decidable and recognizable languages. I'm having difficulty proving undecidability of "is this CFG prefix-free?". M = `` On input 〈𝑮〉, where 𝑮is a CFG: 1. , is there a CFG such that a string ?) is true? Consider the language $A_{CFG}$ = { < $G$, $w$ > | $G$ is CFG that generates $w$}, where < $G$, $w$ > is a string encoding of $G$ and $w$. 15 of the third REX is decidable Theorem The language A REX r–R,w‰¶R is a regular expression that generates the string wx is decidable. opt A: They are saying about Halting Turing Machine which is decidable because it halts within a certain amount of steps. All that said, I think that the table has a typo. EXPLANATION: Option 1: The emptiness By definition, a language is decidable if there exists a Turing machine that accepts it, that is, halts on all inputs, and answers "Yes" on words in the language, "No" on words not $\begingroup$ Convert the CFG to a PDA, use the product construction, and convert back. 113 3 3 bronze badges. We Provide Services CFG’s 3. In this case, we say that the Language is recognized by the Turing Machine. You can also easily simulate a PDA on a TM. Proving languages are not regular, using the pumping lemma for regular languages. language of pairs (DFA, string) A DFA = fhB; Check emptiness CFL Membership Test A CFG = fhG;w i j G is a CFG that generates string w g Could we enumerate all derivations and check them in turn ? Emptiness Problem for CFG Problem: "Given a CFG G , is L(G ) empty? Language Formulation (Emptiness Problem for CFG) E CFG def= fh G i j G is a CFG and L(G ) = ;g Theorem The language E CFG is decidable. However, in the case of an NFA, due to the absence of stack, there can only be a finite set of possible states before reading any input symbol, even Warm-Up: Some Decidable Languages Show that the following languages are decidable by describing (at a high level) an algorithm that deci des them (see more in Sipser 4. In other words, show that $\left\{\langle G\rangle \mid G\right. $\endgroup$ – 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 . 1k 12 12 gold badges 169 169 silver badges 311 311 bronze badges. L1 and L2 decidable ==> INTERLACE(L1, L2) decidable. Given the string encoding of a CFG and some string , a TM can decide whether can derive . R = “On input <G>, where G is a CFG: Mark all terminals of G Mark empty string symbol Repeat until no more variables get marked Mark any variable A where A!U 1U 2U n and all U i The emptiness problem for regular languages asks whether a given regular language is empty, i. - However, the process may not stop; - Would only prove Turing recognizable for A CFG. We can nd out which non-terminals of G can derive a terminal string: i. Maybe you're confusing two different problems. Cite. Log in Join. 1 st) Membership Problems ( Whether the given rammer is CFL ) 2 nd) Emptiness problem ( The given language is empty or not) 3 rd ) Finitness Problem( Means the given language is finte or not). Equivalence Problem for CFGs BBM401 Automata Theory and Formal Languages 16 accepts w} is a decidable language. What are Undecidable Problems? The Is EQCFG is a decidable language? How about, if we are given a CFG, can we find a TM that decides the same language? The answer is YES! Is it decidable whether a given CFG accepts a non-empty language? Yes, it is. Language Regular Context-Free Decidable Increasing generality (Chomsky also studied context-sensitive languages (CSLs, e. FL] 28 Mar 2017 Decision problems of CFG Decidable 1)Emptiness 2)finiteness 3)membership Undecidable 1)equivalence L1 and L2 CFG then is L1=L2? 2)Inclusion or containment A language is decidable if and only if some NTM decides it Halting problem recognizer are more powerful than decider CFG Does a CFG (B) accept some string (w)? – E CFG Is the language accepted by some CFG (B) the empty language (the empty set of strings). Repeat until no new variables get marked: a) Mark any variable A where G has a rule A-->U 1,. Repeat until no new variable is marked 1. We can. * E−DFA = “On input & [IDEA: Check Emptiness of a CFL By checking the production rules of the CFL we can easily state whether the language generates any strings or not. Proof: R = “On input , where is a CFG: 1. A decision problem P is decidable if the language L of all yes instances to P is decidable. However not all languages are decidable by PDA. ) S = “On input <G,w>, where G is a CFG and w is a string: 1. Follow edited Jul 18, 2014 at 21:20. Mark any variable A if G contains →𝑈1𝑈𝑘 and 𝑈1,,𝑈𝑘 have all been marked 3. Formulate this problem as a language and show that it is decidable. In other words, given a language L, we can always find a TM M where M is a recognizer for L. A recursive language is a formal language for which there exists a Turing machine that, when presented with any finite input string, halts and accepts if the string is in the language, and Nevertheless the concept "empty language" might be very confusing to some. Although it might take a staggeringly long time, M will eventually accept or reject w. , Show that A(NFX) is decidable. We can compare two DCFG (DPDA) but we cannot Compare two The Emptiness Problem for the Language of a Finite Automaton. A That is, we transform an instance Mof the emptiness problem into the instance (M;M;) of the equivalence problem. 3 Formal Languages, F. Further, for a given CFG, there This is interesting because the empty language and any language containing just 1 (or any nite number) of strings must be regular and therefore context-free. The intersection of a regular language and a context-free language is context-free, and you can construct a CFG for it. About reduction: for as long as you can reduce this problem to some other decidable problem, its fine. Visit Stack Exchange Turing decidable problems are the problem for which the Turing machine always halts on every input, accepting or rejecting it. Rice's theorem shows only when a language is not decidable, hence it can't be used. If you start to doubt if the language is context-free, you should use a so-called "pumping lemma for context-free languages". (%): from above, decidable languages are closed under We can easily prove this language is semi-decidable. Definition 1. If this problem is decidable, we'd expect the problem to be decidable for any arbitrary string. question asked if there is a set of intermediate size between ℕand ℝ. Is L(G) = Φ? This is the emptiness problem of CFLs which is decidable. We will show some specific language 0. Convert G into Chomsky normal form. That is why this breaks on semi-decidable languages where not both the Yes-Answer and the No-Answer are guaranteed to be delivered in finite time (I always imagine that as a "return" behind an infinite loop). Perfect languages|a term coined by Esparza, Ganty, and Majumdar|are the classes of languages that are closed under Boolean operations and enjoy decidable emptiness problem. Decidability 23/43 Proof (1) To determine whether L(G) = ;, the algorithm might try going Myhill–Nerode does indeed provide a characterization of the regular languages but that isn't enough to show that the problem is decidable. 1 . There exists a TM that decides the above language. 2(Recognizable Language). ,U k provided You don't say exactly what you mean by equivalence "structurally (at a very basic level)". So, since regular languages are closed under complement, Is the equivalence problem of a CFG and a FSM decidable? 1. . Convert PEG to BNF. Naive algorithm has O( n 2 ) time complexity where n is the size of G (sum of the lengths of the It can be partially decidable but never decidable. Share. In a recent exam some students noted that $\{a,b,c,d\}$ had $17$ subsets: $2^4$ usual ones, plus the empty set. They can be used to show that it's not possible to convert a CFG to a regular language when possible, but this does not yet mean it's undecidable to merely decide whether a context-free language is regular. Validity of Prove that the finiteness problem for regular and context-free languages is decidable. Consequently, any regular language and its complement are a pair of complementary context-free languages. Let S be the following Turing machine: T= “On input <G>, where G is a CFG: 1. In class. Since the emptiness of CFG is decidable, we can use it to decide whether $\bar{C}$ is empty and therefore whether A problem is decidable if there is a TM which decides instances of the problem. Every decidable language is Turing-Acceptable. L1 says number of a’s should be equal to number of b’s and L2 says number We can easily prove this language is semi-decidable. Is it decidable that G accepts whatever the regular expression does? In other wor It then discusses various decision problems for regular languages, including emptiness, finiteness, membership, and equivalence. The following language is decidable A CFG = fhG;wijG is a CFG;w 2L(G)g Proof. L(M) is a set of strings. Church-Turing thesis; examples of decidable languages. Run M on w. An algorithm is defined by the existence of a TM that implements the algorithm. Using context-free grammars, we can decide the Emptiness Test for TM (2) Proof Idea: By reducing A TM to E TM. A language is decidable if there is a Turing Machine that halts and accepts strings that belong to the language, and halts and rejects strings that do not belong to the language. 5 5 Decidable CFL Properties Finally, we show that two languages related to properties about context-free grammars are decidable. Another thread has the very different question "is this regular language prefix-free". 4: is a decidable language. My question Decidable Languages Consider the following language about DFAs: A dfa = { <M, w> | M is a DFA accepting input w} Is this language decidable? That is, given as input, a DFA and a string, can we decide whether the DFA accepts the string? The answer is yes. r. There CFL Emptiness is Decidable Theorem It is decidable whether a context-free grammar G • A total Turing machine R, given M and w, can prepare a CFG for L. Lemma. The set R is the set of all decidable languages. Problem 3: describe algorithms to test whether an arbitrary string is an element of a context free language, (i. Language Ambiguity | Ambiguous Language Context Free Grammar- Context Free Grammar Normal Forms- Chomsky Normal Form Decision Algorithms of CFG- To decide whether CFL is empty or not. 7 A CFG is a decidable language. Proof is by induction on the string length n. In 1997, Géraud Sénizergues proved that equivalence of DCFLs is decidable, and he was awarded the Gödel Prize for it. Now we have to show that Yes it is decidable. Tur. Undecidable Problems Showed these problems for FA and CFG’s decidable: Acceptance, Emptiness, Equivalence Qu: What about for TM? A = {<M,w> | M is a I have this question for a homework. This is the equivalence problem of regular languages which is decidable. Complementing a machine where all Yes A language is decidable if and only if. But in the end, we are proving that the grammar is ambiguous, not the language. Home; Expert (P, G, w) | w is accepted by PDA P or CFG G} We have an Answer from Expert View Expert Answer. We use a reduction from PCP. Follow answered Oct 13 , 2021 at 18: Maybe you're confusing two different problems. Also, we can consider the Some of these CFL problems are decidable, some are not. Proof outline. $ is a CFG over $\{0,1\}$ and $\left. The Halting Problem Turing-recognizable vs. A variable is "productive" if it can make any string of terminals at all. The set of decidable languages are closed under the same operations, and also under A language is decidable if there is a Turing Machine that halts and accepts strings that belong to the language, and halts and rejects strings that do not belong to the language. is decidable . The classic example of a non-context-free language whose complement is context-free is {ww|w∈{0,1}*}. S →aS |ε This CFG produces either nothing, or itself with an a right before it. Title [] UNDECIDABLE PROBLEMS [] ABOUT Let $\Sigma=\{0,1\} . Rice's Theorem. Recall that it is decidable whether or not the language generated by a given context-free grammar is empty. Turing Machines (May Not Always Halt) • If Always Halt . opt D:CFG is decidable with emptiness problem. Hot Network Questions Is biological stress related to covid lockdown policies a better explanation of excess pandemic deaths than covid infection? } Emptiness test for CFGs E CFG = {<G> | G is a CFG and L(G) = Ø} This language is decidable. so it is undecidable Think about the situation for regular languages, where such a shortcut does exist. I saw that Ambiguous CFGs and Equality of CFGs are considered undecidable. answered CFL Emptiness and Equality • Let E CFG ={<G> | G is a CFG such that L(G) is empty}. (4)Every regular language is Does this suffice to show that the language is decidable and did I make any mistakes? computer-science; formal-languages; Share. If the machine has no accepting states, then it cannot possibly accept any string given to it. Expert Answer . Hence, the emptiness problem for CFGs is decidable. Q. Undecidability means that there exists no general algorithm that solves every instance of this •Decision problems: solvable (decidable, recursive), semi-decidable (recognizable, recursively enumerable/re, generable), non-re •A set is re iffit is semi-decidable •If set is re and complement is also re, set is recursive (decidable) •Halting problem (K 0): diagonalization proof of undecidability •Set K0is re but complement is not CFL Emptiness is Decidable Theorem It is decidable whether a context-free grammar G • A total Turing machine R, given M and w, can prepare a CFG for L. not decidable Proving Undecidability of a Language L Diagonalization (directly) e. CFL Decid. Q: What would happen if we prove that A TM is decidable? If A TM were decidable, the language of any TM would be decidable. A Decidable Languages Robb T. Pay close attention! – EQ CFG Do two CFGs (A and B) accept the same language. answered Jul 18, 2014 at 20:57. Equivalence Problem for CFGs BBM401 Automata Theory and Formal Languages 16 The Emptiness Problem for the Language of a Finite Automaton. • It turns out that EQ CFG is NOT a decidable language. Ali V. For general context-free languages, it is undecidable. ?” is Decidable. A dfa is decidable. b) the empty language is decidable: the answer is always "no, this string is not in the empty set". Consider the language Aε CFG = {<G> : G is a CFG that generates ε} . • This one is quite interesting, and not what one might expect. L(G) is in nite i such a parse tree exists. (So, what can R The set of all context-free languages is identical to the set of languages accepted by pushdown automata, which makes these languages amenable to parsing. The well-known undecidability result concerns what is also called weak equivalence: Two grammars are equivalent if and only if they generate the same language (= set of words). "Decidable" means that there is an algorithm (more formally, a Turing machine) that terminates for every input and, of arXiv:1702. Language A CFG. Firstly assume we have a TM R that decides E TM. Abstract. ADFA is decidable Theorem The language ADFA = { B,w ∣ B is a DFA that accepts the string w} is decidable. Gabriele R oger (University of Basel) Theory of Computer Science April 25/May 2, 2022 5 / 27 C2. I think that if its already known that a language is deterministic then to decide "is the language ambiguous" boils down to deciding "is it empty or not". e. (<G, w>)) Emptiness test for CFG's (E_CFG(<G>)) DFA Deciders. is decidable? a) Because CFGs are generators and DFAs are recognizers. 145 1 1 silver badge 7 7 bronze badges. 83. The procedure we gave for translating an NFA to an equivalent DFA was mechanical and terminating, so a halting Turing CFG Emptiness Is Decidable Theorem. We do not need stronger models. These results imply that tmso + zero has decidable satisfiability. Maintain a set of \marked" non-terminals. Initially N Your task is to prove that the language is decidable. Convert G to an equivalent grammar in CFG is a Decidable Language Proof Idea: – For CFG G and string w we want to decide whether G generates w. The language Language A CFG is Decidable Language. If the start symbol is marked, accept. However, the machine that you describe will actually Here we show that the emptiness problem for context-free grammars (CFGs) is decidable. Base case: the shortest strings ab and aabb are in the language (ab + aabb)+ and are generated by the grammar as shown above. N =“On input string hB;wi, where B is an NFA and w is a string: 1. Let ! DFA = & & is a DFA and ' & = ∅} Theorem: ! DFA. We have an Answer from Expert Buy This Answer $5 Place Order. Consider the problem of determining whether a pushdown automaton has any useless states. We start by Find step-by-step Computer science solutions and your answer to the following textbook question: A useless state in a pushdown automaton is never entered on any input string. I. We want to build a TM M that decides ADFA: M = “On input B,w , 1 Run (or simulate) B on w 2 If B ends in an accept state, accept; otherwise reject” Since the simulation always ends after ∣w∣ steps, M is a decider. Rec. The next theorem states that determining whether two DFAs recognize the same language is decidable. That is straight forward, because an empty language has no strings at all, even the empty string. Mark all terminal symbols in G 2. L ∈ R iff L is decidable –L s iTuring-decidable if there is some TM that decides L. The given language may be CSL --In this case also the above is Decidable. $\endgroup$ – As an example we prove the undecidability of \intersection emptiness" ISECFG. Theorem 5. After a finite number of steps (1)Every language is recognizable. ” Given a CFG in Chomsky normal form, is there an algorithm that solves the emptiness problem in linear runtime? I thought about using depth search here, but I think it's a little bit above linear runtime. EXPLANATION: Option 1: The emptiness problem for Context-free language is decidable. 2 Repeat until no new state is as we know, every thing is decidable for regular languages. If w ∉ L(B), then as we know, every thing is decidable for regular languages. A regular expression for this language is: a∪b b. 185) To define a Turing machine, we could give a. The question stems from the fact that you can determine whether a regular language is empty by using a Turing machine to count the states n in the given FSM. In other words TM are more computationally powerful than PDA, i. babou babou. Is this Question: Question 5 Given the language ECFG={ G ∣G is a CFG with L(G)=∅}, prove that it is decidable (briefly describe the proof idea). To decide membership of CFG | CKY Algorithm language accepted by a CFG(Context free grammar) is nonempty ; Update: In first question I think 1. 1. , Assume E TM is decidable, show A TM is decidable. Expert Help. • But the other direction does not hold---there are languages that are Turing-recognizable but not Turing-decidable. Lang. , it contains no strings. One of the answers in that thread claims to prove undecidability for CFLs (even though that did not explicitly ask for a proof of undecidability for CFLs) using By definition, machine M accepts a string s if it reaches an accepting state when given input s. and more. We don't have to check words with length > n because Every regular language is context-free. I mean, I give you a set S with a finite number of CFG Grammars, how can you even know that two of them generates the same language? Where does that piece of knowledge come from? Examples for incomparable semi-decidable but undecidable languages. Then we run M0on input hM;wi. 1^{*} \cap L(G) \neq \emptyset\right\}$ is a decidable language. Identified Q&As 3. Title [] UNDECIDABLE PROBLEMS [] ABOUT the emptiness problem of nonzero automata is decidable, with complexity np ∩ co-np. CFL Emptiness is Decidable Theorem It is decidable whether a context-free grammar G • A total Turing machine R, given M and w, can prepare a CFG for L. Also by definition, L(M), referred to as the "Language accepted by M", is the set of all strings accepted by M. However, we run into one big problem: the class of context-free languages is not closed under complement! We saw earlier that, if a class of languages is closed under Decidable Languages 18 •Theorem: A CFG is a decidable language •proof (cont. Let A CFG = { G, $\begingroup$ Actually emptiness of context-free languages is decidable (a single one, not the intersection of two). Given two arbitrary CFGs $G_1$ and $G_2$, we know this for certain: either $L(G_1) = L(G_2)$, or Here we show that the E_DFA problem is decidable, which is determining for a given DFA, whether or not its language is empty. I synced with this Hendrik Jan's answer that to prove undecidability of regularity for CFL is usually obtained from two properties of the context-free languages: (1) they are closed under union, and (2) universality is undecidable. Koether Homework Review Decidable Languages Decidable Problems Concerning DFAs and NFAs Decidable Problems Concerning CFGs and PDAs CFG is a decidable language •Let E CFG = {<G> | G is a CFG and L(G) = Φ}, then E CFG is a decidable language (emptiness testing) •Let EQ CFG = {<G, H> | G & H are CFGs and L(G) = The Emptiness Problem for the Language of a Finite Automaton. Run M on hC;wi The proof that regularity of the language generated by a CFG is undecidable is very similar to the proof that universality of the language The latter for instance, because DCFL are closed under complement, and emptiness is decidable (in fact for all CFL). If w ∈ L(B), then M will accept. Assuming that we would have an algorithm to decide If a language decidable by a PDA then it is also decidable by a TM. Is the problem of determining whether a CFG generates a string in the form 0*1* decidable? 1. A CFG, E CFG decidable, ALL CFG, EQ CFG not decidable A TM, HALT TM, E TM, etc. E CFG . This only works for intersecting a CFG with a regular language, not two CFGs the emptiness problem of nonzero automata is decidable, with complexity np ∩ co-np. Concatenation : If L1 and If L2 are two context free languages, their concatenation L1. This will not work, why? • This method a best will yield a TM that is a recognizer, not a decider. For a decidable language, for each input string, the TM halts either at the accept or the reject state as depicted in the Some Decidable/Undecidable problems about CFL’s Problems about CFL’s Problem (a) Is it decidable whether a given CFG accepts a non-empty language? Yes, it is. Commented Jan 31, Hence, the given problem is decidable. Proof: in class Since we now know that the class of context-free languages is not closed under either intersection or com-plement, it seems hopeless to believe that the equivalence problem for context-free languages is decidable. CFG = {<G,H> | G and H are CFGs and L(G) = L(H)} is NOT a decidable language. Let $G_1$ be CFG is decidable. Can we decide EQ CFG = { G,H : G,H are CFGs and L(G) = L(H)}? If no, why does the strategy from Question 6 not work here? Question 8. Decidable languages and unrestricted grammars? 19. Language of Grammar- Language of Grammar. proof idea: Use G to try all possible derivations to see if any of them derive w. The emptiness problem for Context Free Grammars is decidable. Consider all production rules to be simple (I CFG = {<G,H> | G and H are CFGs and L(G) = L(H)} is NOT a decidable language. Solutions available. By simple logic. L2 will also be context free. AI Homework Help. 06858v2 [cs. is not decidable. st . Check if there is a parse tree within a height of 3n, where n is the number of non-terminals in G, that contains a pump. If not, reject. Language A CFG is defined as follows: { (M ,w) : M is a Context Free Grammar (CFG) that accepts the string w } Therefore, in this language DFA is enough. To prove that A CFG is decidable, we note the following two facts: Every CFG has an equivalent Chomsky Normal Form grammar But why finiteness property is decidable inspite having infinite configuration space? Skip to main content. For example, L1 = { a n b n | n >= 0 } and L2 = { c m d m | m >= 0 } L3 = L1. Consider T = “On input hAiwhere A is a DFA: 1 Mark the start state of A. • A DFA accepts some string if and only if reaching an accept state from the start Theorem 6: is a decidable language. The decidability of a problem refers to the Some language is not decidable. 32b in Sipser 3rd edition). Examples of regular and nonregular languages. Gödel and Cohen showed that we cannot answer this question by using the standard axioms of mathematics. In this field, we explore the inherent difficulty of solving computational problems and classify them based on their computational resources required. LANGUAGE CLASS: Closure Properties, Language Problems, NOT in class . Because L and its complement are both semi-decidable, we have an effective procedure for deciding each: run the TMs for L and its complement alternately, and one will eventually list the input. Question 7. Repeat until no new variable is marked: 3. there exists a derivation X ) w for Let G G the CFG and S S its axiom, then: If the axiom S S of G G is marked, then the language is not empty, otherwise it is empty. Proof-For testing a language to be CFL you need to check if it can be accepted by push down automata. S Emptiness testing: for the language of a FA E DFA is a decidable language. In particular, recognizable languages would Video answers for all textbook questions of chapter 4, Decidability, Introduction to the Theory of Computation by Numerade G is CFG. a nb c),a subset of decidable languages recognized by linear 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 Decidable Languages Robb T. WewanttobuildaTMM thatdecidesA REX. Mark all terminals in G. That is why they use the term "inherently" ambiguous; to refer to the CFG but not the language. • A DFA accepts some string if and only if reaching an accept state from the start Context-sensitive languages have context-sensitive grammars, and context-free languages have context-free grammars. Similar questions. A language L is decidable if and only if L¯ is decidable. The class of data languages generated by RCFG are closed If you want another kind of proof, consider the following. Can you guys list the problems that you are aware that are decidable for Context free Language and for Deterministic Context free languages. Proof: Construct a decider M for E CFG: M = " 1. Stack Exchange Network. This brings me to the definition of a Turing Recognizable Language : Def : A Language is called Turing Recognizable if some Turing Machine recognizes it. 7: is a decidable language. I think D is decidable. In second question I think 2 is right, but not sure about the decidability of non emptiness of CFG and DFSA. CFG is decidable. LetR You are on the right track in that if you can construct a TM that always halts for a given language L, then L is decidable. 6: iACFG s a decidable language. 1. It is decidable. , is there a CFG such that a string ?) is true? Problem 2 describe algorithms to test whether the language generated by a CFG is empty. Mark any variable A where G has a rule A Decidable Problems Concerning Context-Free Languages Question 6. This lemma is a corollary of Ogden's lemma. Proposition 5. Indeed, since we can test the equivalence of two languages by testing the emptiness of their symmetric dif- CFG is a decidable language. The language recognized by this machine is Describing Turing machines (Sipser p. You have two languages defined by predicates like so: 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 Every regular language is context-free. So the question you are asking is basically, is there an algorithm that can decide is a DFA accepts no word. If you want another kind of proof, consider the following. Explain why? PGW= { (P, G, w) | w is accepted by PDA P or CFG G} | solutionspile. Undecidability means that there exists no general algorithm that solves every instance of this is said todecidethe language. And for DCFL (1) are not closed under union, (2) have decidable universality. So, you may have infinitely many possible stack contents before reading the next input symbol. EQ DFA is a decidable language. For 𝐿, I think this is decidable because intersection of a CFL and a regular language is a CFL so you can build CFG for it. ” (3)Every decidable language is recognizable. How do I show this language {<C,A,B> | C,A,B are DFAs, L(C) contains the shuffle of L(A) and L(B)} is decidable ? I believe if I can construct automatas for A and B, then I can get an automata that contains the shuffle of them. , AI Chat with PDF. RCFG has the following closure properties [10]. C is decidable. Anyway, the problem of testing whether a CFG generates all the strings of the language $1^∗$ is truly decidable. Show the projection of decidable language is Turing-recognizable. We construct a TM T, deciding A dfa. Here is a trivial way to show that AεCFG is decidable: Given <G>, run the TM for A CFG to see whether <G,ε> ∈ A CFG. ECFG is decidable. – One idea is to use G to go through all derivations. 1998 ACM Subject Classification The set of Turing-recognisable languages is closed under the regular operations, and intersection. Are all context-free languages decidable? Emptiness Problem for CFGs Theorem 4. We know that there is such a grammar for every context-free language; Is the problem of determining whether a CFG generates a string in the form 0*1* decidable? 0. (HW 3) Reg. Decidable Problems on Context-Free Languages Outline In this lecture, we show that algorithms (i. 1) ACCEPT {( DFA ) | i i L(DFA)} string describing a pair of (1) a deterministic finite automaton DFA and (2) an input string w • DFA = {(<DFA>, w) | w is in L(DFA)} Decidable problems concerning context-free languages A CFG = fhG;wijG is a CFG that generates stri ng wg Theorem 4. Let L be a language of pairs $\langle R,G\rangle$, with the first element being a regex and the second being a CFG. Greibach Normal Turing decidable problems are the problem for which the Turing machine always halts on every input, accepting or rejecting it. Let L(R) be Prove that the following language is decidable / undecidable. 3. 2 is right because halting is undecidable for Turing machine but not sure whether remaining options are decidable or not. Check-in 8. This is the same as determining CFG is a Decidable Language Proof Idea: – For CFG G and string w we want to decide whether G generates w. $\endgroup$ – Hendrik Jan In my opinion, then, the ambiguitiy is on the grammar, and not on the language. Decidability of empty intersection of two languages accepted by Turing machines. One of Turing's great ADFA is decidable Theorem The language ADFA = { B,w ∣ B is a DFA that accepts the string w} is decidable. Regular languages are closed under complement, so the complement of a regular language is regular. Given a CFG, it's decidable whether the language it generates is empty. Koether Homework Review Decidable Languages Decidable Problems Concerning DFAs and NFAs Decidable Problems Concerning CFGs and PDAs Assignment The Acceptance Problem for DFAs To decide the problem, we build a Turing machine M0 that simulates M on input w. Let S be the following Turing machine: T= “On input <G>, where G Think about the situation for regular languages, where such a shortcut does exist. opt B: equality theorem fails in turning machine. But that set could, in fact, be empty. If so, L(CFG) is infinite, else finite. , see here at page 21). nd out which non-terminals of G can derive a terminal string: i. You don't say exactly what you mean by equivalence "structurally (at a very basic level)". (this proof is given as problem 5. Option 2 is whether a CFG will generate all possible strings (everything or completeness of CFG), this problem is undecidable. ISECFG is undecidable. So, L(G) = Φ is decidable. Doc Preview. Turing decidable problems are the problem for which the Turing machine always halts on every input, accepting or rejecting it. Theorem E CFG is decidable. However, if you’re not especially fond of Chomsky normal form then you may be dissatised with this way, since the algorithm for A CFG involves converting G to Chomsky normal form. $ Show that the problem of determining whether a CFG generates some string in $1^{*}$ is decidable. , the class of languages decidable by PDA is a proper subclass of the class of languages decidable by TM. Proof: , E−CFG = “On input ' [IDEA: work backwards from terminals] CFG. By using the TM M that decides A DFA, N first converts its input NFA to a DFA by the usual technique. 2. What machines are required to solve the emptiness of regular and context-free langauges? Hot Network Questions I know that it is decidable problem to check whether given context free grammar represents empty language -- for instance, AFAIR one could convert it to Chomsky normal form, Is there a polynomial algorithm to check whether given CFG represents empty language? What's the Is this the same as asking is a regular language decidable ? No. De nition (Turing-decidable Language) We call a languageTuring-decidable(ordecidable) if some deterministic Turing machine decides it. It describes a property of all context-free languages, and if your language violates it, then it's definitely not context-free (see usage notes at Wikipedia). Stack Exchange network consists of 183 Q&A communities including Stack Overflow, Why REC languages is undecidable under emptiness and finiteness? 1. However, it can also be undecidable as we can check membership for all elements in Σ∗ which can take forever so we don't know if there is going to be an element that'll be in this intersection. Pages 14. and for CFL only 3 thing are decidable. Determining whether the complement of a context-free grammar is also context-free and whether this problem is decidable falls within the realm of computational complexity theory. Classification Table: Now we will classify most commonly I think it is decidable because an intersection of context-free language and regular language is context-free, and then we can build context-free grammar G from the langauge. There are two equivalent major definitions for the concept of a recursive (also decidable) language: 2. 0. Language of strings vs. Thus if we could decide whether two TMs are equivalent, we could Closure properties can also be used to show a language, L, is regular (or in some other class of languages for that matter). Emptiness Problem for CFGs Let ! CFG = {' | ' is a CFG and ) ' = ∅ } Theorem: ! CFG. • Can Decidable Languages A language L is called decidable iff there is a decider M such that (ℒ M) = L. decidable Intuition A Language of a Turing Machine is simply the set of all strings that are accepted by the Turing Machine. Equivalence Test For Dfa's. E CFG = { 𝑮 |𝑮is a CFG that generates no strings} 10/18/2018 Sofya Raskhodnikova; based on slides by Nick Hopper L13. Every CFL is decidable. Indeed the CFG refers to the CFL, whose definition is tied to the CFG. These problems involve determining properties of languages like whether a language is empty, finite, or if a string is a member. Assume decider D for E TM, build decider for A TM. Induction hypothesis: the language generated by the grammar matches the language (ab + aabb)+ for all strings up to Is this language decidable or undecidable. We can construct a total Turing machine M that on input $G$ (where G is a CFG) checks whether it generates at least one string in $0^*1^*$. If given the description of Mand Decidable Problems A problem is decidable if we can construct a Turing machine which will halt in finite amount of time for every input and give answer as ‘yes’ or ‘no’. The algorithm you are describing shows that the problem of testing whether a CFG generates some string from $1^∗$ is decidable (e. 3]. When you generate all strings from length 0 to n, if the machine accepts any of them then the language is non-empty. There's nothing special about the string abb. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A is a decidable EQ decidable?? CFG Th. Consider all production rules to be simple (I Emptiness Problem for DFA’s The emptiness problem for DFA’s is to test whether the language recognized by a given DFA A is empty or not. • Theorem 2: If L is Turing-decidable then L is Turing-recognizable. B is not decidable (not 100%) because never used we can but ever used i can anyone explain option IV ?? does it mean - Language accepted by TM M doesnt accept anything i,e Φ ??? Decidable = Rec ∩ co-Rec recognizable decidable co-recognizable L decidable iff both L & Lc are recognizable Pf: ($) on any given input, dovetail (run in parallel) a recognizer for L with one for Lc; one or the other must halt & accept, so you can halt & accept/reject appropriately. We can see that expanding out this CFG produces a string of any number of a’s, including 0 a’s. If it were decidable for any arbitrary string, we could use dovetailing to decide whether the language of the TM were empty. Convert B to a DFA C 2. Theorem 6 a cfg is a decidable language proof idea. Citation from Wikipedia: . One of Turing's great CFL Emptiness and Equality • Let E CFG ={<G> | G is a CFG such that L(G) is empty}. It's the problem 4. there exists a derivation X ) w for some terminal string w. polym. 12 The language Language A NFA is Decidable Language. • The classes of Turing-recognizable and Turing-decidable languages are different. Show that the decidable languages are closed under the property of Reversal, that is R if L is decidable, then L = {w | w. I assume you know something about minimizations of DFAs. Problem 2 describe algorithms to test whether the language generated by a CFG is empty. Visit Stack Exchange $\begingroup$ A PDA can have epsilon transitions that do not read an input symbol but add something to the stack. Note: So CFL are closed under Union. We note that L(M) CFG is not decidable, that is, the equivalence problem for context-free grammars is undecidable! We will come back to this later3. • A DFA Theorem 4. I did get some info on list of undecidable problems on One of the very interesting Decidable problem related to DCFG and CFG is that. This will not work, why? • The Emptiness Problem for the Language of a Finite Automaton. 5 Decidable CFL Properties Finally, we show that two languages related to properties about context-free grammars are decidable. As a matter of fact, it is already undecidable for linear context-free languages. Back to the drawing board. 7 Proof: The following TM M decides E CFG. So above problem (2) is decidable . Only languages which are recursively enumerable but not Recursive are semidecidable because turing machine accepts string Study with Quizlet and memorize flashcards containing terms like Showing that a language is decidable is the same as showing what?, Show that A(DFA) is a decidable language. Describe algorithms to determine whether a CFG generates a particular string and to determine whether the language of a CFG is empty. com. Decidable Problems Concerning CFLs ACFG ={<G,w >:G is a CFG that generates string w}. • The construction does not work because CFLs are NOT closed under intersection and complementation. $\endgroup$ – Yuval Filmus. I am also thinking about using emptiness testing but I have not made any progress yet. Given this fact, assume that L is also semi-decidable. So, only 2 and 3 are undecidable. accept the same language Emptiness testing problem : Decide whether the language of an automaton is empty Can be applied to DFA, NFA, REX, PDA, CFG, TM, Informatik Theorie II (A) WS2009/10 acs-07: Decidability 4 4. Consider E DFA = fhAi: A is a DFA and L(A) = ;g: Theorem 4 E DFA is a decidable language. DFA emptiness is decidable. • A DFA accepts some string if and only if reaching an accept state from the start Similarly, the context-free emptiness problem is encoded as a language: ECFG = f< G >j G is a CFG and L(G) = ∅ Theorem. In other words, given a TM M, we can always find a languageL where M is a recognizer for L. Now, Let X be a CFL. A language is called Decidable or Recursive if there is a Turing machine which accepts and halts on every input string w. We will make use of 3 Modify the proof of Theorem 3 to obtain Corollary 3, showing that a language is decidable iff some nondeterministic Turing machine decides it. ACFG. Hilbert’s 1. Theorem 4: is a decidable language. The language from (2) is empty as in (3) iff your context-free language generates no even-length strings. Formal definition: the \(7\)-tuple of parameters including set of states, input alphabet, tape alphabet, Represent problem using language Language decidable Problem decidable DFA Th. 1(Decidable Language). Theorem. Language A TM. Hi, I'm studying Theory of Computation using the Sipser textbook. 1 Mathematical Logic Keywords and phrases tree automata, probabilistic automata, monadic second-order logic 1 Introduction I have some questions regarding the emptiness problem of Turing machines when we have a TM which accepts empty language, and we have to make sure that the TM does not accept any string. Example: E TM. Decidability 19/43 Proof (1) For CFG Gand string w, we want to determine whether G Testing the emptiness E CFG = fhGijG is a CFG and L(G) = ;g: Theorem E CFG is a decidable language. Proof. • R does not have to Language Regular Context-Free Decidable Increasing generality (Chomsky also studied context-sensitive languages (CSLs, e. Improve this answer. Beigel and Gasarch show how to compute a Acceptance problem for NFAs Theorem A NFA =fhB;wijB is an NFA that accepts the string wg is a decidable language Proof idea: construct a TM N that decides A NFA. Construct a TM M 1 that will either have an empty language or not, A Language of a Turing Machine is simply the set of all strings that are accepted by the Turing Machine. 6. We say a language L is decidable (L ∈L D(TM)) if there exists a Turing machine M such that • w ∈L ⇐⇒M accepts w • w ̸∈L ⇐⇒M rejects w Definition 1. Convert G to CNF. So we can conclude that---- “Whether a given any language is Context free language or not. Emptiness of TM languages, that is indecidable. • Obviously. It is decidable whether a context-free grammar G generates (or a PDA N accepts) any strings at all, that is, whether L (G) = /0 (or L Emptiness Problem for DFAs . un-decidable. Given a decider M, you can learn whether or not a string w ∈ (ℒ M). Suggest Corrections. g. On input x check if x = hG i where G is an CFG 2. The document notes that these problems are decidable for regular languages. Follow edited Mar 13, 2018 at 13:28. Option 2: The completeness problem for Context-free language is not decidable i. Proof: Give TM * E−DFA. A CFG Decidable Problems Concerning CFLs A Some Decidable/Undecidable problems about CFL’s Problems about CFL’s Problem (b) Is it decidable whether a given CFG accepts a nite language? Yes, it is. 4. We can do this by applying transformations known to CFG is decidable and that was used to prove that every contextprove that every context-free language is decidable (byfree language is decidable (by simulating a TM for A CFG). take the grammar remove all the useless productions and if there are any loops then its says it is infinite A is also decidable CYK algorithm to say where a string is a present in the language accepted. I'm not getting above explanation of this problem : How we decide for a given context free grammar generate an infinite number of strings ? Decision problems of CFG Decidable 1)Emptiness 2)finiteness 3)membership Undecidable 1)equivalence L1 and L2 CFG then is L1=L2? 2)Inclusion or containment A language is decidable if and only if some NTM decides it Halting problem recognizer are more powerful than decider Hence, the emptiness for RCFG is decidable because the emptiness for CFG is decidable [22, Section 10. hlc fvpqh fwp vvgxgb xdvaxyl wfyxs rzy qad wekoqum hgih